Distance Between 2 Randomly-chosen Points on a Sphere

What is the most likely straight-line distance between two randomly selected points on a sphere?

Straight line across the centre of the sphere, or are we restricted to the

) )> What is the most likely straight-line distance between two randomly )> selected points on a sphere? ) ) Straight line across the centre of the sphere, or are we restricted to the ) surface.

Across the centre, I would imagine, otherwise the line wouldn't be straight. I could be wrong of course, it's entirely possible that the OP meant 'straight across the surface' when he wrote 'straight', given that he's also asking for something 'most likely' from a continuum, with which he probably means the median, but may also mean the average.

SaSW, Willem

Now assumed...

So, in my head, it looks like being:

dist = radius * sqrt(2)

radius * (4/3)

I don't know, but somewhere between the radius and diam of the sphere?

(remember straight line distance is always through the sphere except when the points are identical).

As always any opinions I may have written above are mine and mine alone.

Looks to me more like the expected value of straight-line distances, not the 'most likely' one...

Agreed, assuming 'most likely' means 'expected value.'

Fix one point at the north pole. Consider the surface area of horizontal strips to see that the most likely lattitude of the second point is on the diameter. Hence, the most likely distance is sqrt(2) times the radius.

We are talking about the mode of a distribution here.

The mean is given by Int 0 to 2 of s^2/2 = 4/3, but this is not the most likely value.

Use (r,theta,z) coordinates. Without loss of generality, let the center of the sphere be (0,0,0) and the first point be (1,0,0).

For z = 0, the average distance is I/pi, where

pi I = integral ( sqrt[ (1-cos(t))^2 + sin(t)^2 ] dt ) t=0

(values of theta from pi to 2*pi are omitted because they're a mirror image of the values from 0 to pi, so they don't change the average)

= integral ( sqrt[ 1 - 2*cos(t) + cos(t)^2 + sin(t)^2 ] dt )

= integral ( sqrt[ 2 - 2*cos(t) ] dt )

Gah, how do you solve this?

Anyway, one possible approach to the 3-D question is to first solve the 2-D question, then integrate that over z = -1 to 1 (remembering to weight each z value in direct proportion to the circumference of the sphere's cross-section at that z value).

Another possible approach is to take (r,theta), rotate it around the x axis, and note that if a given second-point has distse (r,theta,z) coordinates. Without loss of generality, let the center of the sphere be (0,0,0) and the first point be (1,0,0).

For z = 0, the average distance is I/pi, where

pi I = integral ( sqrt[ (1-cos(t))^2 + sin(t)^2 ] dt ) t=0

(values of theta from pi to 2*pi are omitted because they're a mirror image of the values from 0 to pi, so they don't change the average)

= integral ( sqrtse (r,theta,z) coordinates. Without loss of generality, let the center of the sphere be (0,0,0) and the first point be (1,0,0).

For z = 0, the average distance is I/pi, where

pi I = integral ( sqrt[ (1-cos(t))^2 + sin(t)^2 ] dt ) t=0

(values of theta from pi to 2*pi are omitted because they're a mirror image of the values from 0 to pi, so they don't change the average)

= integral ( sqrt[ 1 - 2*cos(t) + cos(t)^2 + sin(t)^2 ] dt )

= integral ( sqrt[ 2 - 2*cos(t) ] dt )

Gah, how do you solve this?

Anyway, one possible approach to the 3-D question is to first solve the 2-D question, then integrate that over z = -1 to 1 (remembering to weight each z value in direct proportion to the circumference of the sphere's cross-section at that z value).

Another possible approach is to take (r,theta), rotate it around the x axis, and note that if a given second-point has distse (r,theta,z) coordinates. Without loss of generality, let the center of the sphere be (0,0,0) and the first point be (1,0,0).

For z = 0, the average distance is I/pi, where

pi I = integral ( sqrt[ (1-cos(t))^2 + sin(t)^2 ] dt ) t=0

(values of theta from pi to 2*pi are omitted because they're a mirror image of the values from 0 to pi, so they don't change the average)

= integral ( sqrt[ 1 - 2*cos(t) + cos(t)^2 + sin(t)^2 ] dt )

= integral ( sqrtse (r,theta,z) coordinates. Without loss of generality, let the center of the sphere be (0,0,0) and the first point be (1,0,0).

For z = 0, the average distance is I/pi, where

pi I = integral ( sqrt[ (1-cos(t))^2 + sin(t)^2 ] dt ) t=0

(values of theta from pi to 2*pi are omitted because they're a mirror image of the values from 0 to pi, so they don't change the average)

= integral ( sqrt[ 1 - 2*cos(t) + cos(t)^2 + sin(t)^2 ] dt )

= integral ( sqrt[ 2 - 2*cos(t) ] dt )

Gah, how do you solve this?

Anyway, one possible approach to the 3-D question is to first solve the 2-D question, then integrate that over z = -1 to 1 (remembering to weight each z value in direct proportion to the circumference of the sphere's cross-section at that z value).

Another possible approach is to take (r,theta), rotate it around the x axis, and note that if a given second-point has distse (r,theta,z) coordinates. Without loss of generality, let the center of the sphere be (0,0,0) and the first point be (1,0,0).

For z = 0, the average distance is I/pi, where

pi I = integral ( sqrt[ (1-cos(t))^2 + sin(t)^2 ] dt ) t=0

(values of theta from pi to 2*pi are omitted because they're a mirror image of the values from 0 to pi, so they don't change the average)

= integral ( sqrtse (r,theta,z) coordinates. Without loss of generality, let the center of the sphere be (0,0,0) and the first point be (1,0,0).

For z = 0, the average distance is I/pi, where

pi I = integral ( sqrt[ (1-cos(t))^2 + sin(t)^2 ] dt ) t=0

(values of theta from pi to 2*pi are omitted because they're a mirror image of the values from 0 to pi, so they don't change the average)

= integral ( sqrt[ 1 - 2*cos(t) + cos(t)^2 + sin(t)^2 ] dt )

= integral ( sqrt[ 2 - 2*cos(t) ] dt )

Gah, how do you solve this?

Anyway, one possible approach to the 3-D question is to first solve the 2-D question, then integrate that over z = -1 to 1 (remembering to weight each z value in direct proportion to the circumference of the sphere's cross-section at that z value).

Another possible approach is to take (r,theta), rotate it around the x axis, and note that if a given second-point has distse (r,theta,z) coordinates. Without loss of generality, let the center of the sphere be (0,0,0) and the first point be (1,0,0).

For z = 0, the average distance is I/pi, where

pi I = integral ( sqrt[ (1-cos(t))^2 + sin(t)^2 ] dt ) t=0

(values of theta from pi to 2*pi are omitted because they're a mirror image of the values from 0 to pi, so they don't change the average)

= integral ( sqrt[ 1 - 2*cos(t) + cos(t)^2 + sin(t)^2 ] dt )

= integral ( sqrt[ 2 - 2*cos(t) ] dt )

Gah, how do you solve this?

Anyway, one possible approach to the 3-D question is to first solve the 2-D question, then integrate that over z = -1 to 1 (remembering to weight each z value in direct proportion to the circumference of the sphere's cross-section at that z value).

Another possible approach is to take (r,theta), rotate it around the x axis, and note that if a given second-point has distance D, then all the points through which that second-point is rotated also have distance D. So integrate (D * circumference of the appropriate circle perpendicular to the x axis) over theta = 0 to pi.