I am trying to place 4 object on a sphere so that they are equidistant. I can do this by placing them at the 4 corners of a tetrahedron, and a tetrahedron is just 4 corners of a cube. (The cube is just a construction object that will go away.)

I put my object at 0,0,0, then rotate it 45 degrees on two axes. (I then translate it to the corner of the cube.)

See screenshot part 1

I know this worked because when I turn the look down the corner of the cube, I can see my object is perfectly aligned.

See screenshot part 2

But when I add the sphere in, the object is not perpendicular to the sphere's surface.

See screenshot part 3

What have I done wrong? I'm pretty sure I'm doing the multiple axis rotation correct. I set the Transformation Orientation to Local so the rotations should be relative to the object, not Global.

(I can fix it manually by rotating X to about 54 degrees rather than 45, but that's pretty sloppy.)

What am I doing wrong? Or better yet, is there an easier way of placing 4 objects at equidistant angles from each other?

3 screenshots of an object on a sphere at four corners of a cube

  • $\begingroup$ See blender.stackexchange.com/questions/10725/… $\endgroup$
    – Ron Jensen
    Sep 5, 2020 at 19:08
  • $\begingroup$ Yes, the tetrahedron is easy. The four corners of a cube form a tetrahedron. The issue I'm having is getting objects to rotate correctly so they are stuck on the apeces of the tetrahedron. $\endgroup$ Sep 5, 2020 at 19:13
  • $\begingroup$ Sorry, thinking out loud, and then I got called away. $\endgroup$
    – Ron Jensen
    Sep 5, 2020 at 22:32

1 Answer 1


Four little monkeys, sitting on a tetrahedron

My first thought was that the vectors from the origin of a regular unit tetrahedron are normal to the plane tangent to the sphere where the vector intercepts, and we must be able to recover the rotation from that.

We see that Wikipedia gives the rather ugly values of $( 0, 0, 1 )$, $(\sqrt{\frac{8}{9}}, 0, -1/3 )$, $( -\sqrt{\frac{2}{9}}, -\sqrt{\frac{2}{3}}, -\frac{1}{3} )$, and $( -\sqrt{\frac{2}{9}}, \sqrt{\frac{2}{3}}, -\frac{1}{3} )$ for these vertices, so I wrote some Python to draw it.

D = bpy.data
from math import sqrt, radians
from mathutils import Vector

radius = 1

bpy.ops.mesh.primitive_solid_add(source='4', size=radius)

tetrahedron = C.active_object

empties = []

theta = radians(120)
vectors = [ Vector( (   0, 0, 0, 1 ) ),
            Vector( ( theta, sqrt(8/9),         0, -1/3 ) ),
            Vector( ( theta,-sqrt(2/9), -sqrt(2/3), -1/3 ) ),
            Vector( ( theta,-sqrt(2/9),  sqrt(2/3), -1/3 ) ),]
hack=[ 0, 2, 3, 1]

for n in range(4):
    bpy.ops.object.empty_add(type='SINGLE_ARROW', align='WORLD', location=(0, 0, 0))

    empties[n].rotation_mode= 'AXIS_ANGLE'
    empties[n].rotation_axis_angle = vectors[n]
    empties[n].location = radius * Vector(vectors[hack[n]][1:4])

for n in range(4):
    bpy.ops.mesh.primitive_monkey_add(size=0.5, enter_editmode=False, align='WORLD', 
        location=radius * Vector(vectors[hack[n]][1:4]) )
    monkey = C.active_object
    monkey.rotation_mode= 'AXIS_ANGLE'
    monkey.rotation_axis_angle = vectors[n]
  • $\begingroup$ Thanks. I ended up doing it manually by orienting the cube so I am looking exactly down a vertex of the cube in ortho mode. I can simply fiddle with the rotation of the smaller object until it's exactly aligned. Looks like 45deg and 35.4deg. I only have to do this once, then I can duplicate x3 it and simply change the +/- signs as-needed. Not my favorite way of doing it, but it worked. $\endgroup$ Sep 29, 2020 at 21:34

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