[filmscanners] Re: Dynamic range
... Of course, the number of bits LIMITS the dynamic range, I've always said that...but BTW, that contradicts Roy's last round, as he claims that 8 bits has the same dynamic range as 16 bits... Yes, I don't agree with Roy on this point. Julian Hi Julian, I'm curious whether we're talking about two different things or that you disagree with what I was actually talking about. It think that your post (in response to Austin) was talking specifically about scanner output. In other words the phrase the number of bits LIMITS the dynamic range was in the context meaning the number of bits in a scanner LIMITS the dynamic range of that scanner. In this context I entirely agree -- a 16-bit scanner has more dynamic range than an 8-bit scanner. My 8-bit versus 16-bit comment was in a very different context. I was talking about a 16-bit Photoshop that was ready to be printed. Thus value 0 was the max black and value 65535 was the max white. At this time the file was converted to 8-bit such that value 0 represents the same max black as 0 in the 16-bit file, and value 255 in 8-bit file represents the same max white as 65535 in the 16-bit file. So both files represent the same black to white range. In this context I say the 8-bit file and the 16-bit file have the same dynamic range because they represent the same tonal range on a output print. The endpoints are the same only real difference is how many levels are in between. So, is there disagreement? If so I'd like to know why and how you look at it. Thanks, Roy Roy Harrington [EMAIL PROTECTED] Black White Photography Gallery http://www.harrington.com Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body
[filmscanners] RE: Dynamic range
Hi Roy, With your arguments a 1 bit file has the same dynamic range as a 16 bits file because value 0 in a 1 bit file represents the same black level as value 0 in a 16 bits file and value 1 in a 1 bit file represents the same white level as value 65335 in a 16 bits file. This is of course not the case. Dynamic range is the ability to distinguish tonal differences. If a certain system is able to distinguish pure black and a slightly less black and at the same time is able to distinguish pure white and a little less white we say that its dynamic range is large. So the number of bits relates to the dynamic range and 8 bits system have, in general, a lower dynamic range then 16 bits systems. It doesn't mater if you are talking about files, scanners, digital cameras or any other system. Thanks, Vincent ... Of course, the number of bits LIMITS the dynamic range, I've always said that...but BTW, that contradicts Roy's last round, as he claims that 8 bits has the same dynamic range as 16 bits... Yes, I don't agree with Roy on this point. Julian Hi Julian, I'm curious whether we're talking about two different things or that you disagree with what I was actually talking about. It think that your post (in response to Austin) was talking specifically about scanner output. In other words the phrase the number of bits LIMITS the dynamic range was in the context meaning the number of bits in a scanner LIMITS the dynamic range of that scanner. In this context I entirely agree -- a 16-bit scanner has more dynamic range than an 8-bit scanner. My 8-bit versus 16-bit comment was in a very different context. I was talking about a 16-bit Photoshop that was ready to be printed. Thus value 0 was the max black and value 65535 was the max white. At this time the file was converted to 8-bit such that value 0 represents the same max black as 0 in the 16-bit file, and value 255 in 8-bit file represents the same max white as 65535 in the 16-bit file. So both files represent the same black to white range. In this context I say the 8-bit file and the 16-bit file have the same dynamic range because they represent the same tonal range on a output print. The endpoints are the same only real difference is how many levels are in between. So, is there disagreement? If so I'd like to know why and how you look at it. Thanks, Roy Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body
[filmscanners] Re: Dynamic range
Hi Roy, I was talking about your context so we are discussing the same thing. You have already got a response from Vincent which puts that case in terms of resolution, here's my quick take from the dynamic range point of view - the two arguments are otherwise essentially the same. The important thing I think to remember about DyR is as always, the definition, and what defines the minimum signal, the MDS, in that definition. The DyR is the range from the max signal down to the MDS, not down to some zero figure. So in the case you are discussing, and if we call black the low end of the range as you have, we can establish the DyR as follows. We know it is max signal / MDS. What is the MDS? It is the minimum signal that can be detected ABOVE whatever corresponds to background or 'zero level'. In this case, and with all digital step limited situations, the MDS is step 1, not step 0. It is the first, lowest signal you have any chance of discerning above the zero signal level, which in this case is pure black. So in the first case, the DyR is : max/MDS = (4096 steps) / (1 step), and in the second 256/1. i,e, DyR 4096 vs 256. Looking at it another way, with an 8 bit file, the bottom step is the same level as step 16 was in the 12-bit case. So when you converted from the 12-bit to the 8-bit, you lost the 16 lowest steps and combined them all into 1, the lowest level of the 8-bit situation. In that conversion you lost the 16 lowest shades of gray, permanently. So all that info is gone and your MDS is now 16 times larger, and correspondingly your DyR has diminished by the same amount, 16 times. If you then converted back to 12-bit, you can't regain those bottom 16 shades, so your picture is permanently degraded. Despite the 12-bit digitisation which implies DyR of 4096, the actual image bottom step- the new minimum discernable signal above black MDS - is actually step 16 of your 4096 and so the DyR is now 4096/16 = 256, the same as the 8-bit case. This must be so because the information content is exactly the same in both cases. Does this make sense? Julian At 17:14 01/09/02, Roy wrote: I'm curious whether we're talking about two different things or that you disagree with what I was actually talking about. It think that your post (in response to Austin) was talking specifically about scanner output. In other words the phrase the number of bits LIMITS the dynamic range was in the context meaning the number of bits in a scanner LIMITS the dynamic range of that scanner. In this context I entirely agree -- a 16-bit scanner has more dynamic range than an 8-bit scanner. My 8-bit versus 16-bit comment was in a very different context. I was talking about a 16-bit Photoshop that was ready to be printed. Thus value 0 was the max black and value 65535 was the max white. At this time the file was converted to 8-bit such that value 0 represents the same max black as 0 in the 16-bit file, and value 255 in 8-bit file represents the same max white as 65535 in the 16-bit file. So both files represent the same black to white range. In this context I say the 8-bit file and the 16-bit file have the same dynamic range because they represent the same tonal range on a output print. The endpoints are the same only real difference is how many levels are in between. So, is there disagreement? If so I'd like to know why and how you look at it. Thanks, Roy Roy Harrington [EMAIL PROTECTED] Black White Photography Gallery http://www.harrington.com Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body
[filmscanners] RE: Dynamic range
Julian, I have never read whatever paper you are talking about, but I GUARANTEE you it does not SAY that dynamic range is a resolution. I am sure that you, Austin, INTERPRET it to say that, but it will not actually say that. You probably should have read the paper before commenting... But no, that is the point. I understand. You believe you know what you're talking about, and I believe you don't. It appears you are holding your hands over your ears and chanting so you can't hear what someone else has to say. Sigh. I am confident that I don't need to look at the paper, because I know that no paper will say what you believe. You are mistaken in what you say and still, to this date, after buckets of wasted electrons and keyboard hours, you have still not produced a single reference that says what you say. That's why no one can argue with you, Julian. When you're shown to be wrong, you simply ignore it, and pretend it never happened, or say that you didn't see it...or whatever. That's disingenuous of you, and certainly not conducive to adult conversation. The fact that you have not done so I think proves the point. The fact that you refuse to read it (or simply don't understand it) surely prove my point. You might not understand why it means what I said anyway. Now as for the rest of the post, I am in a bind. If I respond to all of yours, you and others will accuse me of being interminable. If I only respond to what I think are relevant points, you will accuse me of being selective. This is very funny, Julian. You have previously accused me of ignoring stuff in your posts, because I didn't respond to them in some verbose way... Well, here you go again, Julian...and this is why I get pissy with you. You take things out of context and apply them to something else. I NEVER said the MDS was ALWAYS noise. In the case of the ORIGINAL SIGNAL, it is noise, in the case of the digitized signal, it is NOT noise. Well let me quote one of many interminable exchanges where I was tearing my hair out because you were insisting that MDS was noise. Please note carefully the contradiction, clear and unambiguous between statement A and the statements B1,2,3 : A) I NEVER said the MDS was ALWAYS noise. - from this post B1) The smallest discernable signal IS noise. - from post in June B2) This is a misunderstanding of the concept of dynamic range. It is ALWAYS based on noise. - post in June B3) Noise and smallest discernable signal are EXACTLY the same thing. - post in June. I have struggled for months to get you to agree that noise and MDS are not the same thing, That's not true, Julian. You're making that claim up. During ALL of these discussion, for at least the past two or more years, I have called it MDS, directly FROM THE HIGGINS PAPER and KNOW, and have ALWAYS KNOWN, that noise is only one of the possible limiting factors to MDS, hence why it simply isn't called NOISE all the time. I am the one who introduced the Higgins paper into the discussion. You are taking the comments above COMPLETELY OUT OF CONTEXT, and claiming it means something it does not. If I said the limiting factor was noise in a PARTICULAR case, then it was, but that does not make it the limiting factor in EVERY case. The fact is in MOST EVERY CASE the limiting factor IS noise. The cases when it is NOT noise are not typical, and really have NO bearing on the conversation at all...since we are talking about film scanners SPECIFICALLY. But, leave it to you to seek EVERY POSSIBLE AVENUE to find some angle to somehow claim something I've said is wrong...when it simply isn't, given the context of the conversation. and now you tell me you have always thought this!! Because I have always KNOWN THIS. I've been using the SAME understanding of dynamic range for over 20 years, Julian. I am pleased that you are coming round, but flabbergasted at the same time. This, and a lot of this post, are why I get nasty to you. You deserve it. You're basically making things up that just were not said, or weren't said in the context you present them in. I have ALWAYS, LONG BEFORE YOUR INVOLVEMENT IN THIS CONVERSATION, KNOWN that MDS and noise were NOT the same thing, though they COULD be, and in SOME cases/discussions they ALWAYS are the same thing. All I can guess is that you have to make things up so you feel like you're winning. It's highly dishonest. I've caught you doing it almost a half dozen times now. Why can't you stop doing that? Austin Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body
[filmscanners] Re: Dynamic range
Roy Harrington [EMAIL PROTECTED] wrote: (Julian) So in the first case, the DyR is : max/MDS = (4096 steps) / (1 step), and in the second 256/1. i,e, DyR 4096 vs 256. Looking at it another way, with an 8 bit file, the bottom step is the same level as step 16 was in the 12-bit case. So when you converted from the 12-bit to the 8-bit, you lost the 16 lowest steps and combined them all into 1, the lowest level of the 8-bit situation. In that conversion you lost the 16 lowest shades of gray, permanently. So all that info is gone and your MDS is now 16 times larger, and correspondingly your DyR has diminished by the same amount, 16 times. I'm sticking with 0 in both files producing the same black and 255 or 4095 producing the same white paper. So the ratio of the amount of light from either file will be the same. The problem here is that you are arbitrarily reinterpreting the values 0 and 2**n-1, and then claiming that that's what they originally meant. What any number reported by a scanner means is that the density at the point measured was in the range of density values for which the scanner reports that value. Since that's a rather circular definition, you need to provide specifications for the scanner that allows users to understand what those ranges are. In particular, the range of densities that a (real) 8-bit scanner will report 0 for is different than the range of densities that a (real) 12-bit scanner will report 0 for. (Here real means actually performs as an n-bit scanner, i.e. that the only noise in the digital output is the quantization noise inherent in a digital value of that number of bits.) FWIW, here's my current take on this: From an engineering perspective, it takes (at least) _three_ numbers to characterize a digital (actually, any numerical) measurement: the center value of the range for the minimum value, the center value of the range for the maximum value, and the dynamic range. Scanner types fudge this by only reporting two numbers (Dmax and Dmin) and their ratio. That only works because the range for 0 can be assumed (a) to include true black (obviously, for scanners) and (b) to be the same width as all the other ranges*, i.e., that there is no significant offset at the low end. If the range of densities for which the scanner reports 0 was significantly larger than the range of densities for which it reports other values, the discussion at http://www.scantips.com/basics14.html would be quite dizzy. Since the offset at Dmax can be ignored, it all works out, but leaves people quite confused about dynamic range and density range. *: My intuition here is that this is actually _not_ true for scanners, but I infer from the definitions used at the above web page, that it _is_ a very good approximation. David J. Littleboy [EMAIL PROTECTED] Tokyo, Japan Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body
[filmscanners] dynamic range discussion
Please, enough already with the dynamic range argument. I want to learn about scanners. Thanks. -bruce Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body
[filmscanners] Re: Dynamic range -- noise
on 8/30/02 3:17 PM, Austin Franklin at [EMAIL PROTECTED] wrote: Back to your interpretation of the DynRange definition/formula. You are transforming the denominator from smallest discernible signal into smallest discernible signal increment. Whether it's the increment or not is determined by what ever the limiting factor is. BUT...you (typically) ONLY resolve down to noise, which is why, typically, noise IS the increment. Do you believe you can resolve a scanner signal further than to noise, and that you get useful information from that? Hi Austin and Julian, I've tried to address this in the past. The more I think about it the more this seems to be a crucial issue. With the way digital scanners work I agree that there's no hope of resolving better than the quantization noise (I prefer quantization error). But when it comes to true random noise you can easily get much better resolution than the noise level. Here's a small example: we have a voltage of 9.37 volts. First the quantization situation, we have a digital volt meter that measures to the nearest volt. The quantization error or noise is +/- .5 volts. No matter what we do we'll always read 9 volts and be off by .37 volts. Second we'll try the random noise situation: to be similar to the above let's say random noise of .5 volts is added to the 9.37 volts. So the voltage is at least bouncing around from 8.8 to 9.9. So any particular sample will have a value anywhere in that range. However if you make many samples and average them the result will converge on 9.37 because the errors due to random noise will cancel each other out. I've been claiming a lot of stuff lately so I'm going to try to back it up with a real demonstration. Here's a step wedge file that I based on the 21step wedges that come with Piezography. The top part is the standard wedge with 21 gray steps from 0% K to 100% K in 5% steps. The bottom is a duplicate with lots of noise added. The PS command is Add Noise 12.5% Gaussian if you want to try it yourself. The noise is a lot -- magnify to 400% on screen and see it, marquee a single step and check the histogram. What was 1 grayscale value now spans more than half the entire grayscale. Marquee and Histogram 2 steps and there no obvious steps. However, print the file out on paper and the step wedge shows through loud and clear. Get out the densitometer and the gray tone measurements of each step match very well whether you measure the noise-less step or the noisy step. So the signal here is the 5% wedge, the noise is large enough to span many steps in the wedge, but its easy to resolve densities much closer than the noise level. Download this file, its a TIFF to insure there is no lossy compression. http://www.harrington.com/21step-noise.tif Austin, I hope you are willing to print this out with Piezo and measure some of the steps. Roy Roy Harrington [EMAIL PROTECTED] Black White Photography Gallery http://www.harrington.com Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body
