I liked your intuitive explanation, but I'm not sure I agree with it as it stands.
Yes, when rolling a circle round the outside of a regular polygon like a hexagon one can expect the circle to roll along the complete length of each side, and so make more rotations than it does inside, where its travel is limited by not fitting into the corners. But surely this problem doesn't arise when rolling a circle inside a larger circle. Its roll brings it into contact with the entire circumference of the larger. So failure to fit in everywhere inside doesn't account for the lost rotation.
This also applies to rolling a straight sided polygon which fits snugly into the corners of a larger version of itself, such as a small square into a bigger square. I tried it with a cardboard square inside a square hole of 2x the side length, also cut out of a piece of cardboard. Inside, just 1 rotation when rolled right round, but outside 3. So it's the same principle, rolling inside gets R/r - 1 rotations, and outside R/r + 1 rotations. 1 and 3 respectively when R/r = 2.
I think this formula suggests the simplest explanation. When R/r = 1, that is when they are the same size, then rolling outside results in 1+1 = 2 rotations. but inside gets you 1 - 1 = 0 rotations. The latter is hardly surprising since you can't really roll a circle, or indeed any polygon, inside another of exactly the same dimensions. It fits everywhere too snugly. So there's your lost rotation.
of a regular polygon like a hexagon one can expect the circle to roll along the complete length of each side, and so make more rotations than it does inside, where its travel is limited by not fitting into the corners. But surely this problem doesn't arise when rolling a circle inside a larger circle. Its roll brings it into contact with the entire circumference of the larger. So failure to fit in everywhere inside doesn't account for that lost rotation.
This also applies to rolling a straight sided polygon which fits snugly into the corners of a larger version of itself, such as a small square into a bigger square. I tried it with a cardboard square inside a square hole of 2x the side length, also cut out of a piece of cardboard. Inside, just 1 rotation when rolled right round, but outside 3. So it's the same principle, rolling inside gets R/r - 1 rotations, and outside R/r + 1 rotations. 1 and 3 respectively when R/r = 2.
I think this formula suggests the simplest explanation. When R/r = 1, that is when they are the same size, then rolling outside results in 1+1 = 2 rotations. but inside gets you 1 - 1 = 0 rotations. The latter is hardly surprising since you can't really roll a circle, or indeed any polygon, inside another of exactly the same shape and dimensions. It fits everywhere too snugly.
I liked your intuitive explanation, but I'm not sure I agree with it as it stands.
Yes, when rolling a circle round the outside of a regular polygon like a hexagon one can expect the circle to roll along the complete length of each side, and so make more rotations than it does inside, where its travel is limited by not fitting into the corners. But surely this problem doesn't arise when rolling a circle inside a larger circle. Its roll brings it into contact with the entire circumference of the larger. So failure to fit in everywhere inside doesn't account for the lost rotation.
This also applies to rolling a straight sided polygon which fits snugly into the corners of a larger version of itself, such as a small square into a bigger square. I tried it with a cardboard square inside a square hole of 2x the side length, also cut out of a piece of cardboard. Inside, just 1 rotation when rolled right round, but outside 3. So it's the same principle, rolling inside gets R/r - 1 rotations, and outside R/r + 1 rotations. 1 and 3 respectively when R/r = 2.
I think this formula suggests the simplest explanation. When R/r = 1, that is when they are the same size, then rolling outside results in 1+1 = 2 rotations. but inside gets you 1 - 1 = 0 rotations. The latter is hardly surprising since you can't really roll a circle, or indeed any polygon, inside another of exactly the same dimensions. It fits everywhere too snugly. So there's your lost rotation.
of a regular polygon like a hexagon one can expect the circle to roll along the complete length of each side, and so make more rotations than it does inside, where its travel is limited by not fitting into the corners. But surely this problem doesn't arise when rolling a circle inside a larger circle. Its roll brings it into contact with the entire circumference of the larger. So failure to fit in everywhere inside doesn't account for that lost rotation.
This also applies to rolling a straight sided polygon which fits snugly into the corners of a larger version of itself, such as a small square into a bigger square. I tried it with a cardboard square inside a square hole of 2x the side length, also cut out of a piece of cardboard. Inside, just 1 rotation when rolled right round, but outside 3. So it's the same principle, rolling inside gets R/r - 1 rotations, and outside R/r + 1 rotations. 1 and 3 respectively when R/r = 2.
I think this formula suggests the simplest explanation. When R/r = 1, that is when they are the same size, then rolling outside results in 1+1 = 2 rotations. but inside gets you 1 - 1 = 0 rotations. The latter is hardly surprising since you can't really roll a circle, or indeed any polygon, inside another of exactly the same shape and dimensions. It fits everywhere too snugly.
So there's your lost rotation.