Here I have this loop, made of parts of two different circles with radiuses $r_1$ and $r_2$, joined with two lines intersecting at $90$ degrees and touching the circles only in one point, as shown in the picture below. How to calculate the length of a way from one point on the loop back to the same point?

Thanks in advance! figure 1: loop

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    $\begingroup$ You need to decide how large the angle is. $\endgroup$ – Saucy O'Path Apr 7 at 13:14
  • $\begingroup$ @SaucyO'Path Or, alternatively, how far apart the centers are. $\endgroup$ – Ethan Bolker Apr 7 at 13:16
  • $\begingroup$ @EthanBolker Yes; in fact I was about to write it. $\endgroup$ – Saucy O'Path Apr 7 at 13:16
  • $\begingroup$ @SaucyO'Path 90 degrees, forgot to write that! ty in advance $\endgroup$ – Xxx Ddd Apr 7 at 13:21
  • $\begingroup$ @EthanBolker 90 degrees, forgot to write that! ty in advance $\endgroup$ – Xxx Ddd Apr 7 at 13:22

With a right angle in the middle, connecting the centers of the circles to the points of tangency produces two squares. Then you can see that the distance $s$ all around the figure eight is $3/4$ of each circle plus twice the sum of the radii: $$ s = \left(\frac{3 \pi}{2} + 2 \right)(r_1+r_2) = C(r_1+r_2). $$ If you know $s$ and $r_1$ then $$ r_2 = \frac{s - Cr_1}{Cr_2}. $$

Note how that requires $s > Cr_1$. When $s = Cr_1$ the second circle is a point and there's a right angle corner in the figure.

  • $\begingroup$ so if the length $s$ equals 280 and radius of the bigger circle $r_2$ is 25, how would you calculate the radius of the smaller circle? Like this - $r_1 = s / (3/2 \pi + 1) - r_2$ ? thanks for answering $\endgroup$ – Xxx Ddd Apr 7 at 15:12
  • $\begingroup$ I think you got the algebra wrong. See my edit. $\endgroup$ – Ethan Bolker Apr 7 at 15:32
  • $\begingroup$ oh, stupid mistake, I added the $(r_1 + r_2)$ just once. $\endgroup$ – Xxx Ddd Apr 7 at 15:35

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