Hello bcdave Originally Posted by

**bcdave** ...While this works for my current scenario, where lines WX, AB and CD are all parallel, we do get a several cases where line CD is not parallel to line AB. Would you have any suggestions for when these line are not parallel?...

Suppose that the line $\displaystyle DC$ makes an angle $\displaystyle \theta$ with $\displaystyle BA$, and that the perpendicular distance from $\displaystyle C_1$ to $\displaystyle DC$ is $\displaystyle d$. (I am assuming that this angle and this distance are known.)

Then it's fairly straightforward to show that the radius of the new arc, $\displaystyle R_2$, is given by: $\displaystyle R_2= \frac{d-R_1\sin\theta}{1-\sin\theta}$

( Note that, where $\displaystyle DC$ and $\displaystyle BA$ are parallel, $\displaystyle \theta = 0$, and $\displaystyle R_2 = d$, as we found before.)

So the position of $\displaystyle C_2$ is along the line joining the 'tangent point' on your diagram (the mid-point of the semi-circle $\displaystyle XY$) to $\displaystyle C_1$, at a distance $\displaystyle R_2-R_1$ from $\displaystyle C_1$, where: $\displaystyle R_2-R_1 = \frac{d-R_1}{1-\sin\theta}$

I haven't time now to draw the diagram and show you the working, but if you want to try it, find the point of intersection, $\displaystyle P$ say, of this 'line of centres' with $\displaystyle DC$. Then:$\displaystyle PC_1 = \frac{d}{\sin\theta}$

$\displaystyle \Rightarrow PC_2 = \frac{d}{\sin\theta}+R_2-R_1$

$\displaystyle \Rightarrow R_2 = PC_2\sin\theta$ $\displaystyle =\left(\frac{d}{\sin\theta}+R_2-R_1\right)\sin\theta$

Solving for $\displaystyle R_2$ gives the result above.

Grandad