Horizontal vs vertical field of view - focal length?

Bit of a weird random thought recently but I assume it has to be easy to explain through math, which I can't think of right now.

Assume I am using a 16mm lens on a full frame camera in landscape orientation - any way to know what equivalent focal length would capture the same vertical field of view?

Found myself shooting more zoomed-in and telephoto panoramic landscapes and just wondering if there's an easy enough formula to calculate equivalent coverage at different focal lengths

What is capture horizontally with a lens will be 1.5X the dimension of what your APS-C or FF digital camera (but not 4/3 format) will capture vertically.

1. So with 50mm mount on FF, at 10' distance it captures 7.08' horizontally. But to capture 7.08' vertically would need a 30.5mm lens to be mounted, so it is not a simple 50mm / 1.5 computation as you might think.
   
2. OTOH, if we start with 24mm at 10' it captures 14'9' horizontally. And to capture 14.9' vertically does take 16mm (24mm / 1.5)
   
3. Continuing that line of thinking, 20mm at 10' captures 17.9' horizontally. And to capture 17.9' vertically does take 13.3mm.
   

Wilt, my guess is that something is wrong with your first calculation; X/1.5 should work in any/every case. Even if X/1.63 is the answer here, the discrepancy is so small nobody will ever notice it.

There is no difference in the calculation for horizontal or vertical, you use the same multiplier.
The lens actually transmits a circular image and the camera crops it to a rectangle.

Which has nothing to do with the question at hand. The 3:2 (1.5) comes from the aspect ratio of the sensor.

Wilt wrote in post #18529830
What is capture horizontally with a lens will be 1.5X the dimension of what your APS-C or FF digital camera (but not 4/3 format) will capture vertically.

1. So with 50mm mount on FF, at 10' distance it captures 7.08' horizontally. But to capture 7.08' vertically would need a 30.5mm lens to be mounted, so it is not a simple 50mm / 1.5 computation as you might think.
   
2. OTOH, if we start with 24mm at 10' it captures 14'9' horizontally. And to capture 14.9' vertically does take 16mm (24mm / 1.5)
   
3. Continuing that line of thinking, 20mm at 10' captures 17.9' horizontally. And to capture 17.9' vertically does take 13.3mm.
   

Go figure, why case 2 and 3 does it work to FL/1.5, but not case 1?!

If focal length and angle of view are not "linear" then using a constant factor of 1.5 would not always work.

I've never looked into it, but it seems a reasonable explanation.

Wilt wrote in post #18529830
Go figure, why case 2 and 3 does it work to FL/1.5, but not case 1?!

Because, as usual, your math is wrong. The correct horizontal field of view for a 50mm at 10' is 7.2'.

Don't take this the wrong way, because I generally enjoy your posts, but as soon as you start using math to explain something I've learned to just skip on ahead. You have a pretty strong track record of basic math errors. Just the fact that you could come up with FL/1.5 as the correct answer in 2 out of 3 cases and then fail to recognize that the 3rd case is clearly a math error on your part is pretty strong evidence as to why any numbers you provide should automatically be regarded as suspect.

Anyway, anyone that wants to play around with the numbers without running the risk of making any math errors can use this FoV calculator here: http://www.bobatkins.c​om …hnical/field_of​_view.html

Doesn't seem you have to use complicated math for the OP's question. It's simple relational aspect ratio. If we do use your source, we find the equivalent vertical FOV for 50mm FF at 10' is 33.33mm FF at 10'. This is the same as using proportions:

2/3 = x/50 : 33.3
2/3 = x/24 : 16
2/3 = x/20 : 13.33

Yeah, I know. And yet...

ShadowHillsPhoto wrote in post #18529979
Because, as usual, your math is wrong. The correct horizontal field of view for a 50mm at 10' is 7.2'...You have a pretty strong track record of basic math errors.

Forgive me sir for using a program -- not math -- to calculate the FOV (fcalc v1.14, 2005 by Tangentsoft).
My math is not wrong, but your supposition of my use of math for FOV calculation is sorely wrong.
Lighten up and do not be automatically so abrasive, it will get you farther in the world with less antagonistic responses from others.

You COULD try to answer the OP, with lens FL, rather than merely pointing out my errors.

Folks, let's stick to the issue and knock off the personal bickering. Remember the rule: "Don't be a jerk."

T.D.
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