How do you get the latitude and longitude of the circumference of an arbitrary hemisphere?

Question

I am attempting to draw the three twilights and night on an azimuthal equidistant projection. If the earth is divided by three planes perpendicular to the sub solar point. How would I find the lat and lon of the points along the circumferences to the resulting circles?

Answer 1

Lets denote by $P:S^2\rightarrow D_1$ the azimuthal equidistant projection. Using geographic coordinates for the sphere $S^2$ and polar for the unit disc $D_1$ this is writen as $$P(\phi,\lambda)=\left(\frac{90-\phi}{180},\lambda\right)$$ The inverse is given by $$P^{-1}(\rho,\theta)=(90-180\rho,\theta)$$ We will create a function $F:D_1\rightarrow \mathbb{R}$ wich will give for each point of the disc, the altitude of the sun. Then lets say, nautical twilight is between -6 and -12 degrees. We take the preimage $F^{-1}([-12,-6])$ wich gives the area on the azimuthal projection covered by nautical twilight. Or maybe just the boundary between civil and nautical twilight: $F^{-1}(\{-6\})$

To do that we start with a point on the disc in polar coordinates $x=(\rho, \theta)$, then $P^{-1}(x)=(\phi,\lambda)\in S^2$. Now we need to change coordinates from geographic to cartesian. This is done by $$cart(\phi,\lambda)=(\cos\phi \cos\lambda, \cos\phi \sin\lambda,\sin\phi)$$

We also need the coordinates of the subsolar point in cartesian, let it be $s=(s_1,s_2,s_3)$. Then we calculate the angle between $cart(P^{-1}(x))$ and $s$, which since we use unit vectors is given by $$arccos\left(\langle cart(P^{-1}(x)),s\rangle\right)$$ And the altitude of the sun at the point $p$ is given by $$F(x)=90-arccos\left(\langle cart(P^{-1}(x)),s\rangle\right)$$