11
$\begingroup$

Imagine there is a Goose swimming counter-clockwise at the edge of a circular pond of radius $R$, and a dog (starting at the center of the pond) is paddling to catch it such that the dog is always pointing towards the goose. If the goose is travelling at speed $u$, what is the minimum speed that the dog must travel in order to eventually catch the goose? What is the minimum speed needed for the dog to catch the goose in time $\tau$?

I've seen pursuit problems like this in my Dynamical Systems textbook (indeed, this is very similar to one) and I've never had any luck solving them. I've found some solutions online to certain problems, but they tend to explain it very poorly in my opinion (for the record, all solutions I've seen to circular pursuit problems involve introducing alternate coordinate systems)

My attempt:

Let $x_d$ be the position vector of the dog, $x_g$ be the position vector of the goose, and $v_d$ be the velocity vector of the dog.

Drawing a diagram makes it evident that for any speed $||v_d||$ at fixed time $t$, $\exists \lambda \in \mathbb{R}$ s.t.

$$x_d + \lambda v_d = x_g = (R \cos(u t/R), R \sin(ut/R))$$

Generally, $\lambda = \lambda(t)$, which makes this much more difficult, but we are able to get two differential equations out of it:

$$x_d' + \frac{x_d}{\lambda(t)} = \frac{R}{\lambda(t)} \cos(ut/R) $$

$$y_d' + \frac{y_d}{\lambda(t)} = \frac{R}{\lambda(t)} \sin(ut/R) $$

Which is a first-order linear ODE, but I don't even know if the integrating factor $exp(\int \frac1{\lambda(t)} dt)$ exists, let alone how to find $\lambda(t) $

In order to catch the goose, we'd need $||x_d - x_g|| = 0$ for some $t \in \mathbb {R} $.

I believe the minimum would occur when

$||x_d - x_g|| \ne 0 \space\space \forall t \in \mathbb {R} \space \text { with } \space \lim_{t \to \infty} ||x_d - x_g|| = 0$.

As mentioned, I'm not sure how I'd find $\lambda $ if it's even worthwhile to introduce to the problem (I originally debated trying Lagrange multipliers or two-timing for this part, but am unsure how if it's possible).

I am open to seeing alternative methods as well as seeing if there's any way to actually use my thought process.

I am also aware that pursuit problems often lack a closed solution, but I thought I'd try my hand it nonetheless.

$\endgroup$
  • 1
    $\begingroup$ You consider $\lambda$ as if it were a constant, which cannot be true. $\endgroup$ – Aretino Jun 16 '17 at 21:33
  • $\begingroup$ Yes, good catch. It is fixed now. $\endgroup$ – infinitylord Jun 16 '17 at 22:26
  • 1
    $\begingroup$ Everything is in this article of Morley in ... 1921 that can I can access at (jstor.org/stable/2973034). Can you also have access to it ? See also (digital.library.okstate.edu/OAS/oas_pdf/v34/p164_165.pdf) $\endgroup$ – Jean Marie Jun 16 '17 at 23:43
  • 1
    $\begingroup$ Can I take, in a friendly way, the opportunity to give you an advice ? I just had a quick glance at your profile, and I saw that you have validated say 25% of the (good...) answers that are given to your questions. Consider a maximum delay of a 2 weeks before deciding to check, among the answers given, the one that has brought you the best insight in your issue (if you have had answers... of course) $\endgroup$ – Jean Marie Jun 16 '17 at 23:59
  • 1
    $\begingroup$ @JeanMarie: Yes, of course. Thank you for the recommendation. I typically do not choose a best answer if I do not feel fully satisfied with any of them, but I suppose I should take the prospect of there being a 'best' answer more literally. $\endgroup$ – infinitylord Jun 17 '17 at 0:10
7
$\begingroup$

We can get a simpler understanding of the problem if we look at dog's path in a frame which is rotating counterclockwise with angular velocity $\omega=u/R$ around the circle center. In such a frame the goose is sitting still at $G=(R,0)$, while the velocity of dog $D$ is made up of two contribution: $\vec{v_1}$, directed along $DG$ and of fixed magnitude $v$, and $\vec{v_2}$, due to the frame rotation, perpendicular to line $OD$ and of magnitude $\omega d$, where $d=OD$ is the distance of the dog from the center.

enter image description here

One can then write down at once a pair of coupled differential equation for the position $(x(t),y(t))$ of the dog. To keep them as simple as possible I conveniently set the unit of length so that $R=1$: $$ \begin{align} & x'(t)=\frac{(1-x)}{\sqrt{(1-x)^2+y^2}}v+\omega y,\\ &y'(t)=-\frac{y}{\sqrt{(1-x)^2+y^2}}v-\omega x\\ \end{align} $$ Notice that these are invariant under the scaling $(v,\omega,t)\to(kv,k\omega,t/k)$, so that the shape of dog's path depends only on the ratio $v/\omega$, as one could expect.

If the dog reaches a point where $\vec{v_1}=-\vec{v_2}$, then it stands there forever. That happens if $\omega d=v$, that is $d=v/\omega$, and if at the same time $OD\perp DG$, that is if $D$ belongs to the circle of diameter $OG$ (see picture below). A simple geometrical reasoning allows then to compute the coordinates of this halting point: $x=v^2/\omega^2$, $y=-(v/\omega)\sqrt{1-v^2/\omega^2}$.

enter image description here

I tried to get an exact solution of the above differential equations with Mathematica, but to no avail. But they can be solved numerically: I plotted the resulting dog's path for ten values of $v/\omega$, from $0.1$ to $1$ (picture below). In any case the path reaches asymptotically the corresponding halting point. For small values of $v/\omega$, this occurs after a long spiralling around it. For values near to $1$ there is no spiral.

enter image description here

As one could expect, for $v/\omega<1$ the dog doesn't catch the goose. For $v/\omega>1$ dog's path is very similar to the one shown above for $v/\omega=1$, and numerical solutions show that the dog arrives at $G$ in a finite time (which is quite obvious). For $v/\omega=1$ I expect the dog doesn't arrive there in a finite time, but that is hard to tell from a numerical solution.

In that case, we can try to compute directly the time taken by the dog to catch the goose, based on the evidence that the last part of dog's path, for $v/\omega=1$, is very near to the halting-points circle. If $s$ is the distance $DG$, we know from the above discussion that if $D$ is on that circle then dog's velocity is directed towards $G$ and its magnitude is given by $v-\omega d=v\big(1-\sqrt{1-s^2}\big)$. It follows that $$ {ds\over dt}=-v\big(1-\sqrt{1-s^2}\big), $$ and the time $t_0$ taken to travel a final length $s_0$ is then $$ t_0=\int_0^{t_0}dt=-{1\over v}\int_{s_0}^0{ds\over\big(1-\sqrt{1-s^2}\big)}. $$ But the last integral is divergent, so we can safely conclude that if $v/\omega=1$ the dog won't catch the goose.

EDIT.

The equations of motion become particularly simple if rewritten in polar coordinates centered at $G=(1,0)$. Defining: $$ 1-x=r\cos\theta,\quad -y=r\sin\theta, $$ we obtain: $$ \begin{align} &{dr\over dt}=-v+\omega\sin\theta,\\ &{d\theta\over dt}=-\omega +{\omega\over r}\cos\theta.\\ \end{align} $$ By dividing the first equation by the second, one gets a single equation for the path: $$ {dr\over d\theta}={c-\sin\theta\over 1-(1/r)\cos\theta}, \quad\hbox{where:}\quad c={v\over \omega}. $$ This equation, with the initial condition $r(0)=1$, gives dog's path in the rotating frame.

$\endgroup$
  • $\begingroup$ Interestingly, the transition between going clockwise and counterclockwise at the last leg appears to be at around $v=0.974$. I wonder why that is. $\endgroup$ – Rahul Jun 18 '17 at 19:20
  • $\begingroup$ @ Rahul For $v/\omega$ near to 1, the dog arrives at the halting-points circle and then follows that circle quite closely until it reaches its halting point. For $v/\omega=0.974$ it happens that the dog arrives at the circle just at the position of its halting point, so it stops there. $\endgroup$ – Aretino Jun 21 '17 at 6:02
  • $\begingroup$ After I wrote my answer I read Morley's paper suggested by @JeanMarie. It is difficult to make a direct comparison, but at the end of that paper the author finds an approximate formula for the arc travelled by the duck in the case $c=v/w>1$: $\theta\approx 0.156/\sqrt{c^2-1}$. But for $c\gg1$ I expect $\theta\approx1/c$ (as I get from my solutions). Am I missing something? $\endgroup$ – Aretino Jun 21 '17 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.