3
$\begingroup$

I would like to design an LED sphere, but I am having some trouble deciding on the placement of LEDs evenly across its surface area.

I would like there to be 32 evenly spaced LEDs across the circumference of the sphere. That is the easy part. I am somewhat certain that there is no way to map LEDs on to the rest of the sphere without some approximations.

I noticed that when I designed this image in illustrator side view of sphere and measured the lenghts of the colored lines that there was no obvious pattern. These were the values I got when I divided the lengths of the lines by each other, starting with the yellow line divided by the light red line:

1.96155694377703,
1.451739343459089,
1.272780965237935,
1.175870069605568,
1.111136167174052,
1.061595183331078,
1.019595348096224

These numbers seem totally random to me. Is there any perfect LED count and sphere radius that would eliminate the need to make approximations in LED count along each line of the sphere? With the numbers I have right now, calculating LEDs along each ring results in a decimal; which is a problem, because you can't split LEDs into fractions... If there are no perfect numbers, then how would I get as close to perfect as possible?

$\endgroup$
1
  • $\begingroup$ This could be a way. Scroll down to the links to the case of 32 charges. $\endgroup$
    – OR.
    Jun 19, 2017 at 23:54

1 Answer 1

1
$\begingroup$

It depends somewhat on how many points you want to have on the surface of the sphere.

The most you can have with absolute regularity (in that no point can be distinguished from any other) is 20; that would be the vertices of a dodecahedron or the face centers of an icosahedron.

You can get 32 fairly regular looking points by combining the 12 vertices of the icosahedron with the extensions to the sphere surface of each of its 20 faces.

If you want more points, this turns out to be the second simplest (the dodecahedron faces themselves the simplest) case in a series of Buckminster Fuller domes: Each is made by placing 12 points the face centers of the 5-sided faces of a dodecahedron, but then filling the remaining gaps with circlets of 1, 2, 3, ... hexagonal faces. So for example, the next in this series is the "Bucky ball$ with 60 points.

The math to determine the coordinates of each point is reasonably straighforward, but you can for the three examples discussed just look up the coordinates on wikipedia pages for those figures.

$\endgroup$
6
  • $\begingroup$ Can I assume you mean "the extensions to the sphere surface of the center of each of its 20 faces"? $\endgroup$ Jun 20, 2017 at 0:13
  • $\begingroup$ Thanks! Are those figures symmetrical by any angle? A really cool feature that I though of for this sphere would be to attach a ring of LEDs perpendicular to a motor and spin that motor really fast. A microcontroller would rapidly change the colors displayed on the LEDs depending at which angle the ring is currently rotated at. To the human eye, there would appear to be a solid image across the surface of the sphere. I was thinking that the ring would be a curved diamond/triangles of LEDs, which could be rotated to create a sort of 3D triangle fan on each hemisphere of the sphere, lol.. $\endgroup$
    – Holden
    Jun 20, 2017 at 0:26
  • $\begingroup$ That obviously wasn't working perfectly though, because there would need to be fractions of LEDs on the curved triangles' surface.. $\endgroup$
    – Holden
    Jun 20, 2017 at 0:28
  • $\begingroup$ Hopefully that makes sense... I may have described it poorly. Triangle fan is probably not the best word... $\endgroup$
    – Holden
    Jun 20, 2017 at 0:32
  • $\begingroup$ They are not perfectly symmetric by angle (there are two different values of opening angles between neighbors) but they are very close. $\endgroup$ Jun 20, 2017 at 17:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .