I would like to see the math.

climbng_vine wrote:

You're wrong. You can, in fact, ignore changing viewing distance within normal use case ranges. Dofmaster confirms this, as do other calculators. Enlargement is the second least important factor, in that you generally have to double or triple the image dimensions to create an appreciable DoF difference at normal viewing distances. Changing the aperture diameter by 10% has a much clearer affect on DoF than going from a 4x6 to an 8x10.

I'd appreciate a demonstration of that claim, if you would. For example, according to DOFmaster,
http://www.dofmaster.com/dofjs.html
, the DOF for an 8x10 image viewed at 25cm on FF from a distance of 10 ft at 50mm f/2 is 0.91 ft.

The aperture diameter at 50mm f/2 is 50mm / 2 = 25mm. So, if we change that aperture diameter by 10% (22.5mm or 27.5mm), we get an f-ratio of 50mm / 22.5mm = f/2.2 or 50mm / 27.5mm = f/1.8. The DOFs for the the same scenario at these f-ratios is 0.81 ft for f/1.8 and 1.03 ft for f/2.2, which represents approximately a 10% difference in DOF.

So, if you would, please show me the math for the DOF of a 4x6 print viewed at 25 cm on FF for a subject distance of 10 ft at 50mm and f/2, and show that it has a much smaller effect on the DOF.

The fact that, in a precise mathematical sense, you cannot "ignore changing any of them" is misleading because, in the real world, some of the factors are barely perceptible except at the absolute edges, and among the rest some create difference more quickly than others. Ignoring this is just stupid.

Please, work the math for me that demonstrates your point above.

You should clarify what you mean by "assume a constant output size", because it's not true unless qualified. It assumes a constant
ratio between output size and viewing distance
, precisely because within normal ranges the difference is negligible.

If we scale the display dimensions and viewing distance by the same amount, the DOF will not change, of course. For example, an 8x10 image viewed at 25 cm and the same image displayed at 16x20 and viewed from 50 cm would have the same DOF.

However, what you said above is that changing the aperture diameter by 10% has a much stronger effect on the DOF than changing the viewing dimensions from 8x10 to 4x6, so I am assuming you mean at the same viewing distance, since, by changing our viewing distance, we can completely cancel out any change to DOF that results from changing the display dimensions.

In any event, I would really appreciate you working the math for me.

EDIT: Ah! Saved by the bell, I see -- the thread has hit the 150 post limit. Well, please start a new thread, perhaps with a title like "The mathematics of DOF", and post the results, since I really am interested to see the demonstration.