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Author Topic: re: energy producing experiments  (Read 116376 times)

Offline Kator01

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Re: re: energy producing experiments
« Reply #345 on: September 04, 2022, 01:52:11 AM »
Delburt,

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

I will re-examine your proposals and check some calculations

Mike



« Last Edit: September 04, 2022, 06:01:37 AM by Kator01 »

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #346 on: September 04, 2022, 04:03:01 AM »

So let’s change this into a practical arrangement. The 2 cm radius could be a 4 cm shaft that is attached to a 160 cm beam. On the end of each side of the beam are 1 kg masses. The shaft can be accelerated by a string that is wrapped around the shaft; the string is suspending a 40 kg mass from its end. This would be 3.27 m/sec/sec acceleration at the surface of the shaft.

After a one meter drop of the 40 kg; the 40 kg will be moving (1 m = ½ v²/ 3.27 m/sec/sec) 2.557 m/sec. But the two 1 kg masses will be moving 102.28 m/sec. Their kinetic energy would be KE = ½ * 1 kg * (102.28 m/sec)² = 5230.6 Joules.

I reposted two sentences; the 160 cm is two 80 cm radiuses, with one kg on each end. The dropped mass (or drive mass) is 40 kg. The 40 kg is accelerating itself and two 1 kg masses, but they are on the end of an 80 cm radius.   

Offline Tarsier_79

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Re: re: energy producing experiments
« Reply #347 on: September 04, 2022, 06:07:16 AM »
Quote
Since we have a pivot-point around which the "dead-masses" rotate ( regardless of the angle covered ) moment of inertia is involved.
This system can not be compared to a pedulum since a pendulum does not transfer torque.


A pendulum can transfer torque, just as it can have torque transferred to it. I only used a pendulum to show why I=mr2. There is link between a weight at a specific radius being accelerated and two weights on an arm being accelerated. Anyone can calculate extra energy using incorrect math.

An atwoods is an easy setup to reproduce. Building a very light one is not as easy. In an atwoods, the mass dropping at radius still acts as if it has rotational inertia from the viewpoint of the atwoods hub assembly, and it acts as if it were linear from the viewpoint of masses accelerating in a gravity field. Both calculations agree with each other (,if you account for the inertia of the hub.).. A simple atwoods does not create energy. I look forward to seeing experiments to investigate the theory though.



« Last Edit: September 04, 2022, 10:29:36 AM by Tarsier_79 »

Offline Kator01

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Re: re: energy producing experiments
« Reply #348 on: September 04, 2022, 06:22:02 AM »
Delburt

Quote
This would be 3.27 m/sec/sec acceleration at the surface of the shaft.[/size]
Quote

That is what I am not sure abput

Mike


Tarsier: I removed my comments above...it was 3 o clock in the morning so had a misunderstanding
I agree :  a special construction of a pendulum-bearing can transfer torque but this is not what is commonly known as a pendulum.
You have to be more specific about the bearing in use.

Mike

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #349 on: September 05, 2022, 07:38:28 PM »
You were correct the first time; a pendulum bob cannot apply torque through a bearing.

I am going to give you real engineering problem.

I have a 14ft fishing boat. If you kneel on the dock and grab the side of the boat, you can apply torque to the boat so as to bring the nose of the boat around to the proper angle for loading it onto the trailer.

I was thinking about making this application of torque much much easier. I could do this by placing a (horizontal ish) pipe at the center of mass so that an 8-foot wooden pole can be placed in the pipe.

And how would you do it?

Would you place a bearing at the center of mass, of the boat, and wrap a rope around the bearing? Or tie a rope to a vertical pipe at the center of mass?


A light tube with 5 kg on both ends is flipped into the air; it rotates about its center of mass. One end is increased to 20 kg, and it again is flipped in the air. The 20 kg is four times harder to rotate so the center of mass shifts closer to the 20 kg end.

But you say that the 5 kg at 4 times the radius is harder to rotate. Well, if it were then way does the center of mass not shift to that end? The center of mass shifts back to a balance where the two ends are equal: 20 kg * 1 r = 5 kg * 4 r

I make Atwood’s machines that produce energy. I made an Atwood’s pulley that has three radii (very similar to these numbers): the shaft at about 1 cm r, a 15 cm radius, and a 30 cm radius; all in one pulley. This is three concentric circles all fixed on one pulley: 2 cm diameter, and 30 cm diameter, and 60 cm diameter. 

You can keep the drive mass at 1 cm r.
 
One kg at 30 cm radius is balanced by 2 kg at the 15 cm radius on the other side.
 
One kg at 30 cm radius is balanced by 30 kg at the 1 cm radius. 

The different drops were timed with a photo gate timer; the photo gate flag was on the pulley head. The times of the gate trips were all the same; sometimes within the same ten thousandth of a second.  Any small addition of mass changed the trip time.

All these had the same trip time.

30 kg at 1 cm balancing 1 kg at 30 cm

30 kg at 1cm balancing 2 kg at 15 cm

30 kg at 1 cm balancing 30 kg at 1 cm

2 kg at 15 cm balancing 1 kg at 30 cm

2kg at 15 cm balancing 2 kg at 15 cm

1 kg at 30cm balancing 1 kg at 30 cm

These are all accelerated by the same mass at 1 cm

All had the same trip time; and 6 additional grams would produce a different trip time. So, there is no way that 1 kg at 30 cm is twice as hard to rotate as 2 kg at 15 cm. Or 1 kg at 30 cm is 30 times harder to rotate than 30 kg at 1 cm.

Now let's look at the energy:

½ 30 kg * 1 m/sec * 1 m/sec = 15 joules

½ 2 kg * 15 m/sec * 15 m/sec = 225 joules

½ * 1 kg * 30 m/sec * 30 m/sec = 450 joules

450 is larger than 15

Offline Kator01

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Re: re: energy producing experiments
« Reply #350 on: September 08, 2022, 01:20:19 AM »
Delburt


as I said im my last comment about your accelleration-calculation " That is what I am not sure about"


So I was looking up an old german Physics-Textbook Grimsehl-Tomaschek, dated 1940. In a chapter about the equivalence-condition
of rotating masses I found an experimental setup in which it is demonstrated that the accelleration of a falling mass which
rotates two masses ( see pic Mass-Equiv_02.jpg)  at both ends of a horizontal mounted bar stays always the same if the value of the moment of inertia ( m*r²)
stays the same for different symetrical positions of the two masses on the bar. In order for (m*r²) to be constant, the masses have to be adapted if their
radial posistion changes.


If you have 2 x 1 kg at r = 0.8 m, moment of intertia I = 2(1kg * 0.64 m²)-> I = 1.28 kg m²


Now you want to move the two masses to r = 0.4m then for I=1.28 kgm² to remain constant ( 1.28) the two masses have to be increased ( I use here the
term replacement-mass m') to


                       m'= 1.28 / 0.4² = 1.28/ 0.16 = 8kg in total, 4 kg at each side,


to fullfill the condition that the accelleration of the falling mass stays the same as with 2 * 1 kg masses at r = 0.8m.


The calculations made in this old book are based on the cgs-System ( centimeter-gramm-second )
and only with this system the formula for the resulting accelleration b of weight G is true :


                       G (gr) * 981 (dyn) = b * (m * r² + G) ( see pic Equiv-Mass_03.jpg) 
 
The formula is correct only for r = 1 cm, as the value of r² = 1 ....so that it is save to assume that the replacment-mass m'
experieneces the same accelleration as G  ( falling weigth )


         b = G (gr) * 981 (dyn) / (m (gr) + G (gr))                            ( Formula Nr. 1 )


......but what would be the new total replacement-mass m' on both sides of the bar at r = 1 cm ?


The moment of Inertia for the two 1kg masses ( in gramm ) at r=80 cm


      I = 2000 gr * (80cm)² = 2000 * 6400 = 12.800.000 gr cm²


In order to have the same moment of Inertia at r = 1cm the new mass m' = 12.800.000 / 1
that is two masses of 6.400.000 gr ( 6.400 kg) at each side.


According the Formula Nr.1 above, falling mass = 40 kg = 40.000 gr


                b = 40 000 g * 981 dyn / (12.800.00gr + 40.000gr) = 39.240.000 /12.840.000
               
                b = 3.05 cm/ s² = 0.0305 m/s²   !!!!
______________________________________________________________________________________




Now in order toget an idea the accekeratio of G with an unchanged replacement-mass m' at r=2 cm ( although this not correct )
we end up with b = 0.121 m/s²


From here you calculate the final velocity of 40kg falling down 1 meter : V = Square root of (2 * 0.121 * 1) = 0.492 m/s


Since the axis-radius is 2 cm the circumference of the axix = 0.126 m
With the velocity 0f 0.492 m/s / 0.126  you get 3.9 revolutions per second.


The cirumference at which the two 1kg masses rotate is : 2 * Pi * 0.8 = 5 m


With 3.9 Revolutions / s the rim-velocity of the two 1 kg masses is 19.5 m/s


Kinetic energy Ekin = 2kg/2 * (19.5)² m²/s² = 380.25 Joule


Input was : m * g * h = 40kg * 9.81 m/s² * 1m = 392,4 Joule


The missing 12 Joule are accounted to the falsy used r = 2cm with the unchanges m' in the Formula N.1


So there is no gain in this system. That is the reason why we dont have overunity-bicycles




Mike


Edit: Sorry for the big rotated pics. I tried to remove and replace them ..but to no avail



















Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #351 on: September 08, 2022, 03:41:22 AM »
Surly you do not believe that 1 kg at 80 cm is balance by 6,400 kg at 1 cm.

Shortly after (although this is not correct) a number comes up from nowhere; .121 m/sec/sec and then this mystery number is used for calculation.

My experiments are real.

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #352 on: September 08, 2022, 05:41:33 PM »
A lever arm can’t move up until it is first balanced.

The 80 to 1 lever arm, according to your math, can’t be moved up until 6,400 kg are placed on the opposite short (1 cm) side. 

And: according to your math, then the long side of a 40 to 1 lever arm can’t be moved up until 1,600 kg are placed on the opposite short (1 cm) side. But the one kilogram on the 40 cm side balances with 40 kg on the 1 cm side; and you then add another 40 kg for acceleration.

But (for you) this 80 kg is still (1600 – 80) 1520 kg short of being balanced. If your rule worked, then the 1 kg at 40 cm will accelerate the 80 kg; not the other way around. For your math the 80 kg at 1 cm * 1 cm; should be overpowered by 1 kg * 40 cm* 40 cm.

----
Experiments collect data to establish or support a concept; so an experiment will collect data points. The concept is; If a longer lever arm with a proportional less mass is harder to rotate, then it will be harder to rotate. For example: if one side of a pivot point is 40 kg at 1 cm and the other side is 1 kg at 40 cm; then according to the mr² theory the long side is 40 time harder to rotate than the short side.

The Laws of Levers says that the two sides are equally hard to rotate. So, let’s construct an experiment that separates the true from the false.

If you apply the same force to two 40 kg masses at 1 cm on both sides; and to two 1 kg masses at 40 cm on both sides, then the harder to rotate side will take longer to rotate, or it will have a smaller acceleration.

The Force is held to be the same by using the same suspended mass at the same location. In the 40 to 1 we used an extra 40 kg on the short side.

You find data points to determine if the acceleration is the same. A 26 mm gate trip at the same times after the two Atwood’s (two 40 kg at 1 cm, and two 1 kg at 40 cm) has moved the same distance (real numbers .0628 and .0631 sec for one of many set ups).  Having the same final velocity after moving the same distance assures us that the different Atwood’s have the same acceleration.

Having the same F divided by the same ‘a’ for both Atwood’s (40 kg at 1 cm both side; and 1 kg at 40 cm both sides) assures us that the inertia ‘m’ has to be the same also.

Real numbers for a certain Atwood’s are:  two 20.7 kg at a 1.03 cm r; compared to two .8628 kg at 24.13 cm. Both took 3.3 sec; in another set up. They had the same F and the same 'a' and the same inertia 'm'.

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #353 on: September 08, 2022, 07:48:09 PM »
I caught myself wondering how science made this mr² mistake in the first place. And then I remembered: it is all about the protection of ½ mv² (probably mv² at the time); if 1 kg can be accelerated to 2 m/sec as easily as 2 kg can be accelerated to 1 m/sec then The Law of Conservation of Energy goes kaput. And that is the sacred cow of physics.   

Offline Kator01

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Re: re: energy producing experiments
« Reply #354 on: September 09, 2022, 01:38:39 AM »
Delburt


Quote
The 80 to 1 lever arm, according to your math, can’t be moved up until 6,400 kg are placed on the opposite short (1 cm) side. [/size]And: according to your math, then the long side of a 40 to 1 lever arm can’t be moved up until 1,600 kg are placed on the opposite short (1 cm) side. But the one kilogram on the 40 cm side balances with 40 kg on the 1 cm side; and you then add another 40 kg for acceleration.
Quote

[/size]
No, this is not what is meant please reread my description.[/size]

[/size]
We have to distinguish between a static situation ( Nm)  and a dynamic one-> accelleration..--> mr² ..resistance to rotary accelleration.[/size]

[/size]
"By the way: this is not "my math",  I just try to understand the basics and s[/size]ince you changed to a lever-system ( away form the Cyl & sphers) in order to find an easier way ( compared to the Cyl. & Sph) I was forced to dive into the depth of what was known to this subject back then.[/size]

[/size]
I cannot and will not argue whether this is true of not but I understood the principle of this experiment. [/size]

[/size]
One thing is sure that these guys back then did extensive and very elaborated experiments. I spent a few days searching the internet ...didnt find anything and then looked the subject up in my old book. I have no time to translate all this to english but the experiment was basis in university
teachings and sure enough they demonstrated this in front of students.
If I had the time and the means I would build this experiment.


I am a pragmatic person, i.e. I build, experiment and analyse...like you to. This I did with the Cylinder & Spheres experiment. Something went wrong
with a brass-bearing and I have to re-engineer the setup. However I could see and film that momentum is conserved not energy. I will continue with this small wheel setup and build it according to sound mechanical means and do not let myself distracted by other setups because I have not reached the core-function of the Cyl & Sphere system.


I agree: They do everything to control attention to protect their narrative.

Levine on moment of inertia

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

Mike












Offline Tarsier_79

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Re: re: energy producing experiments
« Reply #355 on: September 09, 2022, 10:57:53 AM »
I=mr2 is easy to prove.

Static balance does not mean equal inertia. You can't pick and chose when you will use 1/2mv2.

I do not know why you got the results you did. I suspect there is a problem with your setup. Probably bearings able to hold 60kg, but able to rotate freely. A bicycle wheel might be a good start for a test such as you suggest.

You do have one good point. I do not know how a balanced offset weight (eg 8Kg at 10cm vs 1Kg at 80cm) doesn't rotate around a virtual inertial center, rather than the static balance center.

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #356 on: September 09, 2022, 03:23:01 PM »
Demonstrating Rotational Inertia (or Moment of Inertia) - YouTube     flipping physics

Question for you:   Why is the torque formula one r and (your) inertia formula is two r s? Because changing the location of torque is exactly the same thing as changing the location of inertia. It is a relationship between the two radii not an actual distance.

I don’t think they (guts back then) did do extensive and elaborate research; and that is why you cannot find the experiments. Just like the flipping physic guys; they don’t do data: just; it is faster it is slower, etc.

You have a 80 cm wheel with a 2 cm shaft and you have 40 kg suspended from the left side of the shaft and 1 kg suspended from the right side of the wheel. It is balanced: but you then add 5 grams to the 1 kg on the right. We will let those 5 grams accelerate the wheel to .1 m/sec. At one tenth meter per sec we add 200 grams to the 40 kg on the shaft left side.    What happens to the acceleration?

Offline Tarsier_79

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Re: re: energy producing experiments
« Reply #357 on: September 10, 2022, 02:29:59 AM »
Quote
Question for you:   Why is the torque formula one r and (your) inertia formula is two r s? Because changing the location of torque is exactly the same thing as changing the location of inertia. It is a relationship between the two radii not an actual distance.

Torque has nothing to do with inertia. Torque is a measurement of applied force. Inertia is a measurement of how difficult something is to accelerate, and mathematically relates to how much energy is gained or lost during rotation.

Getting back to a static balanced 8Kg vs 1Kg rotating in space (or on an ice rink). The 1kg is 8 times further from the point of rotation, and is 8x more difficult to move. Logic does tell me that the assembly should rotate around a point where both sides are equally difficult to rotate comparatively......Diving Deeper: Again if I am at the center of static balance, the 8kg pulls my left arm forward with 8x less energy than the 1kg pulls my right arm backwards. Again, I feel like this is not where the assembly wants to rotate around if it were not constrained on an axle.



Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #358 on: September 10, 2022, 05:29:48 PM »
Energy?? Tarsier; since when does F and m make energy?  F = ma makes linear momentum. “Rotation gained or lost” is acceleration. So you have the three components of Newtonian motion; and Newton never made energy.

Torque is an applied force upon a resistance; if there is no resistance there is no torque. From Newton Third Law we know that the inertia is applying a force upon the torque: equal and opposite but entirely in the same manner. It is a reverse torque upon the applied torque. This inertial resistance is force being applied at a radial distance.

I will give two arguments:

Make a horizontal lever arm with one kilogram attached (at rest) at one meter. You then (perpendicularly) collide 2 kg moving 5 m/sec at the .5 meter location. What is the resultant motion?

Second argument:

In the flipping physics ‘Demonstrating Rotational Inertia (or Moment of Inertia) - YouTube’ lets assume that the torque can be placed at two locations, on the pulley, where one location is twice the radius of the other. And we know that we can place the inertia at twice the distance; one location toward the middle, and then one location on the end of the rods.

Now we have four possible arrangements of the spoke wheel: small pulley torque to inertia position on the end of the rod, small pulley torque to inertia position in the middle of the rod; large pulley torque position to inertia position on the end of the rod; large pulley torque to inertia position in the middle of the rod.

Now ask yourself; is the small to small equal to the large to large? And I hope you say yes.

So if you halve the torque radius you double the inertial resistance.

But (for your law): if you double the inertia radius you quad the inertial resistance.

So you can move from ‘large radius torque and middle radius inertia’: to ‘large r torque to large r inertia’. And you say it will be one fourth the acceleration. 

Or you can move from large radius torque and middle radius inertia: to small r torque to middle r inertia. And it will be one half the acceleration.

But the radius torque / radius inertia is the same in these two changes.

You can’t have two rules.

Offline Tarsier_79

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Re: re: energy producing experiments
« Reply #359 on: September 11, 2022, 03:35:10 AM »
1 newton meter of torque is 1 newton of force at 1 meter, or its equivalent regardless of the radius it is applied. 2 newtons at 1/2 meter etc. It doesn't care if there is very little resistance or very much.

inertia is directly related to energy math. To accelerate a mass till it has 50 joules of energy takes 50 joules of energy. A mass at 2x radius at a fixed rpm will have 4x the energy, just as it is 4x harder to accelerate, because it takes 4 times the energy to accelerate it to that energy level. Putting a mass on a lever doesn't magically contravene physics.

https://www.youtube.com/watch?v=lNx0yPdl960
See the acceleration at 2:40. According to your theory, the masses are around 3x further out, so should be approx 3x harder to rotate. This is clearly not the case. According to conventional physics, it should be around 9x harder to accelerate. The inner weight model moves a full rotation as the outer weight model only moves around 1/8th.

I understand your unwillingness to accept this fact. Accepting it would throw many or all of your "creating energy" theories away.