Solutions to the gear rotation questionsQuestion 1. Can you prove these mathematically? (a) An odd number of gears in a closed loop in a plane will not turn no matter what their size and tooth count. (b) An even number of gears in a closed loop in a plane will turn smoothly no matter what their size and tooth count.
The underlying principle isn't physics, but geometry, and it applies not only to gears, but to smooth friction wheels as well. The figure shows two friction wheels of different diameter. When they turn without slipping, the two circles must turn through the same arc, but in opposite sense of rotation. That is, if the left wheel turns clockwise through arc A, the right wheel turns counterclockwise through arc B, and B =  A.
There are even exotic gears that have noncircular perimeters, such as oval or elliptical. You could make a model with an even number of these in a loop, and the gears in the model would turn smoothly.
Question 2. Find a way to make a model of an odd number real gears interlocked in a series loop (each one meshing with the two adjacent gears) so that they all can freely turn. That's an easy one. Now try to make the gears turn on their own fixed axles. Consider the model as shown at the right. Now rotate the entire model. This is an example of a "trick" question that depends on ambiguity in wording. When we reword it using "turn on their own axles" instead of merely "all can freely turn" it remains an impossible task. At least I've not seen a solution yet. One reader, John Crick, suggested a solution that took advantage of an ambiguity in my earler statement of the problem in which I failed so specify a "series loop" of gears. He proposed a rather exotic gear in which the gear widht was twice as large as the gap between teeth, allowing three gears to mesh there if at least one of them was in another plane inclined to the first two. It still seems that the problem with gears in a series meshing in a closed loop, on separate, fixed axles, cannot freely turn if there's an odd number of them. Now the challenge is to devise a mathematial proof of this assertion. Return to Whoops!
