I have yet to see anyone prove mathematically that the number of rotations of the outer small circle is R/r + 1, where R and r are the radii of the large and small circles respectively and the ratio R/r is an integer. I haven't yet done it but I can see how the math problem should be formulated.
Imagine the large circle with its center at the origin of the x-y axis. Now place the small circle with its center at (0,R+r). Draw an ARROW as the diameter line of the small circle such that the tail of the arrow is at (0,R) and the head of the arrow is at (0, R+2r). This is crucial for counting complete rotations of the small circle. Now imagine rolling the small circle clockwise along the circumference of the large one and STOP when the arrow points STRAIGHT UP again. This corresponds to exactly one rotation of the small circle (orientation of the arrow is the same as at the start). This happens BEFORE the tail point of the arrow again comes into contact with the large circle. But the arc length between consecutive tail points touching the circumference of the large circle is clearly 2*pi*r, so the distance swept along the large circle's circumference when the small circle performs one complete rotation is LESS THAN 2*pi*r. It is 2*pi*r - D, where D is the arc length of the "shortfall".
Because D is not zero, it's already clear that by the time the small circle reaches its starting point back at the top of the large circle it will have completed MORE than R/r rotations. But how many more? To figure that out we need to calculate D.
We KNOW that if R/r = n (an integer) then the small circle will arrive PRECISELY back at its starting point after exactly n rolls of its circumference along the circumference of the large circle. But this also implies that the total number of rotations of the small circle is an integer that is GREATER than n. From cardboard circle experiments we know that the total number of rotations is n+1. This can happen only if n*D = 2*pi*r - D, or
D=2*pi*r^2/(R+r).
If you can derive this formula mathematically, from the geometry, then you've just proven that if R/r = n then the small circle rotates n+1 times as it goes around the big circle.
No, you haven't proven anything that's relevant.
I have yet to see anyone prove mathematically that the number of rotations of the outer small circle is R/r + 1, where R and r are the radii of the large and small circles respectively and the ratio R/r is an integer. I haven't yet done it but I can see how the math problem should be formulated.
Imagine the large circle with its center at the origin of the x-y axis. Now place the small circle with its center at (0,R+r). Draw an ARROW as the diameter line of the small circle such that the tail of the arrow is at (0,R) and the head of the arrow is at (0, R+2r). This is crucial for counting complete rotations of the small circle. Now imagine rolling the small circle clockwise along the circumference of the large one and STOP when the arrow points STRAIGHT UP again. This corresponds to exactly one rotation of the small circle (orientation of the arrow is the same as at the start). This happens BEFORE the tail point of the arrow again comes into contact with the large circle. But the arc length between consecutive tail points touching the circumference of the large circle is clearly 2*pi*r, so the distance swept along the large circle's circumference when the small circle performs one complete rotation is LESS THAN 2*pi*r. It is 2*pi*r - D, where D is the arc length of the "shortfall".
Because D is not zero, it's already clear that by the time the small circle reaches its starting point back at the top of the large circle it will have completed MORE than R/r rotations. But how many more? To figure that out we need to calculate D.
We KNOW that if R/r = n (an integer) then the small circle will arrive PRECISELY back at its starting point after exactly n rolls of its circumference along the circumference of the large circle. But this also implies that the total number of rotations of the small circle is an integer that is GREATER than n. From cardboard circle experiments we know that the total number of rotations is n+1. This can happen only if n*D = 2*pi*r - D, or
D=2*pi*r^2/(R+r).
If you can derive this formula mathematically, from the geometry, then you've just proven that if R/r = n then the small circle rotates n+1 times as it goes around the big circle.
Good luck!!