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Differential geometry

Figure 2: Triangle resulting from infinitesimal latitude change
\scalebox {0.5}{\includegraphics{t1.eps}}

A displacement of $\mathrm{d}\theta$ in parametric latitude along the meridional ellipse is illustrated in Figure 2. The poleward displacement is $\mathrm{d}(b\sin\theta)=b\cos\theta\mathrm{d}\theta$ and the equatorial component is $\mathrm{d}(a\cos\theta)=-a\sin\theta\mathrm{d}\theta$ From this we see that the geodetic and parametric latitudes are related by

\begin{displaymath}
\tan\phi=(a/b)\tan\theta
\end{displaymath} (2)

and that the displacement along the meridian is given by:


$\displaystyle (a^2\sin^2\theta+b^2\cos^2\theta)^{1/2}\mathrm{d}\theta$ $\textstyle =$ $\displaystyle a(1-e^2\cos^2\theta)^{1/2}\mathrm{d}\theta$  
  $\textstyle =$ $\displaystyle a\frac{(1-e^2)}{(1-e^2\sin^2\phi)^{3/2}}\mathrm{d}\phi
%\equiv{}R_\phi \mathrm{d}\phi
=R_\phi \mathrm{d}\phi$ (3)

$R_\phi$ is the radius of curvature of the meridional arc at $P$. The radius of curvature in the perpendicular plane (i.e. in the plane of the parallel), $R_L$ is given by

\begin{displaymath}
OD=a\cos\theta=a\cos\phi/(1-e^2\sin^2\phi)^{1/2}
\end{displaymath} (4)

Figure 3: Triangle resulting from infinitesimal latitude and longitude changes
\scalebox {0.5}{\includegraphics{t2.eps}}

In Figure 3 we illustrate the result of a displacement of $\mathrm{d}\theta$ in parametric latitude and of $\mathrm{d}L$ in longitude, resulting in a Northerly displacement of $a(1-e^2\cos^2\theta)^{1/2}\mathrm{d}\theta$ and an Easterly displacement of $a\cos\theta\mathrm{d}L$, respectively. By Pythagoras' theorem the displacement distance $\mathrm{d}s$ is given by:

\begin{displaymath}
\mathrm{d}s^2 = a^2(\cos^2\theta \mathrm{d}L^2 +
(1-e^2\cos^2\theta) \mathrm{d}\theta^2)
\end{displaymath} (5)

or, using eqn. (2),
\begin{displaymath}
\mathrm{d}s^2 =
a^2(\frac{\cos^2\phi \mathrm{d}L^2}{1-e^2...
...hi}+
\frac{(1-e^2)^2 \mathrm{d}\phi^2}{(1-e^2\sin^2\phi)^3})
\end{displaymath} (6)

The true course $\alpha$ is given by:
\begin{displaymath}
\tan\alpha=\frac{\cos\theta}{(1-e^2\cos^2\theta)^{1/2}}
\f...
...phi(1-e^2\sin^2\phi)}{1-e^2}\frac{\mathrm{d}L}{\mathrm{d}\phi}
\end{displaymath} (7)

where we have again used Eqn. (2) to transform between reduced and geodetic latitude coordinates. Equations (5), (6) and (7) are the fundamental relations relating distances and directions on the spheroid at a point.


next up previous
Next: Rhumb Lines Up: Spheroid Geometry Previous: Spheroid Geometry
Ed Williams
2002-03-21