Discussion of 347(5a): Force Analysis of the Gyroscope

Eq. (18) is:

F sub Z = (v sub X partial v sub Z / partial X + v sub Y partial v sub Z / partial Y
+ v sub Z partial v sub Z / partial Z)

where v = v (r(t), t) is the velocity field of a fluid spacetime, aether or vacuum. So:

v = v(X(t), Y(t), Z(t), t)

and in general the three partials are non zero. One could choose a model flow field for example. In classical dynamics:

v = v(t)

and the three partials vanish.

To: EMyrone@aol.com
Sent: 12/01/2017 23:15:47 GMT Standard Time
Subj: Re: 347(5a): Force Analysis of the Gyroscope

A vanishing force (18) would come out for any non-constant flow v_Z in Z direction. A more realistic case would probably be a circular wave in Z direction. I will try out some examples.

Horst

Am 11.01.2017 um 13:07 schrieb EMyrone:

It is clear from this force analysis that the classical dynamics of the gyroscope never gives a counter gravitational force because of the result (12), the invariant of the galilean transformation. This is everyday observation, because the point of a spinning top for example never lifts off the ground. However in fluid dynamics, there is such a force, given by Eq. (18). It is a vacuum force and is very small (by observation) in a spinning top, but in the Laithwaite or Shipov experiments it could be responsible for their observations. This can be looked in to using computer algebra by precisely defining the configurations using by Laithwaite and Shipov. The force field of Eq. (18) can be plotted with gnuplot. It is is positive valued in the k axis, giving counter gravitation. It is due to the fluid nature of spacetime, aether or vacuum. The video made by Laithwaite needs to be studied very carefully to find exactly what kind of motion he describes. It is clear that no one can hold a forty pound (20 kilo) weight at arms length for very long. Laithwaite does so with ease.

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