Mathematical formula for a lens's representation of what you see?

[MDesigner]

You know how approx 50mm is 1:1 as far as what the eye sees.. at 18mm, things look further out, and 300mm brings you close in (obviously). I'm wondering if there's a mathematical formula for that. Something with trig probably, involving triangles.. and would also let you factor in crop factor for certain DSLRs. To be more clear, a formula that will let me take the size of an object and figure out how large (or small) it would be at a particular focal length.

Anyone know?

 

[Garbz]

1:1 as far as the eye sees means nothing. In reality the human eye sees more than a 180 degree field of view so you would need a 6mm lens to emcompass that. But your fisheyed perspective would not be anything like the eye.

50mm gives the same perspective as the human vision. But after the last thread on this topic I am somewhat confused. I suppose if you have a 300mm lens and make the photo 1/6th the size of the photo from the 50mm then you still have the same perspective??

 

[Helen B]

...To be more clear, a formula that will let me take the size of an object and figure out how large (or small) it would be at a particular focal length.

How large or small on a print, or on screen, or on the sensor/film? The linear size relationship or angular size relationship?

The magnification on the sensor film is easy to formulate for reasonably distant objects. It is


m = f / Do

Where

m is the linear magnification of the object on the film (ie image length divided by real length);

f is the focal length of the lens; and

Do is the distance to the object, in the same units as f.

The rigorous formula is more complicated than that, because the rough and ready Do needs to be replaced by the distance from the lens' front (first) nodal point to the object, and f needs to be replaced by the distance from the lens' rear (second) nodal point to the image plane. If Do is very much greater than f, it doesn't matter.

If you now want the relationship between the object and its size on a print or screen, you simply multiply the two magnifications - ie the magnification of the object to the sensor/film image and the magnification of the sensor/film image to the print or screen.

If you want to get the relationship between the angular size of the object and the angular size of the image (which is what the 1:1 relationship you refer to for the 50 mm lens is all about) it is also quite simple, but it then depends on the distance at which the print is viewed.

If the print is viewed at a distance equal to the diagonal of the print (which is a conventional assumption) then the relationship can be worked out easily: a 1:1 correlation happens when the focal length is equal to the diagonal of the sensor/film image. It's easy to take it from there, if you get that relationship in your mind.

Does any of that come close to answering your question? I see many possibilities...

Best,
Helen

 

[MDesigner]

I guess what I'm looking for is something like this:

Say I have a photo I've already shot, at say 70mm (not counting the x1.6 crop factor). I'm thinking to myself, "I wonder what this shot would've looked like, on my screen, if I had shot with a 300mm lens." That is the formula I'm looking for...

 

[Helen B]

That's easy.

Imagine the diagonal of the image. Divide it by the focal length ratio and you have the new diagonal.

If you are looking at a 9-inch diagonal image taken with a 70 mm lens, then a 300 mm lens would have taken the image that has the centre 2.1-inch diagonal. (9 x 70 / 300 = 2.1). The same ratio applies to the sides, of course.

Best,
Helen

 

[christopher walrath]

Get a piece of cardboard. Cut a rectangle in the middle the size of your neg/CMOS. Then hold it however far from your eye to see the composition at different focal lengths. 70mm (3 inch or so) from your eye, 70mm (3 inch or so) focal length.

 

[MDesigner]

That's easy.

Imagine the diagonal of the image. Divide it by the focal length ratio and you have the new diagonal.

If you are looking at a 9-inch diagonal image taken with a 70 mm lens, then a 300 mm lens would have taken the image that has the centre 2.1-inch diagonal. (9 x 70 / 300 = 2.1). The same ratio applies to the sides, of course.

Best,
Helen

PERFECT. Thank you!

So, the below shot was shot at 50mm.

If I happened to have a telephoto and zoomed in to 300mm, the area in the white rectangle is what I'd see, correct? (I think I did my math right) The handy thing is, I just took the pixel width and multiplied by 50 then divided by 300.. and it works! Now I can go around shooting w/ my 17-50mm and then go home and figure out what the shot would've looked like at 300mm or 250mm, that will help me make my decision on which lens to get.

 

[Garbz]

Just a quick note bear in mind that if this was shot on a 70mm it is unlikely the 300mm will allow you to get this close to the cat. The near focus distance of the lens would likely limit you. So this rule isn't universally true.

 

[MDesigner]

Right, makes sense.

Actually I think the formula is a bit off. I did a simple test by taking two photos, one at 17mm and one at 50mm, then cropping the 17mm (using the formula) so it looks like a 50mm. The objects in the "simulated" 50mm look a bit larger than the actual 50mm. Still trying to figure out why...

 

[MDesigner]

OK I've been doing some math, some tests at various focal lengths (17mm, 35mm, 50mm). SEEMS like this formula yields a 100% perfect result:

cw = pxw * (f1 + c) / f2

Where:

pxw: pixel width of the photo shot at focal length f1
f1: actual focal length of photo
c: crop factor
f2: desired focal length
cw: resulting pixel width to crop the original photo

So for example, if I have a 17mm photo and I want to crop it down to see what it would look like shot at 250mm (my 40D shoots at 3888 x 2592):

3888 * (17 + 1.6) / 250 = 289 (produce 1:1 crop at 289 x 193)

Since I only have a 17-50mm lens, I was only able to test this formula out at a couple diff. focal lengths.. 35mm & 50mm. But with both cases, my "simulated" zoomed photo was an exact match with the actual 35/50mm photos.

Thoughts? Actually if someone with a different lens could try this out, I'd appreciate it. Hold the camera still & take two shots, one at 55mm and one at 80mm or 150mm, etc. Apply the formula above to pick a part of the 55mm and crop it, see if it matches the ACTUAL zoom shot.

EDIT: for those with a killer eye.. does this look right? Examine the below photos as if they were full size shots out of the camera. The first one is real, not simulated.

17mm (actual true shot):

50mm (simulated)

250mm (simulated)

300mm (simulated)

Pretty interesting.. the difference between 17mm and 50mm (33mm) looks huge.. but the 50mm difference between 250mm and 300mm looks pretty small in comparison. So either my formula is way off, or that's how lenses work

 

[Helen B]

How close was the subject in your comparison photos? At close distances the position of the lens' entrance pupil will matter - the lens will be 'seeing' from a different point. The formula I gave is only correct for infinity focus - it would have to be adjusted slightly for close distances, and hence become more complicated. Another factor could be that the focal length markings are slightly incorrect, particularly if it is an IF (internally focusing) lens because they can change their focal length as they focus.

I wonder if it is just coincidence that your formula has worked in this case, because I see no logic behind it - crop factor has nothing to do with the relationship. Take an example of two cameras, one with a crop factor of 2 and one with a crop factor of 1, and compare two pairs of lenses: 12.5 mm and 25 mm for the crop factor 2 camera and 25 mm and 50 mm for the crop factor 1 camera.

(12.5+2)/25 = 0.58

(25+1)/50 = 0.52

There's no reason for those two numbers to be different. In each case the field of view (in terms of distance) of the longer lens is exactly half that of the shorter lens, if the lenses are rectilinear.

Interesting.

Could you show us the images?

Best,
Helen

Edit: That's how lenses work. It isn't the difference in mm that matters, it is the proportional difference.
17/50 = 0.34 (ie 50 mm has 34% of the view of 17 mm)
250/300 = 0.83 (ie 300 mm has 83% of the view of the 250 mm)

 

[MDesigner]

OK, turns out it was a coincidence after all. Damn!

Anyway, here are image comparisons using your formula, pxw * f1 / f2.

Base 17mm image (actual):

The following will look almost the same, you'll probably have to save to disk & flip back & forth to see the differences.

24mm actual:

24mm simulated:

50mm actual:

50mm simulated:

So you can see, it's not quite exact. I wonder if it becomes more inaccurate with longer focal lengths? Probably, because we're taking a 17mm shot (and the distortion that comes with it) which would never properly translate to a 300mm shot.. still though, I can't help but wonder if there's some formula for it.

BTW, I'm grabbing the focal lengths from the EXIF data.. pretty sure those are exact.

 

[Helen B]

As I said, the formula will not be exact at close distances. It would have to be made more complex, because it isn't the focal length of the lens that really matters, it is the distance from the rear nodal point to the image plane, and that increases as you focus more closely. I could write the complex version if you wish, but I'm not sure that this exercise warrants that degree of accuracy. The simple formula I gave is correct for distant subjects, and the theory behind it is sound.

Also, remember that as you zoom, the point at which the lens 'sees from' changes, so that even if you hold the camera stationary, the lens' viewpoint will be moving as you zoom.

Does the EXIF always show the true focal length of an IF lens?

Try repeating the test with a distant subject.

Best,
Helen

 

[Helen B]

It's easy enough to just give the version that has a close-range correction.

Replace the focal length in the formula I gave you with this:

1 / (1/f - 1/Do)

where f is the focal length and Do is the distance from the front nodal point of the lens to the object, which must be the same in both cases. You may not know that exactly, but if you want to know how to find it I could explain...

Best,
Helen