Dynamic Range -- what it is, what it's good for, and how much you 'need'

Started Oct 17, 2011 | Discussions thread
boggis the cat
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In reply to Great Bustard, Oct 21, 2011

Great Bustard wrote:

boggis the cat wrote:

The difference between systems in terms of light collecting ability for lenses using the same f-ratio and photos displayed with the same area is quite simple:

  • 4/3 vs 1.6x: log2 (332/225) = 0.56 stops more light on the 1.6x sensor.

  • 4/3 vs 1.5x: log2 (372/225) = 0.73 stops more light on the 1.5x sensor.

  • 4/3 vs FF: log2 (864/225) = 1.94 stops more light on the FF sensor.

Not quite, because of the efficiency of the aspect ratio.

These calculations ignore the additional aspect ratio efficiency of 4:3 over 3:2, which is roughly 4%. This changes e.g. the Canon APS-C to FT ratio from 0.55 to 0.49 "stops", and the 135 to FT ratio from 1.94 to 1.89 "stops".

Remember that the aperture is the diameter of the image circle, so the closer to 1:1 the aspect ratio is the more of the light from the image circle is being used (efficiency).

This ends up being 4:3 having 4% greater efficiency than 3:2.

Actual calculations for efficiency of the aspect ratios being considered:

  • 4:3 aspect ratio: efficiency = 12 / (25/4) x pi = 0.6112

  • 3:2 aspect ratio: efficiency = 6 / (13/4) x pi = 0.5876

So we get a 4% increase in efficiency from the 4:3 aspect ratio, and this must be factored in to "equivalence" to make the starting conditions correct.

Revising your calculations:

  • 4/3 vs 1.6x: log2 (332/ (225 x 1.04) ) = 0.5 stops more light on the 1.6x sensor.

  • 4/3 vs 1.5x: log2 (372/ (225 x 1.04) ) = 0.67 stops more light on the 1.5x sensor.

  • 4/3 vs FF: log2 (864/ (225 x 1.04) ) = 1.88 stops more light on the FF sensor.

This actually makes comparing FT to the two APS-C variants easier: compared to Canon, FT is down a 1/2 "stop"; compared to 1.5x, FT is down 2/3 "stop".

Likewise, we can repeat for the systems displayed at 4:3 (or more square):

  • 4/3 vs 1.6x: 2 x log2 (14.9 / 13) = 0.39 stops more light on the 1.6x photo.

  • 4/3 vs 1.5x: 2 x log2 (15.7 / 13) = 0.54 stops more light on the 1.5x photo.

  • 4/3 vs FF: 2 x log2 (24 / 13) = 1.77 stops more light on the FF photo.

and for the systems displayed at 3:2 (or wider):

  • 4/3 vs 1.6x: 2 x log2 (22.3 / 17.3) = 0.73 stops more light on the 1.6x photo.

  • 4/3 vs 1.5x: 2 x log2 (23.7 / 17.3) = 0.91 stops more light on the 1.5x photo.

  • 4/3 vs FF: 2 x log2 (36 / 17.3) = 2.11 stops more light on the FF photo.

This is all for cropping , and is not what I am pointing out.

If you want the correct differences you have to take account of the "conversion efficiency" of the sensor aspect ratio with respect to the image circle.

I define "trivial" as any difference less than 1/3 of a stop (0.33 stops), so none of the differences are "trivial", but the difference between 4:3 and 3:2 most certainly has a trivial effect.

But, let's put this into some persective by comparing the f-ratios of lenses on the respective formats that would result in the same light on the sensor for the same shutter speed and the same DOF for the same perspective and framing (equivalent settings):

  • f/2 on 4/3 --> f/2.5 on 1.6x

  • f/2 on 4/3 --> f/2.7 on 1.5x

  • f/2 on 4/3 --> f/4 on FF

These are all slightly out, too.

Is the difference "significant"? Well, that's for each individual to decide for themselves.

My point is simply that if you are going to calculate something out accurately you should correct for the aspect ratio efficiency as well.

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