"Greg Neill" escreveu na mensagem
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"El Enrrabadore-mor" wrote in message


Euler's equation of motion cannot solve the
problem, nor it seams to work for a sphere.
If the gyroscopic mass is a sphere, all its
mass inertia moments are equal and you
end up with nothing.
Physically, the result for a sphere will be that
a sphere won't precess, contrary to evidence.

Why would you say that? The necessary torque to
result in a precession can be about the sphere's
center of mass, and a gravitational field can
provide that for a sphere spinning on a mount
attached to one of its poles -- such a mount is
not coincident with the center of mass.

My point was that, if the only tool you have
to make the analysis are the "Euler's equation of
motion", hence, for a sphere where I1=I2=I3,
you got nothing to work with:

T1 = I1 dw1/dt + (I3-I2) W2 W3
T2 = I2 dw2/dt + (I1-I3) W3 W1
T3 = I3 dw3/dt + (I2-I1) W1 W2
The result will be:
T1 = I1 dw1/dt
T2 = I2 dw2/dt
T3 = I3 dw3/dt
From which nothing can be said about precession.

I've started to work with Euler's equations
and I've noticed that the link you've provided
is an old page updated.
Equate terms like:
dw2/dt = W3 W1
only happen to be true for the "horizontal",
where theta is 90 degrees.
The link is honest about that and no errors
exist (didn't read about earth).

We only miss the other 180 degree spectrum.
(the remaining 180 is symmetric).

The equation for precession w2 due to
torque T1 is:
T1 = I1 dw1/dt + I3 w3 sin(theta) +
+ (I3-I2) w2^2 sin(theta) cos(theta)
(since I3 and I2 don't mix, there are
4-torque terms).
Theta is the angle to the vertical.
Everything else is obvious to you, for sure.


The Earth is another matter, since it is not
a perfect sphere -- the equatorial bulge
provides a "handle" for torques.

I've had an interesting discussion about that
about half a Year ago.