# Maths - Vectors - forum discussion

From: meron@cars3.uchicago.edu (meron@cars3.uchicago.edu)
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-19 12:26:49 PST

In article <14f4f692.0212190309.5639bb54@posting.google.com>, (Martin Baker) writes:
>Todd,
>
>> For instance, here it looks like you're really trying to find the instantaneous
>> linear velocity resulting FROM the angular velocity.
>I was trying to do the inverse of this, to find the instantaneous
>angular velocity resulting FROM the linear velocity of a point. Do you
>think this is a valid thing to do? It seemed to me that it was
>provided one defined a point about which the rotation is measured. Or
>is angular velocity to linear velocity a one-way function?
>
No, why. You've v = w x r, and you've w = (r x v)/|r|^2, where r, v
and w are vectors, of course, and x signifies vector product.

Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"

From: Martin Baker
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-22 10:59:02 PST

Mati,

> No, why. You've v = w x r, and you've w = (r x v)/|r|^2, where r, v
> and w are vectors, of course, and x signifies vector product.

Does this result come from vector algebra, ie if,
a = b x c
b = (c x a)/|c|^2

Or are there additional physical conditions that are needed to derive w = (rx v)/|r|^2 ?

Martin

From: meron@cars3.uchicago.edu (meron@cars3.uchicago.edu)
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-22 14:52:41 PST

In article <au51vv\$1u2\$1@venus.btinternet.com>, "Martin Baker" writes:
>Mati,
>
>> No, why. You've v = w x r, and you've w = (r x v)/|r|^2, where r, v
>> and w are vectors, of course, and x signifies vector product.
>
>
>Does this result come from vector algebra, ie if,
>a = b x c
>b = (c x a)/|c|^2
>
>Or are there additional physical conditions that are needed to derive w = (r
>x v)/|r|^2 ?
>
I was about to answer the e-mail but can just as well do it here
(perhaps somebody else will find it useful).

What is used here is the (very useful) vector identity, valid for
arbitrary 3D vectors:

a x (b x c) = (a.c)b - (a.b)c

where the "." signifies dot product.

So, if you've v = w x r then

r x v = r x (w x r) = (r.r)w - (r.w)r = w|r|^2 - (r.w)r

Now, if you add the *additional* assumption that r and w are
perpendicular (so that r.w = 0), then the second term on the right is
zero and you get

r x v = w|r^2|
i.e.
w = (r x v)/|r^2|

If you don't add this assumption then, rewriting the relationship
above as

r x v = |r^2|(w - ((r.w)r)/r^2)

and noting that the expression in parentheses is just the component of
w *normal* to r, you can still say

w_n = (r x v)/|r^2|

where w_n is the "component of w normal to r" mentioned above.

You can't do bettere than this since to begin with, when you calculate
v as (w x r) you only use the component of w that's normal to r and,
if there is a component parallel to r, the information about it is
lost. Once lost, it can no longer be recovered.

Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"

From: Martin Baker
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-23 03:41:02 PST

Mati,

> I was about to answer the e-mail but can just as well do it here
> (perhaps somebody else will find it useful).

Yes, I think so, I do anyway so I have put a copy at:
https://www.euclideanspace.com/maths/algebra/vectors/meron.htm
if that's alright with you?

> Now, if you add the *additional* assumption that r and w are
> perpendicular (so that r.w = 0), then the second term on the right is
> zero and you get

That's very interesting because this assumption would not be valid if we
translated the frame-of-reference and we were trying to measure the angular
velocity about some point other than its centre of rotation, or if it is
doing some non-circular motion. Perhaps this "component of w normal to r"
that you suggest would be a possible way to allow us to define angular
velocity for non-circular motion.

Thanks,

Martin

From: meron@cars3.uchicago.edu (meron@cars3.uchicago.edu)
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-23 13:54:03 PST

In article <au6smu\$aaj\$1@sparta.btinternet.com>, "Martin Baker" writes:
>Mati,
>
>> I was about to answer the e-mail but can just as well do it here
>> (perhaps somebody else will find it useful).
>
>Yes, I think so, I do anyway so I have put a copy at:
>https://www.euclideanspace.com/maths/algebra/vectors/meron.htm
>if that's alright with you?
>
Sure, no problem.

>> Now, if you add the *additional* assumption that r and w are
>> perpendicular (so that r.w = 0), then the second term on the right is
>> zero and you get
>
>That's very interesting because this assumption would not be valid if we
>translated the frame-of-reference and we were trying to measure the angular
>velocity about some point other than its centre of rotation, or if it is
>doing some non-circular motion. Perhaps this "component of w normal to r"
>that you suggest would be a possible way to allow us to define angular
>velocity for non-circular motion.
>
You've to be careful, because you're confusing different (though
related) concepts here.

First of all, angular velocity *is* perfectly well defined for *any*
motion. The definition is just what the name implies, change of angle
with time. Meaning, graphically, observe a point at a given time t,
draw the line from your observation loction to this point. Repeat the
same at time t + delta_t where dt is a small time increment. Now you've
two lines which in general have some angle delta_theta between them. Now,
the magnitude of the angular velocity is given by delta_theta/delta_t
(or, more exactly by the limit of this expression for delta_t going to
0). As for the direction of the angular velocity, we take it as the
direction orthogonal both to the displacement of the point and to
the line joining the observation location with the moving point.

So, we've a definition that is perfectly OK for any motion, not only
circular and it is a matter of rather simple math to show that the
recipe above is equivalent to saying "given a reference point and
another point moving with velocity v, take the projection of v on the
plane orthogonal to r (the vector joining the moving point and the
reference one) and divide its length by the length of r. This gives
you the magnitude of the angular velocity and the direction is
orthogonal to both v and r". And if you do a bit of algebra you'll
find that all this amounts simply to taking

w = (r x v)/|r^2|

This is an exact result, no assumptions or approximations involved.
An exact result for the angular velocity of the object *relative to
your chosen observation point*. And note that the result depends not
only on the object but on the choice of observation point as well.
Even within the same reference frame the result is *not* a property of
the object alone.

you asked (or appeared to me to be asking) was a subtly different
question. Not "what is the angular velocity relative to the
observation point?" but "assuming that an object moves with a
momentary velocity v resulting from a rotation around some axis with
some angular velocity w, how do I find this angular velocity?" And
here you've to be careful. You can pick any reference point P and
translate the observed velocity to angular velocity relative to P,
using the prescription above, but this result will in general differ
from the angular velocity of rotation that gave rise to the motion in
the first place. Even if you pick your P on the proper rotation axis
you will, in general get a different result. *Only* if P is ion the
rotation axis *and* the line joining P with the object is
perpendicular to v, only then will the w you calculate using the
formula above coincide with the angular velocity of the rotation.

Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"

From: Martin Baker
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-25 01:04:01 PST

Mati,

> So what was the business about "assumptions" I mentioned? Well, what
> you asked (or appeared to me to be asking) was a subtly different
> question. Not "what is the angular velocity relative to the
> observation point?" but "assuming that an object moves with a
> momentary velocity v resulting from a rotation around some axis with
> some angular velocity w, how do I find this angular velocity?" And
> here you've to be careful. You can pick any reference point P and
> translate the observed velocity to angular velocity relative to P,
> using the prescription above, but this result will in general differ
> from the angular velocity of rotation that gave rise to the motion in
> the first place. Even if you pick your P on the proper rotation axis
> you will, in general get a different result. *Only* if P is ion the
> rotation axis *and* the line joining P with the object is
> perpendicular to v, only then will the w you calculate using the
> formula above coincide with the angular velocity of the rotation.

So if we have a rotating solid object, made up of an infinite number of
particles, and we calculate the angular velocity of each of these particles
about an arbitrary point P. Each particle of the solid object will
potentially have a different angular velocity. However I suspect that there
is one particular value of P (I'll call it Pc) where the conditions that you
mention above will be met for all particles and the angular velocity for
each particle about Pc will be the same.
So for a solid object, the definition of angular velocity must be the
However for complex motion, such as planetary motion, this is not so obvious
because this Pc will itself be moving, in fact its track could be a
circular, so do we take this rotation into account? (assuming an absolute
frame-of-reference)

Martin

From: student
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-25 10:45:41 PST

On Wed, 25 Dec 2002 09:03:08 +0000 (UTC), Martin Baker wrote:
>So if we have a rotating solid object, made up of an infinite number of
>particles, and we calculate the angular velocity of each of these particles
>about an arbitrary point P. Each particle of the solid object will
>potentially have a different angular velocity.

Not if it's a rigid body. If it isn't rigid, you'll need some more precise
way to define angular velocity.

>However I suspect that there
>is one particular value of P (I'll call it Pc) where the conditions that you
>mention above will be met for all particles and the angular velocity for
>each particle about Pc will be the same.
>So for a solid object, the definition of angular velocity must be the
>However for complex motion, such as planetary motion, this is not so obvious
>because this Pc will itself be moving, in fact its track could be a
>circular, so do we take this rotation into account? (assuming an absolute
>frame-of-reference)
>
>Martin
>
>

From: Martin Baker
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-25 10:58:01 PST

> Not if it's a rigid body. If it isn't rigid, you'll need some more precise
> way to define angular velocity.

Yes, I mean a rigid body. The angular velocity on each point on the rigid
body will be different, if measured about an arbitrary point which is not on
the axis of rotation. Unless I have misunderstood Mati.

Martin

From: student
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-25 14:01:55 PST

On Wed, 25 Dec 2002 18:57:51 +0000 (UTC), Martin Baker wrote:
>> Not if it's a rigid body. If it isn't rigid, you'll need some more precise
>> way to define angular velocity.
>
>Yes, I mean a rigid body. The angular velocity on each point on the rigid
>body will be different, if measured about an arbitrary point which is not on
>the axis of rotation. Unless I have misunderstood Mati.
>
>Martin

Um, no, I think I misunderstood, pls disregard.

The point I should have been making is that v=wxr (which I think is
what you were taking as a starting point) is not accurate if |r| has
nonzero time derivatives; otherwise v contains some contribution from
dr/dt. For a rigid rotating object, that means r in that formula
should be displacement from axis of rotation.

The point is most clear if you consider the line where v=0.
wxr then has to be zero as well if v=wxr, and if w!=0 then
the component of r perpendicular to w must be zero -- in other words
that line must be the axis of rotation (much as you were stating).

From: Martin Baker
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-25 15:07:01 PST

So, as an example, the earth is rotating on its own axis and also rotating
around the sun. In the frame-of-reference of the sun (ignoring any rotation
of the sun about its own axis) I can calculate the angular velocity of any
particle on the earth relative to the centre of the sun (if my knowledge of
geometry were good enough). This angular velocity will be a function of time
and also a function of its earth coordinates.

So given this function for the angular velocity of any particle of the
earth, how could I get to the angular velocity of the earth as a whole?
(also in the frame-of-reference of the sun and relative to the centre of the
sun) Which, for all I know, might also be a function of time?

Martin

From: Uncle Al (UncleAl0@hate.spam.net)
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-25 15:20:28 PST

Martin Baker wrote:
>
> So, as an example, the earth is rotating on its own axis and also rotating
> around the sun. In the frame-of-reference of the sun (ignoring any rotation
> of the sun about its own axis) I can calculate the angular velocity of any
> particle on the earth relative to the centre of the sun (if my knowledge of
> geometry were good enough). This angular velocity will be a function of time
> and also a function of its earth coordinates.
>
> So given this function for the angular velocity of any particle of the
> earth, how could I get to the angular velocity of the earth as a whole?
> (also in the frame-of-reference of the sun and relative to the centre of the
> sun) Which, for all I know, might also be a function of time?

Assume the Earth is a point mass to calculate its angular velocity
around the sun. How does that differ from integrating all its
infinitesimal volume contributors? Now you can worry about geocentric
(pure Earth) or barycentric (Earth-moon center of mass) coordinates.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!

From: meron@cars3.uchicago.edu (meron@cars3.uchicago.edu)
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-25 18:54:25 PST

In article <auddl9\$816\$1@sparta.btinternet.com>, "Martin Baker" writes:
>So, as an example, the earth is rotating on its own axis and also rotating
>around the sun. In the frame-of-reference of the sun (ignoring any rotation
>of the sun about its own axis) I can calculate the angular velocity of any
>particle on the earth relative to the centre of the sun (if my knowledge of
>geometry were good enough). This angular velocity will be a function of time
>and also a function of its earth coordinates.
>
>So given this function for the angular velocity of any particle of the
>earth, how could I get to the angular velocity of the earth as a whole?
>(also in the frame-of-reference of the sun and relative to the centre of the
>sun) Which, for all I know, might also be a function of time?
>
Measure the angular velocity relative to an inertial frame (fixed
stars, in this case). The motion around the sun is not relevant to
this.

Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"

From: Martin Baker
Subject: Re: Simple question about rotations
Newsgroups: sci.physics
Date: 2002-12-26 03:03:12 PST

I think I have found a way to put together what you are all telling me, am I
on the right track?

I have tried to do this in terms of kinematics, in other words I have tried
to avoid dynamics terms like centre-of-mass. The reason for this is that I
would like to be able to generalise the approach to other situations, to
take one example, the rotation at the end of a robot arm.

So the argument goes like this:

A solid body has 6 degrees of freedom, in other words its position can be
defined by a translation and an orientation. To do this we need to define a
local coordinate system on the body. For example, in the case of the earth,
we could choose to measure relative to the middle of the earths core, or a
place on the surface, or any other location. Now as the earth is moving,
both the translation and orientation are functions of time.

If we choose to measure the local coordinate system relative to a point on
the surface, then the translation will be a complicated function of time,
but if we measure the local coordinate system relative to the centre of the
earth then the translation will be a less complicated function of time. So
in this case it is better, but not necessary, to have a local coordinate
system relative to the centre of the earth. As far as I can see the
orientation as a function of time is independent of the local reference
point, in the case of the earth, one rotation per day.

So the angular velocity of a solid object is just the rate of change of its
orientation. The angular velocity of a particle on the object takes into
account both the translation and the orientation of the object. So the
angular velocity of a solid object and its component particles is different.
This is what was confusing me!

Sorry to be so slow, but I think I am now starting to get it clearer in my
mind.

Thank you all for your help and patience,

Martin

From: meron@cars3.uchicago.edu (meron@cars3.uchicago.edu)
Subject: Re: Simple question about rotations

Newsgroups: sci.physics
Date: 2002-12-25 18:51:32 PST

In article <aubs8b\$h6r\$1@sparta.btinternet.com>, "Martin Baker" writes:
>Mati,
>
>> So what was the business about "assumptions" I mentioned? Well, what
>> you asked (or appeared to me to be asking) was a subtly different
>> question. Not "what is the angular velocity relative to the
>> observation point?" but "assuming that an object moves with a
>> momentary velocity v resulting from a rotation around some axis with
>> some angular velocity w, how do I find this angular velocity?" And
>> here you've to be careful. You can pick any reference point P and
>> translate the observed velocity to angular velocity relative to P,
>> using the prescription above, but this result will in general differ
>> from the angular velocity of rotation that gave rise to the motion in
>> the first place. Even if you pick your P on the proper rotation axis
>> you will, in general get a different result. *Only* if P is ion the
>> rotation axis *and* the line joining P with the object is
>> perpendicular to v, only then will the w you calculate using the
>> formula above coincide with the angular velocity of the rotation.
>
>So if we have a rotating solid object, made up of an infinite number of
>particles, and we calculate the angular velocity of each of these particles
>about an arbitrary point P. Each particle of the solid object will
>potentially have a different angular velocity. However I suspect that there
>is one particular value of P (I'll call it Pc) where the conditions that you
>mention above will be met for all particles and the angular velocity for
>each particle about Pc will be the same.

Yes, with the caveat that you should think not in terms of angular
velocity about a point, but in terms of angular velocity about an
axis. Rotation is about an axis, not about a point. This distinction
is often lost when we make our little drawings in textbooks, since
what we usually draw are 2D systems where the axis is represented as a
point. In 3D, though, this is not the same.

So, with this caveat, for any rotating solid object you've a line (the
axis of rotation) such that the angular velocity of all the points

>So for a solid object, the definition of angular velocity must be the
>However for complex motion, such as planetary motion, this is not so obvious
>because this Pc will itself be moving, in fact its track could be a
>circular, so do we take this rotation into account? (assuming an absolute
>frame-of-reference)

You mean "assuming inertial frame of reference", I guess. If so, then
the motion of any solid object in such frame can be represented as the
combination of motion of the center of mass of the object following
whatever trajectory (straight line if there are no forces acting on
it) and a rotation of the object around an axis passing through the
center of mass.

Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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