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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 $P\phantom{\rule{0.3em}{0ex}}R$ be a parallel of latitude; let $O\phantom{\rule{0.3em}{0ex}}M=x$, $M\phantom{\rule{0.3em}{0ex}}P=y$, $M\phantom{\rule{0.3em}{0ex}}N\prime =dx$, $RQ=dy$, all in radian measure; and let the angle $M\phantom{\rule{0.3em}{0ex}}O\phantom{\rule{0.3em}{0ex}}P=RPQ=\alpha $; then

$$tan\alpha =\frac{RQ}{P\phantom{\rule{0.3em}{0ex}}R}\text{,but}P\phantom{\rule{0.3em}{0ex}}R=M\phantom{\rule{0.3em}{0ex}}NcosM\phantom{\rule{0.3em}{0ex}}P,17$$ |

hence $dx\phantom{\rule{0.3em}{0ex}}tan\alpha =dy\phantom{\rule{0.3em}{0ex}}secy$, and $xtan\alpha ={gd}^{-1}y$, there being no integration-constant since $y$ vanishes with $x$; thus the required equation is

$$y=gd(x\phantom{\rule{0.3em}{0ex}}tan\alpha ).$$ |

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

$$ds=dy\phantom{\rule{0.3em}{0ex}}csc\alpha ,\text{whence}s=y\phantom{\rule{0.3em}{0ex}}csc\alpha .$$ |

To illustrate numerically, suppose a ship sails northeast, from a point on the equator, until her diﬀerence of longitude is $4{5}^{\circ}$, ﬁnd her latitude and distance:

Here $tan\alpha =1$, and $y=gdx=gd\frac{1}{4}\pi =gd(.7854)=.7152$ 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 $({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$ gives

$$({x}_{2}-{x}_{1})tan\alpha ={gd}^{-1}{y}_{2}-{gd}^{-1}{y}_{1};$$ |

hence ${gd}^{-1}{y}_{2}={gd}^{-1}{y}_{1}+({x}_{2}-{x}_{1})tan\alpha $, 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 $({y}_{2}-{y}_{1})csc\alpha $ 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 ($3{0}^{\circ}$ N, $2{0}^{\circ}$ E) and ($3{0}^{\circ}$ S, $4{0}^{\circ}$ E).

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