"For a translating non-rotating body in a medium to experience a
torque, the body must be CHIRAL,"

HK Moffat, "Six lectures on general fluid dynamics and two on
hydromagnetic dynamo theory," pp. 149-234 in R Balian & J-L Peube
(eds), "Fluid Dynamics" (Gordon and Breach, 1977)

Shapes of reduced symmetry can experience differential pressure when
translating through a fluid. The question is whether there will be a
net non-zero torque over time - not merely an altered translation.
The wing can flutter all it wants and still average to zero net torque
over time.

On Mon, 8 Apr 2007, Edward Green wrote:

As for your claim that mirror symmetry => no torque, I find that
surprising: consider a straight section of wing of constant cross
section, with an airfoil flat on bottom, curved on top. Mirror
symmetric, no? Yet it certainly seems plausible that the upper,
curved surface would experience more drag than the lower flat
surface. That looks like a torque to me.

Drag force: F = A v, where F and v are vectors means A is a rank-2 tensor.
Drag torque T = B omega, where T and omega are pseudovectors (ie they
change direction when you go from a right-handed coordinate system to a
left-handed system), and B is a rank-2 tensor.

If translational motion produces a torque, you have something like
T = C v, where T is a pseudovector, and v is a vector. If there is a plane
of mirror symmetry, centre the origin on it and reflect from right-handed
to left-handed.

I think it's pretty clear that the vector components of the torque in this
plane must be zero (after all, both sides are the same, so how can there
be a preferred direction?), and that's 2/3 of the way there. My source
(which, btw, is available online; do read it) claims that all components
are zero, which has a touch of magic about it.

I didn't expect that from experience in electrodynamics - the rules
concerning symmetry are a little different, although mirror symmetry is
special, chiral particles are special, etc:

http://www.arxiv.org/abs/physics/0402046

However, since we were doing the hydrodynamics of cubes in Stokes flow
assuming those things to be true (for convenience, and it would be a good
enough approximation, even if not true), I like the result.

As an aside, the usual path is hydrodynamics -> electrodynamics, since
fluid flow is more concretely visualisable than EM fields (Picture holding
a firehose in front of a classful of students, "This is non-zero
divergence!" They would remember that lesson unto the very end of their
days.) It's cute to be able to get some benefit from the opposite path.

Anyway, think about the classic structures that couple translation and
rotation: propellers and fans. Can you design a propeller or fan that has
a plane of mirror-symmetry? Basically, are the symmetry principles
mentioned above restricted to the linear regime (ie Stokes flow), or are
they general?

BTW -- I take it the tensorial form can be deduced purely from
linearity (though it doesn't quite suffice to invoke the word).

Linearity + it being a vector-to-vector transformation.

I also
see that "lift" can be generated purely by parasitic drag: for an
arbitrary shape, the general angle of attack will generate some force
perpendicular to the direction of motion.

Vector -> vector, no problem. I'm giving serious consideration to getting
a Stokes flow syrup/oil/glycerol tank as a toy. Alas, wall effects are
important. (I was about to do the calcs for the rotational wall effect,
which has an analytical solution if you care to use bi-polar coordinates,
with the real sphere and an image sphere centred on the origins. Of
course, the solution involves various special functions and is known to
converge poorly, N > 200 is needed. I thought I'd have one last look
through the literature just to see if anything useful had come out in the
last 2-3 months since I'd last looked, and there was this paper where
they'd done the calculations, and gave a polynomial fit to the results.
Given that I wanted the numbers, not the functions, superb!)

Btw, isn't "parasitic drag" due to remoras, lice, fleas, barnacles, etc
ruining your streamlining?

On Mon, 8 Apr 2007, Edward Green wrote:

As for your claim that mirror symmetry => no torque, I find that
surprising: consider a straight section of wing of constant cross
section, with an airfoil flat on bottom, curved on top. Mirror
symmetric, no? Yet it certainly seems plausible that the upper,
curved surface would experience more drag than the lower flat
surface. That looks like a torque to me.

Drag force: F = A v, where F and v are vectors means A is a rank-2 tensor.
Drag torque T = B omega, where T and omega are pseudovectors (ie they
change direction when you go from a right-handed coordinate system to a
left-handed system), and B is a rank-2 tensor.

If translational motion produces a torque, you have something like
T = C v, where T is a pseudovector, and v is a vector. If there is a plane
of mirror symmetry, centre the origin on it and reflect from right-handed
to left-handed.

I think it's pretty clear that the vector components of the torque in this
plane must be zero (after all, both sides are the same, so how can there
be a preferred direction?), and that's 2/3 of the way there. My source
(which, btw, is available online; do read it) claims that all components
are zero, which has a touch of magic about it.

I didn't expect that from experience in electrodynamics - the rules
concerning symmetry are a little different, although mirror symmetry is
special, chiral particles are special, etc:

http://www.arxiv.org/abs/physics/0402046

However, since we were doing the hydrodynamics of cubes in Stokes flow
assuming those things to be true (for convenience, and it would be a good
enough approximation, even if not true), I like the result.

As an aside, the usual path is hydrodynamics -> electrodynamics, since
fluid flow is more concretely visualisable than EM fields (Picture holding
a firehose in front of a classful of students, "This is non-zero
divergence!" They would remember that lesson unto the very end of their
days.) It's cute to be able to get some benefit from the opposite path.

Anyway, think about the classic structures that couple translation and
rotation: propellers and fans. Can you design a propeller or fan that has
a plane of mirror-symmetry? Basically, are the symmetry principles
mentioned above restricted to the linear regime (ie Stokes flow), or are
they general?

BTW -- I take it the tensorial form can be deduced purely from
linearity (though it doesn't quite suffice to invoke the word).

Linearity + it being a vector-to-vector transformation.

I also
see that "lift" can be generated purely by parasitic drag: for an
arbitrary shape, the general angle of attack will generate some force
perpendicular to the direction of motion.

Vector -> vector, no problem. I'm giving serious consideration to getting
a Stokes flow syrup/oil/glycerol tank as a toy. Alas, wall effects are
important. (I was about to do the calcs for the rotational wall effect,
which has an analytical solution if you care to use bi-polar coordinates,
with the real sphere and an image sphere centred on the origins. Of
course, the solution involves various special functions and is known to
converge poorly, N > 200 is needed. I thought I'd have one last look
through the literature just to see if anything useful had come out in the
last 2-3 months since I'd last looked, and there was this paper where
they'd done the calculations, and gave a polynomial fit to the results.
Given that I wanted the numbers, not the functions, superb!)

Btw, isn't "parasitic drag" due to remoras, lice, fleas, barnacles, etc
ruining your streamlining?

On Mon, 8 Apr 2007, Edward Green wrote:

As for your claim that mirror symmetry => no torque, I find that
surprising: consider a straight section of wing of constant cross
section, with an airfoil flat on bottom, curved on top. Mirror
symmetric, no? Yet it certainly seems plausible that the upper,
curved surface would experience more drag than the lower flat
surface. That looks like a torque to me.

Drag force: F = A v, where F and v are vectors means A is a rank-2 tensor.
Drag torque T = B omega, where T and omega are pseudovectors (ie they
change direction when you go from a right-handed coordinate system to a
left-handed system), and B is a rank-2 tensor.

If translational motion produces a torque, you have something like
T = C v, where T is a pseudovector, and v is a vector. If there is a plane
of mirror symmetry, centre the origin on it and reflect from right-handed
to left-handed.

I think it's pretty clear that the vector components of the torque in this
plane must be zero (after all, both sides are the same, so how can there
be a preferred direction?), and that's 2/3 of the way there. My source
(which, btw, is available online; do read it) claims that all components
are zero, which has a touch of magic about it.

I didn't expect that from experience in electrodynamics - the rules
concerning symmetry are a little different, although mirror symmetry is
special, chiral particles are special, etc:

http://www.arxiv.org/abs/physics/0402046

However, since we were doing the hydrodynamics of cubes in Stokes flow
assuming those things to be true (for convenience, and it would be a good
enough approximation, even if not true), I like the result.

As an aside, the usual path is hydrodynamics -> electrodynamics, since
fluid flow is more concretely visualisable than EM fields (Picture holding
a firehose in front of a classful of students, "This is non-zero
divergence!" They would remember that lesson unto the very end of their
days.) It's cute to be able to get some benefit from the opposite path.

Anyway, think about the classic structures that couple translation and
rotation: propellers and fans. Can you design a propeller or fan that has
a plane of mirror-symmetry? Basically, are the symmetry principles
mentioned above restricted to the linear regime (ie Stokes flow), or are
they general?

BTW -- I take it the tensorial form can be deduced purely from
linearity (though it doesn't quite suffice to invoke the word).

Linearity + it being a vector-to-vector transformation.

I also
see that "lift" can be generated purely by parasitic drag: for an
arbitrary shape, the general angle of attack will generate some force
perpendicular to the direction of motion.

Vector -> vector, no problem. I'm giving serious consideration to getting
a Stokes flow syrup/oil/glycerol tank as a toy. Alas, wall effects are
important. (I was about to do the calcs for the rotational wall effect,
which has an analytical solution if you care to use bi-polar coordinates,
with the real sphere and an image sphere centred on the origins. Of
course, the solution involves various special functions and is known to
converge poorly, N > 200 is needed. I thought I'd have one last look
through the literature just to see if anything useful had come out in the
last 2-3 months since I'd last looked, and there was this paper where
they'd done the calculations, and gave a polynomial fit to the results.
Given that I wanted the numbers, not the functions, superb!)

Btw, isn't "parasitic drag" due to remoras, lice, fleas, barnacles, etc
ruining your streamlining?

On Apr 2, 7:05 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:

Moffat next gives a very nice result: for a translating non-rotating
particle to experience a torque, the particle must be chiral ie lack
mirror symmetry.

I gave an apparent counter-example for torque about axes perpendicular
to the flow.

Now I'm not sure this theorem is correct for torque
about the flow axis either.

Consider two similar rectagular plates joined at a 90 degree angle at
the center of the short edges.

This object has mirror symmetry in the
plane of either rectangle. Now, consider a flow incident on each
plate at a 45 degree angle -- the thing begins to look like an
impeller, does it not?

In fact, if each plate experiences drag in the
plane of the surfaces, these drags will each have a component
producing a torque in the same sense about the flow axis.

Any bad assumptions?

We might remark that the overall system -- object + boundary
conditions -- is chiral.

We also remark that the propeller works backwards! Instead of being
twisted around towards the leading edge by dynamic forces, it is
pulled towards the trailing edge by drag. If we wanted to "fly" in a
Stokes fluid we would notice the same thing: we must run the propeller
backwards according to the normal scheme of things, to pull the plane
forward. In fact, in the dominance of drag, all control surfaces
would have opposite their usual effect.

On Apr 2, 1:41 am, Timo Nieminen <t...@physics.uq.edu.au> wrote:

In the Stokes limit, it's all linear.

Er, professor (raises hand tentatively)... what'st the Stokes limit?

"John C. Polasek" wrote:

=20
On 29 Mar 2007 22:05:37 -0700,

I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to =

this?

Thank you.

It would be interesting to see how you calculated that fact. But it =

true and can be proved trivially.
=20
J is a 2d rank tensor for a cube J11=3DJ22=3DJ33 so we J is equal to =

scalar j x the identity tensor I.
=20
The similarity transform for a 2d rank tensor T is A*TA
where A and A* are cosine rotation matrices for angle a and -a. Thus
for any rotation a about (any) axis (compound or principal is OK)
=20
J' =3D A*JA =3D A*IA =3D A*A=3D I =3D J
=20
Thus the rotations have no effect on the cube tensor.
=20
John Polasek

=20
What about the right circular cylinder of height=3D(radius)[sqrt(3)]?=20
Get funky!

"John C. Polasek" wrote:

On 29 Mar 2007 22:05:37 -0700,

I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?

Thank you.

It would be interesting to see how you calculated that fact. But it is
true and can be proved trivially.

J is a 2d rank tensor for a cube J11=J22=J33 so we J is equal to the
scalar j x the identity tensor I.

The similarity transform for a 2d rank tensor T is A*TA
where A and A* are cosine rotation matrices for angle a and -a. Thus
for any rotation a about (any) axis (compound or principal is OK)

J' = A*JA = A*IA = A*A= I = J

Thus the rotations have no effect on the cube tensor.

John Polasek

What about the right circular cylinder of height=(radius)[sqrt(3)]?
Get funky!

On Mar 30, 4:34 pm, "PD" <TheDraperFam...@gmail.com> wrote:

On Mar 30, 12:05 am,

I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
Thank you.

I don't think the statement is correct.

The moment of inertia is a symmetric tensor, so takes on the form of a
symmetric matrix. This can be decomposed into its eigenvalues and
eigenvectors. For a solid whose faces are rectangular (including a
cube) the eigenvectors lie along the 3 axes parallel to the directions
of the edges of the solid; the eigenvalues related to the dimensions
of the solid. For a uniformly distributed cube, all 3 eigenvalues are
equal. That makes the matrix diagonal. Therefore, every vector is a
eigenvector with the same eigenvalue and the moment of inertia is
independent of orientation.