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Gravity powered devices => Gravity powered devices => Topic started by: Delburt Phend on February 04, 2017, 03:31:19 PM

https://youtu.be/YaUmzekdxTQ
This experiment produces energy

This experiment is a yoyo despin experiment; except that in this experiment the tethers are left attached and the masses on the end of the strings return all the motion back to the cylinder.
If the Dawn Mission satellite would have left the tethers attached; it too would have had the spheres return all the spinning motion back to the satellite. A roughly similar spinning mass ratio to that of the Dawn Mission would be a one meter diameter 400 kilogram cylinder spinning at 1 m/sec around the arc of the circle; and this would be attached to two .5 kilogram spheres on the end of the two tethers. This would be a 400 to one; cylinder to spheres, mass ratio.
Spinning at one meter per second the 400 kg cylinder would have 400 units of momentum. The spheres will have to have 400 units of momentum to return all that momentum back to the cylinder.
The 1 kilograms of spheres will have to be moving 400 m/sec to have 400 units of momentum. At 400 m/sec the spheres will have 80,000 joules of energy.
The original energy of the 400 kilogram cylinder moving one meter per second was 200 joules.
The experiment in the video proves that the spheres will return all the motion back to the cylinder.
The mass ratio in the videoed experiment is only 4.5 to one. Because of this smaller mass ratio the spheres actually stop the cylinder twice; and they restart it twice. The original arc velocity is 1.2 m/sec
At the first stop of the cylinder in the video: the arc velocity (of the spheres) required for momentum conservation would be 1.2 m/sec *4.5 = 5.4 m/sec. This is the momentum (sphere mass * 5.4 m/sec) required to restore all of the motion back to the cylinder: and only momentum is transferred from small to large.
The arc velocity (of the spheres) required for energy conservation would be only 2.54 m/sec; this is only 47% of the need motion to restore the momentum back to the cylinder; for only momentum can be transferred from the small spheres back to the larger mass cylinder. For energy conservation; there would be only 22% of the motion needed to restart the cylinder after the second stop.

It seems that your calculations were performed using the
Initial velocity and the final radius of the arc
This is not accurate.
The angular velocity drops as you move to a larger radius.
Do you have a way to accurately measure the velocity of the balls?
Or the impact force when they hit the pipe?
Also the input energy ( your hand twisting is hard to measure)

All of your motion cannot be restored to the cylinder.
Most of the energy is in the collision, which approaches
a radial vector, towards the axis. Which actually jolts it
Sideways. Some of this is lost due to the vector the
cylinder is already traveling in. The rest, that does not
translate into linear horizontal motion, is converted to
heat at the surfaces of the ball and the pvc.

Actually the initial motions is not hard to evaluate.
There is a frame by frame mode in my computer that subdivides the motion into 240th of a second.
It takes four frames for the cylinder to cross the distance of the black square; at the beginning; in the middle ; and again at the end. At these three points of highest cylinder rotation; the system is moving 20 mm *240/4 = 1.2 m/sec.
The momentum is the same at these three moments in time.
Only linear Newtonian momentum is conserved when a small object (spheres) give their motion to a large object (cylinder and spheres). This law is true even if the objects are in arc motion. There is no loss of motion; and no need to attribute any loss of motion to heat.
There is however a 450% increase in energy when the spheres have all the motion; for they must be moving 5.4 m/sec. To return (.594 kg *1.2 m/sec ) .7128 units of momentum the spheres must be moving 5.4 m/sec (.132 kg * 5.4 m/sec).
The spheres do not collide with the cylinder. The spheres come very near the surface of the cylinder but they do not collide.

Then what is that loud kerklunk as each of the balls
Smack the side of the pvc in your video?

The string gets shorter and shorter as the ball approaches
Ultimately the arc formed by the string becomes smaller than
the arc of the cylinder. Follow the final trajectory when r=0
And you can see that the ball is moving straight (linear)
Towards the axis. It strikes the pvc at a 90degree vector.
Notice how it bounces straight away from the surface?
If it is truly moving at 1.2m/s ( I cannot verify that)
steel vs pvc is partially inelastic collision. I could prove to you
exactly how much of the impact momentum gets converted to
heat. It's not an amount that is significant, but it was the first
identifiable loss. After further review, I have a few others to
discuss.
What I see from slowing the video down is the ball in front
Closest to the screen hits first, the pvc moves towards the
Other ball, shortening the time it would have hit by a fraction
Of a second, and the second ball strikes a moment later,
Causing the pvc to stop moving in that direction
a portion of the rotational momentum carries back to the pvc
certainly not 100% of it
However, the forward and downward momentum of the test device
Appear to be the dominant momentii throughout the video.
The input from your hand was much more than the returning balls
Put back as rotational momentum to the pvc.
I don't see energy being created here.
I see energy dissipated as heat, sound, physical vibrations
And reversal of angular velocity by momentum exchange.
All of these things are minus energy

It is important to note that the quantity of momentum at the end is equal to the quantity of momentum at the beginning. This means that the quantity of momentum is the same at every point in between the beginning to the end. The total quantity of momentum will not change; without the application of outside force. The cylinder moves 20 mm in four frames at the beginning; in the middle; and at the end. But these three points are not where the energy is at a high point. The energy is at a high point when the cylinder is stopped.
When you search yoyo despin you will get about a dozen sites that throw the tethered spheres off when the cylinder (disk, satellite) is stopped. NASA predicts that these thrown spheres have the same energy of the original spinning energy; this is not true. The spheres have the same momentum as the original spinning momentum. These dozen experimenters are conserving momentum not energy.
You can’t grab momentum out of the blue; and you can’t lose it. There is no 95% heat loss for momentum; it remains the same. The final momentum is the same because the spheres have the same momentum when they have all the motion. And in this experiment; when the spheres have all the motion they have a 450% increase over the original energy.
In most of the dozen sites; the experimenters are producing much more than 450% increases in energy because the mass ratio of system/spheres is larger.
The double stop experiment proves that the motion can go back and forth. Energy allegedly loses heat and therefore cannot go back and forth.
When a 399 kilogram block (at rest) is struck by a 1 kilogram bullet moving 400 m/sec the block will accelerate to 1 m/sec.
When a 399 kilogram rim (with bearing) mass wheel (at rest) is struck tangent by a one kilogram bullet moving 400 m/sec; the rim will accelerate to 1 m/sec.
If we wrap a string around a 399 kilogram rim at rest; and place the bullet moving 400 m/sec on the end of the string and moving perpendicular to the string; would we then accelerate the rim to 20 m/sec or still just 1 m/sec?

I think you are missing the point
While total momentum is conserved, true
This momentum has magnitudes and vectors
At the end, the momentum of the balls is
In a direction away from the cylinder and
Not is the direction the cylinder is moving.
The total momentum of the system is divided
Along 3 vectors (minus that which was lost in
the collision). The primary vector being that
Direction the cylinder is traveling. Each of the two
Balls have their final momentum along the other
Two vectors. Which are actually gravitational based
ballistic events, that arc downwards in a calculable manner.
By the ballistics path, knowing the weight of the balls
You can calculate their velocity
And by knowing the materials constants for the steel
And pvc, a reversing of the equation we can know the
Impact velocity and resulting force, thereby calculating
How much energy was converted into heat, and how much
Went into moving the cylinder 90degrees to its path of motion.
May not have been a true 90, you whirled it kind of lopsided
And I'm only judging it from your camera angle.
But the motion is visibly transverse to the cylinders direction
And by Newtonian mechanics, must have been 180degrees to
The direction of the balls (+/ a factor for relative curvatures)
The standard experiment shows that the balls have the same momentum
as the balls. NOT the momentum of the cylinder OR the momentum
Of the entire contraption cylinder,balls, and string.
Momentum is a quantity defined by mass.
The cylinder had more of it than the balls
In your experiment, the momentum that was transferred to the
cylinder from the balls, cancelled out because the balls were opposite
each other.
Perform the same experiment with both balls on one side
They go around, smack the side and the cylinder goes diagonal
Balls bounce off just like before and if it doesn't touch down first
The balls will give a backwards jolt at the end

Before you attempt to quantify this experiment
You need a consistent launcher, other than your hand
So you know your input energy.
Maybe there with your software and the original footage
You feel comfortable using a digital frame rate for a clock
There are too many unknowns for me to place any accuracy
on that situation.
Conversion rates, frame buffers, storage rates, possible glitches
Software based frame alteration, file conversions,
And all before thebYOUTUBE antishake filter touched it.
You can assign an initial velocity and do the math to see exactly
What will happen. Looks similar to your video, but with more
precision, and elegance in flight.

I don't see any significant software or video anomalies. The technology is getting very good; and it is financially available to about anyone.
When you compare the real results with expected results the momentum conservation becomes an absolute.
Momentum conservation requires that the cylinder motion at the end will be four frames to cover 20 mm.
Energy conservation requires a very low speed for the spheres when the spheres have all the motion. The sphere speed is 2.12 m/sec for the first cylinder stop. Sufficient momentum to restart the motion of the cylinder is gone. By the second restart the motion would be down to 22% if energy were conserved by the spheres.
Only momentum can be given from the spheres to the cylinder. So if you lose a little over half your momentum from cylinder motion to spheres motion; twice; you are down to 22% of the original motion.
Energy conservation would require 4 frames /.22 = 18 frame to cross 20 mm.
At the end of the experiment it appears that the rotation rate is still 4 frame to cross 20 mm.
It does not in any way appear to be 18 frame to cross 20 mm.
You have to pick one or the other; is it 18 or 4.
If you have 100 unbiased observers pick between 4 and 18, they would 100% of the time pick 4.
The fact that the experiment is falling is total inconsequential. The spinning motion and the falling motion are independent of each other.
I still don't think the spheres hit the side of the cylinder; but it does not matter whether they do or not the motion is still 4 frames to cross 20 mm. Nor does the direction of the spheres matter they can still be directed up from any direction.

Hello Delburt Phend... or should I say pequaide ?
we stopped the discussion about 8 years ago:
http://overunity.com/1995/freeenergyfromgravitationusingnewtonianphysic/msg165397/#msg165397 (http://overunity.com/1995/freeenergyfromgravitationusingnewtonianphysic/msg165397/#msg165397)
and as I remember there was good consense about the basics.
Now we better think about finding a technical solution to earn the energy as this is the crux of the matter.
regards
Mike

I am not sure how you are using he word 'earn'; is that making the energy, or using the energy.
The energy is made by making a closed system place all the motion is smaller subset of the mass.
The energy can be used by throwing it up into the air; and holding it as potential energy.
The larger combined mass; at 1.2 m/sec; would only rise .0733945m. For .594 kg *9.81 N/kg * .07339 m = .4277 joules of potential energy.
The smaller .132g mass at 5.4 m/sec would rise 1.486 m. For .132 kg * 9.81 N/kg *1.486 m = 1.924 joules of potential energy.
The energy when the spheres have all the motion is 450% larger 1.925 J /.4277 J.
You would have the same 450% increase if the spheres had a mass of 5 tons and the cylinder 17.5 metric tons.
If the mass ratio was 20 (total) to one (spheres mass) the present increase would be 2000%.

How to make this remarkable concept into the machine?

The Atwood’s machine produces massive amounts of momentum: and it has a wheel or rim.
The cylinder and spheres machine produces massive amounts of energy; and the cylinder can be a wheel or rim.
Bring together the Atwood with a cylinder and spheres concept and you have the machine in question.

oh, I have to apologize .
"earn" is the wrong word.. i fell back into my german thinkig. Of course I mean gaining energy.
As I said the topic was fully discussed in the above mentioned thread, you can go there and reread all post there.
My, telecom and smOky2 questions are related to the practical technical means of energy extraction that is what concept of a machine could be developed for this principle.
More than 8 years have passed and still no solution found
Are you serious in telling the community here that throwing this setup in the air is a practical way to extract energy ?
I even posted a solution in reply #109 here:
http://overunity.com/1995/freeenergyfromgravitationusingnewtonianphysic/105/#.WKEGX1IeCuI (http://overunity.com/1995/freeenergyfromgravitationusingnewtonianphysic/105/#.WKEGX1IeCuI)
The problem of energyextraction was recognized back then and obviously could not be solved ....after 8 years
Mike

For me there is no "question", more so I was attempting to
describe the situation from a physics standpoint.
But apparently physics and mathematics fall to the wayside
When it comes to digitized frame rate of an unscaled distance
Judged by a visual interpretation of a recording.
The cylinders and spheres is an elementary physics lab
Generally covered in the first two years of engineering
Bachelors degrees. This is a simple experiment, and all
Energies and momentii can be accounted for, both physically
And mathematically.
But hey who am I?
This guy says he "created energy"
And his camera frame rate proves it!!
Maybe he can figure out how to get the energy out.....
All I saw from the video was Mr hand putting way more momentum into
The cylinder that the balls were able to overcome in the time it was airborn
You can clearly see the momentum transfer to the cylinder when the balls
Impact the side at 90degrees.
If you don't think this occurs......
Hold the strings in your hands and spin
Then stop spinnning
Make a fist so it's 'round' like the cylinder
Or hold your hands down at your side so your
Hips act like the cylinder
Tell me what happens when the balls run out of string
But are still moving.
Remember TeatherBall?
When the balls gets wrapped it smacks the pole!
Make a big noise and the pole vibrates.
The same thing happens to the pvc, except it is free
to move from the collision.
A steel cylinder will behave slightly different than the pvc
The inertia from the collision is what determines how much
energy is transferred to the cylinder and the how much reflects
With the ball.
Steel vs pvc is not fully elastic, nor is it a fully inelastic collision
But contains properties of both in different degrees, based on the
Materials properties.
Pvc is fairly rigid, but it flexes, bends and heats up more so than
Two pieces of stainless colliding.
Steel pretty much just bounces off like the balls on a pool table
Two masses of same weight steel ball and pvc ball
You can compare to steel vs steel and visibly see this
The two steel balls will behave just as the pool que and a ball.
The pvc collision will absorb some of the energy and both balls
will roll differently
But changing your materials will not change your experiment
The balls and cylinder have two momentums
First being thrown in the direction of travel, this is a linear momentum
Second is the rotational momentum from the twisting action
These two fight each other through the whole experiment
Because their vectors are different.
The linear momentum, being of greater magnitude, is dominant.

The momentum transferred to the cylinder, mostly goes back into the balls
Some is transferred. But each ball cancels out the other ball in this regards
In fact, if you fix the lengths of your strings so they are equal
Such that both balls hit at the same time.
None of the momentum will be transferred to the cylinder
(Except the flexing factor of the materials)
And both balls will repel 180degrees from the collision
With almost all of the momentum they had when they ran
Out of string.
But again this is in a vector opposing the forward momentum
Of the entire device.

If you want to really perform this experiment in the way
You propose:
Perhaps a turntable or stationary rotatable accommodation
Could get rid of the forward toss that prohibits a full examination
Of the forces the balls impose to the cylinder
Of course the cylinder should not be attached to the turntable
So that it may freely respond to changes in velocity in almost every
Available vector.

You seem to be focusing on the last few frames of the experiment. Here the cylinder is moving 20 mm in four frames and we can leave it at that. I record maximum speed for the cylinder of 4 frames per 20 mm right at what you think is a glance. If the sphere and cylinder touch they are moving at the same speed. But the experiment is over at this point.
Maximum energy was achieved about .066 sec before the glance you see. And this is the second time maximum energy is achieved.
The cylinder is stopped 16/240 seconds after release: here the spheres have all the motion. This is where the energy of the system is highest. Most cylinder and spheres experiments do not proceed past this point.
The end frames confirm that all the motion is maintained throughout the experiment. And only linear momentum can be maintained throughout. Because linear momentum is the only thing a ballistic pendulum conserves.
The fist and tether ball are not similar experiments. The fist and steel post are not at liberty to rotate. When the sphere comes close to the cylinder at the end of the experiment both the cylinder and the spheres are moving at the same speed; just like they were at the beginning of the experiment.
All you need to do to extract the energy is point the motion up and cut the string.
Lets make a small machine to see what it looks like.
A one kilogram mass will proceed up 25 meters if it has a velocity of 22.15 m/sec. D = ½ v²/a
This is 22.15 units of momentum.
Make a 25 meter string with 1 kilogram at each meter of length: except there is no mass on one end. Place the string vertically with a one kilogram mass at the top. That would leave you with no mass on the bottom of the string. When the string is lowered one meter that would leave you with one kilogram on the bottom; and a kilogram mass at each meter length in between and no mass at the top.
To restore the one kilogram to the top would require that you accelerate the bottom kilogram to 22.15 m/sec and point the motion up and cut the string.
Place this 25 kilogram string on an Atwood with a balanced mass of 97.67 kilograms; That would be 25 kilograms accelerating 122.669 kilograms for an acceleration rate of 25/122.669 * 9.81 m/sec/sec = 2 m/sec/sec.
After a drop of one meter the entire mass of 122.669 kg would be moving 2 m/sec. From d = ½ v²/a
That is 122.669 kg * 2 m/sec = 245.338 units of momentum.
Place all this momentum in the bottom one kilogram using the cylinder and spheres and it will rise 3067.8 meters, actually air resistance would prevent that much rise but you have vastly more than enough motion to restore the one kilogram at the bottom of the string back up to the top of the string.

Well yes it is exactly like you say
Except for the part about the mass
And the string
And the momentum
25kg moving at 2m/s is going almost straight down
Not up. It will not be moving fast enough to put tension
On the string of 25 meters long
In the earths gravitational field.
Much different than your handtossed pipetoy
You are confused
And this is observable in your descriptions
Taking the linear transform of angular momentum
You must include the vector
Nonparallel vectors subtract from one another
Leaving only difference, at the new vector.
I am focusing on the entire 5 seconds of fussy video
In your original post.
The cylinder is still rotating as it is released. (2s)
There is an angular momentum and associated
Angular velocity, as visible by the markings on
The pvc.
There is also a linear momentum of the system
As visible by the trajectory of the entire apparatus
Two Separate values, two different operations
Let's ignore the linear momentum for now
Even though it is far greater than any other momentum
In this system.
Let us focus on the rotational momentum
As the strings unwind, tension is placed on the strings
As a result of the (vector variant) angular momentum of each ball
At that moment, (2.82.9s on vid) the strings are at their maximum
Length. And as such are following their largest radius, and coincidentally
Traveling at their lowest velocity. Angular momentum at this point is
Subtracting from both the cylinder and the balls. Both balls combined
Are providing the torque on the strings which slows the cylinder and
Causes it to reverse directions.
The momentum of the balls (each) are subtracted by half of the total
Cylinder momentum that the cylinder had at (2s)
It was depleted across almost 1sec of time.
(This is why the balls are released!!!)
Immediately after this, the strings begin to shorten and the angular
velocity of the balls begins to increase.
This is accelerated by the cylinders (now opposite) rotation.
This continues for almost 3 seconds.
Until the strings are again completely wrapped.
Why 1 sec to unwrap, and more than double that to wind back up?
And the impact occurs at (5.9s)
The second reversal of the cylinder has already occurred at this point
Almost half a second before impact.
This is an effect of the linear velocity we are intentionally ignoring.
If gravity were not able to take control, and there was no linear momentum
To deal with, the cylinder and balls would spin back and forth
The cylinder changing directions twice every oscillation of the balls.
Like an odd flying pendulum. Until all of the energy has dissipated.
Which in your inefficient system, would probably occur before the 4th
full oscillation.

They make these things for demonstrating physics labs
It's on a stick with a plastic knob you can spin with a pull string
The pull string is set just shy of one full rotation for the input.
This is meant for demonstrating momentum, not energy.
More importantly, it demonstrates how the momentum is
transferred between the two varying radii.
Via the attached points of the strings
The math is actually just like a pendulum
Or more accurately, like that of a kid on a swing that has
Enough balls to wrap the top post a few times.
Different kind of balls than the ones in your video.
But the momentum transfers just the same.
If the playground isn't bolted down a small child can
Tip it over like that.
Even though it weighs a lot more than the kid and has
Feet spaced for support.
The momentum of the kid is transferred into torque twisting
On the axis. Your "cylinder" in this case, is the upper beam
Holding the kid up. The entire playground, beam and all
Flip right over.
How did the kid build up so much momentum so quickly?
By wrapping the post with his swinging action, he shortened
The chains he was swinging on. Which caused him to move
Much faster.

This type of experiment was done by a student in Ohio.

I use several vertically mounted wheels that throw a single tethered mass. Two meters per second works just fine; that is why I chose it.
The sphere only stay 2 m/sec for a very brief period of time; it is soon 3 m/sec; and 7 ; and 20; and whoosh up it goes. Even in blaze orange you often never see them again.
What you see in the video is about 1/12 actual speed and with a low mass ratio 4.5 to 1. Most throws are much faster.
Angular momentum does not work in the lab because there is no gravitation acceleration of the spheres; which is required for angular momentum conservation to be true.
Further more angular momentum conservation will give lower velocities for the spheres than energy. And energy conservation does not work because there is not enough linear momentum to return the original motion to the cylinder: which clearly happens.
I would like to see a picture of the playground equipment; sound like if you set that up in the U.S. you would be arrested.
The trouble with the playground equipment and the plastic knob stick thing; is that you are adding energy to the system. I can't envision either the knob or the playground thing, much like you can not envision a Atwood’s with a 25 kilogram string on one side.
If 'much faster' (in the last sentence) means linear velocity; then that is not true, linear velocity can not change without the application of outside force.

In the U.S., what I was talking about is a normal playground
Aframes on the ends, an upper support bar, and swings hanging
from it.
I grew up in the 80's , these things were not bolted down back then.
Now we bolt the playgrounds to the ground or concrete the posts to
prevent this from occurring. (And other events that cause tipping)
I learned this the hard way, at a young age. Way before I understood
the physics that make it happen.
as far as your misconception concerning linear velocity of an object
That posses multiple vectored angular momentii......
My inertial propulsion research team got a big laugh out of your statement.
I am not even going to respond to it myself. I think I will let you stew on
That proposal for a moment.
Energy is always put into the system. The hand in your video inputs this energy.
In a controlled experiment, driven by a stepper motor, or even the rudimentary
pullstring this energy is quantized. Meaning it holds an experimental 'value'
From which the data can be calculated.
Without knowing this value, you cannot perform energy conservation analysis.
If you did such analysis, you would probably go back and delete your posts.

Does your propulsion team have any videos that shows energy being conserved and linear momentum being lost? Does linear momentum increase without the application of outside force; as you laughers propose? Please: I am shewing myself sick.
Of course energy is put in but not while the experiment is under way.

@Delburt or whatever your posting name was or is:
Like Kator01 I would like to know whether anything tangible (besides venting too many words and out of context numbers) happened during the last eight years in respect of useful experiments concerning your alleged tether energy production?
You will admit that the table top experiment from your video https://youtu.be/YaUmzekdxTQ (https://youtu.be/YaUmzekdxTQ) is at least inconclusive and very messy. Is that all you can do experiment wise?
May be I am wrong, but the biggest hurdle is that the tether experiment needs to be done in free fall (best in space).
There is the possibility that you do your experiments from a tower or high building (20 meters might be high enough). You mount a good high resolution camera on the ground over viewing the experiment from bottom to top and you throw your gadget (which still seems to be the same from eight years ago) from the tower or high building. A mechanism to always spin the gadget consistently with the same speed at launch would also be good.
You can also throw the device from a bridge. Many bridges have an area underneath which is still land and not water (at the beginning or end of most bridges). In urban areas many bridges lead over land and not over water. You do not need extraordinary height, 10 to 20 meters would be plenty to make a short movie (definitely better than a table).
The best would be a free fall tower like this one https://www.zarm.unibremen.de/droptower.html (https://www.zarm.unibremen.de/droptower.html) , but for some initial experiments a high building or cliff would be good enough. The most important part of the experiment would be a launching mechanism, the camera and the high building or cliff or bridge are trivial.
Yes, this is harsh criticism, but your too many words and strange posting behavior (over eight years) pose serious questions and cause doubt. Nothing has ever been achieved with words and theoretical numbers, the experiment is the mother of all progress. First is the repeatable experiment and then come the words and numbers based on the experiment. That is how useful science is done since ages.
Greetings, Conrad

What I see is a perfectly excellent experiment; there is no why to add motion to the experiment after it is released and it is easy to see what motion it has by counting the frames for a crossing of the 20 mm square. And it comes back to that exact quantity of motion twice. And only linear momentum conservation can do that.
I don't even know why people refer to it as blurry: it is in motion what do you expect. It is a $200 camera not a $4,000 one. And the $4000 camera will give you no more information than the one you see. Because the end motion is four frames not 18 frames to cross the 20 mm. There is zero chance that energy is conserved. These experiments will be near the top of all physics experiments.
Conrade; Did you ever think that no experiment will ever make you happy; and you probably don't ever want to see one that actually proved energy can be made. What the experiment proves can not be any clearer; maybe it is you that refuses to see it.
Any motion experiment has to be a closed system. Once the experiment starts there can be no application of outside force. There can be no energy added; no momentum added; and none of your angular momentum added, it has to be a closed system to be an experiment. This rule disqualifies a person swinging something around on a string; and pulling the string in and out of a tube. It disqualifies the Ice skater. It disqualifies a child on an A frame swing. It disqualifies a person on a swivel chair pulling barbells in and out.
The cylinder and spheres qualifies as a closed system experiment because the first data point occurs after the fingers have released the spheres and cylinder. There is no data point that has anything to do with an outside force being applied. Just after release the fastest speed of 1.2 m/sec is recorded: and that same speed is recorded two other times. This experiment qualifies as a closed system.

so what is the reason that this subject which was discussed in 15 pages 8 years ago
is brought again to the attention of the members here ?
Conrad if you wonder about the strange posting behaviour: maybe this here will help you to understand
https://zeltser.com/botscontrolsocialnetworkingcontent/ (https://zeltser.com/botscontrolsocialnetworkingcontent/)
and no, he can not afford to pay just one hour at ZARM: Only Prof. Szasz of Ungaria had the chance to perform his freee fall experiment here:
https://www.youtube.com/watch?v=jkNjvCmsWOU (https://www.youtube.com/watch?v=jkNjvCmsWOU)
https://www.youtube.com/watch?v=WsyJjxC7SRc (https://www.youtube.com/watch?v=WsyJjxC7SRc)
the whole issue does not make sense besides distracting attention by presenting old stuff again and
of course to cause traffic for obvious reasons...period
Mike

Only Prof. Szasz of Ungaria had the chance to perform his freee fall experiment here:
https://www.youtube.com/watch?v=jkNjvCmsWOU (https://www.youtube.com/watch?v=jkNjvCmsWOU)
https://www.youtube.com/watch?v=WsyJjxC7SRc (https://www.youtube.com/watch?v=WsyJjxC7SRc)
the whole issue does not make sense besides distracting attention by presenting old stuff again and
of course to cause traffic for obvious reasons...period
Mike
@Mike,
thank you for the link to the videos of Prof. Gyula I. Szász. He has his own website with many papers http://atomsz.com/ (http://atomsz.com/) (as you might know). This is very interesting and I try to understand it.
Particle physics is driven by a handful of people who have access to a particle accelerator. And this is the experimental source for thousands of scientists who can never really understand and check what is done with the particle accelerators. The particle accelerators are because of theire enormous cost a political issue. They have to succeed because failure would be a catastrophe. Who can admit that one blew away billions and did useless experiments for decades? My unimportant and humble opinion: the particle accelerators are a terrible distraction and prohibit meaningful research of the microcosm. But who I am to judge that? I only know very little. I derive my opinion from the complexity of the technology used in particle accelerators. Everything is beyond the mensurable, it is a statistics wank. If you have trillions of measurements and you then do statistics long enough you will find something, specially if your career and your money supply depends on it.
Concerning our good man Delburt:
One never knows what motivates people to write endless and senseless rants in forums. Delburt's behaviour is very typical: no facts, no clear answers, deliberate obfuscation. He obviously does not want a meaningful dialogue and he has nothing tangible to show.
Greetings, Conrad

https://youtu.be/YaUmzekdxTQ (https://youtu.be/YaUmzekdxTQ)
This experiment produces energy
Any given effect has a bad tendency to be a result of the cause. In other words, it is very hard, if not impossible, to achieve an effect that does not have any connection to the cause. If it wasn't your calculator would be useless. 11=0 no matter how hard you try to change the result.
The cause is your hand, the effect is two balls having fun around a PVC tube. The mechanical movement is a direct cause of the energy you supply by your hand.

When a force causes a mass to move in a particular direction that is the positive direction for that mass. It does not matter what the actual direction is (N, S, E, W, up, down, left, right). When the same force causes a different mass to move in a different direction that direction is still the positive direct for that mass. And the momentums of the two masses are added.
Proof: An Atwood’s machine is used to prove F = ma. A little over half of the motion in an Atwood’s is going down and a little less than halve is going up; but the two momentums are added.
So you are not going to make vector mistakes again are you. When one force causes different masses to move in different directions all the directions are positive and the momentums are added together.
A 97.63 kilogram Atwood’s with 25 extra kilograms (122.63 total mass) on one side will accelerate to 2 meter per second velocity after the 25 kilograms has dropped 1 meter. This drop will take one second. With 9.81 newtons per kilogram for 25 kilograms applied for one second; this is 245.25 newton seconds.
After this Atwood’s is in motion it can apply 245.25 newtons for one second.
When 9.81 newtons is applied to this Atwood’s it will take 25 seconds to make the Atwood’s stop.
When 9.81 newtons is applied to a one kilogram mass for 25 seconds it will be moving 245.25 m/sec.
A one kilogram mass moving 245.25 m/sec will rise for 25 second. And it will rise 3065.625 meters.
This is 3065.625 m * 9.81 newtons/ kg * 1 kg = 30,073.78 joules of energy.
Twenty five kilograms dropped one meter is 25 kg * 9.81 N/kg * 1 m = 245.25 joules
So with 245.25 joule you can make 30,073.78 joules.

https://www.youtube.com/watch?v=aaYoJjIPAo

https://www.youtube.com/watch?v=aaYoJjIPAo (https://www.youtube.com/watch?v=aaYoJjIPAo)
Excellent!
Now comes the important part
E = (m2m1)gh
Where h is the distance downward the large mass moves
during the experiment.
You see that energy is conserved

That is correct: the total mass of the Atwood’s in the video is 40 kg and the accelerating mass is 20 kilograms. The 20 kg in the balanced portion of the Atwood’s is not dropped; only the extra 20 kg is lowered.
Twenty kilograms at a height of one meter has a potential energy of (1 m * 20 kg * 9.81 N/kg) 196.2 joules.
Twenty kilograms dropped one meter in free fall has a velocity of 4.429 m/sec for an energy of (1/2 * 20 kg * 4.429 m/sec * 4.429 m/sec) = 196.2 joules.
The 20 kg /40 kg Atwood’s has a final velocity of 3.132 m/sec after the 20 kg drops 1 m; this is (1/2 * 40 kg * 3.132²) = 196.2 joules.
A twenty meter long string that has one kilogram at each one meter length has 196.2 joules of energy after it is dropped one meter.
A twenty meter long string that has one kilogram at each one meter length has 196.2 joules of potential energy when it is raised one meter.
A one kilogram mass dropped 20 meters has a final velocity of 19.81 m/sec for 196.2 joules of energy.
So: 196.2 joules for everything; so far.
The 20 kg /40 kg Atwood’s has a final velocity of 3.132 m/sec after a drop of 1 m; this is (40 kg *3.132 m/sec) = 125.28 units of momentum.
As proven by the double stopping cylinder and spheres you can place all of that momentum into one kilogram.
For one kilogram to have 125.28 units of momentum it must be moving 125.28 m/sec.
One kilogram moving 125.28 m/sec will rise (1/2 *125.28² / 9.81 m/sec) = 800 meters.
The potential energy of 1 kg at a height of 800 meters is (800 m * 9.81 N/kg * 1 kg) = 7848 joules of energy.
A 7848 J / 196.2 J = 4000% increase over the original energy.
Please note that there is no mention of radius. Which eliminates what?
It is also important to note that all the 196.2 joules listed are from an original production of motion; they are not transfers of previously existing motion.
If the one kilogram moving 19.81 m/sec were to transfer its motion to the Atwood’s at rest it would only be moving .48317 m/sec for only 4.786 joules of energy not 196.2 J.
If the 20 kg moving 4.429 m/sec were to join its motion to the Atwood’s at rest it would have (20 * 4.429 = 60 * X) = 1.4765 m/sec for 59.1 joules of energy not 196.2.
Momentum is always conserved in the transfer of motion from one objects to another not energy. And radius is never mentioned; so what kind of momentum is it?

Imaginary momentum, represented by a j
The momentum of the Atwood is not 40kg moving down
It is the difference between the two masses.
They each move a momentum in opposite directions
Smaller mass moving up
Larger mass moving down
Gravity accelerates both downward
The net acceleration is on the difference
The momentum is also the difference as both masses are joined
via the string
They subtract from each other because the mass is balanced out
How much does a kid weigh on a teeter totter?
Why can they jump so high?
I admire your enthusiasm, but you have to understand what is being
conserved.
all the momentum is there, it is just not all in the same direction

What is the momentum of two cars tied to a chain
Driving away from each other?

With an acceleration rate of 4.905 m/sec/sec it will take the 20 kilograms .63855 seconds (d = 1/2at²) to accelerate down one meter.
The other twenty kilograms is balanced and its center of mass does not rise or lower.
Twenty kilograms exerts a force of 196.2 newtons. So 196.2 newton applied for .63855 seconds is 125.28 units of momentum.
The final velocity is 3.132 meters per second.
A newton * second = kg * m/sec.
Ten kilogram moving up at 3.132 m/sec is 31.32 units of momentum:
10 kilograms moving down at 3.132 m/sec is 31.32 units of momentum:
Twenty kilograms moving down at 3.132 m/sec is 62.64 units of momentum.
And 62.64 + 31.32 + 31.32 = 125.28. The momentums are added in Newtonian Physics.
The fact is that Newtonian Physics works for every type of motion in the lab; but not everyone will properly use it.
One thing that helps verify that the starting rotational momentum in the double stop cylinder and spheres https://youtu.be/YaUmzekdxTQ experiment is that four frame per second is what you get in all the throws all the time. Of all the throws I have done I don't remember any faster than 3.5 frames to cross the 20 mm black square. And I do not remember any slower than 5 frames to cross the black square. It seems like the wrist is fairly consistent in producing a relatively slow spin. And I have done hundreds of spins.
Massive quantities of energy are observed when the ballistic pendulum is taken as a proof of Newtonian momentum conservation; and when Newtonian math is used appropriately.

Excellent!
Now comes the important part
E = (m2m1)gh
Where h is the distance downward the large mass moves
during the experiment.
You see that energy is conserved
Totally agree sm0ky2. This topic, as "all" other topics, there are a confusion between force and energy. Energy is what we want at the end of the road. Forces tells nothing about the energy unless we apply displacement.
Energy is conserved even if it doesn't look like it when the experiment is only dealing with forces alone.
I can lift a bulldozer with my finger if I use enough pulleys to do it, but I need to displace hundreds of meters of rope with my finger to lift the dozer a few centimeters. 10 000 kg displaced 1cm require the same energy as displacing 1 kg 100 meters.
The confusion about forces and energy is the one and only reason why all gravity or permanent magnet over unity experiments fail.
Conservation of energy is always the partykiller  not the naysayers ;)
Vidar

If you apply 10 newtons for one second to a mass of 10 kilograms you get 5 joules of energy. (½ mv²)
If you apply 10 newtons for one second to a mass of 1 kilograms you get 50 joules of energy.
If you transfer all the motion of 10 kilograms moving 1 m/sec to one kilogram; the energy changes from 5 J to 50 J.
This is exactly what happen in the video; The motion of a massive object is given to a small object and the small object gives the motion all back; twice. The same quantity of motion is contained in the small object as is contained in the large object.
The small object with the same quantity of motion contains significantly more energy.

If you apply 10 newtons for one second to a mass of 10 kilograms you get 5 joules of energy. (½ mv²)
If you apply 10 newtons for one second to a mass of 1 kilograms you get 50 joules of energy.
If you transfer all the motion of 10 kilograms moving 1 m/sec to one kilogram; the energy changes from 5 J to 50 J.
This is exactly what happen in the video; The motion of a massive object is given to a small object and the small object gives the motion all back; twice. The same quantity of motion is contained in the small object as is contained in the large object.
The small object with the same quantity of motion contains significantly more energy.
I can clearly see the confusion here. Your equation of kinetic energy is correct, but I think you have missed something out. Energy IS ALWAYS conserved, so there MUST be a misconception somewhere. I can't point it out for you, but I strongly believe you've missed something out.
I'll digg into my old papers, and maybe I find something that explains it all.

Now I think I got it.
If you use the 10kg mass to transfer energy into the 1kg mass, the heaviest weight does not loose ALL its kinetic energy, but only some is transfered to the lightest weight.
If you have two steel spheres in space. One small and one large that has 10 times more volume.
If the small sphere is stationary, and you take the large sphere and push it with 1m/s head on to the small sphere. What happens in the collision?
Does the large sphere stop completely, and the small sphere shoots away in 10m/s? That is what your idea suggests.
Vidar

That is absolutely correct Q; that is absolutely what happens.
But instead of using one ten kilogram sphere you use two five kilogram spheres at 180° on a light rim. Wrap a string around the rim from each of the 5 kilogram spheres; place one kilogram on the ends of the two equal length strings. We now have 12 kilogram on a very light rim. Spin the rim at one meter per second; and release the two one kilogram masses. When released the two one kilograms spheres will unwrap and the two five kilogram spheres will be quickly stopped. The two kilograms will have all the motion that previously existed in the 12 kilograms. They will have 12 units of momentum; because the 12 kilograms had 12 units of momentum. The two spheres will be moving 6 m/sec; for an energy increase of 600%
The video proves that this is what happens; because the spheres restore all the motion back to the cylinder twice. And the cylinder had been stopped twice. only Newtonian momentum can do this.
I could have stayed with 10 kilograms by using two 4 kilogram masses at 180°.
But instead of using one ten kilogram sphere you use two four kilogram spheres at 180° on a light rim. Wrap a string around the rim from each of the 4 kilogram spheres; place one kilogram on the ends of the two equal length strings. We now have 10 kilogram on a very light rim. Spin the rim at one meter per second; and release the two one kilogram masses. When released the two one kilograms spheres will unwrap and the two four kilogram spheres will be quickly stopped. The two kilograms will have all the motion that previously existed in the 10 kilograms. They will have 10 units of momentum; because the 10 kilograms had 10 units of momentum. The two spheres will be moving 5 m/sec; for an energy increase of 500%

I think you're ready to build a device that is possible to measure all events. Because the example of the balls in space (space balls haha) isn't correct. The large ball will not stop completely, but if it did, the small ball would bounce away with approx 3,162277m/s  not 10m/s.
3,162277 happens to be the square root of 10  square root of the relationship between the two weights of respectively 1kg and 10kg. The formula for kinetic energy is a quadratic equation (or do you say "second degree equation"?), remember?

If you have two steel spheres in space. One small and one large that has 10 times more volume.
If the small sphere is stationary, and you take the large sphere and push it with 1m/s head on to the small sphere. What happens in the collision?
Does the large sphere stop completely, and the small sphere shoots away in 10m/s? That is what your idea suggests.
Vidar
DP makes an arrangement where he ensures transfer of all the momentum when
string unwraps.
In your case, when 2 bodies hit each other, it doesn't happen.

The Dawn Mission is closer to 400 kilogram to one kilogram of spheres; but we know that the small spheres stop the rotation the satellite. We also know from the double stop video; that those small spheres can fully restart the rotation of the satellite. That means that the small spheres in the Dawn mission yoyo despin device have all the Newtonian momentum that was contained in the spin of the satellite. The spheres are moving 400 m/sec (if the spin was 1 m/sec); check the energy increase.

The Dawn Mission is closer to 400 kilogram to one kilogram of spheres; but we know that the small spheres stop the rotation the satellite. We also know from the double stop video; that those small spheres can fully restart the rotation of the satellite. That means that the small spheres in the Dawn mission yoyo despin device have all the Newtonian momentum that was contained in the spin of the satellite. The spheres are moving 400 m/sec (if the spin was 1 m/sec); check the energy increase.
There is no energy increase. You must go through your calculations again.
The velocity of the small balls is not multiplied with the relationship between the large and small ball, but multiplied with the square root of this relationship. 10kg ball at 1m/s use all its kinetic energy to accelerate a 1kg ball at 3.162277m/s (Not 10m/s as you suggest). Energy is conserved, and no gain is achieved. Sorry mate.
Vidar

There is no energy increase. You must go through your calculations again.
The velocity of the small balls is not multiplied with the relationship between the large and small ball, but multiplied with the square root of this relationship. 10kg ball at 1m/s use all its kinetic energy to accelerate a 1kg ball at 3.162277m/s (Not 10m/s as you suggest). Energy is conserved, and no gain is achieved. Sorry mate.
Vidar
Sir Isaac was saying that the momentum is conserved, not energy.
Consevation of the momentum is derived from the 3rd law of Newton.

I just video taped another cylinder and spheres arrangement.
When spun and released the system (cylinder and spheres) took four frames (4/240th sec) for the black square to move from one side to the other side.
The spheres quickly stopped the cylinder and then continued to unwrap. The spheres soon had the cylinder back up to full rational speed. It again took four frames for the black square to cross from side to side.
The mass of the cylinder is 972 grams: the mass of the spheres is 132 grams; for a total mass of 1104 grams.
The cylinder's rotation speed is 1.2 m/sec (from the four frames).
This means that when the spheres contain all of the Newtonian momentum (and the cylinder is stopped) they must be moving 8.36 times (1104 g / 132 g) faster that the original speed = 10.03 m/sec.
Your calculation say the spheres are moving 2.89 times faster than the original speed; 3.47 m/sec; when the cylinder is stopped.
Your calculations say that 65% of the Newtonian momentum is missing when the smaller mass spheres restore the motion back to the cylinder. Ballistic pendulums prove that only Newtonian momentum can be given from small objects to larger objects. So your calculations show that the spheres do not have enough momentum to return the motion to the cylinder.
The motion of the cylinder is completely restored; so your hypothesis is false.
The energy increase is 836%

Great! Now just connect this up to a generator

I added 216 grams to the 972 gram cylinder: so now the cylinder has a mass of 1188 grams. The two spheres have a total mass of 132 grams. This makes the total system mass of 1320 grams. So the spheres are about 1/10th of the total mass.
Now we have the 10 to one ratio mentioned by Q. And lets start with one meter per second arc velocity just before the released of the cylinder and spheres.
Upon release: the spheres soon have the cylinder stopped; and then soon after the stop the spheres have the cylinder fully restarted. This is four frames at the start and four frame at the finish; as stated many times (1.2 m/sec). But lets stay with 1 m/sec just to simplify the math.
If energy is conserved when the spheres have all the motion then they are only moving 3.16 m/sec; if Newtonian momentum is conserved then the spheres will have to be moving 10 m/sec.
Now if energy is conserved there is only 31.6% of the Newtonian momentum remaining for the return of all the motion back to the cylinder. Collisions prove that only Newtonian momentum is conserved when small masses share their motion with larger masses.
This cylinder and spheres returns all the motion to the cylinders; just like all the scores of other cylinder and spheres do. This is a 10 to one mass ratio and I think I will stay with this model for a while.
Telecom; you are correct about Newton and his third Law. The force in the tether is equal to itself and the force is in both directions. So the momentum lost by the cylinder is gained by the spheres. And then when the spheres share the momentum back with the cylinder they have the adequate amount. The spheres can't be short (only 31.6%) of motion because all of the motion is restored back to the cylinder.
In this cylinder and spheres: 90% of the original motion belongs to the cylinder; because it has the same speed and 9 times the mass. For simplicity I let the speed be 1 m/sec around the arc of the circle.
If energy were to be conserved as suggested then 90% of the motion becomes 21.6% of the motion. This is because the spheres start with 10% of the motion; and only 21.6% is needed to achieve 31.6%. If the spheres have only 31.6% of the motion, when the cylinder is stopped, then 90% has become 21.6%. Confusing math to say the least. Nine units of motion are lost by the cylinder and only 2.16 units of motion are gained by the spheres.
F = ma on the other hand is most precise math. With F = ma 9 units of mass decelerates from 1 m/sec to zero; while 1 unit of mass accelerates from 1 m/sec to 10 m/sec. This is an equal quantity of momentum change on both ends of the tether; Newton's third Law.
There is not any experimental evidence for dropping Newtonian physic; but the abolition of Newton has certainly occurred.

OK. Do a simple experiment. The earth and a 2 gram steel ball.
Drop the steel ball to a hard surface and see what happens. The earth is much heavier than the steel ball, but momentum is still valid and conserved. If your theory is correct, the steel ball will bounce off the surface in a velocity greater than the speed of light.
If it does, you have a problem :)
Vidar

This is the cylinder and spheres with a total mass of 1320g; and the spheres have a mass of 132g.
In the photo the cylinder is stopped and the spheres contain all the motion.
About a 1/12th of a second prior to the photo the motion was shared between the spheres and the cylinder. (four frames to cross 20 mm)
About a 1/12th of a second after the photo the rotational motion of the cylinder will be completely restored. (four frames to cross 20 mm)
If the spheres had conserved energy when they had all the motion it would take 12.65 frames to cross the 20 mm after the motion is restored to the cylinder.
The photo is at the high point of the energy;1000% of the original energy.

OK. Do a simple experiment. The earth and a 2 gram steel ball.
Drop the steel ball to a hard surface and see what happens. The earth is much heavier than the steel ball, but momentum is still valid and conserved. If your theory is correct, the steel ball will bounce off the surface in a velocity greater than the speed of light.
If it does, you have a problem :)
Vidar
Vector of speed of the earth is 0 in the direction of the momentum of the ball.

Correct telecom; the only momentum in Q's system is from the 2 grams. In relationship to the 2 grams the earth is at rest. The spinning motion of the cylinder is transferred to the spheres and then back again; but the earth does not do that.

Oops; I was thinking of the ball being thrown; but Q said dropped. That means that the ball accelerates toward the earth and the earth accelerates toward the ball. They have equal and opposite momentum.
The earth's momentum can not be measured but we have to assume that it bounces back away from the point of collision just like the 2 gram ball. The ball keeps its momentum and the earth keeps its momentum. There is no momentum transfer. The only noticeable motion will be that of the ball.

Oops; I was thinking of the ball being thrown; but Q said dropped. That means that the ball accelerates toward the earth and the earth accelerates toward the ball. They have equal and opposite momentum.
The earth's momentum can not be measured but we have to assume that it bounces back away from the point of collision just like the 2 gram ball. The ball keeps its momentum and the earth keeps its momentum. There is no momentum transfer. The only noticeable motion will be that of the ball.
It does not matter if the ball or the earth is in motion. The end result is a total momentum that is conserved. Would it make any difference if you throw the ball into earth, or even throw the ball into a vertical wall that is fixed to the earth, or use a large crane to move a 10000 kg concrete block towards a stationary 2 grams steel ball? Would the steel ball fly away as a projectile? Nope!
If I understood you correctly, you assume that the earth must have motion and hit a stationary small ball, and not the other way around. If so, your assumption is not correct. The energy and momentum is conserved in any case.
and it does not matter if you throw or drop the ball. The ball gains momentum as it accelerate towards the earth, and the earth gains the very same momentum as it accelerate towards the ball. The earths acceleration is however very tiny, but it's there.
Vidar

It does not matter if the ball or the earth is in motion. The end result is a total momentum that is conserved. Would it make any difference if you throw the ball into earth, or even throw the ball into a vertical wall that is fixed to the earth, or use a large crane to move a 10000 kg concrete block towards a stationary 2 grams steel ball? Would the steel ball fly away as a projectile? Nope!
If I understood you correctly, you assume that the earth must have motion and hit a stationary small ball, and not the other way around. If so, your assumption is not correct. The energy and momentum is conserved in any case.
and it does not matter if you throw or drop the ball. The ball gains momentum as it accelerate towards the earth, and the earth gains the very same momentum as it accelerate towards the ball. The earths acceleration is however very tiny, but it's there.
Vidar
The earth doesn't haver any momentum because its linear speed equals 0.
The momentum of the ball is conserved, this is why it reflects back with the same momentum it had before the impact less losses.

Place two masses in deep space, the only gravitational attraction is from each other.
One of the masses is ten kilograms and the other is one kilogram.
From Newton's Third Law we know that the mutual attraction is equal in both directions.
From F = ma we know that the acceleration of the one kilogram will be ten times greater than the acceleration of the 10 kilograms.
After a period of time the one kilogram will be moving 10 times faster than the 10 kilograms. When the one kilogram is moving one meter per second the 10 kilograms will be moving .1m/sec.
Then ½ *10kg *.1 m/sec * .1 m/sec = .05 joules
And ½ * 1 kg * 1 m/sec* 1 m/sec = .5 joules
Energy is not conserved.
If the 2 gram ball had another source for it's velocity the acceleration of the two spheres would not be interdependent and therefor they would not have equal momentum.

Place two masses in deep space, the only gravitational attraction is from each other.
One of the masses is ten kilograms and the other is one kilogram.
From Newton's Third Law we know that the mutual attraction is equal in both directions.
From F = ma we know that the acceleration of the one kilogram will be ten times greater than the acceleration of the 10 kilograms.
After a period of time the one kilogram will be moving 10 times faster than the 10 kilograms. When the one kilogram is moving one meter per second the 10 kilograms will be moving .1m/sec.
Then ½ *10kg *.1 m/sec * .1 m/sec = .05 joules
And ½ * 1 kg * 1 m/sec* 1 m/sec = .5 joules
Energy is not conserved.
If the 2 gram ball had another source for it's velocity the acceleration of the two spheres would not be interdependent and therefor they would not have equal momentum.
Energy conservation does not mean that the two balls must have the same energy. "Conservation" of some quantity in Physics means that the value of the quantity at some time t1 is the same as the value as another time, t2. If you calculate the total energy of the system when it starts moving and at a later time, they will have the same energy. You need to consider potential energy of the system and the sum of the kinetic energies to get the total mechanical energy.
Vidar

The energy of the two different mass objects is not the same; after an application of the same quantity of force for the same quantity of time. This deep space analogy is nearly identical to the tether of the cylinder and spheres. You should not expect equal changes in energy; you should expect equal changes in momentum on the two ends of the tether.

Energy of the system will be conserved no matter what, so does the momentum. Rethink your eksperiment, and figure out where the misconception is hiding.
Practical experiments are useless unless you do accurate measurements. I don't think you have done accurate measurements, but assume that your theory is legit. I have learned that if the practical experiment doesn't fit the theory, the theory is incorrect. Not the other way around.
Vidar

The cylinder and spheres with a mass ratio of 10 to 1 would require 12.6 frames to cross the 20 mm black square for energy conservation; but only 4 frames for Newtonian momentum conservation. I measure four frames to cross at the beginning and the end.
This four frames is the fastest the cylinder rotates. This entire experiment only takes 48 frames. It takes about 24 frames to go from 4 frames to cross the 20 mm black square to the cylinder being stopped. It then it takes another 24 frames to restore maximum rotation of the cylinder. From the stop: it only takes about 8 frames for the cylinder to be rotating faster than 2 mm per frame (10 frames to cross the 20 mm black square). And then you still have 16 more frames to move past the 2 mm / frame speed. You have 16 more frames that are progressively moving faster and faster past the maximum speed for energy conservation. The possibility of the cylinder rotating as slowly as required of energy conservation is zero. There is no way to mistake 4 frames to cross for 12.6 frames to cross.
You might consider that Newton is right.
And no; Newtonian momentum and energy can not both be conserved; it is either 4 frames to cross or it is 12.6 frames to cross. This is 5 mm per frame or 1.6 mm per frame.

Whats frames are you talking about. Maybe I've missed out a video?
Vidar

My middle range high speed camcorder takes 240 frames per second. The software has an application that slows the video and then allows you to advance frame by frame. The videos have to be slowed down and looked at frame by frame: because at normal speed you can't see any thing definitive.
The cylinder and spheres experiment that I am investigating at this time is the 10 (total) to 1 (spheres) mass ratio. I am varying the tether length: the last photo posted had a circumference length tether; the tether length I am using right now is a ½ circumference length.
The spheres and cylinder is released from the hand at about frame 1:03:166.
The black square on the cylinder crosses from side to side from frame 1:03:162 to 166. This is 4/240th of a second.
Just before the cylinder hits the floor the black square on the cylinder crosses from side to side from frame 1:03:231 to 235.
The end rotational speed is the same as the initial rotational speed; and the spheres are up near the cylinder as in the start.
At 1:03:189 the cylinder makes its first stop. The cylinder stops it clockwise rotation and it is sent backward (counterclockwise). This is because the cylinder is stopped before the spheres reach 90° to tangent. The spheres reach 90° to tangent at about 1:03:202 At this point the counterclockwise rotation begins to slow and the cylinder stops again at 1:03:214. From frame 214 onward the cylinder is accelerating clockwise.
By frame 232 the cylinder is back up to its original rotational speed and it is moving in the same direction; clockwise. Even though the cylinder spent some time moving in the opposite direction the final speed is not altered.
NASA predicts that the spheres conserve energy: but this is impossible because they would only have 31.6 percent (½ * 10 kg * 1 m/sec *1 m/sec = ½ *1 kg * 3.16 m/sec * 3.16 m/sec ) of the momentum necessary to return all the motion back to the cylinder. When the spheres have all the motion they are actually moving 10 times as fast as at the original speed: because their momentum is sufficient to restore all the motion back to the cylinder.
If I reduce the tether length a little more the spheres will not stop the cylinder before 90° to tangent. There would be no backward motion.
When the correct tether length is selected the spheres will be at 90° to tangent just as the cylinder is stopped. There would be only one stop. I have done this a few time before: but it is still kind of fun. Back to the lab.
This is much like the video presented on page one. I don't make the slow motion videos myself; it is done by a friend. I don't want to make too many; he might get weary of me asking.

A perfect stop (for the 10 to one mass ratio cylinder and spheres) appears to occur at a tether length of about .834 diameters of the cylinder. The backward motion of the cylinder is gone; it now stops and precedes forward. It is still four frames to cross the black square at beginning and end.
The cylinder stop is also dead center of the complete experiment. The tether at 90° to tangent occurs about half way between the two four frames (start and finish) velocities.

Oops; I forgot to add the radius of the sphere (center of mass), which is 12.75 mm. That would make the tether length .978 diameters instead of .834.

In one of the runs of the 10 to 1 cylinder and spheres it took four frames to cross the black square at the start and four frames to cross at the finish of the experiment. In the middle is a dead stop of the rotation of the cylinder.
Energy conservation predicts that at the end of the experiment it will take 12.6 frames to cross the black square. Because for energy conservation; when the spheres have all the motion, the spheres would have to be moving at the 'square root of ten' instead of ten times faster than the starting motion of the cylinder and spheres. Four frames time 3.16 is 12.6 frames.
Well I was curious: I found the middle frame, of the videoed experiment, where the cylinder is stopped. There are actually about five frames where the cylinder does not appear to move; but I picked the center of these five frames. From this middle frame I clicked off 12 more frames. By the end of the 12th frame one black square had been crossed from side to side. So, from the stop, the average velocity is at least 1 square per 12 frames. That makes the final velocity at least 2 squares per these 12 frames; final velocity is roughly double average velocity if you start from a stop. This is twice as fast as predicted for energy conservation; and this is starting from a stop. These 12 frames start at the slowest portion of the experiment for the cylinder.
In the next twelve frames (that would be 13 24 frames from the stop) there were 2.5 black squares crossed in 12 frames. Now the average speed is 2.5 times faster than the max expected for energy conservation.
In a linear acceleration the final speed would be double the average speed of 2.5 squares; for 5 black squares per 12 frames (1.67 per 4 frames). The graph of this acceleration is not linear because it troughs out at 4 frames per crossing and there are several frames that have almost zero motion. The graph of this acceleration would probably be more like a section of a sine curve.
The point is that 12.6 frames to cross is way too slow to be the correct answer; and energy conservation is eliminated as a possibility. The direct measurement of four frames for momentum conservation fits perfectly.
This troughing out, or little velocity change for several frames on the graph, is logically expected. When the cylinder is stopped in this experiment and the tether is at 90° to tangent little (or no) rotational force can be applied by the tether. And at the end of the experiment the spheres and cylinder are moving at the same speed; the spheres on the tangent tether are moving at the same speed as the rotation of the cylinder. Both these arrangement cause acceleration to cease.

I built 2 more cylinder and spheres machines; I converted the 10 units of total mass to one unit of sphere mass model. I made a 20 to one and a 30 to one by adding 1320 grams and then 1508 additional grams. The 1508 was stainless steel rods so they were left too massive.
I did a careful mass to diameter evaluation of the three cylinder and spheres models. The spheres are assumed to be point masses of 132 grams at a diameter of 114.3 mm; compared to the cylinder diameter of 88.9 mm; and so on for all the other mass. The greater diameter gives you greater speed; and four sets of mass are at different diameters.
So the 10 to one is actually a 9.78 to one.
And the 20 to one is actually a 19.33 to one.
And the 30 to one is actually a 32.87 to one.
I wanted to compare the tether length that causes a perfect stop of the cylinder at full extension; when the tether is at 90° to tangent. I used the tether length that actually touches the cylinder.
I got 72.4 mm of tether for the 9.78 to one.
I got 146.4 mm of tether for the 19.33 to one.
I got 239.7 mm of tether for the 32.87 to one.
There appears to be a one to one relationship between the length of the tether and the mass the tether is able to stop.
I went back to the 4.5 to one total mass to sphere mass, cylinder and spheres, and shortened the length of the tether for a perfect stop; with this one to one relationship in mind. The 4.5 to one is actually 4.55: after the above diameter evaluation. So 72.4 mm tether length times 4.55/9.78 mass ratios would give you about 34 mm tether length.
I found that a tether length of 32.5 mm, for the portion of the tether that actually touches the cylinder, had a perfect stop at 90° to tangent. This remains within the 5% error ranger.
This seems to confirm the one to one relationship between the tether length and the cylinder mass stopped; but lets put tether length in radial length of the cylinder.
A tether length of .731 radii stops 4.55 to one. 4.55/.366 = 6.22
A tether length of 1.629 radii stops 9.78 to one. 9.78/ 1.629 = 6.00
A tether length of 3.294 radii stops 19.33 to one. 19.33/3.294 = 5.87
A tether length of 5.39 radii stops 32.87 to one. 32.87/5.39 = 6.1
So if you want to have a perfect stop at 90° to tangent: for a cylinder that has a mass of 23 times that of the spheres you would use a tether length of about 4 radii. This is for any size cylinder. And the tether length is from the cylinder surface.
All lengths can restart the cylinder to the full initial speed.

Or is the relationship 2 * pi? And why?

I used a larger diameter (129 mm) cylinder and spheres that was a 16.36 to one mass ratio. I used the new formula for determining tether length. That is (mass ratio) / (2 * pi) * r = tether length for a perfect stop at 90° to tangent.
That is 16.36 / 6.28 = 2.60 * 64.5 mm = 168 mm for tether length
By using this length of tether I got a perfect stop when the spheres are at 90° to tangent; first try. Awesome!

In these models: an additional tether length of 15.9% of a radius (of the cylinder) will stop an additional 132 grams of cylinder. It does not matter if the 15.9% is added to a 1 radius length tether; or if the 15.9% is added to a 6 radius tether length.
Each 1/6.28th of a radius tether length stops 132 grams in these models. This is 15.9% of a length of radius of the cylinder. Six radius lengths of tether stop 4973 grams; 6.159 radius lengths of tether stop (4973 +132) 5105 gram. One radius lengths of tether stops (1 r * 6.28 * 132g) 829 grams; 1.159 radius length of tether stop (829 +132) 961 gram.
NASA predicts that extra length added to a longer tether has a greater ability to stop larger quantities of mass: wiki “and their (sphere mass) effect grows as the square of the length of the cables.” But this is not true. The same added length of tether stops the same added mass no matter what the length of tether.
It should be noted that this is true of the restart as well. An additional tether length of 15.9% restarts another 132 grams. When the tether has twelve units of 15.9% of the radius it restarts a 1584 (12 * 132 g total mass) gram cylinder. The thirteenth unit of 15.9% of radius length, added to the tether, will restart an additional 132 grams.
For the 129 mm diameter model this 15.9% is 10.27 mm; for the 88.89 mm diameter model this 15.9% of the radius would be 7.08 mm.
The tether length to mass stopped (at 90° to tangent) is a linear relationship not a square relationship.

In these models: an additional tether length of 15.9% of a radius (of the cylinder) will stop an additional 132 grams of cylinder. It does not matter if the 15.9% is added to a 1 radius length tether; or if the 15.9% is added to a 6 radius tether length.
Each 1/6.28th of a radius tether length stops 132 grams in these models. This is 15.9% of a length of radius of the cylinder. Six radius lengths of tether stop 4973 grams; 6.159 radius lengths of tether stop (4973 +132) 5105 gram. One radius lengths of tether stops (1 r * 6.28 * 132g) 829 grams; 1.159 radius length of tether stop (829 +132) 961 gram.
NASA predicts that extra length added to a longer tether has a greater ability to stop larger quantities of mass: wiki “and their (sphere mass) effect grows as the square of the length of the cables.” But this is not true. The same added length of tether stops the same added mass no matter what the length of tether.
It should be noted that this is true of the restart as well. An additional tether length of 15.9% restarts another 132 grams. When the tether has twelve units of 15.9% of the radius it restarts a 1584 (12 * 132 g total mass) gram cylinder. The thirteenth unit of 15.9% of radius length, added to the tether, will restart an additional 132 grams.
For the 129 mm diameter model this 15.9% is 10.27 mm; for the 88.89 mm diameter model this 15.9% of the radius would be 7.08 mm.
The tether length to mass stopped (at 90° to tangent) is a linear relationship not a square relationship.
I think there are two types of action  when the force is momentary vs the force is
constant.
The work calculations are not applicable to the momentary force action,
but they are applicable to the constant force action.
But by sending the projectile up against the gravity with a momentary force,
it becomes possible to calculate work on the way down, when the gravity force is constant.

I thought I was loading a smaller picture; how do you get the big one back off?

I think you have the concept; but I prefer to look at it as a function of time; telecom.
A one kilogram missile moving 19.81m/sec will rise 20 meters. From d = ½ v²/a. This will take 2.019 second. From d = ½ at².
A kilogram mass applies 9.81 newtons of force. This is 9.81 newton applied for 2.019275 seconds. = 19.81 N seconds; N*s
Form a chain of twenty one kilogram masses; they are one meter apart, 20 meters high. This vertical chain could be dropped one meter. The original configuration can be restored if only one kilogram is raised 20 meters.
Connect this 20 kilogram chain to a 80 kilogram flywheel. It will take 1.009637 seconds for the chain to drop one meter while it spins the wheel. But this is 20 kilograms dropping one meter for (20 kg * 9.81 N/kg) 196.2 N applied for 1.009637 seconds for 198.09 N seconds.
The output momentum is ten times that of the input momentum.
The output momentum is 100 kilograms moving 1.9809 m/sec for 198.09 kg*m/sec.
The input momentum was 1 kg moving 19.81 m/sec for 19.81 kg*m/sec.
The input energy is 196.2 joules.
The output energy is 19,620 joules.
In the 20 kilogram; twenty meter; vertical chain arrangement one kilogram applies its 9.81 newtons for 1.0096 seconds 20 times before it needs to be returned to its original vertical position at the top of the chain. The other 19 kilograms in the chain always assist in the application of force but each one kilogram applies its 9.81 N for 1.0096 seconds 20 times.
The descending mass is 9.81 N applied for 20.19274 seconds. This is 198.09 Ns.
The ascending mass is only 9.81 N applied for 2.0193 second for 19.809 Ns.
This time difference multiplies the momentum by 10 times and the energy by 100 times.

I think you have the concept; but I prefer to look at it as a function of time; telecom.
A one kilogram missile moving 19.81m/sec will rise 20 meters. From d = ½ v²/a. This will take 2.019 second. From d = ½ at².
A kilogram mass applies 9.81 newtons of force. This is 9.81 newton applied for 2.019275 seconds. = 19.81 N seconds; N*s
Form a chain of twenty one kilogram masses; they are one meter apart, 20 meters high. This vertical chain could be dropped one meter. The original configuration can be restored if only one kilogram is raised 20 meters.
Connect this 20 kilogram chain to a 80 kilogram flywheel. It will take 1.009637 seconds for the chain to drop one meter while it spins the wheel. But this is 20 kilograms dropping one meter for (20 kg * 9.81 N/kg) 196.2 N applied for 1.009637 seconds for 198.09 N seconds.
The output momentum is ten times that of the input momentum.
The output momentum is 100 kilograms moving 1.9809 m/sec for 198.09 kg*m/sec.
The input momentum was 1 kg moving 19.81 m/sec for 19.81 kg*m/sec.
The input energy is 196.2 joules.
The output energy is 19,620 joules.
In the 20 kilogram; twenty meter; vertical chain arrangement one kilogram applies its 9.81 newtons for 1.0096 seconds 20 times before it needs to be returned to its original vertical position at the top of the chain. The other 19 kilograms in the chain always assist in the application of force but each one kilogram applies its 9.81 N for 1.0096 seconds 20 times.
The descending mass is 9.81 N applied for 20.19274 seconds. This is 198.09 Ns.
The ascending mass is only 9.81 N applied for 2.0193 second for 19.809 Ns.
This time difference multiplies the momentum by 10 times and the energy by 100 times.
But doesn't it require each of the masses to be lifted 1 m for the original position?
Or I'm missing something?
Regards

Stack 20 dominoes on top of each other on their sides. Hold the second domino from the bottom and slide out the bottom domino. Place the removed domino on the top of the stack.
We now have the starting position; of 20 dominoes that are at a distance of one domino width from the table. We can lower the stack of twenty dominoes by one domino width so that the stack again rests on the table. We can remove the bottom domino again; and place it on the top of the stack. So by raising one domino twenty domino widths we can continually restore the original configuration. (This is quite a trick with twenty; but works easily with six)
All the one kilogram masses in the chain drop one meter; but only one needs to be raised 20 meters to restore the driving chain (stack) to its original configuration. It takes a whole lot less momentum to raise one kg twenty meters than to raise 20 kg one meter. It takes 19.81 (19.81 m/sec * 1 kg) units of momentum to raise one kilogram 20 meters: and 88.59 (4.429 m/sec * 20 kg) units of momentum to raise 20 kilograms one meter. 19.81 / 88.59 = 22.36%
You are actually using 19.81 units of momentum to make 88.59. The momentum increase is 88.59 / 19.81 447%. When you transfer all the motion of the 88.59 units of momentum to one kilogram you get (½ * 1 kg * 88.59 m/sec * 88.59 m/sec) 3924 joules of energy: for an energy increase from 196.2 J to 3924 joules; 2,000%.
If you attach the chain to a 80 kilogram flywheel you are then using 19.81 units of momentum to make 198.1.

Stack 20 dominoes on top of each other on their sides. Hold the second domino from the bottom and slide out the bottom domino. Place the removed domino on the top of the stack.
We now have the starting position; of 20 dominoes that are at a distance of one domino width from the table. We can lower the stack of twenty dominoes by one domino width so that the stack again rests on the table. We can remove the bottom domino again; and place it on the top of the stack. So by raising one domino twenty domino widths we can continually restore the original configuration. (This is quite a trick with twenty; but works easily with six)
All the one kilogram masses in the chain drop one meter; but only one needs to be raised 20 meters to restore the driving chain (stack) to its original configuration. It takes a whole lot less momentum to raise one kg twenty meters than to raise 20 kg one meter. It takes 19.81 (19.81 m/sec * 1 kg) units of momentum to raise one kilogram 20 meters: and 88.59 (4.429 m/sec * 20 kg) units of momentum to raise 20 kilograms one meter. 19.81 / 88.59 = 22.36%
You are actually using 19.81 units of momentum to make 88.59. The momentum increase is 88.59 / 19.81 447%. When you transfer all the motion of the 88.59 units of momentum to one kilogram you get (½ * 1 kg * 88.59 m/sec * 88.59 m/sec) 3924 joules of energy: for an energy increase from 196.2 J to 3924 joules; 2,000%.
If you attach the chain to a 80 kilogram flywheel you are then using 19.81 units of momentum to make 198.1.
I think I finally understood this, but this looks like a perpetual energy making machine to me,
unless I'm hallucinating!
Where is the catch then???

Yes: that is what it is; 'a perpetual energy making machine'

The picture is of another picture off of a monitor; sorry, rather fuzzy. I am just trying to figure this picture stuff out. It is a picture of the cylinder; rotationally stopped.
I did learn how to make my own slow motion videos.

https://www.youtube.com/watch?v=boLk57cKNao

https://www.youtube.com/watch?v=KgT70vSIUgA
It is actually impossible that Newton can be wrong in this energy source:
I have one inch spheres that have a mass of 152 grams, they are tungsten spheres and should weight a little more. But I will do some math with what is real. (152 grams)
If I accelerate one of these spheres to 20 m/sec it will rise 20.387 meters. This 20.387 m is 802.6 inches.
This means I can stack 802 of the 152 grams spheres on top of each other. This is 121.9 kilograms. And I can drop the entire stack one inch (2.54 cm).
When the stack is dropped 2.54 cm it will have a final velocity of: the square root of (.0254 m * 2 * 9.81 m/sec/sec) = .7059 m/sec. This is a momentum of 121.9 kg * .7059 m/sec = 86.054 units
The sphere needs only .152 g * 20 m/sec = 3.04 of these 86 units to travel back up to the top and reconfigure the predrop arrangement.
The 121.9 kilograms is equal to a rim of the same mass; and moving at the same speed of .7059 m/sec around the arch of the circle.
121.9 kg * .7059 m/sec = 86
Or it could be a 60.95 kilogram rim moving 1.4118 m/sec around the arch of the circle.
Or 30.475 kg moving 2.836 m/sec.
Or 15.2375 kg moving 5.647 m/sec.
Or 7.615 kg moving 11.29 m/sec.
Or 3.809 kg moving 22.588 m/sec. = 86.046 units; this 22.588 m/sec will send 3.8 kilograms up 26.0 meters.
When this 22.588 m/sec for 3.8 kilograms is given to .152 kilograms it will send it to the next county.
You would hear a whir and you would never see the sphere again; if you were lucky enough not to get hit.

https://www.youtube.com/watch?v=dHxdmKmAfl8
It takes (19 frames) to go from 1.2 m/sec of rotation of the cylinder and spheres to the first stop of the cylinder. It takes the same amount of time (19 frames) to go from the last stop, of the cylinder's rotation, to the last full return of the cylinder's rotational motion. This motion appears to be the same 1.2 m/sec (from counting the frames needed to cross the black square). The 19 frames confirms that the rotational velocity is indeed the same. The accelerations are of the same magnitude (19 frames) therefore linear Newtonian momentum is conserved.
It would be moving about a fourth that fast (83 frames) if energy was conserved and the alleged heat was lost. It would take 83 frame to go from the last stop to the last restart instead of 19 frames. The entire experiment is only 75 frames.
A mass moving on the end of a string can wrap around a stationary post. The string will become shorter and the radius will be reduced; but the linear Newtonian momentum will remain the same.
In this (two 86 gram bolts) cylinder and sphere experiment the total mass of a spinning object is reduced from 1448g (cylinder and spheres) to 304g (spheres); but the linear Newtonian momentum remains the same.
In this (two 86 gram bolts taped to the cylinder) cylinder and sphere experiment the energy increases from (.5 * 1.448 kg * 1.2 m/sec 1.2 m/sec) = 1.0425 joules to (.5 * .304 kg * 5.716 m/sec * 5.716 m/sec) = 4.966 joules: but the linear Newtonian momentum remains the same.
This is an unlimited source of free energy. Because you can transform a 400 kilogram rim moving 1 m/sec into a 1 kilogram rim moving 400 m/sec; for an increase from 200 J to 80,000 J.

I made a 10 (1296g) to 1 (132 g) cylinder and spheres and it follows the same pattern as the 4.5 to 1 and others cylinder and spheres. It takes 25 frames to stop the cylinder's spin and it takes 25 frames to fully restart the spin of the cylinder. There are three frames needed to cross the black square at the release; and it takes three frames to cross from one side of the black square to the other side after 50 frames (after a full stop and a full restart).
If energy were conserved when the cylinder was stopped it would only have one third of the linear Newtonian momentum needed to return the cylinder to full rotation. That means it would take 75 frames to return the less than one third of the motion. It takes 25.
If energy were conserved when the cylinder was stopped it would only have one third of the linear Newtonian momentum needed to return the cylinder to full rotation. That means it would take 9.5 frames to cross the black square from side to side after the restart. It takes 3.
The Linear Newtonian momentum formula (mv) would be satisfied with a velocity increase of 10. This is an energy increase to 1000%.
The kinetic energy formula (½ * m * v * v) would be satisfied with a velocity increase of the square root of ten: 3.16. Ballistic pendulum experiments prove that only Linear Newtonian Momentum is conserved as the small mass spheres collide with the cylinder; kinetic energy is never conserved.
The cylinder and spheres event keeps its Linear Newtonian Momentum.
This means that a 400 kilogram rim moving 1 m/sec would throw off weighted strings of 40 kilograms moving 10 m/sec. This 40 kilograms could throw off weighted strings of 4 kilograms moving 100 m/sec. This 4 kilograms could throw off 1 kilogram moving 400 m/sec. Now we have ½ * 1 kg * 400 m/sec * 400 m/sec = 80,000 joule and you started with 200 joules.

Is the only way to capture this excess energy by sending the projectile upward?
And then using it as a potential energy?

Is the only way to capture this excess energy by sending the projectile upward?
And then using it as a potential energy?
Just have in mind that the potential energy is not more than the energy you put in to "create" this potential energy.
If you throw a ball upwards, its mass acceleration while the ball is still in the hand, require a given amount of energy to achieve a given velocity.
This velocity is the reason why the ball reach a given hight. The mass is the same all the way, and the velocity of the ball when it comes back and hits your hand, is the same, or less due to air resistance.
So no gain in potential energy.
Vidar

Yes, but there is a gain in kinetic energy which is transferred into potential.
How are you going to harvest kinetic energy?

Telecom's question: How are you going to harvest kinetic energy?
I am going to let the technology used in hydroelectric plants harvest the kinetic energy.
Lets throw a 147 gram ball up into the air 20 meters. That will require a velocity of (d = ½ v²/a) 19.81 m/sec. This is (1/2mv²) 28.83 joules.
We will wrap a 20 meter string around a (147 gram *39) 5733 gram rim mass wheel. We will attach the ball to the string and drop the ball 20 meters.
The ball and rim will accelerate at (9.81m/sec/sec / 40) .24525 m/sec/sec.
The ball and rim will have a final velocity (after the ball has dropped 20 meters) of (d = ½ v²/a) 3.13 m/sec.
This is 5880 grams moving 3.13 m/sec. This is 18.4 units of momentum.
We will now transfer this 18.4 units of momentum to the 147 gram ball by using the cylinder and spheres.
The ball will now be moving 125 m/sec.
This is (1/2mv²) 1151 joules of energy; and you started with 28.8 joules.
And it is better to stack the balls and drop the stack of 20 balls.
The mass of 147 grams is about that of a baseball; so this can be, and is, applied to a real experiment.
At Sault St. Marie the Saint Marys River drops 2000 metric ton of water 7 meters every second. This is 30 MW
So we will increase the drop to double the drop (50 meters) at Niagara: 100 meters.
You could drop 2000 metric tons of mass with a head of 200,000 metric tons every second.
Less than 3% of the energy produced would be needed to reload the system and bring one 2000 metric ton unit back to the 100 meter top. This would generate at least 500 MW.

There a three formulas that describe motion. Modern physics claims that all three formulas are conserved quantities. It is claimed that all three are simultaneously conserved. A mathematical evaluation of the three formulas would quickly reveal that they cannot all be conserved: the evaluation would show that only one of the three formulas could be conserved.
The formulas are mv: 1/2mv²: and mvr. These are linear Newtonian momentum: Kinetic Energy: and Angular Momentum.
So in a closed system where we have motion interactions all three formulas would allegedly remain the same.
First let’s take the ballistic pendulum experiments. A 1 kilogram mass moving 50 m/sec collides with a 19 kilogram mass at rest. The resultant motion is a 20 kilogram mass moving 2.5 m/sec. The mv is conserved. The 1/2mv² loses 95% of the physical motion to unrecoverable heat. Now heat is considered motion but a return to the original condition of 1 kilogram moving 50 m/sec would expose this heat content to be a myth. There is no mechanism for recovery of the heat. The cylinder and spheres proves that the yoyo despin can be returned to its original state; of the slowly spinning satellite. This totally eliminates 1/2mv² as a conserved quantity. A number is rarely equal to its square.
The mvr can be exposed by interrupting the string of a rotating mass on the end of a string. Rotate a soft ball on the end of a string, have someone interrupt the string somewhere down its length. As the ball begins rotating in the smaller circle angular momentum is decreased; linear Newtonian momentum remains the same. Mutilating a conserved number by different r's will not give you the same conserved quantity.
Of the three laws only one is conserved; mv.

https://www.youtube.com/watch?v=oeG7RcSodn8
The mass difference, of cylinder to sphere, is only about 3 to 1.There are not two spheres only one. Half the motion is on the other side of the center of mass still in the mass of the cylinder. The motion is complex.
All the motion comes from gravitational potential energy. The sphere’s tether gets very long, but the sphere is moving very fast.
If angular momentum is conserved the arc velocity of the sphere would become only .375 of the original speed of the rotating cylinder; because of the long radius. It is obvious that the sphere velocity is much higher. And the gravitationally source is not at the center of rotation: as is the situation in space.
For kinetic energy to be conserved the top velocity of the sphere would have an increase from only 1 to 1.73.
If momentum were to be conserved the velocity of the sphere would have an increase from 1 to 3. These higher speeds seem more apparent; especially since the sphere can stop and lift a falling cylinder.
One might ask why the sphere does not return the cylinder to the top starting position. Well the cylinder’s spinning first has to be stopped and that takes time. It takes time for the sphere to restart the spin of the cylinder in the opposite direction. And it would take time to bring the cylinder back up to another stopped position at the top. Going from stop to stop would take a ton of time. And in all of this time the cylinder is under gravitational acceleration. It is amazing that the single sphere gets the cylinder as far back up as it does.
Sorry: that the release position is not in view; it starts just above the viewing area. There are two strings on the cylinder and the sphere is in the center.

https://www.youtube.com/watch?v=oeG7RcSodn8
The mass difference, of cylinder to sphere, is only about 3 to 1.There are not two spheres only one. Half the motion is on the other side of the center of mass still in the mass of the cylinder. The motion is complex.
All the motion comes from gravitational potential energy. The sphere’s tether gets very long, but the sphere is moving very fast.
If angular momentum is conserved the arc velocity of the sphere would become only .375 of the original speed of the rotating cylinder; because of the long radius. It is obvious that the sphere velocity is much higher. And the gravitationally source is not at the center of rotation: as is the situation in space.
For kinetic energy to be conserved the top velocity of the sphere would have an increase from only 1 to 1.73.
If momentum were to be conserved the velocity of the sphere would have an increase from 1 to 3. These higher speeds seem more apparent; especially since the sphere can stop and lift a falling cylinder.
One might ask why the sphere does not return the cylinder to the top starting position. Well the cylinder’s spinning first has to be stopped and that takes time. It takes time for the sphere to restart the spin of the cylinder in the opposite direction. And it would take time to bring the cylinder back up to another stopped position at the top. Going from stop to stop would take a ton of time. And in all of this time the cylinder is under gravitational acceleration. It is amazing that the single sphere gets the cylinder as far back up as it does.
Sorry: that the release position is not in view; it starts just above the viewing area. There are two strings on the cylinder and the sphere is in the center.
I think these kinetic, gravitic, whatevertic ideas are a good thought experiment, but when you have your "analogy" figured out with the balls and strings then you want to convert that idea to something that is not diminished by air drag or friction. Humans think you want to go bigger and bigger, but the correct answer might be to go smaller and smaller. maybe even to molecular levels. If you got problems in transfering the energy back from molecular level then just heat water with it and use a turbine.
Even electricity has inertia so if your idea is kinetic, you might be able to do a solid state version of it.

https://www.youtube.com/watch?v=8Q7L2BOYkjE
The reason this works is because you can apply the same quantity of force for different periods of time. You can apply the force for a short period of time on the side going up; and on the side going down you can apply the same force for a long period of time. This process will produce massive quantities of energy.
By throwing a mass up fast; you can pass units of distance that result in only minimal loss of momentum. By changing the arrangement of the applied force (from the same mass); these passed units of distance can give you larger units of momentum on the way back down. Momentum is a function of time.
An example for a one kilogram mass: The first meter of an upward throw of 100 meters only costs you .222 units of momentum. The same meter of drop on the way back down can give you 4.429 units of momentum. The speed of the upward throw of 100 meters; changes from only (square root of (100 m*2*9.81m/sec/sec) 44.294 m/sec to 44.0724 m/sec in the first meter up. But you get 4.429 units of momentum on each individual meter on the way back down. You gain (4.429 .222) 4.207 units of momentum.
It takes .4515 sec to drop one meter: that means you have applied the force for .4515 second.
If you are moving up at 100 meter per second you can cross 19 meters in .4515 second.
On the way back down ‘each’ of these 19 units of distance can be crossed in .4515 seconds; this is 18 more units of time from the same distance of 19 meters. Gravity does not make you pay for these 18 extra units of ‘time’ over which the force is applied.
In the kilowatt hour you are paying for each unit of time ‘of the applied force’; you won’t get 18 free units. Gravity will apply the force for free.

https://www.youtube.com/watch?v=8Q7L2BOYkjE (https://www.youtube.com/watch?v=8Q7L2BOYkjE)
What were the parameters of this experiment, the mass of the tube and ball? How come did the the tube start rotating when you let it go, was it weighted on one side? Did you calculate the amount of energy that was transferred between the two?

Do not repeat this experiment.
Even after the sphere careens off of a wellpadded floor; it then snaps a knot of a 65 pound test fluorocarbon fishing line; and then bounces off of a cardboard box and crosses back (above the viewing area) to the other side the room and hits the chair I had been sitting in. This one might be unsafe; and all of the experiments have some danger. The video is 1/8th speed.
We can predict the velocity of the sphere from the other experiments.
In several of the posted experiments the original rotational speed is returned to the cylinder and spheres “combination” at the end of the experiment. This means that the motion has to be maintained throughout the entire length of the experiment.
The ballistic pendulum proves that the small spheres can only give Linear Newtonian Momentum to the larger cylinder and spheres combination. Let’s look at the experiments that start with a spinning combination; that then move to the spheres having all the motion; and then back to a spinning combination. These experiments can only conserve Linear Newtonian Momentum; because half of the experiment is the same as a ballistic pendulum experiment. The motion of the small spheres is being transferred to a larger combined mass of the cylinder and spheres.
So the first half (of these motion conserving experiments) has to conserve Linear Newtonian Momentum as well; because that is where the spheres got there motion. If the spheres can give it back then they must have received it.
This experiment is the first half of the motion returning experiments. And this experiment must conserve Linear Newtonian Momentum. All the Linear Newtonian Momentum of the spinning cylinder must be transferred to the sphere; with the exception of the pendulum movement of the cylinder and air resistance.
The cylinder is suspended from only one side by two strings. This causes the cylinder to rotate, unhindered, counterclockwise. The string of the sphere is wrapped in the other direction. The sphere stays put until the cylinder rotates down to it. The input motion is the spinning cylinder; or the total distance dropped at the point where the cylinder is stopped.
The cylinder starts throwing the sphere as soon as contact is made. It takes a while to transfer all the motion to the sphere; but eventually the sphere has the cylinder stopped. The sphere even lifts the cylinder and counter rotates it a bit.
The cylinder has a mass of approximately 2700 grams; the sphere has a mass of 66 grams; about 40 to 1. If we attribute a fourth of the motion to the swinging cylinder: then that means the sphere is moving about 30 times faster than the cylinder was rotating.
This is a proportional kinetic energy increase of about ½ * 2.7 kg * 1 m/sec * 1 m/sec = 1.35 J; to ½ * .066 kg * 30 m/sec *30 m/sec = 29.7 J about 22 times.
If energy were to be conserved the sphere would be rotating about half as fast as the cylinder; how can you unwrap when you are rotating at .50 the rate of rotation. The sphere would have trouble unwrapping at the speeds required for energy conservation. And the experimenter would not be wearing a helmet for fear of getting thumped.
The experimenter can comfortable catch the spinning cylinder with his left arm; but the sphere could put you in the hospital or the funeral parlor.

If we attribute a fourth of the motion to the swinging cylinder: then that means the sphere is moving about 30 times faster than the cylinder was rotating.
Like you say "IF". however "if" is not based on any data you collected from your experiment. I agree your experiment does show a complete stop of the cylinder but nowhere have you measured the actual speeds involved in this experiment. I also agree with the fact that the total energy input is equal to the drop of the tube to where it comes to a complete stop but again, nowhere do you mention the exact height of where this happens relative to the starting height. You just assume the final speed of the sphere has conserved all the linear momentum of the cylinder and go on to explaining how much energy gain this gives you.
Please share the actual data of this experiment as it's hard to deduce from this video without having some calibration points. Or redo the experiment from a different point of view (looking directly into the tube) and adding a calibration stick perhaps at the front/rear of the view.

I would say that the ballistic pendulum is a solid fact; and that only linear Newtonian momentum is conserved when a small object gives its motion to a larger object.
I would also say that, in several of these experiments, there is a complete restoration of the motion to the cylinder and spheres combination. If half of the experiment conserved linear Newtonian momentum then so also does the other half of the experiment. The same motion is maintained throughout.
The current experiment (a gravitational energy source) shows that these experiments are not condemned to avoid speed. High speeds a obtained easily; so much so, that some experiments are to be avoided.

You haven't answered the question at all, in fact you are encouraging me to not even experiment which is hilarious. Again I ask you, what is the measured final velocity of the sphere, what is the height difference of the dropped tube from where it's dropped to where it comes to a halt? If you do not have these numbers how can you make any statement about energy gain/loss?

Let's see I used your latest video and did some calculations from what I can roughly deduce from the video. It was interesting to see that even with the rough estimates the input energy is pretty much equal to the output energy. According to your statements the sphere has to be moving at much greater velocities for your theory to hold.

I think you have calculated the average speed not the top speed.
As far as claiming the production of energy I will sight Issac Newton. When Newtonian momentum is conserved there is a massive increase in energy.
Please consider that someone (Newton, or Leibnitz, or a faulty use of Kepler) has to be right and someone else has to be wrong. There is going to be only one velocity. There are two or three equations under consideration and the same number will not satisfy more than one of these equations. The three formulas are mv for Newtonian momentum: ½ mv² for kinetic energy: and mvr for angular momentum.
The Dawn Mission is a big throwing device so let’s use it; it is about a 400 to one ratio.
So: We have a 400 kilogram (399 kg cylinder + 1kg spheres) thin walled hollow cylinder. The cylinder is spinning in space with an arc velocity of 1 m/sec around the arc of the circle. It can be stopped by two .5 kilogram spheres.
When the cylinder is stopped: what is the arc velocity of the spheres?
The choices are 400 m/sec. for mv; 20 m/sec. for 1/2mv²; or 16 m/sec. for mvr.
There is the concept of verification; but there is also the concept of elimination. How close do we have to be to 400 to eliminate 20? This is not the introduction of a new concept that needs verification: it is picking which one of the three hundred year old concepts is actually applicable to the experiment.
I have experiments that consistently fell within a 5% error range. In one experiment I would make four runs to get four data points: of the four data points two were sometimes the same thousandth of a second (i.e. .643, .642, .640, .642.) But I would not risk my photo gates on a sphere that is a blur. I do not have the capacity to measure blurs. And this brings us back to 400, 20, and 16.
If in the Dawn Mission experiment we got 30 m/sec would you be content to eliminate 400 m/sec. But in the same way if you got 270 m/sec would you be happy to eliminate 20 m/sec and 16 m/sec?
I know for a fact that the two .5 kilograms spheres can return all the motion back to the 399 kilogram cylinder. You simply leave the spheres attached and all the motion returns to the cylinder. You can not do that with 20 units of momentum. It will take 400.

Lets use Broli's chart to see how much momentum he has lost.
The drop is .5 meters. That is a final velocity of 3.13 m/sec. The square root of (.5 m * 2 * 9.81 m/sec/sec)
This gives us an input momentum of 2.7 kg * 3.13 m/sec = 8.451 units
Broli's chart has the final velocity of the sphere (.066 kg) at 20.56 m/sec.
That will give us a output momentum of 20.56 * .066 = 1.35 units.
That will give us 8.45 – 1.35 = 7.1 units of lost momentum.
Which means that 7.1 / 8.45 = 84% of the momentum is lost.
You can't possibly lose this much momentum. The Law of Conservation of Momentum (Newton's Three Laws of Motion) is the only motion law that actually stands up under real physical experiments.
I think these experiments can be conducted with only 5% loss of momentum; which gives you massive increases in energy.

You seem to be contradicting yourself. I'm gathering data from YOUR experiment. Even if I was accounting for half the velocity the final energy should at least be double of what I can deduce from your video. At around 7 sec the tube stops completely and the sphere makes a half turn from bottom to top, in fact your average speed argument does not hold as the speed stays fairly constant because it took around 11 frames for a 90 degree turn from 6 to 3 o'clock and then another 11 frames from 3 to 12 o'clock. In other words 22 frames for 180 degree turn.
This is all from your own video.

But is there a portion in the 22 that is faster?
Is the cylinder moving a few mm per frame? Put that motion in the sphere and energy is way over.

But is there a portion in the 22 that is faster?
Is the cylinder moving a few mm per frame? Put that motion in the sphere and energy is way over.
You will always argue the contrary even if the slow motion was at 1 million fps, you'll argue the excess velocity will be hidden in between frames. If you cannot analytically show an excess of energy then what's the point of claiming there is one?

It takes four frames to cross from one side of the black square to the other side; at the start and at the finish of the video. In https://www.youtube.com/watch?v=YaUmzekdxTQ If energy produced all that heat it would drop to 16 frames by the end of the video. It would be hard to mistake 4 for 16; energy is not conserved. There is no heat given off when the spheres restart the cylinder. Only Newtonian momentum is conserved.

It seems rather silly to make presumptionous acusations of some arbitrary energy
value, based on a partial (visual) analysis of only a small portion of the interactions
taking place. In short, you have yet to even identify all of your variables,
or present an accurate description of the momentum, or the changes in energy
throughout the system.
To make any definitive evaluation of what is going on, you must find a method of
gathering accurate realtime velocity data.
Of not only the cylinder, but each individual weight, and their support strings.
Who’s tension coefficient must also be taken into account.
https://www.wikihow.com/CalculateTensioninPhysics (https://www.wikihow.com/CalculateTensioninPhysics)
https://en.m.wikipedia.org/wiki/Rotational_energy (https://en.m.wikipedia.org/wiki/Rotational_energy)
consistency of release (not by your hand) is essential.
not just height, and mechanism of release.
But also orientation  as release from different orientations around the circumference
leads to different mathematical results.
there are pages of variables which are being ignored here
some are frivolous, while others can be shown to be rather significant
but they must be identified
https://en.m.wikipedia.org/wiki/Scientific_method (https://en.m.wikipedia.org/wiki/Scientific_method)
Accurate data, controlled variables, unbiased observation
there is a reason these things are important
if you are serious about investigating this device
then set up a ‘controlled’ experiment
and collect real data
Identify your variables, your controls, your hypothesis, and demonstrate
your conclusion, in a way that reflects the results of experiment, rather than
the results of your delusion.
the human self cannot make accurate measurement
because of the way our brains dilute the passage of time.
your physical state determines the length of the ‘second’ as you perceive it.
this experiment requires a tool to gather the data.
some way to accurately measure speed and distance, over time.
Video analysis CAN be a useful tool, but it has its limitations
and should not be considered an ‘accurate’ measurement device.
especially when all of the variables are not considered
i.e.: sample rate, processing speed (and relevant duty cycles),
playback speed, buffering delays, compression ratios,
angle and distances of observational perspective,
this is a simplified shortlist, but you get the idea.
A lot of time an effort has been put into this
by its’ inventor, and by others following along.
it’s time to put away the crayons and put on your lab coats.

We know that kinetic energy is not conserved in motion alone. Kinetic energy allegedly gives off heat when small objects strike large object. This excuse of heat is negated in this experiment because there is no motion loss when a small sphere interact with a larger cylinder (twice).
If energy were conserved when the cylinder gives its motion to the spheres only a fraction of the motion would be contained by the spheres. The motion would not be available to return all the motion back to the cylinder and spheres combo. It would actually take 18 frames to cross the black square in this experiment; if energy were conserved.
The crossing of the black square in four frames at the beginning; middle; and end: is consistent with Newtonian Momentum Conservation. And the energy increase is about 450%. But this 450% is very small compared to other arrangements.

A 305 gram mass moving 5.02 m/sec will combine its motion with a 972 gram mass at rest; the combined mass of 1277 grams will be moving 1.2 m/sec. This is not a debatable statement: it is the Law of Conservation of Momentum. If you know the velocity of the combined mass you then know the velocity of the incoming small mass.
You can count the frames as the black square crosses from side to side (four). By measuring the velocity of the spinning cylinder (1.2 m/sec) we then also know the velocity of the spheres (5.02 m/sec); when they contain all the motion.
The energy of the larger combined mass is .919 joules: the energy of the spheres is 3.85 joules. The energy increase is proportional to the mass difference. 1277 g / 305 g = 3.85 J / .919 J
This event has no mass limit; the sphere could have a mass of 305 metric tons. The mass difference can be very large; the spinning wheel mass could be 30,500 metric tons. A tower of dropping masses can increase the time over which the force acts; the energy increase would be 10,000%. The 305 tons can be throw up ever two seconds; and then output would be measured in megawatthours.
The output: construction cost; and maintenance, would be similar to a hydroelectric plant.
http://hyperphysics.phyastr.gsu.edu/hbase/balpen.html
Note the word 'inaccessible'. The energy is inaccessible; it cannot come back. The motion energy would disappear: it is inaccessible.
But the experiments show that there is no loss of motion. Energy conservation is a false concept.

Accelerate a bar so that it is moving perpendicular to its length. Accelerate the bar to 1 m/sec and then catch it on the end so that it must rotate around one of its ends. When the bar starts rotating on one end the center of mass will continue at the same speed; in this case 1 m/sec.
When the center of mass continues at the same speed the energy of the bar increases to 133%. For energy to remain the same (when the rotation starts) the speed of the center of mass would have to decrease by 13.4%. The fact that the speed of the center of mass does not decrease means that energy is not a conserved quantity.
Momentum conservation always overrules energy conservation in real experiments.

A 305 gram mass moving 5.02 m/sec will combine its motion with a 972 gram mass at rest; the combined mass of 1277 grams will be moving 1.2 m/sec. This is not a debatable statement: it is the Law of Conservation of Momentum. ...
You can't combine the momentum or kinetic energy of separate parts to have that of the whole. The analysis of the law of conservation must be made in the same referential frame, preferably an inertial frame, and the speed applies to the center of mass. Energy/momentum depend on the referential.
https://www.youtube.com/watch?v=8Q7L2BOYkjE
...
I see nothing else than energy and momentum conservation.
What makes you think that the potential energy of the cylinder that is used when falling to feed the movements of the ball would be less than the energy acquired by the ball?

Is the frame an excuse? Is the frame something used by the smart people; to explain why the rest of us don't understand things that are perfectly simple. F = ma is perfectly simple; I see no need for it to be framed.
The problem with Q7L2BOYkjE is that the cylinder is still accelerating toward the ground. But I think we could take a snap shot in time where the cylinder is stopped; both in rotation and in falling: and at that point the sphere has all the momentum and or energy. So what is conserved all the momentum or all the energy; does it have all the mv or all the 1/2mv²? The proof of which one of the two is conserved comes from another experiment: the double despin https://www.youtube.com/watch?reload=9&v=YaUmzekdxTQ
At two points all the rotational motion is in the spheres and (twice) the spheres give all the motion back. Two small masses restart a much larger mass: experiments prove that only linear Newtonian momentum is conserved when small masses give their motion to larger masses.
In the Dawn mission yoyo despin; three kilograms stops the rotational motion of 1420 kilograms. If the initial average speed was one meter per second then the initial momentum around the arc of the circle was 1420 units. When the rotation of the satellite was stopped the spheres had 1420 units of momentum; or 473.3 for each kilogram; requiring a velocity of 473.3 m/sec.
For ½mv² to be conserved the initial energy of 710 joules would require that the three kilograms would have a velocity of 21.755 m/sec. But then the 65.27 (3kg * 21.755 m/sec) units of momentum would have to restore the 1420 units if the weighted cables were left attached; as in the double despin.
By restoring the rotational motion; after the weighted string are left attached, the double despin proves that unlimited quantities of energy can be made from gravitation.

Is the frame an excuse? Is the frame something used by the smart people; to explain why the rest of us don't understand things that are perfectly simple. F = ma is perfectly simple; I see no need for it to be framed.
F = ma is "perfectly simple" but false if you use it outside its domain of application.
It is false when m is not constant, you must use F=dp/dt instead where p is the momentum.
It is false when relativistic effects cannot be neglected (therefore surely when we consider currents of electrons).
It is false when you measure "a" in a referential and F in another one.
It is false in general: you forgot that F and a are vectors, not scalar.
The two last points are the reasons why your analysis is as wrong as your calculations.
It is not enough to keep the simplest equation because we understand it, to make correct calculations. Physics is not "perfectly simple".
The "perfectly simple" principle that the analysis by the lagrangian (https://en.wikipedia.org/wiki/Lagrangian_mechanics) of the system is equivalent to the analysis by the forces rules out the possibility of extra energy for purely logical reasons. Hoping to defeat the laws of physics by using the laws of physics is childish because they are internally consistent, it's like searching for a particular case "n" where n+1 would be different from n+1.
I have a reluctance to let things I consider false be said, and in this case I explain why I consider them false, it can help to get out of ignorance as I myself have benefited from others.
That said, I do not prevent anyone from wanting to live in ignorance or illusion.
I even hope that I will be shown that I am the one who lives in illusion, it is enough that those who show illogicality in their analysis of a mechanical system are nevertheless able to build a working looped machine.
So far, I have read hundreds of their claims, but I have not observed the slightest working machine created by their minds.

F = ma: Acceleration (a) is equal to the change in velocity over the change in time v/t. Therefore: Ft = mv. And mv is a conserved quantity and unabated.
Most of those confused about vectors think a spinning rim has no momentum: because each quantity of mass has an equal quantity of mass on the other side moving in the opposite direction. But in fact all the mass is moving in the same ‘vector’ direction.
A ten kilogram rim spinning at 3 meters per second, around the arc of the circle, has 30 units of momentum. Proof of this statement can be obtained by wrapping the spinning rim with a string weighted with a 10 kilogram mass floated on dry ice.
When the string comes taut the 10 kg rim will jerk the 10 kg on dry ice to a velocity of 1.5 m/sec. The 10 kilograms on dry ice will have 15 units of momentum and the rim will still have an arc velocity of 1.5 m/sec. The 1.5 m/sec is half the original velocity of the rim: so half the original velocity of the rim is equal to 15 units of momentum. So the original momentum of the rim was 30 units of momentum.
Furthermore: 20 kilograms (the rim and the block on ice) moving 1.5 m/sec is 22.5 joules of energy. But the 10 kilogram rim had an original energy of 45 joule. You cannot claim a 50% loss of heat energy because the despin cylinder and spheres machines move back and forth: in this case 22.5 J to 45 J and then 45 J to 22.5 J
So this simple experiment proves that a spinning rim has momentum and that the Law of Conservation of Energy is false.
Actually it can get simpler than the rim. Do you remember what the governor of an antique steam engine looks like? It is two counter balance spheres. If you released one sphere and then the other a half rotation later the true momentum of the two spheres would be seen. But the linear momentum they have after release is the same as the momentum they had before release.
The portion of the machine that produces energy (despin technologies) is the only important part; because the other half of the loop is a chain draped over a pulley.
I have experiments that make energy: that should end all the illusions.

F = ma: Acceleration (a) is equal to the change in velocity over the change in time v/t. Therefore: Ft = mv. And mv is a conserved quantity and unabated.
...
Not right. Mv is not necessarily conserved. An accelerating rocket expels its mass of fuel. mv is not a conserved quantity if you look only at the rocket.
The conservation requires that m is the total mass (current rocket mass + expelled mass), and v the speed of the center of mass of the whole, not the speed of the rocket.
The ball is a moving object relative to the cylinder as the expelled fuel relative to the accelerating rocket. You have to consider the whole system cylinder+ball, to calculate where is the center of mass and its speed, it's the only way to prove that mv wouldn't be conserved, otherwise your conclusions are simply irrelevant.

The chemical energy of the fuel is used as an outside force. This force changes the momentum of the burnt fuel mass; and that change in momentum is equal to the momentum change of the mass of the rocket (Newton’s Third Law of motion). It is never that the kinetic energy change in the exhaust fuel mass is equal to the kinetic energy change in the rocket mass. Energy is not conserved it is always momentum conservation. Yet when NASA predicts the velocity of the thrown masses in the Dawn Mission they predict energy conservation. Apparently NASA is not familiar with the universality of a universal law: Newton’s Third Law.
When the spinning Dawn Mission satellite loses 1420 units of momentum: the thrown masses gain those 1420 units of momentum; Newton’s Third Law.
There is no outside force applied to the despin of the Dawn Mission satellite. But there has to be an internal force within the closed system that changes the spin of the satellite from 48 RPM to zero RPM. That force is in the cable that causes equal changes in the (spinning) momentum of the satellite and the momentum of the thrown masses. Absolute proof that it is Linear Newtonian Momentum Conservation is achieved when the Double despin experiment returns all the spinning motion back to the cylinder.
Ballistic pendulums prove that small object can only deliver their linear Newtonian momentum to a larger object. The bullet energy is not conserved in ballistic pendulum experiments.

Statements not supported by evidence. Misinterpretations of elementary physics, outside of any science.
Incantations of ignorance, as irrefutable as any religious dogma.
I see nothing here to discuss that would be compatible with reason and rationality.

Just experiments that make energy.

What is being ignored in the videos is the conservation of angular momentum. I studied that many years ago in college physics. What I am talking about is the same principle used by figure skaters to increase their spin rate to a high speed and then stop spinning very quickly. A large slow spinning object will increase greatly in rpm if the mass can be brought closer to the center of rotation.
A simple and fun experiment will demonstrate this. Find a chair like an office chair that has good bearings so that it will spin easily. Sit in the chair with your arms and legs extended out as far as you can get them. Now have someone give you a good spin. While spinning, pull your arms and legs in as close as you can to your body. You will feel the chair speed up in rotation. When you extend your arms and legs again you will feel the chair slow back down. If the bearings are really good in the chair you should be able to again retract your arms and legs and again feel the chair speed back up. You can repeat these motions until the drag of the bearings and wind resistance slows you down.
There is no extra energy gained but this is an interesting example of the conservation of angular momentum. I don't remember the formula for calculating it but I am pretty sure there was one we used when doing the experiments.
Carroll

The formula for angular momentum (L) can be found in Wikipedia; L = r mv where r is the radius of rotation (sometimes called the moment arm) and mv is of course Linear momentum.
The formula in Wikipedia is just to the left of the vector diagram window.
This formula can be checked by applying it to the appropriate place for which it was invented: planets comets and other satellites. You take the product of the long radius and the slow speed (mv) of Haley's comet at apogee; and compare it to the product of the smaller radius and fast speed of the comet at perigee; and those numbers will be the same. This is conservation of angular momentum.
The speed change of the comet is caused by gravity. But gravity does not cause this kind of change in the laboratory. Gravity does not increase the speed of the barbells as the student; spinning on the chair; pulls them in. Then why does angular momentum conservation work in the laboratory? Well it doesn't. There will be an increase in rotational speed because the radius of rotation has decrease but the new speed will not fit Kepler's formula for satellites.
You will never see; in common use; an angular momentum conservation experiment conducted where the experimenter carefully measures the changing radii of the rotating mass. An experiment that is done well will be ignored by your professors. And the results scorned by the same professors.
One good example of this is Galileo's pendulum conducted over 300 years ago. Galileo used pins to interrupt the string of a simple pendulum in the down swing position; and the radius of rotation changes. The mv of the long and short pendulum sides is the same at the down swing so the changed radius changes the angular momentum. Thus angular momentum is not conserved.

Let’s look at a despin experiment where the cylinder is stopped. In this experiment the attach spheres are 500 grams each and the cylinder; that is at rotational rest; has a mass of 10 kilograms. The radius of rotation of the spheres are 24 times that of the cylinder (cylinder r = 1.75 inches, 4.445 cm). The spheres are moving 1 m/sec; at a radius of 24 times that of the cylinder (42 inches, 106.68 cm); when the cylinder is stopped.
The original angular momentum is therefore 1.0668. If the spheres were to wind around the cylinder and draw into the same radius then you would have 11 kilograms all moving at v meters per second: so that 11kg * v m/sec * .04445 m = 1.0668. That would leave us with a linear velocity of 2.1818 m/sec. for the entire 11 kilogram system. So angular momentum conservation would require that 1 unit of linear momentum would produce 24 units of linear momentum. This production of 24 units of linear momentum from 1 is a clear impossibility. In any closed system (no application of outside force) the linear momentum always remains the same. Angular momentum conservation does not work in the lab.
Another proof would be to place an immovable post in the middle of a frictionless plane; have a puck on the end of a string wind or unwind from the post. The radius of the rotating puck would be constantly changing but the linear Newtonian velocity of the puck would remain constant. As the puck rotates the radius would shorted (or length if unwinding) for each orbit of the post. An infinite number of radii would multiple the same linear momentum (L = r mv); for an infinite number of angular momentum.

If the student on the chair, with barbells, can reduce the radius of the mass to one half then the rate of rotation doubles. This is the increased motion you see; but the linear momentum has remained the same.

A one meter rod can be made to have a mass of 1 kilogram. One hundred such rods stacked on end on top of each other would have a mass of 100 kilograms. Place this 100 kilograms on one side of a 1900 kilogram Atwood's (950 kilograms on each side of a pulley).
The acceleration rate is 9.81 m/sec/sec / 20 = .4905 m/sec/sec.
When the entire stack of rods is dropped only one meter the final velocity of the whole 2000 kilograms would be; .99045 m/sec (d = ½ v²/a). This would be a momentum of 1980.9 units. The single rod at the bottom of the stack could be sent back up to the top of the stack with the consumption of only 44.29 units of momentum. The other 97.76% of the motion can be used to turn an electric generator.

If the student on the chair, with barbells, can reduce the radius of the mass to one half then the rate of rotation doubles. This is the increased motion you see; but the linear momentum has remained the same.
You are confusing angular momentum with linear momentum. They are not the same. In angular momentum the direction of movement is constantly changing. Linear momentum is in a straight line. You are also incorrect when you state the conservation of angular momentum does not work in the lab. The experiment in the chair proves that is does.

Citfa hi can't help but notice as you appear to be a knowledgeable guy who has experience in the zero point
I feel you might already have the knowledge or even a developed device by your comments any chance you can share
anything of interest?

Citfa hi can't help but notice as you appear to be a knowledgeable guy who has experience in the zero point
I feel you might already have the knowledge or even a developed device by your comments any chance you can share
anything of interest?
I am probably not nearly as knowledgeable as you think. But I do have many years of experience working in electronics and on industrial machine tools so I have learned a few things along the way. I do NOT have an OU device as yet, but I know a couple of guys that are getting close. I believe they probably already have something working.
Here is a link to the research and work they have been doing for several years:
http://www.energeticforum.com/renewableenergy/19774basicfreeenergydevice.html
I also have worked with them on and off over the past 7 or 8 years or so. I know from our research that it is definitely possible to run a load for much longer using the 3 battery generating system than if you just ran the load directly off of the three batteries. At least a dozen or so of us have proven that on the bench. Energy CAN be recycled. But most of the naysayers on that forum and this one insist that is not true. They will not take the time to properly set up and run the tests. You must use reasonable size batteries and keep the current down. The reason for that is that lead acid batteries do not take a charge quickly very well. The larger batteries help because they have a lower internal resistance. This allows them to take the charge easier. But you will see most people insist on using small batteries and larger loads and then complain the system doesn't work.
Now if you couple the 3 battery system with a super efficient generator you can then go above a COP of one. I have not gotten to that point yet. Dave (Turion) has posted publicly that he has a generator that is more than 100% efficient. But he has had problems with it overheating and is currently in a major project of renovating his old house so it can be sold and remodeling his new house so he can move into it. So he has no time for research at this time. He has revealed how to wind the coils to make the generator speed up under load and produce some serious power. As I posted, he is still having problems with overheating that need to be worked on.
Here is another link to the original thread where you can read about all the effort and years of research we did before we finally decided how to best use the 3 battery generating system:
http://www.energeticforum.com/renewableenergy/106103batterygeneratingsystem.html
Carroll

If angular momentum conservation in the lab is such a solid law then why does it not have experiments where they measure the mass; they measure the radii; and confirm the speed by other means.
The ballistic pendulum experiments measures the mass of the bullet (steel ball) and the mass of the receiving block (cup). The experimenter only measure the radius to determine rise; and thy confirm the speed by other means. These are real experiments.
The ballistic pendulum can be two pendulums where the incoming mass is also a pendulum bob (such pendulums do exist in the literature). The combined mass of the steel ball and cup will then swing up to indicate the speed of the combined mass. You could use d =1/2v²/a: d = rise a = 9.81 m/sec/sec to determine initial velocity; energy need not come into the experiment.
The incoming pendulum can be much longer than the cup pendulum. The two pendulum lengths become the same when the incoming pendulum string contacts the top of the cup pendulum.
The ballistic pendulum experiment need not be in a vertical plane: it can be done in a horizontal plane. The speed changes would be exactly the same; but of course the cup on a rod would no longer be a pendulum. But the speed could be determined by a photo gate.
There are a large number of videos on the internet that show ballistic pendulums, most are the modern spring gun and cup types. You will not get to far into the video when they tell you that ballistic pendulums conserve linear momentum. Some time they do not mention the word linear; but mv is linear. They never mention radius in the formula so you know it is linear momentum being conserved.
'Linear momentum' when the combined mass is moving in a circle. These experimenters do not believe that linear means a straight line. Because the final momentum is not in a straight line.
The energy of the steel ball is not conserved; typical 80% or more is lost.
Angular momentum is not even mentioned in ballistic pendulums. You could easily construct a collision between pendulums of different lengths. But this would prove that angular momentum conservation is false (in the lab) so this experiment is never done. The nonexperiments of chairs and skaters has caused a mind block that delayed the advancement of knowledge.

To make free energy from gravity you need to utilize two simple machines. We use the Atwood's https://www.youtube.com/watch?v=4ovhEkSIqV0 to increase the quantity of momentum; this is done by increasing the quantity of time over which the force acts. This momentum is then transferred to a small object by using the despin event as seen in a satellite despin procedure https://www.youtube.com/watch?v=uMQnXig2hrg and in the double despin cylinder and spheres machine https://www.youtube.com/watch?v=YaUmzekdxTQ. Or https://www.youtube.com/watch?v=KgT70vSIUgA
https://www.youtube.com/watch?v=w7d66JscI8

I set up an Atwood's machine that has two pulley radii on the same rotational shaft. The one radius is 10 .035 mm; half of the 20.07 mm shaft diameter. The second radius is a wheel on the shaft with a radius of 97.2 mm. 97.195 mm / 10.035 mm = 9.6856
A 650 gram mass suspended on the shaft could be balanced with 68 grams suspended on the wheel.
I placed 650 grams on both ends of a string and draped it over the shaft. I accelerated this Atwood's arrangement with a certain mass on the shaft and had it trip a photo gate after one rotation. The average trip time of the photo gate was .0433 seconds.
I placed 68 grams on both ends of a string and draped the string over the wheel. I accelerated this same Atwood's with the same certain mass on the shaft and arranged it to trip the same photo gate, in the same manner, after one rotation. The average trip time of the photo gate was .04325 seconds.
This proves that a two radius Atwood's produces an F = ma outcome; as long as the longer radius has a proportionally smaller mass. The 68 grams is moving 9.68 time faster than the 650 grams. They have about the same momentum .068 kg times 9.68 is roughly equals to .650 kg times 1.
The two Atwood's arrangements do not produce the same quantity of energy: .068 kg at 9.68 has about 10 times as much energy as .650 kg at 1. But they had the same input energy.
Atwood's can make energy.

They did not have the same input energy.
One had a larger mass to be initially accelerated.
What was this quantity of “ input energy”?

The quantity of mass being accelerated has nothing to do with the input energy.
The Atwood’s was accelerated by the same mass suspended from the shaft. It was only about 62 grams. The photo gate recorded a little less than one rotation. The flag was released just after the 2nd trip gate and it came around one rotation. So the 62 grams dropped (20.07 mm * pi) * 9.81 N/kg = .038 joule for input energy.
The same mass was used to accelerate the two 68 gram masses and the two 650 gram masses. But the exact quantity of input force is not important; because the motion of the system without the added masses is the same for both the 136 grams and the 1300 grams.
The quantity of force needed to overcome bearing friction would be the same for both the 136 gram run and the 1300 gram run.
The quantity of force needed to set the shaft and wheel in motion (to .75 RPS) would be the same for both the 136 gram run and the 1300 gram run.
The quantity of force remaining to set the 136 grams, on the wheel, in motion would be the same as the quantity of force remaining to set the 1300 grams, at the shaft, in motion.
This remaining force accelerates the 1300 grams to .04769 m/sec. Or this same force accelerate 136 grams to .46189 m/sec. The 1.300 kg has .001478 joules of energy; and the .136 kg has .0145 joules of energy. ½ mv²
.0145 J / .00148 J = 9.814 times as must energy for this portion of the Atwood's.
The quantity of momentum is the same 1.300 kg * .0477 m/sec = .062101 units: .136 kg * .46189 m/sec = .06281 units of momentum .06281 /.062101 = 1.01

Take a one meter tube and mount a quality bearing on one end: arrange it so that the tube rotates in a vertical plane. We will start from a horizontal position for the tube.
We place a force gauge at one decimeter from the point of rotation: the gauge reads a downward force.
The experimenter then places 10 kg at the one decimeter position. The scale or force gauge will read 10 kilograms or 98.1 newtons.
The experimenter then removes the ten kilograms and places a one kilogram mass at the 10 decimeter position. The scale or force gauge will read 10 kilograms or 98.1 newtons. The lever arm multiplies the force at the 1 decimeter position.
Now lets place the bearing and tube into a horizontal plane; and place it just above a frictionless plane so that the masses rest on the frictionless plane. In this arrangement gravity no longer has any affect upon the system; the tube is now rotating in a horizontal plane. The one and ten kilogram masses no longer exert force they only have inertia.
Now lets place ten kilograms of mass at the one decimeter position. Now lets apply ten newtons of force, in a clockwise direction, at the one decimeter position. After one second the ten kilograms will be moving 1 meter per second.
Now lets place a one kilogram mass at the ten decimeter position. Then lets apply ten newtons of force in a clockwise direction at the one decimeter position. After one second the one decimeter position will be moving 1 meter per second. The inertia of the one kilogram mass has been multiplied by ten at the one decimeter position just as the force had been multiplied by ten in the vertical experiment above. After one second the ten decimeter position will be moving 10 meter per second.
A 10 kilogram mass moving 1 meter per second has 5 joules of energy and a 1 kilogram mass moving 10 m/sec has 50 joules. But the same amount of force * time can be used to make 5 or 50 joules.
Conclusion: It takes the same amount of force to rotate a tube having different quantities of inertia along the length of the tube: as long as the length of the radius of rotation is inversely proportion to the differing quantities of inertia.
Example: ten kilograms at one decimeter rotates as easily as one kilogram at ten decimeters;
An experimental proof: is that an Atwood's finds it no more difficult to accelerate 136 grams at a 97.195 mm radius than it does to accelerate 1300 grams at 10.035 mm. And the 136 grams has 9.81 times as much energy as the 1300 grams.
Any suggestions on a corporation that would like to make use of this?

Take a 10 N motor and set up your machine.
then use another motor as a generator.
You can then go offgrid and use your monthly
electric bill to finance your advertising campaign

https://www.youtube.com/watch?v=dq5jbPoE_Cc
http://hopsbloghop.blogspot.com/2013/10/whataboutmroberth.html
"...So accelerating a mass already moving fast gives you more kinetic energy for your buck..."

https://play.google.com/books/reader?id=EKY7UvvTOoC&printsec=frontcover&pg=GBS.PP1 (https://play.google.com/books/reader?id=EKY7UvvTOoC&printsec=frontcover&pg=GBS.PP1)
https://files.pearsoned.de/inf/ext/9781292112619 (https://files.pearsoned.de/inf/ext/9781292112619)

sm0ky2, thanks for the books.
I just don’t understand what they have to do with Oberth Effect. :)
experience from the video shows the manifestation of the Oberth Effect.

sm0ky2, thanks for the books.
I just don’t understand what they have to do with Oberth Effect. :)
experience from the video shows the manifestation of the Oberth Effect.
Actually that was for Delburt
Oberth has a much more efficient way to accelerate a rocket than straight up
Not sure what that has to do with energy production. Just less waste.

we calculate the balance of speeds and kinetic energies in Beletsky's experiment:
https://www.youtube.com/watch?v=dq5jbPoE_Cc (https://www.youtube.com/watch?v=dq5jbPoE_Cc)
Mm is the mass of the missile assembly;
Mr is the mass of the rocket body;
Mb is the mass of the ball;
Mm = Mr + Mb = M;
n <1;
Mb = n * M;
Mr = (1  n) * M;
Vm  the speed of the missile assembly;
Vr  rocket hull speed;
Vb  ball speed;
Mm * Vm = M *Vm = Mr * Vr + ( Mb * Vb ) =
= ((1  n) * M) * Vr  n * M * Vb;
((1  n) * M) * Vr = M * Vm + n * M * Vb = M * (Vm + n *Vb);
Vr = (M * (Vm + n *Vb)) / (M * (1  n)) = (Vm + n *Vb) / (1  n);
Ekr  kinetic energy of the rocket body;
Ekb  kinetic energy of the ball;
Ekm  kinetic energy of the rocket assembly;
Ekoe  the addition of kinetic energy from the Obert effect;
Ekr = (0.5 * M * (1  n)) * (((Vm + n *Vb) / (1  n)) ^ 2) =
= (0.5 * M * (Vm + n *Vb)^ 2) / (1  n) =
= (0.5 * M / (1  n)) * (Vm ^ 2 + 2 * n * Vm * Vb + n ^ 2 * Vb ^ 2) =
= (1 / (1  n)) * 0.5 * М * Vm ^ 2 + (n / (1  n)) * 0.5 * (n * М) * Vb ^ 2 + ((2 * n) / (1  n)) * 0.5 * M * Vm * Vb;
Kkm = 1 / (1  n); (fraction of kinetic energy Ekm in Ekr )
Kkb = n / (1  n); (fraction of kinetic energy Ekb in Ekr )
Kkoe = (2 * n) / (1  n); (fraction of the kinetic energy of Ekoe in Ekr )
Ekr = Kkm * 0.5 * M * Vm ^ 2 + Kkb * 0.5 * (n * M) * Vb ^ 2 + Kkoe * 0.5 * M * Vm * Vb;
Ekr = Kkm * Ekm + Kkb * Ekb + Kkoe * Ekoe;
_____________________________________________________________________________
as can be seen from these elementary calculations, the theory is consistent with experience
the speed (Vk) and energy (Ekk) of the body (the body of the "rocket") depends on the speed of the body before the impact of a force pulse ...
just in case, I’ll attach once again a picture where everything is painted for kindergarten

You are forgetting the source of the acceleration.
Which explicitly comes from the gravitational potential well.
Newton already told us it is easier to accelerate a moving object
We have since then learned how to derive the moment of inertia
of the moving mass relative to the applied force.
You consume less fuel accelerating while your car is moving
than you do from a dead stop.
The benefit of that rocket launching method is simply an improvement
upon our current method.
Again, not a “gain” in energy. Just less waste.
Further, if you had to raise the rocket to the elevated starting point
you’re wasting more than you save by using the method.

about the "gravitationalpotential hole" did not understand?
this simple experience shows very definitely  the greater the speed of the "body", the greater the increase in the kinetic energy of the "body" from the same impulse of force to this "body".
it’s not clear why to drag here a “gravitationalpotential hole” and other “difficulties”?
the experience is "simple" and "transparent", where can there be a hoax? everyone can repeat.
p.s. imho, all OU machines where something rotates take OU from mechanics ...

How?/why? Is the rocket moving?
Before the engines fire?
Because it drops from a height
Gravity accelerates the rocket initially
Then the rockets fire and speed it up more.
You have to get the rocket up there to drop it
Or I have a genius idea.... use a 2stage booster
The first stage can get the rocket moving
just like dropping it does in the video
The second stage can boost the already moving rocket
like when the ball fires on the video!!
Wow I should call nasa and let them know

you don’t need to call nasa, they know about the Obert effect.
back to this experience, forget about real rockets.
the person in this video makes a “rocket start” at different points on the trajectory.
in terms of energy conservation, it does not matter at which point on the trajectory the "rocket launch" was.
energy from the gravitational field (potential + kinetic) and energy from the ball should be added up.
but for some reason, the kinetic energy of the rocket hull turns out to be maximum when the "launch of the rocket" occurs at the bottom of the trajectory, when the speed of the entire "system" is maximum. violation of the laws of physics is not here, Obert brought his effect purely theoretically, "at the tip of the pen." this is because the kinetic energy of the body is the square of the speed of the body. no mysteries, no mysticism in this.

I did this experiment quite a while ago; but it is easy enough to remember so I will just present it from memory.
I took a board about ½ inch thick by 2 inches wide and 1.1 meters long. I drilled a hole in the center of the board at 55 cm. I cut the hex head off of a ¼ inch bolt and fastened the bolt to the board through the hole. I placed the smooth end of the bolt in a hand held, variable speed, drill.
I slowly began to rotate the board with the drill: it rotated uniformly and smoothly with no wobble. Now lets make some observations.
One: the mass on both sides of the board are equal.
The torque on both sides of the board are equal.
Third: the energy used to rotate the left side and the energy used to rotate the right side are equal.
Now lets continue the experiment by drilling a hole at 10 cm from the point of rotation on the right side of the board. And then place a half inch bolt, 4 inches long, through the hole fixing the bolt at the center of its mass.
Now slowly begin to rotate the board and bolt; you will quickly notice a distinct wobble.
The mass on both sides of the board are no longer equal.
The torque on both sides of the board are not equal.
The energy used to rotate the left side and the energy used to rotate the right side are (*probably) not equal. *The drill is flying around so who know where the energy goes.
Now lets continue the experiment by drilling a hole at 50 cm from the point of rotation on the left side. And then place a quarter inch bolt through the hole and fasten it at the center of mass. This quarter inch bolt should have one fifth the mass of the ½ inch bolt.
I slowly began to rotate the board with the drill: it rotated uniformly and smoothly with no wobble. Now lets make some observations.
One: the mass on both sides of the board are not equal.
The torque on both sides of the board are equal.
Third: the energy used to rotate the left side and the energy used to rotate the right side are equal. The energy on both sides was equal when we just had the board and we had no wobble; and now we are back again to no wobble so surly the energies are equal again.
The small bolt is moving 5 times as fast as the large bolt and it has five time as much energy; but the drill treats the two sides as if they were identical. The board with no weight has no wobble; and the board with these two weights has no wobble. The same amount of energy is used to accelerate the large and small bolt and the small bolt has 5 times as much energy.
The Law of Conservation of Energy is violated by this experiment.

A 10 kilogram mass moving 2 m/sec can give half of its Newtonian momentum to a 1 kilogram mass moving 10 m/sec. The energy loss for the 10 kg is 15 J and the energy gain for the 1 kilogram is 150 J.
A 10 kilogram mass moving 1 m/sec can give all of its Newtonian momentum to a 1 kilogram mass moving 20 m/sec. The energy loss for the 10 kg is 5 J and the energy gain for the 1 kilogram is 250 J.
Math: ½ mv²: ½ * 10 kg * 2 m/sec * 2m/sec = 20 joules: ½ * 10 kg * 1 m/sec * 1m/sec = 5 joules: ½ * 1 kg * 10 m/sec * 10 m/sec = 50 joules: ½ * 1 kg * 20 m/sec * 20 m/sec = 200 J ½ * 1 kg * 30 m/sec * 30 m/sec = 450 joules:
Math: mv: 10 kg * 2 m/sec = 20; 10 kg * 1 m/sec = 10; 1 kg * 10 m/sec = 10; 1 kg * 20 m/sec = 20 ; 1 kg * 30 m/sec = 30;

I was in my lab and I saw a piece of extruded aluminum 362 mm long with 29 holes evenly spaced along its length. The holes were slightly elongated down the center line of this bar shaped aluminum. So I thought I would put some numbers to the above experiment.
I fixed a 1/4 bolt in the center hole: and then put this bolt in a hand held variable speed drill.
The bar accelerated smoothly and uniformly as I slowly turned the drill on.
Then at two holes out (25mm) from the center; I placed a ¼ bolt nut and washers with a mass of about 115 grams.
The drill wobbled distinctly as I turned the drill on.
Then I balance the bar with the two attached bolts; with a third bolt at 14 holes out (175mm) on the other side. This was to the nearest washer (one nut; two washers and one bolt). This came to about 17 grams. And then I placed this back onto the drill.
The bar accelerated smoothly and uniformly as I slowly turned the drill on.
The drill accelerated 17 grams at 175 mm just as easily as it accelerated 115 grams at 25 mm. The drill did not balk at producing more energy on one side than the other side. And all other parts of the bar being equal: If I had ramped up the drill to 380 rpm you would have had .0575 joules on one side and .4165 joules on the other side.
The point is that real world experiments do not show any trend toward conservation of energy.
And it is important to note that the Newtonian momentum on both sides is equal.
Experiments conserve momentum not energy.
Now it is true that this is not a closed system; but double radius Atwood's experiments perform exactly the same way. They will accelerate smaller objects to higher speeds while conforming to F = ma. Not ½ mv²

Good stuff magneat. So you think Chas figured a way to use this for overunity benefit?

Good stuff magneat. So you think Chas figured a way to use this for overunity benefit?
I am sure that it is.
Did you notice that YouTube is filled with mechanical OU generators made in Asia, Africa, Latin America and other "nonEuropean" countries? there are already tens, if not hundreds of them, on YouTube, despite the fact that they are periodically removed.
imho, this is because the authors either taught physics from other textbooks, or selftaught.
"European" physics textbooks hammered into the heads of people from the 1st grade that the "Law of conservation of energy" is a "sacred cow", no one dares to doubt it. but there are many situations where the "Law of conservation of energy" conflicts with the "Law of conservation of momentum."
Continuing the topic: task
_____________________________________________________________________________
we write down the boundary conditions for the task:
we will consider only uniform, steady movements. transient processes, and hence accelerations at the moments of action of force pulses will not be considered. we are not at all interested in the magnitude of the force F and the duration of the force Δt. we are only interested in the result of the action (F * Δt)  a change in the velocity of the body ΔV.
those. the "installation" operation algorithm is as follows:
we have a “body” of mass M, which moves uniformly with speed V1.
the "mechanism" periodically acts on the "body" by a force impulse (F * Δt = const).
transients (moments of acceleration) skip.
after the action of the force impulse (F * Δt), the "body" again moves uniformly, but with a different speed V2 = V1 + ΔV
i.e., in accordance with Newton’s 1st law, at times when the body moves uniformly, we will have equal inertial reference systems (IRS).
__________________________________________________________________
for simplicity, consider linear motion.
the physics of the process will be the same as for the rotational motion.
there is a body of mass M, there is a certain "mechanism" ("mechanisms)" that is capable of
transmit to the body some impulse of an external force (F * Δt).
to simplify, let all momenta of force be the same (F * Δt = const)
using the “mechanism”, the body accelerates to a certain speed V1, while the body
acquires kinetic energy Ek1 = 0.5 * M * V1 ^ 2
if you paint all the formulas:
body weight  M
initial body speed  V1
final body speed  V2
body speed gain  ΔV = V2  V1
the energy expended by an external force to increase the speed of the body  ΔEst = 0.5 * M * ΔV ^ 2
important point: body speed increased from V1 to V2 (changed by ΔV)
those. for an external force there is no difference with what speed the body moves before and after its impact.
there is such a theorem  Impulse Theorem:
______________________________________________________________________________
The change in the momentum of a mechanical system over a certain period of time is equal to the geometric sum of the elementary impulses of external forces applied to the system over the same period of time.
______________________________________________________________________________
initial kinetic energy of the body  Ek1 = 0.5 * M * V1 ^ 2
kinetic energy of the body after an impulse of an external force  Eк2 = 0.5 * M * V2 ^ 2
increase in kinetic energy of the body:
ΔEk = Ek2  Ek1 = 0.5 * M * V2 ^ 2  0.5 * M * V1 ^ 2 = 0.5 * M * (V2 ^ 2  V1 ^ 2) = 0.5 * M * ((V1 + ΔV) ^ 2  V1 ^ 2)) = 0.5 * M * ((V1 ^ 2 + 2 * V1 * ΔV + ΔV ^ 2)  V1 ^ 2)) = 0.5 * M * (2 * V1 * ΔV + ΔV ^ 2) = 0.5 * M * ΔV ^ 2 + M * V1 * ΔV
write again:
ΔEst = 0.5 * M * ΔV ^ 2
ΔEk = 0.5 * M * ΔV ^ 2 + M * V1 * ΔV
ΔEc> ΔEst
_______________________________________________________________________________________
where is the mistake?
_______________________________
p.s. from the formulas it turns out that for any initial speed V1> 0, it should already be OU :)
in reality, friction losses should be taken into account, losses due to the fact that the impulse of an external force is transmitted with an efficiency of <100%.
but in practice must be V1min, after which OU > 0.
the formula shows OU = M * V1 * ΔV (ideal conditions), in reality minus losses.
Of course, this is all theory :)
need experimentation.
p.p.s. it seems that all the OU generators where something massive rotates get OU from MECHANICS
and all kinds of coils, spirals, nozzles, etc., are purely engineering solutions for implementation
conditions for obtaining OU...
____________________________________________________________________
“Only puny secrets need protection. Big discoveries are protected by public incredulity.”
/ Marshall McLuhan /

I have been at this for 36 years.
Ever since I was old enough to comprehend the
theories about conservation of energy.
(which predate thermodynamics by centuries)
These theories seem illogical at their core.
And I have always known there was something wrong.
During my research I have seen many violations of
these theories. And from my observations, it is not
our physics that are violated, but our understanding of them.
Maxwell made allowance for these violations, we have simply
convinced ourselves that things outside our ‘box’ cannot be “real”.
The negative (non real) solutions are still perfectly valid.
We just have to utilize the system in a way to work with itself,
rather than against itself.
Gravity can be used in a system to do this.
But not gravity alone. There needs another force in opposition.
Then the system can intake synchronous energies from around itself.
Not strictly in terms of vibration, but synchronization, of changes in motion
or changes in force/pressure/etc.
Gravity/air
Gravity/magnet
Gravity/ impulse force
Gravity / gravity (from a second well  tidal, etc)
Do some quick math on the mass of the Earth’s oceans
and how long Tidal activity would take to slow down the moon.
You see the problem
The solution lies in how we define energy itself.
Einstein made a valiant attempt to define mass in terms of energy.
Because on the most basic level, mass itself is a system that works
exactly like we are trying to do.
Satellites orbit their star because the forces pulling outwards balance
the gravity pulling it towards the star.
Eventually it winds down and gravity wins because it is the sole force
acting on the system.
Let’s take a system like mars just a few million years ago.
A giant volcano caused the planet to move away from the sun
and take on a much larger orbit.
The timing of this event placed a force in opposition to the sun’s
gravity in just the right direction that it prolonged the orbit well
beyond the determinable wind down time of the system.
Let’s take a look at another example:
A gravity potential storage in the form of a raised water tank.
And a change in pressure by a constricted nozzle.
Mathematically there comes a solution wherein the pressure and
the height of the water jet violate our theories.
Therefore, the theory must be incorrect.
Or another: a pulsed potato cannon:
Wherein an extended tube is given a series of synchronous pulses
of air. Total volume of air released and total initial pressure of the
air tank are held constant. The pulse cannon violates our understanding
of the original device. (which lies at the heart of ballistics theory)
In about 56 months we will revisit Archer Quinn’s Magnetogravitic Generator.
This device uses a synchronous magnetic impulse to oppose gravity.
Chas Campbell created a device which uses the momentum of a rolling mass
to accelerate a wheel faster than gravity is pulling on it.
Under constant brute force, this would clearly waste energy.
However, with the proper timing and mass*velocity of impact vs wheel mass:
There are solutions which violate our theories.
(Note: i was not a proponent of the Chas Campbell device for many years)
(because I had overlooked it’s true operating principal)
Conclusion:
When we arrive at a more correct definition of Energy
“Overunity” will not be what we see it as today.
But will be accepted by consensus.

everything has been invented before us, for example, John Bedini motorgenerator, Robert Adams motorgenerator, Richard Clem overunity motor, VEGA motorgenerator, etc.
Bedini, for example, began his experiments with a generator Raymond Kromrey.
further excerpts from the book "Bedini SG  The Complete Advanced Handbook "
...The original design of this generator was developed by a Swiss engineer named Raymond Kromrey. Kromrey did extensive research and testing on this concept in the 1960s, and received US Patent #3,374,376 on the basic design in March of 1968...
...Being a firm believer in the "Law of Conservation of Energy", he postulated that the energy gain produced by the machine was coming from an interaction with the Gravitational Field, but he also believed that the specific mass of the rotor was involved in this gain. The US Patent is written in a relatively simple manner, and the patent is granted on a single claim based on the physical design alone. There is no attempt to officially "claim" that the generator does anything unusual...
...So here we see for the first time, in Kromrey's patent, the fundamental components of John's "Free Energy Device" consisting of an electric motor, a low drag generator and a flywheel...
...When John first discovered the Kromrey Patent, he knew it might hold the key to what he was looking for. He immediately attempted to contact the inventor in Switzerland. Unfortunately, by the time John did this, Raymond Kromrey had already died. He was able, however, to track down his surviving widow...
...She told John he could do anything he wanted with it and not to contact her again...
...Left completely on his own, John began to replicate the machine and see if he could duplicate the results Kromrey reported...
Jim Watson Generator (Big Bedini Generator Model)
...After John published his book, Jim Watson contacted him, and they had numerous discussions on the phone. Jim wasn't very familiar with electronics, and asked John to develop an even simpler control circuit. In response to this, John developed this 555 based timer/relay circuit. Using this simpler circuit controller, Jim Watson built his first machine...
...Jim Watson was sufficiently impressed with his first model that he decided to build a bigger one. In the run up to the Tesla Centennial Symposium in Colorado Springs that summer, Jim would only tell John that he was working on a "surprise" for the conference. Jim was an accomplished automobile mechanic, so working on big machines was well within his level of expertise. Here is a picture of what Jim Watson demonstrated for the attendees at the first Tesla Society Conference in the summer of 1984...
...Remarkably, Jim Watson did not let John inspect the machine at close range...
Jim Watson's large model of John's "Free Energy Generator" design was shown at the Tesla Centennial Symposium, held at Colorado College, in Colorado Springs, on August 9th through the 12th, 1984. The published Proceedings of the Tesla Centennial Symposium, in the Light of Modern Physics does not mention either Jim Watson's demonstration or John Bedini's presentation on his "Cigar Box Tesla Switch." Apparently, to be included in the published "Proceedings" document, the presentations had to be written up ahead of time, and Jim's and John's remarks were essentially spontaneous in nature.
While the MACHINE seems to have disappeared immediately after this demonstration, Jim Watson did not...
_____________________________________________________________________________________________
It is noteworthy that in all these machines there is a FLYWHEEL, those MASS OF rotating components is important.
without a FLYWHEEL, these machines will not work
imho, in all these machines works the "Overt effect", described as early as 90 years ago, works.
p.s. John Bedini removed from wiki  https://en.wikipedia.org/w/index.php?search=JOHN+BEDINI&title=Special%3ASearch&go=Go&ns0=1 (https://en.wikipedia.org/w/index.php?search=JOHN+BEDINI&title=Special%3ASearch&go=Go&ns0=1)

Some paid hoaxers people are scamming on top of working technology is what I think.
I say this because of the Earth Engine, I dont think anybody would be ballzy enough to make this without external guarantees.
Its all brown folk networks, not indian also. The other brown folks.
Good old youtube isolate and scam tactic.
Also, I am not sure that Clem got it working, there was an expensive attempt and it was not pursued.

...Also, I am not sure that Clem got it working, there was an expensive attempt and it was not pursued.
It’s just that those who tried to copy Clem, imho, could not correctly calculate the parameters of the generator and insist on its operation.
it’s like, for example, giving the task to an “alien from another planet” by taking a photo of his appearance to make and start the internal combustion engine, without explaining all the “subtleties” of his work.
Continuing the topic:
article in THE WALL STREET JOURNAL about the american generator "Earth Engine" 
https://www.wsj.com/articles/onemansunlikelyquesttopowertheworldwithmagnets11558029179 (https://www.wsj.com/articles/onemansunlikelyquesttopowertheworldwithmagnets11558029179)
full tex can be read there  https://revolutiongreen.com/onemansunlikelyquestpowerworldmagnets/ (https://revolutiongreen.com/onemansunlikelyquestpowerworldmagnets/)
company website  ie.energy/
one of these generators is installed in Las Vegas  https://www.youtube.com/watch?v=XZkuteW_Q9k (https://www.youtube.com/watch?v=XZkuteW_Q9k)
and what we see  the device is more than 2 tons
I’ll express my opinion, quotes from the article from the link above:
"... IEC says its first commercial model, the R32 Earth Engine, hucks two 900kilogram flywheels at speeds between 125 and 250 rpm, generating 240V or 480V at 100 amps...
will not work without flywheels
"...IEC has yet to file any patent paperwork for experts to examine, which Mr. Hinz said was due to fear the technology would be poached in a patent counterclaim..."
Mr. Hinz and Mr. Danzik know that they use other people's ideas
"...IEC isn’t selling, or even leasing, the machines, to keep the secret guts secret. Instead, it will charge by the kilowatthour delivered in the field—and a pittance, too, 8 to 45 cents per kWh..."
They know that the technology is the simplest, and therefore they are afraid
"...The magnets IEC uses are also highly onesided, or «anisotropic», which means their field is stronger on one face than the other—say, 85% North and 15% South..."
from the point of view of physics, nonsense...

https://revolutiongreen.com/earthengine/
See the comments here.

https://revolutiongreen.com/earthengine/ (https://revolutiongreen.com/earthengine/)
See the comments here.
it’s hard to believe that a team of such professionals would look for suckers...

It seems unreal they would dare to pull off such a scam.
Then again, does the work of a paid hoaxer even have a meaning?
Or maybe I am just imagining conspiracies and this guy got inspired for a scam I dunno.
Bummed out because it seems like everywhere you go its a tough road to OU, Wht do you expect LOL?
Ou is like Jedi and Sith status almost, and this is the far flung galaxy.
but whats this what Pierre Cotnoir first showed?
A while ago.
If Chas Camplbell wasnt pulling any tricks and really self looping in that video it was definitly OU.
The video where it start, "were filming"...

https://www.youtube.com/watch?v=2qyvX9j5i3U&t=83s
Actually, someboy should analyze this by video and see the speeds.
It seems like the light was still lit after that as well.

https://www.youtube.com/watch?v=2qyvX9j5i3U&t=83s (https://www.youtube.com/watch?v=2qyvX9j5i3U&t=83s)
Actually, someboy should analyze this by video and see the speeds.
It seems like the light was still lit after that as well.
The Chas Campbell Gravity Wheel may well be working, but to understand the OU principle of a mechanical generator, I like F.M.CHALKALIS ENERGY MULTIPLIER more  https://www.youtube.com/watch?v=iHhZZ9DuzK4&feature=emb_title (https://www.youtube.com/watch?v=iHhZZ9DuzK4&feature=emb_title)
This video shows all the OU classics for a mechanical generator.
only one moment, imho, is not correct for F.M. CHALKALIS  why does it give an impulse to the flywheel at the TOP point of the trajectory, when according to theory the impulse should be applied at the moment when the "body" speed is maximum, i.e. at the BOTTOM of the trajectory?
and also, imho, on rollers that give an impulse to the pendulum, you need to add a flywheel.
then the drive electric motor can be taken of small power, energy will accumulate in the flywheel, and in the pulse will be transmitted to the pendulum.
it is also unclear why the goods are placed in the middle of the radius of the pendulum, it is more logical to place them at the end of the radius, the torque and stored energy will be more ...
then the calculation of the parameters F.M. CHALKALIS ENERGY MULTIPLIER can be done according to elementary formulas of classical physics ...
F.M. CHALKALIS ENERGY MULTIPLIER can serve as a training tool for understanding the OU principle of a mechanical generator.
for the training stand, the design can be greatly simplified  take an inflated bicycle wheel, strengthen one load (several loads?) from the inside of the rim, fasten (stick?) several strips of rubber from the outside to the tire cover through which from the drive roller (with flywheel) a force impulse will be transmitted to the wheel with the load (s).
The Chas Campbell Gravity Wheel, imho, is also a working structure, just not as transparent as F.M. CHALKALIS ENERGY MULTIPLIER.
p.s. for any OU generator, you need the correct calculation of the parameters, and the correct setting, like for any other (ordinary) mechanism. and then later, some "OU seekers" have statements: "this OU is not working" ....

I think that the despin concept is in some way incorporated into any wheel that produces energy. So I will outline the despin concept.
It takes the same amount of force, applied for a period of time, to make a mass move around the arc of a circle as it does to make the mass move in a straight line. So the math used for the arc of the circle is the same as that used for a straight line. Which is F = ma. When a mounted rim has a certain amount of force applied for a certain amount of time the mounted rim has a certain amount of momentum.
A 9 kilogram rim can be accelerated by wrapping a weighted string around the rim; and a one kilogram mass can be placed on the end of the string. After the one kilogram mass has dropped .5097 meters the system will have a velocity of 1 m/sec.
A 10 kilogram rim mass wheel moving one meter per second, in the arc of a circle, has 10 units of momentum. When a 1 kilogram mass swings out on the end of a string (as in the cylinder and spheres) it will have 10 units of momentum when the rim's motion is stopped.
This is proven to be momentum because the 1 kilogram can restore all the motion back to the rim if the 1 kilogram is left attached. Small masses can not give their energy to larger masses.
Ten kilograms moving 1 m/sec (the rim with spheres attached) is 5 joules of energy.
One kilogram moving 10 m/sec (the spheres) is 50 joules.
The one kilogram moving 10 m/sec can be sent upwards 5.097 meters.
So we start over . A 9 kilogram rim can be accelerated by wrapping a weighted string around the rim; and a one kilogram mass can be placed on the end of the string. After the one kilogram mass has dropped only .5097 meters the system will have a velocity of 1 m/sec. This one meter per second, for the 10 kilograms, bring us back to the point where we transferred the motion of the rim to the smaller spheres. The smaller spheres are then sent upward 5.097 meters. And we have 5.097  .5097 = 4.587meters left over.
Now the question for all these flywheels is: How do you gain an energy increase without releasing the spheres and letting them rise. They need a tower for the spheres to rise; and then catch the spheres for a slower driving descent.
Another question I have is: How can all these people be so close and not try wrapping their flywheels with a weighted string and see if it works.

Delburt Phend, your question was for me?
English is not my native language, so I might not understand you correctly.
I don’t understand the expression "despin concept".
it would be more clear if there was a drawing + formulas.
my point of erosion is outlined in the following posts (if you have not read them):
https://overunity.com/17113/reenergyproducingexperiments/msg541516/#msg541516
Obert effect video  https://www.youtube.com/watch?v=dq5jbPoE_Cc
popular explanation of the Obert effect  http://hopsbloghop.blogspot.com/2013/10/whataboutmroberth.html
mathematical presentation of the Obert effect, and assumptions about its presence in OU generators:
https://overunity.com/17113/reenergyproducingexperiments/msg541532/#msg541532
https://overunity.com/17113/reenergyproducingexperiments/msg541788/#msg541788
https://overunity.com/17113/reenergyproducingexperiments/msg541808/#msg541808
https://overunity.com/17113/reenergyproducingexperiments/150/

https://www.youtube.com/watch?v=boLk57cKNao
Please notice that the spheres swing out from the cylinder and the cylinder stops.
The cylinder's mass is a great deal larger than the spheres.
The spheres absorb all the motion of the cylinder. This is the despin effect.
Also note that the rotational motion of the cylinder is restarted. This restart guarantees that this motion transfer is momentum conservation. Because as proven by ballistic pendulums energy is not conserved when a smaller mass transfers its motion to a larger mass. And a great deal of motion would be lost.
The spheres could be released as the cylinder is stopped and they could be directed upward.
The increase in energy is proportional to the mass difference. If the total (cylinder's and spheres') mass is ten time that of the spheres then the energy increase is ten times; when the spheres have all the motion.
For other experiments type delburt phend youtube into the search engine.

...For other experiments type delburt phend youtube into the search engine.
Your experiments are very interesting!
but, how do you see their practical application for obtaining OU? Do you already have thoughts on this?

How does that compare to when the cylinder is on a bearing?
Tetherball?
Bolas?

Absolutely
The one gentleman had a wheel that must have had a mass of one hundred kilograms. And it was probably moving several meters per second; so lets start there. With 100 kilograms moving 2 meters per second. And let transfer that momentum to 5 kilograms.
One hundred kilograms moving two meters per sec is 200 units of momentum.
When we transfer all of the momentum to 5 kilograms the five kilograms will have to be moving 40 meters per second. At 40 meters per second it can rise 81.5 meters. d = 1/2v²/a
After it has risen 81.5 meters we can place it on a stack of eighty 5 kilogram masses 80 meters high. These masses would be spaced 1 meter apart.
You can drop the entire stack one meter. And all you have to do to return the stack to its original configuration is to take the lowest 5 kilograms and throw it back to the top.
You can throw the lowest five kilograms back to the top by accelerating the 100 kilogram rim to 2 meters per second and transfer the momentum into a five kilogram mass.
A stack of 80 five kilogram masses has a mass of 400 kilograms. 400 kilograms can accelerate a 100 kilogram rim to 2 m/sec after the stack has dropped a distance of .2548 meters. But in this drop of one fourth meter not only is the rim moving 2 m/sec but so is the 400 kilograms stack. So now you have 1000 units of momentum and you only need 200 (5 kg * 40 m/sec) units to reload the stack.
200 units of momentum is one fifth of 1000 and the distance dropped is only a fourth of a meter. The stack was dropped one meter. One fifth of the momentum and one fourth of the distance is only 5%.
The stack is reconfigured by using only 5% of the motion produced from the one meter drop of the stack.

E= mgh (only when it falls)
If I have a rock that weighs 10 tons
And I lift one end up (leaving the other on the ground)
I am only lifting part of the rock. This is much easier than
lifting the entire rock. (something less than 10 tons)
Now if I place a fulcrum at its center, it will balance itself out.
Literally lifting the other half of the rock for us.
we have 10 tons in the air we can drop.
Let’s take two 5 ton weights.
Place a raise bar above them. Center pivot. Tie chains
and pull them tight, and use their balance to pull the chains
tighter and get them off the ground. Like scales.
You may have to dig some dirt then bounce them pulling the chains
tighter to get them airborne.
now: you can move these and spin the bar around with little effort
we have 10 tons flying through the air.
do this with another pair in a “T” shape so it pivots on 2 gimbals
we have 20 tons flying through the air on a 4way pivot.
Get them spinning fast
now you can get them all swinging up and outwards on their chains.
Now we have 20 tons higher than before.
catching the momentum of these yields an incredible burst of energy
And the Height at which we catch them is higher than before.
Support the weights raise the supports and do it again.
Each time gaining a burst of impact energy 20 tons slamming into something.
Each time 20 tons higher than before.
Each time just little you spinning a perfectly balanced axle with ease.

Like spinning a heavy barrel

If we want to achieve a sustainable energy footprint
and not be dependent on anything
It is a simple equation.
How much energy do you need?
How many times does your generator need to spin
to make enough power for the rest of your life?
Figure out what kind of gearing this would require to operate a
Gravity dropweight system at a reasonable height.
Meaning if you had weight X at a height you can build a tall enough
support to, geared down to spin your generator for 5060 years.
Calculate the torque on that gear at nominal operating rpm.
And then figure out how much weight you need to get up there.
You can compare that to say..... your water bill.
Which will lift the mass of water to the height of your local tower for free.
(at the same rate you pay for water, and needs to be pre pressure regulator
for max height)

I like to equate energy in terms I can relate to as a country boy like cutting the grass with my lawn mower. So my household requires 1 KW on average neglecting peaks and troughs at an average efficiency of 3040% not unlike my lawn mower. So my real world not BS energy needs are not unlike a 5hp generator/lawn mower screaming away in my back yard 24/7.
Now let's take maybe 4 billion 5hp lawn mowers screaming away as residential and 8 billion as commercial/industrial and now we have a little fucking problem. Think of it this way, a billion is a thousandmillion so were talking about let's say 12 thousand million generators screaming away every day just to keep up with our current demand. Understand this is only our electrical demand and has nothing to do with heating/cooling or HVAC in the rest of the world.
For example where I live in Alberta and it is supposed to hit 35 Celsius with wind chill into 55 the next few days. If I do not have fucking "energy" I die unlike many of these pussies in Florida at maralago who think not being able to charge their phone is the end of the world.
So if were going to talk about energy then lets talk about energy but don't think think anyone just can throw out some bs they cannot justify. I live and breath energy because my life depends on it because I live in a environment designed to kill most people.

I have to disagree magneat, the Chalkalis device, was not overunity.
The Chas Campbell device was on a loose wooden plank, as I have shown before on video but it seems someting happened to that.
It may have have had improper weight distribution.
You are not the first to come here and say these things, I have said all this before.
The problem is that we, me , everybody, wont build a proper test bench because small people are small and stay small.
Its a big endeavour to make it the way I want it, and theres cheaper things that show much more promise.
Dominating factor is still money, economics.

We have two extremes here: how much money does it take to prove the theory; and how much money will it take to make a useful machine? One that will keep you alive in Alberta.
I was raised in Colorado; my uncle lived in the San Luis Valley;  40° was common in the Valley. In a breeze, unprotected skin would freeze in a minute or so. My aunt would freeze dry close on the close line. I remember duck hunting on ten inches of ice. The ducks and geese kept a spot open, over some springs I assume. Back up systems, and good neighbors keep the residents alive.
The actual proof that energy can be produced takes less than one hundred dollars. If a 'cylinder and spheres' has a mass 16 times greater that the spheres then the spheres must be moving 16 times as fast when they have all the motion: or they are only moving 4 times as fast. 16 conserves mv. And 4 conserves ½ mv².
But if the spheres are only moving 4 times as fast then they can not restore the motion to the cylinder if the spheres are left attached. A one kilogram object with only 4 m/sec velocity would only give (upon collision) 16 kilograms a velocity of .25 m/sec. Ballistic pendulums are common and we know what velocity a small object gives a large object.
Plastic tubing used for electrical conduit; 4 inch schedule 80, would work great. You could use a schedule 80 fitting; ten or twelve dollars. Spheres are also available; for only a few dollars each. And a few feet of fishing line. You could probably build a cylinder and spheres for less than 50 dollars. The experimental proof is worth upwards of 5 trillion dollars.
Now about surviving in Alberta. Well: why would a gravity dam cost more than a hydroelectric dam. Glen Canyon Dam has been a cash cow for about fifty years now and it produces no carbon emissions.

Delbert have you done this experiment below? Could you give us a drawing? I don't quite understand it. Or did I miss previous posts? Norman
Delbert said .....
"The actual proof that energy can be produced takes less than one hundred dollars. If a 'cylinder and spheres' has a mass 16 times greater that the spheres then the spheres must be moving 16 times as fast when they have all the motion: or they are only moving 4 times as fast. 16 conserves mv. And 4 conserves ½ mv².
But if the spheres are only moving 4 times as fast then they can not restore the motion to the cylinder if the spheres are left attached. A one kilogram object with only 4 m/sec velocity would only give (upon collision) 16 kilograms a velocity of .25 m/sec. Ballistic pendulums are common and we know what velocity a small object gives a large object.
Plastic tubing used for electrical conduit; 4 inch schedule 80, would work great. You could use a schedule 80 fitting; ten or twelve dollars. Spheres are also available; for only a few dollars each. And a few feet of fishing line. You could probably build a cylinder and spheres for less than 50 dollars. The experimental proof is worth upwards of 5 trillion dollars. "

https://www.youtube.com/watch?v=boLk57cKNao
I went into the lab and found what looks like this experiment. The tape on the cylinder is different but the tape can change any time. The cylinder has 972 written on it; and that no doubt is its mass. The spheres are the steel spheres with a hole through a diameter and they have a mass of 66 each (66 g *2 = 132 grams). So the total mass before the spheres swing out is (972 g + 132 g) 1104.
So the ratio of total mass to spheres is 1104 g /132 g = 8.36.
As you see the rotation of the cylinder is quickly stopped. The string has not come close to the 90° position to a tangent line on the surface of the cylinder. Before the 90° position the spheres will cause the cylinder to reverse the spin; and that is what you see. Before the video stops you can see a spin magnitude that equals the original spin. The sphere motion has restored the motion back to the cylinder.
https://www.youtube.com/watch?v=w7d66JscI8
This looks like the same length of schedule 80 conceit with the addition of a schedule 40 water tubing connecter. These connecters have a mass of about 218 and that brings us to 1320 grams for the total mass.
So the ratio of total mass to spheres is 1320 g /132 g = 10.0.
The spheres find it harder to stop the cylinder; but they do and they also restart the cylinder.
There are other experiments out there if you type in Delburt Phend youtube. One mass ratio stops the cylinder twice; some have more massive spheres.
https://www.youtube.com/watch?v=KgT70vSIUgA

I posted a video of the Atwood's that I collect data from. It is a heavy flywheel that has two radii: the shaft at 10 mm and the outside circumference at 97 mm.
https://www.youtube.com/watch?v=1iM_thYbDlc
In one experiment the accelerating mass of 219.87 g was placed on the shaft radius. Without any added mass on the fly wheel the gate tripped at .0266 seconds. This was an average of 5 runs (.0266, .0264, .0266, .0267, .0267).
I then placed the small masses on the outside of the wheel. This would be about 67.7 grams on each side. The gate was then tripped at .0271 seconds, (.0269, .0272, .0272, .0271, .0272.). By adding 135 grams to the outside of the wheel the gate trip; of the Atwood's; was slowed by .0005 second. This may not seem like much but the flywheel is heavy and 135 grams probably is that proportion of that mass. But the more important experiment was to compare the light outside mass with the balancing heavy mass placed on the inside.
The next experiment had the gates repositioned so the times are not the same as the previous experiment.
I placed the light masses (67.7 g each side)) on the outside circumference of the flywheel and the gate tripped at .0294 seconds (.0294, .0296, .0293, .0295, .0292).
I replaced the light mass with the balancing heavy mass on the inside shaft radius. This would be about 657 grams on both sides. The gate now tripped at .0294 seconds (.0291, .0293, .0296, .0297, .0292).
The Atwood's accelerate the same for 135 grams at the 97 mm radius as it does for 1313 grams at the 10 mm shaft radius.
If the inside shaft circumference was moving 1 m/sec
The energy of the .135 kg is: ½ * .135 kg * 9.7 m/sec * 9.7 m/sec = 6.35 joules
And the energy of the 1.313 kg is: ½ * 1.313 kg * 1 m/sec * 1 m/sec = .6565 joules

What he is talking about is a crudementary handspun version
of an experiment performed by George Atwood in 1790.
He spins it with his hand and drops it, then estimates rpm based
on his camera frame rate.
No experimental controls, no accuracy of velocities, no real data...
Just “blah blah I can spin this cylinder and make the balls spin really
fast, and I think the cylinder would spin faster than it started if the
thing didn’t hit the ground so fast”
This of course doesn’t work with the cylinder on a bearing because
you can see it slow down.
but when it hits the ground just before you can make a determination!
it’s all up to the observer and his/her imagination.
For those that don’t know, Atwood is the guy that standardized
weights and measures by using gravity and a pulley.
(for some reason we ignore Archimedes...)
This experiment was performed by NASA in low g orbit
with much greater precision, and they had nothing of the sort
to say about the experiments. It is about as interesting as a yoyo.
Great way to observe conservation of momentum.
In low g the cylinder despins then spins back up rewrapping the
strings and this continues each time slowing down more and more
until the balls no longer wrap around and instead spin on straight
strings with the rotating cylinder.
Then it stops.

When you say that NASA performed this experiment are you referring to the 'cylinder and spheres' or the Atwood's. Please tell us where we can find that experiment. It is scientifically good etiquette to find the experiment you are referring to and let us look at it.
Frame rates of a camera are not an imagination; they are quite solid data. Does it take 4 frames to pass from one side to the other (of the black square) or does it take 16. Yesterday I collected 215 data points that were to a ten thousandth of a second. In many of the five data point runs several data points were exactly the same as the other runs.
I have never said that the cylinder can spin faster; the energy of the spheres (when they contain all the motion) increases dramatically. There is a ten fold increase in the energy of the system when the spheres have a mass of one tenth. Watching the restart proves that the spheres contain the original momentum.
The energy increase of the cylinder and spheres occurs in a third of a rotation and in a few tenths of a second; so friction is negligible even with bearings. When making energy you release the spheres you do not leave them attached and wait for them to slow down.

Extra energy could perhaps be a possibility, but only in a few ways.
Since static physics is well defined we cant discovr anything, areas to discover have to do with lag of reaction force during impulse.
Now you must finda way tio input a disconnecting flywheel onto a greater flywheel, and find an arrangement where the delay is so delayed
that you actually have time to physically decouple using a system that you make. Thus causing a bug in the system that has a beat frequency.
Using an eccentric wheel, one can translate an intermediate wheel in high tension or no tension zone, and if a squeeze static wheel is there to limit the movement an oscillation can be had between it and the input static wheel wich is attached to the input motor.
Now the magic happens when you communicate with the stuff around you, at the magic frequency,This will cause a vacuum to form with the universe and you will take in energy from a localized orberth effect from the universe just as a space probe. Slowing the universe down.
We must find ways to oscillate things in the varying of the acceleration, pulse, the third derivative wich will interact with the universe. For the goal is to use a machine that has a localized orberth advantage,
You like my theory?

And you cant constantly gain velocity either, at some point you must connect to release the energy and disconnect the output so start back up, and so on. In a sawtooth fashion.
Lets say the third derivative oscillation resonates to soften things, it preps the area for you to suck it in.

In the Atwood's experiment yesterday: I started with a drive mass on the shaft of 127.9 grams. I recorded five rotations of the wheel and made five runs. I then added 12.8 grams of drive mass and did the runs again. I added similar drive masses seven times. On the average each added gram of drive mass resulted in a gate trip time change of .000135 seconds.
The adding of 135 balanced grams to the outside diameter resulted in a time change of .0005. So how many grams of drive mass at the 10 mm radius shaft are needed to compensate for the addition of 135 (67.5 both sides) grams at the circumference radius of 97 mm.
So I set up an experiment to collect real data. (This is a different position for the the gate and the flag so the numbers will again change from the previous set up) I rotated the wheel with no balanced mass on the outside circumference of the wheel. The gate tripped at around .0292 sec.
With the 135 g mass added on the outside the gate tripped at around .0298.
I then added 5.72 grams to the drive mass at the shaft; the trip time returned to around .0292 second.
This data corresponded with the data collected yesterday but this added drive mass (5.72 g) directly compensates for the decrease in motion caused by a balanced mass added on the circumference (135 g).
The gate distance is 27 mm and it took .0292 seconds to cover the distance. That puts the velocity at the gate to be .9246 m/sec. But the wheel surface (where the 135 g is) is inside the gate position by about 90%. So the speed at the surface of the wheel is about .83 m/sec.
So 5.72 grams dropped (2r * pi) 64 mm causes 135 g to accelerate to .83 m/sec.
135 g moving .83 m/sec has: ½ .135 kg * .83 m/sec * .83 m/sec = .04650 joules of energy
5.72 grams (.00572 kg * 9.81 N/kg) has a force of .0561 N times the distance dropped (20.4 mm * pi) of 64 mm (.064 m) so the energy is .0561 N * .064 m = .00359 joules
.00359 joules produced .04650 joules

The experiment is reasonably close to F = ma which would give you .72 m/sec; I could have added another gram of drive mass. The time over which the force acts is almost right on target. I missed my guess by a gram.

How come the space probe gets to gain pulse advantage and not a machine here on earth?
What if there was a way?
What if you can get a localized orberth advantage? Why wouldnt you?
The space probe thing is just as illogical.

The concept being investigated by the Atwood's machine is that: as increments of balanced mass are placed at the circumference of the Atwood's wheel; then increments of drive mass have to be placed at the shaft in order to keep the rate of acceleration constant. That constant acceleration is confirmed by a constant velocity at the gate.
The increment of balanced mass (67.5 g both sides) placed at the wheel circumference is; in this experiment; 135 grams.
F = ma predicts that the increment of mass placed at the drive location on the shaft should be about 7 grams. This 7 grams will compensate for the added mass at the circumference and the acceleration will remain the same. This means that the speed at the gate will remain the same at .83 m/sec.
So now lets do the math for energy conservation.
The energy produced is ½ * .135 kg * .83 m/sec * .83 m/sec = .04650 joules.
We know that the distance that the drive mass drops is only (20.45 mm * pi) 64.24 mm: and potential energy is Nm, so now we need to find the newton needed to produce the final amount of energy (.04650 joules).
So N * .06424 m = .04650: .04650 / .06424 m = .724 newtons
Now we need to change newtons into kg: .724 newtons / X = 9.81 N / 1 kg: X = .0738 kg
So energy conservation predicts that we should add 73.8 grams to the drive mass at the shaft.
I know what will happen if you add 73.8 grams to the drive mass at the shaft: but I guess I should go do it and get real data.
11820 The Atwood's (gate and flag) had to be set up again so the number may not match the last runs.
The gate trip times for a wheel with a 219.87 g drive mass at the 20.04 mm shaft. And no mass at the circumference. (The string has a diameter of .45 mm so I add that to the diameter for calculating drop distance.) .0291, .0291, .0292, .0292, .0291 sec.
The gate trip times for a wheel with a 219.87 g + 7.01 g drive mass at the 20.04 mm shaft. And no mass at the circumference. .0285, .0285, .0287, .0287, .0286 sec.
The gate trip times for a wheel with a 219.87 g + 7.01 g drive mass at the 20.04 mm shaft. And a balanced mass of 135 g at the circumference. .0297, .0293, .0294, .0295, .0295 sec.
The gate trip times for a wheel with a 219.87 g + 72 g drive mass at the 20.04 mm shaft. And no mass at the circumference. .0246, .0247, .0247, .0247, .0248 sec.
The gate trip times for a wheel with a 219.87 g + 72 g drive mass at the 20.04 mm shaft. And a balanced mass of 135 g at the circumference. .0253, .0254, .0254, .0256, .0254 sec.
The drive mass of 7.01 g added was the closest but not on top of .0291 so I then decided to try to hit the added mass that would get us back to .0291. This meant the wheel had to move a little faster so I tried 8.75 g of added drive mass.
The gate trip times for a wheel with a 219.87 g + 8.75 g drive mass at the 20.04 mm shaft. And a balanced mass of 135 g at the circumference. .0291, .0291, .0291, .0291, .0290 sec. Wow: good guess
The gate gives us a speed of (27 mm gate distance / .0291 sec) .9278 m/sec at the gate; and .844 m/sec at the surface of the wheel. And the formulas for F = ma predicts a speed of .886 m/sec.
Whoa that is .886 / .844 = 1.04976 = 5% I'll take it F = ma wins

There was rumors, evidence (internet evidence), that the UFO's had a layered material. With the Schwartz thing popping up I believe that now.
This sort of thing goes on on the electronic side of things.
This is a thread about mechanical things, so we are gonna talk about that.
Extra energy could perhaps be a possibility, but only in a few ways. A physical characteristic that is not well defined is reaction force time.
http://www.halexandria.org/dward133.htm
Since static physics is well defined we cant discover anything, whatever they say, goes, 100%. So lets go where they dont even have formulas for yet, areas to discover have to do with lag of reaction force during impulse.
Now you must find a way to input a disconnecting flywheel onto a greater flywheel, and find an arrangement where the delay is so delayed
that you actually have time to physically decouple using a system that you make. Thus causing a bug in the system and every action or lack of has a reaction, now we are entering the zone, that has a beat frequency. Using an eccentric wheel, one can translate an intermediate wheel in high tension or no tension zone, and if a squeeze static wheel is there to limit the movement an oscillation can be had between it and the input static wheel wich is attached to the input motor.
Now the magic happens when you communicate with the stuff around you, at the magic frequency,This will cause a vacuum to form with the universe and you will take in energy from a localized orberth effect from the universe just as a space probe. Slowing the universe down.
We must find ways to oscillate things in the varying of the acceleration, pulse, the third derivative wich will interact with the universe. For the goal is to use a machine that has a localized orberth advantage,
And you cant constantly gain velocity either, at some point you must connect to release the energy and disconnect the output so start back up, and so on. In a sawtooth fashion.
Lets say the third derivative oscillation resonates to soften things, it preps the area for you to suck it in.

I think F = ma is fairly well understood but it got replaced with Conservation of Energy: about 150 years ago.
The only experiment that I remember NASA doing in the 'cylinder and spheres' area was the Dawn Mission and other such satellites . Energy conservation is a loser in the Atwood's and it is a loser in the ballistic pendulum and all collisions. But when NASA makes predictions (never backed up with data) they chose energy conservation. Go figure: NASA goes to the Moon with Newton and then picks Leibniz to predict the velocity of the extended spheres in the Dawn Mission.
The Dawn mission is a despin satellite that had a mass ratio of cylinder mass to sphere mass of about 400 to one (1420 kg to 2.7 kg). For energy conservation NASA predicted that the spheres would only be traveling about 23 m/sec at the point of release. Well why release the spheres at all? The 2.7 kg moving 23 m/sec would only give the 1420 kg a velocity of about .05 m/sec. That is slower than the 3 rpm back spin that they actually got and they released the spheres. They never released data on the velocity of the spheres. Can you guess why?
When the 2.7 kg moving 23 m/sec returns its motion to the satellite it must comply with the laws of physics. The ballistic pendulum tells us what those Laws are: small object can only give their momentum to a large object, they can not give their energy. And the end rotation of the satellite would only be about 5 cm per second.
If on the other hand the 2.7 kg were moving over 400 m/sec then the rotational motion of the satellite would be restored and you would be back were NASA started. Would anyone think that you would not be back were you started if you left the spheres attached?
It is only that NASA thinks that energy conservation must occur that this 23 m/sec (a 12 year old boy can throw this fast) idea could have arisen.
Newton lost this debate; but he should have won. We would have had free energy long ago.

If only you had a complete motor arrangement to go along with this. ;)
Could you engineer a QMOgen/motor with what I described to you above?

In an Atwood's 8.75 grams can accelerate a balanced mass of 1310 grams to a velocity of .09128 m/sec after the 8.75 grams has dropped .064 meters. This occurs when the accelerated mass of 1310 grams is at the same radius as the 8.75 g drive (the mass that provides the force) mass. In this experiment the drive mass is at a 10 mm radius; and the 1310 grams of balanced mass is at the same location.
In an Atwood's 8.75 grams can accelerate a balanced mass of 135 grams to a velocity of .885 m/sec after the 8.75 grams has dropped .064 meters. This occurs when the accelerated mass of 135 grams is at a radius 97 mm while the 8.75 g drive mass remains at 10 mm. The 8.75 grams is still moving .09128 m/sec and the 135 g is moving 9.7 times faster.
The 1310 grams has a momentum of 1.310 kg * .09128 m/sec = .11958 units
The 135 grams has a momentum of .135 kg * .885 m/sec = .11948 units
The 8.75 grams has dropped .064 m; To restore its original position you would have to throw it upwards .064 m. A velocity of 1.12 m/sec is needed to throw a mass upwards .064 m. So .00875 kg * 1.12 m/sec is the needed momentum to restore the original position of the drive mass.
The momentum needed to restore the original position of the 8.75 grams is .00875 kg * 1.1206 m/sec = .0098005 units
The quantity of momentum available in the 1310 grams or the 135 grams is 12.19 times that which is necessary to recycle the system. .11952 / .009805
This is possible because the time over which the force acts in a free fall of .064 m is .114 sec.; and the time over which the force acts to produce the motion of the 1310 grams or the 135 grams is 1.4023 seconds. 1.4023 sec / .114 sec = 12.33
The 1310 is at a radius of 10 mm. And 1/9.7ths of that mass (135 g) is at 9.7 times that radial distance (at 97 mm).
What would prevent us from placing 1/12.28ths the mass (106.67 g) at 12.28 times (122.8 mm) the distance?
The 106.67 grams would be moving 12.28 times faster than the 8.75 grams moving at .09128 m/sec. You would have 106.67 grams moving 1.12 m/sec.
106.67 grams moving (.09128 m/sec * 12.28) 1.12 m/sec would have .10667 * 1.12 m/sec = .11947 units of momentum.
The 1.12 m/sec velocity is sufficient to reload the system and you have 106.67 grams instead of 8.75 grams.
Or how about 26.2 grams at a radius of 500 mm? 50 * .09128 m/sec = 4.564 m/sec

Delburt,
wow, 12 years have passed by...you remember ?
For those here to better understand how the dynamics of your system works we go back 12 years where you posted the pictures of the evolving motion in series including description: go to reply#116, January 18th
https://overunity.com/1995/freeenergyfromgravitationusingnewtonianphysic/105/ (https://overunity.com/1995/freeenergyfromgravitationusingnewtonianphysic/105/)
I am still scratching my head...to find a technical solution
Mike

A pulley with two radius lengths can be used to make energy. As in a double Atwoods. When the ratio of the length of the long radius to the length of the short drive radius is greater than the square root of the mass ratio between the accelerated mass over the drive mass. The speed of the accelerated mass throws it beyond the distance dropped.

The ratio of accelerated mass to drive mass would be; as in the above experiment 1318.75 g / 8.75 g. And then you place a 1 / 12.28 mass at 12.28 times the distance.

Delburt,
found this simulationapplet, however you can not vary the radius. I myself can not run the simulation , my LinuxSystem misses the Javacode
https://demonstrations.wolfram.com/DoubleAtwoodMachine/ (https://demonstrations.wolfram.com/DoubleAtwoodMachine/)
Now concerning this guy here:
https://www.asc.ohiostate.edu/durkin.2/treb2.htm
Lets do some energy calculations:
Mass of the Brick 1.5 kg, g 0.98 kg*m/sec exp2,
Mass of golfball 0.046 Kg, speed 16 m/s
From the video I only can guess the hight of the falling brick is about 0.8 m
Potential energy of the brick : m * g + h = 1.5 x 9.8 x 0.8 = 11.76 JEnergy of the golfball : 1/2x 0.046 x 16 exp2 = 5.88 J
So this system is not showing energyincrease. I assume it misses the right point in time to release the ball.He mentioned that  accoding to his observation  the brick almost stood still at times while falling.
I would really find more about this Mr. Gordon here : http://garydgordon.com/Despin/ (http://garydgordon.com/Despin/)
especially how he connected the pucs at the rim so they could be catapulted tangentially.
Good find anyway
Mike

now here another puzzle comes along
https://sciencedemonstrations.fas.harvard.edu/presentations/doubleatwoodsmachine (https://sciencedemonstrations.fas.harvard.edu/presentations/doubleatwoodsmachine)
Mike

I need to specify what I mean by a double Atwood's: It is a pulley with two radii, one for the accelerated mass and one for the drive mass. Maybe I should say two radii Atwood's.
I added 245.6 grams at the larger circumference (instead of 135 g) in the two radii Atwood's. I had to add 17.5 grams at the drive mass location of the shaft to get the gate time back to .0291 seconds. This was off of F = ma by 10% instead of the 5% for the 135 grams, and 8.75 grams. Still: to conserve energy I would have had to added 134.7 grams at the shaft location. One or two grams off is a lot closer than 134 grams.
The energy of the 17.5 grams dropped .064 meters is .010987 joules
The energy of 245.6 grams moving .83 m/sec is .08459 joules
.08459 J / .010987 J = 7.699 770%

I set up the double radii Atwood's again; because I intend to extend the legs, so it will be out of service for a while.
In this experiment I placed 200 grams at the drive mass location on the shaft. The gate was tripping at .0299 sec with no mass placed on the circumference.
After I placed 245.6 g of balanced (122.8 grams both sides) mass on the circumference the gate trip time increased to .0319 seconds.
To get the gate trip time back to .0299 seconds I had to add 14.58 grams at the drive mass location on the shaft.
So 14.58 grams can accelerate 245.6 grams to .8217 m/sec after the 14.58 grams has dropped .064 m.
The 245.6 grams at 97 mm radius has the same momentum (245.6 g * .027 m / .0299 sec * (97 mm / gate radius = .8127 m/sec)) as 2382.3 ( 2382.3 g * .8127 m/sec * 10 mm / 97 mm) grams moving at the shaft. Previous experiments showed that the two masses (at their locations) would cause the same rate of rotation. Even though the 245.6 grams is at 97 mm it is the same as 14.58 grams accelerating 2396.88 (2382.3 + 14.58 g) grams at the shaft.
So a perfect F = ma at the shaft would have a velocity of: the square root of (.064 m * 2 * 14.58 g / 2396.88 g * 9.81 m/sec) = .08739 m/sec which would give you: .08739 * 9.7 = .848 m/sec at the circumference.
About 4% error: .848 m/sec / .8127 m/sec
I think this proves that: no matter what the radius of the mass; you will get the same amount of Newtonian momentum produced per unit period of time in a double radius Atwood's. And the rate of rotation will remain the same as long as the quantity of mass is inversely proportional the the length of the radius.
And what is the practical application? To throw this momentum into a smaller mass. Or make the second radius large enough that you do not have to throw at all. That is why the next step is to increase the length of the Atwood's mounts.

I made a mistake. The real speeds (that were determined by the photo gate trip times) are off because I had the Photo gate in the wrong mode. I thought the time started at one photo gate and ended at the other photo gate. This is true in some modes but not in the simple (gate mode) that I had selected.
The gate was recording the time over which the single gate was started when one side of the flag blocked the gate and then it stopped when the other side of the flag passed. But the flag is a 3/8 inch dowel pin.
So in any situation were I used .027 mm for the gate distance will be incorrect. I have to reevaluate everything I have done for the last few weeks. The ratios concerning radius to mass will remain the same; because the relationship of the gate times will remain the same. But raw speed and F = ma %s will be inaccurate. Sorry: I had totally forgotten how the different modes worked.

The newest data confirms that the concepts are correct: That Atwood's make momentum. And that that momentum is controlled by the 'mass to inverse radius length' concept of the wheel or pulley. For example: The driving force will accelerate 1 kilogram at a radius of 100 just as easily as it will accelerate 100 kg at a radius of one
Now lets place the current experiment on a commercial scale by multiplying the mass by around a million; and the linear dimensions by 20. The throwing radius would then be 32.6 m and the drive radius would be 200 mm. This shaft diameter of 400 mm does not need to support the mass of the system; it needs only to accelerate it. The accelerated or thrown mass is now 1 metric tons. But what is the drive mass?
Lets use an evenly spaced stack of 100 one ton masses stacked one hundred meters high.
To obtain a velocity of 44.29 m/sec at the 32.6 meter radius (of the throwing wheel) we will need a shaft surface speed of .2717 m/sec. This 44.29 m/sec will allow the bottom one ton mass to be thrown back up to the top of the stack of one hundred.
We can now employ the stacking concept to obtain the drive mass. The machine is capably of throwing the accelerated mass of 1 metric tons up 100 meters. The stack of one hundred 1 ton masses is reconfigured when you throw one mass from the bottom to the top. The drive mass would then be (1 tons * 100) 100 metric tons, in an evenly placed stack 100 meters high.
When we employ a drive mass of 100 metric ton we then need to have a wheel with significant rotational inertia. It could have a rotational inertia of 1900 metric tons. We then have 100 ton accelerating 2000 ton (100 ton + 1900 ton) for a acceleration rate of .4905 m/sec/sec.
With an acceleration of .4905 m/sec/sec a drop distance of .07525 m gives you a final velocity of .2717 m/sec at the shaft; and 44.287 m/sec at the 32.6 m circumference. At 44.29 m/sec the one ton mass will rise 100 meters. This is a total momentum of (2000 ton *.2717 m/sec) 543,400 units. And the one ton rise 100 meters costs 44,290 units.
So the system can throw one ton from the bottom of the stack back to the top of the stack and it has only dropped 7.525 cm. The drop distance of .07525 m goes into 1 m 13.29 times. You can drop it 13 times and you only need one drop of this distance.
The one ton masses are one meter apart so you have .92475 meters of free energy. Further: after the 1 ton is thrown from the 32.6 m circumference the wheel and stack are still moving .2717 m/sec at the shaft. This is 499,113 units of momentum. As the wheel feeds another one ton mass out to the circumference the speed will drop. But to bring the wheel and stack back up to speed the stack only needs to add 44,290 units of momentum. The stack does not need to restart the wheel every time it throws.
When the second (to be thrown) mass has reaches the circumference at 32.6 meters; the wheel will have slowed to .24955 m/sec at the shaft (before any force has been added). The dropping stack needs only to accelerate the system back to .2717 m/sec at the shaft. This will be achieve with less than 7cm of drop; I would guess around 2cm. You will need giant electric generators to hold back this acceleration.
I began collecting data in a different manner. The flag is a 9.3 mm dowel: the distance between the gates is 27 mm: I place the electronics timer in pulse mode. I am now placing the flag between the gates. The flag comes out from between the gates: the first gate starts the timer as the flag interrupts the gate. After nearly one full rotation: the flag interrupts the second gate and the timer stops.
Now the times are in full seconds: 3.73 seconds is common. The times are the same for a balanced mass of 245.6 grams on the circumference at the 97 mm (radius): and for 2383 grams of balanced mass at the shaft of 10 mm. You can also move the driving force: and that will remain equal if you acknowledge the (one tenth mass at 10 times the radius: or five times the mass at one fifth the radius) radius to mass concept. One ton at 32.6 m is like 163 ton being accelerated at .2 meters.

By changing the dimensions of this experiment we can get a doable measurable experiment. One hundred meters is too high for an amateur to measure but 10 meters is not. You can tell the difference between 30 feet and 60 feet. I have already throw to the top of our trees, which is 80 to 90 feet.
So lets rearrange the above experiment to throw 10 meters upwards. This is also the height of the stack of drive masses.
And lets let the drive distance drop be .0667 meters. Lets place one kilogram in this drive distance of .0667 meters. One meter can be divided by .0667, which shows that 15 one kilogram masses can be evenly spaced along the distance of one meter.
There are 10 meters for such 15 mass units so the drive mass is 150 kilograms.
The rotational inertia of the wheel should be about 19 times (19 * 150 kg = 2850 kg) this is so that the total mass accelerated is 20 times the quantity of the drive mass. This 20 to one will give us an acceleration of .4905 m/sec². 9.81 m/sec² / 20 = .4905 m/sec²
So the speed (at the shaft) after the drive mass has dropped .0667 meters is going to be about .2557 meter/sec. The square root of (.0667 m * 2 * .4905 m/sec²) = .2557 m/sec.
Now this is 3000 kilogram moving .2557 m/sec which is 767.1 units of momentum.
One kilogram moving 14.007 will rise 10 meters. This is 14.007 units of momentum. One kilogram rising ten meter will restore the configuration of the stack; which has dropped .0667 meters.
So the system can reconfigure with 14.007 units of momentum and it produces 767.1 units.
Now we need to apply the radius to mass rule for the rotational inertia of spinning objects. If the accelerated mass (3000 kg) is at ten times the radius of the shaft then the wheel only needs to have a mass of 300 kilograms. But this 300 kg is moving 2.557 m/sec. For the same 767.1 units of momentum.
The thrown mass is 1 kilogram: which has come off the bottom of the 10 meter stack.
So: as in the cylinder and spheres experiment we attach two ½ kilograms masses to the outside of the wheel. We spin the wheel and release the spheres on their tethers and one of them will rise up to the tree tops. And the other sphere will thump into the ground; I have done it many times.
But not with a 300 kilogram rim mass wheel. So lets start downsizing.
I use 152 gram one inch spheres. Two of these are 304 grams instead of the 1000 grams in the above experiment. This means that the 300 kilograms for the one kilogram above becomes 91.2 kg for the .304 grams of thrown mass. .304 * 300 kg = 91.2 kg.
Using a thrown mass of .304 kg give a drive mass of .304 kg * 150 = 45.6 kilograms. And an accelerated mass of 912 kg and a momentum of 912 kg * .2557 m/sec = 233.19 units.
This means that I can walk into the yard and spin a 91.2 kilogram rim up to 2.557 m/sec and see if one .152 kg mass can rise 10 meters up.
Well maybe some of you are 91 kilogram hombres; but I am more like a 3 kilogram guy. But if I use a 3 kilogram wheel does that mean that I can spin it 30 times faster? 91.2 kilograms moving 2.557 m/sec is 233.1 units of momentum. And 3 kilograms moving 76.71 m/sec is also 233.1 units of momentum.
So this means I can take a 3 kilogram rim that has two 152 grams masses attached, in the cylinder and spheres fashion, and I can spin it as fast as I can. And after I release the spheres when one of them reaches the tops of the trees then it proves that the .0667 meter drop can be repeated. The spheres have more energy (½ * 1 kg * 14.007 m/sec * 14.007 m/sec + Extra meters) than (½ * 3000 kg *.2557 m/sec .2557 m/sec) when they rise higher than 10 meters.
Did you notice that the 3 kilogram wheel already has a velocity of over 5 times faster than the spheres would need to rise 10 meters?
Okay lets check some of the relationships to see if I am staying consistent.
The stack of one kilogram masses has 767.1 units of momentum for the one thrown kilogram to work with; and a thrown mass of .304 kg has 233; checks
The one kilogram needs 14.007 units of momentum to cycle and the .304 kg needs 4.26
300 kilogram moving 2.557 m/sec has 767.1 units of momentum: and 3 kilograms moving 77.6 m/sec has 233 units 14.007 / 767.1 and 4.26 / 233 is only 2% to cycle.
Baseballs have a mass of 142 grams and they make great tree ornaments.

The difference between Newtonian momentum and energy is that momentum is Force times time: and energy is force times distance. And the original definition of energy was Liebniz's mv².
So the energy of motion is Fd = mv² and F = ma is Newtonian momentum.
Acceleration is v / t. So F = ma is Ft = mv
Distance is v * t So Fd = mv² is F * v * t = m * v * v We can simplify the equation by dividing both sides by v. And we get Ft = mv which is Newton's equation. So it appears that kinetic energy is a mathematical error: which would explain why it can not be conserved in the lab. Because the correct equation had been multiplied by v.
More importantly the energy of motion needs imaginary heat to be conserved; as in a ballistic pendulum. Distance is merely a consequence of acceleration; and kinetic energy is not a conserved quantity.
I thought I would discuss Newtonian Mechanics in a manner that would be useful. And at the same time discuss the alternate concept that replaced it.
When an object is dropped it gains a speed of 9.81 m/sec in the first second. The object travels 4.905 meters because its average speed is (the starting speed) + (the final speed) / 2 = 4.905 m/sec. This average velocity is applied for 1 second. For a one kilogram mass this is 9.81 units of momentum.
When this same object continues to drop the second second it will travel 14.715 meters in that second. This is because its original speed is 9.81 meters per second and the final speed is 19.62 meters per second for an average speed of (9.81 + 19.62) / 2 = 14.715 meters traveled in one second. For a one kilogram mass this produces 9.81 units of momentum; because the speed has increased by 9.81 m/sec.
When this same object continues to drop the third second it will travel 24.525 meters in that second. This is because its original speed is 19.62 meters per second and the final speed is 29.43 meters per second for an average speed of (19.62 + 29.43) / 2 = 24.525 meters per second; and it traveled for one second. For a one kilogram mass this produces 9.81 units of momentum; because the speed has increased by 9.81 m/sec.
When this same object continues to drop for the twentieth second it will travel 191.295 meters in that second. This is because its original speed is 186.39 meters per second and the final speed is 196.2 meters per second for an average speed of (186.39 + 196.2) / 2 = 191.295 meters per second; and it traveled for one second. For a one kilogram mass this produces 9.81 units of momentum; because the speed has increased by 9.81 m/sec.
So Newtonian Mechanics produces 9.81 units of momentum for each and every second that the one kilograms mass drops.
What replaced Newtonian Mechanical is the Law of Conservation of Energy that says the first second produces 48.11 joules of energy: and the second second produces 144.35 joules: the third 240.59 joules and the twentieth 1876.6 joules.
When you use a 1 kilogram drive mass in a 10 kilograms Atwood’s one second produces 4.811 joules of energy and 9.81 units of momentum.
A 1 kilogram drive mass in a 100 kilogram Atwood's produces 9.81 units of momentum and .4811 joules of energy in one second.
Force applied for one second always produces a uniform quantity of momentum; and the production is independent of how fast the accelerated mass is moving. And momentum is conserved in collisions.
Force times distance on the other hand produces an infinite number of energies. As an example one kilogram dropped for one second can produce: .4811 joules, 4.811 J, 48.11 J, 144.35 J, 240.59 J, 1,876.6 joules. Yet when interacting in a collision most of this energy is lost. Kinetic energy is not conserved by a ballistic pendulum because multiplying by an extra v give massive amounts of energy that can not be found.

A yoyo uses gyroscope techniques as you know.
Rotation of the Earth is a quite involved process. Deep knowledge of certain areas of physics is indispensable for its understanding and researching. It is necessary to clarify the physics of rigid bodies’ rotation because the usage of terms precession and nutation by experts in geosciences are generally not correct, sometimes confused. In physics nutation is the motion of a free gyroscope, and precession is the motion of a heavy one. The torque of an external force causes the precession of a heavy gyroscope, but nutation is a motion of a free gyroscope, if its rotation has not started around its symmetry axis. While the cause of precession of a rotating body always arises from the effect of external mass sources, nutation is exclusively a function of the intrinsic mass distribution of the rotating body. After discussing some important aspects of rotational mechanics (motion of a free and a heavy gyroscope), the Earth’s nutation and certain elements of Earth’s precession (normal precession, lunisolar precession, planetary precession, disturbing precession) are defined and discussed. A new terminology is proposed, e.g. disturbing precession instead of the widely spread deceptive expression of astronomical nutation.
Read more here > https://www.researchgate.net/figure/PrecesionalmotionoftheEarthsrotationaxislunisolarprecession_fig4_250008115 and https://www.researchgate.net/publication/250008115_Physical_backgrounds_of_Earth's_rotation_revision_of_the_terminology/download from that website.
Regarding the gyroscope technique and newton, what stays in motion actually stays in motion, exactly the same with the torque output.
The free energy secret is in toys whey and i bought one for fun, https://www.amazon.co.uk/gp/product/B007FF5PVC/ref=ppx_yo_dt_b_asin_title_o01_s00?ie=UTF8&psc=1
The gyro scope technique is the secret sole key for some free energy inventions so click and read this and come to your own conclusion > https://drmyronevans.files.wordpress.com/2016/12/iufooutlinev23.pdf

More about the discussion on the 12th.
If a one kilogram mass accelerates another 99 kilograms (as in an Atwood's machine) for 20 seconds it will be traveling only 1.962 m/sec. It will have dropped 19.62 meters and the 100 kg system will have 192.47 joules of energy.
If a one kilogram mass accelerates itself in free fall (in a vacuum) for 20 second it will be traveling 196.2 meters per second. It will have fallen 1962 meters and it will have lost gravitational potential energy of 19247.2 joules. If we direct the one kilogram moving 196.2 m/sec upward it will 'ideally' travel back up 1962 meters; and its gravitational potential energy will be restored. But its kinetic energy (19247.2 J) is a math error and it is not there. Restoration of height is a natural consequence of velocity.
The true motion energy for one kilogram moving 196.2 m/sec can be found by collision with a 99 kilogram mass at rest. The combined mass of 100 kilograms will be moving 1.962 m/sec and will have 192.47 joules of energy.
The kinetic energy is not there: it doesn’t disappear as heat (as in a ballistic pendulum) it was simply never there. The kinetic energy scheme was invented to prevent people from seeing how easy it is to make energy.

More about the discussion on the 12th.
If a one kilogram mass accelerates another 99 kilograms (as in an Atwood's machine) for 20 seconds it will be traveling only 1.962 m/sec. It will have dropped 19.62 meters and the 100 kg system will have 192.47 joules of energy.
If a one kilogram mass accelerates itself in free fall (in a vacuum) for 20 second it will be traveling 196.2 meters per second. It will have fallen 1962 meters and it will have lost gravitational potential energy of 19247.2 joules. If we direct the one kilogram moving 196.2 m/sec upward it will 'ideally' travel back up 1962 meters; and its gravitational potential energy will be restored. But its kinetic energy (19247.2 J) is a math error and it is not there. Restoration of height is a natural consequence of velocity.
The true motion energy for one kilogram moving 196.2 m/sec can be found by collision with a 99 kilogram mass at rest. The combined mass of 100 kilograms will be moving 1.962 m/sec and will have 192.47 joules of energy.
The kinetic energy is not there: it doesn’t disappear as heat (as in a ballistic pendulum) it was simply never there. The kinetic energy scheme was invented to prevent people from seeing how easy it is to make energy.
I agree 100% about the kinetic energy scheme that was invented and the way the earth processes work would shatter that concept, but science it self rejects this explanation sadly, if you applied all the known laws of energy to it, it won't answer as to why it works or make sense in that regard but not all of it though, there are some independent more open minded scientists to accept the truth about it.
This torque output effect is better explained using Russian torsion theory regarding what your pondering about and the gyro effect.
Here is something about torsion theory > https://blog.worldmysteries.com/science/torsionthekeytotheoryofeverything/
So view torsion as the magic key which is what your after, the torque manifests as a torsion wave with regards as to what your playing or working out with here.
Also read > https://infinityexplorers.com/teslastorsionfieldswavesbosnianpyramids
https://infinityexplorers.com/everythinglightincredibleinterviewnikolatesla1899
It is also crucial to work out the torque pendulum, here is some demonstration > https://www.youtube.com/watch?v=GcPO2pKaiO0
To keep the torque pendulum running you will need an input (maybe electric motor or whatever) taking advantage of this to keep the whole process going and maintained perhaps instead of manually doing it.
It would have an effect of little power input to bigger output power perhaps with the way this effect works
It would have to be maintained using forced precession methods i would not be surprised if you get more out than in but take that with pinch of salt

I took a video of a 2.5 kilogram cart wheel that can be stopped by two 152 gram spheres. I tried to post it on line but YouTube said the app was temporarily closed.
This cart wheel would be the same basic experiment as the cylinder and spheres (see Delburt Phend; you tube). The mass ration is about 9 to 1. This means that when the spheres have all the motion they have nine times as much energy. ½ *.305 kg * 9 m/sec * 9 m/sec = 12.35 joules; ½ * 2.745 kg *1 m/sec * 1 m/sec = 1.37 joules: momentum is conserved; .305 * 9 roughly equals 2.8 *1
The wheel is roughly the same shape as the gyroscope. And the interesting thing is that the energy transfer to the spheres only takes about one third rotation. Experiments have shown that the stop occurs in the same quantity of rotation no matter what the rate of rotation. The quantity of friction in one third rotation must be very small. So you have a huge quantity of energy produced with very little friction and in very little time.

There should be a very simple experiment on the internet: but I could not find it. He experiment would be a puck, on a frictionless plane, rotating on the end of a string; and then the string comes in contact with an immovable pin. The puck then begins rotating about the immovable pin. Lets say the string length is originally 50 cm; and then after it begin rotating about the pin it has a 10 cm string length. So we have a puck rotating at 50 cm and then at 10 cm.
When we burn through a string that is rotating a puck; the puck will move in a straight line with the same speed as it was moving around the arc of the circle. Therefore we also know that the speed around the smaller 10 cm circle will be the same as the speed of the puck in the 50 cm circle.
Just to get real numbers lets say that the puck has a mass of 80 grams and the speed is 3 m/sec.
So in the large circle the energy is: ½ .080 kg * 3m/sec * 3/sec = .36 joules
In the 10 cm circle the energy is: ½ .080 kg * 3m/sec * 3/sec = .36 joules
The linear momentum in both circles is: .080 kg * 3 m/sec = .24
But what about angular momentum conservation? The angular momentum is L = R * linear momentum.
The angular momentum of the .5 meter circle is: .5 m * .24 = .12
The angular momentum of the .1 meter circle is: .024
So the angular momentum of the system is not conserved.
Shouldn't you be skeptical of all those high school teachers and college professors that told you something that is false.
Angular momentum is not a conserved quantity. Some lecturers will actually tell you that linear momentum changes so that angular momentum can be conserved. These are the people that never back up their conjectures with experiments.

https://www.youtube.com/watch?v=AzgasHgVOy8
The experimenter cuts the string and the momentum is then equal to the linear momentum (velocity * mass) that the puck had while traveling around the arc of the circle.
It doesn't make any difference what length of string the experimenter used, he could have used any length that fits the table. That would of course be an infinite number of lengths for R; the radius. This infinite number of Rs would give you an infinite number of angular momentums: because angular momentum is linear momentum times R.
Further: the experimenter could reattach the puck to a different length of string. That would give you the same linear momentum on a different length of string. That would give you a different angular momentum. Obviously angular momentum is not a conserved quantity.
So what else did all those high school teachers and professors tell you that was false.

You might want to read that:
https://courses.lumenlearning.com/boundlessphysics/chapter/conservationofangularmomentum/ (https://courses.lumenlearning.com/boundlessphysics/chapter/conservationofangularmomentum/)
Give yourself the time to understand. Your high school teacher was not such a bad guy.
Greetings, Conrad

They use the ice skater because there are no real experiments that prove that angular momentum is conserved. There is no way to quantitatively evaluate (at least not in high school, or at most universities) the mass of an ice skater or someone seated on a spinning chair.
Use a pincher to hold the string instead of a scissor, and the puck will proceed with the same arc motion in a smaller circle. Angular momentum is not conserved: but it is used to cover for their other false concepts.
Linear momentum is conserved in rotation. Not all lines are straight.

Now here is a very good one.
I do not know if he would have announced results that would debunk the accepted theory; but he does go to extremes to get the results he wants. He yanks on the string; faster and faster; until he gets the results he wants. Is it possible that yanking on the string adds linear velocity to the ball?
Indeed; does he not say that the linear velocity doubled. He yanks on the string fast enough to increase the linear velocity (tangent velocity) enough to double the original linear velocity. Now how do you increase velocity? How do you increase linear velocity except with the application of outside force? He has added outside force.
He yanks on the string fast enough so that the force in the string is no longer at 90° to the original direction of the motion. It is no longer a balanced force but imbalanced. Thus: he increases the linear velocity of the ball.
If he were willing to accept the correct answer for the experiment he would have conducted the experiment as gently as possible; to make sure that he was not adding external force.
https://www.youtube.com/watch?v=LBeX74AVFgU
Rather than force the wanted answer: why doesn’t he rotate a puck on the end of a string on a frictionless plane? And then interrupt the string with a pin somewhere along its length. The puck would continue with the same tangent velocity but in a smaller circle.
This is an important experiment because it proves that in order to conserve angular momentum you must violate the law of conservation of linear momentum.

Also check the kinetic energy increase in this experiment. Does it quadruple? Why is that not a problem?
https://www.youtube.com/watch?v=LBeX74AVFgU
And then this formula L = mvr is used to predict that the thrown mass; in the Dawn Mission despin event, is moving very slow.

Does the linear velocity remain constant in this experiment, by Paul Nord?
I could not find an address: but look under; circular motion paul nord
It is an unwinding puck on the end of a string.

Lets review the other hoax formula; The Conservation of Energy.
In a ballistic pendulum there is a small amount of motion lost to heat, but the loss is so small that the linear momentum conservation formula hardly notices the change. In the very same ballistic pendulum experiment this same loss of heat almost makes the kinetic energy disappear. In truth: this massive amount of heat, to explain the loss of energy, is made up. It is just an excuse to pretend that the Law of Conservation of Energy is true.
If the heat was lost then the motion could not come back; but the double despin experiment shows that the motion does come back.
AWHS davjohn41 ballistic pendulum you tube This is another experiment with actual data: note 4:10 minutes in. The teacher forgot to tell them the accepted excuse. Styrofoam is not a heat sink.

This is copied from Wikipedia: we are given a set of information and then a false statement. And we are missing one needed piece of information; which is the, rotational mass, diameter of the satellite.
Wiki: “As an example of yoyo despin, on the Dawn Mission, roughly 3 kg of weights, and 12 meter cables, reduce the initial spin rate of 1420 kg of spacecraft from 36 RPM down to 3 RPM in the other direction[1]. The relatively small weights can have such a large effect since they are far from the axis of the spin, and their effect grows as the square of the length of the cables. “
From the picture; the satellite has some equipment extending to give it the 1.7m diameter, but I think its rotation mass radius is more like .4 meters. The twelve meters of extended cable is also a radius.
This gives you an initial rotational velocity of .4 m * 2 = .8 meter diameter * pi * 36 rpm / 60 sec = 1.50 meters per second. So the faulty formula gives you: 1420 kg * 1.5 m/sec *.4 m = 852 units of angular momentum; which is all given to 3 kilogram (I think it was actually 2.6 kg).
So according to the faulty angular momentum conservation formula the new velocity for the released masses is only; 852 = 3kg * V * 12 m = 23.66 m/sec. 54 miles per hour A catcher's mitt can handle 100 mph
So why not let this 72 units of momentum (3 kg * 24 m/sec) rewrap around the satellite; ballistic experiments prove that it could only give the satellite a rotational velocity of (72 = about 1420 kg * v) = about .0507 meters per second.
This .0507 m/sec is below the 3 rpm that they did get; and the masses were released.
For linear velocity to be conserved the 3 kilograms would have to be moving around 700 m/sec. This 700 m/sec is dangerous; and that is why they are released.
The double despin experiment proves that the satellite could be returned to the original rate of rotation by the rewinding masses, 24 m/sec won't restore the spin.
Yes I know: I have said this before; but it is worth 7 trillion dollars.

https://www.bing.com/images/search?view=detailV2&ccid=TMu2hSEA&id=6BB16497F82989ED2B478664F434FFEA6B0CE030&thid=OIP.TMu2hSEAPL8m07FjJA8ggHaFl&mediaurl=https%3A%
https://www.bing.com/videos/search?q=pendulum+with+peg&qpvt=pendulum+with+peg&view=detail&mid=4A152BF53AB0A335A03A4A152BF53AB0A335A03A&&FORM=VRDGAR&ru=%2Fvideos%2Fsearch%3Fq%3Dpendulum%2Bwith%2Bpeg%26qpvt%3Dpendulum%2Bwith%2Bpeg%26FORM%3DVDRE
I don't know where the short names when but here are a few, of what some refer to as Galileo's pendulum.
The energy and therefore speed at the down swing position is the same on both sides; as the upward swing starts. That also means that the magnitude of linear momentum is the same. But what happened to angular momentum conservation. L = mvr
Angular momentum is not conserved because the length, or radius, does not remain the same.
Kinetic energy is not a conserved quantity; so what is the only Law that explains the restoration of spin in the despin.

Pasco has a few experiments where they claim angular momentum is conserved. They do this by not changing the radius. The problem is that to get angular momentum you have to multiply linear momentum by the radius. They drop a disk on top of another disk; of the same dimension (Radius).
When you have a before and after experiment where the before and after radius are the same; then you are merely stating that linear momentum has remained the same. You have not searched out whether or not you can change the radius.
In the interrupted pendulum; also known as Galileo's pendulum, the radius is changed. And angular momentum does not remain the same as the pendulum swings through the down swing position. This experiment is used to prove that energy is a conserved property. Conservation of energy should not be a surprise because nothing (mass and velocity) has change.
Change the mass, as in a ballistic pendulum, and the conservation of energy drops out; with a totally makebelieve quantity of heat. The double despin proves that the motion is still there; it has not been lost as heat.

The wheel is roughly the same shape as the gyroscope. And the interesting thing is that the energy transfer to the spheres only takes about one third rotation. Experiments have shown that the stop occurs in the same quantity of rotation no matter what the rate of rotation. The quantity of friction in one third rotation must be very small. So you have a huge quantity of energy produced with very little friction and in very little time.
Ok so, to understand this i want you to consider the energy transfer along the string:
From our perspective we can view this as a ‘wave’, which has an amplitude and velocity
It’s frequency, however, is predetermined by our analysis.
It is a wave, traversing a string, therefore we know it has a wavelength equal to the free motion of the string. At 1/3 we find the node of maximum change in amplitude, and thus maximum transfer of rotational momentum.
Tension plays a major role in the scalar equation here
This is derived from the force between the two masses which approaches its’ peak at the same node
and subpeaks at subsequent nodes. (as does the remainder of momentum transfer; remember that an object that ‘stops’ and changes direction from our perspective doesn’t necessarily have a 0momentum and to know the true value we would first have to determine our own)
We dont need to apply relativity here, most of that is semantical for our application.
We can now apply a peg at the 1/3 node, such that as the cylinder rotates and the smaller masses apply tension to the string, at 1/3 rotation and at 1/3 the length: the string hits the peg.
Then we can observe the momentum change directions in almost a pure fashion

A disk with a 1.278 meter diameter and a thickness (height) of one decimeter can have a mass of one thousand kilograms.
To throw this disk up 30 meters will require a velocity of 24.26 m/sec. From d = ½ v²/a
This would be 24.26 m/sec * 1000 kg = 24,260 units of linear Newtonian momentum.
A stack of these disks that is 30 meters high would have a mass of 300,000 kilograms.
When this 300,000 kg stack is dropped 5 cm it will have a velocity of .9904 m/sec. This is 297,136 units of linear Newtonian momentum. You only need 24,261 units of the 297,136 units to reconfigure the stack, and the stack has only dropped half of the available distance.
So here is the procedure; you let the entire stack drop 5 cm. You transfer 24,261 units of momentum to the lower disk, you throw the disk back up to the top and the stack is ready to be dropped again.
The remaining 272,859 units of momentum and the remaining drop of 5 cm can be used to spin electric generators.
The 24.261 m/sec velocity of the one thousand kilograms can be achieved by use of the despin event.

John Mandlbaur; successfully argues that angular momentum conservation does not work in lab. You may find it interesting. john@baurresearch.com

Tie a string to a puck, at rest, on a frictionless plane; put the string through a small low friction tube. Pull the puck toward the tube. Does the puck begin accelerating in a clockwise direction? What in the pull would cause the puck to rotate and then accelerate that rotation?
Nor would you lose velocity; the friction in the tube is not upon the puck; in is on the hand. The inward pull will have little whatever to do with the motion of the puck.
Yet the establishment goes nuts when you tell them angular momentum is not conserved.

https://groups.spa.umn.edu/demo/mechanics/movies/1Q4020.mp4
At first the balls are far apart, and I measured the average rotational speed during this section to be 6.6 rad/sec (63 rpm).
Then he squeezes the handle and the balls are pulled nearer. During this section, I measured the the average rotational speed to be 17.5 rad/sec (167 rpm).
Finally, he released the handle and the balls resume their original distance. At this section, the average angular speed was 5.0 rad/sec (48 rpm).
So over the course of about 10 seconds, the speed dropped from 6.6 to 5.0 (24%). Hopefully, we can agree that is due to various losses  probably friction and aerodynamic drag.
But if we take a look at how the speed changes when the radius changes, we should get a good picture of how angular speed is related to radius.

I should say that this is someone else that is doing this evaluation.

Hello,related #211 I reed yesterday in old threads in a discussion between Gustav Pese and Kator01 as given example by Kator/Mich(a)el a similar physical configuration !
Probably he,Kator01, can give good advices/links/help !
Sincere
OCWL

Gustav Pese ... he should Rest in peace... he would love to see what Mr Zündel developed regardimg to gravity. Hello Mr de Lanca .. long time.

Hello,related #211 I reed yesterday in old threads in a discussion between Gustav Pese and Kator01 as given example by Kator/Mich(a)el a similar physical configuration !
Probably he,Kator01, can give good advices/links/help !
Sincere
OCWL
I refound it : https://overunity.com/30/extraelectricpowerpoweramp/45/ (https://overunity.com/30/extraelectricpowerpoweramp/45/) #54 and meaning that translated does more sense ::) :
A DESIGN FOR A PERFECT QUANTUM RESONANT PENDULUM
http://www.unifiedtheory.org.uk/ (http://www.unifiedtheory.org.uk/)
Here the principle of storing and retrieving centrifugal force is explained very well. Nobody had such an idea so far. However, it must be built differently than shown. Centrifugal forces, which are actually inertial forces (from the point of view of a corotating observer) take on enormous dimensions.
Here, too, the solution lies in moving away from the "rigid" concept towards a dynamic one so that the enormous potential can be decoupled.
Example: A weight of 100 grams generates a centrifugal force of approx. 1480 N [kg m / sec exp ^ 2) when rotated 20 cm from the axis at 3000 rpm.
First you have to temporarily store this force, if you let it run against a resistance for 1 second, for example (= impulse braking, the weight remains on the axis, it does not fly away tangentially) you get an impulse = F xt = force times time of 1480 kg meter / sec.
If you let go of the weight (as every hammer thrower does), you get the impulse = F xt = M x V (mass times speed) = 0.1 kg times 47.12 meters / sec = 4.7 kg m / sec, i.e. only approx. 1/300 of the possible centrifugal force impulse (in SI units)
What a difference if done right.
So far, nobody has thought of it because all circular systems are viewed rigidly.
The author claims that he took a DC motor with a similar configuration to 3000 rpm, then switched off the DC input, after which the motor continued to run with a vibrating noise for 20 minutes (autoresonance) before it switched off then had to brake violently.

Gustav Pese ... he should Rest in peace... he would love to see what Mr Zündel developed regardimg to gravity. Hello Mr de Lanca .. long time.
??? Wer sind Sie ? ???
;D Halloechen,helmut und Duisburg !? Oder Wohnort geaendert ? Immer noch politisch links von der Mitte ? ::)
Oder nun doch "eingefleischter" Merkelianer ? ;) ( An Sitzfleisch mangelt es dieser Person ja nicht ! :\ )
Die oide "UeberMutti der Nayssion" geht ja baldigst von Bord ! :'( Neuer bloeder Lotse gesucht !
Sie vielleicht ? 8) Erst Duisburg unter Sparkommissariat dann die ganze Nayssion ! Lockout forever ! :o ;D
Bei overunity.de war doch ein Frankfurter/Main( der mit dem Kaefer,nicht zweibeinig,vierraedrig) mit dem "Ei" (Wasserstoff/bzw.HHOKata/Elektrolyse)
anderes "Ei" https://www.weka.de/einkauflogistik/daswasserstoffeideskolumbusrevolutioninderenergieversorgung/ (https://www.weka.de/einkauflogistik/daswasserstoffeideskolumbusrevolutioninderenergieversorgung/)
https://www.youtube.com/watch?v=kfYiQVzB3Cg (https://www.youtube.com/watch?v=kfYiQVzB3Cg)
individuelle pro/contra Meinung https://www.youtube.com/watch?v=PkbjkXTBsyw
Schon wieder Freitag , also schoenes geruhsames home office  Wochenende wuenschend !
Bis dann mal
OCWL
p.s.: habe den Research nicht verlassen ,aber guenstige TechnologieAlternativen zu Imris/Ferreira Spulen
da spielt ein hiesiger ExPhillipsIngenieur, aktuell als Professor taetig, eine Rolle,wobei wir " uns technologisch" gegenseitig befruchten

Hallo Hallo . Ja , wir kennen einander. Der Käfer aus Frankfurt rollt wohl nicht mehr. Leider hat der Schöpfer sich zurück gezogen. Andere werkeln nun weiter an HHO und meine eigenen Aktivitäten mit OLGA der Trennzelle ruhen noch eine Weile. 300 Tage bis zur Rente. Zwischenzeitlich hatte ich einen Abstecher in die Electromedizin getan und stecke noch fest , bis ich mir genug Wissen über die Frequenzerzeugung im Ghz Bereich angeeignet habe. Aus Duisburg komme ich erst raus , wenn ich die Hütte verkauft habe. Mutti und ihre Schergen giften uns mit ihrer hässlichen Fratze an. Oft denke ich über Auswandern nach. Ich grüsse Sie.

You could construct a strong light wheel made of modern fiber that had a high quality center bearing. The wheel could be attached by a string to a 399 kilogram mass floating on dry ice. This same wheel could have a second string attached that had a 1 kg mass on the end.
The string for the one kilogram is arranged so that is can wrap around the wheel and the 399 kg is just pulled by the wrapping of the one kilogram.
What proponent of COAM are saying is that the one kilogram that is only moving twenty meters per second can wrap around the wheel and cause the 399 kilograms to achieve a velocity of 1 m/sec. The one kilogram only needs a string long enough to give the one kilogram a angular momentum sufficient for the 400 kg * 1 m/sec movement. This is ridiculous 20 units of momentum can not make 400 kg*m/sec.

This is about an Atwood's machine with 199.5 kg on one side and 200.5 on the other.
Yes; the difference between M1 and M2 is 1 kg; and the sum is 400 kg. That means that the acceleration ‘a’ is 1/400 times 9.81 m/sec/sec. a = .024525 m/sec²
Now the formula for time is used; d = ½ at² With a drop distance of one meter and an acceleration of .024525 m/sec² we have a drop time of 9.03 seconds.
Doing the same for free fall we get a drop time of .4515 seconds.
9.03 sec / .4515 sec = 20
The force of 9.81 N is applied for 20 times as long in the (400 kg to 1 kg) Atwood’s machine.
The final velocity of the Atwood’s is found by this formula; d = ½ v²/a; this is just d = ½ at² where v/a is substituted for t.
The final velocity for one meter of free fall is 4.429 m/sec; The final velocity for the 400 to one Atwood’s is .22147 m/sec.
This is 4.429 (1 kg * 4.429 m/sec) units of momentum for free fall and (400 kg * .22147 m/sec) 88.589 units of momentum for the Atwood’s. This is 20 times as much momentum produced; and 20 times as much time over which the force acts.
It takes 4.429 units of momentum to return the one kilogram of imbalance to the original position of the Atwood’s. And you get 88.589 out.
Now this is the quantity of momentum that Newton said was conserved and you put 4.429 units in and you get 88.589 units out. Every cycle; 4.429 in 88.589 out.
All that is needed is to transfer the motion of the 400 kilograms into the one kilogram: and that can be done with a cylinder and spheres machine.
There is also a modified Atwood’s that eliminate the up and down motion; if that bother some.

https://pisrv1.am14.unituebingen.de/~hehl/Drehimpuls.pdf
The last two are of interest; and the labrat disproves his own theory.
You have to click on the graph to make the video play.

This is about a discussion on the cylinder and spheres experiments.
If you were to shoot a 1 kilogram mass moving 20 m into a 399 kilogram block; the block will not start moving 1 m/sec. This is the conservation of energy and it will not happen; as you know. But you are proposing that you put the one kilogram on a string and then it will happen; but it won’t, the string will not make it happen.
You lose it all to heat; right. That is what the theory says, so stick with it.
It takes four frames to cross the black square at the start and four frames after the spheres return the motion. The spheres are much smaller than the cylinder and they can not return the energy to the cylinder. The spheres return the momentum to the cylinder, which means they received the momentum from the cylinder.
The Paul Nord experiment proves that angular momentum conservation does not work in the lab. That leaves you with only linear Newtonian momentum. And if Newton is correct this leaves us with free energy.
Circular Motion  YouTube
https://www.youtube.com/watch?v=zww3IIMRo4U

This is about an Atwood's machine with 199.5 kg on one side and 200.5 on the other.
Yes; the difference between M1 and M2 is 1 kg; and the sum is 400 kg. That means that the acceleration ‘a’ is 1/400 times 9.81 m/sec/sec. a = .024525 m/sec²
Now the formula for time is used; d = ½ at² With a drop distance of one meter and an acceleration of .024525 m/sec² we have a drop time of 9.03 seconds.
Doing the same for free fall we get a drop time of .4515 seconds.
9.03 sec / .4515 sec = 20
The force of 9.81 N is applied for 20 times as long in the (400 kg to 1 kg) Atwood’s machine.
The final velocity of the Atwood’s is found by this formula; d = ½ v²/a; this is just d = ½ at² where v/a is substituted for t.
The final velocity for one meter of free fall is 4.429 m/sec; The final velocity for the 400 to one Atwood’s is .22147 m/sec.
This is 4.429 (1 kg * 4.429 m/sec) units of momentum for free fall and (400 kg * .22147 m/sec) 88.589 units of momentum for the Atwood’s. This is 20 times as much momentum produced; and 20 times as much time over which the force acts.
It takes 4.429 units of momentum to return the one kilogram of imbalance to the original position of the Atwood’s. And you get 88.589 out.
Now this is the quantity of momentum that Newton said was conserved and you put 4.429 units in and you get 88.589 units out. Every cycle; 4.429 in 88.589 out.
All that is needed is to transfer the motion of the 400 kilograms into the one kilogram: and that can be done with a cylinder and spheres machine.
There is also a modified Atwood’s that eliminate the up and down motion; if that bother some.
Can this transfer be done by the different means?
For example using m * v = F * dt
In your case for the t= 1 sec F is equal 80 N
The piston of the hydraulic cylinder can be pressed with 80 N for 1 sec, which can be used to turn the generator...
I can be wrong, though.
Tnx

I did an air table experiment today were I used the center pin of the air table to interrupt the string on a puck with a 108 cm radius. After the string hit the pin; the new radius was 14.9 cm. This is a radius change of 108 cm to 14.9 cm = 7.25. 108 cm is 7.25 times larger than 14.9 cm.
Therefore if angular momentum is to remain the same the linear velocity of the puck will have to increase by a factor of 7.25.
I did a video and I placed this video on my computer; I put the video in a frame by frame mode and counted the frames needed to cross a distance of 2.86 cm on the puck. It took 21 frames to cross this distance on the puck when the radius was 108 cm and when the radius was 14.9 cm. At 240 frames per second this is a speed of .327 m/sec for both the small circle and the large circle. There is no change in velocity when you go from the large circle to the small circle.
I did it in the opposite direction as well; going from the small 14.9 cm circle to the 108 cm circle. There is no change in velocity when you go from the small circle to the large circle.
Now I will try to post the video.

https://www.youtube.com/watch?v=egP0bjVi2ss

Lets use real numbers from commonly done real experiments.
Four hundred kilograms floating on dry ice can be accelerated by a string that is draped over a pulley. The string is weighted with one kilogram that can drop one meter.
After the one kilogram has dropped one meter the velocity is .2212 m/sec and the momentum is 88.7 kg * m / sec.
Using the despin procedure (as demonstrated by the 36 g and 2000 experiment) you can transfer this momentum to a small one kilograms mass.
A one kilograms mass with 88.7 units of momentum will be moving 88.7 m/sec.
The one kilogram moving 88.7 meters per second can rise 401 meters.
At a height of 401 meters one kilogram has 3933.81 joules of potential energy.
This series of events was started with one kilogram at one meter of height which has 9.81 joules of potential energy.

How much momentum is produces by a 12000 kg Atwood with 5985 kg on both sides; and the Atwood’s uses a 30 meter high stack of 30 one kilogram masses as a drive mass. And how does that momentum compare with the momentum needed to reload the stack after a drop of one meter?
input
You have a 30 meter high stack of 1 kilogram masses spaced one meter apart. To increase the height of this stack by one meter; you only need to remove the bottom kilogram and place it on the top of the stack. To do this you need to give the one kilogram a velocity of 24.26 m/sec. This velocity will require the application of 9.81 newtons for 2.473 sec.
Formulas used are: d =1/2 at² and d = ½ v²/a and F = ma
The input force * time is 9.81 N * 2.473 seconds = 24.26 N * sec; for one kilogram this is 24.26 m/sec
The input momentum is 24.26 m/sec * 1 kg = 24.26 kg*m/sec
Real numbers for output momentum;
A stack of 30 kilograms exerts a force of 30 kg * 9.81 N/kg = 294.3 N; upon a mass of 12000 kg (5985 kg + 5985 kg + 30 kg).
For an acceleration of F = ma: 294.3 N = 12000 kg * a = .024525 m/sec²
After a drop of one meter the entire 12000 kilograms will be moving .22147 m/sec.
Formula used: d =1/2 v²/a = the square root of (1 m * 2 * .024525 m/sec²) = .22147 m/sec.
For a momentum to 2657.67 kg * m/sec; 12,000 kg * .22147 m/sec = 2657.64 kg* m/sec.
This takes 9.03047 sec for the stack to drop one meter; the stack has a force of 294.3 N, for 9.03047 sec * 294.3 N = 2657.66 N * sec.
So we put 24.26 N * sec in; and we get 2657.65 N * sec out, 109.5 times more momentum out than in.
The cylinder and sphere proves that this large slow motion can be given to a small object and in fact the small mass can give the motion back to the large mass. So the circle is complete.
You arrange for the large mass to give a small portion of it motion to the small mass and you have a cycling system, worth 7 trillion dollars.

sure glad I didn't want to be a mathematician, this hurts my dendrites just reading it.,

How much momentum is produces by a 12000 kg Atwood with 5985 kg on both sides; and the Atwood’s uses a 30 meter high stack of 30 one kilogram masses as a drive mass. And how does that momentum compare with the momentum needed to reload the stack after a drop of one meter?
input
You have a 30 meter high stack of 1 kilogram masses spaced one meter apart. To increase the height of this stack by one meter; you only need to remove the bottom kilogram and place it on the top of the stack. To do this you need to give the one kilogram a velocity of 24.26 m/sec. This velocity will require the application of 9.81 newtons for 2.473 sec.
Formulas used are: d =1/2 at² and d = ½ v²/a and F = ma
The input force * time is 9.81 N * 2.473 seconds = 24.26 N * sec; for one kilogram this is 24.26 m/sec
The input momentum is 24.26 m/sec * 1 kg = 24.26 kg*m/sec
Real numbers for output momentum;
A stack of 30 kilograms exerts a force of 30 kg * 9.81 N/kg = 294.3 N; upon a mass of 12000 kg (5985 kg + 5985 kg + 30 kg).
For an acceleration of F = ma: 294.3 N = 12000 kg * a = .024525 m/sec²
After a drop of one meter the entire 12000 kilograms will be moving .22147 m/sec.
Formula used: d =1/2 v²/a = the square root of (1 m * 2 * .024525 m/sec²) = .22147 m/sec.
For a momentum to 2657.67 kg * m/sec; 12,000 kg * .22147 m/sec = 2657.64 kg* m/sec.
This takes 9.03047 sec for the stack to drop one meter; the stack has a force of 294.3 N, for 9.03047 sec * 294.3 N = 2657.66 N * sec.
So we put 24.26 N * sec in; and we get 2657.65 N * sec out, 109.5 times more momentum out than in.
The cylinder and sphere proves that this large slow motion can be given to a small object and in fact the small mass can give the motion back to the large mass. So the circle is complete.
You arrange for the large mass to give a small portion of it motion to the small mass and you have a cycling system, worth 7 trillion dollars.
Last time I looked at it, they involve law of the energy conservation when transferring linear momentum between 2 bodies ( elastic).
I'm not saying this is correct, but this is what they use in the physics study books.

Momentum is conserved in all collisions. We don’t even need to discuss elastic collisions because momentum is conserved in those collisions as well.
Opponents to energy production from gravity would have you believe that in certain cases energy conservation overrides momentum conservation. This is an illusion and it is totally false. Momentum is conserved in all collisions.
So you have 12,000 kilograms moving .22147 m/sec and it is going to share its momentum of 2657.7 kg * m/sec with one kilogram. Who in the world would drop the Law of Conservation of Momentum and go to the law of conservation of energy? Are they to pretend that 12 tons moving over 1/5 meters per second can’t give 1 kilogram a velocity of 24.26 m/sec? Are they to pretend that 2633.44 units of momentum disappear?
The experiments like https://pisrv1.am14.unituebingen.de/~hehl/despin2kg.mp4 or the ‘cylinder and spheres’ do not show an inability to transfer the motion of large objects to small objects.

Momentum is conserved in all collisions. We don’t even need to discuss elastic collisions because momentum is conserved in those collisions as well.
Opponents to energy production from gravity would have you believe that in certain cases energy conservation overrides momentum conservation. This is an illusion and it is totally false. Momentum is conserved in all collisions.
So you have 12,000 kilograms moving .22147 m/sec and it is going to share its momentum of 2657.7 kg * m/sec with one kilogram. Who in the world would drop the Law of Conservation of Momentum and go to the law of conservation of energy? Are they to pretend that 12 tons moving over 1/5 meters per second can’t give 1 kilogram a velocity of 24.26 m/sec? Are they to pretend that 2633.44 units of momentum disappear?
The experiments like https://pisrv1.am14.unituebingen.de/~hehl/despin2kg.mp4 or the ‘cylinder and spheres’ do not show an inability to transfer the motion of large objects to small objects.
I wonder if cylinder and spheres is a special case, bc they look at the collision of 2 balls. I will try thinking how to make some kind of the mechanical device based on this.

The kinetic energy for the one kilogram is equivalent to 9.81 N applied for 2.47 seconds. 294.3 J
The kinetic energy of the 12000 kg flywheel would be caused by 294.3 N applied for 9.03 seconds. 294.29 J
When motion is transferred from one object to another; the force in the string is equal in both directions; the force is equal to itself.
When motion is transferred from one object to another the time over which the force acts is equal for both masses; the time is equal.
When the one kilogram mass is unwinding from the flywheel; the force in the string is equal upon the flywheel and upon the one kilogram mass.
The time over which this force is applied is equal upon the flywheel and upon the one kilogram mass.
So as the string applies force between the two; the change in the kinetic energy of the flywheel and the change in the kinetic energy of the one kilogram mass will not be the same. The force of 9.81 N for 2.47 seconds will stop the one kilogram mass but it would hardly faze the flywheel.
Once the flywheel is in motion; that motion that is shared with another mass will not conserve kinetic energy. It will conserve linear Newtonian momentum.

This is long but some may have interest; it produces energy.
What would be the acceleration; if an Atwood’s had a drive mass of 100 g at a 1 cm radius: and a balanced mass of 100 grams on each side, at a 10 cm radius?
100 g at 10 cm would have the same torque as 1000 grams at 1 cm. So you could have 1 kg at 1 cm on one side of the Atwood's with the 100 grams of drive mass. And on the other side you could have 100 g at 10 cm; and the Atwood’s would accelerate at the same rate.
In fact you could have all the mass at the 1 cm radius by trading the remaining 100 grams on the 10 cm radius for 1000 grams at the 1 cm radius. Now you have 2000 grams of balanced mass at the 1 cm radius with the 100 extra grams that would be the source of the accelerating force.
So you would have an accelerating force of: F = ma, .981 N = 2.1 kg * a: .981 N / 2.1 kg = .467 m/sec²
Okay: keep in mind that 1000 g at 1 cm is equivalent to 100 g at 10 cm. Their angular acceleration will be the same.
Also keep in mind that I am purposefully manipulating machines to my advantage; to make energy from the gravitational field. I will shift motion from one machine to another to achieve this purpose.
The acceleration of the Atwood's drive mass of 100 grams at a radius of 1 cm is .467 m/sec².
After a drop of one meter for only this 100 gram drive mass: the drive mass will be moving: .9664 meters per second. d = ½ v²/a
If the 2000 grams at 1 cm were present they would also be moving .9664 m/sec; but the 2000 grams is not present; what is present (as stated in the question) is (2 * 100 g at 10 cm) 200 g at 10 cm.
At 10 cm the two 100 gram masses are moving 9.664 m/sec.
At 9.664 m/sec the two masses have an energy of: ½ *.2 kg * 9.664 m/sec * 9.664 m/sec = 9.339 joules of energy. The drive mass of .1 kg is moving .9664 m/sec for .04669 joules. For a total final energy of 9.3857 joules
The drive mass of .1 kg dropped 1 meter for an input energy of; 9.81 N/ kg * .1 kg * 1 meter = .981 joules. From; Nm; or ½ * .1 kg * 4.429 m/sec * 4.429 m/sec = .981 joules.
So you see that the output energy is 9.3857 joules and the input energy was only .981 joules.
When released both these 100 grams masses, on the end of the tube, will rise 4.76 meters. And you started with 100 grams at one meter.
Mechanical questions: Release the masses at highest velocity while they are moving up. Catch them on a ledge and feed them back down through.
A comment from someone else:
So if your calculations appear to show a net gain in energy you must be making a mistake or there is some mass flow down which you have not accounted for.
It’s not my job to show what the error is; it’s up to you to demonstrate that the machine or mechanism you propose actually works like you claim. That means you have to build it; proving physics is wrong by using its equations to show it is, doesn’t make sense unless  maybe  you can mathematically prove you can create energy.
So until you build one there is no reason to believe you have created what is, in essence, a perpetual motion machine of the first kind (i.e. an effective efficiency larger than 100%).
Phend
This one is 957%. And this one has only one meter of working height; without a stack, or a large flywheel.
“Not your job” Oh you would be more than willing to point out the error; but you can’t find one. You can obviously do the math because you came up with the correct acceleration. You just can’t find an error.
I have tested all the steps in these machines with real experiments; and they all work.
Those that will accept a paradigm shift will become billionaires.

This is from someone of the oppositional view; but it is an excellent description of the cylinder and spheres experiments on youtube.
"Fine, I’ll humor you by doing some theoretical qualitative analysis and comparing it to some of your video evidence with cylinders and spheres.
The cylinder and spheres is a yoyo despin.
Let me tell you what I think will happen in a few conceptually simple cases.
If the wirelength is just right, it will be unwound and perpendicular to the cylinder exactly when the cylinder is no longer spinning.
This is the “ideal” case if you want to despin a satellite completely but hardly ever is achieved because things never happen exactly like you planned.
If the wire is too long, the cylinder will come to a stop before the wire is completely unwound and (completely) perpendicular to the cylinder. This will cause the cylinder to reverse its spinning  because the weights/string cause a torque  to some degree.
It entirely depends on the exact value of some parameters and variables e.g. mass of spheres, moment of inertia of cylinder, position of the wire when the cylinder stops  what will happen next.
The balls could reverse the cylinder again and rewrap, or start rewrapping and then reverse the spin.
We also can’t tell what happens when a wire is too short. There are just too many variables and parameters for anyone to take a guess what will exactly happen.
At least one of your own experiments actually has the cylinder reverse its spin twice before or while the wires appear to rewrap in the opposite direction (“132 grams / 1320 grams”).
Another configuration completes such a rewrap and then appears to start dewrapping again before hitting the floor (“spheres 305 g cylinder 972 gra”/”newtonian momentum energy increases”).
The “restart of despin” actually just shows a reverse in spin, and the “Double yoyo despin” does a rewrap and starts dewrapping again.
I won’t do any calculations here, first of all because you would not care, and second because the exact relationship between the variables and parameters make this much more complicated than the above descriptions might suggest.
It is possible mind you, just not something I am interested in doing if you’re not.
And NASA already did it (e.g. THEORY AND DESIGN CURVES FOR A YOYO DESPIN MECHANISM FOR SATELLITES) so why bother.
So my theoretical qualitative “prediction” matches with most of your video evidence.
It would be interesting if you would try out the same cylinder and spheres but with different wire lengths to check if my intuitions are right."

Were the forces mentioned in the Third Law of Motion the same force that was defined in the Second Law of Motion?
Danny Kodicek
, Educational Software Developer and Teacher
Answered 45m ago
Yes, all three laws of motion are essentially the definition of what we mean by force. Collectively, they also comprise the law of conservation of momentum: the first law tells us that momentum can only be changed by the action of a force; the second tells us the effect of force on momentum; and the third says that whenever a force changes the momentum of one object, another force must change the momentum of some other object by an equal amount in the opposite direction, thus leaving the total momentum unchanged.
Larry Rothstein
, Engineering at SAIC (1981present)
Answered 40m ago
In a way yes, and in a way no. The Second Law of Motion describes the motion of a body subjected to an External force. In this case it it the net Force on the body, or the sum of all External Forces. Note the emphasis on External. For example, you push a ball and the ball accelerates and flies through the air.
The Third Law of Motion describes the reaction to the External Force. For example, when your foot is in contact with the ball, the ball is pushing back on your foot. The ball accelerates and the your foot decelerates a small amount
Delburt Phend
32m ago
Note the answers to the 2nd and 3rd Laws question. What does this mean?
It means that the force in the tether (in the despins) causes an equal change of linear Newtonian momentum for the released mass and the satellite. F = ma is a linear (curved lines also) equation.
Dawn Mission yoyo despin is an example of a despin.

What is the acceleration of a one kg rim mass wheel; that has a one meter diameter and is accelerated by 100 kg suspended by a string from a 1 cm shaft?

A massless one meter beam balance has 4 kg on one end and 1 kg on the other. It is rotating about its center of mass at 6.19 radians/sec. What is the arc speed of the one kg?
The mass relationship is 4 kg to 1 kg so when the beam is rotating about it center of mass: the 4 kg will have a radius of 20 cm and the 1 kg will have a radius of 80 cm.
At 6.19 radians per second the 4 kg is moving .2 m * 6.19 rad/sec = 1.238 m/sec around the arc of the circle.
At 6.19 radians per second the 1 kg is moving .8 m * 6.19 rad/sec = 4.952 m/sec around the arc of the circle.
At 4.952 m/sec the 1 kg can rise 1.25 m.
One kilogram in a modified Atwood’s will accelerate a 4 kg block to 1.98 m/sec after the 1 kg has dropped 1 meter.
Okay lets go through this again: You drop a one kilograms mass to accelerate a 4 kilogram block in a modified Atwood’s.
This gives you a final velocity of: the square root of (1 m * 2 * 9.81 m/sec² * 1/5) = 1.9809 m/sec
And this is 9.9045 units of momentum. Because this is 5 kg moving 1.9809 m/sec
A balanced beam that has 4 kg on one end and 1 kg on the other end; and is moving 6.19 rad/sec also has 9.9045 units of momentum.
The one kilogram on the beam has half the momentum (4.95 units) and it can rise 1.25 m.
It was dropped 1 m.
If the one kilogram had all the momentum it would rise 5 m.
Physics with George 5：Rotation about the center of mass of a system like the MoonEarth  YouTube

2 things dilbert
1  to get your 9 units you had to spend 10 getting the 5kg up to 60rpm
2  to transfer all of the momentum into the 1kg mass:
you have to stop the 4kg mass…..

will anytime any picture here ? :)

smOky2 quote: 1  to get your 9 units you had to spend 10 getting the 5kg up to 60rpm
No: it only cost 4.429 units of momentum. If you drop a one kilogram mass one meter it will have a velocity of 4.429 m/sec. To throw a one kilogram mass up one meter it will cost you 4.429 units of momentum. The one kilogram was dropped one meter.
smOky2 quote: 2  to transfer all of the momentum into the 1kg mass:
you have to stop the 4kg mass…..
This is true. We can place this rotating beam in a horizontal plane. The 4 kilograms will have lost 7.9236 units of momentum: but it is in the same plane as that plane where it started; and it is at rest which is also how it started.
Now we have one kilogram that adds those 7.92 units of momentum to 1.98 and we have one kilogram moving (7.92 = 1.98) 9.9 m/sec and it can rise d = ½ v²/a; 5 meters. It was dropped one meter.
It seems possible that we can place a 4 kg block on one end of a horizontal beam and a 1 kg mass on the other end of a horizontal beam. Even though both masses are moving; I think this is doable.
As for pictures; “Delburt Phend youtube ” shows rotating objects giving all their motion to smaller mass spheres.
And the ‘MIT Atwood’s machine’ has an input momentum of .0443 units and an output of .4624 units of momentum. .01 kg * 4.429 m/sec 1.11 kg * .41666 m/sec
If this MIT Atwood’s was a modified Atwood’s you would have 1.100 kg moving .4166 m/sec (on a plane) and one meter away there would also be .01 kg moving .4166 m/sec. If these two masses were placed on the ends of a balanced beam; and the beam were rotating about its center of mass, then the 1.100 kg would be moving .2102 m/sec and the .01 kg would be moving 23.135 m/sec. At 23.135 m/sec the .01 kg will rise 27.255 m.
The .01 kg will rise 27.255 m and it was dropped 1 meter.
So one way to make energy is to combine a modified Atwood’s with a rotating balanced beam.

is that talk about the same?
mechanical device Kapanadze.

The Administrator is welcome to remove the kolbacict post. I thought I would post again so first timers see something that is on subject.
Look at the last statement: My point is: one way to make energy is to combine a modified Atwood’s with a rotating balanced beam.
You start with a modified Atwood’s. You have a 4 kg block on dry ice. The block is tied to a string that is draped over a pulley that has a 1 kg mass suspended from the other end.
The acceleration is 1 kg / 5 kg * 9.81 m/sec² = 1.962 m/sec²
The one kilogram is allowed to accelerate toward the floor for one meter. This also means that the 4 kilograms on the dry ice will accelerate for one meter at a rate of 1.962 m/sec.
After the one kilogram has dropped one meter you have 4 kilograms moving horizontally on dry ice at 1.9809 m/sec.
And you have 1 kg moving down at 1.9809 m/sec.
We then change the direction of the one kilogram and have it moving on a horizontal plane. Let’s direct the one kilogram so that it is moving in the opposite direction of the 4 kilograms.
So we have a 4 kg mass moving east on a frictionless plane. About a meter under it we have a 1 kilograms mass moving west.
At this point we connect the two moving masses to each end of a massless beam a meter long.
On this beam the two masses will rotate about their center of mass but the ‘tangent velocity’ of each mass will not remain the same. The total ‘tangent’ momentum will remain the same but the individual velocities will not.
The momentum of the one kilogram after it has dropped one meter is 1.98 m/sec * 1 kg = 1.98 units of momentum. The ‘tangent’ momentum of the one kilogram as it enters the rotation on the beam is 1.98 kg m/sec.
The momentum of the 4 kg is (4 kg * 1.98 m/sec = 7.9236) 7.9236 kg m/sec.
The total momentum is 7.9236 kg m/sec + 1.98 kg m/sec = 9.9 kg m/sec.
When an object is rotating about its center of mass it is also rotating about its center of ‘tangent’ momentum. So the 1 kg has half the momentum so it will be moving 4.95 m/sec.
The total momentum of the beam remained 9.9 kg m/sec.
At 4.95 m/sec the one kilogram can rise 1.25 m.
If the 1 kg is given all the momentum, as in the cylinder and spheres, it will rise; d = ½ v²/a; 5 m
So we start with a modified Atwood’s machine and then we place those masses in a massless beam. We then arrange for a cylinder and spheres event and we get a 500 % increase in energy.
I am going to point out your errors as they occur.
A line does not have to be straight; that is the whole point of curvilinear and rectilinear. You have at least two types of lines; curved and straight. You can’t properly look at Newtonian physic and have an improper definition of line. Your side want to set up a difference set of rule for mass moving in a circle; but Newtonian physics is about mass that is often moving in a circle.
Florens quote: “It will also have to spin at a larger radius, so, by the law of levers it will decrease its speed, because v/r is constant.”
Wrong: not if you are talking linear v. This is where your misconception of linear messes you up every time. Linear speed does not care one iota about r. It is the same as changing direction; you can change direction with a long pendulum or a short pendulum. You are not going to change v by using a long or short r.

The common misconception with an Atwood cylinder
is that they think they observe a gain in energy
due to an observation made from the velocity of the
rolling cylinder.
In actuality this has more to do with aerodynamics of rolling cylinders.
and why the cylinderplane flies.
Momentum is conserved, even in an Atwood

I did not know that an Atwood's (machine) had anything to do with a cylinder. Honestly what are you talking about?

Oh… ok so history for the chronologically gifted
Well some time after the death of Atwood,
his knowledge was advanced as far as science could
(a pully is generally cylindrical)
The most advanced form took place in space
in microgravity, using a cylinder and 2 spheres
on strings, that wound around the cylinder (pulley)
Different masses were tested etc.

The results confirmed that momentum is always conserved

What i am talking about is that you can use
the momentum of the large weight to
throw the smaller weight more than it ‘should’
but you are missing the point that you putt that
into the larger weight to begin with

Re: re: energy producing experiments
« Reply #241 on: November 16, 2021, 05:45:57 PM
The Administrator is welcome to remove the kolbacict post.
You are some kind of angry ... :)

I know this is a recent repeat but I want to cover the first paragraph.
And the ‘MIT Atwood’s machine’ has an input momentum of .0443 units and an output of .4624 units of momentum. .01 kg * 4.429 m/sec 1.11 kg * .41666 m/sec
If this MIT Atwood’s was a modified Atwood’s you would have 1.100 kg moving .4166 m/sec (on a plane) and, one meter away, there would also be .01 kg moving .4166 m/sec. If these two masses were placed on the ends of a balanced beam; and the beam were rotating about its center of mass, then the 1.100 kg would be moving .2102 m/sec and the .01 kg would be moving 23.135 m/sec. At 23.135 m/sec the .01 kg will rise 27.255 m.
The .01 kg will rise 27.255 m and it was dropped 1 meter.
This 27.25 times more energy is done without transfer all the motion to the small mass. If you transferred all the motion to the small mass (as in the cylinder and spheres) the energy increase would be 110 times (ideal).
"And the ‘MIT Atwood’s machine’ has an input momentum of .0443 units and an output of .4624 units of momentum. .01 kg * 4.429 m/sec 1.11 kg * .41666 m/sec"
The input momentum (.0443) is the mass times the velocity achieved by dropping a .01 kg mass 1 meter.
Any mass dropped one meter will develop a velocity of 4.429 m/sec. Therefore the momentum needed to throw an object up one meter is 4.429 units of momentum for each kilogram. Ten grams is .01 kg so it only needs .0443 units of momentum in order to rise 1 meter.
The .01 kg was dropped one meter; so to return the system to it original starting configuration .0443 kg m/sec is all that is needed.
The MIT Atwood’s covers one meter in 4.80 seconds for and average speed of .20833 m/sec, and a final speed of .41666 m/sec. The center of mass of the 1.100 kg does not move. But the 1.100 kg is moving .4166 m/sec at the end of the drop of the .01 kg mass.
So the output momentum is 1.110 kg * .41666 m/sec = .4625 kg m/sec
This is an increase of .4625  .0443 = .4182 kg m/sec
We can restart the MIT Atwood’s ten times. That is 1000% of the energy we put in.
smOky2: Could you post links or an address for the Atwood’s experiment conducted in space? Thanks

m1*g*h = 1/2 m1 * v(final)^2 + 1/2 m2 * v(final)^2 + m2 * g * h
———————————————————————————————————
m1*g*h  m2*g*h = 1/2(m1+m2)*v(final)^2
————————————————————————
(m1m2)gh=1/2(m1+m2)*v(final)^2
———————————————————
2[ (m1m2)g/(m1+m2)]h=v(final)^2
it doesnt matter when or where you stop it
when or where you transfer the energy between the 2 masses

You might explain what you are trying to say.
I assume you are saying that the energy of the .01 kg mass dropped 1 meter (v = 4.429 m/sec) is equal the energy of the 1.110 kg Atwood’s moving .41666 m/sec: which is quite correct. ½ m v²
I am not proposing that you give the energy of the 1.110 kg to the .01 kg I am proposing that you give the momentum of the 1.110 kg to the .01 kg
The 1.110 kg has .4625 units of momentum; if you give all of that momentum to the .01 kg it will be moving .01 kg * v = .4625 kg m/sec. v = .4625 kg m/sec / .01 kg = 46.25 m/sec.
At 46.25 m/sec it will rise 109 meters. It was dropped 1 m.
Cylinder and spheres machine give the motion of a massive cylinder to much smaller mass.

Their momenti are not separate, but part of one system.
In 238 years you are not the first person to walk this path

v = (sqrt)[19.6(1.110.1)/(1.11+0.1)]
———————————————————
v = 4.04479055 m/s

Name the others that have walked this path.
To transfer momentum from a large mass to a small mass is incredibly simple.
But; name the others that have walked this path.
v = (sqrt) [19.6 .01 kg /1.110 kg= .4202 m/sec ideal
The experimental value was very close .41666 m/sec

The momenti are one and the same object, they are linked.
It is not two separate momenti (one for each mass),
But rather one single quantity, from both interconnected masses in motion.
If the large mass suddenly becomes disconnected (no transfer of momentum),
the smaller mass is accelerating downwards at 9.8m/s/s
its’ upward momentum being instantly disconnected as well.
Cut the string
If a scenario were present such as an impact lever at the very bottom,
point of maximum velocity, such that the large mass hit and transferred
100% of its’ momentum into an upward thrust, sending the smaller mass skyward.
And you were to catch it at its uppermost peak::::: even with 0 losses from wind or anything
(pay attention here!!)
The Potential Energy in the small mass:
E=m[small] * 9.8m/s/s * height = m[large]*9.8m/s/s* initial height
Do the tests yourself
This has been done, literally, millions of times

show one of your millions of experiments. smokey
you can list it with one of those that has walked before
and with your Atwood's in space
The cylinder and sphere make large amounts of energy. Delburt Phend Youtube

Well, im not going to do your homework for you
This has been discussed on this forum somewhere
inside those threads are other members experiments
The Nasa video was linked a few times (spacelab possibly?)
Of course ignoring physics, and imagining that i would make things up
just to prohibit you from exploring, is your choice to do.
Feel free to learn first hand from your own personal experience:
If you think you can make it do what you think it will do
Then build one.

Or at the very least,
Set up a series of controlled experiments
observe the transfer of a pair of interconnected vectored momenti
to a state outside its frame of reference
The force and acceleration are in the downward vector
for both masses
They are linked by the tension of the string
the larger mass uses an equal amount of its’ own downward momentum
to bring the acceleration into a negative value
The remainder is split between the two
Let’s take an extreme example and allow your Atwood to be 2 meters tall
at that point, would you expect the smaller weight to simply fly up and break off its’ string?
No, it Stops
Just like every other time.

When you are doing construction,
The best way to get something to the top floor
is to put it in a bucket, on a pulley and the other bucket
is filled with trash on the way down
The dumb waiter elevator works the same
i’m pretty sure that has been going on since way before
the scientists wanted to examine the math
As Archimedes describes in the 2nd century B.C.,
mechanical systems were already well developed
Lets say we have a spring in the bottom of the bucket that set and latched
when the heavy weight hit the ground
And the value of the energy stored in this spring
were exactly equal to the impact force of the heavy weight
Then we place that spring underneath the smaller weight
at exactly the moment the large weight hits the ground
(before gravity has a chance to make the downward acceleration positive again)
And fired off the spring
E = m(large)*9.8*h(initial) = m(small)*9.8*h(final)
= m(large)*1 = m(small) * h(final)
h(final)= m(large)/m(small)
= 11.1m
11.1 * 9.8 * .1 = 10.878j
1 * 11.1 * 9.8 = 10.878j

If you are going to cite an experiment then you should be ready to at least provide a location or link; not have some vague memory of it.
The MIT Atwood’s produces enough momentum to make the dropped mass rise 110 times higher than what it was dropped.
And I have experiments that place the momentum of very large masses into very small masses. Delburt Phend youtube This is all that needs to be shown. (Atwood’s – “cylinder and spheres”)

Then show the M.I.T. Version that does not follow the known equations above.
If it does what you say, just use the potential energy of the smaller mass to reset the larger mass
And have (97x excess?) to use for something else
While the larger mass throws another small mass into the sky
(or did you miss a decimal?)

You got it smoky; The MIT Atwood’s produces 1.110 kg * .42042 m/sec = .46662 units of momentum.
When a cylinder and spheres gives all its momentum to a .010 kg mass then the mass will be moving .010 kg * v = .46662 kg m/sec = 46.667 m/sec
At 46.667 m/sec the .010 kg will rise 111 meters.
Only 1 of those 111 meters is needed to restart the Atwood’s for another 111 meters.
So the excess energy is easily 97%.

You might need to ask for a calculator for Christmas.

Can't your calculator find the error; then I won't need a new one. Please show the error.

You don’t show your work, so its hard to tell
But you appear to be off by 1 decimal, twice
Resulting in 100x a value you should have ended up with
I’m not here to correct your work,
I have shown above the mathematical solution.
If you are convinced that it will work
Then build it

BUONGIORNO COME STAI.
https://www.youtube.com/watch?v=jiAZF0pMfyA

BUONGIORNO COME STAI.
https://www.youtube.com/watch?v=jiAZF0pMfyA (https://www.youtube.com/watch?v=jiAZF0pMfyA)
Bene ! Salute !
https://www.youtube.com/watch?v=YLNLePqVaQU&t=310sdispositivo gravi centrifugo per la produzione di energia elettrica
Comments
David Davids vor 2 Jahren (https://www.youtube.com/watch?v=YLNLePqVaQU&lc=UgxFn4ZaXzGz1L5GHIh4AaABAg) some info. at 1:15 (https://www.youtube.com/watch?v=YLNLePqVaQU&t=75s), the input force mechanism, at left, has been correctly conceived; but poorly executed. in it, reductionMA is converted to INHERENT MA; in the largeradius wheel. power is, correctly, taken off at a radius that is smaller than the radius at which at which force was applied. Mechanical Advantage. THEN, the rotary output force is, stoopidly, converted to Reciprocating motion. ultimately, this machine does Nothing. there is NO energy gain in the secondary mechanism; in fact, the asymmetrical orbiting mass is acting as a governor/speed limiter. any attempt to utilize gravity to achieve an energy gain, Will fail. why is this true. any apparent energy gain on the 'downhillside' is ALWAYS offset, by the Energy Required to OVERCOME gravity, on the 'uphillside'. if interested, review my ou viaMA research. googletranslate governor https://www.youtube.com/channel/UCNSWIWz7trnMo4Wdbug0j9A/videos (https://www.youtube.com/channel/UCNSWIWz7trnMo4Wdbug0j9A/videos)
Best greeting to Seychelles their fishermans,in search of Excellence, ::) , or nice mermaid/s !
Adeus
OCWL

@ seychelles
A real work of art
and
an amazing investment of time and materials !
But it looks like the generator / motor is powering it.
There is no principle, of which I am aware, and by which this machine
can operate otherwise.
regards
floor

FAIRLY SIMILAR TO MR SKINNER GRAVITY MACHINE BUT IN REVERSE ROUND
ABOUT WAY.

@ seychelles
A real work of art
and
an amazing investment of time and materials !
But it looks like the generator / motor is powering it.
There is no principle, of which I am aware, and by which this machine
can operate otherwise.
regards
floor
Screwdriver pointed up, box wrench one end on the screwdriver
Tilt the screwdriver slightly one way then back the other.
Just a tiny bit, small small force required, each direction
get the timing right, you will feel it take off.
a small rodent could operate this
But get too close the wrench will take your face off
You’re creating a situation of constant gravitational acceleration,
by tilting the axis of rotation.
depending on where you place your fulcrum,
the energy required to tilt can be less than the change in potential energy
Another example is a small child moving a 3000lb barell of water.
Tilt it on its’ edge and spin it, the barell moves itself.

oops

You’re creating a situation of constant gravitational acceleration,
by tilting the axis of rotation.
Another example is a small child moving a 3000lb barell of water.
Well, in a straight horizontal line, we can move any mass for any distance as long as there is no friction. Without doing work, if you have already given the body the initial acceleration.
Or am I wrong, and this machine nevertheless works? ;)

The physics are the same as rolling a marble around in the interior of a bowl.
One can build up and store kinetic energy in the form of momentum.

Kinetic Energy can disappear; why not store momentum in the form of momentum.

Momentum of an object is
the mass of that object (Kilograms)
multiplied by
that object's speed (meters per second).
The direction of its motion is also a consideration
(speed and direction, given together are the objects velocity).
Momentum is perhaps the most basic form of kinetic energy.
The kinetic energy of an object is given as...
Ek = 1/2 M x V ^2
or
kinetic energy = 1/2 times the mass times the velocity squared.
This is also the kinetic energy required to accelerate an object of
a given mass to a specific velocity.
The kinetic energy the object has (due to its motion), is the same amount of energy as
was needed to cause its acceleration it to that specific velocity.

Incorrect: motion is caused by Force applied for a period of time. Ft
The product of force * time is linear momentum mv. not energy
Newtonian momentum is a conserved quantity; kinetic energy is not conserved.

No acceleration and no mass ?
edit .. momentum = mass x velocity p = mv
and
force = mass x acceleration F = ma
Delburt Phend quote
"motion is caused by Force applied for a period of time."
end of Delburt Phend quote
or
motion is caused by force applied for a distance
like i said
Momentum of an object is
the mass of that object (Kilograms)
multiplied by
that object's speed (meters per second).
The direction of its motion is also a consideration
(speed and direction, given together are the objects velocity).

Floor quote "This is also the kinetic energy required to accelerate an object of
a given mass to a specific velocity."
Motion is made by F applied to a mass not by kinetic energy

Floor quote: "The kinetic energy the object has (due to its motion), is the same amount of energy as
was needed to cause its acceleration it to that specific velocity."
How is this statement consistent with ballistic pendulums?

Floor quote "This is also the kinetic energy required to accelerate an object of
a given mass to a specific velocity."
Motion is made by F applied to a mass not by kinetic energy
You are correct, in so far as motion is caused by force, but forces are the manifestation of
energy.
The action of a force is a transferance of energy. We measure the changes caused
by a force acting. From those changes we calculate the energy of the event. Energy cannot
be measured. Only the changes caused by energy acting as a force, can be measured.
also ...
I should have stated THIS as "This is also the ENERGY required to accelerate an
object of a given mass to a specific velocity" rather than "kinetic" energy.
best wishes with your experiments
floor

Floor quote: "The kinetic energy the object has (due to its motion), is the same amount of energy as
was needed to cause its acceleration it to that specific velocity."
How is this statement consistent with ballistic pendulums?
It is consistent, that is how.
Unless we are examining in terms of other than ordinary physics / mechanics.

A 5 kg mass moving 8 m/sec can collide with a 20 kg mass at rest. The input kinetic energy is 160 J and the output KE is 32 J.
To make 5 kg move 8 m/sec it takes an application of 9.81 N of force applied for 4.077 second.
To make 25 kg move 1.6 m/sec it takes an application of 9.81 N of force applied for 4.077 second.
The conserved quantity is Force * time: Ft; but Ft is not a form of energy.

It would be interesting to see an elastic collision take 4 seconds….
However, in the real world the 20kg mass has a moment of inertia
defined by the mass*radius^2
And the momentum of the smaller mass is transferred in a fraction of a second
Want to see this? Shoot a pool cue at a bowling ball

quote
"It would be interesting to see an elastic collision take 4 seconds…."
end of the quote
! was thinking of an experiment along these lines, (in another topic).
explanation ...
1. Place a section of copper tube or a thick roll of aluminum foil upon a digital weight scales,
with the long axis vertical. Note its weight.
2. Drop a powerful neodymium magnet into the tube and note the weight as
the magnet falls through the tube.
3. Note the weight of both the tube and the magnet when the magnet is at rest
at the bottom of its descent / resting on the scale.
It is probably a good idea to isolate the magnet from contact with or very near
proximity to the weight scales (as it might damage the scales).
I am pretty certain that less weight will register during the magnets descent than
when the magnet is at rest at the bottom of the tube.
What are your opinions ?

4. Next place the magnet upon a vertical, wooden, dowel rod, upon the scales.
5. Drop the the tube around / over the magnet.
6. Use a tubular copper wire coil, first shorted then not shorted.
P.S.
Hope this hasn't drifted too far off topic.
best wishes
floor

Relationship between momentum and kinetic energy
ke = 1/2 * m * V^2 kinetic energy = 1/2 * mass * velocity squared
ke = 1/2 * m * v * v kinetic energy = 1/2 * mass * velocity * velocity
p = m * v momentum = mass * velocity
ke = (m * v) * (1/2 * v) kinetic energy = momentum * 1/2 * velocity
An object's kinetic energy due to its motion is equal to the product of its momentum
and 1/2 its velocity. ke = p * (1/2 * v)
also
An object's momentum is the quotient of the division of its kinetic energy
by 1/2 of its velocity. p = ke / (1/2 * v)

The collision does not take 4 sec; I am pointing out that the quantity of Ft to make the initial motion is equal to the quantity of force * time needed to produce the final motion.
You mention "moment of inertia" is this moving in a circle?

@ Delburt Phend
? 1. What is the momentum of 3 Newtons of force times 25 seconds of time ?
? 2. What is the difference between an apple ?

The apple will have 75 kg m/sec of momentum.

Relationship between momentum and kinetic energy
ke = 1/2 * m * V^2 kinetic energy = 1/2 * mass * velocity squared
ke = 1/2 * m * v * v kinetic energy = 1/2 * mass * velocity * velocity
p = m * v momentum = mass * velocity
ke = (m * v) * (1/2 * v) kinetic energy = momentum * 1/2 * velocity
An object's kinetic energy due to its motion is equal to the product of its momentum
and 1/2 its velocity. ke = p * (1/2 * v)
also
An object's momentum is the quotient of the division of its kinetic energy
by 1/2 of its velocity. p = ke / (1/2 * v)
... ... ... ... ... ... ...
... ... ... ... ... ... ...
scenario 0
force = 3 N
time = 25 seconds
mass = zero
force = mass times acceleration
momentum = mass times velocity
momentum = zero
... ... ... ... ... ... ...
scenario 1
force = 3 N
time = 25 seconds
mass = 15 kg
force = mass times acceleration
momentum = mass times velocity
momentum = 15 kg * 5m/s .. "75 momentum"
kinetic energy = 1/2 * m * v^2
velocity = 5 m/s
because
1 Newtons of force will cause 1 kg to accelerate to 1 meter / second in 1
second (by definition)
and
1 Newton of force will cause 15 kg to accelerate to 1/15 (0.0666666666666666
6666666666666667) meter / second in 1 second
and
3 Newtons of force will cause 15 kg to accelerate to ( 0.0666... times 3) which is
0.2 meter/second in 1 second
and
3 Newtons of force will cause 15 kg to accelerate to, 0.2 times 25, in 25 seconds
and
this is equal to 5 meters per second velocity (0.2 * 25 = 5)
A 15 kg mass moving at 5 m/s has 187.5 joules of kinetic energy
Ek = 1/2 * Mass * Velocity ^2
kinetic energy = 1/2 * 15 kg * 5m/s^2 ... 1/2 * 15 kg * 25 = 187.5 joules
5 m/s ^2 = 25
15 kg * 25 = 375
375 / 2 = 187.5 joules
An object's momentum is the quotient of the division of its kinetic energy
by 1/2 of its velocity. p = ke / (1/2 * v)
187.5 joules / 1/2 * 5 m/s as in 187.5 joules / 2.5 m/s as in "75 momentum"
momentum = mass * velocity
15 kg of mass * 5 m/s = "75 momentum"
Momentum = force * time ???
3 Newtons * 25 seconds = "75 momentum"
... ... ... ... ... ... ...
... ... ... ... ... ... ...
scenario 2
force = 3 N
time = 25 seconds
mass = 4 kg
force = mass times acceleration
momentum = mass times velocity
momentum = zero
... ... ... ... ... ... ...
force = 3 N
time = 25 seconds
mass = 4 kg
momentum = 4 kg * 0.75 m/s = 3
kinetic energy = 1/2 * m * v^2
velocity = 18.75 m/s
because
1 Newtons of force will cause 1 kg to accelerate to 1 meter / second in 1
second (by definition)
and
1 Newton of force will cause 4 kg to accelerate to 1/4 (0.25) meter / second in 1 second
and
3 Newtons of force will cause 4 kg to accelerate to 0.25 times 3 which is
0.75 meter/second in 1 second
and
3 Newtons of force will cause 4 kg to accelerate to 0.75 times 25, in 25 seconds
and
this is equal to 18.75 meters per second velocity (0.75 * 25 = 18.75)
... ... ... ... ...
A 4 kg mass moving at 18.75 m/s has ? joules of kinetic energy
Ek = 1/2 * Mass * Velocity ^2
kinetic energy = 1/2 * 4 kg * 18.75 m/s^2 ... 1/2 * 4 kg * 351.5625 = 703.125 joules
18.75 m/s ^2 = 351.5625
4 kg * 351.5625 = 1406.25
1406.25 / 2 = 703.125 joules
An object's momentum is the quotient of the division of its kinetic energy
by 1/2 of its velocity. p = ke / (1/2 * v)
703.125 joules / 1/2 * 18.75 m/s or stated as 703.125 joules / 9.375 m/s or stated
as in "75 momentum"
momentum = mass * velocity
4 kg of mass * 18.75 m/s = "75 momentum"
Momentum = force * time
3 Newtons * 25 seconds = "75 momentum"
Momentum = force * time
cool !

Floor quote: "The kinetic energy the object has (due to its motion), is the same amount of energy as was needed to cause its acceleration it to that specific velocity."
How is this statement consistent with ballistic pendulums?
Good explanation here
http://hyperphysics.phyastr.gsu.edu/hbase/balpen.html
"The ballistic pendulum is a classic example of a dissipative collision in which conservation
of momentum can be used for analysis, but conservation of energy during the collision
cannot be invoked because the energy goes into inaccessible forms such as internal
energy. After the collision, conservation of energy can be used in the swing of the
combined masses upward, since the gravitational potential energy is conservative."

ALIAS TINMAN AND OR OZSOLAR POWER HAS JUST COME UP
WITH A SUPER GREAT IDEA AND OR INVENTION. CHECK IT OUT.
https://www.youtube.com/watch?v=CmTCj5E8Si4

Thanks syechelles for the link
Guys check it out
https://www.youtube.com/watch?v=CmTCj5E8Si4
It looks promising
A cross between the Flynn brothers ? and the MagMirror engine ?
Kenneth C. Kozeka, Ph.D. KEDRON CORPORATION (EDEN PROJECT) ?
But definitely new, and all his own invention ultimately.
another link and video
https://www.youtube.com/watch?v=fQtJTodw3QQ
P.S.
Apologize / don't mean to be stepping on your topic

Hi, very good. But how high is the COP?

LOTA HOW MANY MILLIONS OF DOLLARS HAVE YOU IN YOUR BANK ACCOUNT.
THEN WE CAN TALK.

Back on topic ...
How much energy is delivered to an object when 1 Newton of force
is applied to it causing that object to accelerate (given that 100% of the
energy applied is transferred as acceleration) ?
This depends upon the length of time the Newton of force is applied for.
See the "Magnets Motion and Measurement part 19 PDF file, attached below,
for detailed explanation.

No it does not.
Force does not produce consistent quantities of energy.
Force produces momentum, F = ma a = v/t Therefore Ft = mv it is not Ft = 1/2 mv²
If 7 N is applied to 3 kg for 4 seconds you will get; 7 N * 4 sec = 3 kg * v = 9.33 m/sec This gives you 28 units of linear momentum and 130.66 J of energy.
If 7 N is applied to 30 kg for 4 seconds you will get; 7 N * 4 sec = 30 kg * v = .933 m/sec This gives you 28 units of linear momentum and 13.066 J of energy.
Force does not produce energy it produces linear Newtonian momentum.

No it does not.
Force does not produce consistent quantities of energy.
Force produces momentum, F = ma a = v/t Therefore Ft = mv it is not Ft = 1/2 mv²
If 7 N is applied to 3 kg for 4 seconds you will get; 7 N * 4 sec = 3 kg * v = 9.33 m/sec This gives you 28 units of linear momentum and 130.66 J of energy.
If 7 N is applied to 30 kg for 4 seconds you will get; 7 N * 4 sec = 30 kg * v = .933 m/sec This gives you 28 units of linear momentum and 13.066 J of energy.
Force does not produce energy it produces linear Newtonian momentum.
It is actually an integral equation. If you want to truly understand this problem,
I suggest learning calculus and trigonometry.
FT = ma = m* (delta)v
That is the change in velocity, integrated over time
Since mass is on both sides, we take that out and we see
that the FT results in a value of acceleration.
Anything off the horizontal constraint (with respect to the gravitational field)
looks something like [FT +{<+\}9.8m/s/s]
Which is a vectored quantity
So the vector is also integrated over time
What we see mathematically, is that both the
vector and the acceleration are changing over time.
4 seconds takes up about a full sheet of notebook paper
if you write it all out in standard academic format.
Then you will understand that every second, you are inputting
different values for acceleration and vector (change in vector changes acceleration)
The summation of which can be estimated at 1/2q^2
This is simple shorthand physics, we do this in most “public equations”
Because the average person can’t perform pages of calculus for a simple problem
in the real world.
To get from one to the other is where trig comes in
A triangulation of the acceleration curve
You are correct in saying that is not the actual equation
It is an average of the change in velocity over time
The results are the same, because one works out the velocity
the other starts with the velocity and works it backwards

Now understand that “one second” is an arbitrary value.
We can instead use a value of 1trillionth of a second.
We see that for every sec/ 1,000,000,000,000 the acceleration and vector change
so 1 trillion sheets of paper later, we realize that the best of our equation are only an approximation.
We could divide the trillionth of a second another trillion times.
and all of this only applies to the “stationary” inertial reference of our position on the earth’s surface
Which moves at 1000mph relative to the sun, which moves at 25Mmph relative to the inertial
reference of our galactic center, which moves at 1 trillion mph with respect to another arbitrary value
representing our sector of expanding spacetime. After which time itself becomes an integral quantity.

Even the 9.8m/s/s of gravity is not entirely accurate
We can integrate gravity using tSecs (trillionths of a second)
and derive many decimals which (were it not for terminal velocity)
would become significant over long drops.
Using this level of accuracy in tests to verify the equation,
gives rise to variances in the true value of gravitational acceleration
from location to location.
Mapping the earth’s gravity for the purposes of physics,
We see that even “sea level” is not the appropriate altitude in every location.
We have no consistency for our standard tests, beyond a certain accuracy.
Not to mention sea level in 1706 is meters under water today.

Im sure you have pulleys, ropes, wood, etc. wherever you are
This shouldn’t be hard to build a series of mechanisms to demonstrate
everything we are discussing here.
It is not the math and theory we are after
It is Energy Producing Experiments
A Lever, the top of the short end is placed at “0”
the Atwood has an H of the large weight, directly above the short end
The long end connects to a chain
Another lever, that when it is fully activated, the long end sits at the
just higher than the bottom of the smaller weight after the Atwood is stopped.
But also such that when at rest the lever swings low enough to provide clearance
for the smaller weight to rise.
The chain goes from the lower lever, under a pulley and up to the short end of
the upper lever, such that when the Atwood drops, the heavy weight hits the lever,
pulling the chain, activating the other lever
Sending the smaller weight upwards, operating the Atwood in an upward vector
and measure the rise of the larger weight as the smaller weight reaches the end
of its string.

If it can lift the larger weight all the way back up,
Then the energy produced would be
The m of the smaller weight, * 9.8 * the length of the string

The RATE here (in the graph attached below) at which the acceleration is occurring, is constant.
(3 meters per second per second in the graph)
This is like when gravity accelerates an object while within a vacuum.
(9.80665 meters per second per second when caused by Earth's gravity)
The kinetic energy in a free falling object increases within each interval of time
within which it free falls. This is so even though the force of gravity is constant
during that free fall.
The RATE AT WHICH THE SPEED IS INCREASED is the same in each given time
interval.
The SPEED ITSELF, that the object is traveling at is greater in each successive interval
of time that passes.
There fore the object is moving faster and travels farther in each successive interval of time,
than it did during the previous interval of time.
The momentum of a given mass or object is greater when its speed is greater.
momentum = mass * velocity
momentum = force * time
The longer the time that a given force acts upon an object that it is ACCELERATING,
the faster that objects moves, the greater is that objects momentum, and the greater is
the kinetic energy of that object due to that motion.
Ek = 1/2 * mass * velocity ^2
But also I agree that actual measurements and experiments are what are probably
most appropriate to the topic so... I'll bow out for now.
Again
best wishes
floor
P.S.
Given that the mass of the object remains constant, the reason the rate of
acceleration is constant ... is because the force applied is constant.
The energy reqired to maintain the force as constant increases through out each
interval of time.

This will hold true in the vertical domain if the force applied is from gravity alone
(or more appropriately stated: at vectors perpendicular to the center of gravity of the source field)
The same if the force applied was constant in the horizontal on a rigid surface
Such that gravity could be negated.
In either case, the second derivative can be negated, as its’ effective value is 0
But the full equation allows for analysis of both an applied force,
and the force of gravity, at any vector.
A ‘ballistics calculator’ does this for us, and can be used in automated trajectory devices.
Or in the field using manually controlled ballistics
The solidstate version of the quivering nerd in the trenches with a notepad
Gravity will always prove to be a conservative field in the perpendicular vector.
And consistent in most locations. (there are anomalies that could be exploited)
What we put in to lift a mass, will always return the same; minus wind resistance,
and other losses.

In the most extreme example, we have a slim 1 meter rod of titanium with pointed ends.
We launch this straight up using a small railgun coil, a series of capacitors and a small power supply
It leaves the rail gun at approximately 10km/sec
And returns in roughly the same general area at about 8 km/sec
Thats mach 23, it destroys the launchsite and the rod,
leaving nothing but a crater.
Consuming 30MW for 5 seconds
Returning with a power of 0.126GJ
And a loss of ~ 15% of our power
And about 20$ from our pockets
Your railgun must be engineered to NOT exceed 10km/sec
or you risk endangering loworbit technologies.

Very large quantities of energy can be produced by placing large quantities of momentum in a small mass.
And very large quantities of momentum can be produced by a rim that is accelerated by a weighted string wrapped around the rims circumference.
A vertical rim will accelerate like a modified Atwood’s; which means it will be an F = ma acceleration. The one kilogram is on a string that is wrapped around the rim. As the 1 kg mass drops one meter it will accelerate the 100 kg vertical rim.
So: The one kilogram has 9.81 N of force. 9.81 N = 101 kg * a = .0971 m/sec/sec. is the acceleration of the rim.
d = ½ v²/a: so: v = sqrt (1 m * 2 *.09871 m/sec/sec) = .4407 m/sec. is the velocity of the rim after it has dropped one meter.
This is .4407 m/sec * 101 kg = 44.51 kg * m/sec units of momentum.
This is 44.51 units of momentum that was produced by one kg dropping one meter. It takes 4.429 units of momentum for a onekilogram mass to be raised one meter. So, you produced 44.51 units of momentum from 4.429 units.
A stack of 30 (1000g / 30) 33.33 gram masses evenly spaced 30 meter high is also 1 kg. This stack can also be dropped one meter. After the one meter drop the stack can be reconfigures by raising one 33.33 gram mass 30 meters.
To raise 33.33 gram 30 meters it needs to be moving 24.26 m/sec. This is 24.26 m/sec * .03333 kg = .8087 units of momentum.
You produce 44.51 units of momentum using .8087 units.

I'm glad to see that you are still exploring this !
best wishes
floor

I would like to see a demonstration of this “energy producing experiments”
We can excite ions with a simple electric field
(Ionized plate particle accelerator)
Even with tiny atomic masses and near relativistic velocities
there have been no anomalies concerning momentum observed
outside the expected quantum factor
as such would alter our view of the equations
Momentum is conserved throughout the entire system being analyzed
Even in the transferring from one mass to another.
If there is any mass at all involved,
you will always have less momentum than before the transfer
If there is no mass, momentum would transfer pure and lossless
[For this reason i argue that a photon cannot be massless,
but rather releases countably infinite energy for a finite time upon impact,
when we set the total energy values to be equivalent the calculated mass of
the photon corresponds to the expected value observed from the physical impact]
Archimedes, Da Vinci, Descartes, Galileo, Newton, Maxwell, Rankine, Einstein, Froude
They all agree on the theory of momentum
The Universe as we understand it would collapse if this theory were to break.

Good research but you may find Nathan Larkin Coppedge's experiments more ambitious.
Nathan's experiments can be investigated here at the rather original https://emporium.quora.com/TheHistoryofPerpetualMotionMachines (https://emporium.quora.com/TheHistoryofPerpetualMotionMachines)

The cylinder and spheres experiments make energy; you would do better to build and experiment with something that works.

I would like to see a demonstration of this “energy producing experiments”
We can excite ions with a simple electric field
(Ionized plate particle accelerator)
Even with tiny atomic masses and near relativistic velocities
there have been no anomalies concerning momentum observed
outside the expected quantum factor
as such would alter our view of the equations
Momentum is conserved throughout the entire system being analyzed
Even in the transferring from one mass to another.
If there is any mass at all involved,
you will always have less momentum than before the transfer
If there is no mass, momentum would transfer pure and lossless
[For this reason i argue that a photon cannot be massless,
but rather releases countably infinite energy for a finite time upon impact,
when we set the total energy values to be equivalent the calculated mass of
the photon corresponds to the expected value observed from the physical impact]
Archimedes, Da Vinci, Descartes, Galileo, Newton, Maxwell, Rankine, Einstein, Froude
They all agree on the theory of momentum
The Universe as we understand it would collapse if this theory were to break.
Mass is a function. Could it be done away with? Yes, partially, if it is converted to energy. The energy in question is a part of spacetime. However, this does not happen easily as something has to facilitate the change, IE the function has to be manipulated by an outside source. Inertia is a resistance to changes of position concerning what we call matter to the surrounding medium, which we currently call spacetime. Momentum is based upon mass, but it is not mass itself, and is a resistance to changes in spatial velocity. Lambda F does equal mc^2 in a particle at rest. Mass is a byproduct of a selfpropagating wave packet, stable or unstable (real or 'virtual'). Light does indeed have mass, which is to be expected, as you cannot separate mass from energy. If there is energy, then there is mass, or vice versa. However, conversion of one to the other can reduce or increase the measurable amount. Momentum is ultimately energy being stored as mass. It is converted back to energy during a collision with something else.
Spacetime itself is somewhat stupid as a notion goes, since adding time to an absence of anything does not yield something which can be bent, warped, torn , tunneled through, or otherwise spatially or physically manipulated. Gag. Many of the supposed effects become iimpossible when viewed in anything but two dimensions, such as warped or 'dented' space causing gravity. (When applied with no frame of reference being given special dispensation, IE in all three dimensions at the same time, since gravity pulls straight towards the center of any mass, the "warp" completely envelopes any object.)
A lot of what we know is still backwards or outdated thinking, unfortunately. When you consider that the medium that we have given the moniker of spacetime is the source for all of the energy we use, and that, according to at least one prominent Physicist, one cup of it could boil off all the oceans of earth concerning the amount of energy stored their, I would say that our notions of efficiency leave something to be desired. If any scientist desired to see how badly off our estimate of the subject is, they should try to heat 100 ml f water 10 degrees, and calculate exactly how much energy is required, then do it. BUT, they should record the whole event with a thermal imaging camera (fleer type) and fail the test if they see any energy radiation in the form of heat into the ambient surroundings, from any source connected to the test, from any angle, including the water or the container it is in, as well as any power sources involved. (That would be the thermal definition of Unity or COP=1)
That also won't happen, let alone measuring for EM loss, EF loss, friction, hysteresis, or other potential forms of energy loss or conversion not accounted for in the factoring as well.
Even if they did all of those things, the question still remains of whether the method used has any real efficiency for increasing the rate of vibration of water molecules (heating water). If any technique is more efficient than merely bashing water molecules with brute force, then the testing method for efficiency is still invalid.
It is time people start calling the BS card.
Paul Andrulis

Disks and rim mass wheels are used to store motion, so it is important to know how they store it.
The moment of inertia for a rim (or ring) is mr²; and for a disk it is 1/2mr². So is a disk, with the same mass and radius, twice as easy to rotate as a rim?
The word inertia implies that The Moment of Inertia has something to do with the difficult of moving the object; or how much force needs to be applied to accelerate the object.
Also, it should be noted that the r in 1/2mr² is the r of the outside circumference, it is not a new r.
For a rim and a disk of the same mass the m is the same. So, the r is the same and the m is the same in 1/2mr² for the disk, and mr² for the rim.
So, the formula is clearly saying that a disk is ½ as difficult to rotate as a rim. So, at the same rate of angular acceleration the mass of the disk is applying half of the inertia; which means the mass is being accelerated half as fast; which means that the average position of the mass is half the distance to the circumference. Or the mass is at ½ r.
But is the average mass of a disk at ½ r?
The disk with its circumference at ½ r has 1/4th the mass of the disk with a radius of r. So, 3/4th of the mass of the disk is beyond ½ r.
Plus, this 3/4th mass is in a position to have a greater portion of the velocity because it is closer to the circumference.
I do not think that if you apply the same force to a disk that it will accelerate twice as fast as an equal mass rim. What does this video show?
https://www.youtube.com/watch?v=NsKIPa4Fnfo

What do we mean by faster ?
Rotations per unit of time ?
or
The numbers of some unit of length around the circumference, that pass a
given point per revolution ?
Given that the disk and the rim have the same r
No real real world, rim has zero thickness toward the center of rotation.
Its mass is not concentrated in an imaginary line. Although as a model we
may consider it as such.
A wheel driven from near its center is a lever. The greater the radius of the
rim in relationship to radius at which the wheel is driven, the greater is the leverage.
The leverage here in, is considered as acting against the resistance to acceleration solely.
Levers trade off, force for distance or distance for force.
... ... ... ... ... ... ... ... ... ...
The mass around a wheel which is being accelerated has, differing amounts of
leverage being applied to it depending upon the r at which that mass is
located (average).
Like wise is the resistance to acceleration against the applied force
of a band of mass at a leverage (r), due to
both
the mass (average) and the leverage factor r.
This is Newton's equal and opposite, acting force and resisting force.
... ... ... ... ... ... ... ... ... ...
A given mass has the resistance to acceleration that it has against a given force
period.
When that force is applied to a lever and from thence to the mass, it does not
change the force needed to accelerate that mass. It changes the force applied.
... ... ... ... ...
A given mass has the resistance to acceleration that it has against a given force
period.
When the resistance of a mass to acceleration is applied to a lever and from
thence to the accelerating force, it does not change the force needed to accelerate
that mass. It changes the force (reactive force) of that resitance to the applied force.
"I do not think that if you apply the same force to a disk that it will accelerate twice
as fast as an equal mass rim."
Incomplete statement. i.e. force is applied at what r ?
and
Do both the rim and the disk have the same diameter ?

1/2mr² is a claim that the disk has ½ as much resistance to motion as a hoop; and should therefore accelerate down the ramp twice as fast. If the disk is rotating down the ramp twice as fast, then it should cover twice the distance in the same period of time. The disk should reach the end of the ramp when the hoop is only halfway down the ramp.
The experiment shows that the disk is harder to rotate than 1/2mr² predicts.

OK, I was looking primarily at why the mass s cancel out.
" The experiment shows that the disk is harder to rotate than 1/2mr² predicts. "
I don't know why with complete certainty but I think it is that...
No real world rim, has zero thickness from the circumference, toward the center of rotation.
Its mass is not concentrated in an imaginary line. Although as a model we
may consider it as such.

" The experiment shows that the disk is harder to rotate than 1/2mr² predicts. "
https://en.wikipedia.org/wiki/Moment_of_inertia (https://en.wikipedia.org/wiki/Moment_of_inertia)
Half way down, a computer simulation clearly showing the disk is not 2x the distance of the ring, a similar result to the youtube vid.
I am no math guru, but there is usually more to the picture than a simple formula.
You have the formula for inertia of both bodies. You then need to take into account angular momentum, rolling vs accelerating around an axle and gravity at an angle.
I found this:

Math typically works that way. :)

Some request models;
Gooble “A scientific model is a physical and/or mathematical and/or conceptual representation of a system of ideas, events or processes. Scientists seek to identify and understand patterns in our world by drawing on their scientific knowledge to offer explanations that enable the patterns to be predicted.”
You say that I have not presented a model, and then you show how the
Law of Conservation of Momentum would work. Apparently, I have presented a model and you are fully aware of it.
35 kg dropped one meter can accelerate 140 kg to 2.21 m/sec. Model: Atwood’s Machine
A 35 meter high stack of evenly spaced 1 kilogram masses exert the same force as a 35 kilogram mass. Model: scale balance
After a drop of one meter the stack can be reconfigured by accelerating one kg to 26.24 m/sec, costing 26.24 units of momentum. The one kilogram moving 26.24 m/sec will rise 35 m. Model: kinematic equations.
140 kg moving 2.21 m/sec equals 310 units of momentum and you only need 26.24 units to recycle the system. Model: kinematic equations.
If the 310 units of momentum is placed in one kg the rise would be 4900 m. Only 35 of those meters is needed to reconfigure the stack. Model: kinematic equations.
The momentum of a large mass can be given to a small mass. Model: cylinder and spheres machine where all the motion is returned to the cylinder; and the Law of Conservation of Momentum.
The stack of 35 1 kg masses could accelerate a 140 kg rim and stack to 2.21 m/sec and then the rim could throw one kg 140 times higher than the original drop. Model: cylinder and spheres machine; the Law of Conservation of Momentum: kinematic equations.
The double yoyo despin ‘cylinder and sphere’ transfers all the motion of the “cylinder and sphere” to only the spheres. Then the spheres transfer the motion back to the cylinder and spheres combination.
The conservation of energy could not return the motion of the spheres back to the cylinder. Because as the conservation of momentum would require a 10 m/sec speed for the spheres the conservation of energy would only require a 3.16 m/sec velocity.
When masses collide, or transfer motion, smaller masses do not and can not give their energy to a larger mass. The smaller masses can and do however give all of their momentum to the larger mass.
So the double yoyo despin demonstrates the transfer of momentum to the spheres and then back again to sharing the motion with a mass that is about nine times more massive.
This means that the double yoyo despin: https://www.youtube.com/watch?v=YaUmzekdxTQ has a ten fold increase in energy twice.

Your oversimplified math does not prove anything. Prove it with a build and a measurable gain in Potential energy. Measurable inputmeasurable output. It should be simple for you, you have many models.
ADD: This is a much better experiment. You can measure the speed of the cylinder and calculate the speed of the ball bearings from where they land.
https://www.youtube.com/watch?v=sWS2Sl5iG0g

You apparently are not familiar with the physics. You have no information about the mass of the spinning (space station shaped) shaft or its distribution of mass. Without that knowledge you cannot predict motion conservation of any type.
The cylinder gives you a known mass at a known radius, for a known momentum. The disk is better than the space station, but not as easy as a cylinder.

You apparently are not familiar with the physics. You have no information about the mass of the spinning (space station shaped) shaft or its distribution of mass. Without that knowledge you cannot predict motion conservation of any type.
Wow, really?...
I was saying this is a more elegant experiment than yours, as is this one: https://www.youtube.com/watch?v=_wsDEOW9DuU&t=0s
Due to the ability to properly measure kinetic energies of the components together, then separately afterwards.
Wild claims require irrefutable proof.

I don’t know why anyone would call these legitimate experiments wild claims. It is only that you (and others) do not like the data they produce.
This disk throw that you posted is a legitimate experiment; if you know the mass of the disk, and you and also know the distribution of mass, then you can evaluate the momentum of the spheres alone and the momentum of the combination of disk and spheres.
You can eliminate the bearing resistance by letting the unwind occur while the system is falling. I have used mechanical releases and video cams can separate the motion into 1/240th of a second. It is easy to tell what is happening.

You can eliminate the bearing resistance by letting the unwind occur while the system is falling. I have used mechanical releases and video cams can separate the motion into 1/240th of a second. It is easy to tell what is happening.
/?
What horizontal movement did the system have? What was the final speed of the yoyos at the point in time the cylinder stopped?
There are excellent bearings available. Your video is from 2017...If it is such a perfect experiment, why has noone duplicated it, or from what I can tell taken much notice of it? This experiment will not produce consistently repeatable results. For me, it doesn't prove a gain in energy. If it doesn't prove it for me, it probably isn't proving it for a number of other people.
If you actually find a way to create energy and prove it, the whole world will sit up and listen.

These problems are on your side of the fence Tarsier; not mine.

#322 : " There are excellent bearings available." price/full load rotations ratio
10 Mio full load rotations means much : 30 000 RPM transformer(m. or g./a.) x 60 min = per hour x 24 = per day
as R.O.I. calculateable friction part : exchange costs per annum,by own skills by external service ?
a good source about ultralong life bearings for recommend ?
Sincere
OCWL

I don't have any problems. I am not the one claiming to have broken energy conservation with a half baked experiment.
35 kg dropped one meter can accelerate 140 kg to 2.21 m/sec. Model: Atwood’s Machine
This isn't energy gain. I do not see a video of this. Your atwoods video is not 140 or 35kg, and does not have a scale for the distance traveled over what time.
A 35 meter high stack of evenly spaced 1 kilogram masses exert the same force as a 35 kilogram mass. Model: scale balance
Balance has very little to do with Potential or kinetic energy.
After a drop of one meter the stack can be reconfigured by accelerating one kg to 26.24 m/sec, costing 26.24 units of momentum. The one kilogram moving 26.24 m/sec will rise 35 m. Model: kinematic equations.
Sounds like energy conservation in an imaginary world with no resistance from your description.
140 kg moving 2.21 m/sec equals 310 units of momentum and you only need 26.24 units to recycle the system. Model: kinematic equations.
If the 310 units of momentum is placed in one kg the rise would be 4900 m. Only 35 of those meters is needed to reconfigure the stack. Model: kinematic equations.
or Naive mathematics...
The momentum of a large mass can be given to a small mass. Model: cylinder and spheres machine where all the motion is returned to the cylinder; and the Law of Conservation of Momentum.
With a claim of 10x energy and an unconvincing proof.
The stack of 35 1 kg masses could accelerate a 140 kg rim and stack to 2.21 m/sec and then the rim could throw one kg 140 times higher than the original drop. Model: cylinder and spheres machine; the Law of Conservation of Momentum: kinematic equations.
More naive maths. Show physical proof.
The double yoyo despin ‘cylinder and sphere’ transfers all the motion of the “cylinder and sphere” to only the spheres. Then the spheres transfer the motion back to the cylinder and spheres combination.
The conservation of energy could not return the motion of the spheres back to the cylinder. Because as the conservation of momentum would require a 10 m/sec speed for the spheres the conservation of energy would only require a 3.16 m/sec velocity.
Again uninformed mathematics.
When masses collide, or transfer motion, smaller masses do not and can not give their energy to a larger mass. The smaller masses can and do however give all of their momentum to the larger mass.
Do they? Show proof and how this helps in energy creation.

Tarsier,
concerning your formula ( relpy#315):
Have you ever made the effort of reading for what a condition this formula is true ?
It needs sharp eyes to read the bottom right sentence
My impression : the discussion is  because of a lot of false assumptions and arguments is rather
exausting and not constructive.
Try first to understand instead of endless arguments and build it...its not so complicated.
@Delburt:
how is this possible that the rotation of the Cylinder is CCW since the steelball is hanging at the right of the cylinder and it seems the steelball remains at the same hight in the first moments ? Can you explain it ? Did you give it a ccwspin before releasing the cylinder ?
https://www.youtube.com/watch?v=8Q7L2BOYkjE (https://www.youtube.com/watch?v=8Q7L2BOYkjE)
Mike
I ask because I am working on a concept of repeating one of the experiments

#325 : " .... More naive maths. ... "
https://en.wikipedia.org/wiki/Mathematics (https://en.wikipedia.org/wiki/Mathematics)
"maths" = ALL spectrum ,non naive/naif
dialectical materialism = theory in images and/or words/numbers about object :
material,non dialectical = without words/numbers/image
#315
FORMULA : as AXIOMATIK part right/wrong ?
https://de.wikipedia.org/wiki/Axiom (https://de.wikipedia.org/wiki/Axiom)
the basic law/Fixpoint : Anti/ Hypo/These
https://de.wikipedia.org/wiki/Kondition_(Mathematik (https://de.wikipedia.org/wiki/Kondition_(Mathematik))
https://de.wikipedia.org/wiki/Vorkonditionierung (https://de.wikipedia.org/wiki/Vorkonditionierung)
: not applyable for this case,conditioning !?
Application and physical result ,between
HubKraft
AUFTRIEBSKRAFT,LASTSCHWERKRAFT,"GRAVITY"
https://www.google.com/search?client=firefoxbd&q=hubkraft (https://www.google.com/search?client=firefoxbd&q=hubkraft)
https://www.google.com/search?client=firefoxbd&q=hubkraft+diagramm (https://www.google.com/search?client=firefoxbd&q=hubkraft+diagramm)
https://www.mathematik.unimuenchen.de/~diening/ws11/numerik/numscript.pdf (https://www.mathematik.unimuenchen.de/~diening/ws11/numerik/numscript.pdf)
and ,point of view :
https://www.google.com/search?client=firefoxbd&q=abtriebskraft (https://www.google.com/search?client=firefoxbd&q=abtriebskraft)
https://de.wikipedia.org/wiki/Normalkraft (https://de.wikipedia.org/wiki/Normalkraft)
ZugKraft
ANTRIEBSKRAFT
https://www.google.com/search?client=firefoxbd&q=zugkraft (https://www.google.com/search?client=firefoxbd&q=zugkraft)
https://studyflix.de/ingenieurwissenschaften/zugkraftdiagramm1171 (https://studyflix.de/ingenieurwissenschaften/zugkraftdiagramm1171)
Force/s their scalar,vectorial,tensorial orientation from source to target
Pressure : https://de.wikipedia.org/wiki/Druck_(Physik (https://de.wikipedia.org/wiki/Druck_(Physik)) https://www.linguee.com/germanenglish/translation/spezifischer+druck.html (https://www.linguee.com/germanenglish/translation/spezifischer+druck.html)
Force,specific :
https://www.linguee.com/germanenglish/translation/spezifische+kraft.html (https://www.linguee.com/germanenglish/translation/spezifische+kraft.html)
The FORMAT https://de.wikipedia.org/wiki/Form_(Philosophie (https://de.wikipedia.org/wiki/Form_(Philosophie))
LOGOS https://en.wikipedia.org/wiki/Tractatus_LogicoPhilosophicus (https://en.wikipedia.org/wiki/Tractatus_LogicoPhilosophicus)
https://de.wikipedia.org/wiki/Logos (https://de.wikipedia.org/wiki/Logos)
ALGORYTHM https://de.wikipedia.org/wiki/Algorithmus (https://de.wikipedia.org/wiki/Algorithmus)
wmbr
OCWL

@Delburt:
how is this possible that the rotation of the Cylinder is CCW since the steelball is hanging at the right of the cylinder and it seems the steelball remains at the same hight in the first moments ? Can you explain it ? Did you give it a ccwspin before releasing the cylinder ?
https://www.youtube.com/watch?v=8Q7L2BOYkjE
This is a much better experiment IMO. Input can be measured, but it is still difficult to measure the output.
I suspect the Ball hangs in the same spot because the tube is wound up on a piece of fishing line/string, its drop causes the unwind and spin. In the initial moments, the ball is being wound up at the same rate the tube unwinds as it is accelerated by gravity.
concerning your formula ( relpy#315):
http://hyperphysics.phyastr.gsu.edu/hbase/hoocyl.html
The formula on the top left is a cylinder rolling without slipping, the formula top right is a hoop rolling without slipping.
The formula at the bottom is either slipping without rolling, the text on the bottom right refers to that bottom formula.
It was in response to:
https://www.youtube.com/watch?v=NsKIPa4Fnfo
1/2mr² is a claim that the disk has ½ as much resistance to motion as a hoop; and should therefore accelerate down the ramp twice as fast. If the disk is rotating down the ramp twice as fast, then it should cover twice the distance in the same period of time. The disk should reach the end of the ramp when the hoop is only halfway down the ramp.
The experiment shows that the disk is harder to rotate than 1/2mr² predicts.
This is the computer calculated race under perfect conditions in a simulation:
https://en.wikipedia.org/wiki/Moment_of_inertia#/media/File:Rolling_Racers__Moment_of_inertia.gif
It closely resembles the outcome of the physical test.
Delberts assumption that the hoop should only be half way down the ramp based on 1/2mrr shows a naive view of the math involved.
Try first to understand instead of endless arguments and build it...its not so complicated.
1. I am not convinced there is enough evidence to justify a build.
2. My build time is currently allocated elsewhere.
3. If I were to build it, I would be complicating it enough to measure input vs output.

IancaIV, what point are you trying to make? Let us look at context:
140 kg moving 2.21 m/sec equals 310 units of momentum and you only need 26.24 units to recycle the system. Model: kinematic equations.
Using online calculators:
https://www.omnicalculator.com/physics/freefall
https://www.calculatorsoup.com/calculators/physics/gravitationalpotential.php
https://www.calculatorsoup.com/calculators/physics/kinetic.php
The 140KG falls from 0.25m to achieve this speed. (343 J)
1kg can rise to 35m at 26.24 kgm/s, which also costs 344 J
There is a difference in momentum, bu where is the energy gain?
If the 310 units of momentum is placed in one kg the rise would be 4900 m. Only 35 of those meters is needed to reconfigure the stack. Model: kinematic equations.
So unless I am missing something, it takes 51,200 J to "place" 310 units of momentum on 1kg. How did we magically transfer 310 units of momentum from a 140kg at 2.21m/s mass to a 1kg mass at 310m/s?

IancaIV, what point are you trying to make? Let us look at context:
Using online calculators:
https://www.omnicalculator.com/physics/freefall (https://www.omnicalculator.com/physics/freefall)
https://www.calculatorsoup.com/calculators/physics/gravitationalpotential.php (https://www.calculatorsoup.com/calculators/physics/gravitationalpotential.php)
https://www.calculatorsoup.com/calculators/physics/kinetic.php (https://www.calculatorsoup.com/calculators/physics/kinetic.php)
The 140KG falls from 0.25m to achieve this speed. (343 J)
1kg can rise to 35m at 26.24 kgm/s, which also costs 344 J
There is a difference in momentum, bu where is the energy gain?
So unless I am missing something, it takes 51,200 J to "place" 310 units of momentum on 1kg. How did we magically transfer 310 units of momentum from a 140kg at 2.21m/s mass to a 1kg mass at 310m/s?
" 310 units of momentum and you only need 26.24 units " : 310/26,24 = ?
https://overunity.com/1763/12timesmoreoutputthaninputdualmechanicaloscillationsystem/msg564829/#new (https://overunity.com/1763/12timesmoreoutputthaninputdualmechanicaloscillationsystem/msg564829/#new)
Short answered about "..... we magically transfer ....." .
magica,mysteria,Magister,Meister,Master M.S.A. : Master Science Administration ~ kybernetes
I can,if You want to introduce Yourself in deeper hyper/deeper physics ,in"QuantumMechanics"Logics explain ,similar ,
but animistic language,less " swollen ":
https://www.google.com/search?q=stephen+hawking&client=firefoxbd&sxsrf=APqWBsymqOZRsoVgTxAIse7P4c6HbEUeA%3A1647771334635&ei=xv42YvapJoj87_UPtYW7sAg&ved=0ahUKEwj2pMnIutT2AhUI_rsIHbXCDoYQ4dUDCA0&oq=stephen+hawking&gs_lcp=Cgdnd3Mtd2l6EAwyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEc6BwgAEEcQsANKBAhBGABKBAhGGABQughYughg3BZoAXACeACAAQCIAQCSAQCYAQCgAQHIAQjAAQE&sclient=gwswiz (https://www.google.com/search?q=stephen+hawking&client=firefoxbd&sxsrf=APqWBsymqOZRsoVgTxAIse7P4c6HbEUeA%3A1647771334635&ei=xv42YvapJoj87_UPtYW7sAg&ved=0ahUKEwj2pMnIutT2AhUI_rsIHbXCDoYQ4dUDCA0&oq=stephen+hawking&gs_lcp=Cgdnd3Mtd2l6EAwyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEcyBAgAEEc6BwgAEEcQsANKBAhBGABKBAhGGABQughYughg3BZoAXACeACAAQCIAQCSAQCYAQCgAQHIAQjAAQE&sclient=gwswiz)
String to PointMatrix
https://en.wikipedia.org/wiki/Urelement
C.Colombo : chicken or egg hyper/deeper physics Dualism: spin or point
or,going 50 years back to its/his,Otto Stein,first publication date :
german,deutscher,Sprache/language : international multilingual traduction available doing ?
https://www.yumpu.com/de/document/view/20184764/ottosteindiezukunftdertechnikpdf (https://www.yumpu.com/de/document/view/20184764/ottosteindiezukunftdertechnikpdf)
Sincerely and a sinfull Sunday wishing
OCWL

What do you think of Destin's evaluation?
https://www.youtube.com/watch?v=AnaASTBn_K4

Delburt
Excellent find.You are a keen observer. I am amazed of the different possibilities to fully transfer momentum of of a big mass to a much smaller one.
Lets calculate:
Mass of the knot including the fluffy part about 2 gr = 0.002 kg
Minimum Velocity according to Wikipedia: v = 342.2 m/sec at T = 20 degree Celsius
E_kin = 1/2 * 0.002 kg * 117 100.2 m_exp2/sec_exp2 = 117.1 Joule.........thats a lot
Remember: The fastest Pitcher can throw a ball 105.5 mph which is about 47 m per second and if you look at the ease with which this woman
throws the whip....
https://sportsnaut.com/fastestpitchevermlbhistory/
There is one peculiar thing: as the wave rolles down to the end it stops its wavelike movement about 10 cm before the knot as then
the knot is accellerated to full speed like the head of a serpent.
Why that ?
Question : would the sound barrier be reached if the wave would roll down to the end ?
There is a small element in the construction of the last part of the whip which makes it work and these "smart three engineers have not figured it out,
they fully missed a key element of the "faster that soundspeedfunction" in their experiment.
I will not reveal it...its a good excercise for sharpening ones observationskills.
I call it "The serpents bite"
Mike_G

There is one point in the video (smarter every day) where they say that the sound barrier is broken before the end; that might put the mass at around 20 grams. Also think of the air resistance you would get at these speeds.

Delburt,
everthing is normal, see here:
https://de.wikipedia.org/wiki/%C3%9Cberschallflug#/media/Datei:Shockwave_pattern_around_a_T38C_observed_with_BackgroundOriented_Schlieren_photography_(1).jpg (https://de.wikipedia.org/wiki/%C3%9Cberschallflug#/media/Datei:Shockwave_pattern_around_a_T38C_observed_with_BackgroundOriented_Schlieren_photography_(1).jpg)
and the compare to first and second picture
Now the third pic : what does this woman hold between their fingers ?
with the fourth pic : what does the corde look like shortly before the bang
Mike_G

also I noticed recurring discussions whether formulas "moment of inertia" or "momentum" should be used . We dont need any further discussion on this as moment of inertia in the cylinder & the spheres experiment is not involved for any further calculations as the spheres  after getting all the momentum  are freely rotating and do not create any torque via a spoke to the axis.
Its just simple as that. I have built a simple setup with a small flywheel from an old record player ( eighties) and can confirm the fullstop of the wheel if
the small mass is out at 90 degress. I used just one small mass. It will take me a longer time to build the complete setup as I need the assistance of a mechanic which has his workshop 25 mile away
Mike_G

At about 2:36 they talk about going supersonics before full extension; so a larger portion of the whip (in addition to the feathered end) must be moving at about the same speed.
Backyard scientist also says they went supersonic with rubber bands and a string. I am still working on that one.

Laws of levers proves you can make energy in the lab.
Let’s apply a force to the lever arm at the same radius as the accelerated mass. Let’s use numbers to keep this clear. The accelerated mass is on a lever arm of .1 m radius; and the force is applied on a lever arm of .1 m radius.
The mass being accelerated is more difficult to accelerate, or more easily accelerated, in accordance with the Laws of Levers. Fr
If the two lever arms are the same .1 m /.1 m them the mechanical advantage of the accelerated mass is 1. In this case you have a direct F = ma situation. If you apply 4 newtons of force to a 1 kg mass it will be moving 4 m/sec after one second. The energy is 8 joules
If you move the mass to .2 m radius; then the accelerated mass has a mechanical advantage of .2 m / .1 m = 2. To keep the torque the same at the .1 m position of the applied force you will have to reduce the mass at .2 m to .5 kg. The torque at the applied force location (.1 m) is the same so you will still get an F = ma acceleration: except now you will get .5 kg moving 8 m/sec.
If you move the mass to 4 m radius; then the accelerated mass has a mechanical advantage of 4 m / .1 m = 40. To keep the torque the same at the .1 m position of the applied force you will have to reduce the mass at 4 m to .025 kg. The torque at the applied force location (.1 m) is the same so you will still get an F = ma acceleration: except now you will get .025 kg moving 160 m/sec. The energy is 320 joules
Remember the second Law is “The change of momentum is proportional to the applied Force.” Note that all these rotating masses have the same momentum.

Delburt
are you talking about a seesaw setup ? Mass on one side...force beeing applied on the other ?
Mike

Consider a seesaw with 40 kg on one side and 1 kg on the other side. To be balanced the one kilogram must be 40 times further from the point of rotation than the radius of rotation of the 40 kilograms.
An extra force could be applied at the same radius as the 40 kg.
First: let us focus on the radius and rotation of the 40 kg. Let’s put the radius at 2 cm. So we have 40 kg at 2 cm on one side. We could also put 40 kg at 2 cm on the outer side of this balanced beam. There would be no rotation because we have no unbalanced force.
When we put an extra 40 kg at 2 cm on one side we will get acceleration. We have (40 kg + 40 kg+ 40 kg) 120 kg accelerated by 40 kg. This is 1/3 standard acceleration; 9.81 m/sec/sec / 3 = 3.27 m/sec/sec. We know this to be true from Atwood’s machine experiments.
Now let us switch out a 40 kg mass on both sides for a 1 kg mass at (2 cm * 40 = 80 cm) 80 cm. The two 1 kg masses at 80 cm are the same rotational resistance as the two 40 kg masses at 2 cm, because they are at forty times the radial distance.
Because the resistance remained the same the acceleration will also remain the same; 3.27 m/sec/sec for the remaining extra 40 kg at 2 cm. But the two one kilogram masses are moving 40 times faster than the 40 kg drive mass. At forty times faster they have forty times the energy. ½ * 40 kg *(2.557 m/sec)² = 130.76 joules ½ * 1 kg * (102.28 m/sec)² = 5230.6 J
So let’s change this into a practical arrangement. The 2 cm radius could be a 4 cm shaft that is attached to a 160 cm beam. On the end of each side of the beam are 1 kg masses. The shaft can be accelerated by a string that is wrapped around the shaft; the string is suspending a 40 kg mass from its end. This would be 3.27 m/sec/sec acceleration at the surface of the shaft.
After a one meter drop of the 40 kg; the 40 kg will be moving (1 m = ½ v²/ 3.27 m/sec/sec) 2.557 m/sec. But the two 1 kg masses will be moving 102.28 m/sec. Their kinetic energy would be KE = ½ * 1 kg * (102.28 m/sec)² = 5230.6 Joules.
The input energy is 40 kg at 1 meter of height which is (40 kg * 9.81 N/kg) * 1 m = 394.4 joules. This is the force produced by the drive mass of 40 kg that was dropped 1 meter.
The output energy is 5230.6 joules * 2 + (½ *40 kg * 2.557 m/sec * 2.557 m/sec) = 10,591.96 Joules.
There were 394.4 joules put in: and 10,591.96 joules came out.

The two 1 kg masses at 80 cm are the same rotational resistance as the two 40 kg masses at 2 cm, because they are at forty times the radial distance.
Except they are not.

You can put the balanced masses at any distance you want. I have not used so large a radial difference (2 cm to 80 cm) but I have confirmed the principal with experiments, with less dramatic differences.
Are you speaking from experimental knowledge; and if so, please sight the experiment.
Fundamentally what you are saying is that in takes more force to rotate the long side of a balanced tube or beam; than the force needed to rotate the short side. Please explain how that might be possible as it is a ridged object. And how were you convinced of this without experiments?

The inertia of a point mass in rotation is I=mr^{2}
The two 1 kg masses at 80 cm will be much harder to accelerate than the two 40 kg masses at 2 cm.
The simple proof is the period of a pendulum. The formula is T^{2}=L/g. Note there is also a squared relationship. The period of a pendulum is derived from its inertia and acceleration. The squared (or square root) part of the formula is from inertial math.
So when um, when you double the length of the pendulum, you see that the period increases by a factor of square root of two.

If your mr² is correct, then a pendulum bob of 2L would be 4 times harder to move than a bob at 1 L; clearly this is a false statement. The inertia of a bob is unchanged no matter what the length of the pendulum. The bob accelerates itself it is not under torque from the point of rotation.
The period of a pendulum has to do with the length of the swing and the sine of the angle; this is not what is being discussed.
MIT Physics Demo  Center of Mass Trajectory
Rotation about the center of mass requires that the linear Newtonian momentum (mv) on one side is equal to the linear momentum on the other side. It is not about angular momentum conservation or energy conservation. The energy of one side is not equal to the energy of the other side and the angular momentum of one side is not equal to the angular momentum of the other side.
If a 40 kg mass is attached by a long tube to a one kilogram mass on the other end, and it is flipped in the air, it will rotate about its center of mass. And 40 kg * r = 1 kg * 40 r.
Explanation of the universe must correspond with what the universe does. And the universe does Laws of Levers and the linear momentum (mv) of one side is equal to the linear Newtonian momentum of the other side.
Do the experiments what do the experiments tell you; this is only worth about 10 trillion dollars.

If your mr² is correct, then a pendulum bob of 2L would be 4 times harder to move than a bob at 1 L; clearly this is a false statement.
It is 4 x harder to move.
The inertia of a bob is unchanged no matter what the length of the pendulum. The bob accelerates itself it is not under torque from the point of rotation.
The reason the pendulum length is not a linear relationship to the period is the same reason inertia doesn't have a linear relationship with mass/radius. A pendulum is the same as your lever. Instead of looking with tunnel vision at the problem, consider the similarities. Gravity provides a constant acceleration F=MA
If I double the length of a pendulum/leverweight, Gravity doubles the torque on the system (leverage), but it doesn't take the same time to rotate. (because it is squared)
The period of a pendulum has to do with the length of the swing and the sine of the angle; this is not what is being discussed.
That is the funny thing with Energy math. Everything relates to eachother, as it should.
The period of a pendulum, gravity accelerates a mass around a pivot point. Your seesaw 2 masses are being accelerated around a pivot point. Regardless of where you apply torque: to the center of rotation, or the furthest mass from the center, the inertia increases exponentially.

Delburt,
Removed my post, because I had some misunderstanding of your proposal.
Since I am a pragmatist I always try to construct a system which I can build in such a way that I can extract the output.
Also I would avoid strong centrifugal forces so that bearings are not destroyed.
I will reexamine your proposals and check some calculations
Mike

So let’s change this into a practical arrangement. The 2 cm radius could be a 4 cm shaft that is attached to a 160 cm beam. On the end of each side of the beam are 1 kg masses. The shaft can be accelerated by a string that is wrapped around the shaft; the string is suspending a 40 kg mass from its end. This would be 3.27 m/sec/sec acceleration at the surface of the shaft.
After a one meter drop of the 40 kg; the 40 kg will be moving (1 m = ½ v²/ 3.27 m/sec/sec) 2.557 m/sec. But the two 1 kg masses will be moving 102.28 m/sec. Their kinetic energy would be KE = ½ * 1 kg * (102.28 m/sec)² = 5230.6 Joules.
I reposted two sentences; the 160 cm is two 80 cm radiuses, with one kg on each end. The dropped mass (or drive mass) is 40 kg. The 40 kg is accelerating itself and two 1 kg masses, but they are on the end of an 80 cm radius.

Since we have a pivotpoint around which the "deadmasses" rotate ( regardless of the angle covered ) moment of inertia is involved.
This system can not be compared to a pedulum since a pendulum does not transfer torque.
A pendulum can transfer torque, just as it can have torque transferred to it. I only used a pendulum to show why I=mr^{2}. There is link between a weight at a specific radius being accelerated and two weights on an arm being accelerated. Anyone can calculate extra energy using incorrect math.
An atwoods is an easy setup to reproduce. Building a very light one is not as easy. In an atwoods, the mass dropping at radius still acts as if it has rotational inertia from the viewpoint of the atwoods hub assembly, and it acts as if it were linear from the viewpoint of masses accelerating in a gravity field. Both calculations agree with each other (,if you account for the inertia of the hub.).. A simple atwoods does not create energy. I look forward to seeing experiments to investigate the theory though.

Delburt
This would be 3.27 m/sec/sec acceleration at the surface of the shaft.[/size]
That is what I am not sure abput
Mike
Tarsier: I removed my comments above...it was 3 o clock in the morning so had a misunderstanding
I agree : a special construction of a pendulumbearing can transfer torque but this is not what is commonly known as a pendulum.
You have to be more specific about the bearing in use.
Mike

You were correct the first time; a pendulum bob cannot apply torque through a bearing.
I am going to give you real engineering problem.
I have a 14ft fishing boat. If you kneel on the dock and grab the side of the boat, you can apply torque to the boat so as to bring the nose of the boat around to the proper angle for loading it onto the trailer.
I was thinking about making this application of torque much much easier. I could do this by placing a (horizontal ish) pipe at the center of mass so that an 8foot wooden pole can be placed in the pipe.
And how would you do it?
Would you place a bearing at the center of mass, of the boat, and wrap a rope around the bearing? Or tie a rope to a vertical pipe at the center of mass?
A light tube with 5 kg on both ends is flipped into the air; it rotates about its center of mass. One end is increased to 20 kg, and it again is flipped in the air. The 20 kg is four times harder to rotate so the center of mass shifts closer to the 20 kg end.
But you say that the 5 kg at 4 times the radius is harder to rotate. Well, if it were then way does the center of mass not shift to that end? The center of mass shifts back to a balance where the two ends are equal: 20 kg * 1 r = 5 kg * 4 r
I make Atwood’s machines that produce energy. I made an Atwood’s pulley that has three radii (very similar to these numbers): the shaft at about 1 cm r, a 15 cm radius, and a 30 cm radius; all in one pulley. This is three concentric circles all fixed on one pulley: 2 cm diameter, and 30 cm diameter, and 60 cm diameter.
You can keep the drive mass at 1 cm r.
One kg at 30 cm radius is balanced by 2 kg at the 15 cm radius on the other side.
One kg at 30 cm radius is balanced by 30 kg at the 1 cm radius.
The different drops were timed with a photo gate timer; the photo gate flag was on the pulley head. The times of the gate trips were all the same; sometimes within the same ten thousandth of a second. Any small addition of mass changed the trip time.
All these had the same trip time.
30 kg at 1 cm balancing 1 kg at 30 cm
30 kg at 1cm balancing 2 kg at 15 cm
30 kg at 1 cm balancing 30 kg at 1 cm
2 kg at 15 cm balancing 1 kg at 30 cm
2kg at 15 cm balancing 2 kg at 15 cm
1 kg at 30cm balancing 1 kg at 30 cm
These are all accelerated by the same mass at 1 cm
All had the same trip time; and 6 additional grams would produce a different trip time. So, there is no way that 1 kg at 30 cm is twice as hard to rotate as 2 kg at 15 cm. Or 1 kg at 30 cm is 30 times harder to rotate than 30 kg at 1 cm.
Now let's look at the energy:
½ 30 kg * 1 m/sec * 1 m/sec = 15 joules
½ 2 kg * 15 m/sec * 15 m/sec = 225 joules
½ * 1 kg * 30 m/sec * 30 m/sec = 450 joules
450 is larger than 15

Delburt
as I said im my last comment about your accellerationcalculation " That is what I am not sure about"
So I was looking up an old german PhysicsTextbook GrimsehlTomaschek, dated 1940. In a chapter about the equivalencecondition
of rotating masses I found an experimental setup in which it is demonstrated that the accelleration of a falling mass which
rotates two masses ( see pic MassEquiv_02.jpg) at both ends of a horizontal mounted bar stays always the same if the value of the moment of inertia ( m*r²)
stays the same for different symetrical positions of the two masses on the bar. In order for (m*r²) to be constant, the masses have to be adapted if their
radial posistion changes.
If you have 2 x 1 kg at r = 0.8 m, moment of intertia I = 2(1kg * 0.64 m²)> I = 1.28 kg m²
Now you want to move the two masses to r = 0.4m then for I=1.28 kgm² to remain constant ( 1.28) the two masses have to be increased ( I use here the
term replacementmass m') to
m'= 1.28 / 0.4² = 1.28/ 0.16 = 8kg in total, 4 kg at each side,
to fullfill the condition that the accelleration of the falling mass stays the same as with 2 * 1 kg masses at r = 0.8m.
The calculations made in this old book are based on the cgsSystem ( centimetergrammsecond )
and only with this system the formula for the resulting accelleration b of weight G is true :
G (gr) * 981 (dyn) = b * (m * r² + G) ( see pic EquivMass_03.jpg)
The formula is correct only for r = 1 cm, as the value of r² = 1 ....so that it is save to assume that the replacmentmass m'
experieneces the same accelleration as G ( falling weigth )
b = G (gr) * 981 (dyn) / (m (gr) + G (gr)) ( Formula Nr. 1 )
......but what would be the new total replacementmass m' on both sides of the bar at r = 1 cm ?
The moment of Inertia for the two 1kg masses ( in gramm ) at r=80 cm
I = 2000 gr * (80cm)² = 2000 * 6400 = 12.800.000 gr cm²
In order to have the same moment of Inertia at r = 1cm the new mass m' = 12.800.000 / 1
that is two masses of 6.400.000 gr ( 6.400 kg) at each side.
According the Formula Nr.1 above, falling mass = 40 kg = 40.000 gr
b = 40 000 g * 981 dyn / (12.800.00gr + 40.000gr) = 39.240.000 /12.840.000
b = 3.05 cm/ s² = 0.0305 m/s² !!!!
______________________________________________________________________________________
Now in order toget an idea the accekeratio of G with an unchanged replacementmass m' at r=2 cm ( although this not correct )
we end up with b = 0.121 m/s²
From here you calculate the final velocity of 40kg falling down 1 meter : V = Square root of (2 * 0.121 * 1) = 0.492 m/s
Since the axisradius is 2 cm the circumference of the axix = 0.126 m
With the velocity 0f 0.492 m/s / 0.126 you get 3.9 revolutions per second.
The cirumference at which the two 1kg masses rotate is : 2 * Pi * 0.8 = 5 m
With 3.9 Revolutions / s the rimvelocity of the two 1 kg masses is 19.5 m/s
Kinetic energy Ekin = 2kg/2 * (19.5)² m²/s² = 380.25 Joule
Input was : m * g * h = 40kg * 9.81 m/s² * 1m = 392,4 Joule
The missing 12 Joule are accounted to the falsy used r = 2cm with the unchanges m' in the Formula N.1
So there is no gain in this system. That is the reason why we dont have overunitybicycles
Mike
Edit: Sorry for the big rotated pics. I tried to remove and replace them ..but to no avail

Surly you do not believe that 1 kg at 80 cm is balance by 6,400 kg at 1 cm.
Shortly after (although this is not correct) a number comes up from nowhere; .121 m/sec/sec and then this mystery number is used for calculation.
My experiments are real.

A lever arm can’t move up until it is first balanced.
The 80 to 1 lever arm, according to your math, can’t be moved up until 6,400 kg are placed on the opposite short (1 cm) side.
And: according to your math, then the long side of a 40 to 1 lever arm can’t be moved up until 1,600 kg are placed on the opposite short (1 cm) side. But the one kilogram on the 40 cm side balances with 40 kg on the 1 cm side; and you then add another 40 kg for acceleration.
But (for you) this 80 kg is still (1600 – 80) 1520 kg short of being balanced. If your rule worked, then the 1 kg at 40 cm will accelerate the 80 kg; not the other way around. For your math the 80 kg at 1 cm * 1 cm; should be overpowered by 1 kg * 40 cm* 40 cm.

Experiments collect data to establish or support a concept; so an experiment will collect data points. The concept is; If a longer lever arm with a proportional less mass is harder to rotate, then it will be harder to rotate. For example: if one side of a pivot point is 40 kg at 1 cm and the other side is 1 kg at 40 cm; then according to the mr² theory the long side is 40 time harder to rotate than the short side.
The Laws of Levers says that the two sides are equally hard to rotate. So, let’s construct an experiment that separates the true from the false.
If you apply the same force to two 40 kg masses at 1 cm on both sides; and to two 1 kg masses at 40 cm on both sides, then the harder to rotate side will take longer to rotate, or it will have a smaller acceleration.
The Force is held to be the same by using the same suspended mass at the same location. In the 40 to 1 we used an extra 40 kg on the short side.
You find data points to determine if the acceleration is the same. A 26 mm gate trip at the same times after the two Atwood’s (two 40 kg at 1 cm, and two 1 kg at 40 cm) has moved the same distance (real numbers .0628 and .0631 sec for one of many set ups). Having the same final velocity after moving the same distance assures us that the different Atwood’s have the same acceleration.
Having the same F divided by the same ‘a’ for both Atwood’s (40 kg at 1 cm both side; and 1 kg at 40 cm both sides) assures us that the inertia ‘m’ has to be the same also.
Real numbers for a certain Atwood’s are: two 20.7 kg at a 1.03 cm r; compared to two .8628 kg at 24.13 cm. Both took 3.3 sec; in another set up. They had the same F and the same 'a' and the same inertia 'm'.

I caught myself wondering how science made this mr² mistake in the first place. And then I remembered: it is all about the protection of ½ mv² (probably mv² at the time); if 1 kg can be accelerated to 2 m/sec as easily as 2 kg can be accelerated to 1 m/sec then The Law of Conservation of Energy goes kaput. And that is the sacred cow of physics.

Delburt
The 80 to 1 lever arm, according to your math, can’t be moved up until 6,400 kg are placed on the opposite short (1 cm) side. [/size]And: according to your math, then the long side of a 40 to 1 lever arm can’t be moved up until 1,600 kg are placed on the opposite short (1 cm) side. But the one kilogram on the 40 cm side balances with 40 kg on the 1 cm side; and you then add another 40 kg for acceleration.
[/size]
No, this is not what is meant please reread my description.[/size]
[/size]
We have to distinguish between a static situation ( Nm) and a dynamic one> accelleration..> mr² ..resistance to rotary accelleration.[/size]
[/size]
"By the way: this is not "my math", I just try to understand the basics and s[/size]ince you changed to a leversystem ( away form the Cyl & sphers) in order to find an easier way ( compared to the Cyl. & Sph) I was forced to dive into the depth of what was known to this subject back then.[/size]
[/size]
I cannot and will not argue whether this is true of not but I understood the principle of this experiment. [/size]
[/size]
One thing is sure that these guys back then did extensive and very elaborated experiments. I spent a few days searching the internet ...didnt find anything and then looked the subject up in my old book. I have no time to translate all this to english but the experiment was basis in university
teachings and sure enough they demonstrated this in front of students.
If I had the time and the means I would build this experiment.
I am a pragmatic person, i.e. I build, experiment and analyse...like you to. This I did with the Cylinder & Spheres experiment. Something went wrong
with a brassbearing and I have to reengineer the setup. However I could see and film that momentum is conserved not energy. I will continue with this small wheel setup and build it according to sound mechanical means and do not let myself distracted by other setups because I have not reached the corefunction of the Cyl & Sphere system.
I agree: They do everything to control attention to protect their narrative.
Levine on moment of inertia
https://www.youtube.com/watch?v=lvfzdibrUFA
Mike

I=mr^{2} is easy to prove.
Static balance does not mean equal inertia. You can't pick and chose when you will use 1/2mv^{2}.
I do not know why you got the results you did. I suspect there is a problem with your setup. Probably bearings able to hold 60kg, but able to rotate freely. A bicycle wheel might be a good start for a test such as you suggest.
You do have one good point. I do not know how a balanced offset weight (eg 8Kg at 10cm vs 1Kg at 80cm) doesn't rotate around a virtual inertial center, rather than the static balance center.

Demonstrating Rotational Inertia (or Moment of Inertia)  YouTube flipping physics
Question for you: Why is the torque formula one r and (your) inertia formula is two r s? Because changing the location of torque is exactly the same thing as changing the location of inertia. It is a relationship between the two radii not an actual distance.
I don’t think they (guts back then) did do extensive and elaborate research; and that is why you cannot find the experiments. Just like the flipping physic guys; they don’t do data: just; it is faster it is slower, etc.
You have a 80 cm wheel with a 2 cm shaft and you have 40 kg suspended from the left side of the shaft and 1 kg suspended from the right side of the wheel. It is balanced: but you then add 5 grams to the 1 kg on the right. We will let those 5 grams accelerate the wheel to .1 m/sec. At one tenth meter per sec we add 200 grams to the 40 kg on the shaft left side. What happens to the acceleration?

Question for you: Why is the torque formula one r and (your) inertia formula is two r s? Because changing the location of torque is exactly the same thing as changing the location of inertia. It is a relationship between the two radii not an actual distance.
Torque has nothing to do with inertia. Torque is a measurement of applied force. Inertia is a measurement of how difficult something is to accelerate, and mathematically relates to how much energy is gained or lost during rotation.
Getting back to a static balanced 8Kg vs 1Kg rotating in space (or on an ice rink). The 1kg is 8 times further from the point of rotation, and is 8x more difficult to move. Logic does tell me that the assembly should rotate around a point where both sides are equally difficult to rotate comparatively......Diving Deeper: Again if I am at the center of static balance, the 8kg pulls my left arm forward with 8x less energy than the 1kg pulls my right arm backwards. Again, I feel like this is not where the assembly wants to rotate around if it were not constrained on an axle.

Energy?? Tarsier; since when does F and m make energy? F = ma makes linear momentum. “Rotation gained or lost” is acceleration. So you have the three components of Newtonian motion; and Newton never made energy.
Torque is an applied force upon a resistance; if there is no resistance there is no torque. From Newton Third Law we know that the inertia is applying a force upon the torque: equal and opposite but entirely in the same manner. It is a reverse torque upon the applied torque. This inertial resistance is force being applied at a radial distance.
I will give two arguments:
Make a horizontal lever arm with one kilogram attached (at rest) at one meter. You then (perpendicularly) collide 2 kg moving 5 m/sec at the .5 meter location. What is the resultant motion?
Second argument:
In the flipping physics ‘Demonstrating Rotational Inertia (or Moment of Inertia)  YouTube’ lets assume that the torque can be placed at two locations, on the pulley, where one location is twice the radius of the other. And we know that we can place the inertia at twice the distance; one location toward the middle, and then one location on the end of the rods.
Now we have four possible arrangements of the spoke wheel: small pulley torque to inertia position on the end of the rod, small pulley torque to inertia position in the middle of the rod; large pulley torque position to inertia position on the end of the rod; large pulley torque to inertia position in the middle of the rod.
Now ask yourself; is the small to small equal to the large to large? And I hope you say yes.
So if you halve the torque radius you double the inertial resistance.
But (for your law): if you double the inertia radius you quad the inertial resistance.
So you can move from ‘large radius torque and middle radius inertia’: to ‘large r torque to large r inertia’. And you say it will be one fourth the acceleration.
Or you can move from large radius torque and middle radius inertia: to small r torque to middle r inertia. And it will be one half the acceleration.
But the radius torque / radius inertia is the same in these two changes.
You can’t have two rules.

1 newton meter of torque is 1 newton of force at 1 meter, or its equivalent regardless of the radius it is applied. 2 newtons at 1/2 meter etc. It doesn't care if there is very little resistance or very much.
inertia is directly related to energy math. To accelerate a mass till it has 50 joules of energy takes 50 joules of energy. A mass at 2x radius at a fixed rpm will have 4x the energy, just as it is 4x harder to accelerate, because it takes 4 times the energy to accelerate it to that energy level. Putting a mass on a lever doesn't magically contravene physics.
https://www.youtube.com/watch?v=lNx0yPdl960
See the acceleration at 2:40. According to your theory, the masses are around 3x further out, so should be approx 3x harder to rotate. This is clearly not the case. According to conventional physics, it should be around 9x harder to accelerate. The inner weight model moves a full rotation as the outer weight model only moves around 1/8th.
I understand your unwillingness to accept this fact. Accepting it would throw many or all of your "creating energy" theories away.

thank you Tarsier
good demonstration .
Levins last sentence..not payed attention to:
https://youtu.be/lvfzdibrUFA?t=817 (https://youtu.be/lvfzdibrUFA?t=817)
I emphasize this because he made the same remark in a electrophysics lecture of induction. Induction very much resembles inertia.
I have to say that I am very grateful to Delburts effort of showing the Cyl. & Spheres system. It certainly was not easy to
analyse the NASA Despin correctly and do all the videodemonstration years ago.
This Cyl. & Spheres System is extraordinary and has a embedded working principle which is not easy to detect/ understand and demonstrates that
energy is not conserved. It contains this pluselement NASA was hiding by blasting away the steelballs at the most interesting point in time concealing the kinetic engergy of the balls
The leversystem is conventional physics and only shows energyconservation.
The whipSystem is another one I regard as suspicious but hard to repeat without complex measurementequipment.
I myself concentrate my efforts on Cyl & Sphere System.
Mike

The cylinder and spheres is interesting. I have not looked at it in depth for a long time. IMO a better experiment needs to be done. I have not had the time or motivation yet to do one., and I have not seen one done the way I want to see one.
I would like to to see a measured input, a calculated kinetic total energy, a complete despin, where the balls are released and their exit KE measured through distance traveled. Not an easy task to successfully test and accurately measure each step. For me it would be ideally on a fixed axle.
I look forward to seeing your tests.

A Beautiful and Engaging Investigation of Angular Motion!  Rotational Inertia Demonstrator  YouTube This difference is not so monstrous 8.4 sec to 3.4 sec for an acceleration difference of 1 to 6.1, which looks very close.
The closer you are to just overcoming the friction; the more friction affects the experiment. The slower the rotation or lower the rotational force, the greater the percent bearing friction plays upon the system. The friction can be significant.
You could conduct the experiment at the same rotational speed; and in that why frictional differences drop out. Such as placing 1/3 the mass at three times the distance. You can accelerate 1/3 the mass at 3 times the distance and it will rotate at the same rate.
You both ignored the negation of your math. The radii ratio of the small ‘torque’ pulley and the near inertia position can equal the radii ratio of the large pulley to the long inertia position.
But when you move from the small inertia position (15 cm) to 3 times that radius; then the inertia for r² is 9 times as great. You can multiply the force by moving the applied torque to the large radius pulley position but that only gets you to three times harder to move.
Let put this in an algebraic description: the small pulley is 2 cm and the large pulley is 6 cm. the small inertia position is at 15 cm on the rod and the large is at 45 cm.
You start with 2 cm / 15 cm ‘torque over inertia’ and end with 6 cm / 45 cm ‘torque over inertia’. But your r² says that 6 / 45 is three times harder to move than 2/ 15. This is impossible because they are the same numerical relationship between force and mass. This is a fatal math error, you can not brush this aside.
If you put the mass at 5 cm it allegedly would be 9 times easier to move; and you could move the applied torque to 2/3 cm. It would still be 3 times easier to move but you are back at the original ratios of radii. This 2/3 cm / 5 cm should not accelerate 3 times faster: 2/3 cm / 5 cm, 2 cm / 15 cm, and 6 cm / 45 cm should all have the same acceleration.
Tarsier quote: 1 newton meter of torque is 1 newton of force at 1 meter, or its equivalent regardless of the radius it is applied. 2 newtons at 1/2 meter etc. It doesn't care if there is very little resistance or very much.
‘It doesn't care if there is very little resistance or very much.”
This simply is not true; If you are tightening a lug nut and it is moving freely, the torque wrench reads zero. Only when the nut tightens against the rim is any torque applied. And the resistance always equals the torque.

The radii ratio of the small ‘torque’ pulley and the near inertia position can equal the radii ratio of the large pulley to the long inertia position.
That is a good point. It is a pity that wasn't shown.
This simply is not true; If you are tightening a lug nut and it is moving freely, the torque wrench reads zero. Only when the nut tightens against the rim is any torque applied. And the resistance always equals the torque.
If a constant torque is applied to a free spinning nut, it will spin very fast. Like what happens to a wheel nut with a pneumatic hammer drill.

You have a one meter low mass tube that has 2 kg attached to one end and 5 kg attached at 30 cm from the other end. The tube is moving sideways at 3 m/sec so that both the 2 kg and 5 kg masses are moving at the same speed (3 m/sec).
The tube is caught on the end near the 5 kg. The tube’s end is then on a bearing and the tube is forced to rotate. The 5 kg now has a 30 cm radius, and the 2 kg has a 100 cm radius.
Where is the center of mass and what is its speed?

From experiments like the hammer or tennis racket toss; the center of mass is the point on the object that travels at the same speed even while the object is put in rotation. The tube is put in rotation by catching its end.
The center of mass of the tube, with a mass of 5 kg at 30 cm and a 2 kg mass at 100 cm, is at 50 cm. We assume the tube has no mass; for easier math.
So, with the center of mass at a 50 cm radius and moving 3 m/sec: then the 5 kg mass with a 30 cm radius would be moving 1.8 m/sec after the tube is caught on the end.
The 2 kg mass has a radius of 100 cm and therefore would be moving 100 cm / 50 cm * 3 m/sec = 6 m/sec.
After the tube is caught on the end: the 5 kg mass would be rotating faster that the 2 kg so it would have to transfer its motion to the 2kg so that they could rotate at the same rate. You could say that the 5 kg is placing torque on the 2 kg. But for you the 2 kg is inertia and would be (2kg * 10 dm * 10 dm) / (5 kg * 3 dm) 200 / 15 harder to move. Then why does the 5 kg lose the same amount of momentum as the 2 kg gains. 5 kg * 3 m/sec = 15 5 kg * 1.8 m/sec = 9 15 9 = 6 and 2 kg * 3 m/sec to 2 kg * 6 m/sec is also 6. The same thing is the same thing. It is not about torque and r² inertia it is about momentum.
The 5 kg’s momentum changes from 15 to 9 kg m/sec; and the 2 kg’s momentum changes from 6 to 12 kg m/sec.
The total initial momentum was 7 kg * 3 m/sec = 21 kg m/sec; and the final linear Newtonian momentum is 5 kg * 1.8 m + 2 kg * 6 m/sec = 21 kg m/sec.
Why does the 2 kg gain the same amount of momentum that the 5 kg loses when the 2 kg is allegedly 200 / 15 harder to move?
When you swing a baseball bat you place torque on it to accelerate it; but if you miss the ball then you place torque on the bat to decelerate it. So the application of torque does not have to be acceleration. Why would it not be legitimate to say that the 2 kg torques the 5 kg. Then the 5 kg would be the inertia and it would be 45 / 20 harder to move than the 2 kg.
Newton’s view is that the 5 kg pushes the 2 kg and the 2 kg pushes the 5 kg; and it all works out beautifully. But your 200 / 15 and 45 / 20 has no math for a realworld event.

Delburt,
[quoteyou have a one meter low mass tube that has 2 kg attached to one end and 5 kg attached at 30 cm from the other end.
The tube is moving sideways at 3 m/sec so that both the 2 kg and 5 kg masses are moving at the same speed (3 m/sec).
The tube is caught on the end near the 5 kg. The tube’s end is then on a bearing and the tube is forced to rotate.
The 5 kg now has a 30 cm radius, and the 2 kg has a 100 cm radius.
You need to be consistent with your arguments. You changed systematics.
We talk about intertia torque which means angular acceleration of masses at a distance from its center of rotation.
We accelerate by applying torque on the axis. Here we experience moment of inertia ( m*r² ) of the masses to be accelerated
Your last example describes a bar with two weights at different position from a center bearing which moves translational then
hits an obstacle and then at almost the same instant is suddenly connected to a bearing around which it rotates.
1) you changed systematics by describing the conversion of a translational moving system to a rotational one.
2) the transition between both systems is irrational in regard to a practical solution
Since there is no practical solution to this transition you need to change the system as follows:
The bar is mounted on the bearing which is fixed to the ground of the earth and another mass with the same mass as the earth hits the system a one end at 3m/s.
Got the point ?
The same irrationality
If we design a system like this which accellerates a rotational system at rest by applying a force ( translational (https://www.dict.cc/?s=translational)[/font][/size] moving mass M hits the system) at a point on the circumference where the mass is,
then we have inertia of the mass involved and not moment of inertia. The masses then rotate freely at the final speed by translational (https://www.dict.cc/?s=translational)[/font][/size] acceleration multiplied (https://www.dict.cc/?s=multiplied)[/font][/size] [/size]by (https://www.dict.cc/?s=by)[/font][/size] the time of physical impact leaving again the system without torque.
Wasn't the topic "pure rotational systems set in motion by angular acceleration" ?
Suddenly we are a lost in a discussion caused by the compulsion to find another counterargument to defend a claim of creating OU.
Mike

It is not irrational and in fact it is deadly simple; and the solution is given.
You could place objects on a cart and drive the cart up against a wall. Fix one end of an object to the edge of the cart. You could test and see if the linear speed of the center of mass of your chosen shapes proceeds at the same rotational speed. You would need a means of reducing friction.
And this event of the tube mentioned does not conserve energy; so, I guess it is a third way of making energy

No,
there are to cases to be considered:
1) elastic impact
2) no elastic impact
3) premise:
configuration must be symmetric, i.e. two identical masses attached to the edge better to 2 bearings on a shaft on two levels ( one above the
other) , shaft mounted in the masscenter of the car.
ad 1)
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car will be repelled > situation develops into chaotic behaviour , behaviour can not be predicted because of timing difference between angular accelleration of the the masses on the edge and the time lapsed during repelling of the car.
ad 2)
momentum of the car will be used up by deformation of the wall and the car ( molecular destruction and heat) . Only initial momentum of the objects attached to the edge of the car is left. No gain possible
Mike

Suppose you were to swing a rod from a 20 meter string as in a physical pendulum. The end of a 1 meter rod is attached to the end of the 20 meter string. Thirty cm down from the attachment of the rod is a 5 kg mass fixed to the rod. And 100 cm down from the string attachment is a 2 kg mass fixed to the rod.
This 20.5 meter pendulum is swung into a pin at the down swing position. The pin is at the point of the attachment of the rod to the 20 meter string.
If the speed of the center of mass of the rod is 1 m/sec before contact with the pin at the downswing position: then what it the speed of the center of mass after contact with the pin?

Premise of 20.5 m PendulumLength ( center of mass ) is not correct. According to my calculation center of mass is at 20.58 m
When the top of the 1 m rod hits the pin the rimvelocity ( at 20.58 m radius) of the center of mass stays the same at the moment of impact.
The angular velocity of the 20.58 m long pendulum however is suddenly increased at the time of impact because angular momentum must be conserved as the radius of rotation is reduced by 20 meters
Shortly after impact however deceleration of the center of mass takes place as it climbs up against 9.81 m/s² gravitational acceleration ( better deceleration) until all momentum is compensated.
I dont do the math but it might happen that the short pendulum loops past the top position ( at 12 o 'clock ) ccwwise and then the behaviour is not
predictable .
I do not see any gain here...standart phisics...the pluselement of the Cyl & spheres experiment is missing
Mike

Well; do the math, angular momentum, cannot remain the same if the velocity of the center of mass remains the same. The velocity of the downswing tells us all we need to know.
The energy cannot remain the same either. What you wish to happen simply doesn't. We are not talking about the 'cylinder and sphere' we are proving that energy can be made in the lab.