# Is it possible to convert an Ellipse into multiple circular arcs?

I would like to know if it is possible to convert/calculate the minimum number of circular arcs*1 which can make up an ellipse*2.

Im not sure if this is even possible but seems like it should be. Bellow is an example of what I expect to do. Also a small margin of error would be fine.

Edits

1. Updated title and body arcs to circular arcs
2. Changed ellipsis to ellipse
• An ellipsis is "..." . I assume you mean "ellipse"? If so, any nontrivial arc of an ellipse is not an arc of a circle, so we'd necessarily be talking about approximations, and in that case, your question is still a bit vague about what you want (but you may want to look up the curvature of the ellipse) Feb 2 at 21:31
• The figures that arent solid are circles? If yes, I dont think is possible to do what you have draw in "Results" because in a circle all points are a the same distance of the center, hovewer in al ellipse that isnt the case. In particular, the green "ARC 1" dont look that cicle arc. Feb 2 at 21:35
• @BrianBritosSimmari I drew this just to represent an approximation. Feb 2 at 21:43
• What is an "arc" here? In many graphics backends (such as svg, postscript, etc.), elliptic arcs are primitives. In this case, an ellipse is exactly one arc. If you are only allowed to use circular arcs, then it's necessary to consider what it means for a circular arc to be "close enough" to an elliptical arc. They'll never be the same, but they can be arbitrarily close. Feb 2 at 21:48
• No portion of an ellipse of positive length is exactly an arc of a circle. Feb 2 at 22:18

I imagine you mean circular arcs.

Here are (Fig. 1) two ovals close to ellipses with (only) four circular arcs (an arc centered in $$A$$ with radius $$AB$$ ; another one centered $$C$$ and radius $$CD$$, and their symmetrical arcs). The left oval with $$\pi/2$$ "openings", the right one with $$\pi/3$$ and $$2 \pi/3$$ "openings".

An important feature of both ovals is that there is a smooth connection of the arcs at meeting points like $$D$$ : the tangent in $$D$$ considered as a endpoint coincides with the tangent in $$D$$ at the beginning of the other arc. This is due to the following property : two circular arcs with centers $$C_1$$ and $$C_2$$ are smoothly joined (in the sense given above) in a point $$P$$ iff $$C_1, \ C_2,$$ and $$P$$ are aligned.

Fig. 1.

These solutions can be traced back to Renaissance, but probably were known since Antiquity. One finds them well described in the works of Serlio, an italian architect working in the 1550s.

The proximity of the first oval with the ellipse represented by a dotted pink line (with foci featured as black points) is very good as one can see on Fig. 2 :

Fog. 2.

One can object that that these constructions are for a fixed eccentricity. A way to palliate this drawback is by extending or shrinking the radii of circular arcs by a same amount in the way described on Fig. 3.

Fig. 3.

See the following references :

• The idea underlying fig. 3 is that you can get in this way all possible ratios $a/b$ (ratios of semiaxes). It will not be always as satisfactory as in fig. 2 ; I agree with you that, depending of the approximations you want, you will have to cope with more than 4 arcs. But, are you obliged to use circular arcs ? Having a certain knowledge of curves approximations, using circular arcs is not at all advisable, I advise you strongly to consider instead curves of the spline family providing means to get very good approximations. Feb 3 at 18:47