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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.)

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Perhaps you should phrase this as a question and not a command? –  Potato Nov 29 '11 at 4:33
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"(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. 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
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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. 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
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3 Answers

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?

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+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
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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.

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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(1+tan^2(g)cos^2(t))dt^2. The corresponding integral is elliptic which I am afraid has no elementary expression.

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