tribalbeeyatch

I'm sure you were. Let's try to stick to the subject at hand, though.

Jayjay

My answer to the opening post would be "yes". However, I notice that a couple of the first responses are "no", made by intelligent people no doubt. So before reading the rest of the thread, I decided to post my disagreement, just so I can feel smug if I was right.

EDIT: Aww shucks. Looks like I was wrong. Can't prove it though.

Jesse

I'm pretty sure the answer is "no"--the top ball will end up with exactly the same position and orientation as when it started (no rotation on the N/S axis or any other axis). If you imagine rolling the top ball by any multiple of 90 degrees along the equator of the bottom ball you can see that it will always work out this way in the 4 possible cases there, and I have a sort of a general proof in mind that this would work for any angle, but it's a visual demonstration which would be difficult to describe in words. If anyone's really interested I'll give it a shot, but first try drawing pictures of various 90-degree rolls along the equator, and maybe a 45-degree roll as well, to see if you can figure it out yourself (and if anyone comes up with a more elegant proof than mine I'd be interested to hear it).

yguy

There is no net rotation of the movable ball's N/S axis because it undergoes equal and opposite rotation during the longitutinal travel. That is not true for the equatorial travel, so that if it travels 90 degrees along the equator, it will rotate 180 degrees about the N/S axis.

Jesse

There is no net rotation along any axis. Try it with fando's initial picture again:

Now, when the red ball rolls down the the equator, the point marked by the stick will turn 180 degrees so it's on the left--it'll now be the point of contact between the red ball and the black ball. If you roll the red ball 90 degrees clockwise along the black ball's equator, so the red ball is in front of the black ball from your point of view, the point marked by the stick will rotate 180 degrees, so it will again be pointing to the right. Roll the red ball back up to the pole and that point will not change orientation, so in the final position it'll still be pointing out to the right. And you already know why the N/S axis does not rotate, so that the same point that's on the bottom of the red ball at the start will be on the bottom at the end; if neither the point on the bottom nor the point marked by the stick has changed, there cannot have been any net rotation along any axis.

attempt at illustration:

It's pretty easy to check that it works out this way for rolling any multiple of 90 degrees along the equator, but a bit trickier to prove it for arbitrary angles.

nermal

Right. Yes. Confusing but exactly right. Still, symmetry is maintained about an arbitrary axis drawn between the balls while the Moveable Ball (MB) is rotated about the equator. Therefore, the relative position of the balls will have been unchanged by the entire exercise. Yep. Um, Yep.

Ed

yguy

That's what I'd like to think, but Jesse's answer seems to compute for 90 degree increments, which contradicts me - but I can't get his example to work in my head for a 45 degree rotation. No way can I see the same stick ending up in the same place except for 90 degree increments.

I suspect we're all missing something.

Jesse

Like I said earlier, I think I have a visual proof that it will work for arbitrary angles. Basically the way it works is, if you roll the red ball x degrees clockwise along the black ball's equator, you know that the stick rotates 2x degrees clockwise from its initial straight-left direction--then the trick is to rotate your mental "camera angle" by x degrees in the same direction, so the red ball is again on the right side of the black ball, and the stick is pointing x degrees from straight-left relative to your new viewing angle. Then rotate the red ball back up to the top, and you'll find that the stick is now pointing at 180-x degrees relative to you, so if you finally rotate your mental "camera angle" back x degrees in the opposite direction the stick will be pointing 180 degrees, or straight out right again. This probably isn't too clear, so I can give a more detailed version if anyone's interested, but try fooling around with the 45-degree case and you might see what I'm talking about.

Principia

Let's see if we can't get some math down:

1) Start with the rolling ball touching the north pole of the fixed ball.
Start with any point xb in the rolling ball (i.e. |xb| <= rb, where rb is the radius of the rolling ball), relative to its center. After each step below, we calculate a new relative position xb', as follows:

2) Roll it down to the equator along a line of longitude.
Let d represent the vector perpendicular to the plane containing the longitude.

Then rolling a ball along the longitude corresponds to rolling it by some angle theta_d about the axis represented by the vector d.

xb' = R_d(theta_d) * xb, where
R_d(theta_d) is a rotation matrix (3x3), with the property that R_d(theta_d) * R_d(-theta_d) = I. The notational convention is R_v(theta), where v is the axis of rotation and theta is the angle.

3) Then roll it along the equator for an arbitrary distance.
Let z represent the vector corresponding to the polar axis. Then, we are rotating now about the z axis of the sphere by an arbitrary angle, theta_z.

xb' = R_z(theta_z)* R_d(theta_d) * xb

4) Then roll it back up to the north pole along a line of longitude.
Using the same conventions as before:

xb' = R_d(-theta_d)*R_z(theta_z)*R_d(theta_d) * xb

In general, then, xb' = R * xb, but R is not the identity matrix I unless R_z(theta_z) = I (which is true if theta_z is a multiple of 2*pi). So xb' is not guaranteed to be xb.

The answer then is yes, the ball can be rotated relative to its original position.

EDIT: ugh

Jesse

Principia, does your math take into account that we are not rolling it an arbitrary distance down a line of longitude, but always 90 degrees, down to the equator? Also, did you take into account that rolling will change the ball's orientation by twice the angle that sliding the same distance would?