Find length of intersection between 2 points and a sphere

I have a sphere and 2 points. The points have (x,y,z) coordinates and the sphere is defined by its centre (0,0,0) and radius R. I am trying to find the length between the 2 points which intersects the sphere. How can I script this out in Matlab?
See below, my objective is Length, L:

2 Commenti

Are you looking for a symbolic or numeric solution?
BenL
BenL il 25 Gen 2017
Modificato: BenL il 25 Gen 2017
im actually looking for a symbolic solution, but need a script for this. I will try to code this from the comments provided. Thanks Jan

Accedi per commentare.

Risposte (3)

Let P1 = [x1,y1,z1], P2 = [x2,y2,z2], and P0 the sphere center.
v = P1-P0-dot(P1-P0,P2-P1)/dot(P2-P1,P2-P1)*(P2-P1);
L = 2*sqrt(R^2-dot(v,v));

3 Commenti

Hi Roger, thank you for the advice, I will try to script this out with my limited knowledge.

Accedi per commentare.

https://en.wikipedia.org/wiki/Line%E2%80%93sphere_intersection
The length L is simply abs(d1-d2) where d1, d2 are the two solutions of the quadratic equation ad^2+bd+c=0.
Best wishes
Torsten.
Since this problem is in three dimensions, you can also make use of the cross product function to compute L as follows. Again, we have: P1 = [x1,y1,z1], P2 = [x2,y2,z2], and P0 is the center of the sphere of radius R.
u = P2-P1;
v = cross(P0-P1,u);
L = 2*sqrt(R^2-dot(v,v)/dot(u,u));

3 Commenti

Hi Roger
Thank you for the advice, can P0 be [0,0,0] ?
I need it to be [0,0,0] (something like center of earth)
Yes, P0 can be any three-element vector, including [0,0,0], in both of the methods I have described. The essential property that is required is that all three vectors P1, P2, and P0 should be such that the infinite straight line through P1 and P2 will intersect a sphere of radius R about P0. Otherwise, in both methods the final code line would be taking the square root of a negative value, which will yield an imaginary number.

Accedi per commentare.

Tag

Richiesto:

il 25 Gen 2017

Commentato:

il 31 Gen 2017

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by