Free energy from gravitation using Newtonian Physic

[pequaide]

Once you are on line type: free_energy, in your Google search engine. Choose: free energy claims, then files, then posted by pequaide, or 8-19-08 four frames. The Yahoo engine does not bring you to the same site. But it is a Yahoo group site.   

[pequaide]

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.

[pequaide]

See the picture labeled 8-25-08 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 √(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 √(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.

[hartiberlin]

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.

[pequaide]

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 Laithewaite’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 â€ââ,¬Å" Everything you need to know, after searching â€ËÅ"Laithwaithe free axis’.   

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. Laithewaite’s data confirms the concepts behind the cylinder and spheres, because energy has been made in the laboratory, in the United Kingdom.

To find Laithwaite’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 A1-A2 (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 = ÂÃ,½*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 Laithwaite’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 1/2mvÂÃ,²

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 don’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 Laithwaite’s experiment should be easy to perform. Are we scientists enough to repeat it?


I have repeated Laithwaite’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 mass’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.