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Particle $A$ traces a smooth closed curve $\mathcal{C}$ in $\mathbb{R}^2$. Particle $B$ is at a constant distance $d > 0$ from the particle $A$ while being orthogonal to the curve $\mathcal{C}$, tracing the curve $\mathcal{C'}$. If the length of the curve $\mathcal{C}$ is $L$, what is the length of the curve $\mathcal{C'}$?

The curves are depicted here:

enter image description here

The problem was stated at a physics class, so some assumptions about $\mathcal{C}$ and $\mathcal{C'}$ might be missing.

My attempt:

Let $\alpha : [0,1]\to\mathbb{R}^2$ be the parametrization of $\mathcal{C}$. Let's determine the parametrization $\beta : [0,1]\to\mathbb{R}^2$ of $\mathcal{C'}$.

The unit normal vector at the point $\alpha(t) = \begin{pmatrix} x(t) \\ y(t)\end{pmatrix} \in\mathcal{C}$ is $\vec{n} = \frac{1}{\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}}\begin{pmatrix} \dot{y}(t) \\ -\dot{x}(t)\end{pmatrix}$, with $\dot{x}, \dot{y}$ being the derivatives of $x,y$. Another assumption seems to be that $\mathcal{C}$ is the inner curve, and $\mathcal{C'}$ the outer one so $\vec{n}$ has to be directed outwards. I think that is the case here, but how to check it?

Anyway, it should be $\beta(t) = \alpha(t) + d\vec{n}$:

$$\beta(t) = \left(x(t) + d\frac{\dot{y}(t)}{\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}}, y(t) - d\frac{\dot{x}(t)}{\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}}\right)$$

The length of $\mathcal{C'}$ is $\int_0^1 \|\beta'(t)\|\,dt$, so we need $\|\beta'\|$:

$$\beta' = \left(\dot{x} + d\frac{\dot{x}\left(\ddot{y}\dot{x}-\dot{y}\ddot{x}\right)}{\left(\dot{x}^2+\dot{y}^2\right)^\frac{3}{2}}, \dot{y} - d\frac{\dot{y}\left(\ddot{x}\dot{y}-\dot{x}\ddot{y}\right)}{\left(\dot{x}^2+\dot{y}^2\right)^\frac{3}{2}}\right)$$

$$\|\beta'\| = \dot{x}^2+\dot{y}^2+ 2d\frac{\dot{x}\left(\ddot{y}\dot{x}-\dot{y}\ddot{x}\right) - \dot{y}\left(\ddot{x}\dot{y}-\dot{x}\ddot{y}\right)}{\left(\dot{x}^2+\dot{y}^2\right)^\frac{3}{2}} + d^2\frac{\dot{x}^2\left(\ddot{y}\dot{x}-\dot{y}\ddot{x}\right)^2-\dot{y}^2\left(\ddot{x}\dot{y}-\dot{x}\ddot{y}\right)^2}{\left(\dot{x}^2+\dot{y}^2\right)^3}$$

Nothing seems to cancel here. Have I made a mistake somewhere? Will I be able to connect the result with $L$?

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    $\begingroup$ This should be related to the curvature: $(\gamma+dN)\dot{} = \dot{\gamma}+d\dot{N} = T - dkT $ where $k$ is the signed curvature. So it should be something like $ \int \lvert 1-dk \rvert \lVert \alpha' \rVert \, dt $. $\endgroup$
    – Chappers
    Sep 8, 2017 at 17:32

3 Answers 3

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Assume the containing space is $\mathbb{R}^2$.

Assume the trajectory of the inner particle is a smooth, simple closed curve $C$, and let $C'$ be the trajectory of the outer particle.

Let $L,L'$ be the lengths of $C,C'$, respectively.

For the case where the region bounded by $C$ is convex, then $L'=L+2\pi d$.

Explanation:$\;$Appoximate $C$ by a convex polygon $P$ with rounded corners. By elementary geometry, the result is true for $P$, hence, taking the limit as $P$ approaches $C$, it's true for $C$ as well.

For the case where there are portions of $C$ which bend inwards, then, restricted to those portions, we get the inverse relationship $L' = L-2\pi d$.

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  • $\begingroup$ How to prove that the circumference of the polygons converges to the curve? This is addressed in one of my earlier questions, as it is not enough to prove that the parametrization of the polygon converges uniformly to the parametrization of the curve. $\endgroup$ Sep 9, 2017 at 8:51
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    $\begingroup$ Hmmm . . .yes, I see the issue. As long as the edges of the appoximating polygon stay "close" to a chord of the curve, then by the definition of arc length as a limit of chord length sums, I think the arc length of the rounded corner polygons will converge to the arc length of the curve, thus avoiding the paradox in your link (the $\pi=4$ example). But my claim that staying close to chords of the curve leads to a correct limit is just an informal, hand-waved claim. To make it rigorous would require a lot more detail. $\endgroup$
    – quasi
    Sep 9, 2017 at 9:17
  • $\begingroup$ Thank you, intuitively it is clear. Do you happen to have a reference for this? $\endgroup$ Sep 9, 2017 at 16:31
  • $\begingroup$ Any Calculus book will define the arc length of a sufficiently smooth curve as a limit of a sum of chord lengths, so if the polygons converge, edges to chords, to the piecewise linear curve given by the chords, then the limit should work out correctly. But as I said, it would need a lot more work to make it rigorous. $\endgroup$
    – quasi
    Sep 9, 2017 at 23:45
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Approximate the curve as a polygon and construct the offset curve. The latter is made of line segments at distance $d$ of the original segments and of the same length, connected by circular arcs. The cumulated length of the arcs is $d$ times the angle swept by the normal, which is an integer number of revolutions (depending on the curve).

By letting the approximation be finer and finer, you converge to the true lengths and the difference in lengths is $2k\pi d$ for some $k$ ($\pm1$ for a non self-intersecting curve).

enter image description here

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  • $\begingroup$ How do you see that the sum of lengths of circular parts is $2k\pi d$, for some $k\in\mathbb{N}$? Why is it $\pm 1$ for a non self-intersecting curve? $\endgroup$ Sep 9, 2017 at 8:48
  • $\begingroup$ Also, how to prove that the circumference of the polygons converges to the length of the curve? See one of my earlier questions. $\endgroup$ Sep 9, 2017 at 8:53
  • $\begingroup$ @mechanodroid: you are perforce turning full turns. $\endgroup$
    – user65203
    Sep 10, 2017 at 10:52
  • $\begingroup$ @YvesDaoust Yves, what happens if the original curve has a concave section? There might turn out to be a loop in the displaced curve. $\endgroup$ Sep 10, 2017 at 17:55
  • $\begingroup$ @CyeWaldman: for the last time, you turn an integer number of full turns. Unavoidably. $\endgroup$
    – user65203
    Sep 10, 2017 at 18:42
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The figure in the OP is somewhat misleading. I consider a random closed form in the complex plane (see the figure below, blue line), call it $z$. A curve with a uniform displacement, say $d$, normal to the curve is desired. The angle of the displacement from the curve is the tangent angle of the curve, i.e., $\alpha=\arg \dot z$ minus $\pi/2$. Therefore, the displace curve, $w$ is given by

$$w=z+de^{i(\alpha-\pi/2)}$$

The displaced curve is shown in red in the figure below. Notice that it loops in regions where the curve is concave.

Finally, the lengths of the two curves are given by

$$ s_z=\int |\dot z|~du\\ s_w=\int |\dot w|~du $$

enter image description here

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  • $\begingroup$ That is a neat way to get the parametrization of the displaced curve. How to calculate $\dot{w}$, though? It contains $\arg$, which is not a differentiable function in the complex plane. Btw how did you create this image? $\endgroup$ Sep 11, 2017 at 18:28
  • $\begingroup$ I'm happy to answer your questions. First of all, I use Matlab almost exclusively. It handles complex variables seamlessly. So the plots are just plot(z) and plot(w), where $w$ is calculated just as I show in the equation. Insofar as I did this numerically I didn't have to worry about how to calculate $\dot w$, but just to look at it, $\alpha$ is only a function of $\theta$, so we can write $\dot w=\dot z+ide^{i(\alpha-\pi/2)}\frac{d\alpha}{d\theta}$. $\endgroup$ Sep 11, 2017 at 20:29
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    $\begingroup$ And let me add that you can run the displaced curve inside the original by just changing the sign of $d$. $\endgroup$ Sep 11, 2017 at 20:31
  • $\begingroup$ $\frac{d\alpha}{d\theta}$ is the part which troubles me. I believe it cannot be calculated. $\endgroup$ Sep 11, 2017 at 21:00
  • $\begingroup$ I understand what you are saying, but think of it this way. We have a vector of $\theta$ values. For each one we have value of $z, \dot z$ and then $\alpha=\arg \dot{z}$. In other words, You could plot $\alpha$ vs. $\theta$. What is preventing you from calculating $\frac{d\alpha}{d\theta}$? $\endgroup$ Sep 11, 2017 at 21:21

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