What curve will a kayak describe if the paddler aims her bow at an object on a distant shore ahead and keeps the bow pointing to that object as she paddles toward it with constant velocity, in the presence of sideways drift due to constant current or wind? It's definitely not a straight line. Is it a common curve? If not, has it got a name?
Name that curve!
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1Change to a coordinate frame moving with the constant current, and you have a kayak moving directly towards an object moving with constant velocity; the kayak follows a [pursuit curve](http://en.wikipedia.org/wiki/Pursuit_curve). MathWorld has [a derivation of the equation of the curve](http://mathworld.wolfram.com/PursuitCurve.html) in the special case that the speed of the current is the same as that of the kayak. (Unfortunately, this is not the same as the path of the kayak seen in a static frame of reference.) – 2012-06-10
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2Strongly related is [this](http://math.stackexchange.com/questions/106912/applied-odes-in-trajectory-problem) – 2012-06-10
1 Answers
$\def\r{{\bf r}} \def\v{{\bf v}}$ As mentioned by Peter Tamaroff, this problem is a slight generalization of this one.
Without loss of generality, assume the shore is at the origin, the initial $x$ coordinate is $x_0=1$, and the current is in the $y$ direction.
The relevant differential equation is coupled and nonlinear, $$\begin{eqnarray*} \frac{d\r}{dt} &=& \v_c + \v_r \\ &=& \v_c - v_r \frac{\r}{r}, \end{eqnarray*}$$ where $\r$ is the position of the paddler, $\v_c$ is the velocity of the current, and $\v_r$ the velocity of the rower. The condition $\v_r = - v_r \frac{\r}{r}$ is imposed since the rower is always rowing towards the origin. Note that $\v_c$ and $v_r$ are constant, but that $\r$ and $r$ (and thus $\v_r$) vary with time.
In components, $$\begin{eqnarray*} \frac{d x}{d t} &=& - v_r \frac{x}{r} \\ \frac{d y}{d t} &=& v_c - v_r \frac{y}{r}. \end{eqnarray*}$$ Thus, $$\begin{equation*} \frac{d y}{d x} = \frac{y - \rho r}{x},\tag{1} \end{equation*}$$ where $\rho = v_c/v_r$. Recall that $r = \sqrt{x^2+y^2}$. With a clever change of variables, this differential equation is not too hard to solve. (Let $y(x) = x u(x)$ and integrate.) See here, for example. The result is $$y(x) = x \sinh(\sinh^{-1}(y_0) - \rho\log x).$$ This can be massaged somewhat. We find $$\begin{equation*} y(x) = \frac{x}{2} \left(\sqrt{1+y_0^2}\left(x^{-\rho}-x^\rho\right) + y_0\left(x^{-\rho}+x^\rho\right)\right).\tag{2} \end{equation*}$$ As mentioned by Rahul Narain in the comments, this is a simple example of a pursuit curve.
Some key facts: $$\begin{eqnarray*} y(1) &=& y_0 \\ y(x) &=& y_0 x, \textrm{ if }\rho=0 \textrm{ (we row in a straight line if } v_c = 0) \\ y(0) &=& 0, \textrm{ if } |\rho| < 1 \textrm{ (we make it to the shore if } v_c
Below we give some sample trajectories. The current is the the positive $y$ direction.
Figure 1. $y_0 = 0$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$
Figure 2. $y_0 = 0.15$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$
Figure 3. $y_0 = 2$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$
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0I really like the simplicity of that last result. – 2012-06-10
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0Thanks @oenamen -- nice answer. This is a common question in kayaking circles and has not, to my knowledge, been answered publicly before. But one question -- does it matter that kayak paddling speed, which I said is constant, is relative to the *water, not to the ground* underneath. Of course, my new question assumes we are dealing with a cross *current* rather than a cross *wind*. BTW, I'm guessing that is of mainly theoretical interest, since it will change the result relatively little in practice. But it is (maybe) an interesting math question. – 2012-06-10
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0@DavidLewis: Thanks, glad to help. The very first differential equation says the kayaker's speed is constant relative to the water and in the direction of the origin. I have not tried to take into account cross winds. For strong currents and light winds the result here should be enough. For heavy winds we would need to consider a much harder differential equation. – 2012-06-10
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1@DavidLewis: One simple approach would be to let ${\bf v}_c \to {\bf v}_c + \alpha {\bf v}_w$, where ${\bf v}_w$ is the wind velocity and $\alpha$ is a parameter telling you how effective the wind is at pushing a kayaker. I expect $\alpha$ to be about $10$-$20\%$. (That is, if you are not paddling and are blown by a 15 mph wind on a lake you will move in the direction of the wind at about 1.5-3 mph.) With these assumptions, the problem reduces to the one above with ${\bf v}_c$ replaced by ${\bf v}_c + \alpha {\bf v}_w$. – 2012-06-10
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0@RahulNarain: Thanks, Rahul. :-) – 2012-06-10
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1@oenamen -- good analysis of the wind issue -- in effect, you are paddling in a composite fluid. I think that 20% is about right. A novice (2-4 knot paddler) would be dead against a 20 kt wind, but an intermediate, 3-6 kt paddler would make headway up to 25-30 kts of wind. The figure is published in a number of places, so I could look it up. – 2012-06-10
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1@oenamen -- question: I am also interested in energy expenditure on a pursuit track as you analyze vs a straight track achieved by paddling upwind/upcurrent at just the correct, varying angle, called the *ferry angle* by paddlers. Experienced paddlers take the straight track using a ferry angle and consider the pursuit track a noobie error. But I've heard speculation that the pursuit track might require less energy, or at least, not more. Or put another way, what's the optimum track, energy-wise? Is that analysis an easy byproduct of this one? If not, I will post it as a separate question. – 2012-06-10
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0@DavidLewis: Your estimates seem quite reasonable. In this model we assume the magnitude of the force applied by the rower is constant and in the direction of travel. Thus, the energy expended will go like the length of the path taken. The "ferry path" will be the most efficient. – 2012-06-10
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0@oenamen -- Yes of course -- constant force ==> shortest path consumes the least energy. Duh and thanks! – 2012-06-11
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0@DavidLewis: One subtlety. ${\bf v}_w$ in the substitution ${\bf v}_c \to {\bf v}_c + \alpha {\bf v}_w$ is the velocity of the apparent wind. If the speed of the current and rower are small compared to the wind speed with respect to the ground, ${\bf v}_{wg}$, then ${\bf v}_c \to {\bf v}_c + \alpha {\bf v}_{wg}$. – 2012-06-11