Spherical trig - sphere radius from 6 lengths

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Discussion Overview

The discussion revolves around the problem of calculating the radius of a sphere given the six distances between four points located on its surface. The scope includes theoretical considerations and mathematical reasoning related to spherical geometry.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that the solution likely does not require advanced concepts from differential geometry and may involve simpler relationships, possibly including the Haversine Rule.
  • Another participant calculates that there are 13 unknowns (coordinates of the points and the radius) and proposes that the system can be constrained by the distances and rotational symmetries, leading to a solvable system of equations.
  • A different viewpoint emphasizes that only two coordinates (latitude and longitude) are needed per point, leading to a similar conclusion about the number of equations and unknowns.
  • One participant questions whether the six distances are Euclidean or along the geodesics of the sphere, clarifying that they are great circle distances.
  • Another participant raises a scenario where all points lie on the same great circle, suggesting that in this case, the radius could be arbitrary.
  • A further challenge is posed regarding the situation where the points are degenerate, implying that the problem may not have a unique solution under certain conditions.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the problem and its potential solutions, with no consensus reached on a definitive method or outcome. The discussion remains unresolved regarding the implications of specific configurations of the points.

Contextual Notes

Limitations include the assumptions about the nature of the distances (Euclidean vs. geodesic) and the implications of degenerate cases, which may affect the solvability of the problem.

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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.
 
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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."

Mentors, please close this thread. I don't need the aggravation.
 
  • Skeptical
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Thread closed for Moderation...
 
At V50's request, the thread is now closed.
 

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