The vertical projection of a chord of a circle?

Question

I was wondering if anyone could help me with the problem below (finding x):

So we are given t_i (the initial tangent angle to the circle), t_o (the exiting angle of the tangent of the circle), the radius R and the x projection of the arc L. We want to find x, the vertical projection of the arc.

I included the case where from t_i we subtend an angle until t_o is 90 degress with respect to the vertical. The x (vertical projection of the arc) is easy to find: R(1-cos(theta)).

But I am having a lot of trouble finding a simple solution for when t_o is not 90 wrt the vertical. I have a solution in mind but it is not elegant at all:

1. Find the chord between t_i and t_o under the assumption (strong) that the angle subtended by the are is simply t_o minus t_i.
2. Use pythagorean theorem with chord length and L to find x.

Any ideas?

Answer 1

If I understand your question correctly it looks like $$R\left(\cos(t_o)-\cos(t_i)\right)$$ as the difference between the $y$-coordinates of the two tangential points.