[normhead]

No, but what you did claim was that FF sees it's image SNR advantage because the pixels were bigger - right?

What I was illustrating was that it still holds an advantage even if the pixels are the same size, demonstrated by the D800 vs D7000 - because the vast majority of the delta between FF and aps-c of the same gen is not about the pixel size, it's about the (drum roll) Total Light. (How about that, I'm staying on-topic with the thread title, not always a guarantee with me )

Anyway - what 'correct answer' are you looking for specifically, the exact sizes of the photodiodes under the microlens? I might be able to find that.

Does anyone have a clue why that is? Why should a D810 have such an advantage over a K-5 when they have practically the same pixel pitch. There's obviously way more to the technology than just pixel size. It has to be the chemistry of the light sensitive materials or the construction of the sensors.. It's a mystery Charlie Brown. Using a pixel level image, 1:1, they should be the same, but they aren't.

Total light does not explain this. Because the amount of light on same sized pixels is the same.

[jsherman999]

Does anyone have a clue why that is?

Why should a D810 have such an advantage over a K-5 when they have practically the same pixel pitch.

Because of Total Light. The larger sensor sees more light for the same exposure. In an image displayed at the same size, that results in a less noisy image.

Using a pixel level image, 1:1, they should be the same, but they aren't.

Well... yes they are, pretty much. Why do you think they're not?

Below is the 'screen' tab from DXO - this basically shows you the per-pixel performance, Total Light taken out of the equation:



It also describes what equal-area crops of images from each sensor would look like, or 100% views from each.

Total light does not explain this.

Total Light explains this:



That's the 'print' tab, and represents how the whole image displayed on screen or printed at the same size would compare.

[ElJamoquio]

Does anyone have a clue why that is? Why should a D810 have such an advantage over a K-5 when they have practically the same pixel pitch. There's obviously way more to the technology than just pixel size. It has to be the chemistry of the light sensitive materials or the construction of the sensors.. It's a mystery Charlie Brown. Using a pixel level image, 1:1, they should be the same, but they aren't.

Total light does not explain this. Because the amount of light on same sized pixels is the same.

There's zero mystery. It's completely explained.

[KevinR]

Because of Total Light. The larger sensor sees more light for the same exposure. In an image displayed at the same size, that results in a less noisy image.

Well... yes they are, pretty much. Why do you think they're not?

Below is the 'screen' tab from DXO - this basically shows you the per-pixel performance, Total Light taken out of the equation:
..

Well done with this detail. Very emphatic data. If people don't get this IMO they never will. I tried a simple principles approach in post 17 explaining almost exactly the same effects, but wow did the thread take on a life of its own. As a saying goes, "everyone is entitled to their own opinion, but not their own facts".
Read more at: https://www.pentaxforums.com/forums/169-pentax-full-frame/275535-total-light-...#ixzz3IRvL7ysZ

Watching this thread mushroom out of all proportion to the complexity of the principles involve had me humorously comparing in my mind to trying to explain fractions and percentages to my kids several year ago before they finally "got it"..

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Last edited by KevinR; 11-07-2014 at 09:55 PM.

[Ian Stuart Forsyth]

Because of Total Light. The larger sensor sees more light for the same exposure. In an image displayed at the same size, that results in a less noisy image.



Well... yes they are, pretty much. Why do you think they're not?

Below is the 'screen' tab from DXO - this basically shows you the per-pixel performance, Total Light taken out of the equation:



It also describes what equal-area crops of images from each sensor would look like, or 100% views from each.



Total Light explains this:



That's the 'print' tab, and represents how the whole image displayed on screen or printed at the same size would compare.

I was late to the thread so I'll let my good friend Porky Pig sum it up

You Tube

You Tube

[LensBeginner]

Because of Total Light. The larger sensor sees more light for the same exposure. In an image displayed at the same size, that results in a less noisy image.



Well... yes they are, pretty much. Why do you think they're not?

Below is the 'screen' tab from DXO - this basically shows you the per-pixel performance, Total Light taken out of the equation:

Well.. it makes perfect sense... same-generation cameras perform almost the same (except for - mostly unwanted - features like before-RAW-NR)...
Thanks for posting the graphs!

[normhead]

As I thought, I must have been lost in the semantics somewhere.

[bxf]

Why do we see references or comparisons of D800/D7000 rather than D800 vs D800 crop-mode? Wouldn't the latter be the obvious comparison?

And, all the explanations, that appear obvious to those writing them, do not explain why more light over a larger area improves the quality of a given unit of area, that common sense dictates should not be affected. I'm not disputing the evidence, but I still don't understand. To say "FF is better because of more Total Light" does not tell me how the distribution of this larger amount of light is different from the distribution of a smaller quantity of light over a smaller area. Having said that. perhaps this is very simple to someone who understands light physics.

[normhead]

OK, here's how I understand it....it's not that the crop sensor of a D810 is better than a K-5, it's that if you shoot the image with a D810 and the same FoV with a K-5 (or D810 in crop mode) ... remember, it's the same field of view, there is twice as much light on the D810's sensor to make the same image...

So say you have noise in your K-5 image... the image has to be twice the size.. so the noise will be twice as apparent.

My guess is that 400 ISO or below this makes no difference at all. But once you go say to 1600 ISO taking 800 as the cutoff for acceptable noise in APS-c, because you've used twice the light, you have half the noise in the FF image.

Obviously some of these small backlit sensor can go against the grain her somewhat, and different sensors have different sensitivities there are all kinds of other variables but essentially... a sensor like the Sony A7s should be (and is) pretty much noiseless to a very high ISOs, because it's a large amount of light on a very large sensor site.

[Rondec]

OK, here's how I understand it....it's not that the crop sensor of a D810 is better than a K-5, it's that if you shoot the image with a D810 and the same FoV with a K-5 (or D810 in crop mode) ... remember, it's the same field of view, there is twice as much light on the D810's sensor to make the same image...

So say you have noise in your K-5 image... the image has to be twice the size.. so the noise will be twice as apparent.

My guess is that 400 ISO or below this makes no difference at all. But once you go say to 1600 ISO taking 800 as the cutoff for acceptable noise in APS-c, because you've used twice the light, you have half the noise in the FF image.

Obviously some of these small backlit sensor can go against the grain her somewhat, and different sensors have different sensitivities there are all kinds of other variables but essentially... a sensor like the Sony A7s should be (and is) pretty much noiseless to a very high ISOs, because it's a large amount of light on a very large sensor site.

I prefer to think of it as a magnification issue. The problem with the K5 is not that it is noisier, per se, it is that you see its pixels quicker when you enlarge the photo/put it on a bigger monitor.

If you aren't magnifying your image much, then the differences will not be visible.

[Ian Stuart Forsyth]

Why do we see references or comparisons of D800/D7000 rather than D800 vs D800 crop-mode? Wouldn't the latter be the obvious comparison?

And, all the explanations, that appear obvious to those writing them, do not explain why more light over a larger area improves the quality of a given unit of area, that common sense dictates should not be affected. I'm not disputing the evidence, but I still don't understand. To say "FF is better because of more Total Light" does not tell me how the distribution of this larger amount of light is different from the distribution of a smaller quantity of light over a smaller area. Having said that. perhaps this is very simple to someone who understands light physics.

I think the best way to show how total light affects the image quality is to see what happens if we shoot 2 images that capture the same total light between 2 different crops
These are using a crop factor of 2 times but it will work for our intended proposes, both images are scaled to the same output size.
The first photo was taken with a D800 140mm iso 3200 F10 1/20 sec

The second D800 70mm iso 800 F5 1/20 cropped to 140mm FOV

With the first image FF when we reduce the light intensity to ( F/10) the total amount of light that the surface area of the sensor captures will be made up of the same total light as crop image using a higher light intensity of (F/5) but using a sensor ¼ the size.

If we take a look at the images as you can see that the appearance of the noise closely matches one and other even when we decrease the light intensity being projected on to the larger surface area. This also shows that iso across formats is completely irrelevant and that total light that is captured in the image is what determines the shot noise we see in the image.

[bxf]

It's too late here to try to digest the last two comments - I will do that tomorrow. But meanwhile, can anyone say if the following makes sense?

To get the same FOV with the crop sensor, you have to move further away from the subject. Hence the reflected light has longer distance to travel to the camera, which must(?) result in lower intensity, no? But then at the same time, since the FOV is exactly the same, I'd expect the exposure to be the same with crop vs FF. Is this a contradiction, or does the greater distance actually result in "less Total Light"? Or is this thought totally irrelevant crap?

[normhead]

That's one way to look at it if that appeals to you. You're further away if you project say an image from a 70 mm lens onto an FF sensor and then move back until the image fills an APS-c sensor. The image will be just as bright, but it will be half the size, so half the light. As you move away from an object, it will appear to get smaller and smaller.

The other way of doing it is to use a wider angle lens which would spread light of the same intensity over twice as much surface area, with only the middle portion being used by the APS-c sensor.

At 100 or 200 ISO this makes almost no difference, the higher you go in ISO the more difference it makes.

If you are a person who really needs excellent low light performance... go for a Sony A7s. But almost any FF is better than APS-c from about 800 to1600 ISO up. That just what you give up to get what APS-c has to offer.

[jsherman999]

Why do we see references or comparisons of D800/D7000 rather than D800 vs D800 crop-mode? Wouldn't the latter be the obvious comparison?

You are right, and I always talk about D800 in crop mode, but sometimes people have a tin ear to that argument - they think it's tangential to "aps-c vs FF" or whatever the main discussion thread is about. A good % of the time when I bring it up someone says something like "I'm not talking about cropping here" and then the discussion goes off on a different tack as we try to discuss why cropping is pertinent.

And, all the explanations, that appear obvious to those writing them, do not explain why more light over a larger area improves the quality of a given unit of area, that common sense dictates should not be affected.

It doesn't improve the quality of a given unit of area, it improves the entire image, because more photons were captured for the same exposure. The resulting image was created with more light.

Keep in mind that when someone says 'larger sensor gets more total light than smaller,' the same FOV and distance to subject - same "framing" - is being assumed in that comparison. It's assumed that the photographer is shooting the same subject from the same distance at the same exposure.

Ex: Take a portrait at 50mm 1/100s f/2.8 (FF) and one at 35mm 1/100s f/2.8 (aps-c) from the same distance, and the FF image has been created using a larger physical aperture (17.8mm) vs. the aps-c image (12.5mm.) The FF image was created with more total light as a result of that, the volume of photons that was collected in the frame over the same period of time was higher, and (if the sensors are similar-gen) that image will be less noisy.



.

---------- Post added 11-09-14 at 10:19 PM ----------

To get the same FOV with the crop sensor, you have to move further away from the subject. Hence the reflected light has longer distance to travel to the camera, which must(?) result in lower intensity, no? But then at the same time, since the FOV is exactly the same, I'd expect the exposure to be the same with crop vs FF. Is this a contradiction, or does the greater distance actually result in "less Total Light"? Or is this thought totally irrelevant crap?

You're introducing other things now - if you move further away from the subject you're reducing the light intensity. That's not what we're talking about.

The same FOV is achieved simply by using a different FL from the same distance. (You can move away too but you change the light intensity and the comparison can really handicap the aps-c combo, and it's not what's being described in the first place.)

[Ian Stuart Forsyth]

It's too late here to try to digest the last two comments - I will do that tomorrow. But meanwhile, can anyone say if the following makes sense?

To get the same FOV with the crop sensor, you have to move further away from the subject. Hence the reflected light has longer distance to travel to the camera, which must(?) result in lower intensity, no? But then at the same time, since the FOV is exactly the same, I'd expect the exposure to be the same with crop vs FF. Is this a contradiction, or does the greater distance actually result in "less Total Light"? Or is this thought totally irrelevant crap?

FOV can only be altered by cropping or changing focal lengths, while moving the location where the photograph was taken from the subject changes framing and perspective.

If you look closely at the details of how the images are captured you would see that the FF image was taken at 140mm lens and the 2 X crop was taken with a 70mm lens the camera did not change its location. The intensity of the light that is being projected onto the sensor was altered using a different F stop F/10 for the FF and F/5 for the 2 xcrop.

This brings us to what determines the total amount of light that falls onto the sensor the virtual aperture.
We can calculate the virtual aperture using the formula that is found on the lens F/5 (f-ratio) is the quotient of the focal length and the virtual aperture

we know the F in the formula is the focal length divided by 5= virtual aperture. For the first image using 140mm divided 10 = 14mm dia virtual Ap. With the image we are using 70mm divided by 5 = 14mm dia virtual Ap. This is why the 2 image have the same appearance of shot noise. The same amount of total light was projected by both lenses, on one format the light intensity was greater but was projected on to smaller medium while the other had less light intensity projected onto a larger medium.