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Let's say I have a sphere (determined by its center and radius) and two planes which cut individually the sphere. Individually, there will be to spherical caps. Let's suppose that both spherical caps overlap partially (i.e. part of the surface of one of the caps corresponds to a part of the other cap). There will be a body with two planar faces and a spherical face, which corresponds to the sphere. How can one calculate analytically the resulting surface area that corresponds to the spherical face of this overlapped cap?

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2 Answers 2

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I would compute it as a different of total area $A = 4 \pi r^2$ and the areas of two caps.

Let $d$ denote the maximal distance from a point on the cap to the plane. Then the area of a cap equals $A_\text{cap} = 2 \pi r d$. Therefore the result would be $2 \pi r ( 2 r - d_1 - d_2)$, assuming caps planes do not intersect within the sphere.

Note also that $d = r - \text{distance from sphere's origin to the plane}$. Denote that distance by letter $h$. Then the formula for the area between two spherical caps becomes $A_\text{btw} = 2 \pi r (h_1 + h_2)$


Added: Let a plane be specified by there equations $a_1 x + b_1 y + c_1 z + d_1 =0$, and $a_2 x + b_2 y + c_2 z + d_2 =0$. The distances to these planes from the center of the sphere (assuming it sits at the origin) are $h_1 = \left| \frac{d_1}{\sqrt{a_1^2+b_1^2+c_1^2}} \right|$ and $h_2 = \left| \frac{d_2}{\sqrt{a_2^2+b_2^2+c_2^2}} \right|$ respectively.

The following numerical calculations confirm the formula and provide an illustration: enter image description here

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this is excellent! the only thing, in your calculations, both caps do not overlap with each other, which is my main problem –  flow Oct 6 '11 at 14:23
    
I have just updated the question to make it more clear –  flow Oct 6 '11 at 14:24

A warning to begin with: The morphology of the situation can vary; so it might be difficult to automatize the computing of such areas. In the following I shall deal with the simplest case: The two caps $C_i$ are less than a hemisphere, and the two midpoints $M_i$ do not belong to the intersection $S:=C_1\cap C_2$. The bounding circles $\gamma_i$ intersect in two points $P$, $Q$ (otherwise $S$ would be empty) which together with $M_1$, $M_2$ form a spherical quadrangle which is symmetric with respect to the diagonal $M_1\vee M_2$. The other diagonal $d:=P\vee Q$ divides $S$ into two pieces $S_i$ which can be treated separately. The part $S_1$ bounded by $d$ and an arc of $\gamma_1$ can be viewed as a sector of $C_1$ minus an isosceles spherical triangle. Since all relevant angles and sidelengths can be computed from the given data, the area of $S_1$ can be found by elementary means, and the same holds for $S_2$.

This means you can forget about the highbrow procedure I proposed yesterday.

$\bigl[$The two planes $H_i$ intersect in a line $\ell$ which hits the sphere in two points $P$ and $Q$. Your piece of surface $S$ is bounded by two circular arcs $\gamma_i$ which connect $P$ and $Q$. From the given data you can compute the radii $r_i$, the constant geodesic curvatures $\kappa_i$ and the length $s_i$ of these arcs as well as the inner angle $\theta$ they enclose at $P$ and at $Q$. Given all this you can compute the area of $S$ by means of the Gauss-Bonnet theorem. You will have to take care of the signs: It makes a difference whether the shape $S$ is "convex" or is a lunula.$\bigr]$

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