## Circular arcs 4 - extrema

One basic operation upon shapes is to compute bounding regions. For a shape made up of pieces of simple curves, this amounts to computing directional extrema on the curve. In other words, given a curve and a direction vector $$d$$ (without loss of generality we assume it is normalized), determine the point that maximizes the dot product of $$d$$ and the vector from the origin to the point. In addition to the point, we typically would also like to know the parameter value of the extremal point.

For a circular arc, there are two cases: the extremum may be an endpoint, or it may be interior to the arc. For any parameter value $$s$$, we shall call the aforementioned dot product $$P(s) = \left[p(s)-0\right]\cdot d$$, where $$0$$ is the origin. Clearly, it is trivial to compute $$P(0)$$ and $$P(1)$$ and we shall compare the value at any other interior points against the extremal point.

For $$g$$ sufficiently far from zero, one may compute the center of the circle $$C$$ and its radius $$r$$, and any interior extremum must be at $$E = C+rd$$. The only task is determining if $$E$$ lies on the arc. For the smaller half of a circle when $$g<1$$, $$E$$ lies on the arc if $$(A-C)\times(E-C) > 0$$ and $$(E-C)\times(B-C)> 0$$. Otherwise for $$g>1$$, then we require that then "and" is replaced by "or". The parameter value can be obtained by any parameterization method discussed in the previous article.

For $$g$$ near zero, the situation is slightly trickier. From the algorithm for evaluation, differentiation provides a method to compute the tangents to the arc, particularly at the endpoints. Therefore, the outward normals to the arc are readily computable, and let us denote them by $$n_A$$ and $$n_B$$. Since $$g$$ is small in magnitude, we can determine inclusion of the extremum within the arc again using cross products. We simply require than $$n_A \times d > 0$$ and $$d \times n_B > 0$$. The harder task is determining the point itself, and it is perhaps easiest to compute the parameter value directly through trigonometry:

$$s = \frac{\tan^{-1}\frac{n_A\times d}{n_A\cdot d}}{2\theta} = \frac{\tan^{-1}\frac{n_A\times d}{n_A\cdot d}}{4\tan^{-1}g} \in [0,1]$$

In implementation, there are a few additional annoying details, such as correctly handling the sign of $$g$$, and ensuring that the arctangents are computed on the right branch, &c.