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

Started Oct 17, 2011 | Discussions thread
boggis the cat Veteran Member • Posts: 6,329
Only 96% correct

Great Bustard wrote:

boggis the cat wrote:

Great Bustard wrote:

  • 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.

That, and the rest of my post, was 100% correct.

No, it was 96% correct.

Add in the aspect ratio correction to get to 100%.

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".

They ignore nothing. You simply don't understand what they are saying (a familiar story). Here -- I'll spell it out to you. For the figures above, the assumption is that we are taking a pic of the same scene with the same exposure, so that the area of the scene in the focal plane is the same for both systems.

4:3 covers a greater area than 3:2 for the same diagonal. It is the sensor diagonal, and thus the required image circle, that relates to relative aperture.

Remember that the aperture is the diameter of the image circle...

That's wrong, right from the start. The aperture is most certainly not the diameter of the image circle.

No, it isn't. Sorry for any confusion. I was trying to make a jump to the important consideration and worded that poorly. (An explanation of how the actual image circle -- which may not be circular due to internal baffles or lens design -- projected at a given f-number is calibrated to yield the correct exposure for the sensor -- circumscribed by the "image circle" -- seemed needlessly complicated, and I assumed that you'd make the same logical leap.)

The f-number relates to the amount of light per second reaching the image circle required to circumscribe the sensor. The important part is that it is the circle circumscribing the sensor .

Varying the aperture varies the amount of light reaching the sensor. A given f-number on a system that can capture more of the light by area is more efficient in "equivalence" terms, is it not?

If you believe it isn't, then explain why. (It is capturing more of the "total light" projected through a given aperture.)

...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.

No. A 4:3 rectangle inscribed in a circle with the same diameter as a 3:2 rectangle will have 4% more area.

It covers 4% more area of the image circle. The aperture setting adjusts the amount of light projected onto that circle.

This does not mean "4% greater efficiency" unless the final photo is displayed at 4:3 or more square. If the photo is displayed 3:2 or more wide, then the 3:2 rectangle is "more efficient". This has been explained to you multiple times, but you still lack the cognitive capacity to understand it.

That's a crop .

Cropping is quite different in effect, because it is not part of the exposure process.

My point is...

...simply incorrect. My post was completely correct, as written. Whether you understand that or not, however, is another matter entirely.

I seem to recall having this exact same argument with you years ago, where you finally agreed that there is a 4% difference but claimed that it isn't significant.

My argument is that if you want to go to the trouble of making calculations then you should correct for this, too.

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