You can say fact all you want, but I don't see it in practice.
Would you like me to demonstrate it?
IF the "fact" is that a totally blown out image with little usable data has less noise...
Whether there is blown out data or not depends entirely on the scene. And in any case, the decision to choose between blown highlights and noisy shadows should be left to the photographer.
For example, autoexposure on Canon cameras tries to get about 3 stops of headroom above middle grey before clipping. If a photographer must use lower light intensity (e.g. f/5.6 instead of f/2.8 ) for whatever reason (DOF, motion blur, low light, etc.), then he will now have *5* stops of highlight headroom at ISO 100. For some scenes, this is just right: any increase in ISO would blow important highlights, and the photographer may choose to live with the increased noise to just to keep them. Other scenes have much less contrast. The highlights are all contained within just a single stop. That allows the photographer to trade *four* full stops of highlight headroom to get less noise by changing from ISO 100 to ISO 1600. On the high contrast scene, that would have blown 4 stops of highlights. But on this scene, nothing is lost, and the photographer benefits greatly from reduced shadow noise.
This is the essence of "ETTR then ITTR". Increase light intensity as much as you can, without blowing important highlights, then increase ISO as much as you can, without blowing highlights. That technique will always result in the lowest amount of noise.
Well that's not of use to anyone as in practice that image is far less useful than one with a little noise.
For some people, such as those that never shoot in low light, it will not be of any use. They'll never need to learn the implications of ISO, since they'll never go above ISO 100.
Even if one is are already doing the "ETTR then ITTR" technique, and understanding all this changes nothing for a person, at least now that person understands it correctly. To me, that's important.
It does.
If one is 4 stops blown that is not what I mean by same exposure,. I'm referring to keeping the exposure = re: metering taking the boosted ISO into account.
FWIW, that's not how photography books (and especially textbooks) define exposure. (Speak of the angel, look at the next post)
i hate to break out the dictionary, but that's as plain as it gets. the most scientifically correct definition of an exposure refers to the total amount of light, which includes duration or period of illumination.
That's correct. The dictionary is correct in this case, but a photography textbook would be more authoritative (as wel as being clearer and more specific).
Yes it is. And it is equivalent in meaning to exposure. The dictionary does not explain that the context is per-area, though it is. That is "total light" means "the total light per area". (The reason it says "total light" is to include effects of *everything*, not just shutter and f-number, but including ND filters, scene luminance, lens transmissivity, etc.)
you could argue that my usage of it by referring to the resulting image is false since technically the sensitivity of the medium the exposure is collected on does not technically qualify as being part of the collective term of "exposure",
Precisely.
but the last line clearly shows how using the term in this manner is acceptable.
No it does not. It says "the image resulting from the effects...". It does not say "the brightness of the image resulting from the effects...", which is what it would say if it included sensitivity.
are you sure you're not just trying to refit the word into what YOU perceive it to be?
Yes, I am sure.
..are you serious? in a properly exposed photo, a higher ISO will garner higher visible noise. nobody cares about a theoretical frame where you're overexposed three stops of light to prove that ISO1600 renders less noise per electron than a well-exposed ISO200 shot.
We certainly do care. If we didn't, we should shoot ISO 100 and just push +4 EC in post. But that results in far more noise than shooting ISO 1600. The reason is that ISO 1600 causes less read noise.
My understanding is that digital sensors have a native level of light sensitivity that corresponds to an ISO.
Yes. (I would use different vocabulary here and elsewhere in your post, but let's put that aside.)
I believe most canon DSLRs are enginered so that ISO 100 is the native light sensitivity of the sensor.
Agreed.
Higher ISO's increase light sensitivity by amplifing the electrical signal received from the sensor.
OK, here I have to take issue with the vocabulary. Higher ISO does amplify the signal. But it does *not* increase the light sensitivity. For a given sensor, the light sensitivity *never* changes. The only thing that can change is how much electronic read noise gets *added* to the signal by the sensor. At high ISO, it adds less read noise.
The noise gets amplified also. However when high ISO's are traditionally used, low light, there is less signal to the same amount of noise.
Close. When the light intensity is reduced (such as when you're shooting in manual, and you're going to change to high ISO, but you haven't yet, all you did is stop down), there is less signal, which increases photon shot noise and reduces the distance between signal and electronic read noise. This results in a *very* noisy image. But after you increase ISO to compensate for the reduced light intensity, the Canon sensor adds less electronic read noise. This improves the signal to noise ratio, but not nearly enough to compensate for the drop in light intensity.
Both the signal and noise get amplified but because there is less signal the noise appears to be greater.
Correct.
There are different ways to compare ISO's and noise.
Setup 1:
There is the same amount of ambient light on the photo subject.
Shot 1: ISO=100, Shutter=1/60th, f/2.8
Shot 2: ISO=800, Shutter=1/1000th, f/2.8
In this setup the second shot will have more noise.
Agreed. It has lower light intensity.
Setup 2:
There is the same amount of ambient light on the photo subject.
Shot 1: ISO=100, Shutter=1/60th, f/2.8
Shot 2: ISO=800, Shutter=1/60th, f/11
In this setup the second shot will have more noise.
Agreed. It has lower light intensity.
Setup 3:
The first shot has 4-stops more light on the photo subject than the second has.
Shot 1: ISO=100, Shutter=1/60th, f/2.8
Shot 2: ISO=800, Shutter=1/60th, f/2.8
In this setup the second shot will have more noise.
Agreed. In the next quote you refer to the (incorrect) definition of exposure, which is "brightness". I am going to change that for you to make it clearer for the reader:
Setup 4:
There is the same amount of ambient light on the photo subject.
Shot 1: ISO=100, Shutter=1/60th, f/2.8
Shot 2: ISO=800, Shutter=1/60th, f/2.8
In this setup one of the images will either be [darker] or [LIghter] or both will not have the same [Brightness].
They both have the same light intensity. But shot 2 will be brighter. Shot 2 will have less noise. (Try it if you don't believe me). Shot 2 will also have less highlight headroom (i.e. greater chance of blown highlights).
Try it. It has less noise and less highlight headroom. (Which *probably* means blown highlights, but not always.)
I point this all out to say that higher ISOs do have more noise If you have proper exposure on both images.
It's not.
how do you draw the conclusion that the lower measurable noise is due to the electronics/mechanics of higher iso setting?
By demonstrating that it is the same even when *nothing* is blown.
The cause of lower noise could be ( likely is IMHO) the loss of image data.
This is disproven by doing the same comparison in low light, where highlights are not blown in either image and the ISO 1600 shot still has less noise than the ISO 100 shot when both have the same light intensity.
I hope you guys can look back and laugh at all this after you realize I'm correct. 
