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\subsection*{Navigational Mathematics}



{\bf 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? This
turns out to be a nice calculus exercise.\vs

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 $(1/\cos L) = \sec \lambda$. 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, $dh/d\lambda = 
\sec \lambda$.
\vs

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 $3^{\circ}42'= 0.658$,
the other at latitude $34^{\circ}50'= 0.608$, and
separated by $97^{\circ}03'= 1.694$ in longitude.\vs


Calculating the height $h$ on the map corresponding to latitude $\lambda$
means integrating the function $\sec \lambda$. This integral, the bane of
generations of Freshmen, has a nice application! As they learned, 
$\int \sec x~ dx = 
\ln |\sec x + \tan x| + C$; in this
case $C=0$. So latitude $\lambda=.658$ corresponds to height $h=.711$, and
$\lambda=.608$ to $h=.649$.\vs


The function $h= \ln|\sec x + \tan x|$ 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 $G(h)$.
 \vs

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
 
$$\lambda = G(.711-.062t),~~ L = 1.694t,~~ 0\leq t\leq 1.$$

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

$$\int_0^1 \sqrt{(d\lambda/dt)^2 + 1.694^2 \cos^2\lambda}~ dt,$$ 
with $\lambda=G(.711-.062t).$\vs

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

The length integral is now 
$$3.39 \int_0^1 e^{.711-.062t})/(1+e^{2(.711-.062t)})~ dt.$$ 
Using the substitution $u = e^{.711-.062t},~~ du =
-.062 e^{.711-.062t} ~dt$, the integral transforms to 
$$-54.68 \int du/(1+u^2) = -54.68 \arctan u
= -54.68 \arctan (e^{.711-.062t}),$$ 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.\vs

The moral of the story is that choosing the great circle track from
the Farallones to Tokyo over the rhumb line saves 234 nautical
miles. 






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