Jayjay

I think what's missing here is a realization that when the ball first rolls "down" to the equator, it rotates along a different axis than later when it rotates "up" again. Thus, they don't cancel each other out entirely.

Principia

I was making more general assumptions, so I actually forgot this restriction:
Suppose we have two balls of the same radius.

In that case, then, you are right that theta_d is always pi/2.

Hmm... I don't see this. theta * r = arc length transversed. So if length traversed is pi*r/2, then theta is pi/2.

Now my head hurts.

Jesse

Doesn't matter, my proof didn't assume it rotates around the same axis when it goes up as when it went down.

Principia

Good point, and I forgot about this too. So in reality:

xb' = R_d2(theta_d2)*R_z(theta_z)*R_d1(theta_d1) * xb

where the first axis of rotation d1 (going down to the equator) is not the same as the rotation axis d2 (going up from the equator).

Hmm...

Jayjay

I wasn't commenting your proof. Heck, I can't even comprehend your proof... I'm really bad at visualizing things in my head.

Silent Acorns

I think the answer is "yes". You can prove it pretty easily by using 2 dice and remembering that for every degree that the ball moves along the surface of the fixed ball it rotates 2 degrees. In other words when the moving ball has gone 90 degrees along the fixed ball it has rotated 180 degrees (with respect to the fixed ball).

Consider two dice with the following orientation:
(red die can move)

..5..
1-4-6 (thats 5 on top, 1 on left, and 4 up front)
..2..

-----
|---|
-----

Step 1: die moves to the front along a line of longitude:


..2..
1-3-6 (fixed die is behind)
..5..

Step 2: die moves 90 deg to the right

----- ..2..
|---| 3-6-4
----- ..5..

step 3: die keeps moving 270 more degrees and returns to the front:


..2..
1-3-6 (same as after step 1)
..5..

step 4: roll the die back to the top:


..5..
1-4-6
..2..

-----
|---|
-----

Same orientation as at the start.

Principia

Let's see if I can't confuse (or clear up) the matter more.

To roll down a longitude to the equator is essentially to roll down the x axis by theta/2 (i.e. 90 degrees). If the longitude is arbitrary, then we can rotate the whole system (i.e. moving ball and fixed ball) about the z-axis, rotate about the x axis, and finally rotate back about the z-axis. So:
R_d1(theta_d1) = R_z(a1)*R_x(pi/2)*R_z(-a1);

So, by the end of step 3) we should have

xb' = R_z(theta_z)*R_z(a1)*R_x(pi/2)*R_z(-a1)*xb;

The final rotation is also representable as 2 rotations about the z axes and one about the x:
R_d2(theta_d2) = R_z(a2)*R_x(pi/2)*R_z(-a2);

But a2 = theta_z + a1;

So by the end of step 4)

xb' = R_z(a2)*R_x(-pi/2)*R_z(-a2)*R_z(theta_z)*R_z(a1)*R_x(pi/2)*R_z(-a1)*xb;

We can simplify. R_v(t1)*R_v(t2)*...*R_v(tn) = R_v(t1 + t2 + ... + tn);

So:

xb' = R_z(a2)*R_x(-pi/2)*R_z(-a2)*R_z(theta_z)*R_z(a1)*R_x(pi/2)*R_z(-a1)*xb;

xb' = R_z(a2)*R_x(-pi/2)*R_z(-a2 + theta_z + a1)*R_x(pi/2)*R_z(-a1)*xb;

xb' = R_z(a2)*R_x(-pi/2)*R_x(pi/2)*R_z(-a1)*xb;

xb' = R_z(a2)*R_z(-a1)*xb;

xb' = R_z(theta_z)*xb;

So the final position is rotated about the z-axis by the amount of rotation about the equator...

Where did I go wrong?

EDIT: fixed sign errors

Jayjay

Uh, wouldn't that make the answer "no"?

Silent Acorns

Opps, I misread the question. I thought it was aking the much less obvious "can the ball return to its original orientation?". What I showed was the only way that it can (i.e. after a whole number of rotations around the equator). Any other trip will return the ball in a different orientation.

To this question my answer is "yes"

Using my previous notation:


..5..
1-4-6
..2..

-----
|---|
-----

Step 1: die moves to the front along a line of longitude:


..2..
1-3-6 (fixed die is behind)
..5..

Step 2: die moves 90 deg to the right

----- ..2..
|---| 3-6-4
----- ..5..

step 3: roll the die back to the top:

..5..
4-6-3
..2..

-----
|---|
-----

NOT the same orientation as at the start.

yguy

Using his model, with the equatorial leg of the journey being 45 degrees, the stick ends up pointing directly away from you before the ball rolls back up. When it does, its axis of rotation is 45 degrees away from what it was on the first leg, so that the 180 degree rotation about that axis will find the stick pointing again to the right.

The trick is that the final leg rotates the ball such that the stick undergoes an effective angular displacement of 2x degrees, where x is the equatorial travel, cancelling out the rotation about the N/S axis produced on the equatorial leg.

Dammit.