# Right n-sided pyramid solid angle

Have:
N-faces right pyramid
value of apex solid angle
value of N

Need to find:
value of flat angle (C on image)

• Anything else is known? The angle will depend on the height of the pyramid. Oct 10, 2017 at 13:32
• @Theorem I thought it meant solid angle. Oct 10, 2017 at 13:32
• @Vasya We do know something else: "value of apex solid angle" Oct 10, 2017 at 13:33
• @Vasya solid angle, number of faces and the fact that pyramid is right is enough to find angle C Oct 10, 2017 at 13:54
• Do you need an exact result, or an approximate solution? Oct 10, 2017 at 14:08

Hint:

Assuming given apex solid angle is defined in the same way as it is on this Wikipedia entry. Assuming the polygon at the base is regular, so that $\angle AOB = 2 \pi / N$. Assume the radius of the circumcircle of base polygon is 1 (w.l.o.g).

Possible steps to a solution are:

• Find $AB$ in the base polygon.
• Find $OF$ in the base polygon.
• Find the height $OS$ of the pyramid using the formula for the solid angle of a right $N$-gonal pyramid with a regular base
• Find $SF$ using $OF$ and $OS$
• Find $c / 2$ using $SF$ and $BF = AB / 2$

The formula referred to on the Wikipedia page for the solid angle of a $n$-gonal right-pyramid with height $h$ and with $r$ as the radius of the circle circumscribing the base is: $$\Omega = 2\pi - 2n \arctan\left(\frac {\tan \left({\pi\over n}\right)}{\sqrt{1 + {r^2 \over h^2}}} \right)$$

Below is an example of the use of this formula for a particular pyramid. Let the base of the pyramid be $A=(1,0,0)$, $B=(0,1,0)$ and $C=(0,0,1)$ and let the apex be $O=(0,0,0)$. This pyramid occupies one eighth of a full sphere. The solid angle of a full sphere is $4 \pi$ so we expect the solid angle for the pyramid to be $\frac{\pi}{2}$.

The point at the centre of the base is $P=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$. The height $h$ of the pyramid is given by $h = |OP| = \frac{1}{\sqrt{3}}$. The radius $r$ of the circumcircle of the base can be obtained from $r = |AP| =\frac{\sqrt{2}}{\sqrt{3}}$. We can then calculate $\sqrt{1 + \frac{r^2}{h^2}} = \sqrt{3}$.

For this pyramid, $n=3$ so $\tan\left(\frac{\pi}{n}\right)= \sqrt{3}$. Putting this together with the above, we can calculate $\Omega$ $$\Omega = 2\pi - 2n \arctan\left( \frac{\sqrt{3}}{\sqrt{3}} \right) = 2\pi - 2 \times 3 \times \frac{\pi}{4} = \frac{\pi}{2}$$ confirming the value we expect for the solid angle.

• ∠OSF is not solid angle. en.wikipedia.org/wiki/Solid_angle Oct 10, 2017 at 13:48
• Thanks Petr, I've adjusted the suggested approach to use that. Oct 10, 2017 at 14:11
• yes, it looks like a solution. It remains only to compile this into one expression. Oct 10, 2017 at 14:20

Complete formula:
$$t = \frac{tan(\pi/n)^2}{tan(\frac{\Omega-2\pi}{2n})^2}-1$$ $$C = ∠ASB = 2arcsin(\frac{\sqrt t * sin(\pi/n)}{\sqrt {t+1}})$$