# Drawing an approximation to a circle in isometric projection

A circle viewed from from the side is an ellipse.

A common approximation can be found on the web (eg do a google image search for isometric circle). This produces something like (with arc centers T,U,A and N):

Alternatively, make a 4 Bezier Curves with control points of each arc being the mid points of a side of the parallelogram and the included corner.

Here is the ellipse which is the image of circle: $x^2+3y^2=3/2$.

Here is a picture of the ellipse (black), the composite Bezier curve(red/green) and the 4 circular curves(blue/orange):

How is this justified? Why do the arcs join up smoothly?

EDIT: I would like to:

• Predict the error (for a bezier approximation to a circle here is an error estimate: wikipedia).

• Generalise either of these methods, say, include more Bezier control points, include more composite arcs, find more circular arc centers ...

• Is that not just an ellipse? – muaddib Jun 12 '15 at 1:00
• True, the true isometric projection is an ellipse. But circular arcs are easier to draw. Hence, the common approximation. – pdmclean Jun 12 '15 at 1:01
• Ah, I see. 4321 Comment is now long enough :) – muaddib Jun 12 '15 at 1:06

## 1 Answer

"Justification" for an approximation may be in the eye of the beholder, but:

1. The true projection is an ellipse inscribed in the indicated quadrilateral.

2. The circular arcs have their centers chosen so that (by elementary geometry) each arc is tangent to two sides of the parallelogram. In particular, each arc is tangent to its "neighboring" arcs, i.e., the arcs join up smoothly.

3. An arc of the ellipse "near" the major axis or minor axis is "approximately circular", since it's the image of a circular arc under a linear transformation $T$ and the axes of the ellipse are the eigenspaces of $T$.

4. The circles' centers are easily found with a straightedge, given the isometric grid.

• 2. I suppose this uniquely defines the circles. – pdmclean Jun 25 '15 at 6:11
• Actually, not. :) If you pick an arbitrary point on one edge of the rhombus, construct the three "corresponding" points under reflection across the diagonals, and draw the perpendiculars at those points, the perpendiculars intersect in pairs, and each intersection is a suitable center of a circle. For the arcs in your diagram, however, the centers are "isometric grid points", i.e., vertices of the pre-existing equilateral triangles. – Andrew D. Hwang Jun 25 '15 at 11:55