How could you get the latitude and longitude of four points (equal distance apart) on a line from $(27,-82)$ to $(28,-81)$? The four points should split the line into 5 parts.
[Math] Latitude and longitude of points on a line
spherical coordinatesspherical-geometry
Related Solutions
For the corners of the square, we have this diagram
To get $r$ from $D$ (converted to an angle by dividing the distance by the radius of the sphere), we use $\cos(\frac{\pi}{4})=\frac{\tan(D)}{\tan(r)}$ to get $$ \tan(r)=\sqrt{2}\tan(D)\tag{1} $$ Then use the Law of Cosines to get the latitudes of the top and bottom corners $$ \sin(lat_1)=\sin(lat)\cos(r)+\cos(lat)\sin(r)/\sqrt{2}\tag{2} $$ $$ \sin(lat_2)=\sin(lat)\cos(r)-\cos(lat)\sin(r)/\sqrt{2}\tag{3} $$ Then solve the Law of Cosines to get the differences in longitude of the top and bottom corners $$ \cos(\Delta long_1)=\frac{\cos(r)-\sin(lat_1)\sin(lat)}{\cos(lat_1)\cos(lat)}\tag{4} $$ $$ \cos(\Delta long_2)=\frac{\cos(r)-\sin(lat_2)\sin(lat)}{\cos(lat_2)\cos(lat)}\tag{5} $$
The area of a spherical polygon is proportional to its excess angle. In the plane, a $n$-gon would have an interior angle sum of $(n-2)\pi$. On the sphere, the angle sum $\alpha$ is larger than that. Compute the difference, multiply with a suitable constant (which is $r^2$ as I'll show below) and you have the area.
$$A = \bigl(\alpha-(n-2)\pi\bigr)r^2$$
The fact that the proportionality factor has to be $r^2$ can be deduced like this: On the unit sphere, the hemisphere can be interpreted as a $2$-gon with two angles of $\pi$, stretching from pole to pole. Therefore you get $A=2\pi$, which is exactly half of the surface area of the complete sphere, as expected. Scale your sphere by a factor of $r$, and all areas will scale by $r^2$.
To compute these interior angles from latitude and longitude, you can have a look at how to compute the initial course for great-circle navigation. The difference between bearings towards adjacent corners will give you the angle at the current corner.
Best Answer
Given the latitudes of two points, $\beta_1$ and $\beta_2$, and their difference in longitude, $\Delta\lambda=\lambda_2-\lambda_1$, compute $\delta\in[0,\pi]$ with $$ \cos(\delta)=\sin(\beta_1)\sin(\beta_2)+\cos(\beta_1)\cos(\beta_2)\cos(\Delta\lambda)\tag{1} $$ Then compute $\alpha$ using $$ \tan(\alpha/2)=\frac{\sin(\Delta\lambda)\cos(\beta_1)\cos(\beta_2)}{\sin{\beta_2}-\sin(\delta-\beta_1)}\tag{2} $$
After computing $\delta$ and $\alpha$, compute $\beta'\in[-\frac{\pi}{2},\frac{\pi}{2}]$ with $$ \sin(\beta')=\sin(\beta_1)\cos(k\delta/5)+\cos(\beta_1)\sin(k\delta/5)\cos(\alpha)\tag{3} $$ Then compute $\Delta\lambda'=\lambda'-\lambda_1$ using $$ \tan(\Delta\lambda'/2)=\frac{\sin(k\delta/5)\sin(\alpha)\cos(\beta_1)}{\cos(k\delta/5)+\cos(\beta_1+\beta')}\tag{4} $$ Where $k$ in $(3)$ and $(4)$ ranges in $\{1,2,3,4\}$.