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i am sure there is a simple answer, but i have seem to have forgen it. How do you calculate the relative apparent size of the subject

i have seen a formula with the field of view and tangent, but i am looking at a simple way to estimate on the fly what lens to use to make the subject 3X larger or smaller than where it is at, with a 50mm for instance.

i seem to remember halfing and doubling focal lenght, but that doesn't seem to work, 400mm is a lot more magnified than X3 (50 to 100 to 200 to 400mm)

Since "super zooms" like 24-240mm are 10X optical zooms, is it based on the starting 24mm lenght, which gives 10 increments, the relative X2 magnification would be 24-48-72-96-120-144-168-192-216-240mm

but that logic seems to break with other applications...

i though of angle of view, but 10° increments of fov, gives only 7 steps from 24 to 240mm (84-74-64-54-44-34-24-14° but 14° diagonal angle of view is 130mm and the next logical step 4° is 500mm

any simple way to estimates?

  • thank you both this and the duplicate giving the (lenght/sensor)*subject size= distance where helpfull. i guess i got confused with simple arithmetic seeing 50 to 400 as 3X increments rather than X8 – Reed Feb 16 '17 at 0:21
  • 400mm is 50mm raised by a factor of 2^3 (that is by a factor of 8). 50mm raised by a factor of 3 is 150mm. – Michael C Feb 16 '17 at 19:20
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The greater the focal length the more the image is magnified. A 100mm lens will project an image that that is 2X greater in size as compared to a 50mm. Thus a 400mm produces an image that is 400 ÷ 50 = 8X larger than a 50mm. Astronomers label the 50mm as magnification 1. Thus if they image using a 1000mm lens, they state the magnification as 1000 ÷ 50 = 20X.

I believe the basis is the diagonal measure of the format. If you are using a compact digital (Dx) the basis is 30mm which is the diagonal measure. Thus if your camera is a Dx and you mount a 400mm, then the magnification is 400 ÷ 30 = 13.3X.

If you mount a lens with a focal length approximately equal to the diagonal measure of the sensor, the angle of view will be about 45° for cameras with rectangular format and held horizontal (landscape). The published angle of view will likely be 53°. This is the diagonal angle of view, not much value but the one most commonly quoted, like buying a TV by the diagonal measure.

Hope this helps

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The apparent size seen in two lenses (all else the same) is simply the focal length ratio.

A 200mm lens will enlarge objects 2x what a 100mm does, and 4x of what a 50 mm lens does. Field size is the reciprocal, 1/2 or 1/4 dimension.

For 3x larger, use 3x longer, or 150mm instead of 50mm. Field size dimension becomes 1/3.

There is a field of view calculator at http://www.scantips.com/lights/fieldofview.html
Enter a distance and focal length, and then it will show field size. It's easy to verify that doubling focal length causes a 1/2x field size.

  • If you saw my earlier comment, please ignore it - I was conflating the angle of view with the field size. It's interesting how the non-linearity of focal length to field of view is perfectly cancelled out by the non-linearity of field of view to field size. Makes perfect sense though if you draw a diagram with similar triangles. – Mark Ransom Feb 15 '17 at 22:26
  • IMO, only the angle of view is not linear (angle is a trig tangent function). The focal length and sensor size, vs distance and field size are easy linear relationships, making similar triangles. – WayneF Feb 15 '17 at 23:54
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According to this web site the formula to calculate angle of view is:

angle=2*arctan(d/2/f)

where

d represents the size of the film (or sensor) in the direction measured

and

f represent focal length

And as you can see the relation is not linear

  • But to quote from the Wikipedia link you gave: "However, except for wide-angle lenses, it is reasonable to approximate α ≈ d/f radians or 180d/πf degrees." That's nearly linear for normal to telephoto focal lengths. – Mark Ransom Feb 15 '17 at 20:56
  • @MarkRansom, as OP ask for quite big difference in focal length I prefer to provide the precise function – Romeo Ninov Feb 16 '17 at 5:12
  • As pointed out by another answer, it's irrelevant anyway. The apparent size is governed by the tan of the angle of view, which cancels out the arctan of your formula leaving a perfectly linear relationship. – Mark Ransom Feb 16 '17 at 13:47

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