Another focal length question

[questions_uk]

Hi. If using:

f (focal length) = a / ( 2 x arctan (theta / 2)) where a = 13 and is the Field of View

the value I get for the focal length is 0.03 and this seems too small. Unless the calculation has not been done right?

Thanks.

 

[Redbelly98]

Something is weird about your equation. If theta is an angle, it makes no sense to take the arctan of an angle.

the value I get for the focal length is 0.03 and this seems too small.

That depends. 0.03 mm would be too small. 0.03 m would actually be reasonable. 0.03 miles could be too large.

Moral: units are an important part of the answer.