I am trying the find the distance between two different sized circles, both centred on the horizontal plane. I know the diameter of each circle, and the length around both circles if wrapped like a bike chain around the front and rear sprockets.

I've tried searching for the formula's but everything seems to work with a known distance to find the external tangent length's etc.

What are the steps to calculate this?

Thanks, Matt

Update: Found a couple of formulas from timing pulley sites that seem to do the job. Would still like to understand how they get built just out of interest.

Formula #1

$C = A + \sqrt{ A^2 + B }$


$A = \frac L4 - \pi \frac{D + d}{8}$

$B = \frac{\left(D - d\right)^2}{8}$

Formula #2

$C = \frac{A + \sqrt{A - 32 \left(D - d\right)^2} }{16}$


$A = 4L - 2\pi \left(D + d\right)$

Both come out with the same answer. Which isn't really a surprise as they look very similar.

  • 1
    $\begingroup$ Could you be more specific about the information you have? When you say "the length... if wrapped like a bike chain", do you just mean the perimeter of each circle, or the actual length of the entire chain? And do you know what the centers of the circles are in coordinates, or is that all the information you have? $\endgroup$ – Jack M May 4 '14 at 12:26
  • $\begingroup$ Calculate their centers' distance and substract the sum of their radiuses... $\endgroup$ – DonAntonio May 4 '14 at 12:32
  • $\begingroup$ @DonAntonio You have to distinguish the case where one circle is inside of the other circle frome the one where it is outside. $\endgroup$ – Thomas May 4 '14 at 12:39
  • $\begingroup$ Indeed so, @Thomas: in the first case it'd be $\;R-d-r\;$ , with the letter denoting the big circle's radius, the centers' distance and the little, inner circle radius, resp. $\endgroup$ – DonAntonio May 4 '14 at 12:44
  • $\begingroup$ I think it's reasonable to assume that each circle is outside the other, or else the bicycle chain measurement doesn't make sense. $\endgroup$ – bubba May 4 '14 at 12:51

enter image description here

Due to similar triangles and the Pythagorean Theorem we have:

$$\frac{h_1 + h_2}{R} = \frac{h_2}{r} \tag{1}$$ $$\frac{D + y}{R} = \frac{y}{r} \tag{2}$$ $$(D + y)^2 = R^2 + (h_1 + h_2)^2 \tag{3}$$

Combining the 3 we get :

$$h_1^2 = D^2 - (R - r)^2 \tag{4}$$

Arcs $A_1$ and $A_2$ are two parts of a semicircle, by scaling, so:

$$\frac{A_1}{R} + \frac{A_2}{r} = \pi \tag{5}$$

The Length of $A_2$ is $A_2 = \theta\,r$, with $\theta$ in radians. $\tan(\theta) = \frac{ h_2} { r } = \frac{h_1}{R - r} \text{(by applying (1))}$, so

$$\tan(\theta) = \frac{\sqrt{D^2 - (R - r)^2}}{R - r} \tag{6}$$

The length of your chain, $L$, is $$L = 2(A_1 + A_2 + h_1) \tag{7}$$

Applying (4), (5), and (6) to (7) you get: $$L = 2\left(\tan^{-1}\left(\frac{\sqrt{D^2 - (R - r)^2}}{R - r}\right)(r - R) + \pi R + \sqrt{D^2 - (R - r)^2}\right) \tag{7}$$

You won't be able to solve this for $D$ with elementary relations.

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Let $r, R$ be the radiuses, $L$ the length of chain, $D$ the distance and $\alpha$ the angle between the line joining centers and the radius of the bigger circle in radians. Then we have $$\frac L 2 = \alpha r + (\pi - \alpha)R + (R-r)\tan \alpha$$ and $$R-r = \cos \alpha D$$

These equations are sufficient to find $D$, but the calculation itself might be unpleasant.

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  • $\begingroup$ I wish there was a simple way to draw diagrams here... $\endgroup$ – Karolis Juodelė May 4 '14 at 12:39
  • $\begingroup$ Which radius of the bigger circle? There's quite a few of them. $\endgroup$ – Jack M May 4 '14 at 12:39
  • $\begingroup$ @JackM, What do you mean? There are two circles. $\endgroup$ – Karolis Juodelė May 4 '14 at 12:41
  • $\begingroup$ "Let $\alpha$ be the angle between the line joining centers and the radius of the bigger circle" $\endgroup$ – Jack M May 4 '14 at 12:47
  • $\begingroup$ By "radius" he means the radial line through the point of tangency. $\endgroup$ – bubba May 4 '14 at 13:02

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