Case study: the ellipse

Now, here's a familiar curve, an ellipse.

[Graphics:../Images/index_gr_20.gif]
[Graphics:../Images/index_gr_21.gif]
[Graphics:../Images/index_gr_22.gif]
[Graphics:../Images/index_gr_23.gif]

[Graphics:../Images/index_gr_24.gif]

[Graphics:../Images/index_gr_25.gif]

The inner and outer curves also look like ellipses. Are they? Try a fatter path.

[Graphics:../Images/index_gr_26.gif]

[Graphics:../Images/index_gr_27.gif]

[Graphics:../Images/index_gr_28.gif]

Not likely! The inner curve is certainly no ellipse. Well, I suppose it is conceivable that it is piecewise elliptical, but to see that this is not so is pretty easy. We'll look at that later.

Here are the same curves, with normals shown to the inner path.

Notice that on the inner curve in the figure above, there are four points where the curve is not smooth, but on the previous picture, the inner curve appears smooth. The outer curves appear smooth in both  pictures. In fact, it seems possible to find the "onset of nonsmoothness":

[Graphics:../Images/index_gr_32.gif]

[Graphics:../Images/index_gr_33.gif]

[Graphics:../Images/index_gr_34.gif]

See this link for an animation showing various widths.

Given a fixed ellipse, let's see if we can find the widths necessary to bring this condition about, and the points at which the cusps occur.

Let [Graphics:../Images/index_gr_35.gif] and let [Graphics:../Images/index_gr_36.gif] denote the (signed) distance of our path. We'll assume, without loss of generality, that [Graphics:../Images/index_gr_37.gif]. It is an easy calculation to obtain the parameterization of the parallel path at a distance [Graphics:../Images/index_gr_38.gif] to be

[Graphics:../Images/index_gr_39.gif].

This is clearly differentiable for all values of [Graphics:../Images/index_gr_40.gif] and it is easy to show that

[Graphics:../Images/index_gr_41.gif].

The denominators are positive for all [Graphics:../Images/index_gr_42.gif], so the only way this will fail to be smooth is if [Graphics:../Images/index_gr_43.gif] vanishes. Setting the above expression equal to [Graphics:../Images/index_gr_44.gif] we are delighted to see that setting each coordinate equal to zero separately gives the same result, namely that

[Graphics:../Images/index_gr_45.gif].

Notice that a lack of smoothness is impossible for negative [Graphics:../Images/index_gr_46.gif], and that this corresponds to the case of the "outer" curves, which did indeed appear smooth.
Also, if [Graphics:../Images/index_gr_47.gif], then the above requires that [Graphics:../Images/index_gr_48.gif], which is just as we expect, for this is the case of a circle, for which any other value of [Graphics:../Images/index_gr_49.gif] gives a (smooth) circle.

A quick trig identity turns our last equality into

[Graphics:../Images/index_gr_50.gif],

hence one solution is

[Graphics:../Images/index_gr_51.gif],

with three other solutions being [Graphics:../Images/index_gr_52.gif]. Of course, there are no such solutions if the argument of the arcsin function isn't suitable. In fact, we need that

[Graphics:../Images/index_gr_53.gif].

This implies that

[Graphics:../Images/index_gr_54.gif].

[Graphics:../Images/index_gr_55.gif]
[Graphics:../Images/index_gr_56.gif]

[Graphics:../Images/index_gr_57.gif]

[Graphics:../Images/index_gr_58.gif]


Converted by Mathematica      November 7, 2001