Outrunning a Dog Four Times Faster, on the Margin Between Pi and Three

Running straight out from the centre loses, because a radius costs you 1 while half the fence costs the dog pi over 4. Inside a quarter of the radius your angular speed beats his, so you can orbit until he is diametrically opposite and then sprint three quarters of a radius against his pi over 4, and the whole escape reduces to 3 being less than pi with a margin of 0.0354 R. That two-phase plan works only up to a speed ratio of pi + 1, while the best known strategy for the problem reaches 4.60334.

Lambdia
August 18, 2026 · 9 min read

You are standing at the exact centre of a circular field of radius $R$ . A dog patrols the fence at four times your speed. He cannot come inside; you cannot leave except across the fence. Can you get out before he reaches you?

Yes, with about three and a half percent of a radius to spare. The margin is that thin because the escape rests on nothing more than $\pi$ being larger than 3.

Why running for it fails

Measure distances in radii and times in units of $R/v$ , so your own speed is 1 and the dog's is 4. Sprinting straight outward costs you a full radius, so time 1. The dog runs the shorter way round to meet you, at worst half the fence:

t_{\text{dog}} = \frac{\pi R}{4v} = \frac{\pi}{4} = 0.7854 \;<\; 1 = t_{\text{you}}

(1)

He is waiting when you arrive, with a quarter of a second of a lead in these units. Any straight line from the centre loses by the same margin, since every radius is the same length and the dog only ever has half a fence to cover. Direction is not the variable to optimise over.

The variable that matters is angle

The dog's position is described by one number, his angle. Yours needs two, and the second one is the lever. On a circle of radius $r$ your angular speed is $v/r$ , which grows without bound as $r$ shrinks, while his is fixed at $4v/R$ .

\frac{v}{r} > \frac{4v}{R} \iff r < \frac{R}{4}

(2)

Inside a quarter of the radius you turn faster than he does. Not slightly faster: at $r = 0.24R$ your angular speed is $4.167\,v/R$ against his 4, so you gain angle on him at $0.167\,v/R$ and can accumulate any lead you like, including the half turn that puts him diametrically opposite you.

Angular speed on a circle

A point moving at speed $v$ along a circle of radius $r$ sweeps $v/r$ radians per unit time. Two runners with fixed speeds on different radii have fixed angular speeds, so their relative angle changes linearly, which is what makes any required lead reachable in finite time.

The dash, from the right place

Now you are on the circle of radius $R/4 - \varepsilon$ with the dog opposite. Run straight out. You cover $3R/4 + \varepsilon$ and he covers half the fence:

\frac34 + \frac{\varepsilon}{R} \;<\; \frac{\pi}{4} \iff \varepsilon < \frac{(\pi - 3)R}{4} = 0.0354\,R

(3)

Both sides of that inequality carry the same factor of a quarter, so after dividing it through, the entire escape is the statement $3 < \pi$ . Nothing else is doing any work.

Fig. 1 — The two runs that decide the problem. Divide both by the dog's speed of four and the comparison becomes 0.750 against 0.785.

Fig. 2 — The same dog, the same fence, two starting radii. The orbit does not make you faster; it removes a quarter of a radius from the run that counts.

The orbit is slow, and that is not a detail

The phrase "circle until he is opposite" hides a real cost. At $r = 0.24R$ the angular gain is $0.167\,v/R$ per unit time, so clearing a half turn takes up to $\pi/0.167 = 18.85$ units of $R/v$ . The dash that decides everything takes 0.75. The escape is 96 percent waiting.

Pushing $r$ closer to $R/4$ lengthens the orbit without limit, since the angular gain tends to zero, and pushing it well below $R/4$ shortens the orbit but lengthens the dash. There is a genuine trade-off here, and equation (3) says the dash is the binding constraint: any $\varepsilon$ below $0.0354R$ works, so the sensible choice is a radius comfortably inside $R/4$ and an orbit you are willing to sit through.

How fast a dog defeats this plan

Run the argument with a general ratio $k$ . Your orbit works below $R/k$ , your dash costs $1 - 1/k$ , his half fence costs $\pi/k$ , and the escape condition becomes

1 - \frac1k < \frac{\pi}{k} \iff k < \pi + 1 = 4.1416

(4)

So $k = 4$ escapes and $k = 4.2$ does not. That threshold belongs to this particular plan and not to the problem, which is a distinction worth keeping. Running radially at the end is the obvious move and it is not the best one: a curved escape path, chosen so that the dog is always losing ground rather than merely starting behind, does better. The best known threshold for the problem is the root of

\sqrt{k^2 - 1} - \arccos\!\left(\tfrac1k\right) = \pi, \qquad k^{\star} = 4.60334\ldots

(5)

which is half a unit above equation (4). A reader who takes "four times faster" away from this problem as a hard boundary has learned the wrong thing. What the two-phase plan buys is a proof you can carry out loud in a minute.

What is being assumed

The dog stays on the fence. If he can cut across the field, the angular game is pointless and nothing here survives. He also runs the shorter way toward wherever you are headed, which is the worst case for you, so the plan is not being helped by a lazy opponent.

Both runners turn instantly and hold constant speed. That matters more than it sounds, because your orbit is a circle of radius $R/4$ traversed at full speed, needing a centripetal acceleration of $4v^2/R$ . On a field of radius 50 metres at 6 metres per second that is about 2.9 metres per second squared, which is survivable. On a smaller field it is not.

And the model ends at the fence. Reaching it a hair before the dog does is the whole of the claim, and what happens on the other side is somebody else's problem.

Sources and further reading

The general family this problem belongs to — Pursuit curve
The quantity in equation (2) — Angular velocity
Where the $\pi R$ in equation (1) comes from — Circumference
The acceleration remark — Centripetal force

The escape was verified by stepping the chase forward at intervals of $10^{-5}$ rather than by comparing formulas: the orbit needed 1,884,956 steps to accumulate its half turn, the dash then measured 0.76 against the dog's 0.785, and the dog was still short of the exit point by a positive angle at the moment of crossing. A bisection over the speed ratio, trying a ladder of orbit radii instead of assuming any formula, returned 4.14159 for the two-phase plan.

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