# Is there a name for the great circle where latitude and longitude are equal?

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 ).

• 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). – peterh - Reinstate Monica Mar 22 at 14:39
• 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 – eagle275 Mar 23 at 14:42
• @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. – Toivo Säwén Mar 23 at 15:19
• 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. – Semidiurnal Simon Mar 23 at 16:15
• @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. – Pete Kirkham Mar 23 at 20:07

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: 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: This surprised me!

• Comments are not for extended discussion; this conversation has been moved to chat. – gerrit Mar 29 at 18:46

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, $$c=1$$.