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Wednesday, March 12, 2014

••◊ Calculating Depth of Field

I was on a shoot this last weekend and we were having massive focus problems due to equipment issues.  Another cinematographer on the shoot suggested that I just give up and close down the aperture to get more depth of field.  Well..."how much?"  So today I'm setting out in an endeavour to further understand depth of field/focus depth.

The first factor that you need to take into account is the circle of confusion (CoC).  This factor is determined using a number of assumptions related to viewing conditions and determines the amount of "acceptable" blur given a sensor size.  For instance, most cameras I deal with now days have a sensor size approximately the same as APS-C photography sensors.  So given the generally accepted d/1500 standard, that gives us a circle of confusion of 0.018mm.  HOWEVER...and this is a big however...there are different suggested rules of thumb for CoC and even different CoC's suggested by brand of lens.  For instance, Zeiss suggests 0.025mm CoC for their Master Prime lenses.  Some people estimate 0.02mm and some estimate as much as 0.05mm for 35mm movie film.  That's a great deal different!  So I'm just going to show example calculations using the d/1500 standard so I don't start an argument on who's right!  Perhaps the ASC has something to say on this(?)

Given the CoC figure, there are a few formulas to calculate the depth of field.  We first calculate the hyperfocal distance.

H = f^2/(N * CoC) + f

H is the hyperfocal distance
f is the focal length of the lens (i.e. 50mm...etc)
N is the lens number (i.e. f/1.4 or f/11...etc)
CoC is the circle of confusion

So let's say we are photographing a person with a 50mm lens set to f/4.0.  That gives us H = 2500/(4*0.018) + 50 = 34772 millimeters or approximately 34.8 meters.

Next we'll calculate the near (Dn) and far (Df) distances that are sufficiently in focus given our chosen circle of confusion (CoC).  The quick formulas assume practical values of H are larger than the subject distance.

Dn ≈ (H * s)/(H + (s-f))

With the actual formula being...

Dn = (s*f*f)/(f*f + N*c*(s - f))...if you need to be more exact for whatever reason.

H is the hyperfocal distance
f is the focal length of the lens
N is the lens number
s is the distance from the camera to the subject being photographed.

Again, let's choose an arbitrary distance of 3 meters (3,000 mm) for the subject distance.  This gives us Dn ≈ (34772*3000)/(34772+3000) ≈ 2765 millimeters or about 2.77 meters.

Df  ≈ (H * s)/(H - (s-f)) given that s < H

With the actual formula being...

Df = (s*f*f)/(f*f - N*c*(s - f))...if you want to geek out.

At 3 meter subject distance that means Df ≈ (34772*3000)/(34772-3000) ≈ 3278 millimeters or 3.28m.  In all that gives us a depth of field of about 0.52 meter.

There are some other interesting findings from this formula.  Let's say the person is standing exactly at the hyperfocal distance - 34.7m in this example.  Then the Dn ≈ H/2 ≈ 17.4 meters and Df =∞.  Also if the person stands beyond the hyperfocal distance then Df will continue to be infinity, however closer objects are still subject to the focus limits of Dn.

I applied these formulas to the Arri Master Prime depth of field chart and found a great deal of discrepancy at 14mm and at 50mm.  But then I remember that these are T-numbers that take into account the efficiency of the lens as well as aperture.  So at T/4 the 50mm lens has an approximate f-number of f/3.3.  Makes sense.  Something else must be going on with wider lenses because I couldn't get the formulas close to their chart numbers. 

Total geek out over.

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