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Gravity powered devices => Gravity powered devices => Topic started by: pequaide on February 17, 2007, 07:39:49 PM

The cylinder and spheres machines
Tethered spheres imbedded (at 180?) in the surface of a hollow cylinder can stop the spinning cylinder when the spheres are allowed to feed out on the end of the tethers. The only quantities of motion that I see might be conserved are kinetic energy (1/2mv?) or Newton?s linear momentum (mv); but not both, for the same v can not satisfy both equations.
I would like to get a few people to make and video tape these cylinder and spheres machines, I would very much like to know which type of motion is conserved. My data shows that it is Newtonian motion. I make my models out of PVC pipe and couplings; fishing string; and one inch spheres from scientific supply houses.
I am not looking for people to tell me what Laws I have broken, or what concept problems I have. I build these machines and they work; I am looking for people to repeat the experiment. I get output energy of about 350% the input energy.
I have video tapes of several models, and the mechanism used to spin them, on DVD. If you think the DVD would be helpful in building your own machines; just post your address and I will send you one free. There are pictures on yahoo/groups/madscientist

Hello pequaide.
can you post a small video to see it please?
thank you.
slncspkr

I get output energy of about 350% the input energy.
What method do you use to measure the output energy and input energy?

How it is that, "the same v can not satisfy both 'equations;'" are we setting linear momentum equal to kinetic energy?
It sounds like you have a complex system of twirling masses with tethers causing discontinuities. Which v are we talking about?
v_{ar1} initial rotational velocity, (should be Ãâ€°_{ar1},) assembly at sphere 1 release.
v_{ar2} initial rotational velocity, (should be Ãâ€°_{ar2},) assembly at sphere 2 release.
v_{a1} initial rotational velocity, (should be Ãâ€°_{a1},) assembly at assembly release.
(It is unclear without seeing the video if the tube is released from the hand before of after the spheres are released from the tube.)
v_{s1} initial linear velocity of sphere 1.
v_{s2} initial linear velocity of sphere 2.
There will be more velocity changes but the speculation grows exponentially without seeing the video.
My guess is that the spheres go around the tube in some cool pattern due to the tethers and the rotational mass of the cylinder. Energy is transferred from rotational to linear and back again. An overall gyroscopic effect may prevent the tube from toppling over, maybe throughout the entire event. Then the whole thing finally comes to rest with the spheres just clearing the tabletop?
Or, if what you are saying is true, you might want to warn people to take safety precautions. At 350% out/in, the acceleration must stress the tethers to failure sending the spheres out as projectiles? :o

Measurements were made by video taping the models that have slits in them. The slits come back from the small hole that allows the string of the bolas to go through a diameter of the cylinder. The portion of the bolas (string with two spheres on each end) that protrudes (the tether) through the cylinder can be varied, and the mass of the cylinder can be varied. These variations allow the experimenter to arrange for the cylinder to stop when the tether is at 90? to a tangent of the cylinder.
Before the tether is at 90? (to the tangent) it will counter rotate the cylinder, after 90? it will pull the cylinder in the original direction. By having a slit at the entrance hole of the tether it allows the video camera to take a few frames while the cylinder is stopped. The movement of the string along the slit allows you to calculate the spheres velocity at the end of the bolas when the cylinder is stopped. The original velocity is determined by a photo tachometer (3.25 rps). Some strobe light photography was also used, with the strobe flashing at 100 flashes per second, which yielded the same results, about 350% output from the original input energy.
All velocities used were Newtonian linear, of course you have to know rps (rotations per second) and radius. But all motion was in the linear frame, which would be the distance traveled around the circumference of the circle in a unit period of time. The end objective would be to release the spheres, and then they would be traveling in a line. Whenever an object is released from a circular motion it will travel in a straight line equal to the quantity of travel around the circumference.
Keeping it simple the original velocity is about, 3.25 rps * 4.81 in * (.0254 m /in) * 3.14 = 1.25 m/sec, and the final velocities of the spheres are about 4.5 m/sec. I admit it is rough data but it is well over kinetic energy conservation, which is our only other choice. I think momentum is conserved.
As far as safety; yes the spheres will on occasion break the string, these are not toys they are somewhat dangerous. Proceed to build at your own risk, and please don?t let children play with them.
I have videos but I can not put them on my computer from D or E drive, I don?t know what the problem is. I might try loading from the camera itself. I can send about any picture you want. The cylinder in the background in the pictures is one with a slit, numerically marked to calculate distance traveled by the string.

Tethered spheres imbedded (at 180?) in the surface of a hollow cylinder can stop the spinning cylinder when the spheres are allowed to feed out on the end of the tethers.
Isn't this conservation of angular momentum? Similar to an ice skater extending or retracting their arms to alter the speed of a spin?
I may have missed some important point but I don't quite get the purpose of a device that stops spinning, isn't the goal to get one to keep spinning?
Could you please share a few more details so I can understand this better?

Hello pequaide
Your development is very impressiv . But to be honest,it is, at least to me, very difficult to emagine,how the Parts move ,and how the Energy is to move out.
It would be very very helpfull to see either a Video,or an animatet gif.
Please give ita push,and spend more Information,so that we can learn from you,how the things work in function.
Regards from Duisburg
Helmut
.....................
All velocities used were Newtonian linear, of course you have to know rps (rotations per second) and radius. But all motion was in the linear frame, which would be the distance traveled around the circumference of the circle in a unit period of time. The end objective would be to release the spheres, and then they would be traveling in a line. Whenever an object is released from a circular motion it will travel in a straight line equal to the quantity of travel around the circumference.
Keeping it simple the original velocity is about, 3.25 rps * 4.81 in * (.0254 m /in) * 3.14 = 1.25 m/sec, and the final velocities of the spheres are about 4.5 m/sec. I admit it is rough data but it is well over kinetic energy conservation, which is our only other choice. I think momentum is conserved...............

I am also having a hard time visualising this setup. So I maybe way off but here is a concepts. When it come to any potential OU device you must break it down to its simplest equations and find all the input and out put energy. To keep it simple like the ice skater spinning with there arms out at a slower rpm and then PULLING there arms in to accelerate the rpm. So the work done to pull there arms in is then stored in the rotational energy of the skater. So, when they open there arms again they are spinning at a higher rpm then originally. I believe the energy is being added from the work done to pull the arms/balls to a closer radios?
Tim

In order to make a perpetual motion machine there must be a portion of the cycle of the machine that makes energy. The cylinder and spheres machines can quadruple (or more) the quantity of energy in less than a second; this makes them applicable to be the impetus for what is envisioned as the classic perpetual motion machines.
In some portion of the cycle of the machine momentum must be increased as well, but this is already proven technology; which is the Atwood?s Machine. This is force times time being used to make momentum so this is not a violation of the Law of Conservation of Momentum.
A 9 kg Atwood, with one kg of imbalance, can produce 14.007 units of momentum from what costs 4.429 units of momentum to achieve. One kg in a 1 meter freefall has a final velocity of 4.429 m/sec, but after it has fallen the same distance on the machine it has a velocity of 1.4007 m/sec (* 10 kg) for 14.007 units of momentum. It only takes 4.429 units of momentum to return the one kilogram mass to the top, to start over.
The Law of Conservation of Momentum is one of the most respected Laws in physics. All three of Newton?s Laws of Motion are contained in this Law. Newton?s Laws of Motion are not restricted to objects that only travel and interact in straight lines, the Laws would be worthless if this were true. Newton used the word oblique to describe the interactions of objects at angles; he also dealt with gravity and planetary motion so he was familiar with forces working at angles.
If the cylinder and spheres do not conserve momentum I think it is the first and only violation of Newtonian Physics.
When the spheres are seated in the surface of the cylinder there mass is moving at roughly the same speed as the cylinder. Let?s say that the three kg cylinder with the imbedded 1 kg spheres is moving at one meter per second around the circumference (center of rotational mass actually) of the cylinder, that would give them 4 units of momentum and 2 units of kinetic energy. To maintain the 4 units of momentum the spheres would have to be moving 4 meters per second when the cylinder is stopped, and that would give them 8 units of kinetic energy. 8 is four times bigger than 2. Formulas used were 1/2mv? and mv.
Now this kinetic energy is real, for the 4 kg can only rise .050968m * 4 kg = one kg at .2038m and the spheres (one kilogram) can rise .8155m. .8155m is four times as high as .2038m. Put this rise into an Atwood and we have a perpetual motion machine. Formulas used were d = 1/2at? or d = ? v?/a
Experiments have been conducted on a frictionless plane using a disk and pucks, with the same results.
So the perpetual motion machine would consist of a horizontally mounted wheel driven by an Atwood machine or an overbalanced wheel. The horizontal wheel would act as a cylinder and spheres machine and would flip the driving mass back up to the top.

It might just be me but I can't understand the workings of the device. You are throwing numbers around but I don't see the mechanical design in which you are speaking. Could you post a video or schematics please?
Tim

Hello Pequaide
I am not really not shure to got the Point.So i discribe it as i see it in my vision as a pendulum.
Therefor i imagin a swinging pendulum.
The axes in the middel is surroundet be two counterrotating masses(weights but imbalanced).The counterrotating Masses are also part of a Pendulum,wich is mountet on the axes from the mainpendulum.Both Pendulums are forced to move in the same direktion.
The speed of the counterrotating weights is according to the speed of the Amplitude,that the Pendulum is acting.As there can place resonanz.
But this is just imagency.
I would like to take part at the Offer ,that you have made,to send Pics and or,a DVD.
Therefore see here the Adress info@satundedv.de
The email adress ist working.
Please writre details for Kontakt
the Page is deactivatet because of reconstruktion.
so thanks in advance
Helmut

I can understand your inquiry for drawings of a functional perpetual motion machine. We are all a little curious as to what the first perpetual motion machine will look like. I think it will be an overbalanced wheel or Atwood?s machine that drives a smaller cylinder or disk that flips masses back to the top like the cylinder and spheres machine.
I made drawing of such a machine that I estimated would have the output of Glen Canyon Dam, on the Colorado. I sent the drawing to the Patent Office: boy was I naive. They not only had a good laugh they also keep my money.
I have determined that to discuss any such machines is a waste of time (which I am doing now). If I can?t get you to repeat a simple $25 experiment, what good is discussion of one that will cost $1,000.00 or more. There is not much sense in posting on sites where all they wanted to do was talk (and make junior high jokes). To my knowledge none would actually build the $25 machines, but they were sure willing to berate them. Plus: who would move on to the $1,000.00 machines before confirming the $25 versions?
The cylinder and spheres machine is in and of itself a huge challenge to the field of physics. If Newtonian physics is true the Law of Conservation of Energy is false. If the cylinder and spheres machine works as proposed the machine itself is one of the most important inventions in history. If it does not make energy it will have to have lost Newtonian momentum, and this too would be a significant event.
Linear momentum can be caught and released from circular motion, with no change in linear momentum. If the rearrangement of motion as in the cylinder and spheres does not conserve circumference momentum then it does not conserve linear momentum. And this too would be a first in the word of physics. So you see that the cylinder and spheres experiment places the world of physics between a rock and a hard place, either the Law of Conservation of Momentum is false or the Law of Conservation of Energy is false.

Hello Pequaide
First i want to thank you,that you are kontinuisly passing by and keep
the conversation alive.
I think,if we follow the law of Newton in all aspekts,then we put our Minds in chains.
Thats wy i prefer,to try a search after new knowledges.
Stefan has made a diskription to show us ,were we do a further search.
Analog to the energy, of flowing water,which we take out by a wheel,we have to find a prossedure to guide out the energy from aether.
By electric energy:
I think,that the key is hidden in either a electric field,like it is present in a capacitor,to capture electrons between a plate and a dielektrikum.Or electrons are Captured in a coil
as a result of a collaborated magnetic field,that was caused of a travel from a moving field and masses throu a existing field.
Perhaps we have to build a stator,which has both of it.Once the Coil or Magnet and/or
a flat capacitor. And the Rotor might also be a sort of multicapturabe construktion.
And the Power from Gravitation need also a prossedure to be captured and to be placed on another point to feed a secondary aktion.Without a need of palarisation.
Just to create a differenz in energy level and keep this situation as long as there is need for energy.
I am so shure,that the plans for free energy are all ready in our Minds.But to find the key to open our Minds and disclosure the fakts,is our greatest job.
Helmut

The following is a description of a simple experiment that proves that the Law of Conservation of Energy is false. It is composed of an upper and lower section. The top is a 208.6g cylinder (3 in. PVC pipe coupler with a 3.5 in. I.D.; both the pipe and the coupler have a ? inch wall) with two 66g spheres seated in the coupler?s surface at 180?. It has a 3 in. I.D. pipe inside that has a mass of 22.2g to help seat the spheres. The top portion of the cylinder (see pictures under: ZPEnergy; FreeEnergy; files; pequaide) and spheres experiment can be connected to different lengths and diameters of pipe by using the bottom opening of the coupler.
The spheres stop the top cylinder from spinning, as the spheres swing out on the end of tethers, before they reach 90? to tangent if there is no additional mass added to the bottom. Only by adding mass to the cylinder can you force the spheres to stop the cylinder when the spheres are at 90? to tangent. Mass is added by placing a straight pipe in the bottom of the top coupler. A slit is placed in the cylinder behind the entry hole of the string (that tethers the spheres) to allow you to evaluate the motion of the cylinder when the spheres reach 90?; the cylinder will maintain the same rate of motion for about 1/10 of a second as the string (tethering the spheres) moves across the slit. One and only one length (mass) of pipe makes the spheres stop the cylinder at 90?, less than adequate mass and the cylinder will stop before 90? and the cylinder is being force backward while the string, holding the spheres, is in the slit at 90?. Too much added mass and the cylinder will still be moving forward while the string is in the slit.
For this particular cylinder arrangement and tether length the added mass that stops the top (and bottom) cylinder at 90? is about 364.9g of a 3 in. I.D. PVC pipe.
If you replace a portion of the added pipe with a larger diameter coupler; you can make calculations and construct a system that keeps the spinning momentum of the added mass the same. But if the rotational momentum is kept the same the kinetic energy of the spinning larger diameter pipe is not the same.
So if the entire system maintains its 90? stop it is reasonable to assume that nothing has changed and that the momentum of the system is being conserved since the input momentum had not changed and the apparent output had not changed either.
Now; one could claim that the 90? stop still occurs but that the spheres are moving faster or slower, but under these conditions the Law of Conservation of Momentum would be false. You simply can?t pretend that both these two formulas (mv and 1/2mv?) are conserved.
This discrepancy between momentum and kinetic energy exists in ballistic pendulums, there kinetic energy is assigned an imaginary friend (heat) to hide the formula?s (1/2 mv?) incompetence.
In the cylinder and spheres experiment it is totally impossible to have this phantom friend of heat come to the rescue, because kinetic energy is increasing not decreasing.
The straight 3 in. pipe is replaced with a combination of a coupler and a 3 in. connecting pipe. If the connecting pipe for the bottom coupler is cut at 73.2g; the momentum change from the 3 in. (I.D.) 364.9g pipe length would be 12% low. The momentum would be the same for this lower portion if the connecting pipe is cut to a length that gives you 112.8g.
If the connecting pipe for the coupler is cut at 112.8g; the kinetic energy change from the 3 in. (I.D.) 364.9g pipe length would be 11% high. The kinetic energy would be the same for this lower portion if the connecting pipe is cut to a length that gives you 73.2g.
This11% and 12% change for this lower portion doesn?t mean that the spheres motion will be going up or down by 11% or 12% that has to be calculated separately because the spheres are going to absorb all the motion of the connecting pipe whether it has a mass of 73.2g or 112.8g. The 11% only predicts that the 90? stop will be lost if you use the wrong one (73.2g or 112.8g) of the two masses. The spheres themselves would ideally have about 500% of the original energy at the 90? stop.
The 112.8g connecting pipe was used and the 90? stop was maintained, that means that kinetic energy is not what the spheres are responding to. The spheres are responding to momentum, and since there was no change in the rotational momentum there was no change in the 90? stop.
This 11% is well outside of the accuracy of the 90? stop method. I think I could detect a 6g change in the connecting pipe mass, and the 73.2g pipe is 39.6 grams off. If the connecting pipe only had a mass of 73.2g the method would be telling you to add more mass to get it to the stop at 90?, but at 112.8g the 90? stop has been achieved and the method is not telling you to add or subtract any mass. The cylinder and spheres experiment responds to momentum.
Another experiment was conducted using the results from the 3 in. pipe addon as a base line, and comparing the results collected by connecting a larger diameter cylinder onto the bottom of the top coupler. After being spun the top cylinder is forced into a stop when the spheres swing out to 90?. This is accomplished when a mass of 364.9 g is added to the top cylinder in the form of a 3.0 in. I.D.(.25 in. side wall) PVC pipe placed in the bottom portion of the coupler. There is one and only one mass that will cause a perfect 90? (to tangent) stop.
I then replaced the 3.0 in. pipe with a 4.0 I.D. PVC pipe. I calculated its relative velocity (at the same RPS) to be about 4.25/3.25 that of the 3.0 pipe. Therefore to maintain the same momentum its mass will be 3.25/4.25 times 364.9, which is 279g. I then placed a 279.1g pipe on the bottom of the top cylinder and began video taping it while spinning and releasing it by hand. Once the appropriate mass is determined the cylinder will stop at 90? at any RPS, being hand held offers no real problem, in fact it proves that the RPS does not matter. And the hand held models are much more readily available for physics class demonstrations.
At 279.1g the cylinder had a slight backward motion, while the string was in the slit at 90?, that indicates that it is a little light. I added 7.5g and the forward motion was greater than the previous backward motion. That means that the 279.1g 4.0 pipe was about 3g light, 3/279 is about 1% off. This is the third object (added to the bottom of the top coupler) with different diameters that have the same rotational momentum, and they all cause the same 90? stop. If graphed this would be three points on the same line, a line that indicates that the cylinder and spheres experiment is driven by the Law of Conservation of Momentum.
To conserve kinetic energy the 4.0 in. pipe would have to have a mass of 213g, about 30% off. I see no way that kinetic energy is being conserved in this experiment. That makes the Law of Conservation of Energy false.
My original estimations of a 350% energy increase for the cylinder and spheres experiments were based upon attempts to actually measure the velocity of the cylinder with embedded spheres (using a mechanical release) and then measure the spheres alone at the 90? stop. I used strobe light photograph, video tapes, and photo gates. This method of adding different diameters is an entirely different and second method that confirms in my mind that the cylinder and spheres conserves momentum and can increase energy by 300% or more. This will be the driving force of perpetual motion machines.
These 90? stop experiments prove that the quantity of motion lost by the cylinder is momentum. Newton?s Three Laws of Motion require that the momentum lost by the cylinder must be gained by the spheres, because the force in the tether string must be equal in both directions.
The cylinder stops with the spheres at 90? with any R.P.S. used if the proper amount of mass at a certain diameter has been added to the bottom of the top coupler. Three proper amounts of mass at a certain diameter have been used to stop the top cylinder when the spheres are at 90? to tangent.
The cylinder will stop with any initial R.P.S. given it by the hand release method. I am going to use 2 rotations per second to get some actual numbers. I think 2 R.P.S. is within the ability of the hand release method.
A 3.0 inch inside diameter PVC pipe (1/4 in. side wall) with a length just under 10 inches and with a mass of 364.9g gives the cylinder a nice stop when the spheres are at 90?. The effective rotational mass of the cylinder is at about a 3.25 in. diameter. Here is the math used for the quantities of motion mentioned.
Mass: 364.9g (3in. I.D. pipe)
Momentum: (mv) 364.9g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .18926
Kinetic energy: (1/2mv?) .5 * 364.9g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .04908
A 3.5 inch inside diameter PVC pipe coupler with a mass of 218.9g combined with a 112.8g connecting 3.0 in. pipe gives the cylinder a nice stop when the spheres are at 90?. The effect rotational mass of the coupler is at about 3.75 in. and the effective rotational mass of the connecting pipe at a 3.25 in. diameter. Here is the math used for the quantities of motion mentioned.
Mass: 218.9 for the coupler and 112.8 for the connecting pipe
Momentum: (mv) 218.9g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .13101
Momentum: (mv) 112.8g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .058506 Total equals .189516
Kinetic energy: (1/2mv?) .5 * 218.9g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .039202
Kinetic energy: (1/2mv?) .5 * 112.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .015173 Total = .05437
Momentum change from the 3.0 in. pipe is .189516/.18926 = 0.1%
Kinetic energy change from the 3 in. pipe is .05437/.04908 = 11%
A 4.0 inch inside diameter PVC pipe with a mass of 283g gives the cylinder a nice stop when the spheres are at 90?. The effect rotational mass of the cylinder is at about 4.25 in. Here is the math used for the quantities of motion mentioned.
Mass: 283g (4inch added cylinder)
Momentum: (mv) 283g * 1kg/1000g* 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .19195
Kinetic energy: (1/2mv?) .5 * 283g * 1kg/1000g * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .06510
Momentum changed from the 3.0 pipe was .19195/.18926 = 1%
The kinetic energy change from the 3.0 in. pipe was .06510/.04908 = 33%
When the tether achieves 90? it enters a slit, if the cylinder is stopped it remains stopped as the tether string crosses the slit. If the spheres stop the cylinder before 90? they will have started moving the cylinder backward and the cylinder will be moving backward while the string is in the slit. If the spheres have not yet stopped the cylinder before they reach 90? then the cylinder will be moving forward while the string is in the slit.
The cylinder will be stopped at 90? only if it has a certain mass, add 6g and the cylinder will still be moving forward in the slit, subtract 6g from the proper mass and the cylinder will be moving backward. It is accurate within 2% of the added mass. Momentum conservation falls within this 2%, kinetic energy conservation fall outside the 2% at 11% and 33%. Kinetic energy is not conserved by the cylinder and spheres experiment.
If the initial momentum of the cylinder and spheres is conserved in the motion of the spheres alone (when the cylinder is stopped) then the systems has over 400% of the original energy.
Momentum: (mv) 208.6g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .12484 (top 3.5 in. I.D. coupler with holes for the string and holes to seat the spheres)
Momentum: (mv) 22.2 g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .011515 (small piece of 3 inch I.D. pipe used to seat spheres)
Momentum: (mv) 364.9g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .18926 (3 in. PVC pipe about 9.8 inch long)
Momentum: (mv) 132g * 1kg/1000g* 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .084265 (spheres seated on the surface of the coupler, 66g each)
Total initial momentum = .40988
Final momentum will be .40988 for a velocity of 3.1 m/sec .40988/.132 = 3.1 for the spheres only
Kinetic energy: (1/2mv?) .5 * 208.6g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .037356
Kinetic energy: (1/2mv?) .5 * 22.2g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .002986
Kinetic energy: (1/2mv?) .5 * 364.9g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .04908
Kinetic energy: (1/2mv?) .5 * 132g * 1kg/1000g * 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps =.026896
Total Initial Kinetic energy = .11632
Final Kinetic energy will be .5 times .132kg times 3.1 m/sec times 3.1 m/sec (1/2mv?) = .63426
This is a kinetic energy change of .63426/.11632 545%
The change in achievable rise (d =1/2 v?/a) of the initial moving objects (spheres, coupler, pipe) to the final moving objects (spheres) will also be proportional to the change in kinetic energy and is called potential energy.

pequaide,
please don't take this the wrong way. i am just a dumb ironworker with too much time on my hands. however much i would like to do your experiment, it looks fascinating, i am having trouble figuring out, how all these parts are supposed to come together. i would be more than willing to duplicate your $25 experiment as all it would cost me is time. i would also like to see your $1,000 experiment because alot of the materials may be available to me from scrap on the construction sites that i work at. and if i need special machineing of parts, that might also be available from friends, in there spare time.
but i am having trouble grasping the really intense concept that you have put out so far with my intelectual ability. can you make it any simpilar for people that have the physical and mechanical skills to get the job done, but don't have the physics and math capabilities, along with the engineering ability that you have?
sorry to be a dumbass
lol
sam
ps. maybe this is over my head conceptually, but if you can just draw me a print i can build you the world.

I am duplicating my current model so that I have an extra for demonstration purposes. I will write a step by step procedure as if we were starting up manufacturing. It will take a while, but here is what you need for materials.
About a 12 inch length of 3 inch I.D. PVC pipe, schedule 40, that is with a ? inch side wall.
A PVC pipe coupler for the 3 in. pipe, they have a 3.5 in. I.D. and a 4 in. pipe will fit on the outside of the coupler.
About a 4 in. length of 4 in. I.D. PVC pipe schedule 40. (not in the picture)
Two steel spheres WLS448020E Sargent ?Welch; These spheres are drilled, they are much cheaper if you can fix nondrilled spheres to a string,
I make wire loops to fit inside the drilled hole in the spheres, the hole is larger on one side than the other, this will stop the wire crimp half way through.
I use fishing wire and crimps, I will check this: my source is old.
Then I use fishing clips and 30 lb fluorocarbon fishing line, to connect it through the holes to the center.
I will submit a detailed description in the not to distant future. Thanks for your help. These machines have energy increases of over 300%.

pequaide,
i've got all that on hand, except for the steel sphaere's. the ones i saw in the photos looked to be about 3/4" steel balls. is this about right? if so the bulldozer guy on my jobsite had to replace something the other day and i know where maybe a hundred of these are laying on the ground getting ready to be buried or salvaged. i also know a machinist that can drill me any kind of hole imaginable for free. so i guess i am sitting on go right now. can't wait for construction details to replicate.
lol
sam

The spheres are 1 inch in diameter; they have a mass of 66g, they have a 1/16 in. hole through the center and it is enlarged to 1/8 in. half way through.
If I did the math correctly a ? inch sphere would only have a mass of 28g. They probably would not stop the cylinder.
You could buy 1 inch spheres from MSC industrial supplies and have your friend drill the holes. They are cheaper if you can buy them solid and get them drilled (free). I think I can have the details tomorrow.

pequaid,
the 66gram figure seems to be the most important part to me. ??? 8) is that 66grams after or before the drilling takes place. i can do it either way. 1" spheres, no problem.
lol
sam ??? 8)

If you are about 66 grams (each) you will be fine, you will have to add or subtract the added mass of the pipe, no problem. Most important is the placment of the holes.
Materials list; for models with two masses added (3 in. pipe and 4 in. pipe)
1. A length of 3 in. I.D. PVC pipe (about 10 in. per machine)
2. A 3 in. PVC pipe coupler (one per device)
3. A length of 4 in. inside diameter PVC pipe (about 4 in. per device)
4. 30 lb fluorocarbon fishing line (only a few inches per device but it is constantly breaking)
5. 20 lb stranded fishing wire with appropriate diameter crimps (.047 in. dia. leader sleeves)
6. fishing connecters (two per device) interlock snaps size 3
7. two spheres 1 in. (two per device)WLS448020E Sargent Welch
Tools list:
1. #60 wire gauge drill bit .0400 in.
2. 1/16 in. drill bit
3. 1/8 in. drill bit
4. 3/16 in. drill bit
5. ? in. drill bit
6. saber saw blade
7. saber saw
8. drill press or drill motor
9. band saw or hack saw
10. hack saw blade
11. calipers (you might use a compass if you have no calipers)
12. metal scribe or awl
13. 7/8 in hole saw
14. sandpaper; wet dry work good
15. fibered tape to reinforce the sand paper
16. rat tail file
17. 1 ? in. dremal cutting disk and dremal saw
Procedure:
1. Find the smooth end of the 3 in. coupler. (one end has Genova made in USA etc. with raised printing on it)
2. Scratch a line (with the calipers) 21.35 mm down from the smooth top end, along the side of the cylinder, all the way around the cylinder; all holes will be drilled on this line. The particular dimension (21.35 mm) is not important but that dimension needs to remain constant (always exactly the same, most calipers have a screw that will hold the dimension chosen)
3. Make a very light indention on the line at any point, just large enough so that the caliper tip can locate it. This is indention 1
4. Set the calipers to 2.000 in. and mark off 2 inch intervals on the scribed line around the cylinder from indention 1, the third mark takes you to the other side (180?). Mark off three 2 in. marks around the other side of the cylinder going the other direction from indention 1, this will leave two third marks very close together (at 180?), half way between these two marks place an indention, this is where you (later) drill the second string hole. This is indention 2
5. Place light indentions along the scribe line 2.538 in. counter clockwise (looking from the top) from the indentions 1 and 2. We will call these indentions 1A and 2A.
6. Place indentions along the scribe line 3/16 in. and 1.5 in. clockwise from indentions 1 and 2. These indention will be referred to as 1B (3/16 clockwise of 1) and 1C (1.5 in. clockwise of 1), and 2B and 2C
7. Drill1/16 in. hole toward the center line of the cylinder at indentions 1A, 1B, 1C; 2A, 2B, 2C, be careful to center the holes; 1A, and 2A are most critical
8. Drill.0400 inch holes at 1 and 2; be careful to center the holes.
9. Enlarge holes at 1A and 2A to 1/8 in., then 3/16 in., then ? in. This step drilling is necessary to keep the hole from floating off center.
10. Cut a 7/8 in. hole using the hole saw at the ? in. hole at 1A and 2A
11. Enlarge holes at 1B, 1C and 2B, 2C to 1/8 in., then 3/16 in.
12. Cut a slit between the holes at 1B and 1C, and 2B and 2C with a 1.25in. dremal disk
13. Enlarge that slit with the saber saw so that it is 3/16 wide from hole to hole.
14. carefully extend 1B into hole1 being careful not to damage the working side of 1
15. carefully extend 2B into 2 being careful not to damage the working side of hole 2
16. Seat a ? in. slice of 3 in. pipe on the inside center stop ring of the coupler, this will be behind the 7/8 inch holes. This will have to be filed to seat the spheres.
17. Place a thin sheet of plastic 7/8 in. wide 1/8 in. thick, with a centered hole, on top of the slice in step 16, the top must be level with the scribed line, and 90? to the bolt in step 18.
18. Place a 1/16 in. (enlarge to 3.11 mm) hole between and above the 7/8 in. hole and the trailing end of the slit (1C and 2C) on the smooth top side of the cylinder for a 1/8 (by ? in.) Dia. stove bolt placement, this is used to clamp the end of the string, See photo)
19. Make two 5/8 inch loops using the wire and leader sleeves and place them through the large side of the spheres, they will extend through about 1/8 inch.
20. Sand and file until the spheres seat in holes 1A and 2A in the same manner.
21. Sand and file until the slit does not catch the string.
22. Connect the snap to the loop, connect the string to the snap, seat the sphere and feed the string through the slit, and then through the center hole of the plastic sheet seated on the plastic slice, wrap it once or twice and feed it up to the small bolt in step 18 and clamp the string under the bolt
Pictured is a finished and an unfinished top cylinder, with a 3 in. pipe and a 4 in. pipe. I though you wanted to get started so I gave you this material; tools; and steps. I will probably change them over and over, and if you need help let me know

I know it's asking a lot and you have written a lot of detail already, but would you be able to provide even the shortest video clip to show us what is happening.
Thanks
DrWhat

Lab. notes and comments
I was trying to build what I thought was a duplication of my last 4 inch O.D. coupler, which had a 90? stop with a 364.9g 3 in. I.D. lower pipe, added. To my amazement it took a 390g pipe to make the new cylinder stop at 90?. This is about a 4% mass increase (233.6g cylinder + 364.9g added mass; and 234.2g + 390g). Then I checked the tether length and I had inadvertently increased it by about 6%. So: the longer the tether the more mass the spheres are capable of stopping, even when the spheres mass remains constant. I had known his but had not put numbers on it before. The longer tether length apparently allows a longer period of time for the force to act.
The top coupler (with seats for the embedded spheres at 180?) has a mass of about 234g; when it has no pipe added to its bottom opening, and the spheres masses are not added. The spheres have a mass of 132.8g. With this mass ratio of 132.8g to 234g the spheres stop the cylinder within about 4 cm from the seat. This means that the tether tightens immediately; as soon as the spheres leave the seat, and all of the force before the stop is tangent to the cylinder, because I don?t think the string (tether) has come away from the cylinder this early in the swing. The cylinder is moving strongly backwards while the string is in the slit (behind the point where the string enters the cylinder).
A 3 kilogram cylinder with two 1/2 kilogram spheres moving one meter per second has 4 units of momentum; before the spheres are released.
After the cylinder is stopped the spheres have 4 units on momentum and a velocity of 4 m/sec. The spheres can rise .8155 meters. d = ? v?/a
If one kg (of spheres) is placed in a ten kg Attwood?s machine (with 9 kilograms balanced and the one kg of imbalance) and the spheres? mass is allowed to fall .8155 m, it will give you 12.65 units of momentum. The spheres are now back to their original level and ready to be reloaded into the cylinder and spheres machine, but now the system has 316% the original momentum.
So you see; it doesn?t matter if the kinetic energy formula is defective or not, this system can make energy with or without the kinetic energy formula.
By using the distance formula: we know that the potential rise of 4 kg moving 1 m/sec is only .051, and the rise of 1 kg moving 4 m/sec is .8155, this is 4 times the rise as is indicated by the kinetic energy formula which predicts 4 times the energy.
Whatever the usefulness of the kinetic energy formula; the energy is still there, it is only a matter of utilizing it.
Videos: people can't get them off the DVDs I make. Can anyone help?

Hi,
for converting your DVD files to MPEGs,
use VOB2MPEG, see:
http://www.videohelp.com/tools/VOB2MPG
After this you can use
virtualdubmod
http://sourceforge.net/project/showfiles.php?group_id=65889
to convert MPEGs to AVIs, e.g.with DIVX.com codec.
Or you can use also DVD2AVI for all in one in ONE step.
http://www.protectedsoft.com/download_dvd2avi.php
Regards, Stefan.

Let me introduce this concept for your constructive comments.
Divide a ring (thin walled hollow cylinder) into 360 equal masses distributed evenly around the circumference of the ring. The circumference of the ring is a smooth rigid circle; let?s say it is made of light modern fibers.
Let?s roll the ring down a smooth rigid incline so that it is moving across a smooth horizontal surface at one meter per second. That means the center of mass is moving one meter per second. Lets mark all the masses from 1 to 360; which is of course the degrees of the circle, but the marks are now rolling with the circumference of the circle.
When the mass marked 360 is on the bottom it is moving zero meters per second (in relation to the surface). At the moment when 360 is on the bottom 180 is on the top moving 2 meters per second. Suppose at this moment we release both the 180 mass and the 360 mass. The mass marked 360 will be at rest on the surface. The mass marked 180 can be caught on the end of a pendulum string. The mass marked 180 is now a pendulum bob and it will rise (d = ? at? or d = ? v?/a) .2039 meters. Now the center of mass of the masses marked 360 and 180 is at .1019 meters above the center of mass of the rim (180 can raise 360 .1019 meters off the surface and 180 is still .1019 meters higher than the top of the rim).
Now let?s do the same with the masses at 359 and 179, and 358 and 178, and repeat the process around the circle. This leaves the entire mass .1019 meters above the center of mass of the rolling ring; that now exists of only light modern fibers.
The question arises: How far did the ring roll down the incline to attain a velocity of one meter per second?
An object in freefall needs to only drop .051 meters to attain one meter per second velocity.
A puck on a frictionless plane need only drop .051 meters to attain one meter per second velocity.
A pendulum bob need only drop .051 meters to attain one meter per second velocity.
I would guess it is the same for a cart.
Is the ring different; or is this a free energy source?

Is the ring different; or is this a free energy source?
The ring is different, because it rolls down the incline and ends up spinning at the bottom. Some of the work done by gravity as it descends must be spent to spin it, in addition to the portion spent to send it forward.
You have shown above that you pretty much know how to calculate how much further the ring must fall in order to end up rolling at 1m/s.
Cheers,
Mr. Entropy

All the motion is in the top mass; whether it is spinning or moving forward.

No portion of the rim is being lifted, because for every point of mass at say 23? there is an equal mass at 203?. If 23 is going up 203 is going down, and the same is true for all points around the balanced ring or rim. Take the example of a modern wind turbine; the blades are not being lifted by the wind they are only turning. Nothing has to be lifted to achieve spin.
The center of mass is dropping because the slope allows gravity to act upon it. And I believe you are correct; the acceleration is in relationship to the sine of the angle. Sine is of course in proportional to the length (rolling surface) of the ramp.
Classic physics would expect the rim to roll up the opposing ramp to the very same height if it were not for the existence of rolling friction and air resistance, the same rise in height that we see in pendulums.
And again: it does not matter whether the rim is spinning or translating, all the motion of two opposing masses is held in the top mass when the other is on the bottom. If both masses have accelerated according to the relationship of F = ma, then all the momentum is held by the one mass and energy has been made.

Force is applied to the pendulum frame (at the point of rotation) just like force is applied to the ramp. Tilt an air table a few degrees and slide pucks on it, force is still applied to the table but it has no affect upon the experiment. These are called balanced forces and F = ma works even though these forces exist.

When a kid leans back on a swing, they go higher and higher. They're increasing their acceleration.
And if a weight attached to the pendulum is doing the same thing, why would it be different ?
I reckon the kid is using back muscles to impart energy to the seat of the swing.

You have a fundamental Law available to you that predicts free energy, stay within that Law and you have a very good chance of being successful; the Law is the Law of Conservation of Momentum.
For example: say you have two pendulums side by side that drop .051 meters, both pendulum bobs are moving 1 m/sec at the bottom. Suppose you devise a way of giving all the motion to one of the two bobs, the bob in motion will not rise .102 meters, it will rise .2038 m. You have doubled the energy. d = 1/2v?/a
I once saw a torsion pendulum, in a German clock, that dropped vertically of course but rotated horizontally. If you then applied the cylinder and spheres phenomenon (previous posts) you would have a simple way to make energy.

For example: say you have two pendulums side by side that drop .051 meters, both pendulum bobs are moving 1 m/sec at the bottom. Suppose you devise a way of giving all the motion to one of the two bobs, the bob in motion will not rise .102 meters, it will rise .2038 m. You have doubled the energy. d = 1/2v?/a
This analogy is wrong. Yes in a perfect enviroment the bob in motion would reach .2038m...double the original height. However there is only ONE bob now at the high point, ie HALF the original mass so the potential energy at this point is EXACTLY the same as the kinetic energy at the low point when one of the bobs magically passes its energy to the other one. End result is ZERO change on total energy.

G'day all,
The kid on the swing changes the center of gravity in relation to the fulcrum when it leans back. This shift requires energy.
If this shift in gravity is in a harmonic relationship with the natural frequency of the pendulum (swing) as far as timing and direction goes, this energy gets imparted to the pendulum.
Simple as this
Hans von Lieven

Hans and @ everyone,
the kid on the swing, will have a real hard time getting any momentum to accelerate, if he/she doesn't have arms and hands, that effectively, with energy, shorten the distance from the fulcrum. thus providing additional accelration that coresponds to more height. not applicable. just jmho.
lol
sam

The kid inputs energy into the 'pendulum' swing through his arms and bum.
Have you ever seen a swing with rigid supports instead of flexible rope or chain?.
No, because that would eliminate the energy transfer mechanism.

Pasted from an elementary physics site:
(http://www.hkphy.org/articles/swing/swing_cm.gif)
Let's talk about playing swing again. As the child grows up, he won't be satisfied with the push and will want to learn how to play swing by himself. Please refer to Fig. 2, the broken line represents the route of the centre of mass of the child. Why can he swing with an increasing amplitude? When the boy swings to point A, he rises his body. Assuming that he rises his centre of mass by a distance and the tension of the rope is , he does a positive work which supplies an energy of into the system. When the boy swings to point B, he letdowns his body to lower his centre of mass. This time he does a negative work. If the tension of the rope is , then the system loses of energy. When the boy is at point B, his velocity is zero, so is just the vertical component of his weight (, see Fig. 3). But when he is at point A, his velocity is the greatest, and the difference between the tension and his weight equals the centripetal force (). Hence we conclude that . Therefore, his positive work is greater than the negative one and energy is fed to the swing. If he supplies energy rhythmically, just like his father's push, resonance is produced and the boy swings with an increasing amplitude.
Hans von Lieven

Hello Hans
Today I wend to the Garden to try the Swing.
After this experience I think it is worth, to do more test on it.
It is a play with Inertia.
helmut

The kid inputs energy into the 'pendulum' swing through his arms and bum.
Have you ever seen a swing with rigid supports instead of flexible rope or chain?.
No, because that would eliminate the energy transfer mechanism.
Not true Roy, this works with rigid arms as well.
Hans von Lieven

Hans,
I would love to see an example of a swing with rigid supports and a non pivoting base.

(http://www.johnlewis.com/jl_assets/product/230176252.jpg)
The swings on both ends have rigid arms, no ropes, no chains.
Hans von Lieven

Nice one Hans, but I said with NONPIVOT bases.
The bases the kids are sitting on pivot on the rigid support.

When a kid leans back on a swing, they go higher and higher. They're increasing their acceleration.
And if a weight attached to the pendulum is doing the same thing, why would it be different ?
Kids on swings would seem to be very clever... It's not about leaning back and forth.
They lean away from the swing's pivot point near the top of their swing, and pull themselves back up towards the pivot near the bottom. When they pull back up, they work against more gravity and centrifugal force than are acting on them when they lean away, because they are close to the bottommost, fastest part of their swing. The extra work they do there adds kinetic energy to their swinging motion.
The interesting thing is that kids don't know this, or need to know it. If you're squirming around on a swing, there's really nowhere that any work you do can go, except into your swinging motion, so all you have to do is find something that's hard to do, and it will make you swing faster. If kids practice swinging for a while, their bodies learn this all by themselves. They learn to lean up, raise their feet, and bend the chains to raise their bottoms, because they can tell when they're working hard and they intuitively come to understand that work  any kind of work  produces speed.
In other words, kids on swings understand the conservation of energy :)
Cheers,
Mr. Entropy

Roy; .2038 meters is not twice the original height it is quadruple the original height of .051m. So: half the mass four times as high is double the original energy.

Sorry Roy,
even if they sit on a rigid bar welded to the arms it will still work, a little uncomfortable perhaps, but it will still work.
Hans von Lieven

Roy; .2038 meters is not twice the original height it is quadruple the original height of .051m. So: half the mass four times as high is double the original energy.
Ah yes, I misread your original post. But how did you use d = 1/2v?/a to calculate the finish height of .2038 instead of .102?
Hans: I guess we have to agree to disagree until I get some kid to try it!

Roy: two pendulum bobs that swing down from .051 m have a velocity of 1 m/sec. If these two bobs are on the opposites sides (and moving the same direction around the circle) of a horizontal circle they can be used as a cylinder and spheres apparatus. As the spheres swing out on the end of tethers all the motion is transferred to a small portion of the mass.
d = 1/2v?/a; if velocity is 1 m/sec rise (d) is .051 m, if velocity is 2 m/sec rise is .2038 m. (a) is acceleration 9.81 m/sec d = 1/2v?/a or Ã¢Ë†Å¡2ad = v
If by any means you can transfer the momentum of a massive object to a smaller portion of itself you have made energy. The cylinder and spheres is one means and the rolling ring is probably another.

d = 1/2v?/a; if velocity is 1 m/sec rise (d) is .051 m, if velocity is 2 m/sec rise is .2038 m. (a) is acceleration 9.81 m/sec d = 1/2v?/a or Ã¢Ë†Å¡2ad = v
If by any means you can transfer the momentum of a massive object to a smaller portion of itself you have made energy. The cylinder and spheres is one means and the rolling ring is probably another.
Yes I see your theory now. The problem is that momentum is a property of the moving mass. Whereever the mass goes, the momentum must go with it. A moving body cannot simply lose mass. The best a body can do is split up into two masses in which case each will have its own momentum based on the velocity at the time of split and the mass of each body.
Therefore the presumption that the velocity of a body will double if it loses half its mass is false.

You don't need to change mass when you want to change velocity.
It's enough to convert a part of linear momentum into angular or angular to linear .
You can convert linear motion into 2 circular with opposite directions, and so you have no change in angular momentum of the whole system, but you have change in linear momentum, because part of the energy of the moving body is converted in energy of angular motion.

Sorry Roy,
even if they sit on a rigid bar welded to the arms it will still work, a little uncomfortable perhaps, but it will still work.
Hans von Lieven
Hi Hans,
I know this is trivial but thought you may be interested. I sent two of my junior staff to the park this afternoon armed with two nice big sturdy G Clamps. Their instructions were to clamp the seats ( face to face type) to the two rigid supports so that the seats could not pivot. And then to try and get a swing going.
The result....total failure! No swing possible. I suspect not even with a good jazz player in the background if I had one.
As I said, trivial....but interesting.
ERS

@ Pmotion
Where in the sphere would you mount the fulcrum and how would you keep it swinging perpendicular to the ground for gravity to act upon?
Hans von Lieven

G'day Pmotion,
I am having difficulty picturing what you are telling me. Aassuming you mount a number of pendula inside a sphere where would they be and what would they do when at "rest".
Please draw me a diagram, I am at a loss.
Hans von Lieven

Let?s review the steps to a free energy machine.
A 1 kg pendulum bob that swings down .051 m (vertical drop) will have a velocity of 1 meter per and a momentum of 1.
Ten 1 kg pendulum bobs that swing down .051 m will have a velocity of 1 meter per and a momentum of 10.
A pendulum bob can swing in any direction; north, south, east, or west.
Ten pendulum bobs can swing down into the same circle and all could be moving in a clockwise direction in that circle. By releasing each bob at the bottom of the swing the ten bobs could be moving at one meter per second clockwise in the same circle; one at 360?, 324?, 288?, 252?, 216?, 180?, 144?, 108 ?, 72?, and 36?. The total momentum of the circle will still be ten because each bob could be released at the 360? point (in the rotation of the circle) and then the ten would be moving in a straight line. We will not release them; but we will transfer their motion to only two of them.
Using the cylinder and spheres experiment concept we will transfer all their motion to the masses at 360? and 180?. Two kilograms now have all the momentum (10). They must be moving 5 m/sec.
At 5 m/sec the two 1 kg masses will rise (d = ? v?/a) 1.27 m. 2 * 1.27 = 2.548
The system started with ten kg at .051 m. 10 * .051 = .51
2.548 /.5097 = 5 the energy of the system has increased by 500%.
It is quite possible that the circle with the bobs could have been a ring dropped as a torsion pendulum.
Or the circle could have been started as an Atwood?s machine using one kilogram accelerating 10 other kilograms. Or start by dropping one kilogram over a frictionless pulley accelerating a 10 kilogram block (ring) on a frictionless plane. These methods would also yield 500% the original energy.

PMotion: I am encouraged to see so many still trying to make free energy with the overbalanced wheel, but others are making a fundamental mistake. The mass that causes the overbalance (or imbalance) must be given the momentum of the more massive wheel and released, preferably, at the high point of the energy production. This is because overbalanced wheels make momentum not energy; you must engineer a momentum transfer that is then an energy increase. You can produce energy with an overbalanced wheel but you have to arrange a momentum transfer.
Here is what I mean. Construct a 9 kg rim mass wheel and place a 1 kg (overbalanced or imbalance) mass on the edge. Let the mass accelerate (drop vertically) .509 meters, and the wheel and overbalanced mass will have a momentum of 10 and an energy content of 5.
Drop the same 1 kg mass in freefall .509 m and you will get 3.16 units of momentum and 5 units of energy. There is a momentum increase between freefall and the overbalance wheel but not an energy increase. Formulas used 1/2mv?: mv: d = 1/2v?/a or Ã¢Ë†Å¡(2*a*d) = v : (a) in the wheel is .981 m/sec (a) in the freefall is 9.81 m/sec
If the overbalanced mass (1kg) had 10 units of momentum it would have 50 units of energy.
So if you find a way to transfer all the momentum of the overbalanced wheel to the smaller overbalance mass you would have an energy producing machine.
I have transferred the momentum using the same principle in three different arrangements; the cylinder and spheres device, a large center disk stopped by two air table pucks, and a heavy puck trailed by a lighter puck that swings out and stops the more massive leading puck.
So I am encouraging people to continue experimenting with overbalanced wheels but you must make the momentum transfer to the smaller imbalanced mass, then let the smaller mass rise (as in a pendulum) and then reconnect the mass to the larger wheel after it has risen. You will then have more energy than when you started, because the imbalance mass will be higher than when you started.

pequaide,
if i am hearing you right, then if we have an axis with two weights on tethers rotating, then all we have to do is shove the weights off of the perpendicular path that the centrifugal force wants to hold them to at the bottom of the rotaion and let the centripital force pull them back into line, for maximum gain in momentum, on the downswing.
in other words if we have two bodies of mass rotateing on an axis and we were able, at the bottom of a swing, to shove the one body aside say thirty degrees, it's actual mass would be decreased significantly, because it will be effectively alot closer to the axis, and therefore alot eaisier to lift, by the other weight that is now in it's fully stretched out maximum radius swing. therefore generating excess power.
at what rotational speed will the first weight be pulled back into the line of maximum radius for it's downward and aroundward voyage? i think i can think of a way to shove or skew the weight at the bottom position so that the effective radius will be smaller, and therefore the actual weight. i am worried about hysterisies in an actual model, but that is something for experimentation.
lol
sam
ps: if this is possible we better figure out how to stop it or, what to do with the extra power. and i mean quick!!!

A ten kilogram mass moving one meter per second in a straight line can be caught on the end of a string and it will then be moving one meter per second around the circumference of the circle. After traveling 360? around the circle it can be release, and it will again be traveling one meter per second in the same original direction. Both momentum (mv) and kinetic energy (1/2mv?) are conserved.
A ten kilogram mass moving one meter per second in a straight line can be caught on the end of a string and it will then be moving one meter per second around the circumference of the circle. The mass can then be subdivided into nine kilograms and one kilogram and all the motion can be given to the one kilogram mass. The one kilogram can be released into the same direct of travel as the original line of motion of the ten kilograms. The one kilogram is now moving in the same original direction and the nine kilograms is stopped; what is the velocity of the one kilogram.

pmotion and pequaide,
thanks for your responses. the idea i have is to take a simple two spoke wheel, around an axis, with the ability to shift the spokes latterally at the bottom of the spin and then bring the spoke back into alighnment at the top of the rotation.
lets say that we use magnets, oriented the same way as weights, on the end of our spokes. at the bottom or 6 o'clock position we place another magnet oriented to repel our weight as it swings by, therefore disorting it's rotation laterally, moving it closer to the axis, at the same time the opposing weight is starting to gain momentum from gravity therefore unbalanceing the wheel.
at what speed will the weight that has been shoved laterally be pulled back into line right as the other weight is shoved latterl,y by the magnet?
lol
sam
ps:i hope i am getting the concept across. i am still trying to get a picture of it in my simple mind.

Hello pequade and all,
may I draw your attention to the following Website :
http://www.unifiedtheory.org.uk/ (http://www.unifiedtheory.org.uk/)
Scroll down to Diagramm 13 :
INTRODUCTION TO STEEL INERTIA QUANTUM DRIVE
Hope you enjoy this.
Kator

Sam; I think you are making this to complex. A rim mass wheel can be accelerated according to Newton?s Second Law of Motion (F = ma). You could wrap a vertically mounted wheel with a string and hang a mass from the string to accelerate the wheel. Or you could mount the wheel in a horizontal plane and drape a string, with a mass on the end, over a frictionless pulley to accelerate the wheel. Both methods should produce an F = ma relationship.
The cylinder and spheres experiment has given all the motion of from up to 8 units of mass to 1 unit of mass. A large center disk is also stopped by two unwrapping pucks (on a frictionless plane).

Hello pequade and all,
may I draw your attention to the following Website :
http://www.unifiedtheory.org.uk/ (http://www.unifiedtheory.org.uk/)
Scroll down to Diagramm 13 :
INTRODUCTION TO STEEL INERTIA QUANTUM DRIVE
Hope you enjoy this.
Kator
Thanks for the link! I like the way this guy thinks. I haven't laughed so hard in quite a while, he's got a good sense of humor. ;D

PMotion: I would make these alterations to your machine. Change length and mass to the metric system, the science world is much more familiar with metric. For example; you rarely see 32 ft/sec/sec as the acceleration of gravity, it is usually 9.81 m/sec/sec.
With this change you have 4.5 kg on each end of a one meter vertically mounted bar (the bar is very light weight) with a high quality center bearing. Add an extra 4.5 kg to one of the end masses so that the overbalance 4.5 kg can rotates one half meter to the bottom. That will mean the extra mass will move from 90? to 180? (or from 3 O'Clock to 6 O'Clock). The acceleration rate should be one third that of gravitation 9.81* (4.5/13.5) = 3.27 m/sec/sec. This is from F = ma 44.145 newtons / 13.5 kilograms = 3.27 m/sec
At the end of the .5 meter drop (of the extra 4.5 kg) all the 13.5 kg will be moving 1.808 m/sec. v = Ã¢Ë†Å¡ (2* .5m * 3.27 m/sec/sec) = 1.808 m/sec
As the extra (overbalanced) mass reaches the bottom release it into a light weight, .1 m radius, horizontally mounted wheel with a high quality bearing point. Now we have 9.0 balanced kg moving 1.808 m/sec in a vertically mounted wheel and 4.5 kg moving 1.808 m/sec in a horizontally mounted wheel. Connect a string from the vertically mounted wheel to the horizontally mounted wheel so that it is winding up on the vertical wheel and unwinding from the horizontal wheel. Now release the overbalanced mass from the horizontal wheel but keep it attached to another string that has been wrapped around the horizontal wheel. While the overbalanced mass unwraps from the horizontal wheel it will absorb the momentum of the two 4.5 kg masses on the vertical wheel. This is the same phenomenon as the cylinder and spheres experiment.
Newtonian Physics predicts that the overbalanced mass will now be moving 1.808 m/sec * 3 = 5.42 m/sec (from the Law of Conservation of Momentum) and it will rise 1.5 m. d = ? v?/a. It was dropped only .5 m.
Add the extra mass back to the vertical wheel at 90? after you have transfer 2/3 of its energy to another system. You can still start over because you have three times the original energy to work with.

Hello all,
I followed this with very much interst and I like to add a bit to this discussion :
I have understood that pequaide has seperated 1/3 of momentum and changed the orientation of
the momentum axis so that both axis are perpendicuar to each other. This is a keyfeature in this system.
After this he transfers the momentom of the biggermomentumsystem to the spunoffsystem.
Regards
Kator

Hello pequade and all,
may I draw your attention to the following Website :
http://www.unifiedtheory.org.uk/ (http://www.unifiedtheory.org.uk/)
Scroll down to Diagramm 13 :
Try liking some more "think"
INTRODUCTION TO STEEL INERTIA QUANTUM DRIVE
Hope you enjoy this.
Kator
Thanks for the link! I like the way this guy thinks. I haven't laughed so hard in quite a while, he's got a good sense of humor.Ã‚Â ;D
Have some more laughs with me. Be careful not to THINK

Thanks for the link! I like the way this guy thinks. I haven't laughed so hard in quite a while, he's got a good sense of humor. ;D
[/quote]
Have some more laughs with me. Be careful not to THINK
[/quote]Hi Janus. I wasn't laughing derisively at Alan's work, which I think is very good, but rather at his social commentary. I find the dry wit of the British exceedingly amusing. ;D

Katore: The advantage of having two wheels with perpendicular axes is that while the overbalanced mass is in the vertical wheel gravitational acceleration causes a great deal of momentum, and when the mass that causes the overbalance has been transferred to the horizontal wheel the momentum of the system is constant because the overbalanced mass is no longer under gravitational acceleration.
Another advantage of the horizontal wheel is that it is smaller and close to the range of apparatuses that I have made and evaluated. The disk that I used on a frictionless plane was (if I remember correctly) 7 inches in diameter: a little smaller than the .1 m radius proposed for the horizontal wheel. The PVC pipes used to make cylinder and spheres machines were four and five inches. I would imagine a large cylinder (say 5 meters) would work but the momentum transfer would occur more slowly because the rate of rotation is slower. In the smaller cylinders the motion has to be slowed by video tape or film photographs using strobe lights in order to see the cylinder stop, because if the string is not released it will restart the cylinder?s motion.
The vertical wheel can have a mass much greater than the 3 (total mass) to 1 (overbalanced mass) proposed. I have PVC pipe models that go up to 8 to 1, and the total momentum transfer is still in the blink of an eye (well maybe two blinks). I assume there is no upper limit to the mass relationship except that the transfer of momentum to the smaller object will occur more slowly as the small mass unwraps.
I should revise the first paragraph of this entire thread by crossing out the words ?feed out?. I don?t let out the line the string simply unwraps. The string gets longer but it is simply because it is unwrapping. Some have expressed confusion about this and it was a bad choice of words, sorry. The spheres unwrap from the cylinder and absorb all of its motion.
I will try to post a few more pictures. My computer connection speed is so slow I have a hard time doing this.

Hello pequaide,
is it possible that you please post here the exact link to the madscientist(yahoogoup) ?
I think the construction of your proposed configuration is really not easy if its based on pure mechanical means . I am just trying to figure out an electronic solution for the momentumtransfer. But the main difficulty I have is with imagination of the forceconnection between the two systems. Might take some time to ponder all this.
Kator

mad_scientist@yahoogroups.com
I have a few pictures posted in files with this group but I have not posted there for quite some time.

I posted a strobe light photograph on mad_scientist. It is a cylinder and spheres device being dropped out of mechanical arms. The center bolt of the arms can be seen (at the top) as the cylinder drops.
Two spheres can be seen; one in front of the cylinder and one behind. Just to the left of lower center there is a line of black holes that are photos of a hole in the cylinder wall. This hole is where the sphere was seated before release. These holes appear to still be moving forward slightly, and it also appears that the mechanical arms throw the cylinder a little off to the left. The releasing of the spheres and the dropping of the cylinder and spheres occurs at the same time.
It can be easily seen that the cylinder is stopping. I have photographs of the hole in the cylinder moving forward and then backward and then forward again.
So the questions a scientist must ask are: where did the motion go, and what is the quantity of that motion.
The photo is fuzzy because I resized it to VGA for my slow computer to send, but portions of the 4M copy are very clear.

@ all,
as pequaides picture points out, it is not a two dimensional problem!!!! you must think at least in three dimensions if a a gravity wheel is ever to be achieved that actually works at over 100% efficiency. the main problem that i see with this idea is that everyone wants to continue to explain why this can't work as a basic wheel, in two dimensions. when it is basically pr oven by mathematics for years that it can't. however i believe that if you expand the mathematics and the experiments into at least three dimensions you will realise that it is entirely possible to have a gravity machine that is at the very least highly efficient.
Lol
SAM

Hello all,
I am refering to post #66 of pequaide.
Assume we have two masses m1 and m2 balanced on a vertical wheel. This system has a spininertia
Is = m x r exp2 and is at rest at the beginning. An additional Mass m3 is put on this wheel at lets say 2 o clockposition in an height H = 0.5 meter above bottom dead point ( bdp at 6 pm ) and let go.
Each mass m1 , m2 m3 is 1 kg.
Now potential energy of mass m3 = m x g x h is the only energy available which can be converted to kinetic energy at dbp. Ekin = 1 x 9,81 x 0.5 [kg x m/sec exp2 x m ) = 4.9 [kg m exp2 / sec exp2].
At bdp this value 4.9 true for the whole masssystem ( balance masses will be accelared ),
Now the formula Ekin = m x v exp2 / 2 = 4.9 is dissovled by V = Ekin = Squareroot of {4.9 / 3 x 1kg}
= 1.2 m/sec ( please note that all 3 masses have to enter the formula, since part of the potential energyof mass m3 is used up to accelerate the spinintertia of the balanced masses m1, m2 )
The resulting velocity of the whole system is much slower than the mass m3 alone falling free this height H = 0.5 meter at bdp ( without beeing attached to the balanced masssystem)
Momentum does not create energy. It just represent the torque = Force x r ( radius).
The assumed velocity in post #66 is calculated wrong.
Sorry, but I can not see any possibility here to have a gain in energy at the end of the process described.
Regards
Kator

Kator01 An energy of 4.9 comes from a velocity of 1.808 m/sec (1/2 * 3 *1.808 *1.808, 1/2mv?); three kilograms moving 1.808 m/sec is a momentum of 5.424 units. One kilogram moving 5.424 m/sec is also 5.424 units of momentum and the one kilogram moving 5.424 m/sec can rise 1.5 meters. You only dropped it .5 meters.

Hello pequaide,
sorry, it was very late in the mornig ( about 2 oclock I guess ). I forgot to multiply 4.9 by 2 and then devide by 3.
Now we have some hard work ahead to think it over or better do sound calculation :
System1 consiting just of one mass m3 = 1 kg dropping vertically down from height 0.5 meter :
Energy of mass m3 at bdp ( bottom dead point or 6 pm) will be :
m x g x h = 1kg x 9.81 x 0.5 m = 4.9 [kg m exp2/ sec exp2] ( 1)
therefore :
velocity at bdp = squareroot of g x 2 / m3 = of 3.13 meter/sec. (2)
Momentum or better impulse of mass m3 at v = 3.13 m/sec
impulse = f x t = mass x velocity exp2 = 1 kg x 3.13 exp2 = 9.79 [kg m sec exp2] (3)
System2 consisting of mass m3 attached to a vertical balanced masssystem m1m2 :
Energy = 4.9 ( 1) available for System2 results in
velocity = squareroot of g x 2 / m1+m2+m3 = 1.807 m/sec at bdp (4)
Momentum or better impulse of mass m1+m2+m3 at v = 1.807 m/sec :
impulse = f x t = mass x velocity exp2 = 3 kg x 1.807 exp2 = 9.79 [kg m sec exp2] ( 5 )
Not only is the energy the same but also the impuls or as you say momentum of system2 equals momentum of system1 ( 3 ) = ( 5 )
Now, I stay with my previous analysis :
If you split mass m3 of system2 at bdp to form a new system3 consiting of one vertical balanced masses m1+m1and one horizontal wheel with one mass m3 and radius 0.5 m ( both of them at v = 1.807 at the moment of seperation) there will be no way to increase the velocity of mass m3 beyond 3.13 m/sec by letting the mass m3 fly off ( for example on a flexible lever from 0.5 to 1 meter )
Indeed you would have measured a higher velocity ( 3.13 m/sec) of mass m3 spinning of system3 ( after transferring of energy of vertical system m1+m2 to m3horizontal system if you compare it with resulting system2  velocity :
3.13 : 1.8
which is correct but not true in the sense of an energygain of 174 % as this 3.13 m/sec is the same velocity mass m3 can reach alone if it drops down 0.5 meter without sharing its energy with m1+m2system.
Sorry to say, no way of gaining energy with this system
Kator

Three 1 kg masses moving 1.808 m/sec have 4.9 joules of energy. 1/2mv?
A one 1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum.
1 kg of overbalance placed on a two kilogram balanced wheel will develop a velocity of 1.808 m/sec for all three kilograms after being dropped .5 meters. The kilogram moving 5.424 m/sec will rise 1.5 m.

Greetings
You will find, when it happens, that math will not have solved it but a dupication of a natural efect in nature. Then and only then will the math be figured out, and then math will be able to improve it.
This is my prediction ;)

Hallo pequaide,
I had an error in calculation ( 2 ). I wrongly typed in the letter g instead of Epot = 4.9. But calculation was done with the value 4.9 correctly
your statement :
"A one 1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum"
is wrong because 1 kg mass falling from height 0.5 has a calculated velocity = 3.13 m/sec and
not 3 times 1.8 m/sec.
and therefore energy is :
Epot = m x g x h = 1kg x 9.81 x 0.5 m = [color=blue]4.9 [/color][kg m exp2/ sec exp2] ( 1)
and not 14,71.
I would suggest you go back and study the relevant kinematic formulas in your sciencebooks.
This is a basic condition for any further serious discussion. Otherwise we waste our time
I will stop here with any further explanations. Please come back with a clear cut explanation why you calculate velocity of mass m3 = 3 times 1.8 meter / second falling from height 0,5 meter.
Explain in detail how you want to achieve this 5.424 m/s of mass m3.
I have given the scientific formulas to do the correct calculation of velocity of a mass m3 falling down 0.5 m.
Velocity is calculated by the formula :
velocity at bdp = squareroot of Epot x 2 / m3 = of 3.13 meter/sec. (2)
= squareroot of 4.9 x 2 / 1 kg = 3.13 m/s
Regards
Kator

Hallo pequaide,
I had an error in calculation ( 2 ). I wrongly typed in the letter g instead of Epot = 4.9. But calculation was done with the value 4.9 correctly
your statement :
"A one 1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum"
is wrong because 1 kg mass falling from height 0.5 has a calculated velocity = 3.13 m/sec and
not 3 times 1.8 m/sec.
and therefore energy is :
Epot = m x g x h = 1kg x 9.81 x 0.5 m = [color=blue]4.9 [/color][kg m exp2/ sec exp2] ( 1)
and not 14,71.
I would suggest you go back and study the relevant kinematic formulas in your sciencebooks.
This is a basic condition for any further serious discussion. Otherwise we waste our time
I will stop here with any further explanations. Please come back with a clear cut explanation why you calculate velocity of mass m3 = 3 times 1.8 meter / second falling from height 0,5 meter.
Explain in detail how you want to achieve this 5.424 m/s of mass m3.
I have given the scientific formulas to do the correct calculation of velocity of a mass m3 falling down 0.5 m.
Velocity is calculated by the formula :
velocity at bdp = squareroot of Epot x 2 / m3 = of 3.13 meter/sec. (2)
= squareroot of 4.9 x 2 / 1 kg = 3.13 m/s
Regards
Kator

1 kg of overbalance placed on a two kilogram balanced wheel will develop a velocity of 1.808 m/sec for all three kilograms after only the one kilogram of imbalance has been dropped .5 meters. You have three kilograms moving 1.808 m/sec. That is 5.424 units of momentum. Thats 3 * 1.808 m/sec. This is after only one kilogram has dropped .5 meters in a three to 1 over balanced wheel .
I have a machine that transfers the motion of several kg (three kg) into one kilogram (1 kg). If the motion of three kilograms moving 1.808 m/sec (3 * 1.808 = 5.424) is transfered to one kilogram the one kilogram must be moving 5.424 m/sec if it conserves momentum. 1 * 5.424 = 5.424 m/sec.
A one 1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy. 1/2mv?
If you don't like the overbalance wheel concept, let each 1 kg mass drop on its own. Each 1 kg will need to drop .1666 meters to be moving 1.808 m/sec. That is a momentum of (3 * 1.808) 5.424. Give that momentum to 1 kg and it will rise 1.5 meters. That is three times as high as .1666 m* 3.

Hello pequaide,,
it seems you have a different understanding of the term momentum.
Please define what you understnad by the term "momentum"
According to kinematics and SIUnits :
Momentum = Impuls = force x time = mass x velocity exp2 [kg m sec exp2]
This would give all three masses at speed 1.8 momentum of 9.79 ( see formula 3) and not 5.4
m3 alone with speed 1.8 ( spinning off the system ) will have 1/3 of this = 1kg x ( 1,8 * 1.8) = 3.24
leaving the rest of the momentum = 6.55 to the balanced masse m1 and m2 at speed 1.8.
There is no more momentum available than 6.55 which you can transfer to m3 which will end up by then having the momentum 9.79
Simply repeating wrong data or calculations does not have any effect on physical facts.
Regards
Kator

Last I knew (mv) 3 kg * 1.808 m/sec = 5.425 units of momentum. It is not three times the acceleration it is three times the velocity. Or how do you come up with 9.79? Try this next example.
A 4 kg mass that is dropped from .051 m will have a velocity of 1.00 m/sec, if it is dropped as a pendulum bob it will be moving 1.00 m/sec horizontally. If it is caught into a horizontal circle it will continue this 1.00 m/sec velocity around the circumference of the circle. If you then release the mass from that circle and catch it in a pendulum it will rise (as a pendulum bob).051m.
A four kilogram mass moving 1m/sec in a circle can give all of its motion to one of its four kilograms, as demonstrated by the cylinder and spheres experiment. A good scientist should ask themselves, what must be the velocity of that one kilogram when it has all the motion? This transfer of motion will leave 3 kilograms at rest, and Newton?s Three Laws of Motion requires that the momentum remain constant. Some complain that the experiment is moving in a circle; but that is a ridiculous complaint because ballistic pendulums move in circles and they are used to establish Newtonian Physics. Others sight direction of motion as a problem, but Atwood?s machines are used to prove F=ma and nearly all the mass in motion of an Atwood is moving in opposite directions (almost half the mass is going up and a little over half is going down).
When a one kilogram mass moving 4 m/sec strikes a 3 kilogram mass at rest the new velocity for the 4 kilogram combination must be 1 m/sec to comply with Newtonian Physics.
Shouldn?t we be able to return to the point from which we came? Which was 4 kilogram moving 1.00 m/sec. When the four kilograms gives all its motion to the one kilogram and then the one kilogram is directed to slide back into the 3 kilograms at rest, shouldn?t that return us to the original motion. The only velocity that could do that is 4 m/sec for the one kilogram mass when it has all the motion.
Mathematically we could state it like this: 4 kg * 1.00 m/sec = 1 kg * 4 m/sec = 4 kg * 1.00 m/sec.
Only 4 m/sec velocity for the one kilogram will bring you back to the point of origin, which was 4 kg moving 1.00 m/sec.
Four kilograms moving 1 m/sec has 2 joules of energy. 1/2mv?
One kilogram moving 4 m/sec has 8 joules of energy. The one kilogram moving 4 m/sec can rise to .8155 meters, four times higher than the total height of the four kilograms at .051 meters. .051 m * 4 = .204 m
Note that I have not even mentioned radius, I don?t need to, velocity is independent of radius. Ballistic pendulums always conserve linear momentum no mater what the length of the pendulum. Galileo proved that velocity and radius are independent of each other.
Formulas are valueless when they require a quantity that we don?t even need to know. The angular momentum formula requires radius and we don?t even need to know radius.
Energy can be made in the laboratory for $25, which is the cost of a cylinder and spheres machine.

We have a pendulum where I work that swings for the better part of an hour. The energy transfer in the cylinder and spheres experiment occurs in less than a half second. Put these two facts together and bearing friction percentage becomes minimal. We are looking at about 300% increases in energy in a 3 or 4 second cycle.

Hallo pequaide,
I apologize. I was totally on the wrong track. I have done myself what I was suggesting to you and looked in my physicsbooks.
I had confusion with english physical terms because momentum is impulse in german language.I understand now what you found. I had another concept in the back of my mind ( Alans discovery m*v exp2) and was mixing up the formulas.
Now, the correct way to calculate the final energy of m3 ( after m1+m2 fully stopped) must me done via the momentumchange
F*t = m * Delta V ( change of velocity of mass m3 ) and not by direct energycalculation since this is what actually happens :
 momentum ( or impuls ) transfer.
This formula is the correct link of calculating the final energy of m3. The mistake lies in simply splitting the energy of masssystem m1+m2+m3 into m1+m2 and m3.
Now the momentum of m1+m2 after m3 spun off is 3.6.
If this system is stopped lets say in 1 second the m1+m2 * Delta Velocity is  3.6 adding as a
+3.6 to mass m3 resulting in v = 5.4 m/sec thus delivering energy = 14.58.
As I said earlier : I think the key lies in changing the direction of momentumvector of spinning m3 by 90 deg to m1+m2system. When this is done there cannot be transfered any information from m3system back to m1+m2system if
m3 is moving to a bigger radius. The momentumvectors of both systems are made independant.Now then you can slow down m3angularvelocity in order to transfer momentum
of m1+m2system.
The challange lies in finding a practical and simple system. I think of water running down on an peltonlike turbine where
the water leaves at 6 pm entering another horizontal turbinesystem and flowing then out radially ( both system impulswise coupled for momentumtransfer )
This is hard engineeringwork and it takes some time. Alan is a good person for enginneringideas. Write him about this.
If you have systems like a pendulum you mentioned why not show a picture including the principle here ?
Have a look to this pendulum Alan describes here : http://www.unifiedtheory.org.uk/
( I am sorry the hyperlinkbutton does not work with my onlinewatchsystem)
at diagramm 16 : A PERPETUALLY RESONANT DAMPED AND FORCED SPRING
Kator

Quote from PMotion It would need to be shown how 3 weights falling could increase the potential of an opposing mass above the potential of the falling weights. A math calculation does not show how the increase occurs.
From the math above, the 3 weights could only cause the opposing weight to rise about .5 meters.
All three one kilogram masses will eventually be raised to .5 meters, but they were only dropped .1666 meters
Velocity of an object dropped equals the square root of the product of 2 times acceleration times the distance dropped. In freefall that would be
V = sqrt of 2 * 9.81 m/sec * .1666 m = 1.808 m/sec
This (1.808 m/sec) is the velocity of each of the three one kilogram objects that was dropped (by using a pendulum) .1666 m. The total momentum of the three is (3 kg * 1.808 m/sec) = 5.425 units.
If this momentum is then transferred to one (using the cylinder and spheres device) of these three kilograms the one kilogram will have to have 5.425 m/sec velocity. This will conserve the momentum of the three. This will leave two kilograms at rest. Rest means they have no ability to rise and therefore are at zero elevation. On the other hand the one kilogram is able to rise 1.50 meters. This places the center of mass of the three objects at .50 meters. d = ? v?/a
The center of mass of the three one kilogram objects started at .1666 meters elevation and ends with the center of mass of all three at .50 m elevation. An energy increase to 300%
You could also use a 2 kg balance wheel (with the mass concentrated on the rim of the wheel) to achieve 5.425 units of momentum. Place 1 kg on the edge of the 2 kg rim. This will give an acceleration of 9.81/3 = 3.27 m/sec. The center of mass of the mounted rim will not change upon acceleration; only the elevation of the overbalanced 1 kg will change.
After dropping the overbalanced 1 kg mass .5 m the three kilograms will have 1.808 m/sec velocity each. So an elevation of .5 m for all three can be achieved by dropping only one .5 m.
You will have to keep in mind whether you are starting by dropping three kg .1666 m in a pendulum or by dropping one kilogram in a 3 kg (total mass) overbalanced wheel.

Hello PMotion,
yes, I understand the problem you describe. The main problems are techical ones. Once the single mass ( in our discussion m3 ) leaves the balanced system at 6 pm at v = 1.808 m/sec no momentum can be transfered by the balancedweight momentum because all masses have the same speed.
But before I get deeper in technical discussions I would like to find in this discussion a simple way to proof the principle. I am still sceptic about the formulas although the conservation of momentum is clear.Some of the formulastuff is directing our attention to wrong interconnections But, as you described it : How do we transfer momentum back to m3 ?
One idea came up but I am not sure it works : lets assume m3momentum ( including m1m2 momentum ) will hit a spring at 6 pm in full elastic collision, compress it and then stick to it by some sort of mechanism ( detached from m1,m2 ) while the spring is locked in compressed state by another mechanism. We then have all the time to set up a testrig where we can release the spring and measure m3 climbing the rig or directly measuring the velocity of m3 and have not to deal with more kinematic problems
The question is : does this setup deliver the answer to the momentumtransfer ? Or do we need free moving masses to do so ?
@pequaide : what you you think ?
Kator

A spring is unlike gravity. Gravity has a uniform quantity of force along the entire freefall (and rise) distance; this gives you uniform acceleration (and deceleration).
In a spring: if you double the distance of compression you also double the force applied to the object performing the compression.
If you then allow the spring to unload on a less massive object, the acceleration would be very large at first. This would move the lighter object out of the compression range in a very short period of time.
Since momentum is a function of time; F = ma, a = Ã¢Ë†â€ v/Ã¢Ë†â€ t, the quantity of momentum used to load the spring would be greater than the quantity of momentum delivered when the spring unloads.
It would be the opposite of the overbalanced wheel, because the overbalanced wheel takes longer to load than to reload (the time it takes to cast the overbalance mass back up in reverse freefall).
I think the key is not the force itself but the time over which the force acts.
In a 10kg (total mass with 1 kg overbalance and 9 kg balanced) wheel with 1kg of overbalance it takes 1.43 seconds for the one kilogram to drop one meter. In reverse freefall the 1 kg can rise 1 meter (back to the place from which it was dropped) in only .45 seconds.
This 1.4278/.4515 difference in time also gives you a proportional momentum difference of 14.007/4.429. And the energy increase is the square of this difference. 14.007/4.429 = 3.16; 3.16? = 10.00
Here is how it works: Transfer the 14.007 units of momentum that is developed by the overbalanced wheel back to the one kilogram of overbalance and it will rise 10.0 meters. And it had only been dropped one meter. Formula used d = ? v?/a; v = Ã¢Ë†Å¡ (2*a*d); d = ? at? or t = Ã¢Ë†Å¡ (2 * d / a) (these three are the same formula of course because t = v/a), F = ma was used to obtain acceleration, mv, and 1/2mv? or 1/2mt?a? because v? = a?t?.
I think the vertical and horizontal wheels would be a good choice of designs, I think this was post #77 this site (masses changed).
In the example you have 1 kg on each end (2 kg) of a one meter vertically mounted bar (the bar is very light weight, this could also be a wheel with evenly distributed mass) with a high quality center bearing. Add an extra 1 kg to one of the end masses so that the overbalance 1 kg can rotates one half meter to the bottom. That will mean the extra mass will move from 90? to 180? (or from 3 o?clock to 6 o?clock). The acceleration rate should be one third that of gravitation 9.81* (1/3) = 3.27 m/sec/sec. This is from F = ma, a = F/m 9.81 newtons / 3 kg = 3.27 m/sec/sec
At the end of the .5 meter drop (of the extra 1 kg) all 3 kg will be moving 1.808 m/sec. v = Ã¢Ë†Å¡ (2* .5m * 3.27 m/sec/sec) = 1.808 m/sec
As the extra (overbalanced) mass reaches the bottom release it into a light weight, .1 m radius, horizontally mounted wheel with a high quality bearing point. Now we have 2 kg moving 1.808 m/sec in a vertically mounted balanced wheel and 1 kg moving 1.808 m/sec in a horizontally mounted wheel. Connect a string from the vertically mounted wheel to the horizontally mounted wheel so that the string is winding up on the vertical wheel and unwinding from the horizontal wheel. Now release the overbalanced mass from the horizontal wheel but keep it attached to another string that has been wrapped around the horizontal wheel (the cylinder and spheres experiment). While the overbalanced mass unwraps from the horizontal wheel it will absorb the momentum of the two 1 kg masses on the vertical wheel. This is the same phenomenon as the cylinder and spheres experiment.
Newtonian Physics predicts that it will now be moving 1.808 * 3 = 5.42 m/sec and it will rise 1.5 m. d = ? v?/a. It was dropped only .5 m.
Add the extra mass back to the top at 90? after you have transfer 2/3 of it energy to another system. You can still start over because you have three times the original energy to work with.

Here is my definition of an overbalanced wheel.
Take a bicycle and turn it upside down, onto its seat and handle bars. The front tire of the bike is now a vertically mounted balanced wheel. Duct tape the padlock to the tire and now you have an overbalanced wheel.
If the vertically mounted balanced wheel has a mass of 9 kg and the out of balance or overbalanced mass is one kilogram then you have a 10 kilogram overbalanced wheel with an overbalance of one kilogram. The radius of rotation of all ten kilograms would be (roughly) the same. This would give you (F = ma) 9.81 newtons of force applied to 10 kilograms for a .981 m/sec/sec acceleration. There is no need to consider lever arm length because they are all the same.
Thank you for allowing me to clarify this.

Hello PMotion,
searching for a technical solution I found the following website  unfortunately in german language.
http://doerler.gmxhome.de/
I find it helpful to visualize necessary kinematic basics in order to locate the technical problem with this approach.
Please go on the left naviarea to Impulserhaltung ( conservation of momentum ) and then to
Newtons's Wiege ( sesaw ). Here you can simulate momentumexchange with 1,.. to 4 balls. Just change the number in the small window.
Insert number 3 and watch what happens. In this setup it is not possible to stopp all 3 masses and accelerate the left single mass to the same momentum of the tree masses swinging in from the right side.. So this is the problem to solve.
But this setup solves another problem you brought to our attention in one of your posts :
If the one weight goes around in a circle and then returns to the pendulum, it will be on the side opposite of the other 2 weights.
How would the pendulum be set up to catch it ? And how would it know when to release it ?
Choose number 1(ball) this will give you an idea how you can transfer the momentum of mass m3 leaving the wheel at 6 pm at velocity 1.8 m/s. By this you can put the horizontal wheel to a given distance away from the 6 pm  position.
The second topic is momentumchange
Left naviarea go to : Kraftstoss and then to Schau's dir an
Here you see the simulation of a car decellerated with different forces.
1) big force  very short time ( to get the car to stop )
to
3) very small force  big time ( to get the car to stop )
I hope this helps to understand some of the basics here.
Please understand I still have not found a technical solution.
Regards
Kator

Pictures: note the slit and the extention of the string. The spheres have most or all the motion, they are much lighter than the cylinder.

@ all,
i think i know for sure what is cleared up. the fact that pequaide has the ability to post a video of his experiment!
lol
sam

PMotion Please be more specific as to what you think doesn?t work; I thought everything works just fine.
Both the last two pictures are of the same cylinder and spheres devices in freefall. It was spun by hand and released when the spheres were seated in the holes. The spheres unwind and swing out on the end of the strings, when they swing out the cylinder stops after about a quarter turn. The string is seen on both sides of the slit showing that the cylinder is not moving and that the spheres are. Video tapes can break it down into 1/30 of a second, the cylinder stops.
The mass of this particular cylinder is about 630g; the spheres have a mass of 67g each. 630+134 = 764, 764/134 = 5.7. This would be like 5.7 kg being stopped by transferring the motion to 1 kg.

PMotion: You asked if it is overunity.
Answer; You start with 1 kg at .5 m elevation and end with 1 kg at 1.5 m elevation. I assume that this is one definition of overunity.
PMotion: What I wonder about is the ability of the spheres to transfer force to the cylinder or the other sphere.
Answer: The strings have tension in them, which means force is in the string, Newton?s Third Law states that the force in the string will be equal in both directions, and the Second Law states that the momentum change will be equal on both ends of the string.
PMotion: It seems the set up transfers force/potential from the cylinder to the spheres.
Answer: The setup transfers momentum from the cylinder to the spheres.
PMotion: For the spheres to be able to effect each other, they would have to be on an arm attached to the cylinder.
Answer: Are we concerned if the spheres affect each other. The spheres remove momentum from the cylinder, I am unconcerned if they affect each other.
PMotion: And the cylinder would have to be designed to allow for spin/velocity to be increased by gravity.
Answer: The vertical overbalance wheel is ?designed to allow for spin/velocity to be increased by gravity?. The cylinder and spheres portion is mounted horizontally and is unaffected by gravity.
PMotion: In other words, it would have to allow for linear momentum to be converted to angular momentum.
Answer: Motion changes from linear to circular easily; often; and is 100% efficient in both directions.
PMotion: But with the 2 spheres being attached by a string, the spin necessary for the cylinder to create enough inertia would be high.
Answer: Inertia occurs in the lowest degrees of slowness. When released the spheres attempt to maintain a straight line motion, whether they are moving slow or fast. The cylinder pulls the spheres from this straight line motion and the force in the string begins to increase, at any velocity.
PMotion: Since any overunity design is dependent upon 9.8m/s/s, and with a changing of the mechanics of a device, its' spin will be limited.
Answer: The momentum transfer of the cylinder and spheres device is not limited by any spin rate that I know of. The effect is the same; slow or fast. The overbalanced wheel portion is of course limited by the acceleration caused by gravity.

Hello all,
I wonder if anybody has looked to my post #110, page 8, because I try to find analogies of impulstransfer either in nature or in manmade system we are not aware of.
PMotion, I also like this thread because it is a challenge to figure out a way to transfer impuls in a way nature does not show us  or should I say scientific or technical education we went through has been conducted in a tunnelview manner.
I looked at the pendulums ( animation I was posting in my reply #110, and to the formulas again and again. I asked myself why the heck is the conservationprocess of momentumtransfer from two 1kg masses hitting a single 1kg  mass happening only in a manner focussed on the targetmass and not
on the velocity ? More clearly expressed : Why does the process not solely increase the velocity of the one kgmass and stop the 2 kg masses in order to fullfill the equilibrium in this equation p = 2 kg * V1 = 1kg * 2*v1 ?
The answer is of course : intertia of masses prevail. The inertia of a 1kg mass cannot stop a 2 kgmass.
I have not yet figure out how to technically do this besides what pequaide has presented here. I only have grasped this one idea of spininertia because here spininertia of a 2 kg mass on a wheel can be made equal in value by placing the reactionpartnermass of 1kg on a wheel with double the radius of the two kgmasswheel. Now both systems having the same spinintertia might provide a chance that velocity of target mass is increased in a way the 2 kg mass is stopped.
Maybe that you all here have already mentally worked this trough. I myself concentrate on different way to do this momentumtransfer.
I hope I have been able to bring across my point here. As an enginneer in applies carconstruction I have learned tunnelview realities and I always was struggeling to overcome this.
Another idea came to my mind : a flowing watermass beeing stopped suddenly by a valve  a technique which is state of the art since almost 200 years  by name of
hydraulic ram pump
principle : Waterhammershockwave
have a look here
http://virtual.clemson.edu/groups/irrig/Equip/ram.htm (http://virtual.clemson.edu/groups/irrig/Equip/ram.htm)
Can this technique be changed so it fits in in this momentumtransferproblem ?
The hydraulicramppumpprincipe increases pressure in a tank.Gas is compressible, water is almost incomressible and thus suitable for the elasticalbouncecase.
This pump does not transfer the momentum of a very long slow moving watercolumn beeing stopped> to a small watermass at high velocity.
What do you think ?
Best wishes for year 2008
Kator

Hello PMotion,
I have a few questions. I simply do miss some steps in your description.
 two weights : weigth at left is in fiixed position ? I assume both masses are of the same weight.
 why does the center of the rotating arm has two positions ? Is the center ( axis ) not fixed ?
Does it wobble ?
I simply do not understand the whole process because I do not understand the basic setup especially this :
In the schematic, when the arm has rotated to the desired angle, if the over balanced weight changes its' course, it can return to a balanced position.
 ... What is the desired angle and at which step in the process ?
 ... Changes its course at what position or angle ?
 ... It can return to a balanced position at what angle ?
Can you please give me a more specific description if you have the time ?
Kator

Hi PMotion,
more explanations > new questions. Please bear with me.
I have to apologize but my english misses the meaning of the term ramp. I like to be more precise : what does this ramp look like ?
Is this a device fixed against ground which catches the overbalanced mass pushing it back upwards in the guiding socket ?
I understood the process but not the technical detail of this ramp
Very intersting idea
Kator

PMotion: You must transfer the momentum of the entire wheel to a subset of itself, and then release it. Since your ramps and lever arms don?t do this; I predict that your wheel will not work.
A very light one meter bar with a quality bearing point at .5 meters has six kilograms on one end and five kilograms on the other end. The six kilogram side (the six kilogram lever arm of .5 m length) will rotate to the bottom with an acceleration in accordance with F = ma. If all eleven kilograms remains attached to the bar the six kilogram side will rise to the same level from which it was dropped. That leaves you with no change in energy.
If the momentum of the system (11 kg at .9444 m/sec velocity) is transferred to the one extra kilogram, when it is at the bottom of a .5 meter drop, it will have 10.38 units of momentum and 10.38 m/sec velocity. When released this velocity will allow it to rise to 5.499 meters and it was only dropped .5 meters.
Please don?t ignore and avoid this fundamental difference between the laws of levers and The Law of Conservation of Momentum.
What the lengthening of the lever arm gains in leverage it loses in distance. The side that has the longer lever arm but equal mass will indeed rotate to the bottom, but when it gets there you are stuck with an attempt to rotate it back to the top. If you leave it in one piece and don't consolidate the momentum, it will cost you the same going up as that which you gained going down.

Hello PMotion,
did you calculate in a simulation the kinetic energy of the overbalancemass at the point where it is beginning to move up ( to be pushed up ) by the ramp and compare this with m x g x h ( 3 cm ) ?
As far as I understand this process cannot be calculated with the impulsformulas because the time for the upwardmovement of the overbalanced mass must be measured first. As your mass is in a continous movement all along one cycle, I do not see a impulstransfer so that the wheelmass is fully stopped. According to the formulas that exactly is necessary in order to get excessenergy. In your desing finally the mass consumes m x g x h ( 3 cm ) in a slowmotionfashion compared to full impulstransfer. If there is Ekin >> Epot you will win.
I concentrate on getting rid of the standardtunnelview for finding unconventional ( new ) physical/technical means to achieve what the formula promises.Some iddeas have come to my mind which I will discuss here soon.
I wish you the best and hope you will proove me wrong.
@peqaide : what is the mechanism you use to keep the spheres in the openings while accellerating the spin of the system and then release it ? As I understand it you have to first reach a certain speed so a certain amount of centrifugal force can develop which then can be released.
What setup did you use doing this with gliding masses on a almost frictionles spinning CDDisk ?
Kator

Kator01 question: What is the mechanism you use to keep the spheres in the openings while accellerating the spin of the system and then release it ? As I understand it you have to first reach a certain speed so a certain amount of centrifugal force can develop which then can be released.
What setup did you use doing this with gliding masses on a almost frictionless spinning CDDisk ?
Pequaide answer: In the mechanically dropped and released models the spheres are held up against the cylinder with arms made of plywood (the arms were painted black for photographic reasons). The seats in the plywood are padded with hard rubber. Strong springs are triggered that jerk the arms away from the spheres. This arrangement releases both the spheres and the cylinder at the same time. The cylinder is held square with a piece of plywood mounted on the spinning rod that holds the arms and springs.
The rod spins at 3.25 rotations per second. The diameter used for the 5 in. O.D. PVC pipe was 4.81 in. (distance to the center of mass of the cylinder). The diameter used for the spheres was 5.00 in. (their center of mass seats on the outside diameter of the cylinder. This gives you an initial velocity of 1.297 m/sec for the spheres and 1.247 m/sec for the cylinder.
In the most recently posted pictures the spheres are held in place with the side of the thumb and the side of the second finger. In the hand held models the RPS is unknown; the mass of the inserted pipe is altered to get the cylinder to stop just as the strings on the spheres are entering the slit in the cylinder. These, as all models, are video taped and clearly show the cylinder (or center disk in the air table model) stopping.
In the air table models the pucks were held up against the large center disk with a string that is pinched up against a raise wall in the center of the large disk. The raised wall is used to accelerate the large disk, while the fingers (or spring loaded padded wooden dowels) are pinching the strings that hold the pucks in position at 180?. In this way the pucks and disk are released at the same time. The strings are then dragged by the pucks across the air table, causing a small amount of friction. The strings that allow the pucks to swing out are of course above the table.
The RPS of these models is also unknown, if held by hand. Some calculations were made using the mechanically released models (the spring loaded padded wooden dowels). My measurements and calculations showed energy increases in all models, as the spheres reached maximum velocity. I can post a few pictures of the arms and center disk, but I will have to do it tomarrow.

Kator01 one 1M picture

one more 1M picture with cylinder and spheres loaded

Hello PMotion, Hello pequaide,
I have to say beforehand that in case you do not get immediate response by me the reason is that this forum has becom so slow in reaction especially during the last few days that some posts I did in another thread simply got lost.
@PMotion : I fully agree with what you say concerning the practical experience and your statement "hands on work". I grew up as a son of a gold and silversmith after Wordwar II and since I was 5 years old I would stand on the tip of my toes to get my nose above the level of my father`s workingplace to watch him working. I was educated naturally with all this practical workmanshiptechniques and still today I do this kind of work in order to relax after my consultantjob. So you are talking to an engineer for applied carconstruction who now gets interested to dive again into an new the field of kinematics.
Anyway, at the present I do not have the room and equipment to do this and I am patient till you will have the time to build this, it is your idea and there is no one who can better do this design.
What pequaide has brought up here fascinates me because I have a special way of thinking or should I better express this a "imagine something an whatch it run" like Tesla did it. I have also have learned to have a close look at formulas when there is a possibility to estimate results. but I do not mainly rely on this.
@pequaide : this last picture really shows a big device or am I wrong here. At the present I work out a technical solution in my imagination ( the way I explained it to Pmotion before) for another testrigdesign of the balland cylindersystem. It will take some time and will post a sketch here for you to comment on.
Is gravitation around the sun the carrier to transfer momentum of the sun to the planets to keep them circling around ? Just a thought.
Kator

Kator01 The cylinder is a 5 inch O.D. PVC pipe coupler, it is a coupler for a four inch I.D. PVC pipe (both coupler and pipe have ? inch side walls). The mouse trap gives you a good idea of the scale. The hand held models take up very little room, and also show the cylinder stopping.
I could not send two 1M pictures at the same time, so I posted twice. Yes the site does seem to be slow this weekend.
Here is how the arms work. The center rod of the arms is in a drill press chuck that has a belt and pulley reduction to a second drill press motor. It rotates at 3.25 RPS. The arms are held closed with a thin brass rod. The rod latch is opened with the impact from the spring of a mouse trap; a pad trips the trap at a certain portion of the rotation. Upon release the arms are pulled away from the spheres with the top mounted springs. The spheres were holding the cylinder in position so both the cylinder and spheres are released at the same time.
Within about a quarter rotation the cylinder is stopped; according to Newton?s Three Laws of Motion the spheres must have all the momentum. The energy increases I have measured range around 300%, depending on the model.

Hello pequaide,
thank you for the explanation. I think you have made is clear. What puzzles me most it the fact that the steelrope under tension by the centrifugal force of the spheres can take all the impact from the cylindermomentum. Is this steelrope made from pianostrings ? Can you please give some numbers on the cylinder/ and sphereweights and the distance the sphere can move away from the centeraxis, so I can do a calculation on this ?
Thank you. Very good work.
As I said I am working on another setupdesign which involves a flexible lever with the spheres attached to it instead of steelstrings and a second one using a moving watercolumn beeing partly stopped. This last design would involve losses, but these losses are tolerable since it is very simple to build.
Regards
Kator

By request from P Motion.
having glanced (note glanced) at this thread and the apparent 2 designs in question, I noted the last answer at post 117 was incorrect, in fact the reverse is true, spinning momentum of any horizontal motion is more limited than vertical motion, as it must contend with the loss at the outer extremities of the wheel/cylinder that is the outer weight of the centrifugal force, then must contend with its own friction on the center bearing, and on that note can we please have no more posts containing the words "frictionless gears or bearings" We are after all creating perpetual motion for the sake of energy in machinery that will need to be quite large and have gravitation weight of friction on all parts.
Then the horizontal mech must contend with the forces of gravity pulling downward creating a greater friction on the base of the vertical bearings or bearings placed beneath in support.
Vertical wheels have all of those except the horizontal problem. This is not to say a horizontal centrifuge design will not work, merely that the last answer in post 117 saying it did not have limitations of speed such as the gravitational speed, it has many more limitations.
On both projects and for all projects for that matter, most posts here miss some very basic physics aspects to their designs and most certainly in their criticism of others when citing gravitational speed velocity.
1) gravitational velocity (whilst i disagree with Newton on this point also) I will use your hero's base to illustrate a primary conversation flaw on this site.
an SLR 7.62 rifle fires a bullet at 2700 feet per second, now point the rifle at the ground from the top 100 foot of a building, it is still pretty close to 2700 feet per second not the very slow gravitational velocity speed. Many of the devices here use momentum or projectile firing mechanisms to move objects to fall points within the wheel or armature, this instantly removes the use of gravitational fall as a measurement in every science lab in the world, with only one exception, and that is during the velocity calculation the questions is often asked at what speed does friction (usually from air pressure as with a bullet) slow the speed to max velocity (Newton?s max equation for gravitational fall)
and is the applied pressure a secondary variant in reaching this slower speed (the kite effect same weight as the small apple but fall slow due to aerodynamics).
So many of the calculations on this board keep quoting gravitational fall which is incorrect physics the instant momentum (a simple baseball in a stocking can beat gravitational fall math for weight by momentum) is applied or a projectile system (even a spring) is used in the design.
As for the secondary device this comes under number two.
2)the pendulum in conjunction with the wheel was always a difficult task, 2 sets of friction for a start, primarily the weight of gravity at the top of the bearing that holds the pendulum, but secondary and most importantly , and have not seen it mentioned on any thread, is "racking" one of the greatest shock absorbers of energy. Pendulums create racking force, in fact the energy ratio can be as much as 75percent to racking, sad but true, take a piece of A4 paper hold one hand on each side and rotate you hands in opposite directions, this twist is racking, for the everyday version of a racking pendulum, that is where a set of children?s swings is not fixed to the ground and overbalances when someone is pushed too hard, now image the pressure required to pin downs those legs at that racking point, that?s a lot of energy, now imagine you are not building your machine buy digging holes for wooden posts to go into the ground (i hope your not) and that the 4 legs are joined and a rectangular frame at the base, this energy now travels right through the frame to disperse, this is racking and where most energy from pendulums goes.
But I'm am open minded enough to assume your individual genius has already considered this and your pendulum create huge energy and you only need a small portion of it to help in over balancing your wheel
No pendulum keeps going so it can not help indefinitely, secondly any pendulum that does run for long periods will be slow by the transfer of energy to help the wheel, the wheel will suffer some effect of the racking through stress and vibration. regardless of wheel design. pendulums will only have a negative effect on the wheel, and are probably the poorest of energy transfer devices for an impact design, although may provide some benefit as a wind up and release device that releases a weight to assist in an overbalance application, but most preferable would like be the reverse where a falling weight swings the pendulum. Unfortunately the only design that worked absolutely that i ever came up with was to combine the silly ocean wave energy system with this practice (bloody surprised no one else ever though of it or did it) simply use the ocean for a float that rises and falls with the machine on land, that?s my freebie for those who want free energy fast if you live on a cliff near the ocean, or have a non stupid local government who would build it for you.
But for what it's worth gravity is certainly key as it is the only free "constant" source of energy and my own device was design to manipulate it.
The perfect example of continuous work is simple, two ring magnets north facing north on a stick, the energy from the two magnets constantly repelling whilst the gravity constantly fights to bring the top ring down, in both cases the expenditure of energy is perpetual until the magnet or planet dies. So this is where i first saw perpetual motion and realized it only hade to be manipulated.
There are many ways to get free energy that are not used, not perpetual motion, but perpetual in nature if manipulated correctly. The problem is not discovery, it has usually been cost.
If you want my first free energy device it was simple, the greatest power on earth is not gravity but heat, solar radiation lifts billions of tons of water against gravity every second of the day and at night, my design was simple and works well, based on 8 liters per hour during the day and 3 liters per hour at night average per square meter, this is free lift of tons of weight, simply use 3 square miles of plastic covered ground (a decent commercial nursery size at the bas of a mesa ( a protruding plateau in the desert) and simply have the base a foot deep and cheap roofing and supports like a nursery the slowly inclines up the base of the mesa to a domed condenser probably about half a mile of the mesa top surface area, (a large tent basically with a white roof and giant heat fins aluminum sunk into the ground) simply run back over a single waterfall to the pond below.
Don?t laugh that is 4 liters per day per square meter solar pond basic evaporation, that equates to ?1.3 million liters per hour?, now get that again per bloody ?hour?, that?s one hell of a self cycling hydroelectric plant. Free energy was never tough to discover or design, I just wanted one everyone could build and afford, so the poor could say up yours to the oil gas and coal companies and government taxes on energy.
Whilst having already built the sword and knowing I am the first to genuinely succeed, I do not believe I will be the only one to have a working design in the future, so I guess the best to say for both of these devices, is stay with what you Know Works !!!! Not what science or naysayers tell you won?t work (and that does include myself, always take on board what I have said, but do not set it in stone. You may know or find something that contradicts my opinion, just don?t quote me Newton?s laws that I have already beaten)

The tension in the string builds slowly; so they don?t need to be as strong as you may think.
I use 20 lb and 30 lb test Berkley Fireline. The strings are eventually worn by roughness in the slit; they need to be replaced after about 100 tests. The wire loops that are inserted in the drilled spheres are made of braided fishing wire in about the same 20 lb test range. They break also at about 200 tests.
The open diameter of the spheres (distance between the spheres center of mass) of the black cylinder seated in the arms is 10 inches and the closed diameter is 5 inches. This double radius was done by design, because the spheres stop the cylinder with or without a slit. The slit is only there to allow time for the video tape to catch the stopped motion of the cylinder.
If the open diameter of the spheres is double the closed diameter the initial and final radius of rotation of the sphere remains the same, because when the cylinder is stopped the radius of rotation is not the original center of the cylinder, the new radius of rotation when the cylinder is stopped is the side of the cylinder were the string comes through a hole. In this way the initial radius of rotation of the spheres is 2.5 inches, and the final radius of rotation of the sphere is 2.5 inches. I did this for those that were using angular momentum formulas; this made the application of angular momentum useless because radius of rotation did not change. It would be perfectly legitimate to secure the cylinder in place when it is stopped and force the radius of rotation of the sphere to remain 2.5 inches, locking the cylinder when it is stopped removes no momentum from the system. And this jamming of the stopped cylinder would have no affect upon the momentum of the spheres either. I have not locked the stopped cylinder and the spheres will start it back up again.
I myself, like Galileo, know that the radius of rotation has no affect upon momentum. In this particular example the momentum of the sphere does not change when the string enters the slit in the cylinder, but the radius changed from 2.5 in. to 5 inches.
When the string that connects the spheres is in the slit you must use the 5 inch radius for calculations. For example: when the string moves 2.54 cm in the slit the sphere itself has moved 5.08 cm.
The sphere in this model (G in the log book) has a mass of 66.2 g. Two spheres will be 132.4 g.
The 5 inch coupler has a mass of 367.2 g at a diameter of 4.81 (center of mass/momentum) inches.
Inside the coupler is a 4 inch (inside diameter) pipe that has a mass of 52.4 g at a diameter of 4.32 inches. This added mass is how you get the cylinder to stop (with a fixed radius of 2.5 inches) just as the string enters the slit. You can adjust the radius or adjust the mass.
In the hand held models previously pictured the closed radius is 4 inches and the open radius is 9.5 inches. This is a proportionally larger open radius, but it also stops a larger mass with roughly the same mass spheres. 395 +208 +20 about 623 grams instead of about 367.2 + 52.4 g about 419.6 g.
An engineering friend from GM placed friction at 3 % per gear, with a 300% increase in one fourth turn I think we can handle it.

the design is clever, of that there is no question, and i applaud your design as exception drive length from an initial motion, however there are 2 problems which none can deny, like yourself i built many a device that could not be used in a commercial sense, as this to could never be built on a large scale to produce quantifiable energy as you rely on specific materials that change in abilities the the size increases, you have built a fly line for trout, you can not catch a larger trout with a larger line as it will as you increase, become weighty and sink, and to the point where even casting becomes impossible, there are many anomalies in physics overlooked by the very best, material choice is a common error, and equipment will often not function when it has to be changed.
Many a thread here is testament to that very thing in conductors, they all carry current, but they all have limits and purpose.
Your greatest error was Galileo, He understood that momentum was not affected by radius, unlike yourself, he understood it and taught it in respect to planetary movement. He was of course refferring to planets drifting out into greater orbits, and to that end was the main derivative for Newton to calculate (steal someone else's discovery) that an object in motion will continue forever if uninterrupted by another force or action, you fail to consider you rotating sphere is connected to your machine and is affected by other forces, both gravitational weight, friction from bearing/bush as not applied to Galileo's thoughts on planetary movement, it was always noted that this was not applicable on earth without consideration of these factors.
Lastly i seam to be missing the generator?? the part where this generates power? the primary friction component, or is this an experiment in self perpetuity and not overunity or perpetual motion, go back to the ring magnets on the pole, two ring magnets north facing north on a post, the gravity endlessly trying to pull the upper magnet down, whilst the lower constantly fights to hold it up, until the planet or magnet dies, self perpetuity exists right there.
But I'm sure you have it in hand

OK guys...how viable is this:
http://youtube.com/watch?v=y9wktSQdyaE

for your device velocity can only be measured by percentage against each other.
EG when you finish building the device (I am granting you that it will work)
It does not just start turning correct?! because it is not a version of an unbalanced wheel that cannot stay balanced (although no design i ever saw claimed that, they had to be held heavy end up first and released) therefore it must have a thrust start point to get it going, all engines must be started even an orbiting planet has to get its first push.
This is the difficulty, for instance my device goes at what ever speed i propell it to up to any intrusive vibration point, it does not get any faster, and has a minimal speed requirement to set off replication, as does yours (in other words a gentle breath of air or push with you finger would not work. so velocity cannot be measure from an operational point without knowing speed of wheel friction etc,
However the velocity of travel in your design is fairly basic except the creation of correctional inertia which is a high drag problem for the machine.
For the others reading correctional inertia is where an object traveling in a perfect circle has equal inertia applied at all points equally during travel, image a car at the circular velodrome style tracks like your old circle electric car track, now whilst the car at set speed is travelling around your 100 foot circle (i made it that big to actually change the track during a circuit) the car goes past, you unclip the track and make the curve tighter going inside the circle, as the car comes round again it will likely roll on the corner or massive extra g forces will apply, huge friction!
The design of this device has a similar structure using an inward spiraling track design for want of better terminology, thus creating a shorter route for some componets to shift counter weights than others, Exceptionally clever design as i said earlier, but this addition friction is equal to the cheat short cut you gain by cutting the corner, the velocity of each of the 3 items is equally lower as the distance is shortened due to the increase in friction the tighter the turn, the old porche whips faster around the sharp corner due to shorter distance and that is how you arrived at your figures, weight over time and distance, but you left out the massive increase in g force at the corner reduction, and the massive torque against the tyres of the porche.
Sorry, love it though, very inventive and clever, it reminds me of an old design i played with, you may wish to think about this (i can't build your invention for you but i can give you a few pointers)
take a bullet, place it in a wide arc circular track and then another and another, the arc being wide enough so as not to affect the flight of the round from the bullets placed close to each other.
detonate the first, each bullet striking the pin in front of the other until all have fired, now use the same concept with a fixed in place recharging shell and spring return for the bullet/pin. use the momentum of the bullet round as your power source for electricity, now you only need to work out what is the power that forces out the pin to strike and release this pin/bullet in front, compressed air from the generator?? who knows it may already be part of your design, just remember, when you fire a semi automatic it reloads itself out of a minimal byproduct of the main carriage.
It's not a pendulum if you pull it back and push it forward, no momentum required.

Hello,
due to a trojanvirus entering my system while I was trying to upload 2 pictures I will not post here as long as the moderator gets active to eliminate this shit.
@pequaide : I have a simple modification of your cylinder & spheres setup which will show the energygain ( if there is any ) in a simply, certain and direct way without complicated videotechnique, but I will wait til this forum is clean of this trojanshit and the responsible advertisingwebsite embedded here is disclosed.
Regards
Kator

For the last post, i think do to the number of threads you should have probably used a general announcement so everyone could see, I have had them sunce being on this site also, however, conspiracy theories nuts aside, if the government was ever going to activly target a website, which websites do you think that would be??? I doubt the webmaster can do too much or is responsible

Hello Eskimo Quinn,
problem is already known to stefan. But it seems to consist, as another member reported trojanattack as he was just opening a german thread.
We may have a pattern. Maybe some threads are selectively choosen to get attacked True conspiracy  all nuts
Since some of these viruses are based on JavaSkript simply turn off Java and JavaSript if you read and post here
Regards
Kator

The (Kogel, Leipzig) pendulum at work is about 8.5 inches (21.6 cm) in length. It has a period of a little less than a second (55 cycles per minute). It is on a bearing and the bob is square (not aerodynamic). The bob has a mass of about .8 kilograms. We don?t even keep a dust cover on the open bearing. It will swing back and forth for at least 31 minutes. That is 55 * 4 * 31 = 6840 quarter cycles. So in the first down swing the pendulum bob will have approximately 99.985 % (1/6840) of the total energy that would be available to it.
Quinn: You quote huge energy losses for pendulums. I have no idea what you are talking about. I think your racking (post145) is an excuse.
Let?s assume that the (Kogel, Leipzig) pendulum drops .216 meters; that would give the bob a maximum low point velocity of 2.058 m/sec, but only if it was in a vacuum and if the bearing point was friction free. Air resistant is probably greater than bearing friction, but let?s give half the resistance to each. So in a vacuum the pendulum bob would be moving 2.058 m/sec * 99.9925% = 2.0578 m/sec. This means that the resistance in the first down swing of the pendulum bob is not measurable with the equipment we are using. And the resistance is insignificant. So I think this alleged 75% loss to racking (Quinn) is an excuse some are using to pretend that energy can?t be made from gravity. 00.0075% doesn?t seem like an impediment to me.
Kator01 All that is needed for a cylinder and spheres device is a horizontally mounted wheel with a center bearing (a pulley), and low friction carts. The pulley has to be wrapped with a string that then has a mass (cart) attached to the end. When the mass on the moving pulley is released the mass will unwrap the string and apply a force in the opposite direction of the motion of the wheel. If the pulley has a mass or is being pulled by a second mass, the mass of the wheel (pulley) or the second mass will stops (unless it is greatly larger than the unwrapping mass).
Let me give you this scenario. A nine kilogram block is moving in a straight line; it is followed by a tenth kilogram a short distance away. All ten kilograms are moving 1 m/sec on dry ice. The nine kilograms has a string attached that is unwrapping from a pulley or wheel, but the pulley is at first providing no resistance. The trailing kilogram is caught by the pulley and is placed on a string that is wrapped around the wheel as in paragraph one. A string from the 9 kg block applies its motion to the pulley but only the one trailing kilogram is caught in the rotation of the wheel of the pulley. As the one kilogram swings out and unwraps the string on the pulley the nine kilograms stops.
You started with 10 kilograms moving one meter per second in a straight line, you end with one kilogram moving in a straight line, and if you choose you can direct it to travel in the same direction.
If your machine that you propose to measure the velocity of the spheres, or swinging out mass, does not comply with The Law of Conservation of Momentum then isn?t the machine worthy of the Nobel Prize in Physics?
If you start with 10 units of momentum (in the above scenario) and end with say 3.2 units then hasn?t Newton?s Three Laws of Motion been violated. These Laws require that the momentum of a closed system remain constant. If the one kilogram object that is now moving only 3.16 m/sec slams into the 9 kilograms at rest what will the new momentum of the 10 combined kilograms be, and where did the 6.84 units of momentum go?
I am not trying to be an antagonist of your work; I earnestly await your results. I am merely pointing out that if your results show that The Law of Conservation of Momentum is false then the cylinder and spheres phenomenon is a significant violation in the world of physics. If on the other hand the one kilogram object in the above scenario moves away at 10 m/sec, and complies with The Law of Conservation of Momentum, the concept would be a free energy source.

Hello pequaide,
Kator01 told me in a German thread about your setup and your ideas of interchanging momentum between different sizes fo masses.
Contrary to Kator01 I'm still not convinced to achieve any additional energy with your stated setup ideas therefore it would be nice if you may enlighten me in some open issues.
Kator01 All that is needed for a cylinder and spheres device is a horizontally mounted wheel with a center bearing (a pulley), and low friction carts. The pulley has to be wrapped with a string that then has a mass (cart) attached to the end. When the mass on the moving pulley is released the mass will unwrap the string and apply a force in the opposite direction of the motion of the wheel. If the pulley has a mass or is being pulled by a second mass, the mass of the wheel (pulley) or the second mass will stops (unless it is greatly larger than the unwrapping mass).
1. How will the cart(s) move outwards? In my opinion if there isn't any friction between cart and surface, the released cart (mass) will move tangential away with circumferential velocity and unwrap the string. Do you agree?
2. If you agree with my assumption under (1) how can the cart (mass) apply a force in the opposite direction of the motion of the wheel? Please can you explain this assumption more deeper?
3. Why do you assume that the mass of the wheel (pulley) will stop spinning after releasing the mass (cart) placed at the circumference of the wheel? I can not see any reason for such a behavior.
I think that are key questions about a possible functionality of your stated test setup and respectively advanced ideas. Therefore I hope you can provide satisfactory answers.
Regards,
Homer

Hello pequaide,
I have attached two picture which show the principe I have in mind. In pic Drillbohrercrankhaft.jpg you
see the main principle how rotating steelspheres attached via strings to a spindlenut will wind up if the spindleaxis is in a fixed position. After momentum is transfered to the spheres and the cylinder has come to rest the balls wind up the spindlenut to a position by which you directly can measure Epot = m x g x h.
If there is a gain of 200 to 300 % in energy than frictionlosses  even if they are 50 %  is not the problem here.
The second picture shows a cylinder ( red ) with a toothed rim. In this case the rim should only have two ( 2 ) tooths at which the rope spins around at momentumtransfer. Not shown in the pic are two electromagnets attached to the spindle.Only the spindle will be accellerated by a motor.These two electromagnets hold the spheres in the cylinderholes from inside the cylinder and by doing this will take the cylinder with them  meaning the cylinder will also be accellerated. At the desired speed you cut of the supply of both the motor and the electromagnet and do a fullbreak at the spindleexis. The spindle has to come to rest immediately so that the spindlenut can windup vertically reaching in the end a position with Epot = m x g x h.
Due to new businessactivities I will not be able to build this but with this setup you can easily proof your concept thus avoiding complicated videostrobemeasurements.
Tell me what you think about this.
Kator

Homer; The released sphere (or cart) would travel tangent to the circumference of the spinning cylinder (or wheel) but there is a string attached to the sphere and it is wrapped forward of the sphere?s release point (or wheel's release point) along the circumference of the cylinder (or wheel). This string prevents the tangent course of travel for the sphere.
Draw a circle with a tangent line. If you draw a diagram with equal distances marked off along the circumference of the moving circle, these marks would represent the motion of the point of release. Equal distances marked off along the tangent line would represent the new position of the sphere after equal units of time from the release. This shows how the point of release is moving away from the tangent line of travel of the sphere and the distance between the sphere and the point of release becomes larger and larger.
The point of attachment of the string to the cylinder moves away from the sphere?s tangent line of travel as well; the string creates a second tangent line between the point of contact of the string with the cylinder and the sphere. The string connected to the sphere remains tangent, as it unwraps, until the point of attachment of the string with the cylinder is reached. Since the distance from the sphere, on its tangent line, to the point of contact of the string with the cylinder is greater than the corresponding length of the string; the string will pull the sphere off the tangent line. The sphere will obtain a new direction.
A portion of the force in the string is in the same direction as the line of travel of the sphere, this force will accelerate the sphere. The sphere will obtain a new velocity, as will the cylinder. The remaining forces in the string are balanced centripetal and centrifugal forces.
With the new velocity of the component parts and new angles, the force applied in the direction of the sphere?s travel will also change. The sphere will obtain a new direction a new velocity and a new force, this new direction new velocity and new force will happen for every subdivision of time.
The force in the string is in the opposite direction of the line of travel (rotation) of the mass in the cylinder, this force will slow the rotation of the cylinder. The force in the string is equal in both directions (Newton?s Third Law of Motion).
You get a complex arrangement of angles and forces but the fact that the cylinder stops has been proven by experiments.
According to Newton?s Second Law of Motion (F = ma) the equal forces applied to the cylinder and the spheres will cause equal momentum changes on both ends of the string. The momentum change of the cylinder will equal the momentum change in the spheres.
Some scenarios are thought experiments, many are real experiments that have been video taped, strobe light photographed, and photo gate timed. Some experiments are hand held some are in spinning mechanical arms and some were conducted on a frictionless plane. All experimental data shows that Newtonian physics is correct, and that the cylinder and spheres phenomenon (the transfer of momentum from a large mass to a smaller mass) is a means of energy production.

Homer; The released sphere (or cart) would travel tangent to the circumference of the spinning cylinder (or wheel) but there is a string attached to the sphere and it is wrapped forward of the sphere?s release point (or wheel's release point) along the circumference of the cylinder (or wheel). This string prevents the tangent course of travel for the sphere.
Pequaide, I have another question: is the string wrapped in the same direction as the rotation direction of the cylinder or in opposite direction?
However, regardless the winding direction of the string, the string cannot apply a force to the sphere (mass) as long as the string will be unwrapped by the cylinder and tangential moving sphere because string and sphere have the same value of velocity.
I hope you agree so far?
Regards,
Homer

In the mechanical arm model the spheres are moving 1.29 m/sec; if not for the force in the string restraining the spheres after release they would move tangent and be over 1 meter away from the cylinder in less than a second. This forced restraint cost the cylinder all of its motion. Yes; there is force in the string. They are moving at the same speed but they are not moving in the same direction, the string must force the sphere back into a circular path.

Good day pequaide,
It's a pity that you didn't answer my question about the wrapping direction of the string in relation to the rotation direction of the cylinder. Hence I guess wrapping and rotation direction are equal!?
I have attached a sketch which shows stills of motion of the released spheres how I would expect it. However, gravitation forces are not taken into account at all.
Here my remarks to the single points:
1. Fixed spheres rotating around the circumference of the cylinder at velocity value Vu.
2. Spheres are released and moving away in tangential direction. The connected string is unwrapping from the cylinder. The velocity of cylinder's circumference, spheres and connected strings is identical and unaltered at velocity value Vu.
3. Strings are still unwrapping from the cylinder. Same conditions as stated under No. 2.
4. Unwrapping of the strings is finished. Tangential movement of the spheres is abrupt stopped by the strings. The translation energy of the spheres as well as a accordant part of the rotation energy of the cylinder was changed into stretch energie (elastic force) and at last into heat.
That means the formely translation energy of the spheres is no more available.
5. The remaining rotation energy of the cylinder (if there is any left) will be partly transferred to the spheres and because of the tight strings the spheres begin to rotate around the center of mass of cylinder and spheres.
The remaining rotation energy is much smaller as at the begin with fixed spheres at the circumference of the cylinder. Thus V1u will be much smaller than Vu, depending on the mass of cylinder, spheres and string length.
Sorry, but how I understand it you have built more or less a break instead of a energy device. ;)
Regards,
Homer

Well Homer ?how you would expect it? is wrong.
The tangent direction is immediately altered by immediate force in the string; it doesn?t wait to jerk the sphere back it happens immediately.
The string goes forward of the sphere,

Hello,
I had a look at your drawing, HomerS. Upon release of the spheres at 12 h the stings will at no time hang around so loose. They will be at all time streched by the continuously rotating cylinder. The misunderstandig you have here is the position of the rope in the slit and the direction of the rotation. One end of the slit is in permanent contact with the string up to the point the cylinder stops and has transferred its momentum to the spheres.If this has happened the spheres rotate around the centre axis whereby the rope moves along the slit to the other end of it and then the rope takes the cylinder again with it.
Measuring the transitiontime along the length of the slit will give you the anglevelocity and by the total radius of the rope you will get the temporary circumferencevelocity during transitiontime along the length of the slit.
If you look at the handdrawing of pequaide the cylinder is rotating counterclockwise and the slit begins at number 1 extending somewhere along to number 5 position. In this way the beginning of the slit is alway in contact to the rope and thus stretching it until momentum is transfered.
I can not follow your far fetched claims of loosing all energy in form of heat or elastic stretch of the steelstring at this low velocities. The elastic stretch is no energyloss.
This is the reason why I proposed another setup in my last post. I hope I will have the time during the next 6 month to build it or pequaide accepts this for another try for himself. I still miss his comment on this.
As I understand pequaide he simply asks people to replicate and find out themselves.
You repeat the pattern I have noticed in our german discussion. You start with your basic assumption of the conservation of energy and develop your arguments from this point of view.
In this way you will never ever come to new knowledge. You should know better after reading the german booklet of Otto Stein. It is all written in there.
Regards
Kator

I thought I should make diagrams to demonstrate how it really works, but I thought simulated pictures would take less time.
I put in a video tape of a hand held model to see what the positions of the spheres, that Homer proposed, really look like. I am using a cut shell to hold the spheres where they appear to be in the four frames in which the video tape shows the spheres opening.
Frames in a video tape are 1/30 of a second apart. In only three frames (3/30 of a second: in one frame the spheres are still closed) the spheres are fully open. The spheres open in 1/10 of a second.
The first simulated picture replicates the video tape frame just at the (approximate) release point.
The second simulated pictures shows that about 1/30 of a second later the sphere is just clear of the seat. The seat has moved about its own length. This is my best guess at the second frame in the video tape.
The third simulated picture shows the sphere and string at about 45? from a tangent to the point where the string attaches to the cylinder. The seat moved about half its length. This is my best guess at the third frame in the video tape.
The fourth simulated picture shows the sphere and string at 90? from a tangent to the point where the string attaches to the cylinder. The seat moved about a fourth of its length. This is my best guess at the fourth frame in the video tape. The cylinder is stopped while the string is in the slit.
I was surprised myself how quickly the spheres stop the cylinder; I was guessing one fourth rotation. In this video tape it looks closer to one eighth.
Kator01 you have a grasp of what I am doing, and that I am asking people to replicate the experiment. I also think your ideas are fascinating, but right now I am thinking of doing more work with the mechanical arms, and trying to go to a more overhead view with the video tapes.

VGAs are pretty blurry, I will go back to 1M next time.

Hello Kator,
I agree that if the length of the string is short enough that the strings will allways be stretched by the cylinder . Sorry but I didn't know anything about the length of the strings. If the strings are longer, the behavior in the very split seconds after releasing the spheres will be similar as stated in my drawing.
Anyway, after releasing the spheres there are acting two forces at each sphere:
1. force in tangential direction
2. force applied by the string
The resulting force will change the direction of the sphere into rotation direction. This "spherecatching" activity needs physical work and cause the cylinder to slow down or stop its rotation, depending of the cylinder's mass and origin angular velocity, i.e. its origin momentum directly after releasing the spheres.
That's the reason why the spheres "overtake" the cylinder rotation in the slit. In this sense you are right if you state momentum of the cylinder was transferred to the spheres, however the overall momentum cannot be greater than the origin momentum with fixed spheres at the circumference of the cylinder.
And please accept the fact that the velocitiy of cylinder's circumference and tangential speed of the spheres are equal! I'm afraid you're not aware of this fact although I have stated this several times in the German SARAthread.
Hence the spheres are able to overtake the cylinder rotation in the slit because the cylinder rotation slowed down and not because the spheres rotation have increased!
In the German thread you asked to post directly here in this thread if there are any questions about this setup idea. But now you state here that I better keep my arguments and assumptions under my hat. All right I will accept again your new position.
And by the way, I do not agree in any case to the statements in the booklet of Otto Stein. His assumptions are wrong. For instance you cannot calculate complex asymmetric rotational movements by separating them into two or more single symmetric rotational movements. Or should I better unbalance all my car tires to create energy? ;D
Hey...das ist wirklich haneb?chener Unsinn! Das glaubst Du doch nicht??? :o
I can recommend a good German book: Dorn Bader Physik Oberstufe. ;)
That are my 2 cents regarding this issue.
@pequaide
Thank you for your pictures for better imagination. I have posted my point of few above and I'm convienced I'm right. I will check this thread about news of your break setup. Anyway, have much fun! ;)
Regards,
Homer

Hello pequaide,
thank you very much for theses pictures. It was a bit surprising to me me but finally one can clearly see how the process works. Very good simulation. The spheres a accellerated outwards along a 90 degree arc too the final position. During this transitionphase the spininertia and the speed of the spheressystem increases to its maximum value and this consumes all the momentum of the cylinder.
Of course I understand that you have your own plans to follow through with further steps. I also was thinking about a mechanical arms, but then you will have less degree of freedom for the spheres to move. But I think it will work also. I would rather think of bycyclechains if one plans to start with higher speedtests.
But then the question remains : How do you harvest the energy gained ? This was the reason for my proposal whis can be modified also with mechanical arms, because the gain has to be stored and I found this the simplest way to do and at the same time it will give you a direct measure of m x g x h.
Can it be done in another more simple way ?
regards
Kator
@homerS : I wonder about your statement that Otto Steins calculation deal with unbalanced masses.
Have a look at a normal planetary gear ( (Planetengetriebe), all parts are well balanced. In the case of otto steins first Drehmotorproposal it only has 2 planetary toothed wheels rotating around the circumference of the center toothed wheel but nevertheless balanced and without the outer toothed ring which makes up for this type of gear.This also is true for differential gear ( Differenzialgetriebe )
Believe me : it is difficult but engineers have done it. I myself had to calculate all that stuff during my engineering education for applied carconstruction ( I forgot it all by now )
I simply give you another example here: During the 60`ths no none of the established engineers at BMW, Daimer etc believed it to be possible to construct a 5 cylinderottomotor because they believed it to be an impossible unbalanced system to build. Somewhere in the years 1972 to 1976 ( where I graduated as a engineer ) AUDI prooved them all wrong and created this 5 cylinder engine.
Your quote "Or should I better unbalance all my car tires to create energy? Grin " is simply a very bad example and does not fit here.

The velocities are roughly equal (actually the spheres are going slightly faster because their center of mass is further from the point of rotation) at release, but they immediately become unequal. In an eighth rotation (1/10 sec) or so the horizontal motion of the cylinder is gone. The momentum of the cylinder has disappeared, and there is scant little friction in the system. So where did the momentum go?
The momentum had to be transferred to the spheres, but their mass is constant.
The spheres? velocity had to change.
When the string is touching the end of the slit the radius of rotation of the spheres is the length of the string to the slit, when the cylinder is stopped. I often make that length equal to the radius of the cylinder. The initial and final radii are equal.
Look at the strobe light pictures. How much does the horizontal distance between the pictures of the sphere increase? Before release the sphere?s pictures were about a half overlapping each other. Now they are how much apart?
It all makes sense; the momentum has been transferred to the spheres.
If the cylinder has a mass three times greater than the spheres, the spheres at max are going four times as fast. ? * 4 kg * 1 m/sec * 1 m/sec is less than ? * 1 kg * 4 m/sec * 4 m/sec (but the momentum is the same). The energy of the system quadrupled.
What you can not see in the pictures is that there is a hole in the center of the black plastic strip in the middle of the cylinder. The strings going up to the top of the cylinder are the ends of the strings that come from the spheres. After being fed through the hole the strings are secured with the small screws. This is a quick and efficient way to change a broken string. In the test the string comes away from the cylinder with nearly no friction, and while in the slit the strings don?t even touch the slit. But when the bottom of the cylinder strikes the bed or padded floor the interaction is violent.

@Kator
So you state a differential gear will produce additonal energy?  Very interesting! I always thought a gear will transfer a part of energy into heat by friction loss? :D
...and by the way I'm an mechanical engineer as well (Konstruktion allgemeiner Maschinenbau). ;)
@pequaide
The velocities are roughly equal (actually the spheres are going slightly faster because their center of mass is further from the point of rotation) at release, but they immediately become unequal.
Please can you explain why a mass will going faster if its center of mass is further from the point of rotation? That is not true! The velocity of a rotating mass is always constant independent from its rotation radius. Only the angular rate (or rotary speed) will change according the value of the rotation radius! I believe that fact is the main reason for your misunderstandings.
In an eighth rotation (1/10 sec) or so the horizontal motion of the cylinder is gone. The momentum of the cylinder has disappeared, and there is scant little friction in the system. So where did the momentum go?
The momentum was used to stretch the strings and to transfer the spheres from translation direction into rotation direction.
The spheres? velocity had to change.
Yes the velocity is smaller than directly after releasing from the cylinder because the spheres were slowed down by the strings fixed at the cylinder.
When the string is touching the end of the slit the radius of rotation of the spheres is the length of the string to the slit, when the cylinder is stopped. I often make that length equal to the radius of the cylinder. The initial and final radii are equal.
Look at the strobe light pictures. How much does the horizontal distance between the pictures of the sphere increase? Before release the sphere?s pictures were about a half overlapping each other. Now they are how much apart?
It all makes sense; the momentum has been transferred to the spheres.
You cannot definitely assume that the velocity of the spheres has increased, the cylinder's velocity can slow down as well.
If the cylinder has a mass three times greater than the spheres, the spheres at max are going four times as fast. ? * 4 kg * 1 m/sec * 1 m/sec is less than ? * 1 kg * 4 m/sec * 4 m/sec (but the momentum is the same). The energy of the system quadrupled.
Nope, that's only your assumption founded in misunderstood physics...
This is a quick and efficient way to change a broken string.
Very interesting! Guess why a string may brake? And of course you think there is no need of energy to break a string made of steel???
Regards,
Homer

The strings are made of Berkley Fireline, I think this is a fluorocarbon line that has a very low stretch modulus. It is very strong but after the cylinder hits the bed the line drags across the outer edge of the slit. You can see the line slowly being cut by the slit after a few dozen tests. The string won?t be cut until over one hundred tests so I am not too bothered by it.
Homer; You cannot definitely assume that the velocity of the spheres has increased, the cylinder's velocity can slow down as well.
Pequaide; The strobe light photo shows the sphere pictures becoming further apart, what has that got to do with what the cylinder is doing. As the sphere pictures separate the cylinder is shown stopping.
Homer quote; Yes the velocity is smaller than directly after releasing from the cylinder because the spheres were slowed down by the strings fixed at the cylinder.
Pequaide; How can both the cylinder and the spheres slow down, I have pictures where the cylinder is stopped and the spheres are a blur of motion.
Homer quote: The velocity of a rotating mass is always constant independent from its rotation radius.
Pequaide; This is true only if there is no unbalanced force being applied to the internal parts of the system. The cylinder shows that there is unbalanced force by stopping. The force also affects the spheres; it will speed them up There is no outside unbalanced force being applied to the system but the internal parts of the system experience unbalanced force.

I'll give up because that's wasted time and time is running too fast. Make your trials and setups and become happy!
Bye
Homer

Iacob alex has a link to an Atwood simulation in his Pulsatory gravitational avalanche post. The simulation shows that the momentum of the pulley has to be incorporated in the F = ma equation to determine acceleration. They do not however state the mass distribution of the pulley: which is necessary to determine if their calculations are correct. Their calculations give the average mass, in motion, of the pulley to be at half the velocity of the circumference.
If the pulley was a ring mounted on dry ice, the calculations would be simple. The entire mass m1 + m2 + m3 is accelerated by the difference between m1 and m2. F = ma; (m2 ? m1) * 9.81 / (m1 + m2 + m3) = a. This is true because m3 (the rim shaped pulley) is moving at the same velocity as m1 and m2. If m3 is not a ring the calculations becomes more complex, but the moving mass m3 still has to be incorporated in the F = ma equation.
This simulation has value because m3 can be used as a cylinder and spheres experiment.
Make the pulley (mass 3) massive (9 kg), and place nearly all the mass of the pulley in the rim. From their calculations I think they are assuming a more or less uniform (from center to rim as in a solid disk) distribution of mass.
Place the pulley in a horizontal plane. Drape the string to mass 2 over a frictionless pulley.
Make mass 1 equal to zero and mass 2 equal to 1 kg. With m1 zero there can be no more friction in the string, so you will have to wrap and tie the string to the horizontal pulley.
This will give you 1 kg accelerating 10 kilograms and an acceleration of .981 m/sec.
When the ten kilograms achieves a velocity of 1 m/sec; mass 2 has dropped .5096 meters. d = ? v?/a
Transfer the nine units of momentum of the horizontally mounted rim (pulley) into one of those nine kilograms as in the cylinder and spheres device and the one kilogram will rise (as in a pendulum) 4.128 meters. This is an energy increase to 810%
Mechanically arrange the kilogram at 4.128 meters to lift mass 2 back to where it started and you still have 3.618 meters left over.
I was reviewing my video tapes and I came across the ten to one model; the134 grams of spheres stopped 1258 grams of cylinder. In that model the slit was extended 3/16 in. from the sphere?s seat hole in the cylinder; the string is wrapped from one seated sphere nearly half way around the cylinder and entered the cylinder through the slit extended from the seat of the other sphere. The string length to the sphere?s center of mass of this model is about 2.8 times the cylinder?s radius.
Let?s say the mass of the cylinder with spheres (1392 g) was moving 1 m/sec around the circumference of the circle before release. That would mean that it would take one newton 1.392 seconds to bring the cylinder and seated spheres to a stop. F = ma, a = Ã¢Ë†â€ v/Ã¢Ë†â€ t, 1N = 1.392 kg *1 m/sec / time; time = 1.392 kg * 1 m/sec / 1 newton.
After the spheres have all the motion it will still take one newton 1.392 seconds to bring the spheres to a stop. F = m * Ã¢Ë†â€ v/Ã¢Ë†â€ t; Ã¢Ë†â€ v = F * Ã¢Ë†â€ t/m; Ã¢Ë†â€ v = 1 newton * 1.392 sec / .134 kg = 10.38 m/sec. The spheres will have to be moving 10.38 m/sec for the one newton to stop them in 1.392 seconds
So Newtonian Physics predicts that the energy change in the system will be; initial energy ? * 1.392 kg * 1 m/sec * 1 m/sec, to final KE, ? * .134 kg * 10.38 m/sec * 10.38 m/sec; which is an increase to 7.22 / .696 = 1038%.
I have scavenged this model to make another model that has slits in the normal positions about half way between the seats, this will change the stopping capacity of the original model.

I think that those that constructed the Atwood simulation assumed that the mass distribution of the pulley was like that of spokes. The spoke arrangement would allow equal mass at every distance from the point of rotation. This uniformity of mass allows momentum (of the concentric hollow cylinders) to be a linear function as the radius is increased from the point of rotation to the circumference. And the average momentum is at half the distance from the center to the circumference. If the mass distribution of the pulley is like that of a solid disk this (average momentum location) is not true, because mass and velocity would increase as the spinning concentric rings (being evaluated) approach the circumference (subdivide the disk into a hundred concentric rings). The average momentum of the solid disk is at about 58% of the distance from the center to the circumference.

www.msu.edu/user/brechtjo/physics/atwood/atwood.html I was hoping that typing in the site would post a link to the site. ???
Put 1000 (average momentum is half way to the circumference, effectively 500 kg) kg in for the mass of the pulley, and 2 kg and 3 kg for the masses of mass 1 and 2. Then calculate the momentum of the pulley after a 1 meter drop. d = 1/2v?/a: a = 1/505 * 9.81
And then calculate the momentum of one kilogram that has dropped one meter. The difference between mass one and mass two is only one kilogram.
And then remember that there is a Law of Conservation of Momentum.
Transfer the momentum of the pulley to the one kilogram and you have a huge amount of energy.

Hello pequaide,
thank your very much for your effort in continoiusly explaining more and more of some important details.
Looks I am the only one left at the time as a discussionpartner.
I plan to build a simple version with a disc and spheres. The spheres in my model are attached to a ballbearing fastened to a centeraxis.
This is different mechanically in comparison to your experimental setups in the sense that
the centerpoint of the tethers can rotate freely and do not wind up on the axis.
My question is this : What would you expect to happen according to your experience ?
I will read this attwoodpage and come back to this thread if I have new ideas/insights/ etc.
Regards
Kator

A disc is fine; I use one for the air table model. But you must wrap the tether of the spheres around the circumference of the disc at least one disc radius. I have no model much shorter than one cylinder radius, but they might work depending on the difference in the spheres' mass to cylinder mass.
As the tether unwraps from the disc the force in the tether forces the disc to accelerate (decelerate; reduce the rate of spin) in the opposite direction, the disc will slow down and stop and then begin rotating in the opposite direction. If you release the tethered spheres when the disc is stopped the disc will of course remain stopped, and the spheres would be at maximum velocity.
I am going to insert a picture of an air table model, since a photograph is worth a thousand words.
The screwdriver goes through a hole that captures a loop on the end of each string. In operation there is a pin in this hole that is tied to a string that is looped around the operator?s finger.
The pucks on the end of the strings must unwrap from the disc about one radius. The loose strings are held up against the plastic wall, by the fingers, and they hold the pucks in place as the disc is accelerated. When the disc is released the strings are released and the pucks will unwrap from the disc. If you pull the pin when the disc is stopped the disc will not be restarted, in the opposite direction. The system works very well.

Let me review something in this model. If the pucks stop the disc before the string is at 90? (to a line tangent to the circle at the point where the string enters the circle) then the disc will be restarted in the opposite direction (unless of course the string is released).
By adding mass to the disc the disc can be stopped just as the string becomes straight. In this model the string leaves the disc at 90?. The 30 pound test line was used for photographic reasons. 10 lb test would work for the air table.
Kator01 Possibly a better answer to your question about whether your design will work or not would be this; how are you going to stop the momentum of the spinning disc? How are you going to apply force to the disc in the opposite direction of the spin?

Hello pequaide,
assume the same setup as you have tested and presented here  but the only difference is that the tethers of both pucs are fastened to a ballbearing on a centeraxis. The disc also can rotate around the
centeraxis with another ballbearing. So the difference wound be that the tethers do not wind up on the axis thus becoming shorter at each turn but can rotate free araound the axis and keep their length.
I have no idea of the effects of this additional degree of freedom. There are some keyvariations to be tested to fully understand the principle which make the fullstop of the disc or cylinder possible.
Anyway your last description ( disc rotates backwards at 90 degreeposition of the pucs = bouncing back ) indicates to the elasticimpactcase if the mass of the pucs is too big and the halfelasticcase if the masses are in the right proportion.
I have to build it. There is no other way to find out.
Regards
Kator

I will send a diagram.
Point D is where the string enters the cylinder or drapes over a pin on the disc.
Line CA is tangent to the disc or cylinder at point D.
Line DB is perpendicular to line CA
If the disc was moving counterclockwise, a puck on the end of a string in quadrant ADB would force the disc to accelerate clockwise. A puck on the end of a string in quadrant CDB would force the disc to accelerate counterclockwise.
So you might see the puck in quadrant ADB slow the counterclockwise rotation of the disc; stop it; and restart it in a clockwise direction. As long as the puck is in quadrant ADB all acceleration is in the clockwise direction. If the disc is too massive the puck will not stop the counterclockwise rotation of the disc. I have stopped cylinders nine times heavier than the spheres.
While the puck and string are on line DB the puck places no rotational force upon the disc.
While the puck, on the end of a string, is in quadrant CDB (and still attached to the disc) it will accelerate the disc counterclockwise. If the disc is stopped the puck will accelerate it counterclockwise. If the disc is still moving counterclockwise it will accelerate counterclockwise. If the disc is moving clockwise it will accelerate counterclockwise. If the disc is to light there will be two stops of the disc; the motion will be counterclockwise (original motion); clockwise (stopped in ADB and restarted); counterclockwise (puck in CDB), I have seen it many many times.
The challenge is to add just enough mass to the disc so that the disc stops while the string and puck are on line DB. If the string is then released from the disc the disc will remain stopped. Or if the string enters a slit in the cylinder it will remain stopped until the string reaches the other end of the slit and accelerates it counterclockwise.
Fireline is extremely inelastic; especially with very light pucks on 10 lb test. I envision no elastic activity.
Okay; lets say your two bearing (one axis) design is placed on an air table. The bearing for the string is just above the bearing for the disc. The bearing for the disc is below the surface of the disc. The string of the puck is fastened to its bearing just above the disc and the string proceeds out to the circumference of the disc; the string then drapes (horizontally) over a pin and proceeds along the circumference clockwise about one disc radius.
You will probably need to hold the pucks in place until just at release. How are you going to accelerate the disc?
It should work. If the disc is about three or four time the mass of the puck it should get you in range of adding just enough mass to the disc to make the disc stop while the puck is on line DB. I can?t give you exact mass numbers because I don?t know the length of the string that wraps around the disc. If the string of the puck is clear of the disc (just above) it should work great.

www.msu.edu/user/brechtjo/physics/atwood/atwood.html
On the Atwood site the Green S in the upper left corner of the page identifies the site as originating from Michigan State University, I would imagine the site was made by students but checked over by their professor before release. Therefore the concepts presented by the site represent those of the University. It should not be surprising that one of the United States? major universities thinks F = ma.
This gives us a source of extra momentum. The spinning wheel (pulley) has a much larger amount of momentum than the momentum developed from the freefall for the same distance of the same mass that gave the momentum to the wheel.
A five kilogram rim mass pulley accelerated by dropping a one kilogram mass one meter (as in an only one suspended mass Atwood) has 9.04 units of momentum; and the one kilogram that could have freefell the same distance has only 4.43 units. When the 9.04 units of momentum are given to one kilogram it will rise 4.167 meters, 4.167 times higher than the dropped kilogram.
We know from ballistic pendulums that linear motion and circular motion are completely interchangeable and experience no loss of motion in either direction. So the rotating five kg rim (above) has 9.04 units of momentum. v = Sqrt (1 m * 2 * 1kg/6kg * 9.81m/sec; mv = 1.808 m/sec * 5 kg = 9.04 units
We know that a one kilogram mass moving 5 m/sec will combine with a four kilogram mass at rest and the combined 5 kilograms will be moving 1 m/sec., for momentum conservation. Can going from 5 kg moving 1 m/sec to one kilogram in motion do anything but also conserve momentum?

Hello pequaide,
thank your for your detailled explanations, I think I have all information necessary for building the setup I have in mind.
I will use spheres held in place by two electromagnets powered by lithiumpolymerbatteries. The have enough amperage ( 1.5 Ampere/h ) to feet the electromagnets. Electromagents and batteries are mounted on the bottomside of the disc and thus rotate with it. There are 2 vertical holes in the disc. Two iron pins in these holes are magentically powered from the electromagnet below thus holding the spheres in place on top at the rim of the disc. Via remotecontrol ( highfrequencysender ) and the receiverboard ( also rotating on the disc ) the electromagnets are shut off at definite speed thus releasing the spheres.
By chance I lately was getting a ceilingmounted horizontalworking fan. The rotating disc ( with the fanblades ) is powered by an electromotor ( AC) and has quite a good amount of mass. Motor and the shaft ( central axis) are mounted together. So the shaft is fixed on the ceiling. Motor and shaft do not move, just the disc with the blade is rotated by an inner gear. This is exactly what is needed for this experimental setup I have in mind.
You also can use such a fan for your experiments. The only thing to change here is a ballbearing to be mounted on the upper shaft with the tethers and spheres attached to it.
I will post a picture of this fan. You will understand it better then.
Regards
Kator

Hello pequaide,,
concerning the atwoodcalculations can you please give more details as to the basic assumptions of the rimmasspulley ?
The massinertia of a rimmass depends of the massdistribution and the radius.
How did you calculate the spinmomentum of the pulley ?
Using 1 kg for mass m1 and 5 kg for m3 ( pulley ) I get 2.8028572 m/sec exp^2 in the simulation.
Now you have to assume a radius first before you can calculate the angular momentum.
Thank you
Kator

Hello pequaide,
in the AtwoodSimulation you have to switch on the frictionButton and enter 5 for mass m3. The text explaines that only if you use the frictiononsimulation that you will see the influence of mass m3.
The term friction used here is misleading. What they mean is the massspininertiaresistance of m3 which consumes part of the energy generated be freefall of m1.
For the case m3 = 5 and m1 = 1 and friction "on" you will get an accelleration of 2.8028572.
After 1 second of free fall mass m1 has momentum P = 2.8028572 kg m/s. Circumference of m3 will deliver the angularvelocity via rot/ s = 2.8028572 / 2 x pi x r .
The spinmomentum then depends on the radius of mass m3.
In order to calculate the spinmomentum of mass m3 you have to use the relevant formulas for :
SpinInertia Is = m3 x r exp^2
SpinMomentum = Is x omega ( angularvelocity )
Now the spiinertia and the angularvelocity as well can only be calculated if you have the radius of the pulleymass m3.
It^s not so easy to calculate this. I have done it by the assumption of r = 0.1 meter and have to look over this calculation again because I am not sure if I made a mistake.
I think that this AtwoodExample is no good for any explanations of what you found.
Regards
Kator

In reference to: www.msu.edu/user/brechtjo/physics/atwood/atwood.html
Indeed the word friction in the friction button is confusing, the word adhesion may be more appropriate. In physics we almost always view friction as something lost. From their own calculations they are not viewing this friction as something lost.
If: m1 is 2 kg, and m2 is 3 kg, and the friction button is not selected, you get an acceleration of 1.962 m/sec/sec.
If m1 is 2 kg, and m2 is 3 kg, and m3 is zero, and the friction button is selected you get an acceleration of 1.962 m/sec/sec.
The acceleration is the same with the friction button on and the pulley moving, as it is with the button off and the pulley not moving. Their calculations assume no loss for this quantity they call friction. And certainly we can build very high quality bearings where this is almost true. But it is incorrect to think that there needs to be a frictional loss on the surface of the pulley.
The strings on the ascending and descending sides of the pulley can be independently wrapped and fastened to the pulley; you really don?t need friction to make a pulley work. If the circumference of the pulley was coated with dry ice, you could still use the pulley by wrapping the string from both sides a few times around the pulley in the appropriate direction, and attaching the other end of the string to the pulley. You would get the same results.
Suppose we use a long steel pipe mounted horizontally on dry ice for the pulley. The acceleration of the pipe?s mass, as it spins, is roughly equal to the acceleration of the masses m1 and m2. It makes a difference what the mass of the pipe is. But if we use a larger radius pipe with the same mass; you will get the same acceleration.
I don?t think it was an oversight that the MSU Atwood site makes no reference to the radius of the pulley. It does not matter what size wheel you use. Therefore F = ma is not in reference to angular momentum changes. F = ma makes linear momentum.
I think that those that constructed the Atwood simulation assumed that the mass distribution of the pulley was like that of spokes. The spoke arrangement would allow equal mass at every distance (radius) from the point of rotation. This uniformity of mass allows the average momentum to be half the distance from the center to the circumference (1/2R). So adding 4 kg to m3 (the spokes have a mass of 6 kg instead of 2 kg for example) would be the same as adding 1 kg each to m1 and m2.
The following is an Email to MSU
Your MSU site on Atwood?s machine has been used as a verification of an experiment that is circulating in Europe and the U.S.
Your Atwood calculations guarantee that momentum can be placed and stored in a spinning wheel. The momentum is placed in according to the relationship of F = ma. If it is placed in according to F = ma then most assuredly it must be taken out according to the same relationship.
Your Atwood site is unusual and useful because it makes prediction concerning the mass of the pulley. Many do not deal with the mass of the pulley and assume it to be without mass.
On your site: if mass 1 and mass 2 are not equal, the unequal portion of the mass accelerates the total mass (m1 + m2) according to the relationship of F = ma. But the unequal mass also accelerates the pulley or wheel, as you have noted.
According to your site calculations the Atwood accelerates the pulley as if the average mass of the pulley is half way between the center axis and the circumference. In a spoke wheel this would be correct. So the mass of the pulley is accelerated twice as easily as the same mass equally distributed between m1 and m2. This is because the average mass of the spoke pulley is moving half as fast as the circumference velocity, and m1 and m2 are moving at the same velocity as the circumference.
For example: #1 let?s start with 2 kg in m1, and 3 kg in m2, and 0 kg in m3. Then adding 1 kg to each of m1 and m2 would be equal to adding 4 kg to m3. Since the hanging masses accelerate according to F = ma then the wheel must be accelerating according to F = ma as well.
When m1 = 2kg; m2 = 3 kg; m3 = 0 kg then acceleration equals 1.962 m/sec.
When m1 = 2kg; m2 = 3 kg; m3 = 4 kg then acceleration equals 1.4014 m/sec.
When m1 = 3kg; m2 = 4 kg; m3 = 0 kg then acceleration equals 1.4014 m/sec.
Example #2 m1 with a mass of 20 kg; and m2 with a mass of 21 kg; and with m3 zero:
Should be equal to m1 equal to zero; m2 with a mass of 1 kg and m3 with a mass of 80 kg.
When m1 = 20kg; m2 = 21 kg; m3 = 0 kg then acceleration equals .2393 m/sec/sec.
When m1 = 0kg; m2 = 1 kg; m3 = 80 kg then acceleration equals .2393 m/sec/sec.
Example #3 when force remains constant the doubling of the quantity of mass that is accelerated should reduce the acceleration to half.
When m1 = 0 kg; m2 = 1 kg; m3 = 80 kg then acceleration equals .2393 m/sec/sec this is 1 kg accelerating 41 kg. 80 kg at half the velocity (1/2 R) would be equal to 40 kilograms on the circumference.
When m1 = 0 kg; m2 = 1 kg; m3 = 162 kg then acceleration equals .2393/2 = .1196 m/sec/sec (from the MSU site). This is 1 kg accelerating 82 kg. 162 kg at half the velocity (1/2 R) would be equal to 81 kilograms on the circumference.
Twice the accelerated mass 82kg / 41kg causes half the acceleration .1196/.2393 because force remained constant ( 9.81N ).
When the cylinder and spheres machine takes the momentum back out of the wheel it can place all the motion into an object with a mass as small as the mass (difference between m1 and m2) that accelerated the wheel in the first place. This is where it gets interesting.
This MSU site confirms a source of extra momentum. The spinning wheel (pulley) has a much larger amount of momentum than the momentum developed from the freefall for the same distance of the same mass that gave the momentum to the wheel. When the momentum of the wheel is given to the mass used to accelerate the wheel, the mass can rise much higher than the distance it was dropped.

Hello pequaide
I have done some research on basics :
Moment of inertia ( can bee accesses only one or two times. Then you must register )
[url]http://hyperphysics.phyastr.gsu.edu/hbase/hframe.html][http://www.efunda.com/math/solids/IndexSolid.cfm/url]
[url]http://hyperphysics.phyastr.gsu.edu/hbase/hframe.html (http://[http://www.efunda.com/math/solids/IndexSolid.cfm/url)
Mechanics>Rotation> Angular momentum
Rotational kinetic energy
Moments of Intertia
Here you find the basic formulas.
I found a german scientific paper about calculating the accelleration of two ( or one ) falling mass ( 1 kg ) in the atwoodexample using a pulley ( full cylinder, no rimmass ) of 5 kg. Your assumtions are not true.
See here the formula right above Paragraph 4) Ein Fallversuch (3)
http://wwwik.fzk.de/~drexlin/Mechanik0607/L8.pdf (http://wwwik.fzk.de/~drexlin/Mechanik0607/L8.pdf)
Assuming the radius of the pulley to be 0,1 meter the value of accelleration is in exact accordance with the atwoodsimulation :
accelleration a = 2.8 m/s exp^2 and not 9,81. This means then for just one mass of 1 kg accellerating downwards ( no other upward moving mass used ) that if you let this mass fall for 1 second the velocity will be 2.8 m/s. The distance it has fallen downwards is 1.4 m. ( s = v x t / 2 )
By this you can calculate the InputEnergy ( raise the 1 kg up to 1,4 m ) = m x g x h = 1 kg x 9,81 x 1.4 m = 13.734 Joule.
The SpinInertia of the pulley J = m/2 x r exp^2 = 2.5 x 0,01 = 0.025 kg m exp^2
Angular velocity of the pully at v = 2.8 m/s of the falling mass > Omega = v/r = 2.8 m/s / 0.1 m = 28 1/s
This will give you the SpinMomentum of the pulley L = J x omega exp^2 = 0.025 x 28 = 0.7 kg m exp^2 / s
The falling mass m at v = 2.8 m/s has momentum of 2,8 kg m/s
Rotational energy of the pulley Wrot = J x omega exp^2 / 2 = 0.025 x 28 exp^2 / 2 = 9,8 Joule.
Energy of the falling mass m at v = 2.8 m/s = 1/2 x m x v exp^2 = 1/2 x 1kg x 2.8 exp<^2 = 3.92 Joule
Total energy after 1 second = 9.8 + 3.9 = 13.7 Joule ( see above for Input Energy )
What you have to to is to exactly measure the velocity of the spheres and compare this to the rotational energy of the cylinderandspheres system before you release the balls.
As I said the atwoodmachine is no good example. Spinmomentum has a different dimension ( kg m exp^2 / s ) than
translatory momentum ( kg m/s ), and I have no idea how this small 0.7 SpinMomentum will convert to translatorry momentum, increasing energy.
There is one thing one can find out by experiment : The proposal of Alan Cresswell on his homepage here :
http://www.unifiedtheory.org.uk/ (http://www.unifiedtheory.org.uk/)
See Diagramm 22 ONCE MORE FOR NEWTONIAN APPLEHEADS
Regards
Kator

How much linear momentum does the wheel and falling mass have after the one kilogram has fallen 1.4 meters?
I think it is about 9.8 units of linear momentum.
If these 9.8 units of linear momentum is transfer to the one kilogram that has fallen, then it will rise, 4.895 meters. This is an increase to 4.895m/1.4m = 350%.
A rim mass pulley or wheel would make it easier to calculate the momentum of the pulley. And I think you would agree that such a rim mass pulley could be constructed.
I think a ring vertically mounted on dry ice would give you a nearly perfect F = ma, if you dropped an extra mass on a string wrapped around the pulley ring.
Let the ring have a mass of 5 kg with an extra dropped mass of 1 kg. Now the dropped mass and the ring have the same velocity. Now; do the math again, the system has more momentum than it needs to return to the top. And the cylinder and spheres transfers all the motion of the cylinder and spheres (ring and dropped mass) to the spheres (dropped mass).

Wrap a string around a 5 kilogram rim mass pulley; suspend a 1 kilogram mass from the string as in a one suspended mass Atwood machine. This is one kilogram accelerating 6 kilograms. Acceleration equals 1/6 * 9.81 m/sec/sec. If you let the mass drop one meter the entire system will be moving 1.808 m/sec.
You can use the kinetic energy formula to prove that the energy of this 5 kilogram rim and suspended one kilogram mass has the same energy as one kilogram that has free fallen one meter.
The square root of (1m * 2* 1/6 * 9.81) = 1.808m/sec, ? * 6 kg* 1.808 m/sec * 1.808 m/sec = 9.806 joules. This is the energy of the 6 kilograms after one kilogram has dropped one meter.
The square root of (1 * 2 * 9.81) = 4.43, ? * 1kg * 4.43 m/sec * 4.43 m/sec = 9.81 joules. This is also the energy of a one kilogram mass that has free fallen one meter. Since energy is Force times distance I guess this should not be surprising.
They have the same kinetic energy; but they do not have the same momentum. Momentum is Force times time so this should not be surprising either.
The time over which the force acts in the one kilogram free fall for one meter is Ã¢Ë†Å¡ (2 *d / a) = t = .4515 sec
The time over which the force acts in the one kilogram dropped one meter over a five kilogram rim mass pulley is Ã¢Ë†Å¡ (2 *d / (a/6)) = t = 1.106 sec
6 kg * 1.808 m/sec = 10.848 units of momentum; 1 kg * 4.43 m/sec = 4.43 units of momentum. 10.848/4.43 = 1.106 sec / .4515 sec
A puck circling on the end of a string, which is wrapping the string around the center pin on a frictionless plane, is not conserving angular momentum. It is conserving linear momentum.
Ballistic pendulums always conserve linear momentum not angular momentum. Even if the incoming and final motion is circular, linear momentum is still (always) the quantity conserved not angular momentum.
How can you trust angular momentum conservation when it can?t pass the simplest of test?
When you reverse the phenomenon of the ballistic pendulum and send off a small projectile from a large circling object, why wouldn?t you expect the same quantity to be conserved as that which was conserved in the forward direction?

Here are some experiments that would prove that the angular momentum conservation theory in the laboratory is false.
Swing a 1 kg sphere down .4587 meters on the end of a twenty meter pendulum string; it will be moving 3 m/sec. Have it embed in a 2 kilogram block that is suspended on the end of a one meter pendulum string. The twenty meter string then releases and the 3 kilogram combination will be moving 1 meter per second in a 1 meter pendulum.
Linear Newtonian Momentum is conserved.
Angular momentum is not conserved.
Kinetic energy is not conserved.
Suspend a pendulum bob from a horizontally mounted roller track sled; accelerate both the sled and the bob to one meter per second keeping the string in a vertical position. Have the sled impact a stationary block. The bob should rise .051 meters. Do the same experiment with one, two, and three meter strings, the bob should rise .051 meters in all situations.
Velocity of the bob at impact is maintained in the three different pendulum lengths.
Linear momentum of the bob is maintained in the three different pendulum lengths.
Angular momentum is not maintained in the three different pendulum lengths.
Some use the frame of reference concept to prop up the theory of angular momentum conservation. This frame of reference concept is basically assigning an imaginary axis; usually it is the original axis. One debater equated this assignment of an axis to the frame of references that might be used in linear momentum conservation.
It is true that linear momentum is conserved under different frames of reference. Let?s say a one kilogram mass moving 1 m/sec on a frictionless plane is struck from behind by a 3 kilogram mass moving 3 m/sec. You can use the frame of reference of any point on; the air table, the one kilogram mass, or the 3 kilogram mass and momentum is always conserved from that frame of reference.
But in changing your axis all axes are using the same point of reference as that used by the air table, it is not a change in reference it is merely a means of pretending that the radius has not changed, which of course has to be avoided because if the radius changes your conservation of angular momentum is lost.
Angular momentum is composed of three quantities, mass, linear velocity, and radius. Radius appears in the formula three times.
The distance traveled around the circumference of the circle in a unit period of time is equal to linear velocity. This linear velocity is then divided by radius to obtain radians per sec. The product of mass and radians per sec is then multiplied by radius two more times. Since radians are linear velocity divided by radius and we multiply by radius two more times this makes two radii drop out of the equation. This leaves us with mass times linear velocity times radius; one radius beyond Newtonian Physics.
This is the funny thing about angular momentum, why not just drop the one extra radius and everything would work just fine. The reason is that angular momentum was developed by Kepler for use with satellites and the extra radius was necessary to compensate for a huge increase in linear velocity caused by gravity. For satellites a small radius means huge linear velocity, and a huge radius mean small linear velocity.
This brings us to yet another proof that angular momentum conservation will not work in the lab. It should be obvious to any mathematician that a formula that works for satellites where you have huge increases in linear velocity caused by gravity can not possibly work in situations where there is no increase in linear velocity caused by gravity.
A thought experiment that should clarify the impossibility of angular momentum conservation in the lab would be to put a pin at the position of the Sun and run a string to the comet at apogee. The comet will continue to rotate around the pin at a distance of apogee.
Place a second pin at a distance from the comet that would have been equal to perigee. Place this second pin between the comet and the first pin so that the string comes in contact with the second pin and the comet then begins orbiting the second pin at a distance of perigee. No linear velocity change has occurred because there is no gravitational acceleration caused by the Sun.
The angular momentum formula is: M Ãâ€° R? where Ãâ€° equals radians per sec or linear velocity / r.
M *(linear velocity / r) * r * r = M * linear velocity * r
The two formulas that have to be equal to each other if angular momentum conservation is to be true are:
M * linear velocity * apogee (rotation about pin #1) = M * linear velocity * perigee (rotation about pin #2) Since mass and linear velocity remained constant with the use of pins the equation is false because apogee does not equal perigee. The two equations are equal only if gravity causes a huge increase in linear velocity.
With the Sun back in place; linear velocity at apogee / linear velocity at perigee = perigee / apogee, and the equation (M * linear velocity at apogee * apogee = M * linear velocity at perigee * perigee) is true.
This two pin experiment can be performed on a frictionless plane; only the scale is changed the principle remains the same. If there is no gravitational acceleration angular momentum conservation does not work.
Kepler needed no frame of reference, he got it right.
You could release the comet orbiting pin #2 in the direction of pin #1. Have the comet come within a perigee distance of pin #1 and recapture it on the end of a string. Now your point of reference that proponents of angular momentum conservation want to use is the same point around which the comet is circling, except now the radius is diminished. So it is not an issue of position is space it is only that proponents don?t want to use the real radius for their angular momentum calculations. And no wonder they don?t want to use the real radius because that will prove that angular momentum conservation is false.
Proponents could build a cylinder and spheres device and attempt to prove that angular momentum conservation overrides linear momentum conservation, the spheres can have only one velocity in the open position. Only 2 times the original velocity satisfies angular momentum conservation, but it will take 4 times the original velocity to satisfy linear momentum conservation. This is in a 3 to 1 mass relationship between the cylinder and spheres. The experiment should be awarded the Nobel Prize in Physics.
When an object rotating on the end of a string wraps around a thick center post it does not have a change in linear velocity. If a one kilogram object moving one meter per second starts wrapping its one meter string around a stationary post it will be moving at exactly the same speed when the string shortens to .5 meters or .25 meter. This means that angular momentum is diminishing. Also; if the circling object is unwrapping from a post there will be no change in linear velocity, you could release the circling object and direct it to move in the same direction with the same linear momentum at 10 meters radius as well as at 2.5 meters radius.
Proponents of angular momentum conservation will use the frame of reference notion to keep the radius the same. But by doing this they basically ignore the experiment.
Real scientists don?t ignore reality: the radius is really changing and linear velocity is not. The best way to deal with this reality is to accept Newton?s view; (mv) radius has nothing to do with momentum.
I can change the string length on the cylinder and spheres and it will not change the final maximum velocity of the spheres (according to my measurements). Newtonian Physics works: no frame of reference is necessary.
The reason that this lengthy discussion is necessary is that angular momentum conservation is often used in an attempt to negate the cylinder and spheres experiment, but angular momentum conservation is false in the lab. And there is no such law as the Law of Conservation of Kinetic Energy.
The cylinder and spheres experiment transfers the motion of four units of mass to one unit of mass, by doing this the experiment guarantees that at least one or more of these formulas is false; mv, ? mv?, M Ãâ€° R?. The spheres can not have two velocities.

Two spheres can be given the same velocity with different initial quantities of kinetic energy; this proves that the Law of Conservation of Energy is false. This phenomenon, which is found in the cylinder and spheres experiment, assures us that energy can be made from the gravitational field.
Mechanical arms release the spheres and cylinder at a particular point in the 148 rpm rotation. By adding mass to the cylinder in the form of a 3 inch I.D. PVC pipe the cylinder is stopped by the unwinding spheres just as the tethers of the spheres are at 90? to a tangent line at the surface of the cylinder. If the 3 inch pipe is a few grams too light or a few grams to heavy this 90? stop will not occur.
This 3 inch pipe (366g) is then replaced with a 4 inch I.D. pipe (282g), when spun the pipes have about equal Newtonian momentum and the 90? stop is maintained. The kinetic energy difference between the spinning 3 and 4 inch pipe is about 32%.
The 90? stop of the cylinder can only be maintained if the velocity of the spheres remains constant at that point; the stop is sensitive to momentum changes of the cylinder but is indifferent to the cylinder?s kinetic energy.
The only form of energy that the spinning cylinder and spheres has is motion energy, Kinetic energy is not conserved and there is no way to pretend that energy is coming from some other source; consequently the Law of Conservation of Energy is false.
The top of this cylinder and spheres experiment is a 3.5 in. I.D. 208.4g PVC pipe coupler with two 66g spheres seated in its surface. The top has a short slice of a 3 in. I.D. pipe inside the coupler which has a mass of 22.2g; this helps seat the spheres. Both the PVC pipe and the coupler have a ? inch wall. There is a 7g plastic strip, for the center hole of the tether string, which crosses a diameter of the coupler. The top can be connected to different lengths and diameters of PVC pipe.
Without added mass the spheres stop the top cylinder before the tether lines reach 90? to a tangent line at the surface of the cylinder. Only by adding mass to the top cylinder can you force the spheres to stop the cylinder when the spheres are at 90? (the tether and spheres are straight out from the cylinder) to tangent. Mass is added by placing a straight pipe in the bottom of the top coupler. A slit is placed in the cylinder behind the entry hole of the tether string to allow you to evaluate the motion of the cylinder at that point; the cylinder will maintain the same rate of motion for about 1/10 of a second as the string move across the slit. One and only one length (mass) of pipe makes the spheres stop the cylinder at 90?, less mass and it stops before 90? and the cylinder is being force backward while the string is in the slit at 90?. Too much added mass and the cylinder will still be moving forward while the string is in the slit, because the spheres are unable to stop the cylinder.
For this particular tether length and cylinder arrangement the added mass (mass that is added to the bottom of the top cylinder) that stops the cylinder at 90? is about 366g of a 3 inch I.D. PVC pipe.
You can then replace the added 3 inch PVC pipe with a larger diameter 4 inch I.D. PVC pipe; you can make calculations and construct a system that keeps the spinning momentum of the added mass (4 inch pipe) the same. But if the momentum is kept the same the kinetic energy of the spinning larger diameter pipe is not the same.
So if the entire system maintains its 90? stop it is reasonable to assume that nothing has changed and that the momentum of the system is being conserved since the input momentum had not change and the apparent output had not changed either.
Now; one could claim that the 90? stop still occurs but that the spheres are moving faster or slower, but the spheres get to the same place, and the spheres are stopping the cylinder at 90?, and they are doing all this in the same period of time. Velocity must be the same. And if velocity is the same then The Law of Conservation of Momentum would be true. You simply can?t pretend that both momentum and energy are conserved.
This discrepancy between momentum and kinetic energy exists in ballistic pendulums, there kinetic energy is assigned an imaginary friend (heat) to hide the formula?s (1/2 mv?) conservation problem.
In the cylinder and spheres experiment it is totally impossible to have this phantom friend of heat come to the rescue, because kinetic energy is increasing not decreasing.
Here again is the main point.
Two 66g spheres are embedded in the surface of a 4 inch O.D. PVC pipe coupler. The spheres are connected to the cylinder with a tether that can unwrap from the cylinder. As the tether unwraps all the motion of the spinning system is transferred to the spheres and the cylinder stops rotating. The spheres are at 180? around the cylinder.
The original momentum of this model is proportional to 208.4g * 3.75 (top cylinder coupler) + 22.2g * 3.25 (seating pipe) + 366g * 3.25 (mass added 3 in I.D. pipe) + 132 * 4.28 (spheres seated on the surface of the cylinder) + 7g (plastic strip for center hole of the tether string) * 1.5 = 2619. The final momentum would be the same and is held only by the sphere because the cylinder is stopped 2619 / 132 = 19.84. The final velocity of the spheres has increased 4.62 times
The original kinetic energy of this model is proportional to ? * 208.4g * 3.75 * 3.75 (top cylinder) + ? * 22.2g * 3.25 * 3.25 (seat pipe) + ? * 366g * 3.25 * 3.25 (mass added pipe) + ? * 132 * 4.28 * 4.28 (spheres seated on the surface of the cylinder) + ? * 7 * 2.0 * 2.0 (the center of kinetic energy of the rotating strip is closer to 2.0 in. than 1.5 in.) = 4738.1. The final velocity of the spheres has already been determined by momentum conservation and the increase is proportional to 19.84/4.28. Therefore the final kinetic energy is ? * 132g *19.84 * 19.84 = 25,979 = 548 % 25,979 / 4738.1
After being spun and released the top cylinder is forced into a stop when the spheres swing out to 90?. This is accomplished when a mass of 366g is added to the top cylinder in the form of a 3.0 in. I.D.(.25 in. side wall) PVC pipe placed in the bottom portion of the coupler. There is one and only one mass that will cause a perfect 90? (to tangent) stop.
I then replaced the 3.0 in. pipe with a 4.0 I.D. PVC pipe. I calculated its relative velocity (at the same RPS) to be about 4.25/3.25 that of the 3.0 pipe. Therefore to maintain the same momentum its mass will be 3.25/4.25 time 366g, which is 280g. I then placed a 282g pipe on the bottom of the top cylinder and began video taping it while spinning and releasing it from the mechanical arms. The cylinder stops very nicely just as the spheres achieve 90? to tangent and the string is then entering the slit.
To conserve kinetic energy the 4.0 in. pipe would have to have a mass of 213g, about 30% off. I see no way that kinetic energy is being conserved in this experiment. That makes the Law of Conservation of Energy false.
My original estimations of a 350% energy increase were based upon attempts to actually measure the velocity of the cylinder with embedded spheres (using a mechanical release) and then measure the spheres alone at the 90? stop. I used strobe light photograph, video tapes, and photo gates. This different diameters of added mass method is an entirely different and second method that confirms that the cylinder and spheres conserves momentum and can increase energy by 400% or more. This will be the driving force of perpetual motion machines.
These 90? stop experiments prove that the quantity of motion lost by the cylinder is momentum. Newton?s Three Laws of Motion require that the momentum lost by the cylinder must be gained by the spheres, because the force in the tether string must be equal in both directions.
The cylinder stops with the spheres at 90? with any RPS used if the proper amount of mass at a certain diameter has been added to the bottom of the top coupler. Two proper amounts of mass at a certain diameter have been used to stop the top cylinder when the spheres are at 90? to tangent.
Once the appropriate mass is determined the cylinder will stop with any initial RPS given, the mechanical release spins at 2.5 RPS.
A 3.0 inch inside diameter PVC pipe with a length just under 10 inches and with a mass of 366g gives the cylinder a nice stop when the spheres are at 90?. The effective rotational mass of the cylinder is at about a 3.25 in. diameter. Here is the math used for the quantities of motion mentioned.
Mass: 366g
Momentum: (mv) 366g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .2373
Kinetic energy: (1/2mv?) .5 * 366g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .07692
A 4.0 inch inside diameter PVC pipe with a mass of 282g gives the cylinder a nice stop when the spheres are at 90?. The effect rotational mass of the cylinder is at about 4.25 in. Here is the math used for the quantities of motion mentioned.
Mass: 282g
Momentum: (mv) 282g * 1kg/1000g* 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .2391
Kinetic energy: (1/2mv?) .5 * 282g * 1kg/1000g * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .10135
Momentum changed from the 3.0 pipe was .2373/.2391 = 1%
The kinetic energy change from the 3.0 in. pipe was .10135/ .07692 = 32%
When the tether achieves 90? it enters a slit, if the cylinder is stopped it remains stopped as the tether string crosses the slit. If the spheres stop the cylinder before 90? they will have started moving the cylinder backward and the cylinder will be moving backward while the string is in the slit. If the spheres have not yet stopped the cylinder before they reach 90? then the cylinder will be moving forward while the string is in the slit.
The cylinder will be stopped only if it has a certain mass, add 6g and the cylinder will still be moving forward in the slit, subtract 6g from the proper mass and the cylinder will be moving backward. It is accurate within 2% of the added mass. Momentum conservation falls within this 2%, kinetic energy conservation falls outside the 2% at 32%. Kinetic energy is not conserved by the cylinder and spheres experiment.
If the momentum of the cylinder and spheres is conserved in the motion of the spheres alone (when the cylinder is stopped) then the systems has above a 400% increase in energy.
Momentum: (mv) 208.4g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .1559
Momentum: (mv) 22.2 g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .01439
Momentum: (mv) 366g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .2373
Momentum: (mv) 132g * 1kg/1000g* 4.28in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .1127
Momentum: (mv) 7g * 1kg/1000g* 1.5in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .002095
Total initial momentum = .522385
Final momentum will be .522385 for a velocity of 3.9575 m/sec: .522385/ 132g spheres because the spheres have all the motion.
Kinetic energy: (1/2mv?) .5 * 208.4g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .05831
Kinetic energy: (1/2mv?) .5 * 22.2g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .0046659
Kinetic energy: (1/2mv?) .5 * 366g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .07692
Kinetic energy: (1/2mv?) .5 * 132g * 1kg/1000g * 4.28in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 4.28in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .048114717
Kinetic energy: (1/2mv?) .5 * 7g * 1kg/1000g * 2.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 2.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .0005572
Total Initial Kinetic energy = .1886
Final Kinetic energy will be 132g times .5 * 3.9575 * 3.9575 = 1.03376 / .1886 = an increase to 548%

I replaced the 366 gram 3.0 inch inside diameter PVC pipe (1/4 inch side wall) with a 4.5 inch inside diameter slice of a PVC pipe coupler (with a 9/32 inch side wall) that had a mass of 249.2 grams.
After being spun and released (as described in previous posts) the top cylinder behaved in exactly the same manner with the 249 gram coupler as when the 366 gram pipe was in position. The cylinder made a beautiful stop just as the spheres reached full extension and the string was entering the slot in the top cylinder.
The same results were achieved because the 3.0 inch 366g pipe and the 249.2 gram 4.5 inch pipe have the same Linear Newtonian Momentum while being spun. Linear Newtonian Momentum is what is being conserved.
The kinetic energy of the two spinning pipes is not the same; kinetic energy is about 47% off.
This is also experimental proof that angular momentum conservation in the lab is a false concept. The angular momentum formula is mr?Î¸ were theta is angular motion in radian per sec. Both the 3 inch pipe and the 4.5 inch tube have the same theta when being spun by the mechanical arms; their masses are 366g and 249g as mentioned, Their rotational masses are at a radius of 3.25in./2 and 4.78in./2 respectively. The proportional equivalent of linear momentum is (366 * 3.25/2 * Î¸) / (249 * 4.78/2 * Î¸). Since theta is equal in both experiments, the equation shows that linear momentum has not changed.
Angular momentum is 249 *4.78/2 *4.78/2 * Î¸ / 366* 3.25/2 * 3.25/2 * Î¸ = 1.47, or 147%, or 47% off.
The cylinder stops nicely when the spheres are fully extended and the string is just entering the slot (described in previous posts) when either the 366g 3 inch cylinder is on the bottom or the 249g 4.5 inch cylinder is on the bottom of the top cylinder, with spheres attached. The spheres? motion identifies the 3 and 4.5 inch cylinders as being identical.
The only thing identical about the 3 inch and 4.5 inch cylinder is that they have the same linear Newtonian momentum when spun.
The 3 inch and 4.5 inch cylinder do not have the same Kinetic Energy when spun at the same rate. The cylinder and spheres experiment does not conserve Kinetic Energy, and the Law of Conservation of Energy is false.
The 3 inch and 4.5 inch cylinder do not have the same angular momentum when spun at the same rate. The cylinder and spheres experiment does not conserve angular momentum; the concept of angular momentum conservation (in the laboratory) is false.

@pequaide:
It seems to me that you are neglecting the different moments of inertia for a cylinder and a pendulum. Your linear force equations must be replaced by the angular torque equivalents to properly model the system. The rotational kinetic energy is what you need to calculate; not isolated, linear KE components.
E_{rot} = 1/2 I omega^{2}
zt

Why do you think that I am looking for a formula that works? I have a formula that works; mv.
The spheres? motion is very sensitive to added or subtracted mass, you can?t miss calculate the added or subtracted quantity without getting a distinct change in the behavior of the spheres.
Mass was incrementally added to the 3 inch pipe until the spheres stop the spinning system when the spheres were exactly at full extension. If you add too much mass the spheres can not stop the spinning cylinder, you must have exactly the correct mass.
I then calculated the correct mass (length) of the 4.5 inch I.D. coupler that would give you the exact same linear Newtonian momentum; and hit the correct mass (cut the coupler the right length) first time, dead on.
I had previously used the same (mv) formula to calculate the correct mass (that mass that would make the cylinder stop just as the spheres were fully extended) of a 4 inch I.D. pipe, and I had cut the pipe a few grams short (where upon I added about 4 grams). The spheres behaved the same way for all three added pipes.
Your formula (Erot = 1/2 I omega2 ) will not work because Newton?s formula does work. I am sure you won?t claim that (Erot = 1/2 I omega2 ) = mv.
This is a bona fide experiment with a formula that works not a philosophical conjecture.

pequaide:
It is clear that you do not fully grasp the concept of angular momentum.
The formulae and equations of angular momentum can be derived from their linear counterparts in a straightforward way. The benefit of using the angular formulation is that one does not have to add up a large number of individual linear momenta.
Let's apply this to the example you gave earlier:
Wrap a string around a 5 kilogram rim mass pulley; suspend a 1 kilogram mass from the string as in a one suspended mass Atwood machine. This is one kilogram accelerating 6 kilograms. Acceleration equals 1/6 * 9.81 m/sec/sec. If you let the mass drop one meter the entire system will be moving 1.808 m/sec.
Only this time, instead of a pulley with all of its mass concentrated on the rim (which is rather unphysical), lets substitute a pulley with a uniform mass distribution.
To proceed as you did in the example, we would have to sum the linear momenta of each infinitessimal mass element, for each ring of mass in the pulley, for the duration of the applied force at the rim. This can be done, and once the calculus dust settles, you will find that you have obtained the formulation for the behavior of the system in the language of (*horror*) angular momentum. In fact, that is how we got these equations to begin with.
Instead of going through a process of reinventing the wheel every time, we notice that a general rule can be applied to these kind of systems. It turns out that (in the absence of friction and the like), the total angular momentum is conserved. This makes calculating the behavior of these systems much easier, provided we know or can calculate the distribution of mass in the system.
The systems you cite as exceptions to the rule appear at first glance to violate the conservation law, but upon closer inspection, they all involve varying moments of inertia, and can be modeled perfectly well using the concept of angular momentum.
While there is nothing wrong with using a linear approach to tackling these kind of problems, it is often far easier to use the angular analogs.
Your assertion that angular momentum is not conserved, while linear momentum is conserved demonstrates a deep misunderstanding of the underlying physics: The conservation law for angular momentum can be directly derived from the linear equations.
Of course, when we venture outside of the bounds of Newtonian physics, these concepts need not apply  but I don't think we need to resort to relativity or QM or any other theory to model your machine good old fashioned classical mechanics will do just fine.

Well said.
But there is one major problem with angular momentum conservation; it does not work (in the lab). Linear momentum conservation does work, it?s a pragmatic issue.
My text books give me the formula of mr?Î¸ for angular momentum. If I were to calculate the mass (cut to length) of the 4.5 inch coupler from the mass (that is stopped at full extension by the spheres) of the 3.0 inch pipe I would have cut it to: 366g * 3.25in. * 3.25in. * 1 radian/sec = 3865.9; 3865.9 / 4.78 / 4.78 / 1 radian/sec = 169 grams.
At 169 grams for the 4.5 inch I.D. coupler (added to the bottom of the top cylinder) the spheres would not have stopped the combined cylinders exactly at full extension. The combined cylinder system would have stopped well before full extension of the spheres; and the cylinder would be moving backwards while the string is in the slot.
But the cylinder made a nice clean stop at full extension using the 249 gram length of 4.5 in. I.D. coupler. This is linear Newtonian momentum conservation.
What people assume and what really works are often two different things.
Early physics was founded upon experimentation, now it seems science just assumes everything.

A pulley can change the direction of a force but it does not change the quantity of force. A 12 newton clockwise tangent force at any position on the circle of the pulley can be balanced with a 12 N counterclockwise tangent force at any position on the circle. Six 2 newton clockwise tangent forces at any position on the circle can be balanced with a 12 N counterclockwise tangent force at any position on the circle.
A two kilogram mass moving 1 m/sec clockwise in a circle will need two newtons of force to be applied for one second counterclockwise in a tangent direction to get it to stop. The mass can move around the circle as the force is being applied, and the applied force can move around the circle as well because the position of the mass or the position of the applied tangent force is unimportant. Note: that the size of the circle with a center bearing is unimportant, and a balanced wheel could be mounted vertically or horizontally.
This means that a moving wheel can apply its motion to an object in a linear path without loss of motion to the system, and vise versa.
This means that a massive large diameter moving wheel can apply its motion to a small diameter light wheel without loss of motion to the system. The moving mass of the cylinder in the cylinder and spheres experiment could be contained in the motion of a separate larger wheel; it need not have the same radius as that of the spinning spheres. In fact; linear motion could be used for the moving mass of the cylinder.
The larger wheel or linear motion being used (instead of the same radius that the spheres use) should eliminate the concept of angular momentum conservation. But more importantly it opens a vast array of possible arrangements for the cylinder and spheres experiment, so much so that we might have to call them momentum (linear) consolidation experiments.
So let us review the simplicity of this energy making machine and see if Newtonian Physics has been violated.
A one kilogram mass moving five meters per second collides with a four kilogram mass at rest. The combination proceeds at 1 meter per second. The combined mass is then caught on the end of a string (of any length) at 90?, the mass will proceed around the circumference of the circle at one meter per second. The circle then acts as a cylinder and spheres experiment and throws all the motion back into one of the five kilograms. If the original motion is to be conserved as Newton would predict the motion must now again be five meter per second.
Momentum is conserved so the event in the proceeding paragraph should sound logical to you, now calculate the energy changes. E = 1/2mv?
My data confirms Newton?s predictions, and the Law of Conservation of Energy is false.

Hello pequaide, I?m very impressed with your experiments and knowledgment.. can you send me your experiment video please ?!
my email is felipe.bit@gmail.com I would like if you can add me on msn messenger.. my contact is felipexz@hotmail.com
Congratulations for your great work
Thank you
Felipe

I think we have narrowed our inability to place video on the internet to DVD formatting problems. I will probably buy a new DVD recorder next week. And one of my constituent uses hotmail so I will ask him about that. Thanks

spyblue: I went to the computer store and told them about not being able to post videos (on the internet) from the DVDs my recorder makes. They said all the DVD recorders (they have) format the same way. They sold me a Dazzle that was supposed to reformat and make a DVD that was computer friendly. I guess my computer is to slow because that did not work either.
Rather than continuing to try to post videos I have decide to go to several less favorable options. I can send free DVDs in the mail; just give me your land address. These DVDs play fine but no one has been able to post from these DVDs. Also; I think I shall take frame by frame pictures of the monitor as the DVD plays and post those. My player has frame by frame capabilities. A picture of a monitor is poor quality but better than nothing I suppose. You can see the cylinder clearly stopped for about two frames, 2/30 sec.

This site would not let me post a picture so I posted over in: free_energy; Files; 81508 cylinder and spheres. This is a picture of a monitor and is a little dark. The cylinder is stopped and appears to be as clear as the background. The cylinder has a mass about four times that of the spheres. The spheres are a blur.

I posted four more pictures on the free_energy site

@pequaide
Which free energy sight?

Type (free_energy) in your search engine. Ideally there is about five times more energy after the spheres open, than if the spheres had remained up against the cylinder.

Once you are on line type: free_energy, in your Google search engine. Choose: free energy claims, then files, then posted by pequaide, or 81908 four frames. The Yahoo engine does not bring you to the same site. But it is a Yahoo group site.

I posted ten more pictures (in order) from a video (under files) on the free_energy site.
If the spheres had been glued to the cylinder the circumference velocity of all the parts (after they were spun and dropped) would have remained about one meter per second. That would be roughly 690 grams moving 1 m/sec.
In actuality the spheres swing out and absorb all the motion. That will give us 133.6 g that now must have (.690 kg * 1 m/sec) .690 units of momentum. .1336 kg * 5.16 m/sec = .690 units of momentum
This is an energy increase of Ke = ? mv?: .5 * .690 kg * 1 m/sec * 1 m/sec = .345 joules to .5 * .1336 * 5.16 * 5.16 = 1.778 joules. 515%
You can make a slideshow of the ten frames, energy has been made in the lab.

See the picture labeled 82508 drawing, in free_energy, under files, gravity wheel
Only mass A is under gravitational acceleration. Gravity will exert a force of 9.81 newtons upon mass A and then, by design, upon the entire system.
All the parts of the system are moving at approximately the same speed. Therefore it is obvious that the acceleration of all the parts is the same. The total mass of the system is ten kilograms. Since F = ma; 9.81 N = 10 kg * a, for an acceleration of .981 m/sec?.
d = ? at? or âˆš(d * 2 / a) = t: So mass A will travel a distance of one meter in 1.4278 seconds.
d = ? at?, or d = ? v?/a because t = v/a, or âˆš(d * 2 * a) = v: So after mass A travels one meter the whole system (10 kg) will have a velocity of 1.4007 m/sec. This is 14.007 units of momentum or 9.81 newtons applied for 1.4278 seconds.
To get mass B to stop you must apply 9.81 newtons for 1.4278 seconds or the equivalent. In 1.4278 seconds mass B can travel straight up 10 meters.

Hi,
if you have a DVD,you can easily use the freeware program:
VOB2MPG
http://software.badgerit.com/VOB2MPG.html
to extract the whole DVD to single MPEG files.
Then you can upload these to youtube or
use
Virtualdubmod
http://virtualdubmod.sourceforge.net/
to recompress them to smaller AVIs
with DIVX.com codec for instance and
MP3 audio.
Hope this helps.
Let usknow, when you have uploaded some videos
to youtube.com or anywhere else.
Many thanks.

Thanks; hartiberlin, I will work on it. I am also repeating Laithwaite's figure 7 experiments so I would like to get that out there also.
Confirmation of the data from the cylinder and spheres experiment has been obtained from the LaithewaiteÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s Mass Displacement experiments
Similar experiments to the cylinder and spheres have been performed by Prof. E.R Laithwaithe at the University of Sussex. I found it under Gyroscopes ÃƒÂ¢Ã¢â€šÂ¬Ã¢â‚¬Å“ Everything you need to know, after searching ÃƒÂ¢Ã¢â€šÂ¬Ã‹Å“Laithwaithe free axisÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢.
Of particular interest is figure 7, elsewhere Laithewaite states that momentum is conserved in his experiments but energy is not conserved. I have performed experiments very similar to his, but the cylinder and spheres yield more energy. LaithewaiteÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s data confirms the concepts behind the cylinder and spheres, because energy has been made in the laboratory, in the United Kingdom.
To find LaithwaiteÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s energy producing experiment click on the link (www.gyroscopes.org/masstran.asp  22k) and go down to Mass Displacement by Circular Motion. Figure seven is an overhead view (I assume) of three objects on a frictionless plane.
Lets give the object A1A2 (the sliding center mass, Laithwaite calls it the centre pivot anchor block) a mass of 2 kg and M1 and M2 a mass of 1 kilogram each.
Lets swing M1 and M2 down from .2039 meters so they both have a velocity of 2 m/sec. d = Ãƒâ€šÃ‚Â½*a*t*t or d = .5 * v * v / a
In figure seven that would give us one kilogram going north at 2 m/sec and one kilogram going south at 2 m/sec. They both transfer some of their motion to the center sled (A1, A2). Let us assume that at some point they are all moving at the same velocity, which would mean that all three objects would be moving one meter per second. This would be the conservation of linear Newtonian momentum; 4 kg * 1 m/sec equals 2 kg *2 m/sec.
In LaithwaiteÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s figure 7 he starts at a high energy point and goes to a low energy point and then back to a high energy point. At first M1 and M2 have all the motion after they are accelerated by a spring, this is a high energy point. Then the motion is shared with the center sliding mass that contains A1 and A2, this is the low energy point. The lowest energy point is obtained when the velocity of the center mass is closest to the velocity of M1 and M2. The energy is highest again when all the motion returns to M1 and M2.
Lets start the motion at the low energy point by dropping the sliding center(A1 and A2) mass and M1 and M2 .051 meters. Arrange that the masses go in the correct direction of course. Let the sliding mass be equal to the sum of M1 and M2. The original velocity of all masses will be one meter per second. The original energy will be 2 joules.
After the system proceeds from the low energy point to the high energy point M1 and M2 will have all the motion. They will have to be moving 2 m/sec to conserve momentum and the energy content will be 4 joules. With velocities of 2 m/sec M1 and M2 can rise .2034 meters: 2 kg at .2034 m is twice the energy of 4 kg at .051 meters.
Now you may ask: how do you know that linear Newtonian momentum is conserved; because the objects are moving in a circular path?
First: Well what if you arrange a head on collision between M1 (moving 2 m/sec) and the center sled at rest, the three kilograms would move away at .667 m/sec. Then if you (in line) collide M2 at 2 m/sec into the combination you would have four kilograms moving one meter per second. Why would you expect different results from the arrangement in figure seven? Would anyone claim that Newtonian physics does not apply to objects moving in a circular path?
Second: The arrangement in figure seven returns to having only M1 and M2 in motion and the center sled is stopped. Ideally that motion should again be 2 m/sec. How can you have 4 units of momentum at the end of the experiment unless you have 4 units of momentum all the way in between; from start to finish?
Third: Laithwaite said that he observed that momentum was conserved not kinetic energy. If kinetic energy was conserved when the 3 objects are moving at the same velocity then that velocity would have to be 1.4 m/sec. This would mean that 4 units of momentum would give 5.6 units and 5.6 units would yield only 4: a clear violation of Newtonian physics. From mv and 1/2mvÃƒâ€šÃ‚Â²
Here is the importance of this experiment. Ballistic pendulums conserve linear Newtonian momentum. They conserve it when the incoming projectile is a pendulum bob and the final motion of the block projectile combination is linear, on a frictionless plane. They conserve it when the incoming motion is linear and the final motion is a pendulum. They conserve it when both incoming and outgoing motion is a pendulum. Depending on the mass distribution of the projectile to block they can lose 50%, 80%, or 95% etc. of the energy of motion. Always this energy loss is blamed on friction that makes heat. How can heat be blamed on the energy losses in the sliding center experiment when the objects donÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢t even touch?
If heat is to blame for the loss of energy (when the motion is shared) then how does the heat come back when the energy is restored; when the motion is again only in M1 and M2?
Laithwaite pays special interest in the fact that the center of mass does not move. This would mean that when M1 and M2 are directly above and below the center sled and the total mass of both M1 and M2 equal the mass of the center sled, that their velocities must be equal. Because half the mass is moving to the right and half the mass is moving to the left, they both must be moving at the same velocity in order that the center of mass holds its position.
If you have access to a frictionless plane LaithwaiteÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s experiment should be easy to perform. Are we scientists enough to repeat it?
I have repeated LaithwaiteÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s figure seven experiments and it appears that he is correct in that the center of mass seems to stay in the same place. In a since this is also the center of momentum, in that if the center sled has twice the mass (of M1 and M2 combined) it has half the velocity (when they are all three in a straight line). This would mean that M1 and M2 always give the center sled half the momentum no matter what mass the sled has.
When M1 and the center sled (with a mass twice that of the combined mass of M1 and M2) and M2 are in a straight line the sled must have half the velocity. If M1 and M2 move 1 cm left then the center sled must move .5 cm right. It seems like this would mean that the experiment proceeds at about the same rate (given the same original velocities) no matter what the mass of the center sled. Because if the M1 and M2 pucks always give half their motion to the center they will always have half left for themselves. But the shape of the oval changes with changing center mass, and maybe that would change the rate at which the experiment proceeds. At any rate this is a very interesting experiment.
If the center of mass stays in position this allows for different quantities of energy to be produced. Because a center mass of 4 kg moving .5 m/sec is not the same energy as 2 kg moving 1 m/sec. In fact if kinetic energy would be conserved the center of mass could not remain is position, because the velocities that conserve momentum retain the center of massÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s position, any other velocities would not move the appropriate quantities of mass to the appropriate position necessary to maintain the position of the center of mass.

See http://groups.yahoo.com/group/free_energy/ then go to files, then to pictures titled 92708 LaithwaiteÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s figure 7.
3 photographs of LaithwaiteÃƒÂ¢Ã¢â€šÂ¬Ã¢â€žÂ¢s Figure 7
From position one (photograph one) a small double lever is used to accelerate the pucks on the end of the string; one is accelerated north and one south (relative directions). The center puck is initially at rest and has a mass of 76g; the end pucks have a mass of about 32 grams each. That means the center of mass is closer to the center puck than to the end pucks.
As soon as the end pucks begin moving north and south the center puck begins moving right. When the center puck reaches the center of mass (which is about the center of the table) the end pucks are in the positions of photograph two, directly above and below the center puck. If the center of mass is to remain in position as Laithwaite stated then the velocity of the end pucks will be moving with half their original velocity, and the center puck will be moving with 64/76 *1/2 the original velocity of the end pucks. These velocities conserve linear Newtonian momentum, and the velocities also conserve the position of the center of mass. This is what the experimenter will observe upon doing the experiment.
All this motion is then returned to the end pucks upon moving from photograph 2 to photograph 3. In photograph three the center puck (now on the right side) is again at rest. The energy change from photograph 2 to photograph 3 is 217%. Start the motion in the center (photograph 2) and you can make energy.

Hi Pequaide,
I find your writings and pics interesting. Hans got me looking at your posts.
Yahoo Tech groups is a low memory messy site.
You should post on somewhere better like Besslerwheel.com
At least your posts will sit together better. And the images can be posted there.
Just friendly suggestions,
Damian

If you can put them to a DVD you can place them on a computer...
Simply put.
If you do not have one already then buy a dvd burner off the internet affordable for anyone now!: http://www.newegg.com/Store/SubCategory.aspx?SubCategory=5&name=CDDVDBurners
First download DVDFabHDDecrypter FREE: http://www.dvdfab.com/free.htm
Then download autogk: http://www.autogk.me.uk/
Encode the video using Xvid Quality and encode at the maximal amount that youtube will allow. http://www.youtube.com
Viola you have video for the masses!
The instuctions are simple if you can and have been posting to this forum it should take you no longer then one day to accomplish this and finally get your message out to people in video form.
If you do not have highspeed internet which may be the case as I have not read the whole thread send a copy to someone and have them do it for you or do the encoding and have a friend do it for you this would be the fastest.
infringer

Thanks: DrWhat, infringer and hartiberlin, all good advice. I will work on it.

You may wish to view BesslerWheel.com, energy producting experiments, pequaide. There are some interesting developments there.

Hello pequaide,
can you please be more precise as this link is just the mainentry. What section are you refering to ?
Regards
Kator01

Sorry; when you use favorites you forget, all the sub groups. It would be BesslerWheel.com, (in my search it was the top choice) after that opens go to discussion groups, then general discussion, then â€˜energy producing experimentsâ€™ (author pequaide).
Of specific interest to me is NASAâ€™s yoyo despin device. I knew this had been proposed to NASA but I did not know that they actually used it. Also of interests is that they could not get the math right, and adequate performance evaded them, and thus they dropped the procedure from their list of things to do. (my interpretation)
And possibly of interest to others is that I am shifting to more massive machines with bearings. To produce useful quantities of energy you will have to make use of bearings.
Thanks for your interest.

This is a 3400 grams spinning wheel that can be stopped with a 456 grams disk as it swings out on the white string. The white disk has a mass well over 200 g (guessing) but every thing stops spinning just as soon as the gray disk swings out. What form of motion is conserved?

I don't quite understand. Can you post a video of the action?

This is a 3400 grams spinning wheel that can be stopped with a 456 grams disk as it swings out on the white string. The white disk has a mass well over 200 g (guessing) but every thing stops spinning just as soon as the gray disk swings out. What form of motion is conserved?
To model the dynamics of this kind of setup, one can start with the wellverified assumption that total angular momentum is conserved. Since the small mass changes its position, the moment of inertia varies as a function of time. The resulting differential equations are therefore nonlinear, so numerical approximation methods (stepwise simulation) would probably work best for modeling this system's dynamics in detail. Otherwise, just looking at the total angular momentum at the start and at the end should yield a rough equality, (with a small amount wasted to friction and such...)

I think the size of the drive wheel eliminates the possibility of angular momentum conservation.
First; when the small gray disk has stopped the white disk and drive wheel it is not rotating around the center bearings of the white disk, it is (instantaneously) rotating around a point on the circumference of the (stopped) white disk. This may be hard to envision so I will allow you to believe that the gray disk is rotating about the center bearing of the white disk, which is impossible because the white disk is stopped.
The white disk has a radius of 3.5 inches. The gray disk has swung or flared out about 3.5 inches (from previous experiments and visual observation) when the white disk and drive wheel are stopped. This gives you about 7.0 inches for the radius of the flared out gray disk and the radius of the larger drive wheel is 9 inches. The effective rotational inertia of the wheel is about 30% inside its circumference so the radius of rotation of the drive wheel and the flared out gray mass are roughly similar. This should make this arrangement behave very differently than when all the mass is in the white disk, an experiment which has been done many many times. But the behavior of the two experiments is the same. Moderate motion of the wheel is turned into violent motion in the gray disk; the gray disk pulls loose from the white disk and slams into the back wall.
Let me ask you two questions. What if I replace the 18 inch drive wheel with a 1 meter drive wheel that has it mass concentrated near the rim?
What happens if I replace the 18 inch drive wheel with a similar mass sled on a bearing track?

Hello pequaide,
this is an interesting setup.
Just one question : The drivewheel is the vertical running wheel on the right side ?
And motion is transferred to the white plasticwheel via the red rope ? So the axis of both wheels are perpendicular to each other, is that so ?
Now if you have this experiment that far it should be not to complicated to establish a speedmeter for measuring
the gray disc and calculate its kinetic energy.
If I would be in your position I would not argue with others about angularmomentum but concentrate on speedmeasurements.
Good luck
Kator01

It is red ribbon; and you are correct on other points.

Sometimes I have access to my computer but I am not in the lab, so let me continue to argue against angular momentum conservation (in the lab).
An average size new construction home could be installed with a seven meter rim mass wheel. The cylinder or white disk could be .25 meters (dia.). So, according to the angular momentum conservationists, the thrown gray disk could then be given all the â€œangular momentumâ€ of the seven meter rim. Wow, you could launch bullets.
The angular momentum formula states that a 100 kg rim with a circumference velocity of 1 m/sec and having a radius of 3.5 meters will have seven times as much angular momentum as a 100 kg rim with a circumference velocity of 1 m/sec and having a radius of .5 meters.
Angular momentum conservation would require this sequence of events. When all the motion of the seven meter rim has been given to the gray disk (corresponding to the above pictured experiment) in paragraph two and the disk has a mass 1/10th that of the seven meter rim, then the gray disk will have to have a velocity 140 times larger than the original velocity of the seven meter rim.
I am expecting little league baseball speeds not bullet speeds, but ha who am I to complain.
As far as collecting data from the above experiment, that is always the objective of a good experiment. But the data has already been collected. There is a substantial quantity of data collected from the cylinder and spheres. The experiment with the detached wheel and connecting ribbon only shows that the rotating mass need not be contained in the cylinder, and of course that the radius of the rotating mass has no effect upon the ultimate velocity of the final mass in motion. All data collected from the cylinder and spheres experiment shows that linear Newtonian momentum is conserved and that energy has been made in the laboratory. The external wheel and red ribbon experiment points to even more useful applications.

I don't quite understand. Can you post a video of the action?
Yes, please post a video or more pictures from each step by step
and post a better planation what is going on.
This needs to be documented on video.
Many thanks.

Drive wheel, White disk with embedded gray disk, will explain later.

In the picture of the white disk you will see a white nylon cord attached to the gray disk. The cord was fastened using a slip knot and secured to the gray puck with gray duck tape. As the system begins spinning the gray puck tries to leave the system on a line tangent to the white disk. The white nylon cord tries to prevent the gray disk from leaving the system and soon all the motion is in the gray disk.
The experiment is arranged so that the spinning motion of the 18 inch drive wheel is transferred to the white disk on the table. The red ribbon combines the momentum of the drive wheel to the momentum of the white disk. The white disk is horizontally mounted and has a mass of 485 grams. The velocity at the circumference times its mass (which is distributed throughout the disk) gives it a rotational momentum. This momentum of the white disk is about equivalent to 300 grams that were located on the circumference. The rotational momentum of the 3,400g drive wheel is roughly equivalent to a 2,200g rim times its circumference velocity. The center of mass of the 456g gray disk is on the circumference of the white disk so its rotational momentum is roughly equal to 456g times the circumference velocity.
Letâ€™s use 1 m/sec for the ribbon velocity; this gives you 1 m/sec circumference velocity for the wheel and white disk and 1 m/sec velocity for the center of mass of the gray disk.
This gives you an initial linear Newtonian momentum of 2.956 newtons (2,200g + 300g + 456g * 1m/sec)
This gives you an initial kinetic energy of 1.478 joules (1/2 *(2.200kg + .300kg +.456kg) * 1m/sec * 1m/sec)
This gives you and initial angular momentum of .5701 (2.200 * .2286 m + .300kg * .0889m + .456kg * .0889m * 1 m/sec)
For conservation of linear Newtonian momentum the gray disk will be moving 6.48 m/sec when it has all the motion.
For conservation kinetic energy the gray disk will be moving 2.54 m/sec when it has all the motion.
For conservation of angular momentum the gray disk will be moving 7.03 m/sec when it has all the motion.
If you used a larger drive wheel angular momentum conservation would give you a final velocity that would be off the charts.
If Newtonian linear momentum is conserved at 6.48 m/sec the final kinetic energy is 9.57 joules.

Correction: a newton is a unit of force not a unit of momentum, Sorry. I have to watch my terms.

Hello pequaide,
I can not watch your experiement without my mind beeing triggered with new ideas for proof of your concept.
The following idea I would like to call : "The magnetandspheres breakawayforceproof".
Referring back to your first cylinder and the spheressetup the following principle could be used for proof of concept.
Let us assume you attach two ringmagnets to the end of the flyawaytethers and attach the steelspheres to the magnets then the accelleration of magnetspheresmass will give you the breakawayforce F = mass x accelleration ( In calculation with the formula M must be the mass of the spheres only ).> accelleration = F / masssphere which will give you the instant velocity at the breakawaypoint. This velocity will be of identical value as the calculated accelleration and at the max value since there will be no additional accelleration after breakloosepoint
By putting a plasticdiskwasher of variable thickness between the sphere and the magnet you can control the breakawayforce which you can test before each run in a static testrig. In this testrig you will measure with a some forcemeter the spherebreakwayforce from the magnet with distancewashers of different thickness.
By increasing the thickness of the washer in 0,3 mm steps ( just a guess ) you will reach a point at which the sphere will break loose which will give you the force and thus accelleration>velocity>energy.
What do you think ?
Regards
Kator

Keep using thicker and thicker washers until the spheres break away exactly when the white disk and drive wheel are stopped. That should be the point where the centrifugal force is the highest. Brilliant; your idea is absolutely brilliant.
Or you could use electromagnets that open at the exact millisecond of your choice. Good thinking Kator. Go for it.

Hello Pequaide,
well, I would love to do it myself but I have no technical means ( mechanic ) to produce the whole setup.
One thing I had in mind was to use a ceilingventilatormotor ( low rot/min) but I have to adapt the mechanics and have to find a way to figure out the inertiamoment of the rotor.
Any way  using electromagnets would make it more complicated because of the wiring and the external triggerelectronics for switching the electromagnets including the powersupply ( batteries ) which must all be mounted on the rotorplane.
I think using permanent magnet would be the simplest solution  although the preliminary breakawayforcetest is more timeconsuming.
Then I was thinking about security.
Question : How much energy is left when the spheres break away ? Is it dangerous ?
My quess is that if you will hit the exact force there will be not much energy left as it is all consumed up in the breakawayprocess.
I even think that you can calculate the force before and than use ( by the prelimninary static test ) the corresponding washermagnetconfiguration for the real test.
I think that even in you present discsetup on the table you can use this idea. There are so many forms of magnets awailable you can find one form which fits your gray disk.
http://www.monstermagnete.de/catalog/index.php?language=en&osCsid=31a1972c8fc03bc8ae5efdcdac512f8f (http://www.monstermagnete.de/catalog/index.php?language=en&osCsid=31a1972c8fc03bc8ae5efdcdac512f8f)
http://www.supermagnete.de/eng/index.php?switch_lang=1 (http://www.supermagnete.de/eng/index.php?switch_lang=1)
Or look at these halfshellmagnets on the long bar ( uppermiddle )
http://www.kundelmagnetics.com/pages/Kundel_Motor_Parts.html (http://www.kundelmagnetics.com/pages/Kundel_Motor_Parts.html)
Unfortunately I will not be able to do it myself.
Regards
Kator01

I switched sides, now the fishing line (the white nylon cord was on the left) is on the right side of the gray puck, and of course the white disk spins the opposite direction.
I changed to a narrower ribbon.
I drilled a hole in the white disk and placed the fishing line through it, this holds the tether to a length of 4 3/32 inches to the center of mass of the gray puck.
I also added 700g at 7.75 inches from the center of the 9 inch radius drive wheel. This is roughly equal to 600 additional grams moving at the same speed as the ribbon. The white disk still appeared to stop; this I think puts us at 7.8 times the mass of the gray puck. This means that the gray puck is capable of finishing with 7.8 times the original energy. Final 1/2 * 456g * 7.8m/sec * 7.8m/sec; original Â½ * 3556g *1m/sec * 1 m/sec

I replaced the drive wheel with a sled of approximately equal inertia. The sled and the white disk were stopped as the gray puck swung out on the end of the tether (fishing line). You just quickly see the stop; it was not quite as distinct as the wheel. I found out why when I measured the mass of the sled. The sled had a mass of around 3.4 kg, much greater than the estimated inertial of the wheel (2.8kg). Video tapes will reveal the true nature of the observed stops. The wheel stop may actually be reversing, or the sled stop may not be quite coming to a complete stop (but it is close).
This should be a fairly good proof that linear motion and circular motion is one and the same thing. How are the angular momentum conservationists going to calculate angular momentum conservation when most of the original momentum is linear?

"How are the angular momentum conservationists going to calculate angular momentum conservation when most of the original momentum is linear?"
Using standard vector calculus and their fingers?
It's a bit of a more complicated problem than is usually found in freshman engineering dynamics textbooks, true. But it reminds me very much of that damn sophomore class I took, once long ago, in a galaxy far far away...
Once all the variables are made constant, and all the constants set to zero, the problem reduces to one already solved.
We just have to figure out which one.
Seriously, momentum conservation calculations can be difficult and misleading, particularly when there is impact between objects of widely differing velocities and/or masses. Throw in rotating elements and it's easy to get misled down the wrong garden trail. If one recalls that momentum IS conserved, at least in this universe, the analysis may become a little easier, because at least you'll know when you made a mistake somewhere.

Good mathematics allows you to make predictions, so knowing the original motion you should be able to predict the final motion.
So tell me the original angular momentum of 3 kg moving 1m/sec in a straight line.

I don't know how you are planning to maneuver me into following your garden path, so I'm not even going to start. I can't see how your device moves, so I don't know how or where straight paths become curved ones, nor do I know the degree of curvature, nor a multitude of other things that are needed for a correct analysis. I'm sure we could arrive at one, with some work. But kinematic analyses are not really my forte, so maybe I shouldn't even have started replying here. But it sounds like you are trying to claim a violation of conservation of momentum, and that's even less likely than a buoyancy drive or a gravity wheel, just on first principles, that much I do know.
If it's not spinning as it translates, it has zero angular momentum, and its linear momentum is mv, or 3 kgmeters/sec. As you well know.
But as soon as it is forced into a curved path, by a centripetal acceleration, it acquires angular momentum. I think.

Newton said F = ma. Newtonâ€™s momentum conservation was linear. You absolutely can not conserve both linear and angular and Newton did enough experiments to know this.
Construct two equal mass vertically mounted rim mass wheels with different radii. If you hang the same mass from each rim the force of the hanging mass will accelerate each rim to the same circumference velocity in the same period of time. This will be linear Newtonian momentum; it is not angular momentum being produced. The angular momentums of the two rims are not equal. F = ma does not make angular momentum, and in fact no such angular momentum is ever conserved in the laboratory.

Hello TinselKoala,
I also had a difficult time to visualize this process.
For better understanding have a look at the pics way back at the pages 9, 12 and 17. Especially the three phases of the spheresmovement until they are fully swung out ( 90 degrees to tangent) are very well shown on page 17.
The only difference is that here instead of the puc steelballs are used and a cylinder instead of the white disc.
The prime mover for the spinnig ot the complete setup is not shown there as this can be done in very different ways.
Regards
Kator01

From Wikipedia:
"Definition
Angular momentum of a particle about a given origin is defined as:
\mathbf{L}=\mathbf{r}\times\mathbf{p}
where:
\mathbf{L} is the angular momentum of the particle,
\mathbf{r} is the position vector of the particle relative to the origin,
\mathbf{p} is the linear momentum of the particle, and
\times\, is the vector cross product.
As seen from the definition, the derived SI units of angular momentum are newton metre seconds (NÂ·mÂ·s or kgÂ·m2s1 or joule seconds). Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the righthand rule."
Well, the math symbols didn't translate, but you can see from the definition on Wikipedia
http://en.wikipedia.org/wiki/Angular_momentum
that angular momentum is defined as the vector cross product of the particle's (or object's) position vector and its LINEAR MOMENTUM.
Linear momentum is conserved; angular momentum is conserved; and the two are related by vector mechanics, just as I have said.
An excellent basic text on these matters is
Statics and Dynamics
Vector Mechanics for Engineers
By Johnston and Beers
(now in its at least 12th edition)
Wherein you will find your problem and many many other similar ones analyzed with excruciating thoroughness.

Newton said F = ma. Newtonâ€™s momentum conservation was linear. You absolutely can not conserve both linear and angular and Newton did enough experiments to know this.
Construct two equal mass vertically mounted rim mass wheels with different radii. If you hang the same mass from each rim the force of the hanging mass will accelerate each rim to the same circumference velocity in the same period of time. This will be linear Newtonian momentum; it is not angular momentum being produced. The angular momentums of the two rims are not equal. F = ma does not make angular momentum, and in fact no such angular momentum is ever conserved in the laboratory.
Sorry, you are wrong about your assertion re the wheels and masses. Please don't make me construct an apparatus to prove it.
You have said nothing about moment arms in your thought experiment. If one wheel's mass is concentrated at the rim, and the other's is at the hub, there will be differences in the circumferential velocity as the linear momentum of the falling mass is coupled to the angular momentum of the wheels. If the (PEKE) of the mass as it leaves the wheels is the same, then the angular momentum of the wheels will be the same, but if the mass distributions are different the circumferential velocities will be differenthence the velocity of the weight when it drops off the wheel will be differenthence the (PEKE) of the weight will be different
you can see how the complexities multiply.
But you may rest assured, momentum, whether angular or linear or some combination, is conserved. There are ample experiments, all around you, that confirm this fact.
Ever ride in an automobile, or an aircraft? Then you can be glad of CofM.

Hello TinselKoala,
I also had a difficult time to visualize this process.
For better understanding have a look at the pics way back at the pages 9, 12 and 17. Especially the three phases of the spheresmovement until they are fully swung out ( 90 degrees to tangent) are very well shown on page 17.
The only difference is that here instead of the puc steelballs are used and a cylinder instead of the white disc.
The prime mover for the spinnig ot the complete setup is not shown there as this can be done in very different ways.
Regards
Kator01
If one picture is worth a thousand words, then one video is worth a thousand still pictures.

TinselKoala quote; As seen from the definition, the derived SI units of angular momentum are newton metre seconds (Nâ€¢mâ€¢s or kgâ€¢m2s1 or joule seconds).Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the righthand rule."
Keplerâ€™s formula worked without all this hocus pocus. A joule is a unit of energy. A â€œpseudovectorâ€ they have got to be joking. Are you sure this isnâ€™t pseudoscience.
Take a 9 kg rim (or ring) that has a 2 meter diameter and place it vertically on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/secÂ². The speed of the string and the speed of the particles in the ring will be roughly equal; you could place the string in a grove in the rim to make the speeds even closer. After one second all parts will be moving .981 m/sec. After one second everything will have moved .4905 m, the dangling mass will have moved down .4905 m and the mass in the ring will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum of the rim would be (mass * radians/sec * radius * radius) 9kg * .981 radians/sec *1m * 1m = 8.829
Take a 9 kg thin wall pipe that has a .1 meter diameter and place its length horizontally so that it rotates in a vertically plane on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/secÂ². The speed of the string and the speed of the particles in the pipe will be roughly equal; you could place the string in a grove in the pipe to make the speeds even closer. After one second all particles will be moving .981 m/sec. After one second everything will have moved .4905 m, the hanging mass will have moved down .4905 m and the mass in the pipe will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum would be (mass * radians/sec * radius * radius) 9kg * 19.62 radians/sec *.05m * .05m = .4414
I did this experiment with about a 40 kg wheel 1.05 m in diameter. The calculations were a little more complex because the wheel had bearings with a small steel hub and steel spokes. The majority of the mass was in the 2 inch by 11/4 inch steel rim. The resulting data proved that F = ma; I probably did not even keep the data because the results where what I expected.
I am sure you could find similar experiments on the internet if you knew what the experimenter called it.
It is similar to an Atwoodâ€™s machine.
TinselKoala quote; Please don't make me construct an apparatus to prove it.
What is you construct one and it proves you are wrong; would that be worth the effort.
In the picture: I cut away the lower circle (forward of the entry hole) that allowed the fishing line to enter inside the circumference which determined tether length. This slows the reacceleration because now the line swings free and is attached closer to the point of rotation. This slowed reacceleration allows the experimenter to more easily observe the motion of the system as it comes to a stop.
I also lowered the center of mass of the sled by using rusty bars instead of the lead anchor ball. The sled has a mass of 2758g; the 456g gray puck easily stops the sled and white disk (300g rotational mass).

TinselKoala quote; As seen from the definition, the derived SI units of angular momentum are newton metre seconds (Nâ€¢mâ€¢s or kgâ€¢m2s1 or joule seconds).Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the righthand rule."
Keplerâ€™s formula worked without all this hocus pocus. A joule is a unit of energy. A â€œpseudovectorâ€ they have got to be joking. Are you sure this isnâ€™t pseudoscience.
Take a 9 kg rim (or ring) that has a 2 meter diameter and place it vertically on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/secÂ². The speed of the string and the speed of the particles in the ring will be roughly equal; you could place the string in a grove in the rim to make the speeds even closer. After one second all parts will be moving .981 m/sec. After one second everything will have moved .4905 m, the dangling mass will have moved down .4905 m and the mass in the ring will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum of the rim would be (mass * radians/sec * radius * radius) 9kg * .981 radians/sec *1m * 1m = 8.829
Take a 9 kg thin wall pipe that has a .1 meter diameter and place its length horizontally so that it rotates in a vertically plane on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/secÂ². The speed of the string and the speed of the particles in the pipe will be roughly equal; you could place the string in a grove in the pipe to make the speeds even closer. After one second all particles will be moving .981 m/sec. After one second everything will have moved .4905 m, the hanging mass will have moved down .4905 m and the mass in the pipe will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum would be (mass * radians/sec * radius * radius) 9kg * 19.62 radians/sec *.05m * .05m = .4414
I did this experiment with about a 40 kg wheel 1.05 m in diameter. The calculations were a little more complex because the wheel had bearings with a small steel hub and steel spokes. The majority of the mass was in the 2 inch by 11/4 inch steel rim. The resulting data proved that F = ma; I probably did not even keep the data because the results where what I expected.
I am sure you could find similar experiments on the internet if you knew what the experimenter called it.
It is similar to an Atwoodâ€™s machine.
TinselKoala quote; Please don't make me construct an apparatus to prove it.
What is you construct one and it proves you are wrong; would that be worth the effort.
In the picture: I cut away the lower circle (forward of the entry hole) that allowed the fishing line to enter inside the circumference which determined tether length. This slows the reacceleration because now the line swings free and is attached closer to the point of rotation. This slowed reacceleration allows the experimenter to more easily observe the motion of the system as it comes to a stop.
I also lowered the center of mass of the sled by using rusty bars instead of the lead anchor ball. The sled has a mass of 2758g; the 456g gray puck easily stops the sled and white disk (300g rotational mass).
First, the quote is from Wikipedia, as I clearly state. Second, it's not hocuspocus, it's recognized vector mechanics. It's how we landed a robot probe on Titan, for one thing. Third, note that the units of angular momentum are kilogrammeters squaredper second, that is, (kg*m*m)/sec, which is exactly the units you have in your calculations, which would be correct, if only they were, well, correct. F does indeed = MA, because that's how they are defined. Acceleration is the response of mass to a force. Force is what it takes to accelerate a mass. Mass responds to force by acceleration. Mass is that which resists acceleration by force. And so forth.
In your thought experiment you have made some incorrect assumptions, and without examining the details of your confirmatory "experiment" I cannot evaluate it.
I can imagine your thought experiment as outlined above. But I still cannot figure out how the device in the pictures is supposed to move. You'll just have to show a video, for me to get it, I'm afraid.
And yes, the Joule is a unit of energy, and angular momentum is a means of energy " storage " which is why it can also be expressed in Joule seconds. It's called a "pseudovector" because, by convention, it points out the top axis of whatever is spinning or curving, according to a righthand rule. By convention. Whereas a "real" vector looks like it points in the direction it's going. Again, by convention.

Pardon me: the formula is correct. It is Keplerâ€™s formula. You have my apologies.
\mathbf{r} is the position vector of the particle relative to the origin,
...is Radius
\mathbf{p} is the linear momentum of the particle,
...is p or (m * radians/sec * Radius)
\times\, is the vector cross product.
This is where I got mixed up; I thought they were multiplying by yet another cross produce. But they are merely saying r * p, which is correct.
Thank you for sticking to your guns and not getting irritated with me.
But I will reiterate; this is a formula that works for satellites because in space radius has an effect upon (p) linear velocity. In the laboratory radius has no effect upon p.
TinselKoala quote: In your thought experiment you have made some incorrect assumptions,
Pequaide: Please be more specific.
I accelerate the sled and white disk and gray puck by tying a string to the opposite end of the sled. I drape the string over a pulley and suspend a mass from it. That brings the red ribbon up tight against the white disk and the (duck tape) flap that holds the gray puck up against the white disk. I hold one of the upright bolts until every thing is ready. After release and acceleration the suspended mass hits the floor about the same time that the ribbon releases the gray puck. I assume the puck holds its position up against the white disk until the ribbon feeds past the flap as the system spins. Once the puck begins to swing out on the end of the fishing line (or tether) it absorbs the motion of the white disk and the sled. The puck has a mass of 456g and the system has a total mass of 3514g.
The consolidation of motion in the puck has been used by NASA, what is in question is â€œdoes the puck have the same linear Newtonian momentum as the original systemâ€?

The MSU Atwoodâ€™s site (page 18 reply 175 this thread) allows the pulley wheel to have mass without throwing off their calculations. So apparently MSU thinks that the pulley itself accelerates in an F = ma relationship.
I searched the web for a while and did not come up with many vertical wheels being accelerated by a string wrapped around the circumference with a mass on the end. Maybe I will have to repeat the experiment and keep the data this time. I never thought I would be required to prove that F = ma.
Does anyone expect the rim to accelerate under some other Law? The block on a frictionless plane, attached to a string draped over a pulley with a mass on the end, accelerates according to F = ma. How is the rim different?
If the block on the frictionless plane is accelerated linearly and then is captured into a circular path by a string does it lose its linear quantity of motion when it enters the circle? If it does then how does it gain the linear momentum back after the string is cut?
Most texts say angular momentum is conserved in the laboratory and then give the ludicrous experiment of the ice skater as an example.

Hello pequaide,
I repost here what we have found some time ago concernig the AtwoodSimulation :
https://www.msu.edu/user/brechtjo/physics/atwood/atwood.html (https://www.msu.edu/user/brechtjo/physics/atwood/atwood.html)
Now I used 1kg for m1, 0 kg for m2 and 9 kg for m3 ( pulley ) which returns ( friction must be set, meaning pulleymass influence is taken into consideration ) a accelleration of 1.7836363 m/ sec exp2 which is double the
value you have calculated in your above post.
I have to do the math again according to the correct formulas I had found in the net way back then.
There is a simple answer to tinselkoala concerning this angularmomentumthingy which seems to get stuck im many peoples head :
Once the drivewheel and the whitewheel has stopped there is no angularmomentum any more. This moment in time is a transitionpoint and the only thing one can use then is the velocity of the rimmass which when suddenly brought to zero will mean a negativeaccellation ( value is the velocity) which will fetch the Force = M x (  a )
Regards
Kator01

Kator01: a acceleration of 1.7836363 m/ sec exp2 which is double the value you have calculated in your above post.
That is correct: MSU assumes that the pulley wheel does not have all the mass in the rim (which without doubt is true) but is something more like a solid disk. They assume that the average rotational mass is half way from the center of rotation to the circumference. This would be true if the pulley were only spokes with a very light circumference mass.

The MSU Atwoodâ€™s site accepts the acceleration of the pulley wheel as an F = ma relationship. If they thought there was a significantly different type of momentum in the pulley they would have set up a different category for the motion to fall into, but they didnâ€™t. If they thought there was a significantly different type of momentum in the pulley they would have probably mentioned the radius of the pulley for that would be necessary to evaluate the output angular momentum, but they didnâ€™t. If they thought there was a significantly different type of momentum in the pulley they would have had another window for angular momentum when the friction was turned on, and the quantity in the window would change as the mass of the pulley was changed, but they didnâ€™t. All of their motion falls into one category: Linear Newtonian momentum.
If you tripled the radius of the pulley (but kept the mass distribution of the pulley wheel the same) what effect would there be on the analysis?
I think there would be no effect. But with triple the radius angular momentum would be significantly different.
Take a 10 kg rim (or ring) that has a 2 meter diameter that has been placed vertically on dry ice. After accelerating the rim to 1 m/sec connect its spinning circumference to a string that is wrapped around the circumference of a 10 kg thin wall pipe, at rest, that has a .1 meter diameter. Place the pipe on its length horizontally so that it rotates in a vertically plane on dry ice. What kind of motion will the large rim share with the pipe, and what will be the final circumference velocity of the larger rim.

Pictured: Four masses (brass bushings) were used to accelerate the wheel and a flag (hexagonal wrench) past two photo gates held at a uniform distance. This should be enough information to determine if F = ma. Do you agree?
I took four consecutive readings that were within 2/10,000th of a second from each other, amazing isnâ€™t it.

F = ma is a mathematical relationship between force, mass, and acceleration. To prove that the statement is true you must show that a proportional change in force causes a proportional change acceleration. For example if force increases by 20% then acceleration increases by 20%, etc.
The experiment; four different descending masses were hung from the ribbon, The force they produce is proportional to their mass, this mass is only a small portion of the accelerated mass. The four descending masses were a brass bushing or a combination of several brass bushings. Bushing 1 was 37.4g, bushing two was 26.7g, three was 26.5g, and the fourth was 73.0g
Descending mass 1 was 37.4g, Descending mass 2 was 64.1g (37.4 + 26.7), descending mass 3 was 90.6g (37.4 + 26.7 + 26.5g), descending mass 4 was163.6g (37.4 + 26.7 + 26.5g + 73.0g),
The time interval for each force to make the flag cross the same distance between the photo gates (from the same starting point) was; F1 .6957 sec, F2 .5336 sec, F3 .4556 sec, and F4 .3420 sec.
All forces worked over the same distance but they did not take the same amount of time to do it. To find acceleration from time we can us the formula d = Â½ a * t * t because all the distance are the same the only variable are time and acceleration. Rearranging the formula to solve for a we get 2 * d / t * t = a, again distances are all the same so the proportional acceleration of F1 is 1 / .6957 *.6957 = 2.078, F2 is 1 / .5336 * .5336 = 3.512, F3 is 1 / .4556 *.4556 = 4.818, F4 is 1 / .3420 * .3420 = 8.55.
The proportion of force to the proportion of acceleration is close to as follows.
F1 37.4g 2.078
F2 64.1g 3.512
F3 90.6 g 4.818
F4 163.6g 8.55
But there is one more correction to make. The fourth force F4 is accelerating a larger mass than the first force. This is in a proportion to its total mass and the total mass accelerated by the smaller force F1, that proportion is the rotational mass (inertia of the wheel, 2600g) of the wheel plus each mass that caused the force. F4 accelerates the larger mass slower in a proportion of 2637.4g / 2763.6g = .954; this means that F4 would have accelerated the smaller mass 2637.4 faster giving it a higher acceleration by 1 / .954 = 1.048 * 8.55 = 8.96.
The proportion of force to the proportion of acceleration is as follows.
F1 37.4g 2.078
F2 64.1g 3.512 corrected for the greater mass being accelerated 2664.1g / 2637.4g to 3.55
F3 90.6 g 4.818 corrected for the greater mass being accelerated 2690.6g / 2637.4 to 4.92
F4 163.6g 8.55 corrected for the greater mass being accelerated 2763.6 / 2637.4 to 8.96
So does a proportional change in force (which equals a proportional change in descending mass) equal a proportional change in acceleration? Or does: 37.4g / 163.6g = 2.078 / 8.96
37.4g / 163.6g = .2286, 2.078 / 8.96 = .2319 .2319 / .2286 = 1.0145
37.4g / 64.1g = .5834, 2.078/3.55 = .5853 .5853/.5834 = 1.0033
37.4g / 90.6g = .4128, 2.078/4.92 = .4223 .4223/.4128 = 1.023
These are within two percent of a perfect F = ma relationship. So yes; F = ma is true for tangent forces working on the circumference of a circle. This is true even though the wheel is not a rim mass wheel, and this is Linear Newtonian Momentum.

I have been using the ribbon to drop different masses from the edge of the wheel. By using the photo gates to record their acceleration rate I can determine the rotational inertia of the wheel. The rotational inertia of the wheel is about 2500g. That means the dropping mass accelerates as if it were accelerating a sled with a mass of 2500g.
For example; I dropped a 137g mass .5715m. This mass and the average mass of the wheel had a velocity of about .7692 m/sec. This is an acceleration rate of .5176m/secÂ². 9.81 /.5176 tells us the relative mass of the wheel to dropped mass.
So after a 137g mass has dropped only .5715m we have 2.02 (2630g * .7692 m/sec) units of momentum.
By using the cylinder and spheres principal we can give all the motion to the 137g. For .137 kilograms to have 2.03 unit of momentum it will have to be moving 14.8 m/sec
The 137g object can rise 11.1 meters with a velocity of 14.8m. d = 1/2vÂ²/a.
The 137g mass was only dropped .5715m. This is an energy increase of 19.4 times the original energy. 11.1m / .5715m.

@pequaid
Why dont you build a setup so that will cause the 137g object to rise 11.1 meters ?

The acceleration of the dropped mass divided by the standard acceleration of gravity is proportional to the mass dropped divided by the total mass being accelerated. The total mass being accelerated is equal to the dropped mass plus the rotational inertial of the wheel.
We know the quantity of mass being dropped (tenth column) and we know standard gravitational acceleration: and the acceleration of the dropped mass can be determined from the final velocity given by the photo gates.
Knowing the distance dropped and the final velocity given by the photo gates we can use the rearranged distance formula (d = 1/2vÂ²/a) to determine the acceleration of the dropped mass (seventh column). We now have all the components necessary to determine the rotational inertia of the wheel; which is in the twelfth column and is called total mass minus dropped mass.

Fred: Note that an increase in the dropped mass gives you a proportional increase in the acceleration of the wheel and dropped mass. The ribbon is of course tangent to the surface of the wheel. So this means that a given force F pulling tangent to the surface of the wheel gives you an F = ma relationship for the velocity changes in the wheel.
So if you apply a tangent force for a certain period of time you will get a certain velocity for the particles in the wheel. And if you are to get that velocity back out of the wheel and return the wheel to rest you are going to have to apply equal force (in the opposite direction) for an equal period of time, or twice the force for half the time, or half the force for twice the time etc..
When the steel puck is unwinding from the white disk it is a line tangent to the wheel with a force F that is sufficient to stop both the white disk and the rotating wheel. The force times time relationship in the fishing line must be equal to the force times time relationship in the ribbon that started the motion. But the force in the fishing line is not just an F = ma relationship that is working on the wheel, it is also working on the puck.
Newtonâ€™s Third Law of Motion tells us that the momentum change in the wheel must be equal to the momentum change in the puck. The above experiment proves that the momentum change in the wheel is linear Newtonian momentum, and the momentum change in the puck must be linear Newtonian momentum. If the wheel and white disk and the puck have six times as much inertia as the puck then the puck must be moving six times as fast when the puck has all the motion. This is an energy increase. All data collected thus far has confirmed that this energy increase does occur and that Newtonian physics is correct.
So the answer to your question â€œwhy donâ€™t you build a setupâ€œ is that I already have. I already have such a machine. We know how high objects will rise given a certain horizontal velocity. Catching an object on the end of a cable and allowing it to rise as in a pendulum is something for an engineering department not for a research Lab.

The only visual equipment that works well, for now, is my camera.
I used a suspended dropping mass off of the big wheel for a uniform velocity; the suspended mass hits the floor about the time that the ribbon releases the tab on the gray puck.
In picture one the white disk and gray puck are achieving maximum velocity. It appears that the tab on the puck has already pulled out from underneath the ribbon. I can see I need to improve this release system. The suspended mass on the big wheel will soon hit the floor and the disk and puck and wheel will gain no more velocity.
The second picture is a different run of the same set up. I just take a lot of pictures and then try to arrange them in the appropriate order. But this is a different event and not near as good as a frame but frame evaluation that can be achieved with a video, I will have to replace my DVD recorder.
The second picture shows the puck (with glaring white tab) coming clear of its seat but it has not yet passed under the ribbon. The ribbon is not loose because the momentum of the big wheel is pulling on it. The white disk is slowing down because the gray puck is pulling on it. As the white disk slows down the momentum of the wheel pulls on the disk through the ribbon. There is a dark line between the two layers of plastic; this is a carpenterâ€™s mark not the fishing line. The fishing line is harder to see.
In picture three the puck is headed under the ribbon, the ribbon is still taut. Note the black square on the white disk, there is a small angular increase of this square between picture three and picture four. The tightness of the ribbon and the angular advancement of the square means the big wheel is still turning.
Shortly after picture four the big wheel and the ribbon and the white disk are stopped. The fishing line enters an open area and is attached nearer to the bearing. This closer attachment of the line causes the momentum exchanges to occur more slowly. Since the gray puck is not released it will reaccelerate the white disk in the same direction.
By picture five there is a large angular change in the black square as the puck reaccelerates the white disk. It will not restart the wheel however because the ribbon is limp, it can not push the wheel. There is not a uniform period of time between photos.
I removed 448g from the big wheel because the white disk and wheel could not be stopped when the gray puck was fully extended. The length of line is a determining factor in how much mass the puck can stop. Longer lines can stop greater quantities of mass because the time over which the force acts has increase. This is not because of angular momentum conservation; remember you could use a twenty meter big wheel and then the puck would have to be moving a bullet speed to conserve angular momentum.
A few obvious improvements should be made in the experiment. I hope to build two electronic releases; one for a pinpoint release position of the puck from the seat of the white disk. And then I would like to release the puck from the rest of the system when the white disk and big wheel are stopped. This would leave the puck moving in a straight line (apparently) headed for the back wall. I also think the dropping mass can be incorporated in the mass of the wheel itself. Then the challenge would be to release the gray puck toward the back wall at precisely the same time that the white disk is stopped and the wheel is stopped with the extra mass at six oâ€™clock. And then of course I should place the photo gates between the release point and the back wall.
I would be delighted if people would repeat any of the experiments. I endeavor to give enough detail to make replication possible. I would not see replication as a form of distrust, but only as good science. And the simpler experiments of the â€˜cylinder and spheresâ€™ or the â€˜disk and pucks on the air tableâ€™ should not be passed up. Their lack of bearings makes them deadly accurate, and they are inexpensive. Their low mass makes them susceptible to air resistance, but you design the experiments different ways: slow and small with no bearings, faster and bigger while using bearing, etc.

I donâ€™t think making energy in the lab is an insignificant advancement. It would be one the most valuable scientific achievement in history; it would have to rank up there with the invention of the wheel and the control of fire. Of course it must be followed by functional machines, but you could not prevent that if you tried. You will have to bolt the doors to keep from getting trampled to death.
A one kilogram mass that is dropped straight down in freefall (9.81m/secÂ²) for one meter will have a velocity of 4.429 m/sec and it will have a momentum of 4.429. A one kilogram pendulum bob that has dropped down one meter has a 4.429 m/sec velocity and 4.429 units of momentum.
Near 6 oâ€™clock the force applied to the bob (in the direction of travel) in a pendulum is small but the time over which the force acts is proportionally greater. So in a simple pendulum you end up with the same velocity as that of the freefall mass. Much like suspending the mass near the axis, the time over which the force acts is longer but the force itself is proportionally small, and you will be getting the same overall velocity change.
The angles over which the force acts in a pendulum are identical to the working angles of the force in the wheel.
A one kilogram mass on a string that is suspended over a frictionless pulley (just off the edge of the table) with the string attached to a nine kilogram block on a frictionless plane will accelerate at a rate of .981m/secÂ² (1/10th of 9.81). This is because only 1 out of 10 kilograms is under gravitational acceleration.
The same is true for the balanced nine kilogram wheel with one kilogram of overbalance. The same quantity of force that was available to you in the simple pendulum is now available to you in the wheel. And the force is going to accelerate the other nine kilogram just like the suspended mass over the pulley accelerated the block. Now we have 1/10 the acceleration rate that got us 4.429 meters per second which will now give use 1.4007 m/sec. This is accepted Newtonian physics. Even though the acceleration is 1/10 the final velocity is greater than 1/10 and the final momentum is greater than 4.429. The velocity is determined by this formula d = 1/2vÂ²/a, where v is the square root of (2 * d * a) which in this case is (2 * 1m * .981m/secÂ²) = 1.4007m/sec. That puts momentum at 10kg (all the overbalance wheel is moving 1.4007 m/sec) times 1.4007 m/sec for 14.007 units of momentum.
Now you put the 14.007 units of momentum into a yoyo despin device and transfer all of the motion to the small overbalance mass of 1 kilogram. If (a very small if) Newtonâ€™s Three Laws of Motion apply to this event then the one kilogram must be moving 14.007 m/sec. At 14.007 m/sec the one kilogram will rise (d = 1/2vÂ²/a) 10 meters, and it was only dropped one meter. All measurements that I have taken confirm Newtonian physics, and that the Laws apply to this event.
I intend to do an experiment of mass attached directly to the wheel, but I am quite sure I will be checking bearing resistance not Newtonian physics. Well; but indeed, I will be checking both.
I placed a mass on the end of the red ribbon which was wrapped around the big wheel
I taped a photo gate flag (hex wrench) to the big wheel. I placed the two photo gates at a fixed distance from each other and in a position that the flag (after acceleration) interrupted the photo gate beams.
I raised the mass 296mm above the point where the flag was half way between the photo gates. After being dropped the mass accelerated the wheel and the flag which crossed the photo gates in .0272, and .0272 seconds, in two different runs.
I then taped the same mass to the inside of the rim of the wheel and raised its position 296 mm above the same photo gate position. After release the flag crossed the distance between the photo gates in .0269 and .0269 seconds.
So if the mass is suspended from the circumference with a ribbon or fixed to the wheel, it is the distance the mass drops that determines the final velocity of the wheel. The velocity is also dependent upon the wheel inertia and the dropped mass of course.
If a mass is dropped the same distance off the circumference or fixed inside the rim the final velocity will be the same.
I donâ€™t see anything that is left to speculation, NASA and RCA both performed the despin stops. And I have accomplished stops of disks, cylinders, and wheels. I have timed the spheres and pucks in these stops and have confirmed Newtonian physics. The wheel wrapped with a ribbon will give Newtonian accelerations and an inside (the circumference) wheel attachment will also give Newtonian results. So we know that the original input motion is achievable.
The distance (or displacement formula; s) formula is s = 1/2atÂ². I think I have seen d used instead of s so I went with that, who would think of s being distance. So now we have d = 1/2atÂ², v = at or t = v/a and tÂ² = vÂ²/aÂ², substituting vÂ²/aÂ² for tÂ² the distance formula is now d = Â½ vÂ²/a. So lets check and see if it is correct. After one second in free fall the distance dropped is 4.9 meters (s = 1/2atÂ²) and the velocity is 9.81m/sec (v = at) so lets plug in 9.81m/sec velocity in (d = Â½ vÂ²/a) instead of time in (s = 1/2atÂ²): d = Â½ * 9.81m/sec * 9.81m/sec / 9.81m/secÂ² = 4.9m, yep it works.
Lets double check it at two seconds. After two second in free fall the distance dropped is 19.62 meters (s = 1/2atÂ²) and the velocity is 19.62m/sec (v = at) so lets plug in 19.62m/sec velocity in (d = Â½ vÂ²/a) instead of time in (s = 1/2atÂ²): d = Â½ * 19.62m/sec * 19.62m/sec / 9.81m/secÂ² = 19.62m, yep it works.
If you leave the one kilogram attached to the wheel â€˜noâ€™ the 14 units of momentum will not be enough to return the one kilogram to12 oâ€™clock, but if you separate the one kilogram from the wheel and give it all the momentum then it will rise 10 meters. Separate the overbalanced mass from the wheel and transfer all the momentum to it. The cylinder and spheres (or the yoyo despin device) separates the mass and transfers all the motion to the small mass, and energy is made.
A question: If I graphed time on the X axis, and velocity on the Y axis  is distance traveled the area under the graph?
Pequaide answer: Yes.
If a suspended mass of 1 kg accelerates a 0 kg rim mass balanced wheel the acceleration will be 1/1 * 9.81 m/secÂ² or 9.81m/secÂ². Just brainstorming here, this is free fall. At the end of a one meter drop the velocity will be 4.429 m/sec and its momentum will be 4.429 units of momentum, and its energy will be (1/2mvÂ²) 9.81 joules, and the mass will rise 1 meter. This is all the energy that is needed to reload the system, less a little friction.
If a suspended mass (on the end of a ribbon wrapped around the circumference of a wheel) of 1 kg accelerates a 4 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/5 * 9.81 m/secÂ² or 1.962m/secÂ². At the end of a one meter drop the velocity will be 1.98 m/sec and its momentum will be 9.90 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 9.90m/sec its energy will be (1/2mvÂ²) 49.05 joules, and the mass will rise 5 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 40 kg rim mass wheel the acceleration will be 1/41 * 9.81 m/secÂ² or .239m/secÂ². At the end of a one meter drop the velocity will be .6917 m/sec and its momentum will be 20.05 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 28.36m/sec its energy will be (1/2mvÂ²) 402.1 joules, and the mass will rise 41 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 1 kg rim mass balanced wheel the acceleration will be 1/2 * 9.81 m/secÂ² or 4.9m/secÂ². At the end of a one meter drop the velocity will be 3.13 m/sec and its momentum will be 6.26 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 6.26m/sec its energy will be (1/2mvÂ²) 19.62 joules, and the mass will rise 2 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 1000 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/1001 * 9.81 m/secÂ² or .00980m/secÂ². At the end of a one meter drop the velocity will be .14000 m/sec and its momentum will be 140.14 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 140m/sec its energy will be (1/2mvÂ²) 9819 joules, and the mass will rise 1001 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
This math is F = ma. And d =1/2 vÂ²/a, and (1/2mvÂ²)

Hi,
sorry for bumping in in such a dumb manner,
but may i ask you, where you have got this flywheel from?
I really would like to follow your experiments handson,
but sometimes reality bites hardly.
Getting the stuff together is sometimes harder than to figure out theories... :/
Thanks,
sushimoto

I found the wheel in a trash container at work. One of its two bearing was missing so I assume that it had smoked a bearing and instead of maintenance buying a new bearing they bought a new wheel. That may have been their best choice but I bought a bearing for $6 or so and the wheel is very accurate. It was one of two wheels from an upright band saw; I would hate to think what it would cost new. It would probably be cheaper to find a used band saw and strip the wheels out of it. You could go to a dealer and ask him what he wants for a replacement band saw wheel. I have been wishing I had two, and then I could attach ribbons to the white disk from both sides.

Someone once said that ice skaters appear to conserve angular momentum. Well if they do it is because they use their muscles to increase their angular velocity. The increased linear velocity of the parts as they are pulled in is transferred to angular velocity.
Here is an experiment that proves that inanimate objects can not conserve angular momentum in the lab.
I was in a warehouse once that the builder stated that the entire floor surface did not vary by one millimeter. The floor was hardened to support fork lift trucks that would stack skids in racks six high. So the 200 meter by 200 meter floor had to be free of vibration.
On such a floor you could take a one ton cart and place it on a 100 meter horizontal cable fixed on the other end. After accelerating the cart to one meter per second you could have the cable come in contact with an immovable post that was 3.33 meters from the center of mass of the cart. The Law of Conservation of Angular Momentum requires that the cart must increase in linear velocity by 30 times the original linear motion. To maintain the original angular momentum the radians per second must increase by 30 times; which can only be accomplished by the linear velocity increasing by 30 times. A radian per second is the travel distanced of one radius around the circumference in one seconds.
1000 kg * 1/100 radian per second * 100 meters * 100 meters = 100,000
1000kg * 1/3.33 radians per second * 3.33 meters * 3.33 meters = 3,333
3,333.4 * 30 = 100,000 the linear velocity must increase by 30 times to maintain angular momentum conservation.
If angular momentum conservation is true then now we have a 1000 kg cart that is moving 30 m/sec. At 30 m/sec we can catch the cart on the end of a pendulum cable and it will rise 45.87 meters. An object need only drop .051 m to achieve a velocity of one meter per second, so after we patch the hole in the roof we can start making energy.
Of course the truth is that angular momentum conservation in the lab is false. It may appear that a skater or a person on a spinning stool conserve angular momentum, but appearances can be deceiving or they do it with their muscles. Probably a little of both.

http://www.youtube.com/watch?v=YzwC6Oa8X0
youtube.com science and technology Dspin

The newest model of the cylinder and spheres is pictured with its three attachable pipes. The tall pipe is a 3 inch inside diameter pipe that is stopped by the spheres, just as the tether string enters the slit, when the pipe has a mass of 712g. The widest pipe is a 4 inch inside diameter pipe that has a mass of 541g. The third pipe is a combination of 215g of 3 inch pipe and 428.4g of 3 inch pipe couplers. All three of these addon attachments have about the same Newtonian momentum when rotated. All three addon pipe attachments will be stopped by the spheres just as the tether enters the slit in the cylinder, the top cylinder is also stopped of course.
First I determined the necessary quantity of 3 inch pipe mass that is needed to have the cylinder stop just as the tether enters the slit in the cylinder side wall. This is done by trial and error; I just kept taping more and more dimes and pennies to the side wall of the 3 inch pipe until I found the mass that allows the pipe to stop at the appropriate position when the rotation stops. You add more mass if the cylinder is moving backward when the string is in the slit and you subtract mass if the cylinder is moving forward when the tether is in the slit. When you determine the appropriate mass you then cut a new 3 inch pipe to length with the appropriate mass (without the pennies).
This trial and error determination of mass for the 3 inch I.D. pipe took about 20 attempts, but once that that mass is determined I can cut the second two pipes (4 inch and 3 inch with couplers) to the appropriate mass the very first time. Why is that possible that I no longer need trail and error to determine the mass of the second two pipes?
One of the variable factors in the quantity of mass stopped by the spheres in the cylinder and spheres (top coupler), just as the tether enters the slit, is tether length. But once the mass of the 3 inch pipe is determined you need only match its Newtonian momentum when replacing it with other addon pipes. A 4 inch I.D. pipe has about 4/3 the rotational velocity of a 3 inch pipe and therefore needs only Â¾ of the mass to have equal Newtonian momentum.
The three addon pipes do not have equal angular momentum when rotated and they do not have equal rotational kinetic energy. This not only proves that angular momentum is not conserved in the lab; it also proves that energy has been made in the lab.
This latest cylinder and spheres experiment takes 977g (total mass of cylinder with spheres) of mass moving in a circular path and gives all that motion to 133g (the spheres). Newtonian physics requires that that mass (133g) must then be moving 7.35 times as fast.
With the spheres moving 7.35 times as fast the energy content of the system has increased to 735% of the original energy. 1/2mvÂ² (1/2 * .977 * 1 * 1 = .4885 joules, Â½ .133 * 7.35 * 7.35 = 3.524 joules; 3.592/.4885 = 735%
This is the experiment that is worth the Nobel Prize in Physics.