Angular radius of a sphere
Given a sphere with radius $r$ about a point $c$, what's the apparent angular radius $\alpha$ of that sphere from point $P$? In other words, if $\vec{o} = c - P$, what's the maximum angle another vector $\vec{v}$ may make with $\vec{o}$ to intersect the sphere?
At first I thought of a simple right triangle, with sides:
- $o = |\vec{o}|$
- $m = r$, perpendicular to $o$, from the sphere's center to its boundary
- $l$, closing the triangle.
Then, as long as $m \leq o$, $\alpha = \arctan(m / o) = \arctan(r / |c - P|)$. However, this answer states it's $\arcsin$ instead of $\arctan$. Now I'm lost.
$\endgroup$2 Answers
$\begingroup$Take a point $P$ outside the sphere, at distance $d$ from the centre $O$ of the sphere. The sphere subtends a circle at the eye. Let $A$ and $B$ be diametrically opposite points on that circle. The lines $PA$ and $PB$ are tangent to the sphere. Note that $OA$ and $OB$ are each perpendicular to $PA$.
Triangle $OPA$ is right-angled at $A$, and has hypotenuse $PO$. Thus $\angle OPA$ has sine equal to $r/PO=r/d$, and is clearly less than $\pi/2$. The same is true of $\angle OPB$. It follows that $\angle APB$ is equal to $2\arcsin(r/d)$.
$\endgroup$ $\begingroup$Your right triangle has hypotenuse the tangent from $P$ to the sphere. You can draw a 2D picture to see that the right angle is between the tangent and the radius vector to the tangent point. This means that the hypotenuse is really the segment from $P$ to $c$. Then the sine of the angle is the opposite over hypotenuse, which is $\frac r{|\vec o|}$
$\endgroup$
