View Full Version : Calculating image circle closer than infinity

Paul Ewins

28-Dec-2005, 14:44

Does anybody know of a formula for calculating the image circle of a lens at closer distances? I'm much more interested in portraits than landscapes, so for me knowing the IC at 2m - 3m is more important than at infinity. It looks like it should be a fairly straight forward calculation, but it has just been too long since High School physics for me to be certain that I could derive the correct formula myself.

thanks,

Paul

Nick_3536

28-Dec-2005, 15:02

You've got a triangle. One side is the distance from the lens to the film plane. Second side is the from the middle of the film plane out to the edge of the image circle. The third side is from lens out to the edge of the film plane. You've got three angles. 90 degrees is at the middle of the film plane. The angle at the lens is 1/2 the coverage angle [I think] The third angle is 90 - 1/2*coverage angle. All three should add up to 180.

Okay you know all three angles. You can figure out the length of the line from the lens to the film plane easy enough.

You then cheat and go here:

http://www.trig.ionichost.com/trig.php

A angle of coverage

B 90 degrees

C 90 -A

Leave side a blank

Side b equals the distance to the film plane

leave side c blank

hit solve. Then double the result for side A.

I'll leave it to the people that remember highschool math to point all the mistakes I'm sure to have made -)

Nick_3536

28-Dec-2005, 15:03

Ooops. A is half the angle of coverage.

Nick_3536

28-Dec-2005, 15:14

Okay let me oops again. Leave b blank to. c is the distance to the film plane.

Michael Gudzinowicz

28-Dec-2005, 15:47

How about...

IC@Infinity x ( 1 + Magnification )

Paul Ewins

28-Dec-2005, 16:53

Ok, that makes it fairly clear. The extra coverage is calculated by dividing the focal length by the subject distance. For a 210mm lens at 2m this is approx 10% extra. Not a huge difference, but will increase with focal length and might be important for ULF work.

thanks,

Paul

Michael Gudzinowicz

28-Dec-2005, 17:22

You don't get much more coverage other than at macro distances. For "far" subjects, you can use your estimate for the magnification, but for close distances (d), the focal length (f) should not be ignored:

IC@Infinity x ( 1 + Magnification ) = IC@Infinity x ( 1 + f/(d-f))

adrian tyler

29-Dec-2005, 07:21

i was using very short focal lengths last summer, 80 and 58mm on 4x5, in enclosed spaces and i was astounded at the amount of movement i could get away with, espacially the 58mm which has almost no movement at infinity, i was getting away with 3-4cm, the biggest problem was the bellows getting in the way.

so it may be worthwile to ignore the math and try it out...

Leonard Evens

29-Dec-2005, 09:39

Several people have given the answer, but let me give it in words. You multiply the image circle diameter at infinity by the ratio of the image distance to the focal length. That ratio is also one plus the magnification. Finaly the ratio is also the subject distance divided by the difference between the subject distance and the focal length. Let me do an example using the last form. Suppose you are using a 300 mm lens and the subject plane is at 3 meters or 3000 mm. The ratio is

3000 divided by (3000 minus 300) = 3000/2700 which is approximately 1.11.

So you increase the image circle diameter in this example by about 11 percent.

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