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Spherical Geometry

The surface area of a sphere of radius $r$ in three dimensions is $4\pi r^2$ — a familiar result. But do you know how to compute the area of a spherical triangle, a spherical cap, or the intersection of two caps? This page addresses these somewhat advanced topics in spherical geometry.

Spherical Polygons and Spherical Triangles

A great circle is the circle formed by the intersection of a sphere with a plane passing through the centre of the sphere. A region on the sphere enclosed by several great circles is called a spherical polygon; the intersections of the great circles serve as vertices, and the arcs along the great circles serve as edges.

Consider the sphere shown in the figure². The blue region is a figure on the sphere bounded by three great circles. Let the three vertices (intersections of the great circles) be $A, B, C$, the side lengths be $a, b, c$, and the interior angles be $\alpha, \beta, \gamma$. The side lengths $a, b, c$ are arc lengths along the sphere (not straight-line distances between vertices); for a unit sphere, they equal the angles subtended at the centre between the respective vertices. Such a triangle is called a spherical triangle, and the associated theory is spherical trigonometry. For a unit sphere, the following theorems hold (shown alongside their plane analogues for comparison).

Theorem	Spherical (unit sphere)	Plane
Law of cosines	$\cos a = \cos b \cos c + \sin b \sin c \cos \alpha$ $\cos b = \cos c \cos a + \sin c \sin a \cos \beta$ $\cos c = \cos a \cos b + \sin a \sin b \cos \gamma$	$a^2=b^2+c^2-2bc\cos \alpha$ $b^2=c^2+a^2-2bc\cos \beta$ $c^2=a^2+b^2-2bc\cos \gamma$
Law of sines	$\displaystyle \frac{\sin a}{\sin \alpha} = \frac{\sin b}{\sin \beta}= \frac{\sin c}{\sin \gamma}$	$\displaystyle \frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$
Area	$\displaystyle 2 \sin^{-1} \left[ \frac{\sqrt{ \sin s \sin (s-a)\sin (s-b)\sin (s-c)} }{2 \cos(a/2)\cos(b/2)\cos(c/2)} \right]$ where $2s = a+b+c$ or equivalently $\alpha+\beta+\gamma-\pi$	$\sqrt{s (s-a)(s-b)(s-c)}$ where $2s = a+b+c$ For a triangle inscribed in the unit circle: $2 \sin\alpha \sin\beta \sin\gamma$

Unlike the plane case, $a,b,c$ have the character of angles, making the formulae look quite different at first glance. However, when $a,b,c$ become small (i.e., the spherical triangle shrinks), one has $\sin a\rightarrow a$ and $\cos a \rightarrow 1-a^2/2$³, and the spherical theorems reduce to their plane counterparts. Finally, with $2s = a+b+c$ and $2S = \alpha+\beta+\gamma$, the following identities also hold:

$$\begin{array}{l}
\sin^2 (\alpha/2) = [\sin(s-b)\sin(s-c)] / [\sin b \sin c] \\
\cos^2 (\alpha/2) = [\sin s \sin(s-a)] / [\sin b \sin c]\\
\sin^2 (a/2) = -[\cos S \cos(S-\alpha)] / [\sin \beta \sin \gamma]\\
\cos^2 (a/2) = [\cos(S-\beta)\cos(S-\gamma)]/[\sin \beta \sin \gamma]
\end{array}$$

Spherical Excess = Area

The sum of the interior angles of a spherical triangle ($\alpha+\beta+\gamma$) always exceeds $\pi$. The amount by which it exceeds $\pi$ is called the spherical excess:
$$E= \alpha+\beta+\gamma-\pi.$$
For a unit sphere, $E$ equals the area of the spherical triangle — a remarkable fact, which can be demonstrated as follows.

The region enclosed between two great circles is a spherical polygon with two vertices, called a lune (digon)⁴. For a unit sphere, the area of a lune with vertex angle $\theta$ is $2\theta$.

Label the three great circles defining the spherical triangle as $G_1, G_2, G_3$, and define the areas of the resulting regions as $A_t, A_1, A_2, A_3$ as shown in the figure. We wish to find $A_t$. The lune $L_1$ bounded by $G_2$ and $G_3$ and containing $A_t$ has area $2\alpha = A_t+A_1$. Similarly, $2\beta = A_t+A_2$ and $2\gamma=A_t+A_3$. Adding these gives $3A_t+A_t+A_1+A_2+A_3=2(\alpha+\beta+\gamma)$. Since $A_t+A_1+A_2+A_3$ equals the area of a hemisphere $2\pi$, we obtain:
$$A_t = \alpha+\beta+\gamma-\pi=E.$$

Area of a Spherical Cap

A spherical cap is the bowl-shaped solid formed when a sphere is cut by a plane. As shown in the figures, a sphere of radius $r$ is cut by a plane at distance $r-h$ from the centre, forming a cap (blue) with base radius $a$, height $h$, and half-angle $\Theta$. We place the cap centre at the north pole $N$.

The surface area $A$ of the cap is obtained by integrating over the polar angle $\theta$ from the cap centre⁵:
$$A=\int_0^\Theta {2 \pi r^2 \sin\theta\ d\theta} = 2\pi r^2 (1-\cos\Theta).$$
Since $r(1-\cos\Theta)=h$, this can be written $A=2\pi r h$; and since $r^2=a^2+(r-h)^2$ implies $2rh=a^2+h^2$, it can also be written $A=\pi(a^2+h^2)$.

Intersection Area of Two Caps

We now consider the situation shown in the figures⁶: a blue cap and a green cap with half-angles $\Theta_1$ and $\Theta_2$ respectively, whose centres subtend an angle $\alpha$. We wish to find the area of the yellow intersection region. The blue cap centre coincides with the north pole $N$, and the green cap centre lies on the zero-longitude meridian. From this point on we set $r=1$ (for general $r$, multiply all areas by $r^2$).

Approach 1: As an Integration Problem

A point $P$ on the unit sphere is expressed as
$$P(\theta, \varphi)= (\sin\theta\cos\varphi, \sin\theta\sin\varphi, \cos\theta).$$
The blue cap centre is $C_1=(0,0,1)$ and the green cap centre is $C_2=(\sin\alpha,0,\cos\alpha)$. The condition for $P$ to lie in the blue cap is $\theta \le \Theta_1$; the condition for it to lie in the green cap is
$$P\cdot C_2 = \sin\theta\cos\varphi\sin\alpha + \cos\theta\cos\alpha \ge \cos\Theta_2.$$
Rearranging in terms of $\varphi$:
$$\cos\varphi \ge \frac{\cos\Theta_2-\cos\theta \cos\alpha}{\sin\theta \sin\alpha}.$$
We define the right-hand side (which appears repeatedly) as
$$k(\theta) = \frac{\cos\Theta_2-\cos\theta \cos\alpha}{\sin\theta \sin\alpha}.$$

From the figure, the yellow region exists for $\theta$ in the range $\alpha-\Theta_2$ to $\Theta_1$⁸. For a fixed $\theta$ in this range, $\varphi$ ranges over
$$-\cos^{-1}\{k(\theta)\} \le \varphi \le +\cos^{-1}\{k(\theta)\}.$$
The area of the yellow region is therefore given by the integral (noting that arc length along a latitude circle is proportional to $\sin\theta$):
$$\int_{\alpha-\Theta_2}^{\Theta_1} 2 \cos^{-1}\left[\frac{\cos\Theta_2-\cos\theta \cos\alpha}{\sin\theta \sin\alpha}\right] \sin\theta\ d\theta.$$
Unfortunately, $\cos^{-1}\{k(\theta)\}$ has no elementary antiderivative, so no further closed-form simplification is possible.

Approach 2: As a Spherical Triangle Problem

An alternative approach reduces the problem to spherical triangles. Let $C_1, C_2$ be the cap centres and $P, P’$ their intersection points. The spherical triangles $C_1 C_2 P$ and $C_1 C_2 P’$ are congruent, so we consider only the former. Its side lengths are $\Theta_1, \Theta_2, \alpha$ and its interior angles are $\beta_1, \beta_2, \beta_3$. Let $Q_1, Q_2$ be the points where the side $C_1 C_2$ meets the boundary of the yellow region. Then the yellow area equals twice the sum of the sector areas of $C_1 PQ_1$ and $C_2 PQ_2$ minus the area of the spherical triangle $C_1 C_2 P$.

The sector areas $A_1, A_2$ and the triangle area $A_t$ are:
$$A_1=\beta_1(1-\cos\Theta_1),\quad
A_2= \beta_2(1-\cos\Theta_2),\quad
A_t=\beta_1+\beta_2+\beta_3-\pi.$$

The angles $\beta_1,\beta_2, \beta_3$ follow from the spherical law of cosines:
$$\beta_1 = \cos^{-1}\left[ \frac{\cos\Theta_2 -\cos \Theta_1 \cos\alpha}{\sin\Theta_1 \sin\alpha} \right],\quad
\beta_2 = \cos^{-1} \left[ \frac{\cos\Theta_1 -\cos \Theta_2 \cos\alpha}{\sin\Theta_2 \sin\alpha} \right],\quad
\beta_3 =\cos^{-1} \left[ \frac{\cos\alpha -\cos \Theta_1 \cos\Theta_2}{\sin\Theta_1 \sin\Theta_2}\right].$$
Using $\cos^{-1}(-\omega) = \pi-\cos^{-1}(\omega)$:
$$\pi-\beta_3 =\cos^{-1} \left[ \frac{\cos \Theta_1 \cos\Theta_2 -\cos\alpha}{\sin\Theta_1 \sin\Theta_2}\right].$$
The intersection area $S$ is therefore:
$$\frac{S}{2} = A_1+A_2-A_t = (\pi-\beta_3) -\cos\Theta_1 \beta_1 -\cos\Theta_2 \beta_2,$$
$$\frac{S}{2} = \cos^{-1}\left[ \frac{\cos \Theta_1 \cos\Theta_2 -\cos\alpha}{\sin\Theta_1 \sin\Theta_2}\right]
-\cos\Theta_1 \cos^{-1}\left[ \frac{\cos\Theta_2 -\cos \Theta_1 \cos\alpha}{\sin\Theta_1 \sin\alpha} \right]
-\cos\Theta_2 \cos^{-1} \left[ \frac{\cos\Theta_1 -\cos\Theta_2 \cos\alpha}{\sin\Theta_2 \sin\alpha} \right].$$
Although complex, the integral form has been eliminated and the result is computable with standard tools.

Special Case 1: Equal Half-Angles

Setting $\Theta=\Theta_1=\Theta_2$:
$$\beta_1=\beta_2 = \cos^{-1}\left[ \frac{1 -\cos\alpha}{\tan\Theta \sin\alpha} \right],\quad
\pi-\beta_3 =\cos^{-1}\left[ \frac{\cos^2 \Theta -\cos\alpha}{\sin^2\Theta}\right].$$
The intersection area is:
$$S= 2\cos^{-1} \left[ \frac{\cos^2 \Theta -\cos\alpha}{\sin^2\Theta}\right]
-4\cos\Theta \cos^{-1}\left[ \frac{1 -\cos\alpha}{\tan\Theta \sin\alpha} \right].$$

Special Case 2: Orthogonal Caps

Setting $\alpha=\pi/2$:
$$\beta_1 = \cos^{-1}\left[ \frac{\cos\Theta_2 }{\sin\Theta_1} \right],\quad
\beta_2 = \cos^{-1} \left[ \frac{\cos\Theta_1}{\sin\Theta_2} \right],\quad
\pi-\beta_3 =\cos^{-1}(\cot \Theta_1 \cot \Theta_2).$$
The intersection area is:
$$S=2 \cos^{-1} (\cot \Theta_1 \cot \Theta_2) -2 \cos\Theta_1 \cos^{-1}\left[ \frac{\cos\Theta_2 }{\sin\Theta_1} \right]
-2 \cos\Theta_2 \cos^{-1} \left[ \frac{\cos\Theta_1}{\sin\Theta_2} \right].$$

Special Case 3: Orthogonal Caps with Equal Half-Angles

Setting $\Theta=\Theta_1=\Theta_2$ and $\alpha=\pi/2$:
$$\beta_1 =\beta_2 =\cos^{-1}(\cot \Theta),\quad
\pi-\beta_3 =\cos^{-1} (\cot^2 \Theta).$$
The intersection area is:
$$S=2\cos^{-1}( \cot^2 \Theta) -4\cos\Theta \cos^{-1}(\cot \Theta).$$

Special Case 4: Caps Intersecting at 120°, Equal Half-Angles

$$S=2\cos^{-1} \left[ \frac{1 + 2 \cos^2 \Theta}{2\sin^2\Theta}\right] -4\cos\Theta \cos^{-1}(\sqrt{3}\cot\Theta).$$

Special Case 5: Caps Intersecting at 60°, Equal Half-Angles

$$S= 2\cos^{-1}\left[ \frac{-1 + 2 \cos^2 \Theta}{2\sin^2\Theta}\right] -4\cos\Theta \cos^{-1}(\cot\Theta/\sqrt{3}).$$

Special Case 6: Caps Intersecting at 45°, Equal Half-Angles

$$S=-2\cos^{-1} \left[ \frac{-1 +\sqrt{2} \cos^2 \Theta}{\sqrt{2}\sin^2\Theta}\right] -4\cos\Theta \cos^{-1}( [\sqrt{2}-1] \cot\Theta).$$

Also called a spherical cap or calotte. ↩︎
Mathematica code: P1 = Normalize[{2, 0, 1}]; P2 = Normalize[{0, 0, 1}]; P3 = Normalize[{1, -2.5, 1}]; gcNormal[p_, q_] := Normalize[Cross[p, q]]; n12 = gcNormal[P1, P2]; s12 = Sign[n12 . P3]; n23 = gcNormal[P2, P3]; s23 = Sign[n23 . P1]; n31 = gcNormal[P3, P1]; s31 = Sign[n31 . P2]; eps = 10^-10; insideTri[p_] := And[s12 Dot[n12, p] >= -eps, s23 Dot[n23, p] >= -eps, s31 Dot[n31, p] >= -eps]; sphere = ParametricPlot3D[{Sin[th] Cos[ph], Sin[th] Sin[ph], Cos[th]}, {th, 0, Pi}, {ph, 0, 2 Pi}, Mesh -> None, ColorFunctionScaling -> False, ColorFunction -> Function[{x, y, z, th, ph}, If[insideTri[{x, y, z}], Blue, Red]], PlotPoints -> {800, 800}, MaxRecursion -> 1, PerformanceGoal -> “Quality”, Lighting -> “Neutral”, PlotRange -> All]; ↩︎
Taylor expansion retaining terms up to second order. ↩︎
From Latin luna (moon). ↩︎
The band between $\phi$ and $\phi+d\phi$ has circumference $2\pi r\sin\phi$ and width $r\,d\phi$. ↩︎
Mathematica code: r = 1.2; sphere = ParametricPlot3D[r {Sin[th] Cos[ph], Sin[th] Sin[ph], Cos[th]}, {th, 0, Pi}, {ph, 0, 2 Pi}, …] ↩︎
Geographers measure latitude from the equatorial plane, whereas spherical polar coordinates conventionally measure the polar angle from the north pole (the z-axis direction). The polar angle is sometimes called the colatitude to avoid confusion. ↩︎
We restrict to the case $0<\alpha-\Theta_2 < \Theta_1$. Cases where the green cap contains the centre of the blue cap, or where the blue cap entirely contains the green cap, are not considered. ↩︎

Theorem	Spherical (unit sphere)	Plane
Law of cosines	\(\cos a = \cos b \cos c + \sin b \sin c \cos \alpha\) \(\cos b = \cos c \cos a + \sin c \sin a \cos \beta\) \(\cos c = \cos a \cos b + \sin a \sin b \cos \gamma\)	\(a^2=b^2+c^2-2bc\cos \alpha\) \(b^2=c^2+a^2-2bc\cos \beta\) \(c^2=a^2+b^2-2bc\cos \gamma\)
Law of sines	\(\displaystyle \frac{\sin a}{\sin \alpha} = \frac{\sin b}{\sin \beta}= \frac{\sin c}{\sin \gamma}\)	\(\displaystyle \frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}\)
Area	\(\displaystyle 2 \sin^{-1} \left[ \frac{\sqrt{ \sin s \sin (s-a)\sin (s-b)\sin (s-c)} }{2 \cos(a/2)\cos(b/2)\cos(c/2)} \right]\) where \(2s = a+b+c\) or equivalently \(\alpha+\beta+\gamma-\pi\)	\(\sqrt{s (s-a)(s-b)(s-c)}\) where \(2s = a+b+c\) For a triangle inscribed in the unit circle: \(2 \sin\alpha \sin\beta \sin\gamma\)