[Sandy Hancock]

But the DOF and bokeh (of a a lens) are still a function of the crop factor (and therefore sensor size) because they are impacted by the distance you can frame a photo with a focal length from.

No. Only the field of view of the recorded image changes.
The depth of field and bokeh are unchanged. You just appreciate them differently because of the crop factor.

[redrockcoulee]

Luckily the mental gymnastics are quite a bit easier than that. As has been covered many times on this forum; 77mm f/1.8 on 1.5 crop APS-C camera is the equivalent to shooting with a 116mm f/2.7 on FF assuming the sensor tech (QE) is the same (eg D7000 & D800). You really do just multiply the focal length and aperture to get an equivalent lens.

There's no light gathering advantage on APS-C because the FF sensor has a 1 stop ISO advantage owing to it's larger sensor suface area.

I do not understand your last statement. in the film world there is no difference in exposure when I move my 120mm from 4X5 to 5X7 or to Whole Plate. It does go from being a mild wide to a regular wide to a very wide lens in terms of field of view but the exposure will not change even though the whole plate has almost three times the film area. If I use an external light meter the exposure is the same on my digital as my film Pentax camera or my medium or large format cameras and I would suspect on medium format digital as well. Why would a 1.8 lens stop being a 1.8 lens on a full frame digital camera? Guess I should borrow my friend's D3 and try it out with the same lens and exposure on my D200 at the same ISO.

[aurele]

no, what he just meant was that FF have a better high iso skills. When APS-C are "good" @3200 iso, FF can give the same "grainy" result @6400iso.

[redrockcoulee]

no, what he just meant was that FF have a better high iso skills. When APS-C are "good" @3200 iso, FF can give the same "grainy" result @6400iso.

Ok that I understand but that is a function of the sensor. Will not today's newst APS-C be better than yesterdays FF at high ISO. In addition my understanding is the D800 is not going to be as good at high ISO as say the D7000 and certainly not the D4. Medium format digital have larger sensors and not as good high ISO. Just seems a very weird way of comparing lenses or describing lenses. The speed of my 50F 1.7 is the same for example on my K-r, my MZ5n or my wife's K10D and yet at high ISO the grainy results would be different on all three cameras, the MZ5n dependent on which film/developer combination I choose and the K-r much better than the K10D at above 1600 which the later cannot even do.

Seems much clearer to say a lens is its particular focal length, that you would get an equivalent focal length on a different film/sensor size and that certain cameras are better at high ISOs. In comparing APS-C to FF at least the aspect ratios are the same whereas in my film world we go from a 2to 1 to a 1 to 1 aspect ratio. The only reason I see for using the term equivalent is to enable some one to visualize what focal length they may want to buy or use if they are moving from one format to another or to compare what is available to two formats in helping decide if either one covers the range they require.

[Marc Sabatella]

Ok that I understand but that is a function of the sensor. Will not today's newst APS-C be better than yesterdays FF at high ISO.

Perhaps eventually, but so far, if you check actual statistics, you'll see that physics does tend to win out - differences *within* the APS-C class tend to be relatively small, as do differences *within* the FF class, whereas the difference *between* the averages - whether measured historically or just using the current models - is much larger. Omly within the last year or so have we finally gotten to the poont where the very best APS-C sensor made approaches the quality of the very worst FF sensor ever made. But concisdering that all sensors made *today* are all similar in quality, size really is the biggest determining factor.

the K-r much better than the K10D at above 1600 which the later cannot even do.

The K10D is perfectly capable of being pushed past ISO 1600 - just not via a camera dial, but via PP. this is no a significant differences, since even the K-r is just performing digital manipulation to get the sensitivty past 1600 - the same exact form of digital maniuplation you can do in PP.

And this comparison is very illustrative, as the difference between actual noise performance of the K10D sensor versus the K-r sensor is *well* under one stop (the K10D being toward the bottom of the APS-C range, the K-r near the top), and this completely pales next to the gap from the K-r to the average FF camera.

Seems much clearer to say a lens is its particular focal length, that you would get an equivalent focal length on a different film/sensor size and that certain cameras are better at high ISOs.

Clearer in what way? It makes you guess at something the actual equivalence formula is able to be quite specific about - the inherent effect of sensor size on image. The sensor *size* - independent of any other sensor attribute - has a measurable effect on FOV, on DOF, and on noise. We can measure and state this quite precisely and succinctly. Removing the precise and succinct statement with a wordier and yet less precise statement makes is less, not more clear.

[Marc Sabatella]

I'll try to explain it one last way, then I'm out. The sensor records the image that is created by the lens. But the lens will create an image, whatever the sensor behind it. It will create an image even if there is no sensor. The size of the sensor means that the (circular) image created by the lens is cropped more or less importantly. Even a FF sensor crops the image. But whatever the crop factor you use, the lens doesn't give a tiny rat's bottom. The image that is created by the lens already has all its characteristics (DOF, bokeh, etc) by the time it reaches the sensor.

Basically true, of course, except for one important clarification: DOF as actually measured in a print or on-screen image ends up being affected by sensor size since one of the components of any DOF equation is "enlargement ratio" - the degree to which the captured image is enlarged when forming the display/printed image. That makes comparison of DOF between formats more complicated.

[bdery]

But the DOF and bokeh [of a a lens] are still a function of the crop factor (and therefore sensor size) because they are impacted by the distance you can frame a photo with a focal length from.

Bokeh has no relation to the sensor.

for DOF, there's only one aspect that will affect it, but the effect is minimal in our case, I'll explain below.

what he just meant was that FF have a better high iso skills. When APS-C are "good" @3200 iso, FF can give the same "grainy" result @6400iso.

That's a simplification I don't enjoy reading.

All things being equal (similar technologies and fabrication techniques, etc), a larger pixel will yield better signal to noise performances, lowering the perceived noise. So if the pixels on a full frame camera are larger, high ISO performances will be better. But it's NOT because the sensor is full frame, it would only be because the pixels are larger.

Consider that with the large sensor resolutions being pushed, pixel size i diminishing, and the "edge" that FF sensors might have is being lowered. Also consider that there is much more R&D being done on APS-C sensors. Also consider that a larger sensor generates more noise, and you begin to see that all those energies spent wishing for FF are not that well spent

DOF as actually measured in a print or on-screen image ends up being affected by sensor size since one of the components of any DOF equation is "enlargement ratio" - the degree to which the captured image is enlarged when forming the display/printed image. That makes comparison of DOF between formats more complicated.

The only aspect influencing the DOF is the circle of confusion, dictated by the sensor resolution. Not its size. And the differences will be minimal, much less significant than all the other imaging factors.

[aurele]

it would only be because the pixels are larger.

and why between two DSLR from the same year would have such different pixel size ? Maybe because the size of the sensor is bigger. It's a snake eating his tail ...

Anyway, we are far from the original question.

[Marc Sabatella]

All things being equal (similar technologies and fabrication techniques, etc), a larger pixel will yield better signal to noise performances, lowering the perceived noise. So if the pixels on a full frame camera are larger, high ISO performances will be better. But it's NOT because the sensor is full frame, it would only be because the pixels are larger.

That's actually not true. I'm talking about identical sensor technologies, same pixel size, etc. The total surface of the light collected ends up being the determining factor. Pixel size doesn't play into nearly as much as was once commonly believed. To some extent, it does work out that more space is wasted in the distance between pixels when you have many smaller pixels as opposed to fewer larger ones, but in practice, this effect is not as significant as the effct of total surface area.

You can work through the math to see this, or you can simply perform an experiment with a single camera and cropping your images. Shoot an image at both 50mm and 35mm at a variety of different ISO settings, crop the 35 images to the same AOV as the 50, then compare. You'll find the cropped 35 compares well with the 50 at a difference of one stop of ISO. Repeat the same experiment with aperture as an additional variable and you'll find the cropped image at 35/2.8/800 matches in virtually way the image at 50/4/1600.

Also consider that there is much more R&D being done on APS-C sensors. Also consider that a larger sensor generates more noise, and you begin to see that all those energies spent wishing for FF are not that well spent

That much I certainly agree with.

The only aspect influencing the DOF is the circle of confusion, dictated by the sensor resolution. Not its size. And the differences will be minimal, much less significant than all the other imaging factors.

Not sure which technical articles on DOF you have been reading, but this also is simply not true. DOF as we see it in the final printed/displayed image is most definitely affected by enlargement ratio. Most DoF calculators allow for this by allowing you to input sensor size as one parameter, and they use that as part of the calculation of what the effective CoC actually is. That is, the effective CoC for a smaller sensor is different from what it is for a larger one, precisely because it is going to be enlarged more.

Pretty much any technical article on DOF will explain this, but this too is trivially easy to see empirically. I don't care how shallow you get the DOF to appear in a print or full screen display of an image taken with your fastest lens; show me a 20x30 thumbnail of that image and your perception of how much is in focus changes dramatically. For that matter, it is the basis behind pixel peeping. Look at something at one size it looks sharp; look at bigger it looks less so. Increased enlargement translates into decreased perception of sharpness, which in turns makes DOF seem smaller, as less of the image is now "acceptably" sharo.

[bdery]

You can work through the math to see this, or you can simply perform an experiment with a single camera and cropping your images. Shoot an image at both 50mm and 35mm at a variety of different ISO settings, crop the 35 images to the same AOV as the 50, then compare. You'll find the cropped 35 compares well with the 50 at a difference of one stop of ISO. Repeat the same experiment with aperture as an additional variable and you'll find the cropped image at 35/2.8/800 matches in virtually way the image at 50/4/1600.

In that test, you're not comparing ISO performances. You are comparing "effective pixels" (for lack of a better term) that are of different sizes. Or, put differently, you're pixel peeping with different test conditions. You cannot compare noise in that way, it's like comparing two sensors. Maybe I'm not understanding the test you propose correctly, but if I do then there are too many variables implicated for it to be conclusive.

The noise level of a single pixel is in large part determined by its size. A larger pixel will have more dark current, thus a higher level of noise offset. However, it will collect more light, which will in turn improve its measurement. Since our sensors do not operate in photon counting mode, they are essentially determining if there was zero or "more than zero" photons that hit it. Since the amount of photons actually hitting the pixel is ludicrously high in absolute numbers, a pretty good averaging can be performed. By decreasing the pixel size, you lower the amount of photons hitting it and the averaging is more prone to errors.

Not sure which technical articles on DOF you have been reading, but this also is simply not true. DOF as we see it in the final printed/displayed image is most definitely affected by enlargement ratio. Most DoF calculators allow for this by allowing you to input sensor size as one parameter, and they use that as part of the calculation of what the effective CoC actually is. That is, the effective CoC for a smaller sensor is different from what it is for a larger one, precisely because it is going to be enlarged more.

Pretty much any technical article on DOF will explain this, but this too is trivially easy to see empirically. I don't care how shallow you get the DOF to appear in a print or full screen display of an image taken with your fastest lens; show me a 20x30 thumbnail of that image and your perception of how much is in focus changes dramatically. For that matter, it is the basis behind pixel peeping. Look at something at one size it looks sharp; look at bigger it looks less so. Increased enlargement translates into decreased perception of sharpness, which in turns makes DOF seem smaller, as less of the image is now "acceptably" sharo.

I'm not trying to prove anything, but for what it's worth : I haven't "read technical articles", I have completed a PhD in physics, specialized in optics. That doesn't make me all-knowing, far from it, but I've got some basics covered

I understand what you mean. Indeed when you lower the resolution enough, everything starts to look evenly sharp. And I understand that it can be useful for a photographer to quantify this in some situations. But in that case, you're not measuring the DOF of your lens/sensor, you're measuring the resolution of your printer and your eye.

Maybe the gap here is caused by the fact that photographers and physics students don't always use the same language and metrics. For instance, f-numbers isn't used all that often in optics, many physics majors will never have used it. Because you can do optics without doing imaging science. So my metric for DOF might be tinted by my background. To me, it is a property of the sensor system. Maybe it's actually more common for photographers to quantify it on a printed standard, and that's why we're (politely ) arguing.

[Marc Sabatella]

In that test, you're not comparing ISO performances. You are comparing "effective pixels" (for lack of a better term) that are of different sizes. Or, put differently, you're pixel peeping with different test conditions. You cannot compare noise in that way, it's like comparing two sensors. Maybe I'm not understanding the test you propose correctly, but if I do then there are too many variables implicated for it to be conclusive.

I think perhaps I didn't explain sufficiently well.

The idea is to compare image where there are only two relevant variables - size of sensor and amount of noise at a given sensitivity - and to show there is a very simple relationship between those two variables.

The idea is that the actual pixels - the physical photon collecting sites on the sensor - are the same size in this comparison. They are also exactly the same distance aprt, the same physical composition, the same associated signal reading and processing hardware, etc. The *only* thing different is the size of the sensor. It's a *perfect* comparison if your goal is to isolate the effect of sensor size independent of those other factors. We seldom get such perfect comparisons in real life, but we get to see one here.

The end result will be that the image from the "smaller" sensor will show more noise for a given ISO. Yes, obviously, this is because of the greater enlargement being applied, which doesn't disporve the notion - it proves how simple the concept actually is. This is a case where literally the only difference is total surface area of the sensor. In the real world, it is not usual to find two sensors whose only diffeence is total surface area - there are usually other technogical difference. But this demonstration illustrates that the total surface area component alone has a meaurable effect, and it works out that the difference between APS-C and FF in terms of noise - all else equal - is about one stop.

. By decreasing the pixel size, you lower the amount of photons hitting it and the averaging is more prone to errors

Yes, of course. But if you reduce pixel size *without changing total surface area*, then you are obviously adding more pixels (I am discounting the fact that some sensor technologies may have larger gaps between pixels than other and assuming this too is held constant). Two sensors, same total surface area, but one has smaller pixels, means that one also has more pixels. The increased per pixel noise from the smaller wells is more or less exactly counteracted by the fact that we now have more pxiels to average together. The total amount of light collected by both sensors will be the same, hence the same total noise.

I understand what you mean. Indeed when you lower the resolution enough, everything starts to look evenly sharp. And I understand that it can be useful for a photographer to quantify this in some situations. But in that case, you're not measuring the DOF of your lens/sensor, you're measuring the resolution of your printer and your eye.

True, but that's exactly how DOF has always been measured In photography - using "typical" values for print size, viewing distance, and visual acuity to yield an appropriate CoC value to plug in to the the rest of the formula.

Maybe the gap here is caused by the fact that photographers and physics students don't always use the same language and metrics. For instance, f-numbers isn't used all that often in optics, many physics majors will never have used it. Because you can do optics without doing imaging science. So my metric for DOF might be tinted by my background. To me, it is a property of the sensor system. Maybe it's actually more common for photographers to quantify it on a printed standard, and that's why we're (politely ) arguing.

I think you have completely nailed it here, so I consider that aspect of the matter closed.

[bdery]

The end result will be that the image from the "smaller" sensor will show more noise for a given ISO.

It will show the exact same noise level per pixel. It will simply be more visible because in your test, you enlarge it more.

But this demonstration illustrates that the total surface area component alone has a meaurable effect, and it works out that the difference between APS-C and FF in terms of noise - all else equal - is about one stop.

Two sensors, same total surface area, but one has smaller pixels, means that one also has more pixels. The increased per pixel noise from the smaller wells is more or less exactly counteracted by the fact that we now have more pxiels to average together. The total amount of light collected by both sensors will be the same, hence the same total noise.

We agree that smaller pixels generate more noise per pixel.

Pixels on a sensor are not averaged together, else we'd loose resolution. Noise processing algorithms will indeed take into account neighbouring pixels, but if we take that into account we're discussing processing, not sensor technology, so I'll disregard this element.

Pixels will also be more prone to reading errors when their dimensions come near the wavelength of light. For instance, the Canon A1300 (I took a camera randomly) has a 16 MP sensor, and a quick calculation shows that pixel size is less than 1,3 microns on each side. Now, the light we see is between 0.4 and 0.7 microns. The K20D, at the other end, has pixel dimensions around 5 microns. That dimension is much larger than the other one, improving its light gathering capabilities and lowering the reading errors. A pixel smaller than the wavelength would be incapable of collecting it, for reference.

If you include processing in the equation, then not much changes, because the same processing algorithms can be applied to any sensor.

A few years ago, people were requesting full frame sensors. Today, we get performances on APS-C that are much better than FF was even 4-5 years ago. But we still wish for full frame...

[aurele]

i wish for FF not (only) for a better sensor, but mainly for bigger VF (and thus easier MF), cheap wide angle

[Marc Sabatella]

It will show the exact same noise level per pixel. It will simply be more visible because in your test, you enlarge it more.

Precisely. And in the real world, that,s how noise is perceived and measured: after enlargement. Comparing same-sized prints.

Pixels on a sensor are not averaged together, else we'd loose resolution.

Not literally, true. I meant this in a more colloquial sense - the overall aggregate effect. It really does work out that way in practice.

A few years ago, people were requesting full frame sensors. Today, we get performances on APS-C that are much better than FF was even 4-5 years ago.

Not according to Dxomark stats. We're only just barely beginning to approach where FF was 4-5 years ago. But yes, slowly but surely, technology improves. Still, *at any given point in time, comparing sensors of roighly the same technology* (or, as in my example, comparing sensors of *exactly* the same technology), sensor size has exactly the effect I have been describing.