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Navigational Mathematics


5. Calculation of rhumb line distance

How far would we have to travel if we followed the straight line on the Mercator map from the Farallones to Tokyo? Under the assumption that the earth's surface is a sphere, this turns out to be a nice calculus exercise.

Let us go back to the properties of the Mercator projection: latitudes and longitudes go to an orthogonal grid, and the projection is conformal. Conformality means that at any point the vertical and horizontal stretching are the same. If we use longitude L and height h (measured in the same units as longitude) as coordinates on the map, then the equator is not stretched at all: it and its image have length $360^{\circ}$. But a circle of latitude at latitude $\lambda$ has length $360^{\circ} \cos \lambda$ on the sphere, and the same length as the equator in the Mercator projection. It has been stretched by a factor $(1/\cos\lambda) = \sec\lambda$. For conformality, the meridians must be stretching by $\sec \lambda$ as they pass through latitude $\lambda$, more and more in higher and higher latitudes. In terms of calculus, $dh/d\lambda = \sec \lambda$.

In order to apply calculus to the problem it is most convenient to work entirely in radian measure, and convert to degrees at the end. In radians, we have two points on the sphere, one at latitude $37^{\circ}42'=0.658$ the other at latitude $34^{\circ}50'= 0.608$, and separated by $97^{\circ}03'= 1.694$ in longitude.

The height h on the map corresponding to latitude $\lambda$ is the integral $\int_0^{\lambda}\sec x ~dx$. This integral, the bane of generations of Freshmen, has a useful application! As they learned, $\int \sec x~ dx = \ln \vert\sec x + \tan x\vert + C$. So latitude $\lambda=.658$ corresponds to height h=.711, and $\lambda=.608$ to h=.649.

The function $h= \ln\vert\sec x + \tan x\vert$ is invertible; in fact the formula can be explicitly inverted. First write $\sec x + \tan x$ as $\tan(\pi/4+x/2)$, and note that the function is positive on the sphere so the absolute value signs can be discarded. Then solve for x to get $x = 2 \arctan e^h - \pi/2$. Call this function G(h).

On the Mercator map, the straight line from A (h=.711, L=0) to B (h=.649, L=1.694) is $t\rightarrow (.711-.062t, 1.694t), 0\leq t\leq 1$. The rhumb line on the sphere is therefore

\begin{displaymath}\lambda = G(.711-.062t),~~ L = 1.694t,~~ 0\leq t\leq 1.\end{displaymath}

The metric on the sphere is $ds^2= d\lambda^2 + \cos^2\lambda dL^2$, so the length of the rhumb line is


\begin{displaymath}\int_0^1 \sqrt{(d\lambda/dt)^2 + 1.694^2 \cos^2\lambda}~ dt,\end{displaymath}

with $\lambda=G(.711-.062t).$

Since G is the inverse of a function with derivative $\sec \lambda$, the derivative of G is $G'(h) = 1/\sec(G(h)) = \cos(G(h))$, so the first term inside the radical is $.062^2 \cos^2(G(.711-.062t))$, while the second term is $1.694^2 \cos^2(G(.711-.062t))$. The radical thus simplifies to

\begin{displaymath}\begin{array}{l}1.695 \cos(G(.711-.062t)) \\ = 1.695 \cos(2 \... ...2t}) \\ = 3.39 e^{.711-.062t}/(1+e^{2(.711-.062t)}).\end{array}\end{displaymath}



The length integral is now

\begin{displaymath}3.39 \int_0^1 e^{.711-.062t}/(1+e^{2(.711-.062t)})~ dt.\end{displaymath}

The substitution u = e.711-.062t,   du = -.062 e.711-.062t  dt transforms the integral to

\begin{displaymath}-54.68 \int du/(1+u^2) = -54.68 \arctan u = -54.68 \arctan (e^{.711-.062t}),\end{displaymath}

giving the length as $-54.68 (\arctan e^{.649}-\arctan e^{.711}) = 1.366$. Converting to degrees gives 78.27 degrees, or 4696 nautical miles.

The moral of the story is that the great circle track from the Farallones to Tokyo is 234 nautical miles shorter than the straight line between them on the Mercator map, as we have computed them.


A great circle track plotted on a Mercator chart. From Dutton's "Navigation and Nautical Astronomy", 7th Ed. Reproduced by permission from the U.S. Naval Institute.

Dutton's rhumb line calculation is different from mine, and takes into account the earth's eccentricity e = .082483399. The tables used in Mercator sailing, and the java applets available on the web, take this factor into account: instead of $h = \ln[\tan(\pi/4+x/2)]$ they use

\begin{displaymath}h = \ln[\frac{\tan(\pi/4+x/2)}{(\frac{\textstyle 1+e\sin x}{\textstyle 1-e\sin x})^{\textstyle e/2}}]\end{displaymath}


which is not as simple to invert.


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