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Is there a name for the great circle where latitude and longitude are equal? I have attempted a google search but only the equator and the prime meridian are defined in the sources I can find. ( It is of relevance in developing a map application which keeps track of latitude and longitude ).

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    $\begingroup$ I think the answer is probably not, but maybe others have some idea. I doubt if it would be useful for anything (maybe some satellites might fly on such an orbit). $\endgroup$ – peterh - Reinstate Monica Mar 22 '20 at 14:39
  • $\begingroup$ Your question seems to be missing vital info .. if latitude and longitude are equal = non-changing .. how can there be a circle .. that defines only a point $\endgroup$ – eagle275 Mar 23 '20 at 14:42
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    $\begingroup$ @eagle275 as latitude $\phi\in[-\pi/2,\pi/2]$, and longitude $\lambda\in[-\pi,\pi]$, there should be a curve formed by the points where $\phi = \lambda$. I hade made the (erroneous, as pointed out by tfb) assumption that this curve would be a great circle. $\endgroup$ – Toivo Säwén Mar 23 '20 at 15:19
  • $\begingroup$ There's a good answer. But it's also worth mentioning that this line doesn't mean anything - because longitude is arbitrary. There's no good reason for any particular meridian to be defined as zero. $\endgroup$ – Semidiurnal Simon Mar 23 '20 at 16:15
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    $\begingroup$ @SemidiurnalSimon the 'Prime meridian' or 'Greenwich meridian' is the name of a great circle (or at least half a circle). That longitude it is arbitrary doesn't mean you can't have a great circle with a name defined by it. $\endgroup$ – Pete Kirkham Mar 23 '20 at 20:07
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The curve where latitude and longitude are equal is not a great circle. But as joe khool writes in his excellent answer, it's called the curve of Viviani! It's easy to see that the curve is not a great circle, because, using naïve spherical coordinates (in radians) $(\phi,\lambda)$ with $\lambda$ being longitude and $\phi$ being latitude (zero at equator), this curve passes through $(0,0)$, and also through $(\pi/2,\pi/2)$ which is the north pole ($(\pi/2,\lambda)$ is the north pole for any $\lambda$), But it also passes through, say, $(1,1)$ which is not on the great circle the between previous two points.

In fact the curve you get looks like this:

not a great circle

Note. I plotted this by defining Cartesian coordinates in the obvious way:

$$ \begin{align} x &= R\cos\phi\cos\lambda\\ y &= R\cos\phi\sin\lambda\\ z &= R\sin\phi \end{align}$$

and then plotting $(x,y,z)$ for $\phi = \lambda$ and $\lambda\in[-\pi/2,\pi/2]$.

An earlier version of this answer plotted $(x,y,z)$ for $\phi = \lambda$ and $\lambda\in[-\pi,\pi]$. This means that $\phi$ takes values which are not in $[-\pi/2,\pi/2]$ of course. I had assumed that these points would end up around the back of the planet: that you'd get a kind of 'S' which wraps around the planet, but in fact it ends up around the front of it again:

also not a great circle

This surprised me!

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – gerrit♦ Mar 29 '20 at 18:46
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If the Earth was a sphere, then the curve in the last picture of tfb's answer is the curve of Viviani; otherwise, if you make the oblate spheroid assumption, you get a slightly distorted version of this curve.

More generally, a clélie is the name given to any spherical curve where the longitude $\varphi$ and colatitude $\theta$ have the relationship $\varphi=c\theta,\quad c>0$, and the curve of Viviani corresponds to the locus of a geosynchronous orbit, $c=1$.

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