If you want to avoid the double root you could also solve for sin(2t). The derivative of the arcsin is simpler if anything.
You may need some special handeling for sin(2t)=1 still. You could pretend the funtion continues there (just join it together with asin(z - 1) + z - 1 + pi/2). Or maybe some other transformation.
> If you were looking at the earth from a distance, the face you’re looking at would correspond to a circle in the middle of the Mollweide map.
That doesn't look like the image. If you're looking at Earth from a distance, there will be foreshortening that squishes the meridians together. But the meridians look almost evenly spaced except for the far left and right edges of the ellipse.
It sounded very weird to me too. Perspective projection isn’t equal area. But I think he’s probably not trying to say that the projection would be identical, just that a circle in the center corresponds to one half of the globe.
If you want to avoid the double root you could also solve for sin(2t). The derivative of the arcsin is simpler if anything.
You may need some special handeling for sin(2t)=1 still. You could pretend the funtion continues there (just join it together with asin(z - 1) + z - 1 + pi/2). Or maybe some other transformation.
> If you were looking at the earth from a distance, the face you’re looking at would correspond to a circle in the middle of the Mollweide map.
That doesn't look like the image. If you're looking at Earth from a distance, there will be foreshortening that squishes the meridians together. But the meridians look almost evenly spaced except for the far left and right edges of the ellipse.
It sounded very weird to me too. Perspective projection isn’t equal area. But I think he’s probably not trying to say that the projection would be identical, just that a circle in the center corresponds to one half of the globe.
Yes. Thanks.
Unfortunately I don't know how to think about the Mollweide projection https://xkcd.com/977/
That comic is never not funny. Classic XKCD.