Re: Tonal gradation in shadows - The $67 Question?
Anthony Farr wrote: My point is that the profile needs to be applied before the data enters the digital domain, early in the a/d conversion. It needn't matter that the voltage rise is squared (if that's what you mean) as the brightness rises, as long as it's a constant and predictable relationship between photons in and voltage out. All that matters is that the profile is applied while it's still a voltage readout, not after it's become bits and bytes. Yup- that would suck and would make the pictures worse. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
Oh. I see. Why? Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Friday, 18 August 2006 11:37 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? Anthony Farr wrote: My point is that the profile needs to be applied before the data enters the digital domain, early in the a/d conversion. It needn't matter that the voltage rise is squared (if that's what you mean) as the brightness rises, as long as it's a constant and predictable relationship between photons in and voltage out. All that matters is that the profile is applied while it's still a voltage readout, not after it's become bits and bytes. Yup- that would suck and would make the pictures worse. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
Anthony Farr wrote: Oh. I see. Why? Because, you've obviously figure out something that electrical engineers and physicists have missed for 20 years. Congrats! -Ryan -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
Not at all. In fact I don't know the working of an a/d converter beyond that they take a stream of voltage readouts from the sensor and perform some voodoo ceremony upon it, whereupon the voltage levels at each photosite are assigned brightness levels of 0 - 255 for 8 bit files, or 0 - 4095 for 12 bit files. That's the sum of my understanding of the subject. If someone who seems to be an expert makes an assertion without an explanation I can: (a) totally and naively trust their opinion; or (b) ask for some explanation. Option (a) has let me down in the past, so I choose (b). So my question is; when the a/d converter assigns arbitrary brightness levels to voltage levels, why can't they be assigned on a logarithmic scale rather than a linear scale? Can't it have a lookup table to adapt the numbers, or even just recalculate them on-the-fly? Hint - the correct answer is not because it's not done that way or because it won't work. Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Saturday, 19 August 2006 12:32 AM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? Anthony Farr wrote: Oh. I see. Why? Because, you've obviously figure out something that electrical engineers and physicists have missed for 20 years. Congrats! -Ryan -- -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
D'oh, I'm apparently receiving posts out of order. After sending this please explain to Ryan, THEN I find his references in my inbox. No wonder his response seemed a little sharp. Sorry Ryan, I'll go and get my explanations from those links now. Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Anthony Farr Sent: Saturday, 19 August 2006 1:22 AM To: 'Pentax-Discuss Mail List' Subject: RE: Tonal gradation in shadows - The $67 Question? Not at all. In fact I don't know the working of an a/d converter beyond that they take a stream of voltage readouts from the sensor and perform some voodoo ceremony upon it, whereupon the voltage levels at each photosite are assigned brightness levels of 0 - 255 for 8 bit files, or 0 - 4095 for 12 bit files. That's the sum of my understanding of the subject. If someone who seems to be an expert makes an assertion without an explanation I can: (a) totally and naively trust their opinion; or (b) ask for some explanation. Option (a) has let me down in the past, so I choose (b). So my question is; when the a/d converter assigns arbitrary brightness levels to voltage levels, why can't they be assigned on a logarithmic scale rather than a linear scale? Can't it have a lookup table to adapt the numbers, or even just recalculate them on-the-fly? Hint - the correct answer is not because it's not done that way or because it won't work. Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Saturday, 19 August 2006 12:32 AM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? Anthony Farr wrote: Oh. I see. Why? Because, you've obviously figure out something that electrical engineers and physicists have missed for 20 years. Congrats! -Ryan -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
At 10:31 PM 8/17/2006, you wrote: I definitely agree that tonal gradation is funky on cameras like the *istDS, when compared to what the human eye sees. Last year, I traveled to Pittsburgh for their annual Light Up Night festival. (One night a year, they turn on almost every light in almost every downtown building. It creates a beautiful city skyline.) While I was shooting this impressive skyline at night, I could never get a tonal gradation and consistency of color that approached what I saw with my eyes. I tried every combination of contrast and saturation the camera had, and nothing was close to the impressive skyline itself. When I adjusted for reasonable highlight exposure and color balance, the shadows were always far too dark. I could never capture the full range of tones I could see with my eyes. I even tried playing with the RAW files in post production, but nothing came close to the original scene. It's a real pity. The original scene is much more impressive to the naked eye than any of my photos will ever show. I'd love to have a camera that handled tonal range better than the *istDS. Don't misunderstand, I love my *istDS, but it's quite far from being the perfect camera. take care, Glen -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
I don't know what these grayscales and step wedges are supposed to be showing me. G On Aug 17, 2006, at 7:31 PM, Anthony Farr wrote: Perhaps this post didn't get through the first time so I'm resending it but with an altered subject line. I can understand how some of us might be binning the Ho** Cr** thread out of sheer irritation with its futility. The $67 Question? refers to my suspicion that the big advantage of 6x7 that means so much to Aaron may be the creamy smooth tonal rendition and ability to render subtle nuances of tone and detail in shadows possessed by medium format and larger cameras. which is much reduced in 35mm film cameras, and absent in digital cameras without resorting to multi- image stacks. -- -- My original message: This should demonstrate why digital photography needs more RD to correct some deficiencies of shadow rendition. The examples are 8 bit for the web of course, so 12 or more bit samples should be somewhat better. http://www.photo.net/photodb/photo?photo_id=4816986 Regards, Anthony Farr -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
Glad it's not just me. I thought I was being dense. Dave On 8/18/06, Godfrey DiGiorgi [EMAIL PROTECTED] wrote: I don't know what these grayscales and step wedges are supposed to be showing me. G -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
My takeaway was that it was underexposed. Other than that, I got nothng. Paul On Aug 17, 2006, at 10:47 PM, David Savage wrote: Glad it's not just me. I thought I was being dense. Dave On 8/18/06, Godfrey DiGiorgi [EMAIL PROTECTED] wrote: I don't know what these grayscales and step wedges are supposed to be showing me. G -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
You need to click the DETAILS tab under the image to show the caption. PhotoNet would rather show you an advertisement when the page first loads. What the scales show is that an unmanipulated linear greyscale produced by a digital camera has considerably darker shadows than a non-linear greyscale produced by film. Correcting the difference in levels leads to nasty posterisation bands in the shadow tones. These are 8bit scales and each step is as near to true as I could get (one or less levels error) so 12 or more bits will be better, but the correction of the linear greyscales to anything like the evenly spaced tonal rendition of the non-linear greyscales will always be an inferior option unless considerably greater bit depth is offered. It would be much better if digital cameras could output files with a non-linear characteristic curve. Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Godfrey DiGiorgi Sent: Friday, 18 August 2006 12:44 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? I don't know what these grayscales and step wedges are supposed to be showing me. G -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
On Aug 17, 2006, at 10:31 PM, Anthony Farr wrote: Perhaps this post didn't get through the first time so I'm resending it but with an altered subject line. I didn't reply because there was no context to what I was looking at, so it didn't make any sense to me. -Aaron -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
Anthony Farr wrote: You need to click the DETAILS tab under the image to show the caption. PhotoNet would rather show you an advertisement when the page first loads. What the scales show is that an unmanipulated linear greyscale produced by a digital camera has considerably darker shadows than a non-linear greyscale produced by film. Correcting the difference in levels leads to nasty posterisation bands in the shadow tones. These are 8bit scales and each step is as near to true as I could get (one or less levels error) so 12 or more bits will be better, but the correction of the linear greyscales to anything like the evenly spaced tonal rendition of the non-linear greyscales will always be an inferior option unless considerably greater bit depth is offered. It would be much better if digital cameras could output files with a non-linear characteristic curve. Except that the sensor is linear- if it's a CCD anyway. It's a photo (okay, electron) counter. If you digitize the output in a non-linear space, you're not getting as much (good) information as if it was a linear digitization. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
PhotoNet's doing, I'm afraid. They hide the details panel under an advertisement, you have to dig it out :-( Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Aaron Reynolds Sent: Friday, 18 August 2006 1:14 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? On Aug 17, 2006, at 10:31 PM, Anthony Farr wrote: Perhaps this post didn't get through the first time so I'm resending it but with an altered subject line. I didn't reply because there was no context to what I was looking at, so it didn't make any sense to me. -Aaron -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
It's not underexposed, because it's not exposed. The greyscales are artwork, and the brightness levels of each step are what they theoretically should be, to one or less levels. The uncorrected linear greyscale at the bottom shows what should be obvious, that half the steps are below the threshold of visibility, the middle step is just barely visible, and only the brightest 4 steps are readily visible. That's only 4 steps out of 9 that are visible without making levels adjustments which introduce posterisation effects (that's what the middle greyscale illustrates). Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Paul Stenquist Sent: Friday, 18 August 2006 12:55 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? My takeaway was that it was underexposed. Other than that, I got nothng. Paul -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
Backwards thinking on your part. Linear may be technically correct, but non-linear is better, it's what film has done for years and years. The irony is that linear capture produces a non-linear greyscale, the brightness of each step is double or half of the neighbouring steps. The steps starting at 0,0,0 on the left take half the bit depth to just get out of the blacks, the suddenly leap up to saturation in just the top 4 stops. The reason we are fooled is that the greyscale horizontal axis is progressively compressed, not only should each step be double the brightness of its predecessor, it should also be double the length of its predecessor. But it's not; it's NON-linear, because it was captured lineally. Film produces a constant increase in brightness from each stop to the next whether they are shadow or highlight stops. A difference of a few lumens in the shadows will give the same increase in brightness (negative density in this case) as will an increase of hundreds or more lumens in the highlights. With linear (digital) capture, a shadow stop may represent, let's say, 4 lumens increase over the previous stop, while a 5 stops brighter highlight will represent 128 lumens increase over the previous stop. That's 32 times more illumination, but it's still just one stop! And guess what? The digital sensor will record that as a 32 times higher increase in brightness in the output file. IOW a shadow stop is a severely discounted commodity. It's all about how our eyes see versus how sensors see. The sensor may technically be correct, but I want to photograph things the way that I see. Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Friday, 18 August 2006 1:22 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? Except that the sensor is linear- if it's a CCD anyway. It's a photo (okay, electron) counter. If you digitize the output in a non-linear space, you're not getting as much (good) information as if it was a linear digitization. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
It's all about how our eyes see versus how sensors see. The sensor may technically be correct, but I want to photograph things the way that I see. Me too. But the data coming from the _physics_ of the CCD is linear. Nothing you can do about that unless you change the sensor technology. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
Another thought. Why would converting the linear CCD output to non-linear A/D output have not as much (good) information? I could understand this if the CCD was outputting digital information and arbitrarily reassigning the values caused stairstepping type errors, but the CCD is analogue. What it's outputting is voltages. Can't the conversion be profiled? Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Friday, 18 August 2006 1:22 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? Except that the sensor is linear- if it's a CCD anyway. It's a photo (okay, electron) counter. If you digitize the output in a non-linear space, you're not getting as much (good) information as if it was a linear digitization. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
Not the sensor technology, but the A/D convertor. It needs to do less of a literal translation and more of a weighted interpretation. Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Friday, 18 August 2006 2:31 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? It's all about how our eyes see versus how sensors see. The sensor may technically be correct, but I want to photograph things the way that I see. Me too. But the data coming from the _physics_ of the CCD is linear. Nothing you can do about that unless you change the sensor technology. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
Anthony Farr wrote: Another thought. Why would converting the linear CCD output to non-linear A/D output have not as much (good) information? I could understand this if the CCD was outputting digital information and arbitrarily reassigning the values caused stairstepping type errors, but the CCD is analogue. What it's outputting is voltages. Can't the conversion be profiled? Regards, Anthony Farr Sure... but a log/log representation would start to have big rounding effects at each higher stop; in other words, it would be stair stepping right past data as it moved up the curve. If the CCD recorded an apparent doubling in brightness as a doubling in voltage your scheme would work great (like film). But it doesn't. A doubling in brightness represents a four-fold increase in charge. (whereas your film, our eyes, etc. see this as 2x) I don't like the word brightness. It makes this discussion difficult. I really dont think Pentax has solved this problem. Even if they do have some sort of built-in gamma a/d conversion- what does it matter if it's in hardware or software once you're in the digital domain?This was all figured out years ago w/ video. References: http://www.normankoren.com/digital_tonality.html http://en.wikipedia.org/wiki/Gamma_correction -R -R -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Friday, 18 August 2006 1:22 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? Except that the sensor is linear- if it's a CCD anyway. It's a photo (okay, electron) counter. If you digitize the output in a non-linear space, you're not getting as much (good) information as if it was a linear digitization. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
Yet another observation. Every time anyone adjusts curves or tweaks the levels in Photoshop, what they are really doing is changing their digital image from a linear rendition to a non-linear rendition. Let he who is without sin cast the first stone. Regards, Anthony Farr -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: Tonal gradation in shadows - The $67 Question?
Anthony Farr wrote: Yet another observation. Every time anyone adjusts curves or tweaks the levels in Photoshop, what they are really doing is changing their digital image from a linear rendition to a non-linear rendition. Let he who is without sin cast the first stone. Even before that! Actually as soon as you load your RAW data (or have your camera save to JPG), you're converting to log/log. -R Regards, Anthony Farr -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: Tonal gradation in shadows - The $67 Question?
My point is that the profile needs to be applied before the data enters the digital domain, early in the a/d conversion. It needn't matter that the voltage rise is squared (if that's what you mean) as the brightness rises, as long as it's a constant and predictable relationship between photons in and voltage out. All that matters is that the profile is applied while it's still a voltage readout, not after it's become bits and bytes. I am using brightness because that's what the light levels are commonly named in image editors. In Sensitometry/Densitometry class it was simply called exposure and density. I still find myself referring to density even where digital images are concerned. Regards, Anthony Farr -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Friday, 18 August 2006 2:55 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? Anthony Farr wrote: Another thought. Why would converting the linear CCD output to non-linear A/D output have not as much (good) information? I could understand this if the CCD was outputting digital information and arbitrarily reassigning the values caused stairstepping type errors, but the CCD is analogue. What it's outputting is voltages. Can't the conversion be profiled? Regards, Anthony Farr Sure... but a log/log representation would start to have big rounding effects at each higher stop; in other words, it would be stair stepping right past data as it moved up the curve. If the CCD recorded an apparent doubling in brightness as a doubling in voltage your scheme would work great (like film). But it doesn't. A doubling in brightness represents a four-fold increase in charge. (whereas your film, our eyes, etc. see this as 2x) I don't like the word brightness. It makes this discussion difficult. I really dont think Pentax has solved this problem. Even if they do have some sort of built-in gamma a/d conversion- what does it matter if it's in hardware or software once you're in the digital domain?This was all figured out years ago w/ video. References: http://www.normankoren.com/digital_tonality.html http://en.wikipedia.org/wiki/Gamma_correction -R -R -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ryan Brooks Sent: Friday, 18 August 2006 1:22 PM To: Pentax-Discuss Mail List Subject: Re: Tonal gradation in shadows - The $67 Question? Except that the sensor is linear- if it's a CCD anyway. It's a photo (okay, electron) counter. If you digitize the output in a non-linear space, you're not getting as much (good) information as if it was a linear digitization. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
