Arc lengths of parallel sections of spherical cap? I am imaging the cross sectional view of an object shaped like a spherical cap:
(see image drawing)
Ideally, I should be imaging the purple arc, which passes through the center of the spherical cap. I also know what "a" should be.
However, it's impossible to align my imaging perfectly, so I end up imaging the orange arc instead -- slightly off center. I know my new a', which is less than "a" by some quantity. I also have an h', and don't know what the ideal h should be.
I would like to calculate the ideal h as well as the purple arc length, given a', a, h', and the measured arc length in orange.
What I am having trouble with:
I figured out that the distance between the two "cutting planes" would be $\sqrt(a^2-a'^2)$
I'm stuck at that point though, and don't know what to do from there. Any sort of advice (or even a nudge in the right direction) would be greatly appreciated!
 A: If you just want $h$ and the length of the purple arc, I would skip the computation of the distance between planes.
You have $a,$ $a'$, and $h'.$
The orange arc is an arc of a circle with radius $r'.$
You have a right triangle with legs $a'$ and $r' - h'$ and hypotenuse $r',$
so $r'^2 = a'^2 + (r' - h')^2,$ from which a little algebra finds that
$$ r' = \frac{a'^2 + h'^2}{2h'}. $$
You also have a right triangle with legs $a$ and $r - h$ and hypotenuse $r.$
But the center of the orange arc's circle is the same distance below the bottom of the cap as the center of the purple arc's circle.
That is, $r' - h' = r - h.$ So
$$ r^2 = a^2 + (r - h)^2 = a^2 + (r' - h')^2. $$
And since you were given $a$ and $h'$ and have computed $r'$ you now can easily find $r.$
Now $h = r - (r - h) = r - (r' - h'),$ and you know $r,$ $r'$, and $h'$, so you can find $h.$
The angle $\theta = \arcsin(a/r),$ and the length of the purple arc is $2r\theta$.
This answer did not use the length of the orange arc. In principle you only need to know two of the three values $a',$ $h',$ and the length of the orange arc, but in practice I think this will get into numeric methods without well-known convenient solutions.
