# Spherical trig - sphere radius from 6 lengths

• I
Staff Emeritus
2023 Award
Four points lie on the surface of a sphere. Given the six distances between the points, calculate the radius of the sphere.

This is (allegedly) an advanced high school level problem. However, it is a remembered problem, so it is possibly misremembered (i.e. there might have been some “bice numbers” so that the general solution is unnecessary). But the solution probably does not involve the full machinery of differential and tensor geometry, and it is possible, but unlikely, that it requires knowllege of the Haversine Rule.

First, is there a solution? If the corners are ABCD, the trianghles ABC, ABD, ACD, and BCD are determined. That give syou the Spgerical Excess, and that in turn gives the radius. Alternatively, one has size numbers. The coordinates of the four points would determine R – one point can be placed at the origin, one along one axis, so we have 5 degres of freesom on where the points go, plus the radius, which makes 6, and we have 6 constraints – 6 equations in 6 unknowns.

It seems to be that the key to this must be a relationship between the two diagonals. Until the second diagonal is drawn, everything could be in a plane. But I have not been able to go from there to a high-school level solution.

By brutal force:

Coordinates of the 4 points and the radius are 13 unknowns. Each point is on the sphere -- 4 equations. Six distances -- 6 equations. The coordinates can be rotated so that one points is on say x-axis -- 2 equations. The coordinates still can be rotated around the x-axis so that another point is in the say x-y-plane -- 1 equation.

So, we get 13 equations with 13 unknowns.

I am sure that this is not what they had in mind.

By the way, there are only two coordinates per point - call them latitude and longitude. Four points gives 8 numbers, plus the radius. Three of those points determine your meridians, so you have 6 remaining, and that's the same situation I described in the OP.

Are the six given distances Euclidean in 3D or along the geodesics on the sphere?

Hill said:
along the geodesics on the sphere?
Yes. Great circle distances.

So, what happens if the points all lie on the same great circle? In this case they can be sorted such that their distances all just add, like they are on a straight line. In this case R is anything you wish.

Sure. And what happens if all the points are degenerate?

This is PF at its worst. "I'm going to quibble instead of helping you."

weirdoguy

At V50's request, the thread is now closed.

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