How can I approach attempting to map a hemisphere onto the unit circle such that the meridians become arcs of a circle through $(0,1)$ and $(0,-1)$ and are evenly spaced around the x-axis (which is mapped from the equator) and the lines of constant latitude also becomes arcs of a circle evenly spaced along the y-axis?
The other two maps of spheres onto a 2-D surface I've looked at are the mercator and stereographic projections but the methods they use don't seem to apply to this problem. For example, the stereographic projection is made by drawing a line from the poles onto a plane. However, I can't think of a way to project meridians (great circles) into arcs of a circle on a plane by drawing lines of projection. I've also considered just projecting the hemisphere onto a plane parallel-wise, as if I was shining a light directly behind the hemisphere onto a sheet of paper but that won't make the meridians evenly spaced around the equator. For the mercator projection, preserving lines of equal compass bearing also made constructing the map straightforward in the sense that it was easy to visualise. However, both these examples work with infinite planes and I'm really struggling on how to come up with a projection method that confines the hemisphere to just the unit circle.
Any direction or ideas would be greatly appreciated, thanks!
Best Answer
First, trying to find the circle passing through $(0,1)$, $(0,-1)$ and $(a,0)$, where $-1\le a \le 1$ and $a\ne 0$. This is the circle which the meridian with longitude $90^\circ a$ would map to.
Let the circle have centre $(p,0)$ and radius $\sqrt{p^2 + 1}$:
$$\begin{align*} (x-p)^2 + y^2 &= p^2 + 1\\ x^2 - 2px + p^2 + y^2 &= p^2 + 1\\ x^2 + y^2 - 2px - 1 &= 0 \end{align*}$$
Substitute $(x,y) = (a,0)$,
$$\begin{align*} a^2 +0^2- 2pa -1 &= 0\\ -2ap &= 1-a^2\\ -2p &= \frac{1-a^2}{a} \end{align*}$$
Then the equation of this circle, in terms of $a$, would be:
$$\begin{align*} x^2 + y^2 + \frac{1-a^2}{a}x - 1 &= 0\\ ax^2 + ay^2 + (1-a^2)x - a &= 0 \tag 1\label 1 \end{align*}$$
If the circle of latitude $90^\circ b$ is mapped to a segment on $y=b$ (a degenerate circle), then to find the $x$-coordinate of the intersection of this line and the above circle:
$$\begin{align*} &ax^2 + (1-a^2)x -a(1-b^2) = 0\\ x &= \frac{-(1-a^2) \pm \sqrt{(1-a^2)^2 - 4a[-a(1-b^2)]}}{2a}\\ &= \frac{-(1-a^2) \pm \sqrt{(1-a^2)^2 + 4a^2(1-b^2)}}{2a}\\ \end{align*}$$
Picking the $+$-root gives the $x$-coordinate that's within $[-1, 1]$ and has the same sign as $a$.
In terms of latitude $\phi$ and longitude $\lambda$ (relative to the centre meridian), where $\phi,\lambda\in [-\pi/2, \pi/2]$, the map from $(\phi, \lambda)$ to $(x,y)$ would be:
$$\begin{align*} y &= \frac{\phi}{\pi/2}\\ x &= \begin{cases} 0, & \lambda = 0\\ \dfrac{-(1-(2\lambda/\pi)^2) + \sqrt{(1-(2\lambda/\pi)^2)^2 + 4(2\lambda/\pi)^2(1-(2\phi/\pi)^2)}}{2\cdot2\lambda/\pi},& \lambda \ne 0\\ \end{cases}\\ &= \begin{cases} 0, & \lambda = 0\\ \dfrac{-((\pi/2)^2-\lambda^2) + \sqrt{((\pi/2)^2-\lambda^2)^2 + 4\lambda^2((\pi/2)^2-\phi^2)}}{2\lambda\cdot \pi/2},& \lambda \ne 0\\ \end{cases} \end{align*}$$
If the circle of latitude $90^\circ b$ is mapped to a circular arc that passes through both the $y$-axis and $x^2+y^2=1$ proportionally between $(0,1)$ and $(0,-1)$ (the two poles), then the arc would pass through
According to WolframAlpha, the circle is centred at $\left(0, \frac{1-b^2}{2\sin \frac{b\pi}2 - 2b}\right)$, and the radius would be
$$\begin{align*} \left|\frac{1-b^2}{2\sin \frac{b\pi}2 - 2b}-b\right| &= \left|\frac{1 - 2b\sin \frac{b\pi}2 + b^2}{2\sin \frac{b\pi}2 - 2b}\right|\\ &= \left|\frac{\cos^2 \frac{b\pi}2 + \left(\sin \frac{b\pi}2 - b\right)^2}{2\left(\sin \frac{b\pi}2 - b\right)}\right| \end{align*}$$
When $b\to\pm 1$, the $y$-coordinate of the centre tends to $\pm 1$, and the radius tends to $0$.
The equation of the circle, when $-1<b<1$ and $b\ne 0$, would be
$$x^2 + \left(y-\frac{1-b^2}{2\sin \frac{b\pi}2 - 2b}\right)^2 = \left(\frac{1 - 2b\sin \frac{b\pi}2 + b^2}{2\sin \frac{b\pi}2 - 2b}\right)^2 \tag2\label2$$
Then given latitude $90^\circ b$ and longitude $90^\circ a$ (relative to the centre meridian) (both non-$0^\circ$, and non-polar latitude), the mapped $(x,y)$ is the intersection of two circles $\eqref 1$ and $\eqref2$ on or inside the unit circle.
There are close forms of the intersection in terms of the two centres and radii.