# Points on surface of spherical cap

I need to be able to choose random points on a spherical cap surface for which the radius and centre of the actual sphere are known.

I'd like to do so by simply restricting the range of possible spherical coordinates (rho,theta, phi as shown on http://mathinsight.org/spherical_coordinates) of points on the cap.

How can I calculate the ranges over which theta and phi can vary for points on the cap, if the cap is specified by the centre of its base and the angle from its normal vector to its base (as depicted on https://en.wikipedia.org/wiki/Spherical_cap)?

• This is equivalent to this previous question: math.stackexchange.com/q/56784/856
– user856
Dec 29, 2016 at 19:02
• Could you expand on this. I don't need uniform distribution. Just cutoffs for the two angles. Dec 29, 2016 at 19:09
• On the face of it, the problem seems trivial: $\phi$ varies from zero to the angle shown in the figure on the Wiki page you linked, $\theta$ varies from zero to $2\pi.$ That assumes you set up your spherical coordinates so the axis passes through the center of the cap. Is there a really good reason why you absolutely cannot do that? Dec 29, 2016 at 19:35
• The reason the axis doesn't pass through the centre of the cap is that the origin is where the measurements are taken from and so needs to be able to be anywhere. I am trying to build test cases simulating this. Dec 29, 2016 at 19:48
• @DavidK Good idea. I could then simply 'move' the points via translation and rotation to the actual cap location. What would the angle of rotation be, and how to rotate a point? Dec 29, 2016 at 21:35

I was considering this question 4 years later, and include the answer for others. We want to generate random $$(\tilde \phi, \tilde \theta)$$ on a spherical cap with angle $$\theta$$. If $$r_1, r_2\in[0,1]$$ are uniform random variables, then

$$\tilde\phi_r = 2 \pi r_1$$ $$\tilde\theta_r = \arccos \left[ \left(1-\cos\theta\right)r_2 + \cos\theta\right]$$ HINT:

To find how $$\rho, \phi$$ get modified, consider trig of triangle including sphere center, south pole and required point.

Angle subtended at center is double angle at south pole. $$\pi/2- ph1 = 2 (\pi/2- ph)$$ $$ph= \pi/4+ ph1/2$$

$$\rho^2 =a^2+a^2 -2a\cdot a \cos ( \pi/2 +ph1)$$ $$\rho= 2 a \sin ( \pi/4 + ph1/2)$$

Note $$\phi, \theta$$ limits I chose in reckoning spherical cap coordinates from south pole:

a = 1; ParametricPlot3D[
2 a Sin[Pi/4 + ph1/2]*{Cos[(Pi/4 + ph1/2)] Cos[t],
Cos[(Pi/4 + ph1/2)] Sin[t], Sin[(Pi/4 + ph1/2)]}, {t, 0,
3 Pi/2}, {ph1, -Pi/2, Pi/4}] Also browse related topic Stereographic Projection.