On May 8, 1:34 pm, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na ...

On May 8, 6:21 am, "El Enrrabadore-mor"
wrote:
I've been claiming for Years that nobody ever fully explained
the top/gyroscope by means of Classic Mechanics, including
in this thread. (relativity not even has a clue).
I think you have to consider the holonomic constraints. The no-
slip condition creates restraints on the type of motion possible.
As far as your experiments go, it may be possible that your
materials are compressible or that there is slip. Under those
conditions, some of your experimental results may not fit models that
assume rigid body conditions with holonomic constraints.
However, your comments concerning spheres makes me suspect that
your main problem is not considering holonomic constraints. Try taking
a rigid ball, and rolling it on a rigid flat plane. Take a bowling
ball, and roll it on your kitchen counter. The holes will mark the
surface of the ball nicely. Try to force a precession or nutation like
motion on the ball with your hand without displacing its position. I
think you will find it can't be done. With an object of spherical
symmetry, rotation about any axis except the vertical axis results in
finite displacement. An oblong top can be precessed or nutated while
leaving the point of contact stationary. However, a completely
spherical top can't behave like that. The shape has to be elongated to
allow room for the motion.
I am talking about a rigid ball on a rigid immovable surface. Of
course, if the material is compressible the behavior gets more
complicated. If the surface starts rolling, you will get complex
behavior.
This is not true. The precession of spinning objects is fully
explained both in Newtonian and relativistic physics. It is you who
don't know the necessary physics involved. It seems to me that you
haven't even tried.

Do you think you know top/gyroscopes?
Do you know eigenvalues and eignevectors? The moment of inertia
is a tensor matrix. If you perform an eigenanalysis on the moment of
inertia, which is called diagonalization, you will get the the
symmetric axes of the object and the moment of inertia around each
body.
For a spherical body, the solution is three orthogonal axis with an
equal moment of inertia. However, this solution is degenerate. You can
chose any three orthogonal axis and it will be an axis of symmetry for
the sphere. I suspect that may be related to your problem. The
equations you are using are ill-conditioned due to the degeneracy of
the eigenvector solution. This is a math problem, not a physics
problem.
Plugging in numbers into an equation where the solution is
degenerate causes problems. When this happens to me, I go back to the
physics. I look for degeneracy, a place where there is too much
symmetry. In your case, I would look at the spherical body more
closely. Try to make a spherical top, or simulate a spherical top
using your hand. You may find a physical restriction that you weren't
aware of.
Read "Classical mechanics" by Goldberg
Read "Classical mechanics" by Korben and Stehele.
This is not Einsteinian relativity, this is Newtonian mechanics. In
relativity, there is no such thing as a completely rigid body. This
creates problems you may not be completely aware of. However, they
have been solved. Look up "Thomas precession" and "right-angled
lever". In any case, the top problem has be solved even under
relativistic conditions. The key is realizing that the "Thomas
precession" and the "right angled lever" are related. The "right-
angled lever" provides the torque for "the Thomas precession." Have
fun.

"Tom Roberts" escreveu na mensagem
...
El Enrrabadore-mor wrote:
The only thing I can guarantee is that the derivative
of the above solution of the integral, gives the function
used by Tom Roberts in first place.[...]
The right solution must be:
\tau = 1/2*(v(t)/c)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)/c)

This cannot possibly be correct. Differentiate your expression by t and
you must get a factor of v'(t) (i.e. dv(t)/dt), but there is no such
factor in the original integral.

One thing is for sure. The derivative of the solution must
give the original function (inside the integral).

So let's do the derivative of your solution (and mine)
and see who got the correct solution.

1 - Your solution:
\tau = sqrt(1-|v|^2/c^2)

2 - My solution:
\tau = 1/2*(v/c)*sqrt(1-v^2/c^2) + 1/2*arcsin(v/c)

The original integral:
\tau = \integral sqrt(1-v(t)^2/c^2) dt

If you are to be correct, then:
d/dt [sqrt(1-|v|^2/c^2)] = sqrt(1-|v|^2/c^2)
which obviously looks wrong.
(the derivative of the function equal to the function
itself sounds automatically bad).

The derivative of your solution is:
d/dt [sqrt(1-|v|^2/c^2)] = d/dt [sqrt(1-u^2)]
being u = v/c
and du/dt = (1/c) dv/dt
d/dt [sqrt(1-u^2)] = d/dt [(1-u^2)^1/2] =
= 1/2 (1-u^2)^-1/2 * du/dt =
= 1/2 (1/sqrt(1-u^2)) * (1/c) dv/dt =
= 1/(2*sqrt(1-u^2) * (1/c) dv/dt
For a constant velocity v:
d/dt sqrt(1-u^2) = 1/[2c*sqrt(1-(v/c)^2]

Now, the derivative of my solution is a little
bit hard.
d/dt [1/2*(v/c)*sqrt(1-v^2/c^2) + 1/2*arcsin(v/c)]
being:
u = v/c
w = sqrt(1-v^2/c^2)
z = arcsin(v/c) = arcsin(u)

Hence:
1/2 * d/dt [uw + z] = 1/2 [u dw/dt + w du/dt + dz/dt]

du/dt = (1/c) dv/dt

dw/dt = 1/(2*sqrt(1-u^2) * (1/c) dv/dt =
= 1/[2c*sqrt(1-(v/c)^2]*dv/dt

dz/dt = d/dt [arcsin(u)] = 1/(sqrt(1-u^2) * du/dt =
= 1/[sqrt(1-(v/c)^2]*(1/c)dv/dt

Hence:
1/2 d/dt [uw + z] = 1/2 [(v/c)*1/[2c*sqrt(1-(v/c)^2]*dv/dt +
+ sqrt(1-v^2/c^2)*(1/c)*dv/dt + 1/[sqrt(1-(v/c)^2]*(1/c)dv/dt] =

= 1/2 (1/c) dv/dt [ v/[2c*sqrt(1-(v/c)^2] +
+ sqrt(1-(v/c)^2) + 1/[sqrt(1-(v/c)^2] ]

Now, multiplying by 1 = sqrt(1-u^2)/sqrt(1-u^2) to take
out the denominator:
= 1/2 dv/dt (1/c) (1/sqrt(1-(v/c)^2) [v/2c + 1 - (v/c)^2 + 1]
= 1/2 dv/dt (1/c) (1/sqrt(1-(v/c)^2) [2 + v/2c - (v/c)^2]
Damn, ...
Can't see what's wrong.

Any help, please?
Yes, I do see the predicted dv/dt term.


Tom Roberts

"Darwin123" escreveu na mensagem
...
On May 8, 1:34 pm, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na
...

On May 8, 6:21 am, "El Enrrabadore-mor"
wrote:
I've been claiming for Years that nobody ever fully explained
the top/gyroscope by means of Classic Mechanics, including
in this thread. (relativity not even has a clue).
This is not true. The precession of spinning objects is fully
explained both in Newtonian and relativistic physics. It is you who
don't know the necessary physics involved. It seems to me that you
haven't even tried.

Do you think you know top/gyroscopes?
I've already faced tens of sharks around here with that
challenge and I'm fishing right now (why you think I've
changed my name - I'm fishing).

Do you want to try some basic general questions yourself?

Wait, you know something about precession.
(And only precession of an horizontal gyroscope for sure).
What about a tilted gyroscope, 360 degrees around?
Mathematically, the periodicity of the trigonometric functions
creates a bit of a problem because the inverse solution of the
equations isn't unique. I can't see what you are doing. However, I'll
bet your problem is with solving the mathematics rather than a real
problem with the physics.
What about nutation?
What about nutation under precession?
What about precession under nutation?
In the case spherical "top" resting on a flat surface, which is
orthogonal to the gravitational field, there is no nutation or
precession. You can't get the axis of rotation to face in a different
direction than the gravitational field, not without breaking the
surface. I am not using mathematics, I am merely resting a ball on a
surface and trying to roll it. If you try to use mathematics to
describe nutation or precession here, you will probably get
pathological solutions.
This is not a problem in physics. You have to pay close attention
to those trigonometric functions. They are tricky. You may be getting
a perfectly good description of a broken surface. Look up "holonomic
trajectories."
I suspect that such behavior is possible on a surface that tilts
with respect to the gravitational field, or with an object that isn't
really rigid. However, that will take real analysis and I don't really
have the time.

Actually I not interested in spheres.
My gyroscope is the following:
http://www.gyroscope.com/d.asp?product=SUPER2
with gimbals:
http://www.gyroscope.com/d.asp?product=GIMBALS

Spheres are just a funny case, that one can use as a gyroscopic
mass, instead of a disk.
Torque is to be applied around the horizontal axis - T1
(a suspended mass for instance) and the gyro will precess
around the vertical - w2.
You can also force the precession motion, by means of
a second external torque T2, and you will get nutation - w1.

A ball on a flat surface cannot have external torques
applied, hence it's just a ball, not a gyroscope.


What about energy conservation?
What about it? Work-energy theorems show when and where mechanical
energy is preserved. As far as I can tell, there is no fundamental
violation of it. Or did you tilt that surface? You know, when a
spinning object falls, potential energy is changed to kinetic energy.

Yes I agree. Energy must be conserved.


I've already spent more then 4,000 hours in the past 3 Years,
looking for gyroscope equations that could predict
basic experimental results.
How sure are you of the experimental measurements? I mean, are
you really sure about the moment of inertia of your rotators? How do
you calculate them?

I'm not doing fine tuned measurements.
I understand the setup, also the forces involved, I know
the masses, radius, torque and inertia moments.
Basically I measure the time it takes to fall and I count the
number of precession turns given.
I've the gyroscope permanently feed by a motor, or else
I leave it free.
Time is in order of tens of seconds, up to 2 minutes.
The number of turns around 10 to 60.
Precession speeds up to 6 rad/s can be achieved.
It is very easy to draw conclusions and see if equations
fit, or not.
I've it mounted on a head of video cassette recorder,
which has a nice inertia moment for torque T2 and
spins with very small friction.
The centrifugal force is clear: m r w^2
When I change the term "m r", that causes force over
the bearings (with constant friction/torque) I got a nutation
that matches a precession speed square term.
So, the centrifugal force is real. You increase it and
the gyro will fall faster, according to *m r w^2*.


And I'm not doing this as a hobby. I really need then.
I got:
- Newton method done (T=dL/dt)
- Lagrangian method done
Already available in classical mechanics textbooks.

Do you have any with the Lagrange method solved ?


- Vectorial method done
You do know that moment of inertia of a nonspherical body is a
matrix. And that the matrix has to be differentiated?

Yes, but I use principal axis.


All them full solutions done by myself, zero simplification.
(I did that because couldn't find a book, link, whatever).
What I need is a genius that could explain me
how to get energy conservation out of it.
Try Goldstein, "Classical Mechanics," or Korben and Stehle,
"Classical Mechanics." A world of wonder awaits you.

I've already discussed Goldstein with Eric Gisse.
It's Euler's equations of motion and it give up nothing.
More on this below.



Euler's equations of motion don't qualify.
Just try them for I1 = I2 = I3 (sphere).
I learned Newtonian mechanics from
"Classical Mechanics" 2nd ed. by H. C. Corben and Philip Stehle
(Krieger 1960). I looked up nutation in the index. In chapter 9, page
149, it qualitatively describes both precession and nutation of a free
rotator.
On page 150, it shows the Euler equations related to a rotator
(equation 52.6). I can't write the equations in the text format, but
they are well known. When I substitute I1=I2=I3, I get a simple spin
with all three axis uncoupled. Basically, it looks like there is no
rotation or nutation for a spherical object.

There is nothing wrong with spheres.
Spheres do precess and nutate like any disk does.
I can make the disk to precess around a circle so that,
more or less, I have I1 = I2 = I3 and nothing strange happens.
I can have I1 = I2 = m r^2 much larger then the main inertia
moment I3 of the spin. Again nothing to notice.

I'm to suppose Euler's equation of motion are good
for nothing. Maybe, not even fully right after all.
At least, for spheres they don't qualify.
Only if I1=I2 much smaller then I3 they can predict anything.


Why should a spherical
object without friction show precession or nutation?

Why?
Why not?
Spheres do precess. Look at Gravity Probe B.


From your complaints, I have a suspicion that you have a problem
both with the physics and the mathematics. Physically: Nutation and
precession are not required in each and every rigid body. They are the
results of different degrees of motion being coupled together. Think
of coupled harmonic oscillators. For some symmetries, the motions are
decoupled. You can't get beats from two pendulums that aren't
connected in some way. Mathematically: you have to analyze
trigonometric functions very carefully. The inverse of a trigonometric
function isn't unique. For instance, consider differentiation. The
derivative of an inverse doesn't have to be the inverse of the
derivative since the function isn't unique.
I don't have my copy of Goldstein available. Someone borrowed it
and didn't return it. Try "Classical Mechanics" by Goldberg. It may
help you.

"Darwin123" escreveu na mensagem
...
On May 8, 1:34 pm, "El Enrrabadore-mor"
wrote:

Do you think you know top/gyroscopes?
Do you know eigenvalues and eignevectors? The moment of inertia
is a tensor matrix. If you perform an eigenanalysis on the moment of
inertia, which is called diagonalization, you will get the the
symmetric axes of the object and the moment of inertia around each
body.
For a spherical body, the solution is three orthogonal axis with an
equal moment of inertia. However, this solution is degenerate. You can
chose any three orthogonal axis and it will be an axis of symmetry for
the sphere. I suspect that may be related to your problem. The
equations you are using are ill-conditioned due to the degeneracy of
the eigenvector solution. This is a math problem, not a physics
problem.

My equations don't have any trouble to predict the sphere
behaviour, for instant (the usual torque/precession for a
gyroscope moment I3w3):
T1 = I1 dw1/dt + I3 w3 w2 sin(theta) +
+ (I3-I2) w2^2 sin(theta) cos(theta)

So, I guess I could fortunately get out of that degeneracy
you've been talking about.
So, you agree that Euler's equations of motion don't
qualify for a sphere?


Plugging in numbers into an equation where the solution is
degenerate causes problems. When this happens to me, I go back to the
physics. I look for degeneracy, a place where there is too much
symmetry. In your case, I would look at the spherical body more
closely. Try to make a spherical top, or simulate a spherical top
using your hand. You may find a physical restriction that you weren't
aware of.

The best gyroscopes are spherical.
Gravity Probe B gyroscope is the most perfect sphere ever made.


Read "Classical mechanics" by Goldberg
Read "Classical mechanics" by Korben and Stehele.
This is not Einsteinian relativity, this is Newtonian mechanics. In
relativity, there is no such thing as a completely rigid body. This
creates problems you may not be completely aware of. However, they
have been solved. Look up "Thomas precession" and "right-angled
lever". In any case, the top problem has be solved even under
relativistic conditions. The key is realizing that the "Thomas
precession" and the "right angled lever" are related. The "right-
angled lever" provides the torque for "the Thomas precession." Have
fun.

Many thanks.
Can you point out any solutions besides Euler's equation
of motion? They look like this:
T1 = I1 dw1/dt + (I3-I2) W2 W3
T2 = I2 dw2/dt + (I1-I3) W3 W1
T3 = I3 dw3/dt + (I2-I1) W1 W2

On May 8, 9:03 pm, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na ...

On May 8, 1:34 pm, "El Enrrabadore-mor"
wrote:
Do you think you know top/gyroscopes?
Do you know eigenvalues and eignevectors? The moment of inertia
is a tensor matrix. If you perform an eigenanalysis on the moment of
inertia, which is called diagonalization, you will get the the
symmetric axes of the object and the moment of inertia around each
body.
For a spherical body, the solution is three orthogonal axis with an
equal moment of inertia. However, this solution is degenerate. You can
chose any three orthogonal axis and it will be an axis of symmetry for
the sphere. I suspect that may be related to your problem. The
equations you are using are ill-conditioned due to the degeneracy of
the eigenvector solution. This is a math problem, not a physics
problem.

My equations don't have any trouble to predict the sphere
behaviour, for instant (the usual torque/precession for a
gyroscope moment I3w3):
T1 = I1 dw1/dt + I3 w3 w2 sin(theta) +
+ (I3-I2) w2^2 sin(theta) cos(theta)
Assuming that you have a "perfect sphere," you probably don't
have a "perfect torque." I mean, this formula seems to imply that only
the torque around the first axis is significant. There are no other
torques. You seem to be implying that:
T2=T3=0.
Furthermore, I have no idea of how you determined T1.
What if the direction of the torque is changing? Usually, the
assumption is that the torque is constant while the moment of inertia
matrix is constant. I am convinced (?) by your certainty (?) that you
know the moment of inertia. However, I am not sure you know where the
torque is coming from.


So, I guess I could fortunately get out of that degeneracy
you've been talking about.
So, you agree that Euler's equations of motion don't
qualify for a sphere?

Plugging in numbers into an equation where the solution is
degenerate causes problems. When this happens to me, I go back to the
physics. I look for degeneracy, a place where there is too much
symmetry. In your case, I would look at the spherical body more
closely. Try to make a spherical top, or simulate a spherical top
using your hand. You may find a physical restriction that you weren't
aware of.

The best gyroscopes are spherical.
Gravity Probe B gyroscope is the most perfect sphere ever made.
Really? How is this gyroscope initially spun? It doesn't have an
axle. Not if its "the most perfect sphere ever made." I was trying to
figure out how an electromotive force could give it a spin. I figure
you must be running electric current through it, and placing it in an
magnetic field.
Note: If you are using electromagnetic induction to spin the
sphere, the torque isn't just caused the mechanical force of your arm
on the axle. It also has a component due to the mechanical force of
the magnet doing the spinning.
I think you may have a weird torque, not a deviation from Euler's
equation. The torque probably depends on time and maybe is a function
of angular velocity. You found an empirical formula that sort of fits
on the assumption that only the mechanical force on the axle causes
torque. However, I'll bet you have a magnet somewhere around that is
spinning the sphere.
You didn't say that your gyroscope is running down. So you are
providing a "constant" flow of energy somewhere. Is your instrument
drawing on electric current? If so, look to the wires. You may find a
wire stressed where you least suspect it.

Many thanks.
Can you point out any solutions besides Euler's equation
of motion? They look like this:
T1 = I1 dw1/dt + (I3-I2) W2 W3
T2 = I2 dw2/dt + (I1-I3) W3 W1
T3 = I3 dw3/dt + (I2-I1) W1 W2
No. I don't know of any. However, I know various transformations
on the Eulers equations. You haven't exhausted the Euler equation yet.
I would suggest going into a deeper analysis of torque. List all
sources of external torque, not just the torque the gyroscope is
resisting. Write a function for T1, T2, and T3 as a function of time
and angular velocity. I don't know your apparatus. However, I really
doubt you fully understand the source of torque.
What if the torque in your apparatus was some matrix times the
angular velocity vector. What matrix would give your empirical result?
I know this would be a fit, not physics. However, your spin inducing
apparatus may be producing a strange torque for reasons that have
nothing to do with Eulers equations.
T1(w1,w2,w3), T2(w1,w2,w3), T3(w1,w2,w3) ?

El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
[...]
1 - Your solution:
\tau = sqrt(1-|v|^2/c^2)

You're not paying attention, and you are MISQUOTING. I said "For
circular motion, v(t) is constant," (TYPO: I of course meant |v(t)| is
constant.)
\tau = sqrt(1-|v|^2/c2) (T2 - T1)

Just MISquoting the formula is not sufficient, you must also read the
words that give the conditions under which the formula is valid.


[...]
Can't see what's wrong.

What is wrong is that it is not possible to perform that integral
without knowing the functional form of v(t). When |v(t)| is constant,
the integral is trivial, and has the value given above. When |v(t)| is
not constant, you must know its functional form in order to do the integral.


Tom Roberts

"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
...
El Enrrabadore-mor wrote:
The only thing I can guarantee is that the derivative
of the above solution of the integral, gives the function
used by Tom Roberts in first place.[...]
The right solution must be:
\tau = 1/2*(v(t)/c)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)/c)

This cannot possibly be correct. Differentiate your expression by t
and you must get a factor of v'(t) (i.e. dv(t)/dt), but there is no
such factor in the original integral.

One thing is for sure. The derivative of the solution must
give the original function (inside the integral).

So let's do the derivative of your solution (and mine)
and see who got the correct solution.

1 - Your solution:
\tau = sqrt(1-|v|^2/c^2)

No, it's \tau = sqrt(1-|v|^2/c^2)*t

where t is the time in the observer frame.
Note that here the sqrt() term is a constant since
you chose a fixed velocity, v.


2 - My solution:
\tau = 1/2*(v/c)*sqrt(1-v^2/c^2) + 1/2*arcsin(v/c)

The original integral:
\tau = \integral sqrt(1-v(t)^2/c^2) dt

If you are to be correct, then:
d/dt [sqrt(1-|v|^2/c^2)] = sqrt(1-|v|^2/c^2)
which obviously looks wrong.

Because it should be:

d/dt [sqrt(1-|v|^2/c^2)]*t = sqrt(1-|v|^2/c^2)

On May 8, 6:50 am, Tom Roberts wrote:

Do you have an equation, dimensionally consistent, where one
could see what you are talking about?

Consider an object moving relative to an inertial frame with a velocity
v(t) (= dr/dt where r is its position 3-vector relative to that frame,
and t is the time coordinate of the frame); v(t) can be an arbitrary
function of time. Its elapsed proper time between t=T1 and t=T2 is:

\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

(This is easily obtained by integrating the metric along the path of the
object.)

One trivially obtains \tau=T2-T1 for an object at rest in this frame.
And one clearly obtains \tauT2-T1 for nonzero v(t). Note that |v(t)| is
constrained to be less than c, so there is never a numerical problem
with the sqrt(.).

Proper time is utter nonsense. It is an abstract excuse to validate
the Loretnz transform. What a person ages, the measure of aging is in
time and not in proper time. It is time for you as an experimental
physicist to do your job. Michelson also an experimental physicist
understood this point. That is why he rejected the Lorentz transform
because the Lorentz transform is just nonsense. It manifests the
twin’s paradox.

"Darwin123" escreveu na mensagem
...
On May 8, 9:03 pm, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na
...

On May 8, 1:34 pm, "El Enrrabadore-mor"
wrote:
Do you think you know top/gyroscopes?
Do you know eigenvalues and eignevectors? The moment of inertia
is a tensor matrix. If you perform an eigenanalysis on the moment of
inertia, which is called diagonalization, you will get the the
symmetric axes of the object and the moment of inertia around each
body.
For a spherical body, the solution is three orthogonal axis with an
equal moment of inertia. However, this solution is degenerate. You can
chose any three orthogonal axis and it will be an axis of symmetry for
the sphere. I suspect that may be related to your problem. The
equations you are using are ill-conditioned due to the degeneracy of
the eigenvector solution. This is a math problem, not a physics
problem.

My equations don't have any trouble to predict the sphere
behaviour, for instant (the usual torque/precession for a
gyroscope moment I3w3):
T1 = I1 dw1/dt + I3 w3 w2 sin(theta) +
+ (I3-I2) w2^2 sin(theta) cos(theta)
Assuming that you have a "perfect sphere," you probably don't
have a "perfect torque." I mean, this formula seems to imply that only
the torque around the first axis is significant. There are no other
torques. You seem to be implying that:
T2=T3=0.

Well, the torque T3=0, theoretically. T3 is the torque around the
spinning axis, assumed to be zero. It is assumed zero friction
around gyro main bearings. Or else, I've a small DC motor
feed by batteries whose voltage/current can be controlled, so
that I can keep a constant angular speed w3 against friction.

Every text book I know considers torque T2 = 0.
T2 = 0 is the only theoretical approach known for gyroscopes.
In real world we have mechanical systems with friction and
torque T2 cannot be zero, ever. Only Gravity Probe B running
on magnetic fields can eventually be considered T2 = 0.
Such torque T2 will cause nutation and, for friction, it causes
a negative torque that causes a positive "falling" of the gyro.
That's why all gyros fall, not because the main gyroscopic
mass I3w3 is running slow with time.

What happens is that all torques T1, T2, T3 and all axes
I1, I2, I3 are orthogonal to each other. Therefore, they are
all independent from each other.
Due to said orthogonality, we go independent equations
for each of every torque. That's what Euler's equation of
motion and all text books say. That's logic too, and I agree.
That's a basic stone of the theory and a hope to be right.

The equation for T2, that causes the nutation w1, is much
more complicated and not covered on text books.
All text books consider that the gyroscope never falls,
hence T2 = 0, and the entire World lives happy with that.

The major challenge required was to define the potential
field V that gives rise to the nutation effect do to torque T:
F = - grad V
T = - grad V


Furthermore, I have no idea of how you determined T1.

T1 and T2 equations have been derived based on 3 methods:
1 - Newton: T = dL/dt
2 - Lagrange L = E - V
3 - Vectorial: T = w x L (cross product)


What if the direction of the torque is changing? Usually, the
assumption is that the torque is constant while the moment of inertia
matrix is constant. I am convinced (?) by your certainty (?) that you
know the moment of inertia. However, I am not sure you know where the
torque is coming from.

Torque changes all the time.
T1 = m g r sin(theta) (for gravity action = most usual)
T2 = user defined


So, I guess I could fortunately get out of that degeneracy
you've been talking about.
So, you agree that Euler's equations of motion don't
qualify for a sphere?

Plugging in numbers into an equation where the solution is
degenerate causes problems. When this happens to me, I go back to the
physics. I look for degeneracy, a place where there is too much
symmetry. In your case, I would look at the spherical body more
closely. Try to make a spherical top, or simulate a spherical top
using your hand. You may find a physical restriction that you weren't
aware of.

The best gyroscopes are spherical.
Gravity Probe B gyroscope is the most perfect sphere ever made.
Really? How is this gyroscope initially spun? It doesn't have an
axle. Not if its "the most perfect sphere ever made." I was trying to
figure out how an electromotive force could give it a spin. I figure
you must be running electric current through it, and placing it in an
magnetic field.

Yes, magnetic field are the source for Gravity Probe B.
No torques are applied over the gyroscope of Gravity Probe.
The aim of Gravity Probe B was to find if there was any possible
torque applied. That is, no precession/nutation should be measured
unless something external exists that causes precession/nutation.


Note: If you are using electromagnetic induction to spin the
sphere, the torque isn't just caused the mechanical force of your arm
on the axle. It also has a component due to the mechanical force of
the magnet doing the spinning.
I think you may have a weird torque, not a deviation from Euler's
equation. The torque probably depends on time and maybe is a function
of angular velocity. You found an empirical formula that sort of fits
on the assumption that only the mechanical force on the axle causes
torque. However, I'll bet you have a magnet somewhere around that is
spinning the sphere.

I've no problem with applied external torques.
Nevertheless, you comment is very important, indeed.
The axis 2 and T2 are not good to work with.
I'm working on the vertical FIXED axis z and a torque Tz and an
angular speed wz have been defined.
Tz = T2 sin(theta)
wz = w2
(simple trigonometry plus a fact seen and explained at least in
one text book - Harvard / Cambridge 2008).


You didn't say that your gyroscope is running down. So you are
providing a "constant" flow of energy somewhere. Is your instrument
drawing on electric current? If so, look to the wires. You may find a
wire stressed where you least suspect it.

Many thanks.
Can you point out any solutions besides Euler's equation
of motion? They look like this:
T1 = I1 dw1/dt + (I3-I2) W2 W3
T2 = I2 dw2/dt + (I1-I3) W3 W1
T3 = I3 dw3/dt + (I2-I1) W1 W2
No. I don't know of any. However, I know various transformations
on the Eulers equations. You haven't exhausted the Euler equation yet.

And what transformations on the Euler's equations are there?
Can you explain please?


I would suggest going into a deeper analysis of torque. List all
sources of external torque, not just the torque the gyroscope is
resisting. Write a function for T1, T2, and T3 as a function of time
and angular velocity. I don't know your apparatus. However, I really
doubt you fully understand the source of torque.
What if the torque in your apparatus was some matrix times the
angular velocity vector. What matrix would give your empirical result?
I know this would be a fit, not physics. However, your spin inducing
apparatus may be producing a strange torque for reasons that have
nothing to do with Eulers equations.
T1(w1,w2,w3), T2(w1,w2,w3), T3(w1,w2,w3) ?

"Darwin123" escreveu na mensagem
...

However, your spin inducing
apparatus may be producing a strange torque for reasons that have
nothing to do with Eulers equations.
T1(w1,w2,w3), T2(w1,w2,w3), T3(w1,w2,w3) ?

The main issue here is "Conservation of Energy".
We are to assume that energy is conserved, that's for sure.

How do we express the conservation of energy?

The truth is that one cannot draw energy conclusions, since
we are working with velocities (w1, w2, w3) and we have
no means to know the angular displacements (theta, phi, psi).
One cannot integrate and, even if we do, the facts are that
the gyro produces tens of turns and in no way an integral
can ever express that number of turns.

My point is that Power is Conserved, instead.
Power conservation follows from Energy conservation
since time is not relativistic.
My point is that: T1 w1 = Tz wz = T2 w2

That is, T3 and w3 is out of the Energy Conservation
problem.

I'm to assume that T3, w3 are an isolated system,
which must hold true since axis 1 and 2 are independent.
If axis 1 and 2 are independent, why should axis 3 be also
independent?

Therefore, we are to assume all axes (1, 2, 3) to be
independent, one from each other.
That is, in the process of precession, nutation and
variable external torques (that we can calculate), the
gyro main angular speed w3 remains constant (except
for bearings friction). Hence, T3 is only due to the main
bearings friction, we can neglect or else compensate
to keep angular velocity w3 constant.

Do you agree with the above picture?

Do you believe it's possible that a torque T3 can arise in
the process, in order to spin up/down the gyro main
angular velocity w3?