T3D f/numbers & DOF (from P3D)

[bercov@xxxxxxxxxxx (John Bercovitz)]

Eric G. writes:

> My old time Rollei IBs state their tables were based on 
> f/1400 for 6 x 6 and f/2000 for rolleikin (35 mm)!!

You've shown the correct approach when the task is to find the DOF for
a stereo camera.  When you say f/1400, you are specifying an angle
of view for the allowable circle of confusion and since we view from 
the center of perspective when we view stereo pairs, we want an angle.

A simple formula isn't difficult; all you do is move the equation to 
object space.  It's simple enough that anyone can put it in a spread- 
sheet.  Again, from Rudolf Kingslake's wonderful book, "Optics in 
Photography" (ISBN 0-8194-0763-1, SPIE) Pg 88:

D1 = s^2/(1500d-s)           D2 = s^2/(1500d+s)

Where:

1500 = the constant for the allowable angle of confusion as in
       1 part in 1500  (the eye is nominally good to 1 part in 3400)

s = the distance to the in-focus object from the entrance pupil
    (essentially the distance at which the camera is focused)

d = the linear aperture of the lens (diameter of entrance pupil)
    (= focal length divided by f/number)

D1 = the range past the in-focus object 
     which is still in acceptable focus

D2 = the range nearer the in-focus object 
     which is still in acceptable focus

Note what happens when s is set to 1500d.  D1 goes to infinity.  
This value of s is the famous hyperfocal distance.  

Kingslake used 1500 in this example but 1400, 1000, and 2000 are just 
as valid depending on what level of resolution your particular optical 
system (from photons in to photons out) will support.

To find the far distance and the near distance, just add D1 to s and 
subtract D2 from s.

As an example, I think we were taking about f/16 on a stereo camera 
with a 36 mm focal length lens.  The f/16 is marked for the case 
wherein the lens is set at infinity (focal length = 36) so we can get 
the exact linear aperture, d, as follows:

d = 36/16 = 2.25 mm

Now let's find the hyperfocal distance:

s = 1500d = 1500*2.25 = 3375 mm = 11.07'

How close can we have an object and still have it in acceptable focus?

s - D2 = s - s^2/(1500d+s) = 3375 - 3375^2/(1500*2.25 + 3375) 
       = 1687.5 mm = 5.54'

Note that this new distance is half the hyperfocal distance.  It always 
will be if the camera is focused at the hyperfocal distance but I thought 
it would be nice to show you how to use the formula in case you didn't 
want to have the back end equal to infinity.  If anyone reads this far, 
please let me know off-list.  And so we see yet again why we should all own 
Dr. K's book.

John B


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