trig help?

Question

I need to know how far apart 40 points are when spread as far apart as possible on a sphere.

think bond angles in chem, but with 40 atoms attached to 1 atom. instead of 4

Answer 1

I'm not sure, but I have an idea. The surface area of a sphere is 4 pi r^2. You have 40 points spread out over that area, so each point is "entitled" to 1/40th of the total area, or (pi r^2)/10 square units.

Now think about how these points, or actually, small areas around points, are going to be packed. They won't be circles ... if you put several pennies together on a table, there's be small gaps wherever three pennies join together.

Hexagons are more efficient. Draw a regular hexagon, and you'll see that you can surround it with six more hexagons, completely filling the space without any gaps.

Now consider just one hexagon -- the space surrounding just one point, which is located in the center. This hexagon, which has an area of (pi r^2)/10, is composed of six equilateral triangles.

Suppose the side of the hexagon is "a" (which is also the side of one of the six small equilateral triangles). Then the height of the small triangle is (a/2) sqrt(3), and the area of the small triangle is 1/2 bh = (a/2)^2 sqrt(3). Since there are six small triangles in the hexagon, the area of the hexagon is 6(a/2)^2 sqrt(3) = 3/2 a^2 sqrt(3).

But that must equal (pi r^2)/10, so

3/2 a^2 sqrt(3) = (pi r^2)/10

Solving for a, we get a = sqrt[pi sqrt(3)/45] r = 0.3477 r

[One cautionary note here: I'm doing plane geometry, and all these hexagons are plane figures. This assumes the sphere is circumscribed around all the vertices of the 40 hexagons ... something like a soccer ball.

[Total area, though, will be equal to the sphere because we divided the sphere's total surface area by 40. Although I don't want to do the work, I think it's possible to include the curvature into this analysis.]

Now we want to know how far apart two adjacent points are located. Each point is at the center of a hexagon, and the distance between two hexagon centers is twice the height of any one of the small triangles.

This height is (a/2) sqrt(3), so the distance between points is a sqrt 3 = sqrt[pi sqrt(3)/15] r = 0.6023 r (and all the angles are 60 degrees). (And that 0.6023 r is probably the answer to your question.)

Now let's get the curvature in there and figure out the central angles. Assume the hexagons are all stretched so that they form part of the sphere. Draw a great circle passing through two adjacent points. We want the central angle of the sector where the arc length is 0.6023 r. But because arc length is just r theta (in radians), the central angle is 0.6023 radians, or 34.5 degrees apart.

Answer 2

There is no simple closed form solution to equipositioning 40 points on a sphere. Equal spacing exisits only for 2, 4, 6, 8, 12, and 20 points. Except for the trivial case of 2, these numbers correspond to the number of sides on the platonic solids.

A reasonable approximation would be to take the average spacing. Each point is assigned 1/40 of the sphere area giving an average area per point of (4*pi*R^2)/40. To figure the radius of a circle with that area:

pi*r^2 = (4*pi*R^2)/40

r = R/sqrt(10)

The average distance (along the sphere) between points is twice the radius:

d = 2*R/sqrt(10)

In terms of angular distance, just divide by R, the angular separation is:

a = 2/sqrt(10)

And the chord (straight line) distance:

s = 2*R*sin(1/sqrt(10)))

Remember, these are not exact values and, in general no exact value can be determined. These are only reasonable approximations.

Answer 3

The size of the sphere is irrelevant and all angles will be the same, in response to the above.

Answer 4

so are you asking how far apart 2 points are from each other?
are you asking the average distance between points?
are you asking the most distance possible between points?
How big is our sphere?