This is a question about solid angles.
According to Wikipedia, the central/binocular field of human vision is about $2\pi/3$ in the horizontal plane, and $\pi/3$ in the vertical axis.
Roughly, this tells me that we can see a "rectangle" that is two-thirds of the horizontal extent and one-third of the vertical extent.
Does this mean that we can see clearly a solid angle of two-ninths of $2\pi$ steradians? Or does something else happen with the geometry because it's spherical.
From Calculus(Robert A. Adams), page 978: Suppose the viewer is at point P. Two rays from P determines an angle at P whose measure in radians is equal to the length of the arc of the circle of radius $1$ with center at P lying between the two rays.Similarly, an arbitrary shaped half-cone K with vertex at P determines a solid angle at P whose measure in steradians is the area of sphere of radius $1$ with center at P lying within K. For example , the first octant of $\mathbb R^3$ is haf- cone with vertex at the origin.It determines a solid angle at origin measuring
$4\pi \times \frac 18=\frac{\pi}2$
since the area of the unit sphere is $4\pi$.
Now, coefficient $c=\frac 18$ represent $\frac{\pi}2$ on x axis and $\frac {\pi}2$ on y axis. We may use simple proportion to find our coefficient:
On x axis: $$\frac {\frac{2\pi}3}{\frac{\pi}2}= \frac 43$$
On y axis: $$\frac {\frac{\pi}3}{\frac{\pi}2}= \frac 23$$
So we may write:
$c= \frac 23\times \frac 43\times\frac 18=\frac 19$
So the measure of the solid angle is:
$\frac{\pi}2\times \frac 19=\frac{4\pi}{9}$.
If you want to consider an Ellipsoid instead of a sphere, you may use the following formula for the surface area:
$$S\approx 4\pi\big(\frac{(ab)^{1.6}+(ac)^{1.6}+(bc)^{1.6}}3\big)^{\frac1{1.6}}$$
What we have is the hight of the viewer which is c. for a and be we take arbitrary values.