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In geometry, a radiodrome is the path followed by a point which is pursuing another point. The term is derived from the Latin word radius (beam) and the Greek word dromos (running). The classical (and best-known) form of a radiodrome is known as the "dog curve"; this is the path a dog follows when it swims across a stream with a current after food it has spotted on the other side. Because the dog drifts downwards with the current, it will have to change its heading; it will also have to swim further than if it had computed the optimal heading. This case was described by Pierre Bouguer in 1732.

It is also the name of a weekly radio show hosted by internet personality Brad Jones and Josh Hadley.

A radiodrome may alternatively be described as the path a dog follows when chasing a hare, assuming that the hare runs in a straight line at a constant velocity. It is illustrated by the following figure:

Graph of a radiodrome, also known as a dog curve
The path of a dog chasing a hare running along a vertical straight line at 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% faster than the speed of the hare.

Mathematical analysis[edit]

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)






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)^2}}






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







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}






one gets

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













From this relation follows that

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






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)






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)






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}) } 






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)






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)






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.