The Englishman Eric Laithwaite and the Russian Gennadi Shipov have something in common : they made, each one, gyroscopes on free axis. Lathwaite, rejected by the Royal Society of London because of its results of non-Newtonian physics, sent into space a gyroscope which has accelerated the movement itself, and Shipov made gyroscopes whose energy increased.

The key is to know what "free axis" means, for physicists.

Hello Mr Pequaide

I just know that the set of Laithwaite turned faster and faster, and took off like an aircraft, to go as a meteor into deep space.
It is true that there are gyroscopes which are made up of several parties, and some of these parts can stop and even go back in the opposite sense ..
In france, during the sixties, we bought "mobiles" to decorate our living rooms, with wires and wooden batons, which turned.

I don't understand when you tell us that there is increased energy in a rotary system ( The density of energy changes, but not the tolal amount ); Lathwaite was excluded from the Royal Society for having challenged the second law of Newton.
The second law says that ,anyway, the energy always comes from elsewhere, from a known part of our universe...To tell the physicists that the energy comes from itself, or from the torsion waves, is sacrilegious...

I have changed the terms " conservation of energy " by " conservation of energy density, " what do you think about that ?

http://www.google.de/search?hl=de&q=Eric+Laithwaite++Gennadi+Shipov&meta=

PESE

i find more in the web and ebooks now the last 5 jears in internet
as i studied 45 years ago , but the old knowledge frim tubes and
transistor, i have knowledges that are hard (and not) to find.
Ask me over Messy, if Informations need.
G.Pese

Confirmation of data has been obtained

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 – Everything you need to know, after searching ‘Laithwaithe free axis’.   

Of particular interest is figure 7, Laithewaite states that momentum is conserved but energy is not conserved. I actually performed an experiment very similar to his, but the cylinder and spheres yield more energy. Laithewaite’s data confirms the concepts behind the cylinder and spheres.

In Laithwaite’s diagram 7 he starts at a high energy point and goes to a low energy point and then back to a point of high energy. Laithwaite clearly states that he sees momentum conserved not kinetic energy conservation. 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 sliding 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 us start the motion at the low energy point by dropping the sliding (A1 and A2) mass and M1 and M2 from a height of .051 meters. Arrange the masses so that they 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.   

Energy has been made in the laboratory, in the United Kingdom.  Maybe the Royal Society needs to reread principia. Momentum does not equal kinetic energy. mv, 1/2mv²

www.gyroscopes.org/masstran.asp - 22k

Okay; click on the above link 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 = 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. 

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, 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 1/2mv²

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?

Does the experiment conserve momentum and make energy, or does it conserve energy and make momentum?

I have repeated Laithwaite’s figure seven experiment 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 it has half the velocity. This would mean that M1 and M2 always give the center sled half the momentum.

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.

The center of mass staying put also 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.