# [optimal] Re: Spherical surface area of a given FOV angle

• From: "Egnatz, Thomas James" <tegnatz@xxxxxxxxx>
• To: optimal@xxxxxxxxxxxxx
• Date: Mon, 30 Apr 2012 15:52:52 -0400
```Thank you, Richard.

```
If a 50 degree photo is about 5 % of a sphere, a 50 degree retinal photo would be about 3.6 % of the retina, since the retinal area is about 72 % of the interior of the globe. Does that sound right?
```
Tom

Quoting "Richard J. Kinch" <kinch@xxxxxxxxxxx>:

```
```There was a question about what proportion of a sphere's surface is seen
by a given field of view angle (FOV).  The formula for a unit sphere is:

0.5 * (1 - cos( 0.5 * FOV)), for FOV = [ 0, 180deg ]

This gives the proportion of the full sphere from an observer in
the center.  For example, a 50-degree FOV into this formula yields a
proportion of a little less than 5 percent, or 1/20 of the whole surface.
An observer at a pupil, such as in the optics of a retinal camera,
instead of the center, sees a bit more due to the different perspective.

And of course the eye is only approximately spherical, but this gives you
an idea of the cosine relationship of FOV to area, not unlike ordinary
photography.

See: http://en.wikipedia.org/wiki/Solid_angle

Richard J. Kinch, PhD
http://www.truetex.com/range.htm

```
```

```