How can I create equally spaced points on the surface of a sphere?

[okie]

I want to create points on the surface of a sphere, all equally spaced. Having lots of difficulty figuring it out.

[f700es]

Don't use a smooth sphere for one. Use a faceted sphere to give you vertex points.

Here's one in Blender...

[fuccaro]

Those vertices are not really equally spaced. I don't think there is a general solution, however in particular cases you can find those points.

Let's consider a regular polyhedron. All the corners are equally spaced on the circumscribed sphere. Now if we place an additional vertex, probable there is no way to rearrange the vertices to met the equally spacing criteria.

[steven-g]

That is an age old problem, ever heard of Trigonometry? It is easy for a low number of points but when you start getting into bigger numbers it becomes more difficult, I beleive that after a certain amount it isn't actually possible, all you can do is go for the greatest distance between points but it won't be equal for all points.

[f700es]

I didn't think you could space them on a sphere evenly to start with. Been wrong before.

[steven-g]

You can only space a certain number of points at exact distances apart eg. the 20 vertices of a dodecahedron one of the 5 platonic solids other than that it is a question of placing points as far away from the next one as possible but distances will vary. So you can do this for 4, 6, 8, 12 or 20 points, and I suppose 2 or 3 points are arguably possible.

[SLW210]

The debate goes on about this problem.

Maybe a platonic solid inscribed in the sphere, 20 vertices and 12 faces on a Dodecahedron.

[SLW210]

I missed your post whilst doing other things and thinking.

I was looking into the possibility of Archimedean solids, not sure if any of those work.

[BIGAL]

I think the answer is in like the soccer ball, a shape repeated there is lisp code around to create such with 5 6 7 8 sided polygons, its a mathematical solution, has a specific name, so say 5 sides you theoretically you could work out the centroid of say the pentagon as xyz then using radius convert it to a true xyz on outside of sphere. 

 

geodesic domes/spheres ?

 

 

 

Edited March 3, 2020 by BIGAL

[lrm]

@BIGAL The first of the three shapes you show at the bottom of your post is an icosahedron.  All its edges are the same length. The other two shapes are geodesic spheres made famous by Buckmister Fuller and the basis of geodesic domes.  Not all the edge lengths are the same. The number of different edge lengths is a function of the frequency.  Check out this link.

[fuccaro]

As I mentioned before, in some particular cases there are solutions, but there is no general solution for any number of vertices.

 Nice link,  lrm!

[paulmcz]

12 vertices. See attached

 

EqLengthEdgesSphere.dwg

Edited March 12, 2020 by paulmcz

[BIGAL]

Nice LRM there was definitely a lisp answer to create the domes.

 

Fuccaro you were mentioned back in 2004 here at cadtutor.

 

This is one I found.

 

 

geodome.lsp

[SLW210]

Looks like the OP disappeared.

[okie]

13 hours ago, SLW210 said:

Looks like the OP disappeared.

Nope, still here.

[steven-g]

And does any of this help? It might help if you told us how many points or was is it just a general question?

[okie]

On 3/5/2020 at 8:30 PM, steven-g said:

And does any of this help? It might help if you told us how many points or was is it just a general question?

It was just an academic question. I was just curious why geodesic domes don't use struts of all the same length and hubs with all the same angles. Maybe someday mathematicians will figure out how to equally distribute more than 20 points on a sphere?

[steven-g]

No they won't, it isn't possible. And that's why geodesic domes are based on the Icosahedron (12 points) it gives you the least number of variations of strut length and hub angles when you start scaling up from a 1v through to a 6v dome

[okie]

8 hours ago, steven-g said:

No they won't, it isn't possible. And that's why geodesic domes are based on the Icosahedron (12 points) it gives you the least number of variations of strut length and hub angles when you start scaling up from a 1v through to a 6v dome

That fascinates me for some reason. I can't wrap my head around why 20 points is the limit.

[steven-g]

Why is a BIG question, I don't know. but imagine a globe (ignoring the fact that the globe isn't an exact sphere). place 10 points equaly spaced around the equator (that's doable) now draw the next ring around the globe above the equator how far above? the circumference is obviously smaller so do you make the circumference to be 9x the distance of the spacing of the points on the equator, then where do you place the 9 points, it would be easy to start with your first point equal distance away from any of the 2 points on the equator, but the second point? that could never be equal distance to the next two points on the equator.

So try the other way round, use a compass (drawing compass not a map compass) set it to the distance between 2 of the points on the equator and draw arcs from each point to somewhere above the equator and where the arcs cross place your new point this gives you 10 new points above the equator in a circle around the globe, but all these 10 points would be closer to each other because the circle they are drawn on is smaller than the circle of the equator (does that make sense).

It turns out the only way to use the second method that works (using a compass to equally space points) , is by placing 4 points around the equator (and when you draw the arc's they cross at the north and the south pole) which gives you a total of 6 points or an Octahedron one of the platonic solids.

If you start your first 'ring' of points ofset above or below the equator then again this only has a few posibilities giving you the Cube (8 points two rings of 3 points plus the north and south poles), the Icosahedron (12 points two rings of 5 points plus the north and south poles), the Dodecahedron (20 points 4 rings of 5 points each and no north or south pole) and finally the Tetrahedron (just 4 points a three sided pyramid).

Trying to divide a sphere with any other number of points using just one single length or compass arc for all the points is impossible.