"Tom Roberts" escreveu na mensagem
t...
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.

I didn't misquote your formula by propose.
I just didn't realise that you've simplified that much the problem.
Such simplification makes your solution ridiculous, that's all.
So, I never realise that one could make such ridiculous simplification:
What you say is that:
1 - The velocity v is constant.
2 - Hence, the sqrt(1-(v/c)^2) is a constant term K.
3 - Therefore, \integral K = K t

Without such ridiculous simplification, to include "t" or not doesn't
make any difference, since it is obvious that dt/dt = 1.
Because dt/dt = 1 is so obvious, I didn't care about it.
Hence, I didn't misquote your formula by propose.




[...]
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.

That's the ridiculous part of relativity approach to the
rigid disk. When |v(t)| is constant, it doesn't matter if it's circular
motion or linear motion, you always got Lorentz equation.

To compare, or to say, that circular motion is the same
thing as linear motion, sounds ridiculous to me.

I'm about to say that, relative to the XXI Century, relativity
theory is in its Stone Age. ...and people stoned with it.


Tom Roberts

"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
t...
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.

I didn't misquote your formula by propose.
I just didn't realise that you've simplified that much the problem.
Such simplification makes your solution ridiculous, that's all.
So, I never realise that one could make such ridiculous
simplification: What you say is that:
1 - The velocity v is constant.
2 - Hence, the sqrt(1-(v/c)^2) is a constant term K.
3 - Therefore, \integral K = K t

Without such ridiculous simplification, to include "t" or not doesn't
make any difference, since it is obvious that dt/dt = 1.
Because dt/dt = 1 is so obvious, I didn't care about it.
Hence, I didn't misquote your formula by propose.

If you want to consider the general case where v(t) does
not have a constant magnitude, then you must integrate
over the actual function. This is obviously more
difficult that assuming constant speed (either linear or
circular motion). In fact, for most functions v(t) there
will not be an easily obtained closed form solution, or
perhaps no closed form solution. In such cases one
usually resorts to numerical integration or an approximation
method.

When you did the integral you made the error of assuming
that dv/dt = dt, which it won't be in general.

If you have an integral such as \Integral _a^b U(t)dt,
you cannot assume that dU(t) = t. You need to have
U(t)du.





[...]
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.

That's the ridiculous part of relativity approach to the
rigid disk.

Absurd. You're miffed because the math for the general case
might be hard.

When |v(t)| is constant, it doesn't matter if it's
circular motion or linear motion, you always got Lorentz equation.

Indeed. What a pleasing result! Wouldn't it be more
strange if, despite the obvious analogs between the
mathematical physics of linear and circular motion
(v ~ w, T ~ F, L ~ momentum, etc., and the functional forms
of rotational and linear energy), that relativity would
predict different forms for the time dilation effect?


To compare, or to say, that circular motion is the same
thing as linear motion, sounds ridiculous to me.

They are clearly not the same thing, as rotational and
linear momenta are separately conserved. Yet the
mathematical form for dealing with them run along parallel
lines.


I'm about to say that, relative to the XXI Century, relativity
theory is in its Stone Age. ...and people stoned with it.

Uh oh.

El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
t...
For circular motion, |v(t)| is constant:
\tau = sqrt(1-|v|^2/c2) (T2 - T1)

I just didn't realise that you've simplified that much the problem.

Look at the subject of this thread. "simplifying" to circular motion is
what is being discussed.


Such simplification makes your solution ridiculous, that's all.

It's not "ridiculous", it is what this thread is about. I gave a general
equation (in response to an inquiry), and then applied it to the subject
of this thread. shrug


[...]
To compare, or to say, that circular motion is the same
thing as linear motion, sounds ridiculous to me.

What is ridiculous is your "sound bite" approach to physics.

Nowhere does anybody claim "circular motion is the same thing as linear
motion". But when the formula in question depends only on |v|, and when
one considers a constant |v| (which can apply to either linear or
circular motion), then THE FORMULA gives THE SAME ANSWER for the two
types of motion. This does not in any way imply the motions themselves
are "the same".

This is not limited to just linear and circular motion
-- ANY motion with constant |v| obtains the same answer
from this formula.


I'm about to say that, relative to the XXI Century, relativity
theory is in its Stone Age. ...and people stoned with it.

Obviously you do not understand it, so your opinion is irrelevant.


Tom Roberts

On May 9, 8:14 am, Tom Roberts wrote:
El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
et...
For circular motion, |v(t)| is constant:
\tau = sqrt(1-|v|^2/c2) (T2 - T1)

I just didn't realise that you've simplified that much the problem.

Look at the subject of this thread. "simplifying" to circular motion is
what is being discussed.

Such simplification makes your solution ridiculous, that's all.

It's not "ridiculous", it is what this thread is about. I gave a general
equation (in response to an inquiry), and then applied it to the subject
of this thread. shrug

[...]
To compare, or to say, that circular motion is the same
thing as linear motion, sounds ridiculous to me.

What is ridiculous is your "sound bite" approach to physics.

Nowhere does anybody claim "circular motion is the same thing as linear
motion". But when the formula in question depends only on |v|, and when
one considers a constant |v| (which can apply to either linear or
circular motion), then THE FORMULA gives THE SAME ANSWER for the two
types of motion. This does not in any way imply the motions themselves
are "the same".

This is not limited to just linear and circular motion
-- ANY motion with constant |v| obtains the same answer
from this formula.

I'm about to say that, relative to the XXI Century, relativity
theory is in its Stone Age. ...and people stoned with it.

Obviously you do not understand it, so your opinion is irrelevant.

Tom Roberts

Yeah, In the past I equated
2x acceleration x radius = velocity squared.
2ar = V^2.
and subbed a=GM/r^2 then V^2 = 2GM/r,
where V is escape Velocity.
Grist's hit me over the head when I done that.
It's close enough ;-), but it really needs to
reveal more subtle effects.
Regards
Ken S. Tucker

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


Do you agree with the above picture?
At this point, I do not have a clear image of your apparatus. I
also don't fully understand the deviation from theory that you claim
to be seeing.
I am a physicist, not an engineer. However, I can provide some
hypotheses based on general engineering considerations. I am not
saying that you are wrong, but have you looked into...

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?
Yes, in principal. I don't know if this is true in your system,
but I will hypothesize some possible sources of T3.
I think the general engineering approach is to assume the
conservation laws and ignore dynamical laws like the Euler equations.
The conservation laws have an amazing ability to police themselves
regardless of mechanical detail. Dynamical laws require detailed
information. However, since the dynamical laws are what you are
questioning, you have to go into some of that detail.
1) Your electric motor. An electric motor is not a simple system
in the sense of Principia because it uses electromagnetic fields. Some
of the angular momentum is stored by the fields, which makes it a bit
more complicated.
No, I don't have direct experience with electric motors. Its one
of the reasons I went into optics.
For example, there is a magnet in your motor that applies a
"static" magnetic field on the coils. What is the torque on that
magnet? I seriously doubt that the third component on the magnet ic
zero. The reason is that the motion of the coils has to drag the
currents with them. If a wire is maintaining an electric current, and
you move the wire in a direction orthogonal to that wire, you have
created a component of current in the direction of motion. The charge
carriers are confined to the wire, so they have to go where the wire
goes. The magnetic field will apply a force to that component. So yes,
I think an electric motor will apply torques where you don't expect
them.
2) Elastic vibrations. One possibility is that T3 is being caused by
elastic vibrations that couple T1 and T2 to your T3. The result may be
beats between vibrations that appear to be precession or nutation,
even if Euler's equations don't predict them.
Euler's Laws are valid for rigid bodies. However, real things bend
and even flow. Furthermore, the larger the object the less rigid the
object becomes. The cube-square applies. So even if you made your
apparatus out of invar-admantium alloy, out of a Marvel comic, if it
is large it will act like a rubber band.
I once was shopping for an optical table. Read a catalogue from a
company that makes optical tables. The catalogues give specifications
as to the resonances and vibrations these optical tables will
generate. You will get to see that it is impossible to completely
ignore vibration. The optical table manufacturers try damp one
resonance and another pops up at a different frequency.
That is one thing you can try. If it is a case of elastic
vibration, you can try to damp particular modes of vibration. You may
not get rid of the problem, but the system may deviate in a completely
different way. If you try to damp a vibrational mode and your
"nutation" changes amplitude, that would be a clue. Of course, you
have to identify the possible modes of vibration first.
3) Displacement. Your apparatus may be bouncing up and down. But this
I will not expand on.

On May 9, 8:03 am, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na ...
Sorry, I took a lunch break. There is a fourth thing I would like
you to consider.
My point is that: T1 w1 = Tz wz = T2 w2

That is, T3 and w3 is out of the Energy Conservation
problem.
Maybe. But what about the power phase?

You have some electrical engineering background. Have you
considered that the is a phase difference between voltage and current?
This is especially important if you use an electric motor. Electric
motors have a power phase that changes with the work done by the
electric motor.
A phase difference will also apply to all mechanical counterparts
to your system. For example, there may be a phase, which I designate
as Ai (i=1,2,3), between your torques and your angular velocities. So
maybe,
T1 w1 cos(A1) = Tz wz cos(A2)= T2 w2 cos(A3)
Maybe you can make an approximation A1=A2=A3. However, I doubt
that Ai=0 in your system for all i. Your time averaged power has to
have a phase correction in it.

Ken S. Tucker wrote:
Yeah, In the past I equated
2x acceleration x radius = velocity squared.
2ar = V^2.

Your "2" looks wrong. In Newtonian mechanics, for any object moving
uniformly in a circle, a=v^2/r. There is no "2".


and subbed a=GM/r^2 then V^2 = 2GM/r,
where V is escape Velocity.

To compute the Newtonian escape velocity V as a function of radius R,
simply set an object's kinetic energy to be equal and opposite to its
gravitational potential energy at radius R from mass M:

0.5*m*V^2 = G*m*M/R = V^2 = 2*G*M/R

Note there is no acceleration involved in this computation. All that is
needed is that the object's total energy at r=infinity be zero.


Grist's hit me over the head when I done that.
It's close enough ;-), but it really needs to
reveal more subtle effects.

I have no idea what you are trying to say. If you mean that GR has more
subtle effects than this Newtonian computation, then sure.


Tom Roberts

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


Do you agree with the above picture?
At this point, I do not have a clear image of your apparatus. I
also don't fully understand the deviation from theory that you claim
to be seeing.

I've the deviation from theory in a 42+20 pages long so far, but
can be made much shorter once I finish the story.
The problem looks very difficult to work here. I need a support
Web page for that, otherwise describe 3 orthogonal coordinate
system, Euler angles and math that most of the equations are
a full line using dots for derivatives, that will be 4 lines long
here.

I've been lucky. On a search for gyroscopes I've found a book
to download - looking for errors. It was a 10 Years old work
from David Morin - Harvard professor - Now available by
Cambridge press since January 2008.
I've digested those pages deep down up to the minimum
detail. The methods I've talked about are all there.
The Newton method is fully solved. The Lagrangian method
provides a clue on the equations, but misses the second
potential T2 = - grad V.
The required potential to solve the secondary axis of
precession (nutation) was engineered by me.
Most of the pages are the work about this, seen by all
possible angles, on the Lagrangian solution.
Both, Newton and Lagrange fully agree.
The vectorial method is a footnote of 2 lines in the book.
Nevertheless, such method pulled out of the hat work
99% perfect. Yes, not 100% perfectly.
That has been a pain in the ass, since the vectorial
method is by far the very best one can have to take
a picture on what is going on. Each term of every equation
is a vector (torque) that one can draw upon the
coordinate system.
I've already ****ed my head on that picture and
I still have the feeling something escapes me.

So, Convervation of Energy was my life save.
Like I told you, I can't see any reason why the torque
T3 should be involved in the process. No vector
exists that can cause it. Only magic can do it.
But gyroscopes are all about magic, and you know that.

I'm very sure that T3 is out of the problem.
The Harvard Book deviation is based on that.
The book says dw3/dt = 0.

Now the funny thing. If dw3/dt = 0 the conservation
of power/energy is the solution I've told you:
T1 w1 = T2 w2 = Tz wz

The vectorial method gives a perfect match for
energy conservation.
The Newton/Lagrange doesn't, because includes the
acceleration dw1/dt and somehow two of the terms
got a number 2 multiplier, one due to the derivative
of a squared term, the other due to the sum (I1 + I2)
and I1 = I2.

What is left out, the difference between Newton and
vectorial methods, is a second order differential
equation I've solved and gives the perfect behaviour
of the secondary axis of precession (the one that is
not usual to be described).
Such new equation was very exciting, but in fact
from the engineering point of view I'm working
with peanuts. That is, the value of the torque that
those terms amount are less then 1% of the big
load involved in the process.

So, I'm ****ed up and puzzled.

The apparatus is a Wave Energy Converter.
I want to run the gyroscope like a gearbox to
speed up motion from Ocean Waves, up and down.


I am a physicist, not an engineer. However, I can provide some
hypotheses based on general engineering considerations. I am not
saying that you are wrong, but have you looked into...

That's very good, since this problem is basically
related to Physics.
The apparatus was investigated from the one University
in the UK connected to the main manufacture of those
devices, based on oil pumps and motors, which is only
25% efficient. They give up do to the lack of understanding
of the gyroscope only, by written. Well, I've pressed then
a little too, but they deserved, specially because they
talk about complex conjugate control and don't have
a clue on impedances.
The gyroscope can mimic whatever impedance I
desire. It can be a damper, a spring or a mass.
Full spectrum control.
Yep, that's a strange gearbox indeed.

Thanks for the remarks, but this is not an
engineering problem. No engineering problems exist
besides the fact that gyroscope bearings can't handle
the load, but I've fixed that too - many gyros in
parallel, what else?

The problem is the trial top / gyroscope problem
from every text book.

The question is, does the top in the process of precession
and nutation, whatever you do to the top, increase or
decrease its main angular speed?
The speed decrease do to friction on the table, or
friction on bearings, is out of the problem.
I mean due to other sources different from what is
trivial. Say, frame-dragging or so (kidding)?

Tom et al.
Best essay I've found on the subject under
discussion is, Pauli's "Theory of Relativity"
pg.150, *53 Simple deductions from the principle
of relativity*.

On May 9, 1:35 pm, Tom Roberts wrote:
Ken S. Tucker wrote:
Yeah, In the past I equated
2x acceleration x radius = velocity squared.
2ar = V^2.

Your "2" looks wrong.

V(e) = sqrt(2GM/r) is escape Velocity.

g_44 = 1 - ( V(e)/c )^2

P= GM/r.

In Newtonian mechanics, for any object moving
uniformly in a circle, a=v^2/r. There is no "2".

In that case use,

P= 1/2 (wr)^2

which is the "centrifugal force potential".
(w is angular velocity).


and subbed a=GM/r^2 then V^2 = 2GM/r,
where V is escape Velocity.

To compute the Newtonian escape velocity V as a function of radius R,
simply set an object's kinetic energy to be equal and opposite to its
gravitational potential energy at radius R from mass M:

0.5*m*V^2 = G*m*M/R = V^2 = 2*G*M/R

Note there is no acceleration involved in this computation. All that is
needed is that the object's total energy at r=infinity be zero.

Grist's hit me over the head when I done that.
It's close enough ;-), but it really needs to
reveal more subtle effects.

I have no idea what you are trying to say. If you mean that GR has more
subtle effects than this Newtonian computation, then sure.

That's right, that's a weak field, small velocity
calculation.
In the quoted article Pauli writes,
"the transverse Doppler effect and the time dilation
produced by gravitation appear as two different modes
of expressing the same fact".

What do you think he means by that?
Regards
Ken S. Tucker

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



When |v(t)| is constant, it doesn't matter if it's
circular motion or linear motion, you always got Lorentz equation.

Indeed. What a pleasing result! Wouldn't it be more
strange if, despite the obvious analogs between the
mathematical physics of linear and circular motion
(v ~ w, T ~ F, L ~ momentum, etc., and the functional forms
of rotational and linear energy), that relativity would
predict different forms for the time dilation effect?

Since time dilatation is something that is very hard for
me to accept, I will say that maybe.

Imagine that the original inquiry about the circular motion
is a 100 seconds-light radius motion around Saturn.
People here on Earth will see that circular motion from an
external point of view of the circle.
It will be like looking to a spinning wheel and compare
clock rates from the point of view of a fixed external
observer. The symmetry of the problem simply cancels
out any possible time dilatation.
Therefore, we have the same circular motion problem and
two different inertial observer (one at the center of the circle,
the other an external observer) that disagree on time dilatation.
Which one is right?


To compare, or to say, that circular motion is the same
thing as linear motion, sounds ridiculous to me.

They are clearly not the same thing, as rotational and
linear momenta are separately conserved. Yet the
mathematical form for dealing with them run along parallel
lines.


I'm about to say that, relative to the XXI Century, relativity
theory is in its Stone Age. ...and people stoned with it.

Uh oh.

Indeed.

My point, that you didn't comment so far, is that effects have
consequences.
Relative to the original inquiry of this thread, you didn't comment
on the main issue. Both observers send electromagnetic signals
to each other. The distance between observers doesn't change
and light speed is isotropic and constant.
Why the moving observer see all signals send by the central
observer and the observer and the central observer can't ever
receive any signal from the moving observer (at the limit)?
Do you see the consequences of a wrong analysis?

The right analysis is that time dilatation is a consequence
of the speed of light and the DISTANCE travelled by light.
That is, time dilatation only is required for observations made
at distance, by means of light, like Einstein said.
Not the actual bull**** that follows because physicists only
have a stone hammer to work with.