solid angle of rectangular pyramid

We calculate the apical solid angle of a rectangular pyramid, as an example of using the http://planetmath.org/node/7266formula of van Oosterom and Strackee for determining the solid angle $\mathrm{\Omega }$ subtended at the origin by a triangle:

$\tan {\displaystyle \frac{\mathrm{\Omega }}{2}}={\displaystyle \frac{{\overrightarrow{r}}_1\times {\overrightarrow{r}}_2\cdot {\overrightarrow{r}}_3}{({\overrightarrow{r}}_1\cdot {\overrightarrow{r}}_2)r_3+({\overrightarrow{r}}_2\cdot {\overrightarrow{r}}_3)r_1+({\overrightarrow{r}}_3\cdot {\overrightarrow{r}}_1)r_2+r_1{r}_2{r}_3}}$ (1)

Here, ${\overrightarrow{r}}_1$, ${\overrightarrow{r}}_2$, ${\overrightarrow{r}}_3$ are the position vectors of the vertices of the triangle and $r_1,r_2,r_3$ their .

Let the apex of the pyramid be in the origin and the vertices of the base rectangle be

$$(\pm a,\pm b,h)$$

where $a$, $b$ and $h$ are positive numbers.  We take the half-triangle of the base determined by the three vertices

$$(a,b,h),(-a,b,h),(a,-b,h),$$

with the position vectors ${\overrightarrow{r}}_1$, ${\overrightarrow{r}}_2$, ${\overrightarrow{r}}_3$, respectively.  Then we have in the numerator of (1) the scalar triple product

The vectors have the common length $\sqrt{a^2+b^2+h^2}$, and the denominator of (1) then attains the value $4h^2\sqrt{a^2+b^2+h^2}$.  Thus the formula (1) gives

$$\tan \frac{\mathrm{\Omega }}{2}=\frac{ab}{h\sqrt{a^2+b^2+h^2}}$$

which result may be reformulated by using the goniometric formula

$$\sin x=\frac{\tan x}{\sqrt{1+{\tan }^{2}x}}$$

as

$\sin {\displaystyle \frac{\mathrm{\Omega }}{2}}={\displaystyle \frac{ab}{\sqrt{(a^2+h^2)(b^2+h^2)}}}.$ (2)

Thus the whole apical solid angle of the http://planetmath.org/node/7357right rectangular pyramid is

$\mathrm{\Omega }=\mathrm{\hspace{0.33em}4}\arcsin {\displaystyle \frac{ab}{\sqrt{(a^2+h^2)(b^2+h^2)}}}.$ (3)

A variant of (3) is found in [3].

In the special case of a regular pyramid we have simply

$\mathrm{\Omega }=\mathrm{\hspace{0.33em}4}\arcsin {\displaystyle \frac{a^2}{a^2+h^2}}$ (4)

where $2a$ is the side (http://planetmath.org/Polygon) of the base square.

Note that in (2), the quotients $\frac{a}{\sqrt{a^2+h^2}}$ and $\frac{b}{\sqrt{b^2+h^2}}$ are sines of certain angles in the pyramid.

Title	solid angle of rectangular pyramid
Canonical name	SolidAngleOfRectangularPyramid
Date of creation	2013-03-22 19:16:02
Last modified on	2013-03-22 19:16:02
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	16
Author	pahio (2872)
Entry type	Example
Classification	msc 15A72
Classification	msc 51M25
Related topic	CyclometricFunctions