Length of loxodrome

On a sphere with radius $R$, find the length of a loxodrome which starts at the equator and makes an angle $\gamma$ with all the meridians.

(No equations for such a loxodrome are given, and should be derived.)

• Perhaps you should phrase this as a question and not a command? – Potato Nov 29 '11 at 4:33
• "(No equations for such a loxodrome are given, and should be derived.)" - cool. Unfortunately you've neglected to say what definition of "loxodrome" you're using. – J. M. is a poor mathematician Nov 29 '11 at 4:37
• @J.M. : I think the definition of "loxodrome" must be a curve that makes an angle $\gamma$ with all the meridians. The way the question is phrased doesn't seem to leave room for another definition. – Michael Hardy Nov 29 '11 at 5:12
• I know perfectly well what a loxodrome is, @Michael; I was subtly nudging OP to show his definitions and hopefully derive the necessary differential equation. Oh well. – J. M. is a poor mathematician Nov 29 '11 at 5:15
• Just a guess: If $\gamma=0$ then clearly the length of the loxodrome is the distance from pole to pole along a meridian, and obviously the length corresponding to $-\gamma$ is the same as the length corresponding to $\gamma$, and it's also obvious that as $\gamma$ approaches a right angle then the length approaches $\infty$. Anyone who's learned trigonometry knows of a function that is equal to $1$ when the angle is $0$, that has the same value at $\gamma$ as at $-\gamma$, and that approaches $\infty$ as $\gamma$ approaches a right angle. Namely the secant function. Therefore..... – Michael Hardy Nov 29 '11 at 5:20

It can be done without calculus. Here is a hint: Consider two latitude circles at latitudes $\theta$ and $\theta+\Delta\theta$ with $0<\Delta\theta \ll 1$. How long is a piece of meridian between these two circles, and how long is a piece of your loxodrome between these two circles?

• +1, but I don't think this is "without calculus". – Michael Hardy Nov 29 '11 at 20:20
• One ends up with the conclusion that the length of the loxodrome between two parallels is $R(\sec\gamma)\;\Delta\theta$ even if one does not assume $\Delta\theta$ is small, and regardless of what latitude one is at. But the argument seems to begin with the idea that it works for infinitely small $\Delta\theta$. If one didn't need that, then it would be truly "without calculus". – Michael Hardy Nov 29 '11 at 20:24
• @Michael Hardy: Of course we have some "intuitive" limit here, but no fundamental theorem of calculus with its primitives etc. – Christian Blatter Nov 29 '11 at 21:23
• It's true that for this problem you don't need the fundament theorem of primitives, because the function being integrated is constant. That's as trivial as an integral can be. But you still need something other than purely discrete math in order to see that it's constant. – Michael Hardy Jun 5 '17 at 16:57

It looks to me as if no "equation" of the loxodrome is needed.

Suppose $d\theta$ is an infinitely small increment of latitude $\theta$. Going from a point at latitude $\theta$ to a point straight north of it at latitude $\theta+d\theta$ means going northward by a distance $R\;d\theta$. Now suppose we are heading $\gamma$ east of north. Thus we go northward by $R\;d\theta$ (the "adjacent" side of a right triangle) and eastward by $R\tan\gamma\;d\theta$ the ("opposite" side), covering a distance of $ds=R\sec\gamma\;d\theta$, the length of the hypotenuse (I'm using "$\sec = \mathrm{hyp}/\mathrm{adj}$").

Then the total length of the loxodrome, from the south pole to the north pole, is $$\int_{\theta=-\pi/2}^{\theta=\pi/2} ds = \int_{-\pi/2}^{\pi/2} R\sec\gamma\;d\theta.$$ The quantity $R\sec\gamma$ is a constant, i.e. it does not change as $\theta$ changes, so this is $$R\sec\gamma \int_{-\pi/2}^{\pi/2}\;d\theta.$$ That's a trivial integral.

Given the change in latitude is $dt$ for a loxodrome the corresponding change in longitude will definitely be $\tan(g)\,dt$. However, the longitudinal contribution to arclength depends on the latitude -- circles of constant latitude are smaller near the poles by $\cos(t)$. So, the actual arclength would be given by $ds^2 = R^2\big(1+\tan^2(g)\cos^2(t)\big)dt^2$. The corresponding integral is elliptic which, I am afraid, has no elementary expression.