| Topic: |
Science > Physics |
| User: |
"Edward Green" |
| Date: |
25 Apr 2006 05:53:20 PM |
| Object: |
Instantaneous axis of rotation? |
Circumstantial evidence leads me to believe that the "instantaneous
axis of rotation" of a rigid body is a meaningful concept. I hear that
a sufficiently asymmetric body will wobble (defined as being what such
a body does under these circumstances) even in the absence of external
torques, and that the explanation of this involves a tensor connecting
the angular momentum and said instantaneous axis.
On the other hand, thinking of ways to describe an arbitrary motion of
a rigid body about its center of mass, I come up with the following:
from the center of mass, draw a arbitrary reference ray fixed relative
to the body. This ray will intersect a sphere drawn around the center
of mass at one point, which we may call a pole. The velocity of the
pole on the sphere has two degrees of freedom. Specifying this
two-dimensional velocity at an instant, we are then left with one more
degree of freedom in the motion, which we may quantify by the following
construction:
Draw a plane containing the center of mass and the reference ray,
perpendicular to the intantaneous velocity vector of the pole, and a
second plane formed by the same construction a time dt later. Consider
an aribtrary body point on the initial plane and off the ray. If the
point remains on the similar plane after dt then we say the body is not
rotating about the ray. If, as will generally be the case, the point
has moved off the plane, we infer from this displacement a value of
w(t), the instantaneous angular velocity of the body about the
reference ray.
Well, this definition of w seems reasonable, does it not. One problem
-- the reference ray was arbitrary! We have described the intantaneous
rate of rotation about an arbitrary body fixed axis, not "the"
instantaneous axis of rotation, should such a thing exist. What
singles out a particular axis when apparently any one will do?
Hmm... body fixed axis... body fixed axis... maybe there's my problem.
We don't want to describe the rotation about an arbitrary axis fixed in
the body, we want to identify an axis _not_ fixed to the body, in
general wandering around the body, which is the best of all possible
axes which the body can be said to be spinning around for that moment.
But how? And what is the mathematical expression of "best" here? Is
there an axis such that the displacement of any body point is given
purely the the distance from that axis and w(t), so that the wandering
of the axis itself may be ignored in determining these displacements to
first order in small quantities?
Intuition flags. Insight invited.
.
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| User: "John C. Polasek" |
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| Title: Re: Instantaneous axis of rotation? |
06 May 2006 09:01:03 PM |
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On 25 Apr 2006 15:53:20 -0700, "Edward Green"
<spamspamspam3@netzero.com> wrote:
Circumstantial evidence leads me to believe that the "instantaneous
axis of rotation" of a rigid body is a meaningful concept. I hear that
a sufficiently asymmetric body will wobble (defined as being what such
a body does under these circumstances) even in the absence of external
torques, and that the explanation of this involves a tensor connecting
the angular momentum and said instantaneous axis.
On the other hand, thinking of ways to describe an arbitrary motion of
a rigid body about its center of mass, I come up with the following:
from the center of mass, draw a arbitrary reference ray fixed relative
to the body. This ray will intersect a sphere drawn around the center
of mass at one point, which we may call a pole. The velocity of the
pole on the sphere has two degrees of freedom. Specifying this
two-dimensional velocity at an instant, we are then left with one more
degree of freedom in the motion, which we may quantify by the following
construction:
Draw a plane containing the center of mass and the reference ray,
perpendicular to the intantaneous velocity vector of the pole, and a
second plane formed by the same construction a time dt later. Consider
an aribtrary body point on the initial plane and off the ray. If the
point remains on the similar plane after dt then we say the body is not
rotating about the ray. If, as will generally be the case, the point
has moved off the plane, we infer from this displacement a value of
w(t), the instantaneous angular velocity of the body about the
reference ray.
Well, this definition of w seems reasonable, does it not. One problem
-- the reference ray was arbitrary! We have described the intantaneous
rate of rotation about an arbitrary body fixed axis, not "the"
instantaneous axis of rotation, should such a thing exist. What
singles out a particular axis when apparently any one will do?
Hmm... body fixed axis... body fixed axis... maybe there's my problem.
We don't want to describe the rotation about an arbitrary axis fixed in
the body, we want to identify an axis _not_ fixed to the body, in
general wandering around the body, which is the best of all possible
axes which the body can be said to be spinning around for that moment.
But how? And what is the mathematical expression of "best" here? Is
there an axis such that the displacement of any body point is given
purely the the distance from that axis and w(t), so that the wandering
of the axis itself may be ignored in determining these displacements to
first order in small quantities?
Intuition flags. Insight invited.
You are fumbling your way to the operation of taking the curl of the
velocity which will be a vector defining the axis of rotation. This
operation may be performed in cartesian, or spherical coordinates.
This is far simpler than assembling rays and normal planes. Look it
up.
John Polasek
http://www.dualspace.net
.
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| User: "Ken S. Tucker" |
|
| Title: Re: Instantaneous axis of rotation? |
07 May 2006 12:10:05 PM |
|
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John C. Polasek wrote:
On 25 Apr 2006 15:53:20 -0700, "Edward Green"
<spamspamspam3@netzero.com> wrote:
Circumstantial evidence leads me to believe that the "instantaneous
axis of rotation" of a rigid body is a meaningful concept. I hear that
a sufficiently asymmetric body will wobble (defined as being what such
a body does under these circumstances) even in the absence of external
torques, and that the explanation of this involves a tensor connecting
the angular momentum and said instantaneous axis.
On the other hand, thinking of ways to describe an arbitrary motion of
a rigid body about its center of mass, I come up with the following:
from the center of mass, draw a arbitrary reference ray fixed relative
to the body. This ray will intersect a sphere drawn around the center
of mass at one point, which we may call a pole. The velocity of the
pole on the sphere has two degrees of freedom. Specifying this
two-dimensional velocity at an instant, we are then left with one more
degree of freedom in the motion, which we may quantify by the following
construction:
Draw a plane containing the center of mass and the reference ray,
perpendicular to the intantaneous velocity vector of the pole, and a
second plane formed by the same construction a time dt later. Consider
an aribtrary body point on the initial plane and off the ray. If the
point remains on the similar plane after dt then we say the body is not
rotating about the ray. If, as will generally be the case, the point
has moved off the plane, we infer from this displacement a value of
w(t), the instantaneous angular velocity of the body about the
reference ray.
Well, this definition of w seems reasonable, does it not. One problem
-- the reference ray was arbitrary! We have described the intantaneous
rate of rotation about an arbitrary body fixed axis, not "the"
instantaneous axis of rotation, should such a thing exist. What
singles out a particular axis when apparently any one will do?
Hmm... body fixed axis... body fixed axis... maybe there's my problem.
We don't want to describe the rotation about an arbitrary axis fixed in
the body, we want to identify an axis _not_ fixed to the body, in
general wandering around the body, which is the best of all possible
axes which the body can be said to be spinning around for that moment.
But how? And what is the mathematical expression of "best" here? Is
there an axis such that the displacement of any body point is given
purely the the distance from that axis and w(t), so that the wandering
of the axis itself may be ignored in determining these displacements to
first order in small quantities?
Intuition flags. Insight invited.
You are fumbling your way to the operation of taking the curl of the
velocity which will be a vector defining the axis of rotation. This
operation may be performed in cartesian, or spherical coordinates.
This is far simpler than assembling rays and normal planes. Look it
up.
John Polasek
John you're using a preferred frame, commonly done in
Newtonian Mechanics, however, in the case of Earth, I
can stand at the North Pole or on a Geosynchronous satellite
and measure the rotation of the Earth to be zero, all three
are relatively stationary.
This leads to the idea that angular momentum like
linear momentum is relative that seems to be part of
why SR was extended to GR. AE put a lot of effort
into explaining that in his preamble to GR1916 by
comparing sphere's and ellipsoids.
More simply, chart the course of a light-ray across
a spinning phonograph record, it will be aberrated
(aka deflected) relatively to the phono frame, but
then one can distort the phonograph by stress and
get the same answer.
Ken
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| User: "John C. Polasek" |
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| Title: Re: Instantaneous axis of rotation? |
08 May 2006 09:01:19 AM |
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On 7 May 2006 10:10:05 -0700, "Ken S. Tucker" <dynamics@vianet.on.ca>
wrote:
John C. Polasek wrote:
On 25 Apr 2006 15:53:20 -0700, "Edward Green"
<spamspamspam3@netzero.com> wrote:
Circumstantial evidence leads me to believe that the "instantaneous
axis of rotation" of a rigid body is a meaningful concept. I hear that
a sufficiently asymmetric body will wobble (defined as being what such
a body does under these circumstances) even in the absence of external
torques, and that the explanation of this involves a tensor connecting
the angular momentum and said instantaneous axis.
On the other hand, thinking of ways to describe an arbitrary motion of
a rigid body about its center of mass, I come up with the following:
from the center of mass, draw a arbitrary reference ray fixed relative
to the body. This ray will intersect a sphere drawn around the center
of mass at one point, which we may call a pole. The velocity of the
pole on the sphere has two degrees of freedom. Specifying this
two-dimensional velocity at an instant, we are then left with one more
degree of freedom in the motion, which we may quantify by the following
construction:
Draw a plane containing the center of mass and the reference ray,
perpendicular to the intantaneous velocity vector of the pole, and a
second plane formed by the same construction a time dt later. Consider
an aribtrary body point on the initial plane and off the ray. If the
point remains on the similar plane after dt then we say the body is not
rotating about the ray. If, as will generally be the case, the point
has moved off the plane, we infer from this displacement a value of
w(t), the instantaneous angular velocity of the body about the
reference ray.
Well, this definition of w seems reasonable, does it not. One problem
-- the reference ray was arbitrary! We have described the intantaneous
rate of rotation about an arbitrary body fixed axis, not "the"
instantaneous axis of rotation, should such a thing exist. What
singles out a particular axis when apparently any one will do?
Hmm... body fixed axis... body fixed axis... maybe there's my problem.
We don't want to describe the rotation about an arbitrary axis fixed in
the body, we want to identify an axis _not_ fixed to the body, in
general wandering around the body, which is the best of all possible
axes which the body can be said to be spinning around for that moment.
But how? And what is the mathematical expression of "best" here? Is
there an axis such that the displacement of any body point is given
purely the the distance from that axis and w(t), so that the wandering
of the axis itself may be ignored in determining these displacements to
first order in small quantities?
Intuition flags. Insight invited.
You are fumbling your way to the operation of taking the curl of the
velocity which will be a vector defining the axis of rotation. This
operation may be performed in cartesian, or spherical coordinates.
This is far simpler than assembling rays and normal planes. Look it
up.
John Polasek
John you're using a preferred frame, commonly done in
Newtonian Mechanics, however, in the case of Earth, I
can stand at the North Pole or on a Geosynchronous satellite
and measure the rotation of the Earth to be zero, all three
are relatively stationary.
This leads to the idea that angular momentum like
linear momentum is relative that seems to be part of
why SR was extended to GR. AE put a lot of effort
into explaining that in his preamble to GR1916 by
comparing sphere's and ellipsoids.
More simply, chart the course of a light-ray across
a spinning phonograph record, it will be aberrated
(aka deflected) relatively to the phono frame, but
then one can distort the phonograph by stress and
get the same answer.
Ken
Well he is saying "The velocity of the
pole on the sphere has two degrees of freedom. Specifying this
two-dimensional velocity at an instant, we are then left with one more
degree of freedom in the motion, which we may quantify by the
following construction:"
So an observation frame stationary to the body is assumed.
The curl of the velocity however measured will be an invariant of the
body. Put dots all over it and spin and take successive photos, so you
can scale off dV_x/dy etc. and the answer will be the same all over, 2
omega.
Even at the pole, make a differentially small area and (let's be
practical, you can measre the velocity or it's a poorly designed
experiment), so you can measure
dV_x/dy - dv_y/dx and come up with 2 omega
It's your choice how you lay out x and y.
John Polasek
http://www.dualspace.net
.
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| User: "Edward Green" |
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| Title: Re: Instantaneous axis of rotation? |
08 May 2006 06:51:05 PM |
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John C. Polasek wrote:
You are fumbling your way to the operation of taking the curl of the
velocity which will be a vector defining the axis of rotation. This
operation may be performed in cartesian, or spherical coordinates.
This is far simpler than assembling rays and normal planes. Look it
up.
Well he is saying "The velocity of the
pole on the sphere has two degrees of freedom. Specifying this
two-dimensional velocity at an instant, we are then left with one more
degree of freedom in the motion, which we may quantify by the
following construction:"
So an observation frame stationary to the body is assumed.
I don't see how that follows from what I wrote. I suppose you mean,
stationary wrt the (arbitrary) center point of the body -- or better,
in which the arbitrary center point of the body is stationary. That's
correct.
On the other hand, you may be onto something about my fumbling my way
towards taking a curl of the velocity field. Interesting. So you are
saying that if we have a rigid body executing its arbitrary tumbling
trajectory, that the concept I want, the instantaneous axis of
rotation, is just given, in magnitude and direction, but taking this
curl? You know, that sounds likely to be true, and further has another
nice property: it hints that this axis is going to be independent of
the choice of center (true). I take it the curl will also be constant
across the body, attributable to its rigidity.
You know, you make some disparaging remark about assembling rays and
normal planes. I don't know. I did arrive at the answer before
anybody replied to my first request for guidance, and in the process
corrected a remaining mental stumbling block. I fully admit and admire
the elegance and power of the methods which consider properties of the
instantaneous transformation between inertial and body fixed frames,
lie groups, and now yours -- taking a curl of the velocity field.
They will all repay study. However, I'm certainly not the poorer by
groping (groping is the prefered word, John, not "fumbling") my way to
the correct idea by considering normal planes and moving poles.
Indeed, I can now, on a pleasant summer day, close my eyes in any
convenient field of grass, and draw the series of pictures which will
lead to the constructive existence of the subject axis. Does knowing
that an anti-symmetric 3x3 matrix has an eigenvalue of zero buy you
_that_?
Well, it would when you finally understood the various arguments to all
be the same argument, expressed in different languages. But that is
really going to take some introspection, I tell you!
The curl of the velocity however measured will be an invariant of the
body. Put dots all over it and spin and take successive photos, so you
can scale off dV_x/dy etc. and the answer will be the same all over, 2
omega.
Even at the pole, make a differentially small area and (let's be
practical, you can measre the velocity or it's a poorly designed
experiment), so you can measure
dV_x/dy - dv_y/dx and come up with 2 omega
It's your choice how you lay out x and y.
You evidently have put your share of labor into this also.
.
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| User: "John C. Polasek" |
|
| Title: Re: Instantaneous axis of rotation? |
08 May 2006 08:43:52 PM |
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On 8 May 2006 16:51:05 -0700, "Edward Green"
<spamspamspam3@netzero.com> wrote:
John C. Polasek wrote:
You are fumbling your way to the operation of taking the curl of the
velocity which will be a vector defining the axis of rotation. This
operation may be performed in cartesian, or spherical coordinates.
This is far simpler than assembling rays and normal planes. Look it
up.
Well he is saying "The velocity of the
pole on the sphere has two degrees of freedom. Specifying this
two-dimensional velocity at an instant, we are then left with one more
degree of freedom in the motion, which we may quantify by the
following construction:"
So an observation frame stationary to the body is assumed.
I don't see how that follows from what I wrote. I suppose you mean,
stationary wrt the (arbitrary) center point of the body -- or better,
in which the arbitrary center point of the body is stationary. That's
correct.
On the other hand, you may be onto something about my fumbling my way
towards taking a curl of the velocity field. Interesting. So you are
saying that if we have a rigid body executing its arbitrary tumbling
trajectory, that the concept I want, the instantaneous axis of
rotation, is just given, in magnitude and direction, but taking this
curl? You know, that sounds likely to be true, and further has another
nice property: it hints that this axis is going to be independent of
the choice of center (true). I take it the curl will also be constant
across the body, attributable to its rigidity.
You know, you make some disparaging remark about assembling rays and
normal planes. I don't know. I did arrive at the answer before
anybody replied to my first request for guidance, and in the process
corrected a remaining mental stumbling block. I fully admit and admire
the elegance and power of the methods which consider properties of the
instantaneous transformation between inertial and body fixed frames,
lie groups, and now yours -- taking a curl of the velocity field.
They will all repay study. However, I'm certainly not the poorer by
groping (groping is the prefered word, John, not "fumbling") my way to
the correct idea by considering normal planes and moving poles.
Indeed, I can now, on a pleasant summer day, close my eyes in any
convenient field of grass, and draw the series of pictures which will
lead to the constructive existence of the subject axis. Does knowing
that an anti-symmetric 3x3 matrix has an eigenvalue of zero buy you
_that_?
Well, it would when you finally understood the various arguments to all
be the same argument, expressed in different languages. But that is
really going to take some introspection, I tell you!
The curl of the velocity however measured will be an invariant of the
body. Put dots all over it and spin and take successive photos, so you
can scale off dV_x/dy etc. and the answer will be the same all over, 2
omega.
Even at the pole, make a differentially small area and (let's be
practical, you can measre the velocity or it's a poorly designed
experiment), so you can measure
dV_x/dy - dv_y/dx and come up with 2 omega
It's your choice how you lay out x and y.
You evidently have put your share of labor into this also.
Of course I simplified it to a 2dimensional region but the 3D curl you
can put anywhere in the body and the output will be the skew symmetric
omega vector always the same value. In tensor notation
W_i = eps_ikm del V_k X_m
where eps_ikm is a 3x3x3 matrix where the elements are +1 for the
sequences 123 231 312 and -1 for 132 213 321, and 0 elsewhereand the
del and X are derivative of Vk with respect to Xm.
If I mentioned it you could put dots on the thing and with successive
photos work out the derivative and find the angular velocity.
John Polasek
http://www.dualspace.net
.
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| User: "Edward Green" |
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| Title: Re: Instantaneous axis of rotation? |
13 May 2006 12:59:23 PM |
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John C. Polasek wrote:
Of course I simplified it to a 2dimensional region but the 3D curl you
can put anywhere in the body and the output will be the skew symmetric
omega vector always the same value. In tensor notation
W_i = eps_ikm del V_k X_m
where eps_ikm is a 3x3x3 matrix where the elements are +1 for the
sequences 123 231 312 and -1 for 132 213 321, and 0 elsewhereand the
del and X are derivative of Vk with respect to Xm.
If I mentioned it you could put dots on the thing and with successive
photos work out the derivative and find the angular velocity.
Nice reply. Thanks.
.
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| User: "John C. Polasek" |
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| Title: Re: Instantaneous axis of rotation? |
04 May 2006 04:07:51 PM |
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On 25 Apr 2006 15:53:20 -0700, "Edward Green"
<spamspamspam3@netzero.com> wrote:
Circumstantial evidence leads me to believe that the "instantaneous
axis of rotation" of a rigid body is a meaningful concept. I hear that
a sufficiently asymmetric body will wobble (defined as being what such
a body does under these circumstances) even in the absence of external
torques, and that the explanation of this involves a tensor connecting
the angular momentum and said instantaneous axis.
On the other hand, thinking of ways to describe an arbitrary motion of
a rigid body about its center of mass, I come up with the following:
from the center of mass, draw a arbitrary reference ray fixed relative
to the body. This ray will intersect a sphere drawn around the center
of mass at one point, which we may call a pole. The velocity of the
pole on the sphere has two degrees of freedom. Specifying this
two-dimensional velocity at an instant, we are then left with one more
degree of freedom in the motion, which we may quantify by the following
construction:
Draw a plane containing the center of mass and the reference ray,
perpendicular to the intantaneous velocity vector of the pole, and a
second plane formed by the same construction a time dt later. Consider
an aribtrary body point on the initial plane and off the ray. If the
point remains on the similar plane after dt then we say the body is not
rotating about the ray. If, as will generally be the case, the point
has moved off the plane, we infer from this displacement a value of
w(t), the instantaneous angular velocity of the body about the
reference ray.
Well, this definition of w seems reasonable, does it not. One problem
-- the reference ray was arbitrary! We have described the intantaneous
rate of rotation about an arbitrary body fixed axis, not "the"
instantaneous axis of rotation, should such a thing exist. What
singles out a particular axis when apparently any one will do?
Hmm... body fixed axis... body fixed axis... maybe there's my problem.
We don't want to describe the rotation about an arbitrary axis fixed in
the body, we want to identify an axis _not_ fixed to the body, in
general wandering around the body, which is the best of all possible
axes which the body can be said to be spinning around for that moment.
But how? And what is the mathematical expression of "best" here? Is
there an axis such that the displacement of any body point is given
purely the the distance from that axis and w(t), so that the wandering
of the axis itself may be ignored in determining these displacements to
first order in small quantities?
Intuition flags. Insight invited.
You have to know about the inertia tensor a 3x3 matrix that when
diagonalized has 3 different moments of inertia J11 J22 J33. The
natural spin, probably minimum energy is about any of these three
axes. If you cand identify the inertia tensor, you can't say much
about it. I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
John Polasek
http://www.dualspace.net
.
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| User: "Edward Green" |
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| Title: Re: Instantaneous axis of rotation? |
05 May 2006 07:03:07 PM |
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John C. Polasek wrote:
You have to know about the inertia tensor a 3x3 matrix that when
diagonalized has 3 different moments of inertia J11 J22 J33. The
natural spin, probably minimum energy is about any of these three
axes. If you cand identify the inertia tensor, you can't say much
about it. I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Just a quick note of disagreement (which are always so much easier to
write than letters of appreciation and statements of awe): the
instantaneous axis of rotation is a kinematical concept, and doesn't
care about the moment of inertia -- or the center of mass for that
matter, though that seems to be a popular association.
(Adding disputatious teasers is another easy way to overcome writers
block ;-).
.
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| User: "John C. Polasek" |
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| Title: Re: Instantaneous axis of rotation? |
06 May 2006 09:53:33 AM |
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On 5 May 2006 17:03:07 -0700, "Edward Green"
<spamspamspam3@netzero.com> wrote:
John C. Polasek wrote:
You have to know about the inertia tensor a 3x3 matrix that when
diagonalized has 3 different moments of inertia J11 J22 J33. The
natural spin, probably minimum energy is about any of these three
axes. If you cand identify the inertia tensor, you can't say much
about it. I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Just a quick note of disagreement (which are always so much easier to
write than letters of appreciation and statements of awe): the
instantaneous axis of rotation is a kinematical concept, and doesn't
care about the moment of inertia -- or the center of mass for that
matter, though that seems to be a popular association.
(Adding disputatious teasers is another easy way to overcome writers
block ;-).
Au contraire my friend. In a free body, spinning, it will spin about
its center of mass that is characterized by the inertia tensor. Every
lump has 3-axis inertia tensor and if it is spinning about the
centroid on a principle axis it will not wobble. If it is spinning off
the proper center then we use the translation theorem J' = J + Mr^2
where r is the offset. Then in time it will adjust to least energy,
but spinning up as J' approaches J. That's if it's near the principle
axis.
If it starts spinning off one of the 3 axes, it will wobble as could
be seen by the momentum resulting from
Mom = A'JA*W
where J is similarity transformed by A' and A rotation matrices so
that the single arbitrary rotation rate W produces Mom with components
on all three axes and not colinear with W.
The kinematical axis on a free body will find itself and its quite
dynamic, not kinematic.
John Polasek
http://www.dualspace.net
.
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| User: "Edward Green" |
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| Title: Re: Instantaneous axis of rotation? |
06 May 2006 11:17:32 AM |
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John C. Polasek wrote:
On 5 May 2006 17:03:07 -0700, "Edward Green"
<spamspamspam3@netzero.com> wrote:
the
instantaneous axis of rotation is a kinematical concept, and doesn't
care about the moment of inertia -- or the center of mass for that
matter, though that seems to be a popular association.
Au contraire my friend.
<snip nice precis of rigid body dynamics>
The kinematical axis on a free body will find itself and its quite
dynamic, not kinematic.
You are talking about what a real body will actually do. My point was
-- born out by two detailed responses -- that we can make sense of the
subject-concept considering a rigid body whose trajectories in space
and orientation are given. Such a description doesn't include an
explanation _why_ the trajectory is observed (dynamics), so this
information cannot be necessary to apply the concept. We may choose to
use the center of mass for future convenience but we may equally well
evaluate the instantaneous axis of rotation through any body point --
fortunately, getting the same answer! (parallel axis theorem)
My method may have been primitive, but at least it gave me a picture --
and once a man has that, he may generate thousands of not-incorrect
words. ;-)
.
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| User: "Ken S. Tucker" |
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| Title: Re: Instantaneous axis of rotation? |
06 May 2006 04:44:04 PM |
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Edward Green wrote:
John C. Polasek wrote:
On 5 May 2006 17:03:07 -0700, "Edward Green"
<spamspamspam3@netzero.com> wrote:
the
instantaneous axis of rotation is a kinematical concept, and doesn't
care about the moment of inertia -- or the center of mass for that
matter, though that seems to be a popular association.
Au contraire my friend.
<snip nice precis of rigid body dynamics>
The kinematical axis on a free body will find itself and its quite
dynamic, not kinematic.
You are talking about what a real body will actually do. My point was
-- born out by two detailed responses -- that we can make sense of the
subject-concept considering a rigid body whose trajectories in space
and orientation are given. Such a description doesn't include an
explanation _why_ the trajectory is observed (dynamics), so this
information cannot be necessary to apply the concept. We may choose to
use the center of mass for future convenience but we may equally well
evaluate the instantaneous axis of rotation through any body point --
fortunately, getting the same answer! (parallel axis theorem)
My method may have been primitive, but at least it gave me a picture --
and once a man has that, he may generate thousands of not-incorrect
words. ;-)
My understanding of that problem is based on GR.
Suppose I construct a semi-rigid body composed
of piezo electrical crystals, that are subject to
deformation that in turn creates an electrical
potential (voltage), that is the basis for electronic
weigh scales, when a deformation (aka stress)
is applied.
You can spin said body, or you can push down
on it, and all the summed piezo electrical energy
can end up with the same energy storage within
said body. In GR that's energy storage by stress,
specifically electrically described, sometimes
that's called pressure if the adiabitic solution is
used, but is easy to convert one to the other.
I think it's good to understand how a piezo-
electrical scale and a spring or balance scale
can provide the same information concerning
centrifugal force, certainly they must agree.
Nifty question
Ken S. Tucker
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| User: "Timo A. Nieminen" |
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| Title: Re: Instantaneous axis of rotation? |
04 May 2006 04:24:02 PM |
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On Thu, 4 May 2006, John C. Polasek wrote:
I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Euler angles suck! Angle--axis parameterisations of rotations rule!
Actually, the one that none can reasonably complain about is to use the
3x3 rotation matrix. Sure, there's redundancy there, but these days, I
think that the extra 6 double precision floating point numbers won't put
an excessive strain on our computers.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "" |
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| Title: Re: Instantaneous axis of rotation? |
04 May 2006 04:52:33 PM |
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In article <Pine.WNT.4.64.0605050721150.1348@serene.st>, "Timo A. Nieminen" <timo@physics.uq.edu.au> writes:
On Thu, 4 May 2006, John C. Polasek wrote:
I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Euler angles suck! Angle--axis parameterisations of rotations rule!
Actually, the one that none can reasonably complain about is to use the
3x3 rotation matrix. Sure, there's redundancy there, but these days, I
think that the extra 6 double precision floating point numbers won't put
an excessive strain on our computers.
I fully agree.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "John C. Polasek" |
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| Title: Re: Instantaneous axis of rotation? |
04 May 2006 08:55:00 PM |
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On Fri, 5 May 2006 07:24:02 +1000, "Timo A. Nieminen"
<timo@physics.uq.edu.au> wrote:
On Thu, 4 May 2006, John C. Polasek wrote:
I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Euler angles suck! Angle--axis parameterisations of rotations rule!
Actually, the one that none can reasonably complain about is to use the
3x3 rotation matrix. Sure, there's redundancy there, but these days, I
think that the extra 6 double precision floating point numbers won't put
an excessive strain on our computers.
I don't know what you're talking about. The Euler angles comprise 3
matrices not one. The similarity transform for the inertia tensor
require pre- and post-multiplication making 6 matrices altogether.
You send the rotation rate in along any axis to the transformed tensor
and get a momentum vector that "sticks out in 3 places" unless it's on
a principle axis.
I'd be interested to learn about axis parametrizations though.
John Polasek
http://www.dualspace.net
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| User: "Timo Nieminen" |
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| Title: Re: Instantaneous axis of rotation? |
04 May 2006 10:20:34 PM |
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On Fri, 5 May 2006, John C. Polasek wrote:
On Fri, 5 May 2006 07:24:02 +1000, "Timo A. Nieminen"
<timo@physics.uq.edu.au> wrote:
On Thu, 4 May 2006, John C. Polasek wrote:
I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Euler angles suck! Angle--axis parameterisations of rotations rule!
Actually, the one that none can reasonably complain about is to use the
3x3 rotation matrix. Sure, there's redundancy there, but these days, I
think that the extra 6 double precision floating point numbers won't put
an excessive strain on our computers.
I don't know what you're talking about. The Euler angles comprise 3
matrices not one.
The Euler angles are 3 _numbers_, not matrices. Yes, one can _calculate_ a
rotation matrix from the Euler angles by multiplying 3 separate rotation
matrices together, but it isn't necessary to do it that way.
I'd be interested to learn about axis parametrizations though.
In general, to describe a rotation in 3D space, you need 3 (real) numbers.
Every n-D rotation matrix can be written as
R = exp(S)
where S is an antisymmetric matrix. In 3D, any antisymmetric S can be
written as 3 parameters and basis matrices:
S = a1 S1 + a2 S2 + a3 S3.
(a1,a2,a3) fully describes the rotation. If we choose
S1 = [ 0 0 0; 0 0 -1; 0 1 0 ]
S2 = [ 0 0 1; 0 0 0; -1 0 0 ]
S3 = [ 0 -1 0; 1 0 0; 0 0 0 ]
then R * (a1,a2,a3) = (a1,a2,a3), so the position (a1,a2,a3) lies on the
axis of rotation. The magnitude of (a1,a2,a3) is the angle of rotation, in
radians.
The are other axis-angle parameterisations, where the magnitude of the
vector parameter is some function of the angle, rather than the angle
itself.
You might like to read:
Olivier A. Bauchau and Lorenzo Trainelli
"The vectorial parameterization of rotation"
Nonlinear Dynamics 32, 71-92 (2003)
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "John C. Polasek" |
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| Title: Re: Instantaneous axis of rotation? |
05 May 2006 04:19:51 PM |
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On Fri, 5 May 2006 13:20:34 +1000, Timo Nieminen
<timo@physics.uq.edu.au> wrote:
On Fri, 5 May 2006, John C. Polasek wrote:
On Fri, 5 May 2006 07:24:02 +1000, "Timo A. Nieminen"
<timo@physics.uq.edu.au> wrote:
On Thu, 4 May 2006, John C. Polasek wrote:
I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Euler angles suck! Angle--axis parameterisations of rotations rule!
Actually, the one that none can reasonably complain about is to use the
3x3 rotation matrix. Sure, there's redundancy there, but these days, I
think that the extra 6 double precision floating point numbers won't put
an excessive strain on our computers.
I don't know what you're talking about. The Euler angles comprise 3
matrices not one.
The Euler angles are 3 _numbers_, not matrices. Yes, one can _calculate_ a
rotation matrix from the Euler angles by multiplying 3 separate rotation
matrices together, but it isn't necessary to do it that way.
The Euler angle methodology already is specifying the gimbal order for
the 3 parameter angles that engender the matrices. These gimbals are
tensors that do away with order of rotation in favor of order of
construction. I can name the 3 variables as angles of rotation.
I'd be interested to learn about axis parametrizations though.
In general, to describe a rotation in 3D space, you need 3 (real) numbers.
Every n-D rotation matrix can be written as
R = exp(S)
exp?
where S is an antisymmetric matrix. In 3D, any antisymmetric S can be
written as 3 parameters and basis matrices:
S = a1 S1 + a2 S2 + a3 S3.
If a1, a2 and a3 are angles, your S implies vector addition which is
as you well know, forbidden. If they are angular velocities, yes.
(a1,a2,a3) fully describes the rotation. If we choose
S1 = [ 0 0 0; 0 0 -1; 0 1 0 ]
S2 = [ 0 0 1; 0 0 0; -1 0 0 ]
S3 = [ 0 -1 0; 1 0 0; 0 0 0 ]
then R * (a1,a2,a3) = (a1,a2,a3), so the position (a1,a2,a3) lies on the
axis of rotation. The magnitude of (a1,a2,a3) is the angle of rotation, in
radians.
The are other axis-angle parameterisations, where the magnitude of the
vector parameter is some function of the angle, rather than the angle
itself.
You might like to read:
Olivier A. Bauchau and Lorenzo Trainelli
"The vectorial parameterization of rotation"
Nonlinear Dynamics 32, 71-92 (2003)
There must be a lot to it that can't be put in a note here, but tell
me, how would you apply this to the case of an inertia tensor being
rotated by an arbitrary rotation vector?
Thank you for your effort to explain.
John Polasek
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| User: "" |
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| Title: Re: Instantaneous axis of rotation? |
14 May 2006 03:36:28 PM |
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John C. Polasek wrote:
On Fri, 5 May 2006 13:20:34 +1000, Timo Nieminen
<timo@physics.uq.edu.au> wrote:
In general, to describe a rotation in 3D space, you need 3 (real) numbers.
Every n-D rotation matrix can be written as
R = exp(S)
exp?
Matrix exponential.
where S is an antisymmetric matrix. In 3D, any antisymmetric S can be
written as 3 parameters and basis matrices:
S = a1 S1 + a2 S2 + a3 S3.
If a1, a2 and a3 are angles, your S implies vector addition which is
as you well know, forbidden. If they are angular velocities, yes.
(a1,a2,a3) fully describes the rotation. If we choose
The magnitude of (a1,a2,a3) is an angle. a1, a2, a3 are not angles.
Note that exp(S) is not equal to exp(a1 S1) exp(a2 S2) exp(a3 S3) in
general.
There must be a lot to it that can't be put in a note here, but tell
me, how would you apply this to the case of an inertia tensor being
rotated by an arbitrary rotation vector?
This is out of my familiarity, since I usually deal with rotations in
the context of transformations between coordinate systems that are
stationary relative to each other, and rotation of objects where
viscous drag is overwhelmingly dominant compared to inertia.
I don't understand your question. It appears to mean "what happens to a
moment-of-inertia tensor under a rotational transformation of
coordinate system", but that's just the usual procedure for rotation of
any tensor under such transformation.
If you mean "what rotation results for an arbitrary object with some
constant angular momentum", that isn't dependent on how the rotation is
parameterised. Since we only do spheres and cubes in creeping flow, for
which the drag tensor is isotropic (such magic of cubes!), I'd have to
look it up in eg Goldstein, which you can do as easily as I. If the
angular velocity is constant, then it's trivial, but Iv being constant
doesn't mean that angular velocity v must be constant if I is not
isotropic.
However, note that a constant angular velocity vector multiplied by
time directly gives you the (a1,a2,a3) rotation vector.
--
Timo
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| User: "Edward Green" |
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| Title: Re: Instantaneous axis of rotation? |
05 May 2006 07:07:58 PM |
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Timo A. Nieminen wrote:
On Thu, 4 May 2006, John C. Polasek wrote:
I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Euler angles suck! Angle--axis parameterisations of rotations rule!
Actually, the one that none can reasonably complain about is to use the
3x3 rotation matrix.
How does that work? 3x3 matrices rotate and stretch vectors.
Rotations rotate, but do not stretch, rigid objects. I presume one
somehow uses the "stretching" degree of freedom to map the extra degree
of freedom in rotating a solid body vs. a vector?
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| User: "Timo A. Nieminen" |
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| Title: Re: Instantaneous axis of rotation? |
05 May 2006 08:26:53 PM |
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On Sat, 5 May 2006, Edward Green wrote:
Timo A. Nieminen wrote:
On Thu, 4 May 2006, John C. Polasek wrote:
I will say the mathematics of rotation and momentum
(involving Euler angles as best choice) is just about insoluble. See
Goldstein Mechanics.
Euler angles suck! Angle--axis parameterisations of rotations rule!
Actually, the one that none can reasonably complain about is to use the
3x3 rotation matrix.
How does that work? 3x3 matrices rotate and stretch vectors.
Rotations rotate, but do not stretch, rigid objects.
Yes. Not just any 3x3 matrix will do. It needs to be orthogonal
(ie R R' = I, R' = transpose of R), and determinant(R)=1.
If you're willing to be a little more flexible in what you call a
rotation, you only need det(R)^2 = 1, which includes transformations
between left- and right-handed coordinate systems (for passive
transformations) and reflections (for active transformations).
But anyway, every 3D rotation matrix is a 3x3 matrix, while every 3x3
matrix is a homogeneous linear transformation, of which rotations are but
a tiny subset.
I presume one
somehow uses the "stretching" degree of freedom to map the extra degree
of freedom in rotating a solid body vs. a vector?
To perform a homogeneous rotation (ie x' = R x, rather than
x' = R x + T), the axis of rotation needs to pass through the origin. But
this is just (!) a matter of choosing an appropriate origin of our
coordinate system. It means we have (gasp!) 3 more parameters we need to
store to describe our transformation.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Henning Makholm" |
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| Title: Re: Instantaneous axis of rotation? |
07 May 2006 07:48:04 AM |
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Scripsit "Timo A. Nieminen" <timo@physics.uq.edu.au>
Yes. Not just any 3x3 matrix will do. It needs to be orthogonal
(ie R R' = I, R' = transpose of R), and determinant(R)=1.
If you're willing to be a little more flexible in what you call a
rotation, you only need det(R)^2 = 1, which includes transformations
between left- and right-handed coordinate systems
You already get (det R)^2 = 1 once you require R R^t = I.
--
Henning Makholm "Apologies if I am repeating obvious
conclusions. My only gateway onto the Net is
very expensive, and I miss many important postings...
Please write to me and tell me what you think. I don't get much mail."
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| User: "" |
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| Title: Re: Instantaneous axis of rotation? |
07 May 2006 02:07:51 PM |
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In article <87ac9u6lx7.fsf@kreon.lan.henning.makholm.net>, Henning Makholm <henning@makholm.net> writes:
Scripsit "Timo A. Nieminen" <timo@physics.uq.edu.au>
Yes. Not just any 3x3 matrix will do. It needs to be orthogonal
(ie R R' = I, R' = transpose of R), and determinant(R)=1.
If you're willing to be a little more flexible in what you call a
rotation, you only need det(R)^2 = 1, which includes transformations
between left- and right-handed coordinate systems
You already get (det R)^2 = 1 once you require R R^t = I.
Certainly, but the opposite isn't true. Thus, (det R)^2 is a weaker
condition, definining a larger set, just as the OP said.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "" |
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| Title: Re: Instantaneous axis of rotation? |
07 May 2006 06:07:02 PM |
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In article <87y7xdfo2v.fsf@kreon.lan.henning.makholm.net>, Henning Makholm <henning@makholm.net> writes:
Scripsit
Henning Makholm <henning@makholm.net> writes:
Scripsit "Timo A. Nieminen" <timo@physics.uq.edu.au>
Yes. Not just any 3x3 matrix will do. It needs to be orthogonal
(ie R R' = I, R' = transpose of R), and determinant(R)=1.
If you're willing to be a little more flexible in what you call a
rotation, you only need det(R)^2 = 1, which includes transformations
between left- and right-handed coordinate systems
You already get (det R)^2 = 1 once you require R R^t = I.
Certainly, but the opposite isn't true. Thus, (det R)^2 is a weaker
condition, definining a larger set, just as the OP said.
If you completely drop the RR^t=I condition, you're not just being "a
little more flexible", but a LOT more flexible. Most determinant ±1
matrices do not do anything that you'd recognize as a form of
rotation. For example, they do not in general preserve distances.
Yes, this is true. There is such thing as being "too flexible":-)
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Henning Makholm" |
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| Title: Re: Instantaneous axis of rotation? |
07 May 2006 05:48:56 PM |
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Scripsit
Henning Makholm <henning@makholm.net> writes:
Scripsit "Timo A. Nieminen" <timo@physics.uq.edu.au>
Yes. Not just any 3x3 matrix will do. It needs to be orthogonal
(ie R R' = I, R' = transpose of R), and determinant(R)=1.
If you're willing to be a little more flexible in what you call a
rotation, you only need det(R)^2 = 1, which includes transformations
between left- and right-handed coordinate systems
You already get (det R)^2 = 1 once you require R R^t = I.
Certainly, but the opposite isn't true. Thus, (det R)^2 is a weaker
condition, definining a larger set, just as the OP said.
If you completely drop the RR^t=I condition, you're not just being "a
little more flexible", but a LOT more flexible. Most determinant ±1
matrices do not do anything that you'd recognize as a form of
rotation. For example, they do not in general preserve distances.
--
Henning Makholm "Also, the letters are printed. That makes the task
of identifying the handwriting much more difficult."
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| User: "Timo Nieminen" |
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| Title: Re: Instantaneous axis of rotation? |
07 May 2006 06:27:55 PM |
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On Mon, 8 May 2006, Henning Makholm wrote:
Scripsit
Henning Makholm <henning@makholm.net> writes:
Scripsit "Timo A. Nieminen" <timo@physics.uq.edu.au>
=20
Yes. Not just any 3x3 matrix will do. It needs to be orthogonal
(ie R R' =3D I, R' =3D transpose of R), and determinant(R)=3D1.
=20
If you're willing to be a little more flexible in what you call a
rotation, you only need det(R)^2 =3D 1, which includes transformation=
s
between left- and right-handed coordinate systems
=20
You already get (det R)^2 =3D 1 once you require R R^t =3D I.
=20
Certainly, but the opposite isn't true. Thus, (det R)^2 is a weaker=20
condition, definining a larger set, just as the OP said.
=20
If you completely drop the RR^t=3DI condition, you're not just being "a
little more flexible", but a LOT more flexible. Most determinant =B11
matrices do not do anything that you'd recognize as a form of
rotation. For example, they do not in general preserve distances.
R R' =3D I gives you 6 independent equations which restrict the possible=20
values of the 9 elements of R, leaving only three independent elements.=20
The determinant condition gives 1 equation, leaving 8 independent elements=
=20
of R, which makes it very clear that most matrices with det() =3D 1, let=20
alone det()^2 =3D 1, can't be rotation matrices.
(btw, I mean to say that by not restricting the transformations to=20
det(R)=3D1, you get reflections. I tried to phrase this in terms of det^2,=
=20
to emphasise that the restriction to det(R)=3D1 is natural, but was=20
unclear.)
--=20
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Henning Makholm" |
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| Title: Re: Instantaneous axis of rotation? |
04 May 2006 03:05:13 PM |
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Scripsit "Edward Green" <spamspamspam3@netzero.com>
Well, this definition of w seems reasonable, does it not. One problem
-- the reference ray was arbitrary! We have described the intantaneous
rate of rotation about an arbitrary body fixed axis, not "the"
instantaneous axis of rotation, should such a thing exist. What
singles out a particular axis when apparently any one will do?
The orthodox mathematical construction goes something like:
Imagine a coordinate system that is fixed in space and one rigidly
attached to the tumbling body. For each point in time, write down
the transformation that goes from the fixed coordinates to tumbling
ones. It will have the general form
X |-> T_t + M_t X
where T_t is a translative term (which we will ignore) and M_t is
a matrix in SO(3).
Let t=0 be the time at which we want to find the instantaneous
rotation axis (and angular velocity). We can have chosen to attach the
tumbling coordinate system in such a way that at t=0 it happens to be
oriented just as the fixed one is, i.e. M_0 = I.
Since everything happens smoothly M_t will have a time derivative
N = dM/dt such that for an infinitesimal dt,
M_dt = I + dtN_0.
It we calculate what it means for I+dtN_0 to be orthogonal (setting
dt˛=0) ge get that N_0 must be an antisymmetric 3×3 matrix. That
means that it has just three independent components, and if we work
through all the algebra of imagining a different orientation for the
_fixed_ coordinate system, we find that those three components of N_0
transforms under such rotations as a pseudovector. That this works is
an accident of 3D geometry; it does not work out that way in other
dimensions.
We now see that we can choose to turn the lab coordinates such that
only one of the three components of N_0 is nonzero, e.g.
( 0 -a 0 ) ( 1 -adt 0 )
N_0 = ( a 0 0 ) M_dt = ( adt 1 0 )
( 0 0 0 ) ( 0 0 1 )
and we then find that the points fixed by M_dt are exactly the ones
on the z-axis. (There is nothing magical about this, because our last
rotation of the lab frame was determined by the movement). Therfore
the direction the z frame ended up in is the instantaneous axis of
rotation of the body at t=0.
In fact, using the matrix exponential, t -> exp(tN_0) defines a
constant rotation about this axis which at t=0 matches the movement of
the original body to the first degree in t. We see that a is the
instantaneous *angular velocity* of the rotation.
Hmm... body fixed axis... body fixed axis... maybe there's my problem.
We don't want to describe the rotation about an arbitrary axis fixed in
the body, we want to identify an axis _not_ fixed to the body, in
general wandering around the body, which is the best of all possible
axes which the body can be said to be spinning around for that moment.
But how? And what is the mathematical expression of "best" here?
The fundamental property is that the rotation axis consist of all
those points that are momentarily at rest relative to the center of
the body. The mathematical development above consitutes a proof sketch
that those points always form a well-defined geometrical line through
the center - except in the degenerate case that N_0=0, in which case
the body is momentarily non-rotating.
--
Henning Makholm Blessed are those with no
shoes, for the earth shall kiss them.
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| User: "Edward Green" |
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| Title: Re: Instantaneous axis of rotation? |
07 May 2006 10:15:01 AM |
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Henning Makholm wrote:
<very nice derivation of instantaneous axis of rotation>
Thank you!
Putting your derivation, mine, and michaeld's together, I think I'm
going to have to keep all three. They all illustrate something --
yours and michaeld's mainly my ignorance of modern mathematics. ;-)
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