- #1

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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.

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- Thread starter questions_uk
- Start date

- #1

- 74

- 0

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.

- #2

- 12,134

- 161

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

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.

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

That depends. 0.03

Moral: units are an important part of the answer.

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