Mercator Projection.

Question

I am trying to derive the metrics for the Mercator Projection using standard spherical coordinates $f(\phi, \theta) = (\cos\phi \cos\theta, \sin\phi \cos\theta, \sin\theta) $. For simplicity, I consider the radius of the Earth is equal to 1. Then I am using a quadratic differential of the first fundamental form: $ds^2=Edu^2+2FDuDv+Gdv^2$; where $E, F, G$ are elements of the matrix representation of the first fundamental form of $f$, that is $E = \cos^2\theta; F =0$, and $G=1$. So the metrics of the sphere is $ds^2=\cos^2\theta d\phi^2+d\theta^2$. I am looking for projections: functions $x=x(\phi,\theta); y=y(\phi,\theta)$. Next, I let loxodrome be a function $\phi=\phi(\theta)$ then the velocity vector will be $v=(\phi',1)$. And meridian has a condition $\phi = const$, then as a parameter, I choose $\theta$, which gives me a velocity vector $m =(0,1)$. Then the loxodrome should make a constant angle $\psi$ with the meridian. So I can write:

$ \cos\psi =\frac{\lt v,m \gt}{\Vert v \Vert \Vert m \Vert} = \frac{1}{ \sqrt{ \cos^2\theta \phi'^2 +1} } $. I understand the last expression. But can someone explain to me the length of the velocity vector $v$? Thank you.

Answer 1

The length of the velocity vector $v=<\phi',1> =<d\phi/dt, d\theta/dt>$ is $ds/dt$. Using the metric $ds^2=\cos^2\theta d\phi ^2 + d\theta^2$, this becomes $(ds/dt)^2= \cos^2\theta (\phi')^2+ 1^2$.