# How does a change in FL relate to a scalar change in FoV?

Suppose I'm thinking about buying a longer lens and I want to visualize the effects of its narrower field of view by cropping-and-upsizing some already taken shots . Then,

- If I have a picture W pixels wide shot with a L-millimeter-long lens, to which W' should I crop the picture to match the FoV of an L' focal length which is n times M?

(Sorry if I'm not trying to figure this formula out myself, but still haven't found the rusty trigonometry that hopefully lies like flotsam in some backwater corner of my mind. Just wondered someone would know it more or less by heart?) Thanks!

edu T wrote:

Suppose I'm thinking about buying a longer lens and I want to visualize the effects of its narrower field of view by cropping-and-upsizing some already taken shots . Then,

- If I have a picture W pixels wide shot with a L-millimeter-long lens, to which W' should I crop the picture to match the FoV of an L' focal length which is n times M?

It's the direct inverse proportion. Twice the FL is equivalent to half the pixel width, three times is equivalent to one third the pixel width. etc ...

Gerry

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I don't quite understand your formula but if you double the focal length the FoV is that which you would get by cropping the shorter focal length shot in half in both directions. If you start with an image that is 3000x2000 pixels you have to crop to 1500x1000.

Chris R

... just as moving forward to half the distance also doubles the linear size of the subject while halving the Field of View.

Regards,

Baz

:

"Ahh... But the thing is, these guys were no ORDINARY time travellers!"

Thanks all, gents.

Well, I see my question was sillier than I thought. It happens that after having seen many times FoV specified in degrees for a given sensor size, I got under the impression that its relationship to a "flat field" wasn't linear and ought to be thought in terms of projection. (Hope that makes sense. :-))

And didn't give it a second thought; as you can see, trigonometry isn't the only rusty thing in between my ears...

.edu

PS: As to what

Chris R-UK wrote:

I don't quite understand your formula but if you double [ . . . ]

under the light of the answers, that formula now simply means

● New linear crop size to match FoV at new FL = (original FL / new FL) x original linear full size

(or W'= W/n where n = L/L' )

You can visual the effect of different focal lengths here:

edu T wrote:

Suppose I'm thinking about buying a longer lens and I want to visualize the effects of its narrower field of view by cropping-and-upsizing some already taken shots . Then,

- If I have a picture W pixels wide shot with a L-millimeter-long lens, to which W' should I crop the picture to match the FoV of an L' focal length which is n times M?
(Sorry if I'm not trying to figure this formula out myself, but still haven't found the rusty trigonometry that hopefully lies like flotsam in some backwater corner of my mind. Just wondered someone would know it more or less by heart?) Thanks!

You should understand that Thales law applies in linear dimensions and another law (I don't remember the name) for areas and focal length.

For linear dimensions to get from a 55 mm to 200 mm FOV you need to crop in such a way that the diagonal is in the same proportion as the focal lengths. So if you have 3000x2000 original photo and 55 mm you get the same FOV as at 200 mm if the longest length would be 825 mm. The result image is 825x550 mm (to keep the proportions) so from 6000000 pixels you are left with 453750 thus the square of the focal lengths ratio. As the ration of focal lengths is 3,6363 the ratio of the area is 13,223 thus the the square of it.

BTW. It is better not to upsize.

Good luck!

Victor

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trekkeruss wrote:

You can visual the effect of different focal lengths here:

Thanks for the tip, but there's a difference here, I'm trying to visualize the effect of longer lenses using my photos as reference...

edu T wrote:

Thanks all, gents.

Well, I see my question was sillier than I thought. It happens that after having seen many times FoV specified in degrees for a given sensor size, I got under the impression that its relationship to a "flat field" wasn't linear and ought to be thought in terms of projection. (Hope that makes sense. :-))

And didn't give it a second thought; as you can see, trigonometry isn't the only rusty thing in between my ears...

Well, it depends on what question you ask. Since you asked about field of view (FoV) then it really is a simple as the answers you've gotten: the change in linear dimensions are proportional to the change in focal length. But if you ask about angle of view (AoV), that's is non-linear and requires trigonometry.

Dave

Yes, I understood that you wanted to use your own images. To me that just seems like a lot of work when you can just look at the tool.

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