BesslerWheel.com

[broli]

Too much text too much talk. Seriously there's a reason why people are ignoring you.

I mean we are even supposed to look for your own threads? Couldn't you just link to them. And stop pointing to the super annoying yahoo group. Put those damn pictures on 1 of 1000's free image hosting websites and directly link them when you mention them instead of annoying the hell out of people by letting them sign up to some yahoo group and so on.

Please be a bit reasonable. If I share ideas I at least spend time in putting together a decent presentation. Have some structure and fullnes to your post.

I'm really interested in what you're saying but I always end up wanting to pull my hair in rage after the whole mix of things. Keep everything in one post and keep it clear!

[rlortie]

[pequaide]

A wheel and an Atwood both have an F = ma relationship when an extra mass is placed on them. If you wish you could use an Atwood for your initial motion. You then transfer all that motion to the spheres in the cylinder and spheres arrangement. The spheres are then allowed to rise upward, and from their elevated position they could then be used to restart the Atwood. But the spheres can rise higher than what is necessary to restart the Atwood, therefore you have free energy.

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.

[DrWhat]

Pequaide I hope you don't mind me putting these images up. If you do let me know. I believe these are the 3 images you are referring to.

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_________________
I only realized too late that life was short.

[rlortie]

DrWhat,

I am not pequaide, but I am rather confident that these are the pictures in question. Sorry I did not consider downloading them earlier. I simply cannot fathom their usage in a continuous revolving wheel.

There are more pictures depicting the construction of his PVC machine in the same folder.

The concept is evaded every time I try to get an answer. The latest thrown-bone is the Atwood simulation, which looks great. Unfortunately I have yet to see a design for a self-sustaining mechanism with an automatic reset button.

Ralph

[pequaide]

Thank you, DrWhat. See two heads are better than one. I never thought of posting from the other site.

I accelerated the two pucks on the right so I went from photo 3 to 2 to photo 1. But of course it does not mater what side you start from. The motion is transferred from two pucks to three pucks and then back to two pucks. When the motion goes from three to two the energy increases. No; this is not a self-sustaining mechanism but Laithwaite does demonstrate that energy can be made in the lab. You could of course start with three pucks in motion and end with two in motion, and the energy content of the system would double. These are the steps that are necessary to make free energy, if I had a model with an automatic reset button would I be posting? One of the photos in the other site is of a drawing that has almost all of the necessary parts for a recycling system; I think it needs a pulley system to reload.

The momentum that is transferred to the spheres need not be in the cylinder itself, it can be in some outside source such as another wheel or an object in linear motion or in an Atwood’s machine. The cylinder itself can be of very light weight material that is then connected with a string to one or more of the other three objects. The force in the sphere’s tether doesn’t know if the force is external or coming from the momentum change of the cylinder.

The reason you would us an Atwood’s machine as a source of momentum is because its physics is unquestioned.

An Atwood’s with a 5 kilogram mass on both sides would accelerate at a rate of 1/11 times 9.81 m/sec² (.8918 m/sec²) if one extra kilogram is placed on one side. After the extra kilogram drops one meter the entire system would be moving 1.3355 m/sec for a momentum of 14.69. You could attach a string from the Atwood’s to a cylinder and spheres machine and the one kilogram of spheres would be moving 14.69 m/sec. At 14.69 m/sec the 1 kg of spheres could rise 11 meters. The 1 kg extra mass was only dropped one meter.

The Atwood’s could have a mass of 10 metric tons with one metric ton of over balance. That would give you one metric ton to be dropped 10 free meters every, (d = ½ at²) (1m = 1/2 *.8918 m/sec² *1.497 m/sec * 1.497 m/sec) 1.497 sec. plus the cycle time of the cylinder which is about a second. Well let’s say 12 cycles a minute. That would be 12 metric tons dropped 10 meters every minute.

[pequaide]

Pictures can be found at yahoo.com groups free_energy, under files, see posting by pequaide 8-19-08 002 is a good one.

The experiment is extremely simple. A spinning hollow cylinder that has been wrapped with two strings will surrender all of its motion to two smaller masses that are on the ends of the unwrapping strings. A four kilogram hollow cylinder whose mass is spinning one meter per second around the circumference can give all of its motion to one kilogram of spheres. Before release the spheres are embedded in the hollow cylinder and have the same velocity. In this example you would initially have 5 kilograms moving 1 m/sec for 5 units of momentum. For the one kilogram of spheres (after they absorb all the motion), to have 5 units of momentum they will have to be moving 5 m/sec.

We know from ballistic pendulums that linear motion and circular motion are completely interchangeable and experience no loss of motion in changing from circular to linear or linear to circular. So the rotating cylinder with spheres (above) has five units of momentum.

The sphere with all the motion will have 5 times as much energy as the cylinder and spheres before release.

½ * 5 kg * 1 m/sec * 1 m/sec = 2.5:

½ * 1 kg * 5 m/sec * 5 m/sec = 12.5

Objects can be released from circular motion, when they are released they travel tangent to the circle with linear motion that has equal displacement to the previous motion around the circumference of the circle. The circular interaction of circumference motion in ballistic pendulums conserves linear motion.

The experiment seeks to prove that circumference motion (in meters per sec) and tangent linear motion are the same and should be covered under the same Laws of Physics (Newton’s Three Laws of Motion; which predicts momentum conservation) and that both should be conserved.

The experiment; A spinning cylinder with spheres embedded at 180° is released to allow the spheres to swing out on the end of tethers. As the spheres swing out they stop the rotation of the cylinder. At this point all the arc motion (tangent linear motion) is held by the spheres. If arc motion is conserved the spheres must be moving faster by a proportion roughly equal to the total mass (spheres and cylinder) divided by the mass of the spheres.

[Jonathan]

Only members of the group can see the files section. Luckily, I understand what you mean, and can assure you that it is not a source of free energy, because they would have noticed that when they use this method to slow the rotation of spacecraft.
The primary problem with your math is that it seems valid no matter what the radius of the 4kg cylinder is, and since this can't be true, the math must not actually be valid. For the sake of argument, suppose that the radius of the hollow cylinder is 1m. This means that the initial angular momentum is roughly
L=2mVr+MVr=2*1kg*1m/s*1m+4kg*1m/s*1m=6 kgmm/s.
Now suppose the strings are 1m long each, so that the spheres end up at a radius of 2m. Then the angular momentum will be
L=2mu(r+l)+Mvr,
where geometry requires
u=(r+l)v/r.
Simplifying, we have
L=2m(r+l)v(r+l)/r+Mvr
=2*1kg*(1m+1m)*v*(1m+1m)/1m+4kg*v*1m
=12v.
Since the net torque on the system is zero, the angular momentum must be conserved:
6=12v,
which means that v, the final tangential speed of the hollow cylinder, will be 1/2 m/s.
The initial energy of the system would be
E=(1/2)(2m+M)VV=(1/2)(2*1kg+4kg)*1m/s*1m/s=3 J.
And the final energy of the system will be
E=(1/2)(2muu+Mvv)
=(1/2)(2*1kg*1m/s*1m/s+4kg*1/2 m/s*1/2 m/s)
=3/2 J.
I can't guarantee that this is all correct because I'm sleepy, but the general conclusion is correct: the momentum exchange occurs in an inelastic manner, so the system ends with less energy than it started with.

_________________
Disclaimer: I reserve the right not to know what I'm talking about and not to mention this possibility in my posts. This disclaimer also applies to sentences I claim are quotes from anybody, including me.

[pequaide]

Quote from Joanthan: because they would have noticed that when they use this method to slow the rotation of spacecraft.

This method was proposed (by RCA if my memory serves me correctly) as a means of stopping the rotation of space craft, but NASA elected to use small rockets instead. To my knowledge the cylinder and spheres phenomenon was never used by NASA. And what makes you think that they would have noticed it? They are mind blocked just like you. How many people did not invent Velcro, how many hunters have picked burrs out of their socks but never thought to use the concept as a fastener.

Quote from Joanthan: The primary problem with your math is that it seems valid no matter what the radius of the 4kg cylinder is,

It is absolutely valid. A one kilogram mass moving one meter per second around the circumference of a 220 meter (radius) circle has exactly the same momentum as if the same mass were moving one meter per second around the circumference of a .2 meter (radius) circle. You can change one circle into the other circle instantaneously without the application of outside force. Place a solid pin to interrupt the string of the 220 meter circle at .2 meters from the mass; you will instantly have a .2 meter circle without the application of outside force. Centrifugal and centripetal forces are not outside forces and they are balanced forces, you have no outside force applied to the system yet your math claims that the momentums are different. Angular momentum conservation in the lab is a bogus concept, there is no application of outside force such as gravity in space, for which Kepler introduced the concept of angular momentum conservation. F = ma or F t = mv if there is no application of force there can be no change in momentum. You tie your own hands with these bogus concepts.

[pequaide]

Down sized I hope.

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[pequaide]

Thanks Scott; once I figured out how to post pictures it was easy. Great site.

In the picture above the wheel that is being used as a pulley has a mass of over 3,400g, but its mass distribution is unknown. Let’s assume that the effective rotational mass of the wheel is equal to a rim that has a mass of 2,300g. The red and blue cylinders on the ends of the strings that drape over the wheel are solid steel and have a mass of 4,210g each. The wheel and red and blue cylinders are a balanced system, they will not move unless a small force is applied to them. The center of mass of the wheel and red and blue cylinders does not move if the system is moving or stationary. The dark disk on the edge of the white disk (on the table) in the picture is steel, and has a mass of 454g.

I placed the 454g disk on the red cylinder and allowed it to drop toward the floor. As the red cylinder (with disk) nearly reached the floor the motion was becoming violent.

Let’s assume that the motion is .6 m/sec. The total mass of the system is 4,210 + 4,210 + 2,300 + 454 = 11,174g. This means the system has a total momentum of 6.70 units. If all that momentum is given to the 454g (by the cylinder and spheres arrangement) it will rise 11 meters. The 454g disk was only dropped about .5 meters.

The Atwood’s is a machine that is used to prove F = ma, (or Ft = mv) the force of .454 * 9.81 N produces a change in momentum of the red and blue cylinders and the wheel, this is classic Newtonian Physics.

[Jonathan]

"NASA elected to use small rockets instead."
They use both: http://en.wikipedia.org/wiki/Yo-yo_de-spin
"what makes you think that they would have noticed it?"
If you were correct, then the math they do would give the wrong answer, which would, for example, prevent cameras and antennas from being aimed appropriately.
"You can change one circle into the other circle instantaneously without the application of outside force."
That's not the same as having weights fly outward, because there is no sudden change in the direction of the weights when the strings catch them. At that moment, the strings must exert a large force to change the direction of the momentum, and in the process a large portion of the momentum is dissipated in the strings as heat, so there is less macroscopic kinetic energy. It's like the weights hit an invisible cylindrical wall, and since some of their momentum is radial at that point, some of it is wasted. I have attached an illustration; the arrows are at the proper angle and relative scale for their respective drawings.

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[Clarkie]

Hi Jonathan,

I have been out of it for nearly a year, very please to see you are back. Going to give you something to test your skills in the new year. Model testing at the moment with good results.

Pete.

[pequaide]

Yo-yo de-spin; reduce the initial spin rate of the 1420 kg spacecraft from 36 RPM down to 3 RPM in the other direction

Doesn’t the, other direction, mean they didn’t know what they were doing. They obviously didn’t have the math right. I see they used the process in the early 60s. Did they drop the process because they could not accurately predict the expected outcome, because they had an inaccurate mathematical explanation?

Johnathan: If you were correct, then the math they do would give the wrong answer, which would, for example, prevent cameras and antennas from being aimed appropriately.

Pequaide; It did give them the wrong answers. Missing by 3 RPM means they had no accurate control of the outcome. Accurate adjustment was done by the rockets. They probably had sensors that released the tether as the rotation slowed, and still missed by 3 RPM.

I can build different and new machines and hit the dead stop dead on, and they missed it by three RPM.

The string don’t come tight and catch the spheres as you propose, the tethers are always taut. The force in the tether may change but they never come thigh or jerk the sphere. Further if it were free the direction of travel of the spheres is always at 90° to that of the force in the tether.

[pequaide]

One thing we know about the yo-yo de-spin device, used by NASA, is that all the rotational motion of the rocket or satellite is transferred to the small quantity of mass (3 kg) on the ends of the tethers. Let’s assume that the rocket’s rotation has been stopped by the tethered masses and the masses have been released. Suppose these two projectiles (simultaneously) then strike the rocket on a tangent line to its circumference at 12 o’clock and 6 o’clock, one projectile moving east and the other moving west. We should be aware by now that the rotational change (in linear Newtonian momentum) in the rocket will be proportional to the linear momentum change in the projectiles. There will be no angular momentum conservation, in fact the projectiles were moving in a straight line (they were released from their tethers) and had no angular momentum. We know that linear Newtonian momentum is conserved in this type of projectile collision because ballistic pendulums always conserve linear Newtonian momentum.

The block of a ballistic pendulum is not moving in a straight line it is rotating. It has a radius and a center of rotation. The block is rotating about its center of rotation. Blocks and projectiles can be made of hardened steel but they always conserve linear Newtonian momentum.

When the tethered masses are released by the tether they must have an adequate quantity of linear Newtonian momentum so that if they were to strike the rocket tangent and at 12 o’clock and 6 o’clock they would be able to return the rocket to its original state of rotation. Otherwise we violate a basic scientific principal of reversibility. This is an argument that NASA was improperly using conservation laws.

I built a cylinder and spheres machine (what NASA called a yo-yo de-spin device, I did not know they had used the concept) out of a three inch PVC pipe coupler. I got the cylinder to stop rotating when the tethered masses were swung out at 90° (to tangent) by adding more and more mass in the form of a 3 inch PVC pipe. I took the mass of the attached 3 inch PVC pipe (that made the cylinder stop rotating when the spheres were at 90°) and calculated it linear Newtonian momentum when rotating. I then cut a length of 4 inch PVC pipe that had the same linear Newtonian momentum (when spun) as the 3 inch pipe. I attached the 4 inch pipe to the outside of the coupler. I then spun the cylinders (3 in. coupler with attached 4 inch pipe) and spheres and released the spheres, as the spheres swung out on the end of their tethers the cylinder was stopped just as the spheres reached 90°. The top coupler with spheres (when spun and released) will behave in the same manner if a 400g 3 inch pipe or a 300g 4 inch pipe is attached. Actually the mass of the 4 in. pipe is at about 307g because of the thickness of the walls, the center of rotational mass of a 3 inch PVC pipe is at about 3.3 in. and the center of rotational mass of the 4 inch pipe is at about 4.3 in.

The linear Newtonian momentum of the 300g 4 inch pipe and the 400 g 3 in. pipe is about the same when being spun. But the angular momentum of the 300g 4 inch pipe and the 400 g 3 in. pipe is not the same when being spun. So the cylinder and spheres is showing that it responds to linear Newtonian momentum conservation not angular momentum conservation. This experiment is yet another argument that NASA got it wrong.

Look at Laithwaite’s experiment again. Are these not masses on the end of tethers? And what did he say was conserved? Linear Newtonian momentum was conserved.

I have independently timed the motion of the spheres as they have all the motion in the cylinder and spheres experiment, and linear Newtonian momentum is conserved.

These are a few reasons I believe that angular momentum was not conserved by the yo-yo de-spin device, as NASA predicted. I predict that NASA will not tell you the actual (measured) velocity of the released masses. They may predict what they though the velocities were, but they cannot even make the cylinder (rocket) come to a dead rotational stop. They obviously did not know what they were doing. They accomplished a task, but they had the math way wrong.