Calculating length of vertical line bisecting parallel arcs

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I have 2 arcs, offset from one another (never intersecting) and a vertical line through them both (NOT at the center of the arcs). Is there a way to calculate the vertical distance between the 2 arcs? It changes depending on where along the arc the vertical line is.

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If the two equations are $f(x)$ and $g(x)$, the obvious answer is $g(x)-f(x)$.

If, as suggested by Monstrous Moonshine, the two curves are the upper half of circles, then the curves are $f(x) =\sqrt{r^2-x^2} $ and $g(x) =\sqrt{s^2-x^2} $, where $s > r > 0$.

The distance is then, doing some fiddling with the expression for the heck of it,

$\begin{array}\\ g(x)-f(x) &= \sqrt{s^2-x^2}-\sqrt{r^2-x^2}\\ &= (\sqrt{s^2-x^2}-\sqrt{r^2-x^2})\dfrac{\sqrt{s^2-x^2}+\sqrt{r^2-x^2}}{\sqrt{s^2-x^2}+\sqrt{r^2-x^2}}\\ &= \dfrac{(s^2-x^2)-(r^2-x^2)}{\sqrt{s^2-x^2}+\sqrt{r^2-x^2}}\\ &= \dfrac{s^2-r^2}{\sqrt{s^2-x^2}+\sqrt{r^2-x^2}}\\ \end{array} $

For other curves, just put in the expressions.

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If the rings are circular and concentric

$$h = \sqrt{R^2- x^2}-\sqrt{r^2- x^2} $$

For a narrow ring width:

$$ w = R-r, h \approx w /\cos \phi $$