Free Energy  searching for free energy and discussing free energy
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.