I'm trying to analytically calculate the angle subtended by a circular segment on an external point (external to the circle corresponding to the circular segment). Could someone please help me out here?
AngleSubtendedByCircularSegment
What I've figured out currently is that the angle subtended by any external point on the circular segment will either
- Go through both the cut off points of the circular segment
- Go through one cutoff point and a tangent to the circle
- Go through tangents to the circle.
I'm kinda lost after this step. Any help will be much appreciated.
(In the problem I have at hand, the external points lie on a circle as well. So I need to figure out the angle subtended by a circular segment on points on an external circle. A more detailed image is here AngleSubtendedByCircularSegmentDetailed)
More content added for context -
This problem comes from another problem I'm trying to solve where I have two concentric spheres, with a fixed outer and inner radius. There are photons that start on the outer sphere in an isotropic direction inwards (i.e., the photons only radiate inwards into the outer sphere, not outwards. Think of every point on the inner side of the outer sphere as a point source radiating in all directions inwards. Image attached for reference here).
The inner sphere is filled with water to a certain height (so the chord is indeed parallel to the x-axis). The problem is to find the fraction of photons that reach the water.
The case where the water completely fills the inner sphere is relatively straightforward to solve. Image here. In this case, I find the solid angle subtended by an arbitrary point on the inner sphere (any arbitrary point works since it is symmetric) and divide that by 2$\pi$ to get the fraction (dividing by $2\pi$ cause the photons only go inside). Using some geometry (solid angle formula for a cone given the angle), I get a value of $1 - \frac{\sqrt{r_2^2-r_1^2}}{r_2}$. So this fraction of photons starting on the outer sphere reaches the inner sphere.
Now the same problem is to be solved but for a partially filled inner sphere. It becomes slightly more complicated because the cone subtended by the inner sphere on the outer sphere has a solid angle that varies with theta. It's symmetrical in phi, so the problem can be reduced to a 2D problem. But the problem still persists.
Say, $R$ and $r$ are radii of the outer and inner circles resp. and $OE = FH = h$.
We want to find angle subtended by point $C$ on a horizontal chord $AB$, which is $\angle BCA$.
$\angle BCA (\alpha + \beta) = \angle BCH (\alpha) + \angle ACH (\beta)$
$AE = BE = \sqrt{OA^2 - OE^2} = \sqrt{r^2-h^2}$
If $\angle COD = \theta, $ then $\angle OCF = \theta$
$CF = R\cos\theta,$ so $CH = R\cos\theta - h$
$OF = R\sin\theta = EH, $
So $AH = \sqrt{r^2-h^2} + R\sin\theta, BH = \sqrt{r^2-h^2} - R\sin\theta$
$\alpha = \tan^{-1} \big(\frac{BH}{CH}\big) = \tan^{-1}\bigg(\frac{\sqrt{r^2-h^2} - R\sin\theta}{R\cos\theta - h}\bigg)$
$\beta = \tan^{-1} \big(\frac{AH}{CH}\big) = \tan^{-1}\bigg(\frac{\sqrt{r^2-h^2} + R\sin\theta}{R\cos\theta - h}\bigg)$
$\angle BCA = \alpha + \beta$