# Spherical cap problem - trigonometry / circle theorems problem / surface area

Graph here

I am trying to derive the following equation from a paper I am studying, which the author has derived from the graph above. The two slightly curved lines here are modelled as the surfaces of spherical caps, with surface area S (known). The equation is

$$\frac{X}{4\sqrt{S/\pi}} = \sin( \phi_0) + \frac{\sqrt{S/ \pi}}{R}\cos( \phi_0).$$

I am thinking that if the spherical cap is flattened out, the curved line can be seen as the diameter (D) of this flattened circle, hence $$D=2\sqrt{S/\pi}$$. I believe this is where the $$\sqrt{S/\pi}$$ term comes from. As for the rest, I have managed to get an equation for $$D$$ in terms of $$\phi$$, $$\alpha$$ and $$D$$, using that $$D=R\theta$$ (as the triangle in the diagram can be seen as a segment of a circle).

My solutions however all seem pretty complicated and I cannot manage to get them to match up to the correct one. Where am I going wrong? Do I need to use a different approach? (I thought about using the volume of revolution but I can't work out how to use that here).

Can anyone help?

This doesn't seem quite right to me. If I take $$S$$ to be half the surface area (i.e., the surface area of one of the "arcs"), then I have $$S=4\pi R^2\sin^2\alpha$$, so $$\sqrt{\frac S\pi} = 2R\sin\alpha$$. By similar triangles, the angle between the arc of the circle and the chord joining the endpoints of that arc is $$\alpha$$ as well, and so $$\frac x2 = 2R\sin\alpha\sin(\phi+\alpha) = 2R\sin\alpha\big(\sin\phi\cos\alpha + \cos\phi\sin\alpha\big).$$ Since $$\alpha$$ is presumably small, we take $$\cos\alpha\approx 1$$, and this becomes $$\frac x{2\sqrt{S/\pi}} \approx \sin\phi + \cos\phi\sin\alpha = \sin\phi+\frac{\sqrt{S/\pi}}{2R}\cos\phi.$$ This is "close" to what you posted, but certainly different. At this point, I'm not sure the author is correct.
• You can find the surface area of the cap by doing the surface integral $\int_0^{2\pi}\int_0^{2\alpha} R^2\sin\phi\,d\phi = 2\pi R^2(1-\cos 2\alpha) = 4\pi R^2 \sin^2\alpha$. For the similar triangles, you have a right triangle (because the tangent to a circle is perpendicular to the radius) with one leg the radius and angle $\alpha$ at the center of the circle (so I'm doing half the chord); now draw the chord, which will be perpendicular to the hypotenuse of the original right triangle. That gives you similar triangles. The angle between the chord and the tangent is $\alpha$. – Ted Shifrin Dec 20 '18 at 2:23
• Sorry, I have a final question! How did you work out that $\frac{x}{2}=2Rsin\alpha sin(\phi +\alpha )$? – maria1991 Dec 20 '18 at 3:00
• Make a right triangle with one leg $x/2$ and hypotenuse the chord joining the endpoints of the arc. – Ted Shifrin Dec 20 '18 at 3:04