2
$\begingroup$

Cone intersects Sphere

An upright (green) cone with opening angle $2a < \pi/10$ has its vertex at point O with cartesian xyz coordinates $(0,0,0)$. The cone axis (dotted line) lies in the plane $y=0$ and is parallel to the z-axis.

A (blue) sphere of radius $r$ is centred at point $P(P_x,0,P_z)$ which may lie inside or outside the cone.

The line segment connecting points $O,P$ has length $p < r$. Thus point $O$ always lies inside the sphere.

The cone surface and the sphere surface intersect in a non-planar curved line loop which includes points $L1,L2$.

From here I have found equations for the loop (where $c$ is the "cone opening parameter" defined by $c^2=(x^2+y^2)/z^2$) :-

$$ (x-P_x)^2+(y-P_y)^2+\frac{x^2+y^2}{c^2}-\frac{2.P_z}{c}\sqrt{x^2+y^2}+P_z^2=r^2 \qquad [1] $$

$$ x^2\left(1+\frac{1}{c^2}\right)-2P_x.x+y^2\left(1+\frac{1}{c^2} \right) -2P_y.y + (P_x^2+P_y^2+P_z^2-r^2)-\frac{2P_z}{c}\sqrt{x^2+y^2}=0. \qquad [2] $$

In this case the value of $P_y$ is zero which simplifies the above equations a little bit.

QUESTION

What (in steradians) is the solid angle $w$ subtended by the "loop" at point $P$ in terms of $a,r,P_x,P_z$?

$\endgroup$
1
$\begingroup$

Let's move everything so that the center of the sphere is in the origin. Then the sphere is simply

$$x^2+y^2+z^2=r^2$$

and the cone becomes

$$\left(x+P_x\right)^2+y^2=c^2\left(z+P_z\right)^2$$

and since you don't want a double cone, you also want

$$z+P_z>0\quad.$$

The $y=0$ plane intersects the cone in two lines, namely

$$x=\pm c(z+P_z)-P_x$$

and these intersect the sphere at

$$x_{1,2}={\frac{\pm c\left({P_z} + \sqrt{{\left(c^{2} + 1\right)} r^{2} - c^2{P_z}^{2} \pm 2 c {P_x} {P_z} - {P_x}^{2}}\right) - {P_x}}{c^{2} + 1}}\quad.$$

For a given $x$ coordinate in that plane, the points on the loop are both at

$$ z(x) = -\frac{{P_z} c^{2} - \sqrt{{\left({P_x}^{2} - {P_z}^{2}\right)} c^{2} + {\left(c^{2} + 1\right)} r^{2} + {P_x}^{2} + 2 \, {\left({P_x} c^{2} + {P_x}\right)} x}}{c^{2} + 1} $$

so the portion of the sphere in that $x$ plane and between the points on the loop will form a circular arc with a length of

$$l(x) = 2\sqrt{r^2-x^2}\arcsin\frac{z(x)}{\sqrt{r^2-x^2}}\quad.$$

To integrate these arcs into an area, you have to multiply them by the arc length element in the $y=0$ plane. This is defined by

$$\mathrm ds^2=\mathrm dx^2+\mathrm dz^2 = \mathrm dx^2\left(1+\left(\frac{\mathrm dz}{\mathrm dx}\right)^2\right)$$

which means we need the derivative

$$\frac{\mathrm dz}{\mathrm dx}=\frac{{P_x}}{\sqrt{{\left({P_x}^{2} - {P_z}^{2}\right)} c^{2} + {\left(c^{2} + 1\right)} r^{2} + {P_x}^{2} + 2 \, {\left({P_x} c^{2} + {P_x}\right)} x}}$$

to obtain

$$\mathrm ds = {\sqrt{\frac{{P_x}^{2}}{{\left({P_x}^{2} - {P_z}^{2}\right)} c^{2} + {\left(c^{2} + 1\right)} r^{2} + {P_x}^{2} + 2 \, {\left({P_x} c^{2} + {P_x}\right)} x} + 1}}\;\mathrm dx\quad.$$

So the portion of the sphere enclosed by the loop has area

$$\int_{x_1}^{x_2}l(x)\;{\sqrt{\frac{{P_x}^{2}}{{\left({P_x}^{2} - {P_z}^{2}\right)} c^{2} + {\left(c^{2} + 1\right)} r^{2} + {P_x}^{2} + 2 \, {\left({P_x} c^{2} + {P_x}\right)} x} + 1}}\;\mathrm dx\quad.$$

At least numerically this should be reasonably simple to integrate. I haven't tried yet to find a closed form for this. It might be simpler if you have explicit numbers instead of all these variables.

In the end, you can divide that area by $r^2$ to obtain the corresponding solid angle.

$\endgroup$
  • $\begingroup$ Many thanks. A closed form solution would be awesome but your formula for numerical evaluation is very helpful :-). $\endgroup$ – steveOw Dec 20 '14 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.