Radiodrome

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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

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

${\dot x}=V_{d}\ {\frac {T_{x}-x}{{\sqrt {(T_{x}-x)^{2}+(T_{y}-y)^{2}}}}}$

(2)

${\dot y}=V_{d}\ {\frac {T_{y}-y}{{\sqrt {(T_{x}-x)^{2}+(T_{y}-y)^{2}}}}}$

(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)

${\frac {y''}{{\sqrt {1+{y'}^{2}}}}}={\frac {V_{t}}{V_{d}(A_{x}-x)}}$

(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.

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