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Rhumb Lines

Rhumb lines are paths of constant true course. They thus satisfy Eqn. (7) with $\alpha$ constant. This is most easily treated in geodetic coordinates. Integrating this equation, we obtain:
$\displaystyle \left.L\right\vert^{L}_{L1}$ $\textstyle =$ $\displaystyle \frac{\tan\alpha}{2}\left.\log\left(\left(\frac{1+\sin\phi}{1-\si...
...\left(\frac{1-e\sin\phi}{1+e\sin\phi}\right)^e\right)\right\vert^{\phi}_{\phi1}$  
  $\textstyle =$ $\displaystyle \tan\alpha \left.\log\left(\tan(\phi/2+\pi/4)
\left(\frac{1-e\sin\phi}{1+e\sin\phi}\right)^{e/2}\right)\right\vert^{\phi}_{\phi1}$ (8)

giving the coordinates of points $(\phi,L)$ on a rhumb line with course $\alpha$ through $(\phi_1,L1)$.

Inverting this relation, to find the geodetic latitude $\phi$ given the longitude $L$, can most readily be done iteratively, using:

\begin{displaymath}
\phi=-\pi/2+2\tan^{-1}\left(\tan(\phi_1/2+\pi/4)\exp\left(\f...
...frac{1+e\sin\phi_1}{1-e\sin\phi_1}\right)}\right)^{e/2}\right)
\end{displaymath} (9)

starting with $\phi=\phi_1$ on the RHS.

Combining equations (6) and (7) with $\alpha$ constant gives us a differential equation for the arc-length along the rhumb line:

\begin{displaymath}
\frac{\mathrm{d}s}{\mathrm{d}\phi}=\frac{a(1-e^2)}{\cos\alpha(1-e^2\sin^2\phi)^{3/2}}
\end{displaymath} (10)

Thus we find that $s$ is given by


\begin{displaymath}
s=(M(\phi)-M(\phi_1))/\cos\alpha
\end{displaymath} (11)

where the function $M(\phi)$ is the distance from the equator to the $\phi$ parallel measured along a meridian, given by:


\begin{displaymath}
M(\phi)=a\int_0^{\phi}\frac{1-e^2}{(1-e^2\sin^2\phi^\prime)^{3/2}}
 \mathrm{d}\phi^\prime
\end{displaymath} (12)

Formally, $M(\phi)$ can be expressed in terms of the elliptic integral of the second kind $E(\phi,e)$ by[1]

\begin{displaymath}
M(\phi)= a(E(\phi,e)-e^2 \sin\phi\cos\phi/(1-e^2\sin^2\phi)^{1/2})
\end{displaymath} (13)

Expanding to $O(e^6)$, $M(\phi)$ is approximately given by[4]:

$\displaystyle M(\phi) =$ $\textstyle a$ $\displaystyle [(1-e^2/4-3e^4/64-5e^6/256-\ldots)\phi$  
  $\textstyle -$ $\displaystyle (3e^2/8+3e^4/32+45e^6/1024
+\ldots)\sin2\phi$  
  $\textstyle +$ $\displaystyle (15e^4/256+45e^6/1024+\ldots)\sin4\phi$  
  $\textstyle -$ $\displaystyle (35e^6/3072+\ldots)\sin6\phi$  
  $\textstyle +$ $\displaystyle \ldots]$ (14)

Along a parallel, which is an E-W rhumb line, Eqns. (8) and (11) diverge, but since $\phi$ is constant, we have from Eqn. (6):

\begin{displaymath}
s=a R_L(L-L_1)=a\cos\phi (L-L_1)/(1-e^2\sin^2\phi)^{1/2}
\end{displaymath} (15)

A map with longitude as the x-axis and $M(\phi)$ as the y-axis has a Mercator[4] projection (with the equator as the standard parallel) on which rhumb lines plot as straight lines with the correct azimuth.


next up previous
Next: Geodesics on a spheroid Up: Navigation on the spheroidal Previous: Differential geometry
Ed Williams
2002-03-21