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

Started Oct 17, 2011 | Discussions thread
Great Bustard
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In reply to boggis the cat, Oct 18, 2011

boggis the cat wrote:

Why doesn't the 5D II -- with all of that extra "total light" that you rave about -- crush the D7000 with its tiny, inferior, sensor?

It's really quite simple, boggis. It comes down to the definition of DR that was given in the OP from three independent sources.

You see, the sensor in the D7000 is far from inferior -- it's the best sensor per area on the retail market in DSLRs at the moment, but larger sensors can outperform it on the basis of sheer size in many important measures, but not DR.

So, indeed the 5D2 sensor will have less noise for a given exposure, but not more DR. And this failure on your part to understand the connection between DR and noise is a stupendous display on ignorance on your part. Not because you don't know, but because all you can do is fling insults at the people who do know.

Come on, give us your best floppy-shoe clown dance.

Well, since we know that neither math, physics, optics, nor reason are your strong suit, I'm not sure how much good it will do for you, but I'll give a quick explanation and work a quick example.

The DR is measured as the number of stops from the noise floor to the pixel saturation. Some use the read noise of the pixel for the noise floor, others (such as myself), use the 100% NSR (Noise-to-Signal Ratio -- where the signal is equal parts noise and information). The difference between the two measures is not significant until the read noise is low (maybe 1-2 electrons or less), and no DSLR has a sensor that efficient yet.

Still, I'll compute the DR for each using both measures at ISO 100.

Using the pixel read noise as the noise floor:

D7000: DR = log2 (49058 / 3.1) = 13.9 stops
5D2: DR = log2 (64600 / 27.8) = 11.2 stops

Using the pixel 100% NSR for the noise floo (DR100):

D7000: DR = log2 (49058 / 3.64) = 13.7 stops
5D2: DR = log2 (64600 / 28.3) = 11.2 stops

Of course, a more meaningful measure is not pixel-level DR, but DR / area -- that is, to compute the DR for the same portion of the photo. DxOMark likes to normalize for 8 MP with their "print" DR, by using one eight-millionth of a photo rather than a pixel, but I prefer to use one millionth of a photo instead -- DR100 / µphoto -- as the "8" just seems cumbersome.

I'm sure you can do the math to see what difference it makes on the score, right?

This advantange of the D7000 is a direct result of the sensor being very efficient at base ISO, likely as a result of the ADC having a DR as wide as the sensor. At higher ISOs, where less light falls on the sensor, the DR bottleneck imposed by the 5D2 becomes less of an issue, and by ISO 1600, the 5D2 surpasses the D7000 in DR.

Of course, you already knew that and were just testing me, right?

Let us know when you've figured out the whole "theory meets reality" concept.

Right from the first link I posted in the OP, showing the 5D2 compared to the D7000:

http://forums.dpreview.com/forums/read.asp?forum=1032&message=38379069

which, not surprisingly, seems to be well represented by the computed numbers. I would call that quite a good example of "the whole 'theory meets reality' concept", and an epic failure on your part, unless, of course, the purpose of your post was nothing but empty self-harming entertainment, in which case you can proudly claim "mission accomplished" once again.

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