Has somebody used Eric Laithwaite 's device ?

[pequaide]

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

[pese]

Hello Mr Pese

From Wikipedia :    Messy est une commune franÃÆ'Ã,§aise, situÃÆ'Ã,©e dans le dÃÆ'Ã,©partement de Seine-et-Marne et la rÃÆ'Ã,©gion ÃÆ'Ã...½le-de-France.
Are you from Messy, France ? Si vous ÃÆ'Ã,ªtes de cette commune, nous ne sommes pas trÃÆ'Ã,¨s loin l'un de l'autre; je suis de Tournai, prÃÆ'Ã,¨s de Lille

And after that, a discovery of John Logie Baird must be reactivated...
I m sure, Mr Pese, that you know a great part of this John Logie Baird's technology...

No i, was know only, that he was television-inventor
ici vous truez quelques livres de lui

http://books.google.de/books?q=John+Logie+Baird&btnG=Nach+B%C3%BCchern+suchen

ci vous etes interesse, cherchez au premier des e-book.

I have ut an longer message in your "box"

Gustave PeseÂÃ,´

[pequaide]

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 = 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. 

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 1/2mvÂÃ,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?

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

[soliris]

Hello Mr Pequaide, it's a short message; I have to study your answers in details.

The first idea coming to me, is that we have not taken in our mind the Time changes around the axis of the gyroscopes..I will try to explain a little bit later.
You follow a 'master-idea', like we say in French; I love that: this is the only way to change this world.

[pequaide]

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.