Determine the bounding curve of a moving circular sector

analytic geometrycurveseuclidean-geometrygeometryproblem solving

Preamble: Consider the curve $\mathcal{C}$ formed by the segments $CD$, $DE$, $EB$ and the arc $BC$ presented below:

Next, let $F$ be a "perspective" point and $l$ a line segment whose other end point is $A$. Suppose that we draw similar curves for each point $T$on the line segment $l$, such that the bottom segment $DE$ in the picture is orthogonal to the ray originating from $F$ and passing through $T$. Below is my best attempt to graph that this looks like at the end points of a segment from $(0, 0)$ to $(2, 0)$. (N.B., the distances are not exact in the picture, as I don't know how to use GeoGebra well).

Question: I would like to determine the bounding curve for the shape which is obtained by the moving/painting area. That is, what is the minimal curve $\mathcal{M}$ which contains all of the "moving" curves $\mathcal{C}$?

It is quite clear that portion of the bounding curve are given by the segments $CD$, $DA$ of the leftmost area, and by the segments $AE$, $EB$ of the righmost area. What I'm struggling with is how to determine the bounding shape of the moving arc given that the line $l$ (the line segment which curve $\mathcal{C}$ moves along) and the perspective point $F$ can be be arbitary?

Best Answer

A point of vocabulary: The generic name of this kind of "bounding curve" is "envelope".

This is not a full answer: it gives, under the parametric representation (1) the biggest part of the equation of the upper envelope.

Take a look at the following picture (obtained with Desmos: you can see an animated version here.

The locus of $C$ is the red curve, the locus of $D$ is the purple curve.

The parameters are

the radius $r$ of the circular arc,
the aperture angle $2a$ of this arc,
the ordinate $b$ of point $F$.

The locus of point $C$ is given under the form of parametric equations that can be seen on the upper left of the figure:

$$\begin{cases}x&=&r\cdot\cos\left(\arctan\left(\frac{-b}{t}\right)+a\right)+t\\y&=&r\cdot\sin\left(\arctan\left(\frac{-b}{t}\right)+a\right)\end{cases} \ \text{where} \ t \ \text{is the abscissa or the arc's center}\tag{1}$$

For point $D$ change $a$ into $-a$.

Explanation about the angles found in (1): let $N$ be the midpoint of arc $LM$. As points $F,G,N$ are aligned, if we denote by $t$ the abscissa of $G$:

$$\text{angle of GN w.r.t.} \ x \ \text{axis} \ = \ \angle AGF = \arctan \frac{AF}{AG}=\arctan \frac{-b}{t}$$

To this angle, we add (resp. subtract) angle $a$ in order to obtain point $M$ (resp. point $L$).

The animation shows that, in the first quadrant, it is essentially this point $C$ that governs the "envelope", with an exception around the origin (in the animation here it is the case for $x<0.5$).

This disruption, calling a specific analysis, slightly increases when $a$ becomes important (say $>\pi/4$). The analysis of this exceptional zone looks possible but more complicated...

Waiting for your reactions.

Related Solutions

Given n line segments and a circular sector, find the leftmost and the rightmost segments

Note: Here we want a line segment $(s_k, e_k)$ to be oriented such that the point $c$ is to the right of the colinear line through $s_k$ to $e_k$.

(BTW, "colinear line" is redundant.) This is a key property. As long as you choose $s_k$ and $e_k$ this way (so $(s_{ky}-c_y)(e_{kx}-c_x) \ge (s_{kx}-c_x)(e_{ky}-c_y)$), the problem has a simple solution. If travelling from $s_k$ to $e_k$ has the circle center on your right, then you are travelling clockwise, and the angle should be decreasing. That is, the argument for $e_k$ should always be less than or equal to the argument for $s_k$.

If this isn't true, there are two ways we can correct it: either add $2\pi$ to the argument of $s_k$ or subtract $2\pi$ from the argument of $e_k$. Which technique makes sense depends on whether we are looking for the rightmost or leftmost segment. And just to be clear, I assume that these mean:

"Rightmost segment" is, among all segments with at least one end in the sector, the one with the most clockwise argument of all ends (include one outside the sector, provided the other end is in).
"Leftmost segment" is, among all segments with at least one end in the sector, the one with the most counter-clockwise argument of all ends (include one outside the sector, provided the other end is in).

Note that if the sector happens to open downward (instead of upward as in your examples), the "Rightmost segment" will be on the left and the "Leftmost segment" will be on the right (this is why "clockwise" and "counter-clockwise" are the better terminology for rotational phenomena).

Initialize variables

MinIndex = -1
MaxIndex = -1
MinAngle = 13   # > 4 * pi
MaxAngle = -7   # < -2 * pi

Pass through your segments, skipping over any segment without at least one endpoint in the sector. For the rest,

If $\arg s_k \ge \arg e_k$, then
- if $\arg s_k >\text{ MaxAngle}$, then $\text{MaxAngle } = \arg s_k, \text{ MaxIndex } = k$.
- if $\arg e_k <\text{ MinAngle}$, then $\text{MinAngle } = \arg e_k, \text{ MinIndex } = k$.
If $\arg s_k < \arg e_k$, then
- if $\arg s_k + 2\pi >\text{ MaxAngle}$, then $\text{MaxAngle } = \arg s_k + 2\pi, \text{ MaxIndex } = k$.
- if $\arg e_k - 2\pi <\text{ MinAngle}$, then $\text{MinAngle } = \arg e_k - 2\pi, \text{ MinIndex } = k$.

When you are finished, MinIndex will point to the Leftmost segment and MaxIndex will point to the Rightmost.

Finding an envelope for a moving circular sector

$\hspace{3cm}$

If $\mathcal{C}$ is a straight line you can always rotate and stretch your coordinate system such that $T$ is at $(0,-1)$ and $\mathcal{C}$ is the $x$-axis. If the center of the arc is at $(k,0)$ and its radius is $r$ the location of $P_1$ and $P_2$ is

$$x = k - r \cos \big( \arctan (k) \pm \alpha\big), \quad y = r \sin \big( \arctan (k) \pm \alpha \big)$$

Solving the second equation for $k$ and substituting that into the first equation yields

$$x = \tan \left( \pm \alpha - \arccos \left(\frac y r \right) \right) - \sqrt{r^2 - y^2}$$

where both branches complement each other beacause of the symmetry.

Additionally, if $|k| < \tan(\alpha)$ the arc reaches higher than its endpoints (also, for $\alpha > \frac \pi 2$ the endpoints never reach the top).

If $\mathcal{C}$ is a path of line segments you may have to add a small portion of the arc to the envelope at the connection points of the segments.

You can find the interactive plot here.

Best Answer

Related Solutions

Given n line segments and a circular sector, find the leftmost and the rightmost segments

Finding an envelope for a moving circular sector

Related Question