I'm trying to calculate the penumbra of the upcoming eclipse, which involves finding the inner tangents of the Sun to the Moon and the seeing how those rays project on the Earth:
I made a simulation in Mathematica with grossly inaccurate radii just to work out the calculations, and got most of the way there. (The Sun is orange, the Moon gray, the earth Green. Yes, I know the Earth isn't larger than the Sun! Just picking radii that fit nicely on the screen.)
Where I'm getting hung up is finding the coordinates of the two points circled in red. I know the distance from the internal homothetic center to the edge of the Earth — the straight red line — as was as the angle from the inner homothetic center relative to the plane (B) and, of course, the radius of the fake Earth.
But for the life of me, I can't figure out how the inner tangent lines intersect with the radius of the Earth (the big green circle). It would be easy if the Earth was flat, but I'm led to understand that's not the case 🙂
EDIT:
I was able to solve this by brute algebra, converting the Earth to x^2 + y^2 = r^2 and finding the intersection with the line:
I was hoping for a geometric solution, but this works!
Best Answer
In the figure:
from the configuartion of the eclipse we know:
$\overline{AE}=r$
$\overline{AD}=l$
$\angle EDA=\alpha$
and we can find $\beta= \angle AED$ using the sin rule: $$ \frac{r}{\sin \alpha}=\frac{l}{\sin\beta} $$
Now we can find $\angle EAD=\pi-\alpha-\beta$ and , from this we find $EF$ and/or the arc $EG$.