DXOMark tests the Canon EOS R image sensor, scores it at 89

Nov 2, 2016
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All true, but it is important to keep in mind that at the end of the day, it is the system that takes pictures, not a bare lens. From that perspective, there is value in reporting results from camera + lens combinations.

Having said that, part of DxO’s implied logic was that it didn’t really matter if the Canon 24-70/2.8 was better as a bare lens than the Nikon counterpart since each lens could only be used on the same brand of camera body. Thus, the higher-scoring sensor of the D850 makes the Nikon combo ‘better’. But mirror less cameras kick that argument to the curb, since in that case you can use a Canon lens on a Nikon MILC or vice versa, or either brand of lens on a Sony MILC. That means the highest possible scores (without getting into whether or not that really means “better“) should be obtained by combining the highest scoring mirrorless sensors with the lenses having the best optical measurements (not scores). Except that DxO only tests within-brand combinations.
The problem is that if you’re testing a component, you need to test that component. So whi,emthe resulting combo will tell you about that, it doesn’t tell you anything about how it will perform as part of another combo. When Canon comes out with a high resolution body, the combo will have to be tested. So the concept tells us nothing.a lens is often sued long after another body is acquired. While testing that lens with recognized methods will give us a good idea how it performed independently, and therefor how it will likely perform on another o]body, just testing it on one body, a lower resolution one doesn’t really help.
 
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The problem is that if you’re testing a component, you need to test that component. So whi,emthe resulting combo will tell you about that, it doesn’t tell you anything about how it will perform as part of another combo. When Canon comes out with a high resolution body, the combo will have to be tested. So the concept tells us nothing.a lens is often sued long after another body is acquired. While testing that lens with recognized methods will give us a good idea how it performed independently, and therefor how it will likely perform on another o]body, just testing it on one body, a lower resolution one doesn’t really help.
I disagree. Most aspects of a lens’ optical performance, including distortion, vignetting, transmission, coma, astigmatism, longitudinal CA, field curvature, etc., do not change from one sensor to another. But perhaps you’re someone for whom sharpness is the only important aspect of lens performance. In that case, if you don’t already know how, I’d recommend learning to interpret MTF curves – they give a good idea of lens sharpness with the caveat that, Zeiss notwithstanding, you’re seeing the ideal/best possible lens performance, which in practice is subject to the vagaries of production.
 
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Shooting wide open the f1.2 L is horrible, I wouldn’t at all be surprised if the new f1.8 stm is sharper wide open, although a full stop slower.
Yes it is horrible at 1.2 but if you close the aperture at 1.8 and compare with the 50 1.8 version, the difference is huge! Especially if you do the same at 2.8 in both lenses. there you can see tremendous difference in sharpness! But sharpness is not everything. For example the ef cinema lenses which will take soon an upgrade to pl mount they will be softer than ef versions to render better skin tones. So in cinema world sharpness is not everything. The 50 1.2 took the nickname of the "magic" lens because the bokeh quality and how it renders a scene is still amazing! SO many things that DXO do not talk about. Its not about the sharpness, its not about the sensor but its the overall experience for me.
 
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jd7

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No, it’s exactly one full stop.
f/1.8 to f/1.2 is about two-thirds of a stop (f/1.8 to f/1.4) plus half a stop (f/1.4 to f/1.2), ie the difference is about one and one-sixth stops. The f-stop series is not linear so it's not exactly one and one-sixth stops, but it's in that ball park. It's definitely more than one stop.
 
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dcm

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Depends on which scale you are using and what your camera supports. My film A-1 and FD lenses supported one-third stop ISO and exposure compensation values on the body, but one-half stop apertures in the lenses and Av values on the body. The exposure compensation was simply an adjustment to the ISO setting for the film. My 1DX2 lets me set 1/3 stop or 1 stop increments for ISO speeds, and 1/3 stop, 1/2 stop, or 1 stop increments for exposure and exposure compensation.

On the one-third stop scale, the difference between f/1.8 and f/1.2 is a full stop, f 1/8 to f1.4 is 2/3 stop and f1/4 to f1.2 is 1/3 stop. You would dial in a full stop of exposure compensation between the two values. This would be the same anywhere on the scale.

On the one-half stop scale you'll see different AVs for the same f/stops, f/1.2 is 0.5 versus 0.7. There is no 1.8 so you'd be forced to choose between 1.7 and 2.0.

from https://en.wikipedia.org/wiki/F-number
183606

Here's a table of the numbers from the computation with 2 decimal digits for both 1/2 and 1/3 stops with N representing the f-numbers. The actual numbers are somewhat different than what you see display with 1 decimal digit. This convention has been around a long time.

183607
 
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That's literally how they test the lenses: on the bodies.

Testing a lens on every compatible body would be rather labor intensive, and also would require them to keep all those bodies on hand.

I believe, though cannot prove, that they test lenses on a few bodies, and then model their performance on the other bodies in their dataset.
 
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Depends on which scale you are using and what your camera supports. My film A-1 and FD lenses supported one-third stop ISO and exposure compensation values on the body, but one-half stop apertures in the lenses and Av values on the body. The exposure compensation was simply an adjustment to the ISO setting for the film. My 1DX2 lets me set 1/3 stop or 1 stop increments for ISO speeds, and 1/3 stop, 1/2 stop, or 1 stop increments for exposure and exposure compensation.

On the one-third stop scale, the difference between f/1.8 and f/1.2 is a full stop, f 1/8 to f1.4 is 2/3 stop and f1/4 to f1.2 is 1/3 stop. You would dial in a full stop of exposure compensation between the two values. This would be the same anywhere on the scale.

On the one-half stop scale you'll see different AVs for the same f/stops, f/1.2 is 0.5 versus 0.7. There is no 1.8 so you'd be forced to choose between 1.7 and 2.0.

from https://en.wikipedia.org/wiki/F-number
View attachment 183606

Here's a table of the numbers from the computation with 2 decimal digits for both 1/2 and 1/3 stops with N representing the f-numbers. The actual numbers are somewhat different than what you see display with 1 decimal digit. This convention has been around a long time.

View attachment 183607

It doesn’t matter what scale you use. There are 1.25 stops between f/1.2 and f/1.8.
 
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AlanF

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I disagree. Most aspects of a lens’ optical performance, including distortion, vignetting, transmission, coma, astigmatism, longitudinal CA, field curvature, etc., do not change from one sensor to another. But perhaps you’re someone for whom sharpness is the only important aspect of lens performance. In that case, if you don’t already know how, I’d recommend learning to interpret MTF curves – they give a good idea of lens sharpness with the caveat that, Zeiss notwithstanding, you’re seeing the ideal/best possible lens performance, which in practice is subject to the vagaries of production.
Agreed about your MTF comments. I like lensrental's optical bench measurements because they are done on multiple production copies and the variation as well as mean values are plotted. And they go as high as 50-100 lp/mm rather than the 10 and 30 lp/mm standard ones from Canon that are in any case computer generated theoretical curves.
 
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Shooting the same greycard with 1/3 or 1/2 setting gives one stop difference in any case between f1.2 and f1.8, so for all intended purposes, there is no more and no less than one stop betweeen them. If you’re going to use a lens on a camera, one stop is the difference.

That likely comes down to the resolution of the meter and how accurate the aperture settings are (probably not very), and how linearly transmission (t-stop) tracks geometry (f-stop).


All else being equal, f/1.2 will have 2.25 times the entrance pupil area as f/1.8.
 
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Don Haines

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It is more like 1.1699 stops. But given how things work in real life, it could function more like one stop than 1 1/3.
And there in lies the heart of the debate..... round it up or round it down to the nearest 1/3 stop when it is almost in the middle :)
 
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dcm

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The markings displayed on the camera or lens are approximate to minimize the space they require and make it easier for us to communicate. The actual lens mechanism and metering uses the more precise increments that are equally spaced in 1/2 or 1/3 stop increments. When your camera is set to 1/3 stop exposure increments and you change the aperture in Av mode, it is 3 steps from 1.2 to 1.8 or 1 stop. This is true throughout the entire range.

You can use the formula f-stop-difference = log(f1/f2) / log(sqrt(2)) for any two f-stops f1 and f2. Using the approximate values gives log(1.8/1.2) / log(sqrt(2)) = 1.17 (1.1699) while the precise measures give log(1.78/1.26) = 1.00 (0.9969).

But who would want to talk about f/1.78 and f/1.26?

Since you introduced entrance pupil, that is also approximate. Are any of Canon's lenses exactly the stated maximum aperture? A look at all of the patent applications suggests no.
 
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AlanF

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The markings displayed on the camera or lens are approximate to minimize the space they require and make it easier for us to communicate. The actual lens mechanism and metering uses the more precise increments that are equally spaced in 1/2 or 1/3 stop increments. When your camera is set to 1/3 stop exposure increments and you change the aperture in Av mode, it is 3 steps from 1.2 to 1.8 or 1 stop. This is true throughout the entire range.

You can use the formula f-stop-difference = log(f1/f2) / log(sqrt(2)) for any two f-stops f1 and f2. Using the approximate values gives log(1.8/1.2) / log(sqrt(2)) = 1.17 (1.1699) while the precise measures give log(1.78/1.26) = 1.00 (0.9969).

But who would want to talk about f/1.78 and f/1.26?

Since you introduced entrance pupil, that is also approximate. Are any of Canon's lenses exactly the stated maximum aperture? A look at all of the patent applications suggests no.

The formula f-stop-difference = log(f1/f2) / log(sqrt(2)) for any two f-stops f1 and f2 is of course correct and has the virtue of being independent of the base of the logarithm. But, I think it is a hangover from the days when log to base 10 was the common one for slide rules and log tables (and Napierian). I prefer the formula f-stop-difference = 2xlog(base2)(f1/f2) since not only does it take one step less in the calculation but the physical reason behind it is clearer - area varies as the diameter of the exit pupil squared.
 
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dcm

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For full disclosure, I used natural logarithms on my calculator to get the irrelevantly accurate results.

log(x) in excel (and many other places) is natural logarithm which I used. log10 is often available, sometimes log2 (but not it excel).
 
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I prefer rolling logarithms.

Key%20Log%20Fall%2014_4004.jpg
 
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