Principia

So does this hold true for any closed path around the surface of the fixed ball?

Silent Acorns

Good question. In the previous case all direction changes were orthagonal. I did a little testing with my "cube" model and found that such movements always bring the ball back to the original position. The cube model is almost useless for non-orthagonal changes in direction.

Tronvillan's mental model also strongly suggests to me that it's true for all paths.

tronvillain

Good question. I am not totally sure, but I think it does. While I am incapable of manipulating my mental model in anything other than longitudinal and latitudinal rolls, any closed path could be approximated by many such rolls, and in that case it would certainly hold. You can make the approximation as close as you like just by making each step smaller and increasing the number of steps, so unless I am missing something the answer would seem to be that it does hold true.

*chuckle* I like that we both needlessly identified it as a good question.

Silent Acorns

Just thought that I'd note that in at least some features of diagonals cannot be thought of as an infinite series of 90 degree turns. Consider the "staircase model" of the Pythagorean triangle.

._
| |_
| . |_
| . . |_
| . . . |_
|_________|

If the stair case climbs a height of H and covers a horizontal distance of L the total length of the carpet with no thickness on the staircase will always be H+L no matter how small the individual steps are.

Principia

I think the implication here is that there is a strict one-to-one mapping for the orientation of the moving ball to any position on the fixed ball, regardless of the path it took to get there...

I can see it...

Jesse

I think it works for any path. Imagine a ball with both lines of latitude and longitude drawn on it, but instead of putting it on another ball, put it on the surface of a mirror. Now keep the center of the ball fixed but allow it to rotate in any way you want. Since its mirror image is always rotating the opposite way, then if you imagine keeping the reflected ball's orientation fixed this would look just like one ball rolling on another with points along the circle the ball is rolling on rotating at twice the rate they would if the ball was sliding, which is just how rolling is supposed to work.

Principia

Very nice!

Jayjay

I think we have a winner! :notworthy

Principia

Just so that you guys can see what other people came up with as solutions (I believe that was the original thread.)

Silent Acorns

Just wanted to add that the total distance travelled using an infinite series of 90 degree turns won't be the same as along a diagonal or curved line. Since distance travelled is directly related to the amount of rotation we can't just assume that the position will always be the same.

However, if the answer to Principia's quetion is "yes" then for any final position on the fixed ball the final orientation of the moving ball is independent of the path taken (i.e. it is a function of the initial and final positions alone). This would suggest that for every point on the fixed ball there is a unique point on the moving ball where the two will touch each other. Obviously, every point on the moving ball can come into contact with the fixed ball, so either the pathway is irrelevent to the final orientation or all orientations are possible at all points.

If we "unroll" the two spheres into two plane surfaces (say two circles with the same area as the surface area of the spheres) it seems clear to me that the rolling of the sphere is analagous to a single point moving over top of two stationary circles (one on top of the other). The moving point represents the contact of the two spheres and the fact that the two circles don't move represents the fact that the orientation is only a function of the current postion of the contact point.