Iliah Borg
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Right, my apologies, I wasn't paying proper attention. Avg on the black is 510 .. 511, means black level subtraction is disabled. Lack of magenta tint threw me off.
Yes. Maybe mention σ too
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Why is DR smaller when FF cameras are used in cropped mode?
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D
Donald B
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Thats a great find ,thanks. wow i always have the default sharpening turned off in in camera raw .didnt realize that it changes when i convert a sony file. Ive been racking my brain over this for weeks. thank you very much.I've got to hand it to you, that was pretty funny.Another case of wanting to give the evidence a helping hand, it seems.Perhaps Don will clarify, but the metadata in the JPEGs posted here indicate that they both had "Adobe Standard" profiles and were converted by "Camera Raw" and "Process 11", so presumably he used either LR or ACR. The metadata also indicates that the Sony image had a sharpness setting of +40 (which is the default in the Adobe converters). On the other hand, the metadata for the Oly shot indicates a sharpness setting of 0, which is definitely NOT the default for the Oly. I agree that the halos in the Sony shot are more aggressive than should be expected from the default settings. As importantly, applying no input sharpening at all to the Oly image is bound to make it look relatively soft to even a "normal" sharpened comparison shot.That's not a small difference. I find it hard to believe that Sony defaults would include quite so much sharpening, to the extent of clearly visible halos.The Oly shot just looks soft. Lots of possible causes, but it's probably not going to come down to one single thing like glare. A couple of things to bear in mind here. First, F8 is farther into diffraction territory for the Oly combo than the Sony combo. The default sharpening for the Sony shot is definitely greater than the Oly shot as can be seen here:
Upsized 400%. View at Full Size in the DPR viewer to easily see the difference in the amount of unsharp masking
Don
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Olympus EM1mk2, Sony A7r2
http://www.dpreview.com/galleries/9412035244
past toys. k100d, k10d,k7,fz5,fz150,500uz,canon G9, Olympus xz1 em5mk1 em5mk2
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Great Bustard
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Would be great if we could just hang around Congress and do the job they should be doing when they don't show up. Of course, we'd probably be spending all of our time undoing what they screwed up when they did show up! ;-)great story. I got into dance photography because the pro didnt show at the photo shoot day of my daughters (she was 5) dance school 130 girls waiting for photos . so i said id give it a go. rushed around bought some extra gear, that was 13 years ago.A long time ago, I used to frequent a rock shop just to look. I couldn't afford them at the time, so one day I approached the owner of the shop and said, "What if I sold some of your rocks on ebay and you let me keep some?" I mean, can't hurt to ask, right? But his reply was unexpected, "OK, sure." Just like that. I didn't even know the guy and that was his answer. "I'll need a digital camera to take photos, though, so you'd have to buy one." I mean, if I couldn't afford the rocks, I couldn't afford the camera, right? May as well go for broke. "OK, sure." And that's how I got into photography. It was the very first time I used a camera.I am more interested in rocks and did you polish them yourself?Both photos were taken with a Canon 6D2 at f/4 1/200 ISO 25600 and converted using the exact same settings:
6D2 + Tamron 35-150 / 2.8-4 VC at 100mm f/4 1/200 ISO 25600 downsampled to the same dimensions of the crop from the 50mm photo below.
6D2 + Tamron 35-150 / 2.8-4 VC at 50mm f/4 1/200 ISO 25600 center crop.
The low exposure was intentionally used to make the differences in noise more visible. The reason the crop of the 50mm photo is so much more noisy than the uncropped 100mm photo is because the latter was made with 4x as much light as the former. Thus it is both less noisy and has greater DR.
Some will argue that it is the downsampling of the 100mm photo that reduced the noise, not the fact that it was made with 4x as much light. However, prints of the same size would show the same thing, and displaying the top photo at full size on an 8K monitor and the crop at full size on a same size 4K monitor would also have the same results.
In short, both noise and DR are *highly* dependent on the the total amount of light that makes up the photo.
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Great Bustard
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When you view at 100%, you are comparing different things. For example, if two photos have the same framing and you are comparing at 100%, you might be looking at the face on one photo and just the nose and mouth on the other -- it simply makes no sense to compare different scenes or compare at different sizes.
The whole point of the comparison is to see which photo looks better. Where is the sense in *artificially* comparing one with greater enlargement than the other? It's like comparing a metal print to a canvas print.
It will still retain a resolution advantage, as my examples downthread clearly demonstrate.
Again, check out the examples I linked.
Great Bustard
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Because at 1:1, you are looking at 1/4 as much of the scene for the 100mm photo as with the 50mm photo, thus the portion you are looking at is made with 1/4 as much light.But when viewed at 1:1, they'll show the exact same amount of noise.Both photos were taken with a Canon 6D2 at f/4 1/200 ISO 25600 and converted using the exact same settings:
6D2 + Tamron 35-150 / 2.8-4 VC at 100mm f/4 1/200 ISO 25600 downsampled to the same dimensions of the crop from the 50mm photo below.
6D2 + Tamron 35-150 / 2.8-4 VC at 50mm f/4 1/200 ISO 25600 center crop.
The low exposure was intentionally used to make the differences in noise more visible. The reason the crop of the 50mm photo is so much more noisy than the uncropped 100mm photo is because the latter was made with 4x as much light as the former. Thus it is both less noisy and has greater DR.
Some will argue that it is the downsampling of the 100mm photo that reduced the noise, not the fact that it was made with 4x as much light. However, prints of the same size would show the same thing, and displaying the top photo at full size on an 8K monitor and the crop at full size on a same size 4K monitor would also have the same results.
That information comes *directly* from the amount of light used to create them.
Like I said, view the 100mm photo full size on an 8K monitor and view the 50mm crop full size on a 4K monitor of the same size. You'll see the exact same thing.
Exposure has all of squat to do with it except inasmuch as exposure is a *component* of the total amount of light making up the photo: Total Light = Exposure x Sensor Area.
BC's PDR metric is not a measure of DR. DR is the number of stops from the noise floor (electronic noise) to the saturation limit over a specified area.
Bottom line: photos made with the same amount of light, all else equal, are equally noisy. In extremely low light, the electronic noise of the sensor becomes significant, so it is possible that two photos made with the same amount of light are not equally noisy because one sensor is more noisy than the other. Likewise for extremely long exposures. But in both these cases, not all else is equal, as was stipulated.
bclaff
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Dynamic range is simply a logarithm of the ratio of a high value over a low value.
The fact that PDR doesn't use the "noise floor" (read noise) doesn't mean it isn't dynamic range. (Same goes for DxOMark landscape dynamic range)
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Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )
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bobn2
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Really, no, Bill. That is not a definition of 'Dynamic range'. By your definition the DIN variant of ISO is a 'dynamic range'. First, it doesn't have to be expressed as a logarithm, it's still a DR if you express it as a plain old number, so long as it gives the range of values over of a detectable . Using log units is a convenience, but it doesn't change what it is. Plus, it really does matter what the 'high number' and 'low number' represent. Not just any old number will do. Dynamic range, in its original (and I would say, only - without some qualification) is a measure of the capacity of a communications channel. It comes from telecommunications engineering. Claude Shannon gave it the direct information theory definition familiar now, but that was always intrinsic in the idea of the capacity of a channel. In the context of the present discussion the camera is the 'channel', communication information about the image projected onto the sensor.
Yes, it does.
DxOMark does not have a 'landscape dynamic range'. They have a 'Landscape Score', for which the criterion is their Dynamic Range measure. The DxOMark dynamic range uses, according to them, an SNR of 1 for the lower bound. This is, I suspect a mistake in description - many people believe that the 'noise floor' is SNR = 1, when it is SNR = 0, the value of the noise with no applied signal. As I understand it, their lower bound is in actuality the level with zero exposure, that is SNR = 0, so it is an error of explanation, not execution.
D
Donald B
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"capacity" perfect
Don
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Olympus EM1mk2, Sony A7r2
http://www.dpreview.com/galleries/9412035244
past toys. k100d, k10d,k7,fz5,fz150,500uz,canon G9, Olympus xz1 em5mk1 em5mk2
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bclaff
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Well, certainly not only read noise but all sources of noise.
Sure DxoMark Landscape Score == DxOMark print dynamic range"capacity" perfect
The SNR = 1 error is as follows.
They determine Noise when Signal is 0 therefore SNR = 0
They set Signal equal to Noise for their dynamic range calculation.
At that point it is SNR = 1 but that isn't the point at which they determined Noise
Kind of a confusing fine point.
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