How to find distance between two different circles 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
$C = A + \sqrt{ A^2 + B }$
where
$A = \frac L4 - \pi \frac{D + d}{8}$
$B = \frac{\left(D - d\right)^2}{8}$

Formula #2
$C = \frac{A + \sqrt{A - 32 \left(D - d\right)^2} }{16}$
where
$A = 4L - 2\pi \left(D + d\right)$

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