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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?

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Change 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. MathWorld has a derivation of the equation of the curve 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.) –  Rahul Jun 10 '12 at 3:22
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Strongly related is this –  Pedro Tamaroff Jun 10 '12 at 3:35

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$\def\r{{\bf r}} \def\v{{\bf v}}$ 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: $$\begin{eqnarray*} y(1) &=& y_0 \\ y(x) &=& y_0 x, \textrm{ if }\rho=0 \textrm{ (we row in a straight line if } v_c = 0) \\ y(0) &=& 0, \textrm{ if } |\rho| < 1 \textrm{ (we make it to the shore if } v_c<v_r) \\ y(0) &=& \frac{1}{2}\left(y_0 + \sqrt{1+y_0^2}\right), \textrm{ if } \rho =1 \textrm{ (we don't make it to the shore if } v_c\ge v_r) \\ y(x) &=& \frac{x}{2} \left(x^{-\rho}-x^\rho\right), \textrm{ if } y_0 = 0. \end{eqnarray*}$$ The last result is equivalent to the one derived by robjohn in his excellent answer to the question referred to above.

Below we give some sample trajectories. The current is the the positive $y$ direction.

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Figure 1. $y_0 = 0$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$


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Figure 2. $y_0 = 0.15$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$


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Figure 3. $y_0 = 2$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$

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I really like the simplicity of that last result. –  Rahul Jun 10 '12 at 9:45
    
Thanks @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. –  David Lewis Jun 10 '12 at 15:25
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@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 ${\bf v}_c + \alpha {\bf v}_w$. –  user26872 Jun 10 '12 at 18:02
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@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. –  David Lewis Jun 10 '12 at 20:11
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@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. –  David Lewis Jun 10 '12 at 20:24

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