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

[BIGAL]

All sounds to hard for me as I watched the spinning disco ball with its little mirrors all over it.

[okie]

6 hours ago, steven-g said:

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.

I'll bet AI solves this sooner rather than later.

[steven-g]

Nothing to solve, just accept.

[Dana W]

don't use the surface as a starting point, and don't think in two dimensions. 

 

A given number of lines radiating out from the center of the sphere, each separated from all the other lines by the same angle, will intersect the surface of the sphere at equally separated points.

 

If you start off  by drawing a line around the sphere at any point you are already hosed.

 

Theoretically one can build a sphere with a surface one helium atom thick.  Then all the atom centers will be exactly the same distance apart.

 

All we have to do is come up with math to find the particular angle for a given number of lines, and their distance apart at the sphere's surface.

 

 

[Dana W]

17 hours ago, BIGAL said:

All sounds to hard for me as I watched the spinning disco ball with its little mirrors all over it.

You are beginning to see the light. >rim shot< 

[Dana W]

Something tells me we will have to vary the space around the helium atoms in order to fit the last one in flush. 

[okie]

2 hours ago, Dana W said:

don't use the surface as a starting point, and don't think in two dimensions. 

 

A given number of lines radiating out from the center of the sphere, each separated from all the other lines by the same angle, will intersect the surface of the sphere at equally separated points.

 

If you start off  by drawing a line around the sphere at any point you are already hosed.

 

Theoretically one can build a sphere with a surface one helium atom thick.  Then all the atom centers will be exactly the same distance apart.

 

All we have to do is come up with math to find the particular angle for a given number of lines, and their distance apart at the sphere's surface.

 

 

That's pretty much how I approached it, but I can't wrap my mind around it. If you have lines all emanating from the same point then the endpoints of those lines would be all equally spaced. It's easy to do that on a flat plane, but you can't do it in 3d.

 

It's interesting because the human mind can conceive of it. Like I can picture it in my head. It's basically a sea urchin. But despite a child being able to picture it, it's this huge mystery apparently. Which just blows my mind.

[steven-g]

Oh well, I give up, when you manage to solve this for just 5 points equally spaced on a sphere, then you can have a go at breaking the 1st law of thermodynamics. And use AutoCAD to design a perpetual motion machine, there are plenty of examples to find on the internet. 

Quote

First law of thermodynamics – Energy can neither be created nor destroyed. It can only change forms. In any process, the total energy of the universe remains the same. For a thermodynamic cycle the net heat supplied to the system equals the net work done by the system.

 

[f700es]

14 hours ago, steven-g said:

Nothing to solve, just accept.

And monumental waste of time.

[Dana W]

Especially since there is theoretically no limit to the number of lines one can radiate out from the center of a sphere.  How many syllables will that xxx-Hedron have if we connect all the surface points?

 

Then there is the question...

Does that result in more points than one can swing a cat at?

[BIGAL]

Almost there. 5 mins work. I give up now.

Edited March 9, 2020 by BIGAL

[steven-g]

The original question was "equally spaced"

[BIGAL]

I know just threw it in there any way, time to give up.

[ammobake]

The only way I can think of to ensure the points are indeed equally spaced would be to ensure each pair of points is antipodal.

So in this sense the most basic primitive you could use to represent this would likely be a cube?  So perhaps you could solve this by rotating a cube, mapping all 8 points at specific rotation angles until you ended up with the solution - just mulling it in my head....

 

The surface area of a circle or a sphere is always going to be irrational (in terms of numbers).
 

-ChriS

[okie]

37 minutes ago, ammobake said:

The only way I can think of to ensure the points are indeed equally spaced would be to ensure each pair of points is antipodal.

So in this sense the most basic primitive you could use to represent this would likely be a cube?  So perhaps you could solve this by rotating a cube, mapping all 8 points at specific rotation angles until you ended up with the solution - just mulling it in my head....

 

The surface area of a circle or a sphere is always going to be irrational (in terms of numbers).
 

-ChriS

Yea that's a fair point that the corners of a cube would qualify. The problem is as you add more points it changes the vectors of the lines that are already drawn to the center.

 

I have a feeling this is something that can't be visualized in 3d. You can visualize the end result, like a sea urchin, but not the way to get to it. Kind of like how you can visualize a hypercube splayed open, but not put back together. So maybe an equilateral bucky ball is possible as a lower dimensional representation of something that can exist in a higher dimension.

[lrm]

Here's an interesting discussion on the related topic of why you cannot create  a sphere from equilateral triangles (except the icosahedron approximation).

[paulmcz]

Why do you say "icosahedron approximation"? Do you think it is not an exact figure?

[Dana W]

22 hours ago, ammobake said:

The only way I can think of to ensure the points are indeed equally spaced would be to ensure each pair of points is antipodal.

So in this sense the most basic primitive you could use to represent this would likely be a cube?  So perhaps you could solve this by rotating a cube, mapping all 8 points at specific rotation angles until you ended up with the solution - just mulling it in my head....

 

The surface area of a circle or a sphere is always going to be irrational (in terms of numbers).
 

-ChriS

Auntie who? 

 

I don't think the distance between two opposite corners of a cube is equal to the distance between two adjacent corners whether measured across the sphere surface or that of the cube.

 

I just remembered why I quit writing code.  It makes my old brain wish for a bass boat and some beer.  This puzzle makes the wish come on faster.

[BIGAL]

Fish like shiny things so chucking the mini disco ball out the back of the boat may work. Only problem is how many sticky mirrors for a tennis ball.

[Dana W]

I think I want to change my cat reference to "... swing Schrodinger's cat at."

Edited March 11, 2020 by Dana W