# Digtal vs 8x10 film

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<p>To elaborate on what I just wrote, I think there is a difference between direct capture of an image by sampling and reconstruction of a sampled image.</p>

<p>To reconstruct the image you would calculate the entire frequency spectrum of the image (including the phase factors) and then add up all the wave forms to reconstruct the image.<br>

On the other hand, if you were directly capturing the image you would say that the image is taken straight off the sensor, without any further signal processing.</p>

<p>Additional discussion is welcome. I am sure there is something we can all learn from this discussion.</p>

<p>Oh, one more thing. I agree that there is no way that a 80 Mb sensor is going to capture all the detail of a good 4x5 image. I agree that there simply aren't enough sampling points in an 80 Mb image to do that. We disagree only in a numerical factor, and depending on how one frames the problem (taken a sampled image directly as the image itself vs. reconstruction of a sampled image using mathematical methods) we may not even be in disagreement with each other.</p>

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<p>Steve, I am easily showing the difference between 745 dpi and 4,000 dpi on TMAX 400. </p>

<p>If the Luminous Landscape claims that they do not have any detail past 745 dpi.</p>

<p>Also, 745dpi scan coincides with the resolution of the 80mp digital; so te crops above represent as well the difference in resolving capacity between the digital camera and the film.</p>

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<blockquote>

<p>(To be strictly correct and include the units the spatial frequency would be 1/inch.)</p>

</blockquote>

<p>Hi Alan,</p>

<p>Your error derives from interpreting each line as a half cycle.</p>

<p>It's not. It's a full cycle. </p>

<p>Each line can have any tonal value from white to black. We're simply assigning discrete values to each for the purpose of discussion. You could just as easily have adjacent white lines, or black lines, or gray lines.</p>

<p>- Leigh</p>

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<p>Leigh, the sampling rate to scan film losslessly is higher than 2x, this is because the grain structure is random. The simplest way to understand it is that if you are scanning a perfectly horizontal line pair, a 2x sampling rate will suffice to reproduce it. If the lines are diagonal and formed by components of random size and distribution, a 2x sampling rate will reproduce a jagged staircase in places where the original was more uniform.</p>

<p>Another way to understand it is by drawing a line on the sand with a 1 millimeter toothpick and then scanning with a 0.5 millimeter sampling rate. You won't get anything like the original.</p>

<p>For practical purposes though 6,000 - 7,000 dpi capture 99 percent of the detail that current film can record. Aside of some jaggies added.</p>

<p>The Luminous Landscape 745 dpi is so ridiculous that can only cater to a brand new generation of novice photographers. Sadly, it will stay on the internet to confuse people for years to come.</p>

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<p>Hi Mauro,</p>

<p>If you'll review my post from 4:51PM, you'll see my comment regarding the 2x factor, thus:<br>

"Even that resolution won't provide a high-quality data set, only a marginal one. An accurate scan would require four to eight times that resolution."</p>

<p>I believe you and I agree on this.</p>

<p>- Leigh</p>

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<p>I definitely agree. </p>
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<p>By the way, if anyone associated with the Luminous Landscape reads this post, my offer to shoot side by side to the 80mp digital with my 6x7 film cameras stands. MF film will show a clear resolution advantage over the 80mp digital (particularity with the same lenses as the crop factor of the digital camera will deteriorate microcontrast significantly).</p>

<p> </p>

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<p>Leigh,</p>

<p>A cycle is, by definition, when a periodic signal has returned to its original state without any additional repeats. If you have a pattern of black-white-black-white-black-white-black-white-... One cycle encompasses a single black-white pair, because that is the minimum amount of pattern that will come back to the original state, from which it can repeat with another black-white cycle, and so forth.</p>

<p>If one tries to define it as half that, say a single black, then it is not a cycle because the next thing to come is white, which is not a repeat of what came just before. In a black-white... pattern a single black line corresponds to a half-cycle (assuming that the black and white lines are of equal width), and a single white line corresponds to a half-cycle.</p>

<p>By the way, a repeating black-white pattern is the combination of a zero-frequency component (the average over one cycle, which is gray) and a pattern containing a fundamental frequency (the basic frequency of the cycle) upon which is superimposed a series of odd-order harmonics whose relative amplitudes are the inverse of the harmonic number. For example, the third order harmonic has a relative amplitude of 1/3, the fifth order harmonic has a relative amplitude of 1/5, etc. However, in a discussion such as we are having it is convenient to ignore the harmonics and pay attention only to the fundamental.</p>

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<p>Yes, Alan,</p>

<p>I know that a square wave is the sum of a fundamental and all odd harmonics.</p>

<p>I also know that the Nyquist limit is the anti-aliasing limit that defines the minimum sampling frequency (or maximum signal frequency) that can be used without introducing spurious responses.</p>

<p>Now go back and re-read my 5:37 PM post regarding sensor/subject alignment and dispute it if you can.</p>

<p>- Leigh</p>

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<p>Leigh, if you are talking about sampling at the nodes and the result coming out gray, I have already acknowledged that point and discussed it.</p>

<p>Now, with respect to the definition of a cycle, here is the definition from that font of all knowledge, Wikidedia:</p>

<p>A process that returns to its beginning and repeats itself in the same sequence...</p>

<p>This is essentially a paraphrase of the definition of cycle I gave in my last post.</p>

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<p>Some additional information on the sampling theorem, again from Wikipedia:</p>

<p>"More recent statements of the theorem are sometimes careful to exclude the equality condition; that is, the condition is if <em>x</em>(<em>t</em>) contains no frequencies higher than <em>or equal to</em> <em>B</em>; this condition is equivalent to Shannon's except when the function includes a steady <a title="Sinusoid" href="http://en.wikipedia.org/wiki/Sinusoid">sinusoidal</a> component at exactly frequency <em>B..."</em></p>

<p><em>The condition you described, which is where the signal was sampled (only) at the nodes, resulting in a uniformly gray set of points, corresponds to the condition where the signal was sampled at exactly twice the signal frequency. According to the more correct version of the sampling theory (quoted above) that sampling rate does not satisfy the sampling theorem. The sampling theorem as usually (and incorrectly) quoted says that the sampling rate must equal or exceed twice signal frequency. However, the more correct version says that the sampling rate must exceed the twice the signal frequency. Sampling at a rate equal to twice the signal frequency is not quite enough.<br /></em></p>

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<p><em>The reason why I take the time to post these things is because unscrupulous test like theirs are posted for personal financial reasons at the expense of hurting an entire community of artists.</em></p>

<p>I am sure that everyone can remember how Luminous Landscape was of the opinion that the 3MP 1.6x crop Canon D30 was equal in quality to medium format film. The site hasn't changed all that much.</p>

<p>Mauro's images are very telling. Thanks for posting them. :-)</p>

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<p>Leigh,</p>

<p>OK, I performed a numerical experiment on the computer. I started with a sinusoidal function with a frequency of 1 and an amplitude offset of 1 unit. (The function is sin(2*pi*f*t)+1, where f=1.) This is to simulate an offset sinusoidal function that varies from perfect dark (a value of 0) to perfect light (which in this simulation is arbitrarily defined to be a value of 2. The function has an average value of 1 (i.e. gray).<br /><br /><br>

I then sampled the function in two ways. The first was to at time points of multiples of 0.03125 seconds. This rate is well in excess of the Nyquist limit (0.50 seconds). The second was at time points of 0.46875 seconds, which is barely within Nyquist limit.<br /><br /><br>

Here is what happened. The first sampling scheme gave a fine grained representation of the original function, i.e. a smooth-looking sinusoidal function with a low of 0, an average of 1, and a maximum of 2. If we take the pattern on the sampling points (the sensor elements) to be a direct representation then the pattern is a visually true result.</p>

<p>The second sampling scheme gave a modulated sinusoidal function. At one point it is gray. Then at later points it starts to modulate, alternating between slightly darker and slightly lighter gray. The modulation builds until it is alternates between full dark and full white. Then the modulation decreases until it is gray again. The pattern repeats itself. This corresponds to a condition I discussed in one of my previous posts. It also corresponds to a strategy where the pattern on the sensors (the sampling points) is taken as if it were a direct representation of the image. However, this pattern is not a visually true result because it does not represent an even and sinusoidally varying pattern of light and dark, but rather it presents a Moire-like pattern, with the sinusoidal pattern being modulated by a function of lower period. Although the sampling rate satisfies the Nyquist sampling criterion, taking the sensor element readings as if it were a direct representation of the pattern does not give a visually true result.</p>

<p>However, In taking the Fourier transform of the two sampled results I get the following results. In the case of fast sampling (0.03125 seconds) I get a peak at zero frequency for the real part, and peaks at frequencies of 1 and -1 for the imaginary part of the function. (The Fourier transform uses complex numbers, which is why there is a real part and an imaginary part.) The two imaginary peaks have amplitudes of equal and opposite sign. The real-number peak at zero frequency corresponds to the average value of the function, which is 1 and which corresponds to gray. The imaginary-number peaks at frequencies of 1 and -1 corresponds to the sinusoidal part of the image, i.e. the alternating part of the function. From this information I can produce an error-free representation of the original offset-sinusoidal image by summing signals containing the three frequency components.</p>

<p>What happens when I do a Fourier transform on the vector obtained by sampling at intervals of 0.46875 seconds? Remember, that sampling interval barely satisfies the Nyquist sampling theorem, and the results of the sampling, if taken as a direct representation of the image, did not produce a true image. Here is the result of the Fourier analysis. I get a peak at zero frequency for the real part, and peaks at frequencies of 1 and -1 for the imaginary part of the function. (The Fourier transform uses complex numbers, which is why there is a real part and an imaginary part.) The two imaginary peaks have amplitudes of equal and opposite sign. The real-number peak at zero frequency corresponds to the average value of the function, which is 1 and which corresponds to gray. The imaginary-number peaks at frequencies of 1 and -1 corresponds to the sinusoidal part of the image, i.e. the alternating part of the function. From this information I can produce an error-free representation of the original offset-sinusoidal image by summing signals containing the three frequency components. In other words, when I perform an appropriate set of mathematical operations I get exactly the same result as when I sampled at the faster sampling rate of 0.03125 seconds.</p>

<p>Thus, the result is as I have previously discussed, namely that if you are sampling a repetitive pattern at a rate just slightly higher than the Nyquist cutoff, and if you take the resulting vector directly as a true representation of the image, you get a false result of a periodic image. However, if you do an appropriate mathematical transformation of the data you can reconstruct an error-free representation of the original image. Thus, sampling at the the Nyquist freequency (actually, to be stricture correct I should say slightly faster than the Nyquist frequency) does not give a true representation of the image if the numbers are taken as a direct representation of the original image, but the data has enough information in it to reconstruct the original image if one performs an appropriate set of mathematical operations.<br /><br /><br>

One result of this analysis is that if you want to take the sampled data as a direct representation of the image then you need to sample at a rate much higher than the Nyquist sampling rate. Your example of using four sampling points over the light-dark line pair is a reasonable approach. However, it is, nevertheless, almost twice the Nyquist rate and therefore is, from a signal processing perspective, somewhat in the realm of overkill. However, I do agree that your sampling rate does produce a good direct representation of the image, and I also agree sampling at the Nyquist rate (or, as just discussed, at slightly faster than the Nyquist rate) does not necessarily produce a good direct visual representation of the image.</p>

<p>If you would like the table of numbers I generated in the simulation then send me a personal message.</p>

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<p>No need to go crazy with theory. Just draw a picture on a square patterned paper. Then try replicating it by randomly placing (position and angle) another square pattern paper on top (against a window) filling only those square that have more black than white. How much smaller the pattern on the second paper needs to be to reproduce the original fairly accurately. </p>

<p>Now, even worse, try the original made of random elements (like grain) instead of squares.</p>

<p>For printing purposes though, as long as you have 300+ dpi you are good.</p>

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<p>Hi Alan,</p>

<p>First of all, a cycle is a time interval. It has nothing to do with amplitude.</p>

<p>Your assertion that consecutive cycles must be amplitude-continuous is invalid. Ask anyone who has ever dealt with binary data. Of course, if there's a discontinuity at the cycle boundary that greatly increases the transmission bandwidth of the signal, but that's relevant to the current discussion.</p>

<p>The white/black dataset constitutes a binary data stream, and must be sampled as such. The Nyquist limit does not relate to accurate reconstruction of a dataset.</p>

<p>- Leigh</p>

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<p>There has been a lot of discussion about the scanning, I can't help but wonder if the samples (especially the Ektar sample) don't have some kind of focus problem or film flatness problem.<br>

Something I am curious about though, if one only wanted to make a 16x20 inch then a 600ppi scan from the 8x10 neg would be enough. Would there be any advantage of say going with a 1200ppi scan would any of the extra detail make it onto the 16x20inch and would we be able to see it looking at the print from about 2 feet away.</p>

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<p>Oversampling the scan and then reducing it, will ensure Nyquist is not limitting you.</p>

<p>To your comment regarding the quality of the picture itself, their scan is so poor that it is impossible to determine whether the picture was also poor. Per their own comments about not finding detail pass 745 dpi you may infer the picture is bad too.</p>

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<p>Leigh,</p>

<p>You said that the Nyquist theorem does not relate to the accurate reproduction of a dataset, and by the context of the discussion I presume you mean a data set composed of a series of light/dark lines. I am not sure exactly what you mean by that statement, but if we take it at face value (i.e. that "the Nyquist theorem does not relate tot he accurate reproduction of a dataset") in the context of the present discussion then your statement goes against all the well established principles of signal processing theory.</p>

<p>Would you care to elaborate on what you mean by your statement?</p>

<p>There is absolutely no question that a digital image is a function that has been sampled and digitized on an equally spaced grid, and there is absolutely no question that the Nyquist theorem applies to the ability of a sampled function to capture the information in the original function. Therefore, the Nyquist theorem has direct application to digital imaging, and in particular it determines the ability to which one can reconstruct the original image from the sampled data set.</p>

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<p>Leigh,</p>

<p>One more thing. Contrary to your assertion, nowhere did I say that cycles must be amplitude-continuous. They must, however, be repetitive because repetitiveness is essentially the definition of cyclical.</p>

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<p>I don't know if anyone noticed but there is a link at the article page stating:</p>

<h6><strong>"Update:</strong> Some people have been critical of this test, saying that the 8X10" scans were not done at a high enough resolution. Markus has responded to this and also posted higher resolution samples <a href="http://www.markuszuber.com/8by10.html">here</a>."</h6>

"Of course you will say that the scanning resolution was not high enough. I have had a close look using a 20x loupe on the negatives and slides to see what I/we all have/had to see that the resolution of the 8x10" films does not have the same richness of details as we can see from the IQ180 shots. When I can see the grain, I know that there is nothing more behind. You can increase microcontrast but this will mainly lead to a more pronounced grain. This is also the reason why I did not go and look for other guys for doing a higher resolution scanning."</p>

<p>What is 20X in comparison to 745dpi?</p>

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<p>Giovanni, there's a number of problems with the authors statements. For starters, I'm not sure how the B&W film was processed....but with a 745ppi scan, there should be no grain visible. Second, his images appear soft from the get go. I don't see such softness in my 4x5 scans even at 2400ppi. SO this points to simply another issue with the capture process.</p>

<p>I think this will become nothing more than the old D30 vs 35mm Provia scans that have been laughed at by experienced photographers for the better part of a decade. All it appears that matters is that web traffic is up at the LL.</p>

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<p>Wow, is this getting extremely technical and I think I am actually understanding it is what is really bugging the crap out of me. However, I wish to ask a few points about scanning and printing. Please excuse my complete ignorance on printing anything above a 120 negative...</p>

<p>One: How does a 8x10 negative get printed? I realize you could contact print, but if you wanted to make that wall sized print, how is that done? Are there enlargers really that big? Is it the same with 4x5?</p>

<p>Two: Scanning. My cheap Epson v500 (I know I should have gotten the 700 or 750.....couldn't throw the money at it then) seems to do fine for the extremely limited 120 and 35mm scanning I have done with it. I will admit it appears that I can scan better than the photo lab I use to develop and scan my film currently (I use a Kodak 6850 to print mostly in 6x8 or smaller). Does a drum scanner simply scan the same point over and over again for better sampling? Is that why they are so much better?</p>

<p>I really would love to see the direct comparison to digital camera to digital print versus film negative to paper print......how about a 4x6, 8x10, 16 x20 and then a couple of monster sized prints from 35mm, 120, 4x5 and 8x10......THAT would be an excellent article for a photography magazine. (yes I realize advertisers would not like that article printed, except for the photo paper divisions of Kodak and Illford.....)</p>

<p>Bob E.</p>

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<blockquote>

<p>Are there enlargers really that big?</p>

</blockquote>

<p>Yes..... and bigger</p>

<p>A lot of people just contact print their 8x10 negatives to make 8x10 prints though. Many of these use alternative techniques which need UV light which will not pass through an enlarger's lens to be effective..</p>

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<p>Enlargers that big used to be commonplace and required in custom labs. We had five of them and the best was a computerized Durst horizontal 10 x 10 complete with pin-registered anti-newton ring glass negative carriers, a 2,000 watt halogen light source, custom lenses and a 15 foot vacuum wall with a vertical paper dispenser. Except for a few parts I could sell, everything was thrown to the metal scrap heap in January. Such is life in the digital world.<br>

You asked about drum scanning, and it remains the gold standard for many reasons. The film is fluid mounted to the drum so it's clean and perfectly flat, therefore sharp and clear corner to corner. Each point is directly sampled in exactly the same way without having to go through a lens or complicated light path so flare is minimal. Since the film is sampled spot for spot by PMT rather than a CCD technology, the results are generally smoother with higher dynamic range.<br>

People think of using a drum scanner when they need large scans, but even small scans are better. It's the not the number of pixels but the quality of the pixels that makes it better. Of course, any scanner is only as good as the operator and his skill in post processing is just as important as the technology.</p>

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<p>Thanks John and Steve. I wish I would have had the opportunity to have seen one of those huge enlargers......must have been an amazing sight to have seen in operation. What a shame to have simply thrown it in the dumpster.....</p>

<p>And now I know exactly how a drum scanner works too. Thanks so much for explaining exactly how it is done. So far, I have only had maybe two photos I would consider drum scanning. All the other "good" ones needed some sort of Photoshoping since I don't think my local lab does much if any real enlarging anymore. I really need to ask. Most of my work centers around close up people photography, not alot of landscape stuff yet....I can barely remember being able to "touch up" photos under the enlarger......that was a long time ago.</p>

<p>Bob E.</p>

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