Radiodrome

In geometry, a radiodrome is the path followed by a point that is pursuing another point. The word is derived from the Latin word radius for "beam" and the Greek word dromos for "running". It describes the motion of a point that is pursuing another point. The classical and most well known form of a radiodrome is known as the "dog curve" as this is the path a dog follows when it swims over a river having a current aiming for some food it has spotted on the other side. Because the dog drifts downwards with the current it will all the time have to change its heading and it will have to swim further than if it had been able to compute the optimal heading. This case was described by Pierre Bouguer already 1732

The "dog curve" can alternatively be interpreted as the path the dog follows when chasing a hare under the assumption that the hare runs along a straight line with constant velocity.

The situation is illustrated in the following figure:

The path of a dog chasing a hare running along the vertical straight line with a constant speed. The dog runs towards the momentary position of the hare and will have to change his heading continuously. The speed of the dog is 20% larger than the speed of the hare

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Mathematical analysis

Introduce a coordinate system with origin at the position of the dog at time zero and with y-axis in the direction the hare is running with the constant speed V_t. The position of the hare at time zero is $(A_x\ ,\ A_y)$ and at time t it is

$(T_x\ ,\ T_y)\ =\ (A_x\ ,\ A_y+V_t t)$

(1)

The dog runs with the constant speed V_d towards the momentary position of the hare. The differential equation corresponding to the movement of the dog, $(x(t)\ ,\ y(t))$, is consequently

$\dot x= V_d\ \frac{T_x-x}{\sqrt{(T_x-x)^2+(T_y-y)}}$

(2)

$\dot y= V_d\ \frac{T_y-y}{\sqrt{(T_x-x)^2+(T_y-y)}}$

(3)

It is possible to obtain a closed form analytical expression y = f(x) for the motion of the dog

From (2) and (3) follows that

$y^'(x)=\frac{T_y-y}{T_x-x}$

(4)

Multiplying both sides with T_x − x and taking the derivative with respect to x using that

$\frac{dT_y}{dx}\ =\ \frac{dT_y}{dt}\ \frac{dt}{dx}\ =\ \frac{V_t}{V_d}\ \sqrt{{y^'}^2+1}$

(5)

one gets

$y^{''}=\frac{V_t\ \sqrt{1+{y^'}^2}}{V_d(A_x-x)}$

(6)

or

$\frac{y^{''}}{\sqrt{1+{y^'}^2}}=\frac{V_t}{V_d(A_x-x)}$

(7)

From this relation follows that

$\sinh^{-1}(y^')=B-\frac{V_t}{V_d}\ \ln(A_x-x)$

(8)

where B is the constant of integration that is determined by the initial value of y^' at time zero, i.e.

$B=\frac{V_t}{V_d}\ \ln(A_x)+\ln\left(y^'(0)+\sqrt{{y^'(0)}^2+1}\right)$

(9)

From (8) and (9) follows after some computations that

$y^'= \frac{1}{2}\left(\frac{y^'(0)+\sqrt{{y^'(0)}^2+1}}{(1-\frac{x}{A_x})^{\frac{V_t}{V_d}}} - \frac{(1-\frac{x}{A_x})^{\frac{V_t}{V_d}}}{y^'(0)+\sqrt{{y^'(0)}^2+1}}\right)$

(10)

If now $V_t \neq V_d$ this relation is integrated as

$y= C - \frac{1}{2}\ A_x\left( \frac{(y^'(0)+\sqrt{{y^'(0)}^2+1})\ (1-\frac{x}{A_x}) ^{1 - \frac{V_t}{V_d}} }{1-\frac{V_t}{V_d}} - \frac{ (1-\frac{x}{A_x}) ^{1 + \frac{V_t}{V_d}} }{ (y^'(0)+\sqrt{{y^'(0)}^2+1})\ (1 + \frac{V_t}{V_d}) } \right)$

(11)

where C is the constant of integration.

If V_t = V_d one gets instead

$y= C -\frac{1}{2}A_x\ \left(\left(y^'(0)+\sqrt{{y^'(0)}^2+1}\right)\ \ln(1-\frac{x}{A_x}) - \frac{ (1-\frac{x}{A_x}) ^2}{(y^'(0)+\sqrt{{y^'(0)}^2+1})\ 2}\right)$

(12)

If V_t < V_d one gets from (11) that

$\lim_{x \to A_x}y(x) = C = \frac{1}{2}\ A_x\left( \frac{y^'(0)+\sqrt{{y^'(0)}^2+1} }{1-\frac{V_t}{V_d}} - \frac{1}{ (y^'(0)+\sqrt{{y^'(0)}^2+1})\ (1 + \frac{V_t}{V_d}) } \right)$

(13)

In the case illustrated in the figure above $\frac{V_t}{V_d} = \frac{1}{1.2}$ and the chase starts with the hare at position $(A_x\ ,\ -0.6\ A_x)$ what means that y^'(0) = − 0.6. From (13) one therefore gets hat the hare is caught at position $(A_x\ ,\ 1.21688\ A_x)$ and consequently that the hare will run the total distance $(1.21688\ +\ 0.6)\ A_x$ before being caught.

If $V_t \geq V_d$ one gets from (11) and (12) that $\lim_{x \to A_x}y(x) = \infty$ what means that the hare never will be caught whenever the chase starts.