Article 35
The Loxodrome.

On the surface of a sphere a curve starts from the equator in a given direction and cuts all the meridians at the same angle. To find its equation in latitude-and-longitude coordinates:

Let the loxodrome cross two consecutive meridians $AM$, $AN$ in the points $P$, $Q$; let $PR$ be a parallel of latitude; let $OM=x$, $MP=y$, $MN\prime =dx$, $RQ=dy$, all in radian measure; and let the angle $MOP=RPQ=\alpha$; then

hence $dxtan\alpha =dysecy$, and $xtan\alpha ={gd}^{-1}y$, there being no integration-constant since $y$ vanishes with $x$; thus the required equation is

To find the length of the arc $OP$: Integrate the equation

To illustrate numerically, suppose a ship sails northeast, from a point on the equator, until her difference of longitude is $45^{\circ }$, find her latitude and distance:

Here $tan\alpha =1$, and radians; $s=y\sqrt{2}=1.0114$ radii. The latitude in degrees is $40.980$.

If the ship set out from latitude $y_1$, the formula must be modified as follows: Integrating the above differential equation between the limits and gives

hence , from which the final latitude can be found when the initial latitude and the difference of longitude are given. The distance sailed is equal to radii, a radius being $60\times \frac{180}{\pi }$ nautical miles.

Mercator’s Chart.—In this projection the meridians are parallel straight lines, and the loxodrome becomes the straight line $y\prime =xtan\alpha$, hence the relations between the coordinates of corresponding points on the plane and sphere are $x\prime =x$, $y\prime ={gd}^{-1}y$. Thus the latitude $y$ is magnified into ${gd}^{-1}y$, which is tabulated under the name of “meridional part for latitude $y$”; the values of $y$ and of $y\prime$ being given in minutes. A chart constructed accurately from the tables can be used to furnish graphical solutions of problems like the one proposed above.

Prob. 103.

Find the distance on a rhumb line between the points ($30^{\circ }$ N, $20^{\circ }$ E) and ($30^{\circ }$ S, $40^{\circ }$ E).