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

Started Oct 17, 2011 | Discussions thread
Rikke Rask
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Re: Understanding the significance of the math
In reply to boggis the cat, Oct 22, 2011

boggis the cat wrote:

Rikke Rask wrote:

We are discussing the relative efficiencies of sensor area. This is not measured in exposure terms.

You really have a bee in your bonnet about this.

Refer to my reply to Joe down-thread, where I lay out the calculations for the "conversion efficiency" of the two different aspect ratios.

Which unfortunately for you displayed your utter ignorance.

Are liters an exponential measure, Boggis?

The exponential nature is irrelevant. One stop appears to double or halve brightness.
It is the effect that is important.

Twice the area means √2 times better S/N ratio. Nothing apparent about that, Boggis.

If you have 1.2 litres of water (or anything else) you have nearly double 0.67 litres.

Yes.

An exposure increase of 1.2 stops is nearly double an exposure increase of 0.67 stop.

No, silly - f/4 is two stops more than f/2; quadrupling.

The number of photons collected is proportional to the area: exp₂(1.2) ≈ 2.3 and exp₂(0.67) ≈ 1.6. And 2.3 is about 44% more than 1.6; that's pretty far from a doubling. Incidentally these factors correspond to differences in S/N ratios of a factor of 1.5 and 1.3, respectively; both figures are close enough to √2 ≈ 1.4 for our purposes.

If we use Nikon DX instead of Canon APS-C, the figure will be exp₂(0.73) ≈ 1.7 and 2.3 is only 35% more than 1.7. The 1.7 figure also corresponds to a change in S/N ratio of 1.3, that's how utterly insignificant the difference between DX and Canon is. Why don't you try to "prove" your silly agenda using Sigma sensors instead?

I repeat my previous questions: What is the accuracy of the shutter speed and aperture labeling? Those are questions that ought to interest a "metrologist".

For practical purposes, DX/APS-C is about one stop smaller than FF and FT is about one stop smaller than DX/APS-C. We don't know the exposure parameters with any degree of accuracy that makes it useful to be more accurate than that.

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Rikke

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