# Drawing in Perspective: Fit sphere into cube

Assuming I have a cube drawn in two-point perspective (or even three-point).

I'd like to draw a sphere inside the cube such that it touches all sides.

A sphere becomes a circle on paper.

• How do I find the center?

• How do I find the radius?

I thought at least finding the center would be trivial (drawing the diagonals) but I'm no longer so sure this gives the right result.

Just to clarify, this is a pen & paper question. Rulers are ok, computers are not.

• Drawing diagonals should absolutely provide both center and radius. Commented Sep 12, 2014 at 4:59
• I'm not so sure because the center of a 3D-circle is not the center of the resulting 2D-ellipse. Maybe the same thing happens with spheres? Commented Sep 12, 2014 at 5:02
• You have to draw diagonals correctly :) Commented Sep 12, 2014 at 5:23

To draw a sphere inside a cube, you first need to find its center. This is indeed quite simple: just draw a straight line from each corner of the cube to the opposite corner. The point where the lines intersect is the midpoint of the cube, and thus also the center of the sphere drawn inside the cube:

(If these lines don't all intersect at the same point, your cube is not actually a cube, or even a general cuboid.)

Now all you need to do is find the radius of the sphere. Unfortunately, in the general case, this is somewhat trickier than finding the center. The first thing you need to do is find the midpoints of the faces, which can also be found by drawing diagonal lines across each face:

These are the points where the sphere will touch the faces of the cube. The problem is that, unless one of the faces happens to be exactly edge-on to your viewpoint (so that it appears as just a line in the 2D projection), these points will not generally lie on the edge of the circle obtained by projecting the inscribed sphere into 2D.

The solution is to first sketch the great circles that connect the contact points on the surface of the sphere. In the 2D projection, these will be ellipses that pass through four of the contact points; they will also be tangent to the lines that divide each face into four smaller squares (not shown), i.e. they will pass approximately halfway between the diagonal lines:

Finally, pick the radius of the sphere so that it is tangent to these ellipses:

Now, if you look carefully at the picture above, you'll notice that the sphere I've drawn is not actually centered on the center point of the cube (and it's not even really a sphere, but an ellipsoid). What gives? Well, the problem is that my "cube" isn't really a regular cube, but just a rectangular cuboid (because I couldn't figure out how to get Inkscape's cube tool to give me an actual regular cube, so I had to eyeball it, and got it a bit wrong). Still, it at least looks like it's approximately nested inside the cube.

• You nailed it. I've read countless instructions on the web, all of them were wrong or only worked under some assumptions (which the authors failed to mention). This is not just correct in all cases, it is also the clearest, best written one. I really appreciate the effort you put into it. Commented Dec 12, 2014 at 12:21
• Generally, an image of a sphere is not necessarily a circle, it may happen to be an ellipse; and the center of the sphere is not necessarily projected into the center of the ellipse. Commented Feb 10, 2016 at 16:11

Once you find the center, the radius will be the distance between the center and the top edge of the Square.

• What exactly do you mean by 'top edge'? Commented Sep 12, 2014 at 15:12
• one of the edges :) I said top just for phylosophical reasons. Commented Sep 12, 2014 at 15:18
• The edges are lines. Distances are between points :) Do you mean the distance between the center of the cube and the center of an edge? Commented Sep 12, 2014 at 15:24
• hmmm...this is not exactly right. You can calculate the distance from a point to a line: en.wikipedia.org/wiki/Distance_from_a_point_to_a_line Commented Sep 12, 2014 at 15:27
• Ok, you got me there. So you mean the shortest distance between the center and any point on the edge. That might be the end of the edge (a corner) though. It totally depends on which edge you select. In isometric perspective the radius could be 0. Commented Sep 12, 2014 at 15:31