# 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 $\Omega$ subtended at the origin by a triangle:

 $\displaystyle\tan\frac{\Omega}{2}\;=\;\frac{\vec{r}_{1}\!\times\!\vec{r}_{2}\!% \cdot\!\vec{r}_{3}}{(\vec{r}_{1}\!\cdot\!\vec{r}_{2})r_{3}+(\vec{r}_{2}\!\cdot% \!\vec{r}_{3})r_{1}+(\vec{r}_{3}\!\cdot\!\vec{r}_{1})r_{2}+r_{1}r_{2}r_{3}}$ (1)

Here, $\vec{r}_{1}$, $\vec{r}_{2}$, $\vec{r}_{3}$ 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 $\vec{r}_{1}$, $\vec{r}_{2}$, $\vec{r}_{3}$, respectively.  Then we have in the numerator of (1) the scalar triple product

 $\vec{r}_{1}\!\times\!\vec{r}_{2}\!\cdot\!\vec{r}_{3}\;=\;\left|\begin{matrix}a% &b&h\\ -a&b&h\\ a&-b&h\end{matrix}\right|\;=\;a\left|\begin{matrix}b&h\\ -b&h\end{matrix}\right|+b\left|\begin{matrix}h&-a\\ h&a\end{matrix}\right|+h\left|\begin{matrix}-a&b\\ a&-b\end{matrix}\right|\;=\;4abh.$

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

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

which result may be reformulated by using the goniometric formula

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

as

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

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

 $\displaystyle\Omega\;=\;4\arcsin\frac{ab}{\sqrt{(a^{2}\!+\!h^{2})(b^{2}\!+\!h^% {2})}}.$ (3)

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

In the special case of a regular pyramid we have simply

 $\displaystyle\Omega\;=\;4\arcsin\frac{a^{2}}{a^{2}\!+\!h^{2}}$ (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.

## References

• 1 A. van Oosterom & J. Strackee:  A solid angle of a plane triangle.  – IEEE Trans. Biomed. Eng. 30:2 (1983); 125–126.
• 2 M. S. Gossman & A. J. Pahikkala & M. B. Rising & P. H. McGinley:  Providing solid angle formalism for skyshine calculations.  – Journal of Applied Clinical Medical Physics 11:4 (2010); 278–282.
• 3 M. S. Gossman & A. J. Pahikkala & M. B. Rising & P. H. McGinley:  Letter to the editor.  – Journal of Applied Clinical Medical Physics 12:1 (2011); 242–243.
• 4 M. S. Gossman & M. B. Rising & P. H. McGinley & A. J. Pahikkala:  Radiation skyshine from a 6 MeV medical accelerator.  – Journal of Applied Clinical Medical Physics 11:3 (2010); 259–264.
Title solid angle of rectangular pyramid SolidAngleOfRectangularPyramid 2013-03-22 19:16:02 2013-03-22 19:16:02 pahio (2872) pahio (2872) 16 pahio (2872) Example msc 15A72 msc 51M25 CyclometricFunctions