# What is the bounding curve for the shape formed by a moving circular sector?

Let $$p_0 = (x_0, y_0)$$ be the center of the sector so that our circular sector is given by $$S(p_0, r, \theta_1, \theta_2) = \{p_0 + r(\cos(\theta), \sin(\theta))\mid 0 \leq \theta_1\leq\theta\leq\theta_2 \leq 2\pi \}$$. Furthermore let $$p_1 \in \mathbb{R}^2\setminus \{p_0\}$$ be a "destination" point, so that $$p_0$$ and $$p_1$$ are connected by a finite segment $$l$$. If our sector $$S(p_0, r, \theta_1, \theta_2)$$ moves along the segment $$l(t) = p_0 + t(p_1 - p_0)$$, what will be the resulting bounding curve for the shape? By "moving along" I mean that the end shape will be given by $$\bigcup_{0 \leq t \leq 1}S(l(t), r, \theta_1, \theta_2)$$.

• You may as well assume $p_0=(0,0)$ and $r=1$, without loss of generality.
– user519413
Commented Nov 18, 2021 at 6:37

I suppose that the vector $$\vec{p_0p_1}$$ does not lie in the cone spanned by the vectors defining the circular sector.
Let $$q_0$$ be the point defined by $$p_0+r(\cos(\alpha), \sin(\alpha))$$ with $$\alpha = (\theta_1+\theta_2)/2$$, i.e. the line $$(p_0q_0)$$ is a symmetry axis for the circular sector. The resulting shape of the sector moved along'' the segment $$[p_0p_1]$$ is obtained by replacing the segment $$(p_0q_0)$$ with the parallelogram with vertices $$p_0$$, $$q_0$$, $$q_1$$, and $$q_0$$, where $$q_1 = p_1 +(q_0 -p_0)$$. The bounding curve has the same geometric description. In case a parametrization is needed, setting $$\gamma(t) = (\cos\theta, \sin\theta)$$, we might take $$p(t) = \begin{cases} p_0 +tr\gamma(\theta_1) & t \in [0, 1] \\ p_0 +r\gamma((t-2)\theta_1 +(t-1)\alpha) & t \in [1, 2] \\ q_0 +(t-2)(p_1 -p_0) & t \in [2, 3] \\ p_1 +r\gamma((t-4)\alpha +(t-3)\theta_2) & t \in [3, 4] \\ p_1 +(5-t)r\gamma(\theta_2) & t \in [4, 5] \\ p_1 +(t-5)(p_0 -p_1) & t \in [5, 6]. \end{cases}$$