# solid angle subtended by an infinitely long circular cylinder?

An infinitely long cylindrical pole, with a diameter $D$, is standing on a plane surface. How to find out the solid angle subtended by the pole at any arbitrary point say P lying on the same plane surface at a distance $x$ from the center O of the base of the pole? (as shown in the diagram above) Is there any analytic formula to calculate the solid angle in terms of $D$ & $x$?

Note: $D$ & $x$ are finite values.

• If you look at the rays from $P$ tangent to the sides of the cylinder, you will notice they lie on two planes. This mean if you project the cylinder onto a unit sphere centered at $P$, the result will be a geodesic triangle and the angles at two of the corners are $90^\circ$ You just need to figure out what the third angle is... – achille hui Apr 22 '15 at 6:32

Let $y$ be measured upwards from zero in the plane shown. If $x$ is large, we can consider the blockage by a rectangle $D \times dy$ at a vertical position $y$. The distance is $\sqrt {x^2+y^2}$, so the rectangle blocks a solid angle $d\Omega=\frac {D\ dy}{x^2+y^2}$ The total solid angle is then $\int_0^\infty \frac {D\ dy}{x^2+y^2}$
If $x$ is not so large you need to find the length of the arc of the circle of radius $\sqrt {x^2+y^2}$ that intersects the cylinder. That will replace $D$ in the integral. You are allowing for the fact that the two tangent lines do not touch the cylinder at diametrically opposite points.