Crazy Circle Illusion: Amaze Your Friends!

  • Topic:
  • Thread starter Thread starter kaliprasad
  • Start date Start date
  • Tags Tags
    Circle Friends
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 2K views · last reply
Mathematics news on Phys.org
Surely this must be tricks of the eye? Anyone up for calculating the real path traversed by the vertices of an octagon when in such a motion?
 
mathbalarka said:
Surely this must be tricks of the eye? Anyone up for calculating the real path traversed by the vertices of an octagon when in such a motion?

You can draw a circle with a cosine horizontally and a sine vertically.
This is how a cosine and a sine are defined on the unit circle.
It makes sense that if you create a whole bunch of sines on straight lines with the proper phase differences, that you'd get a circle.
 
Mesmerizing! :D

I have embedded the video so people can just watch it here.
 
I think the "trick" to this is that the "inner circle" (polygon) has exactly half the radius of the outer circle.

Imagine we trace the path of a point on a circle of radius $r$ as it travels inside a circle of radius $2r$. Since it doesn't really matter "when" we start tracking it (the path is periodic), assume that both circles are touching at the point $(0,2r)$ at $t = 0$, and that the outer circle is centered at the origin.

As the inner circle "rolls" counter-clockwise, the point on the inner circle we are tracking moves CLOCKWISE around a shifting center.

This center is at: $((2r-r)\cos t,(2r-r)\sin t) = (r\cos t,r\sin t)$. Since the outer circle's circumference (which is directly proportional to radius) is twice that of the inner circle, as the center has moved through an angle of $t$, the point we are tracking makes an angle of $2t$ with the point of tangency. Half of this angle is $t$, the other half is the angle our tracked point makes to a horizontal line passing through the center of the inner circle.

It follows our tracked point has coordinates:

$(r\cos(-t),r\sin(-t)) + (r\cos t,r\sin t) = (2r\cos t,0)$.

As $t$ varies, the image $\{(x(t),y(t)): t \in \Bbb R_0^+\}$ is the interval $[-2r,2r]\times \{0\}$, which is a "straight-line" (segment).