- #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.

- 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

- 12,121

- 160

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

**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.

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

Moral: units are an important part of the answer.

Last edited:

- Last Post

- Replies
- 10

- Views
- 2K

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 14

- Views
- 2K

- Last Post

- Replies
- 8

- Views
- 2K

- Replies
- 1

- Views
- 892

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 2K

- Replies
- 2

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Replies
- 0

- Views
- 3K