# Pixel diameter of a given field of view calculation equisolid (equal-area) fisheye lens

I have a 180° equisolid (equal-area) fisheye lens with a focal length of 8 mm. I want to crop images to a circle defined by an angular field of view from the zenith angle of the lens (i.e. the middle of the circle in the original image).

I have already found an equation which I think may be on the right lines:

$$R = 2f\sin\left({\theta\over2}\right)$$

Where $$\\theta\$$ is the angle (in radians) from zenith defining the circle to crop to (i.e. my given field of view), and $$\f\$$ is the focal length of the lens (i.e., 8 mm). $$\R\$$ is the radius of the circle drawn by the field of view (I think this is the radius on the sensor plane?), in mm.

The issue is I can't figure out how to relate $$\R\$$ to a pixel length on my images so I can crop to a circle of that diameter.

I've already looked at these pages:

What is the relationship between field of view and focal length for fisheye?

Calculate angle/field of view from 2D image

• Related (but not directly applicable since your question is about equisolid projection): Diagonal Angle of View given Horizontal or Vertical angles of view for equidistant lenses
– scottbb
Sep 11, 2018 at 14:31
• It would probably be more correct to say "angle of view". Although often used interchangeably (I've been guilty as well, until I 'saw the light' regarding this), "field of view" is normally used to describe a linear distance that appears in the image at a specific distance from the camera/lens. In other words, a 50mm lens in front of a FF sensor may have a diagonal AoV of 46°. That translates to a specific height and width of a target perpendicular to the optical axis of the lens at a specific number of feet in front of the camera. The FoV are the dimensions of that target that can be seen. Sep 11, 2018 at 18:23

So given your image-plane measurement $$\R\$$ (in mm), then conversion to pixels is simply
$$\text{Pixels} = {R\,\mathrm{[mm]} \times {1\over\mathrm{pixel\,pitch\,}\left[{\mathrm{\mu m}\over\mathrm{px}}\right]}} \times{1000\,\left[{\mathrm{\mu m}\over\mathrm{mm}}\right]}$$