Largest print size from medium format ?

<i>"He and Q.G. are still sticking to this story too "</i><br><br>Graham,<br><br>That's a libelous contention.<br><br>I never even attached to that story, and certainly am not sticking to it. What i did was point out the silly things that are used (on whatever side of the argument) again and again, and in this thread too, in efforts to make a point and undermine what the opposite side is putting forward.<br>You're strong determination to do your best to make your point must have made you blind to that. ;-)<br><br>(Blind rage, painting with broad brushes, an inability to grasp the subtleties of an issue: all (it's one and the same thing, really) part of the range of silly things that happen, without fail, in these sorts of arguments. So no surprise. ;-) )<br><br>Back on topic though: that assertion you plucked of the blog is true, of course. Certainly not about every bit of exposed film, but lacking a mention of how big that bit of film is and how it was exposed, it's a given that there will be a bit of film that makes it true. ;-)<br>Subtleties, brushed over because too eager to make a point... A silly point, because it is not ad rem, but no more but an attempt to make someone look bad.<br>See?

<p>Q.G, if you don't agree with those figures then I am glad to hear it, but you appeared to agree twice with Jens' numbers:<br>

"films like the ones mentioned have no problem (in everyday, 'real life' situations) resolving those numbers of lp/mm" and<br /><br />"the length of my experience with film easily match yours, and i can confidently say there's nothing silly about those figures"</p>

<p>Ray - I agree; I browsed Jens's article, but rather than getting into a debate about what this or that film and lens combination may theoretically be capable of (and I <i>have</i> seen some very high resolution monochrome detail resolved by film), I thought it best to restrict myself to the actual image being shown here. As I said, I agree with you about the 4xoversampling - from Jens's other images, he does seem a bit keen on showing a scan that's got more pixels, but not necessarily more detail, than a digital image (there are some point sampled enlargements that are a little misleading).<br />

<br />

I think it may have been a little misleading of Graham to post a medium format digital image (very nice though it was) in response to a 35mm film scan, without stating up front that this was the source. Still, there'e been a lot of putting words into other people's mouths on this thread (especially the whole 10MP thing); I'm sure it would be better if we made sure we were sticking to the facts.<br />

<br />

Anyhoo. I have no intention of dismissing film, I just wanted to make the point that the image Jens presented is not a mind-blowing example that digital isn't up to snuff.<br />

<br />

As Joe said early on, the issue is not the print size, it's the viewing angle. If you work with the maximum angular resolution that the eye can resolve (I worked from <a href="http://www.northlight-images.co.uk/article_pages/print_viewing_distance.html">this</a>), I calculated that the eye can resolve the pixels of my D700 when the field of view of a print matches, roughly, a 60mm lens (in good light). It's a 40mm lens for a D3x. I have a 30x20 print that's fine from a normal viewing distance, but since the image was from a 24mm lens, if you view it at the right distance (20") the camera's resolution is clearly not up to snuff. 35mm film wouldn't be either; I keep meaning to get a 35mm (focal length) lens for my Pentax 645 for this kind of image, or just go large format. Hence I'm genuinely interested in the answers on this thread, in addition to the film/digital debate (which I agree has been done to death).<br />

<br />

I'm mildly curious how much resolution-per-film-area drops as film size increases. The increased thickness of 5x4 film makes matching 35mm resolution difficult. The flatness of 120 film may or may not allow it to keep up with 135. The larger formats have more total resolution, but I doubt they're absolutely identical per unit area; after all, 5x4 can get away with f/64 solely because of the larger size.<br />

<br />

The short answer to "how large can I print 6x7 at maximum quality" is: 6x7, because a contact print will have higher resolution than a projection. After that, it's very much a sliding scale.</p>

Graham,<br><br>There are many figures bandied about in this thread. I have no problem believing (as if it would be a case of 'believing') the figures Zeiss provide concerning the resolution films are capable of recording in every day shooting situations. There's nothing strange or controversial about those. On the contrary: they are part of that well-established, broad knowledge base gathered through many decades of using film and witnessing how film progressed.<br><br>How that translates to MP is the issue hotly debated over and over again. Discussions that are quagmires of misunderstood misinformation, partizan 'factoids' and convoluted ways of reasoning, in which terms are often given new, convenient, meanings, and in which inconvenient factoids are ignored. On both sides of the argument.<br><br>In other words: if you wish to think (as you apparently did) that i attached myself to anything Jens (or anyone else) claims, you of course are free to do so. Doesn't make it true. Just illustrates how silly these debates all too often are. ;-)

<p>Graham. Re the maths of combining MTFs. I made no mistake. The product of percentages is exactly how the combined MTF contrast figures are calculated, and give exactly the same results as Zeiss's graph. To use your example: 70% of 32% is 22.4%, (70/100 * 32/100 *100) which coincidentally is almost the same figure as arrived at by your method of 1/(1/32+1/70) = 21.96. However this is a pure coincidence, and if you check the results using your method with other points on the Zeiss graph, you'll find they're at variance.</p>

<p>Take the 80% (lens) and 52% (film) contrast points at 40 Lp/mm.<br>

80% of 52% is 41.6%, exactly as shown on Zeiss's graph. Using your method we would get 1/(1/80 + 1/52) = 31.5, which doesn't agree with the graph. </p>

<p>Again, if we take the two points that are the same for both film and lens at approximately 92% and 17 Lp/mm.<br>

92% of 92% is 84.64%, which also agrees with Zeiss's graph. But 1/(1/92 + 1/92) = 46. Oops! Sorry, but that's a mile out. And 30% of 30% is indeed 9% , not 15%.</p>

<p>The <a href="http://www.zeiss.de/C12567A8003B8B6F/EmbedTitelIntern/CLN_30_MTF_en/$File/CLN_MTF_Kurven_EN.pdf">full PDF from Zeiss, entitled "How to read MTF curves" is available here.</a></p>

<p>The mistake I did make that no-one has picked up on was to use Line pairs per millimetre as the spatial frequency measure, when the standard for MTF graphs should be Cycles per millimetre. But then Zeiss has made the same boo-boo too.</p>

<p> </p>

<p>Regarding the question of how large can you print with a given film format, there is an experiment that is fairly easy to help decide how large of a print you can make before moving up to a larger format film would show an improvement.</p>

<p>Using any format you wish take a photo with one of your sharpest lenses, in my case I use my 50mm, then change to a longer lens and take the same shot, in my case i would use around a 100mm lens. Make a series of print sizes from the photograph taken with the shorter lens and prints that match the scale with he longer lens. For example if I shoot my two photos with a 50mm and 100mm lens then for a 12x18 inch print from the 50mm shot I would make a 6x9 inch print from the shot using the 100mm lens. The result is that the two photos will have the same scale. When you get to a print size where you can see an improvement in the photo taken with the longer lens you are past the limit for the format the photos were taken with.</p>

<p>Of course the viewing distance you use is going to have a large impact on when you see the differences between the two shots.</p>

<p>This can all be done with a 35mm format and then scale up the results for larger format.</p>

<p>My guess is that if you are viewing the print close up that enlarging past around 8x will be produce a print that be noticeably poorer compared to going to a larger format and enlarging less, again viewing the print up close. Clearly the exact number depends on the film, lens and scanner (if use) to make the print.</p>

<p>Another way to look at this is some people are pretty happy with their 4000 ppi scans for film, but if they took the same shot with a format twice and large and scanned at 2000 ppi scan the larger format scanned at 2000 ppi would look far better.</p>

<p> </p>

<p>Joe, the same formula that I posted appears here: http://www.largeformatphotography.info/forum/archive/index.php/t-44093.html<br>

and here: http://www.apug.org/forums/forum44/87068-large-format-lens-test-results-perez-thalmann-realistic.html<br>

and page 6 of this presentation: www.light-and-shadow.org/Download/RettnerSharpAsATack-07-2005.ppt<br>

It also appears in this link, but with a cautionary note which I have quoted below: http://www.largeformatphotography.info/fstop.html<br>

"Several sources use the formula <samp>1/r_final = 1/r_lens + 1/r_film</samp> to compute the final resolution. This formula is an approximation to the exact calculation consisting of the convolution of the response of the film and the response of the lens. This approximation is most valid when both the film and lens are being used near their resolution limits (spatial frequencies with very low contrast). This corresponds roughly to f-stops up to f16. Thus the formula is pretty good for 35mm and MF work. However, when the frequencies involved are nowhere near the film limits, the formula is a poor approximation which predits a worse <samp>1/r_final</samp> than what you actual get. For f-stops of f32 and higher, what you get on film is in fact practically equal to the aerial resolution, and the formula shouldn't be used in that case."<br>

Norman Koren writes:<br>

"The response of a component or system to a signal in time or space can be calculated by the following procedure.</p>

<ol>

<li>Convert the signal into <strong >frequency</strong> <strong >domain</strong> using a mathematical operation known as the <em>Fourier transform</em>, which is fast and easy to perform on modern computers using the FFT ( Fast Fourier Transform) algorithm. The result of the transform is called the <em><strong >frequency</strong> components</em> or <em>FFT</em> of the signal. Images differ from time functions like sound in that they are <em>two dimensional</em>. Film has the same <strong >MTF</strong> in any direction, but not lenses.</li>

<li>Multiply the <strong >frequency</strong> components of the signal by the <strong >frequency</strong> response (or <strong >MTF</strong>) of the component or system.</li>

<li>Inverse transform the signal back into time or spatial <strong >domain</strong>."</li>

</ol>

<p>Source: http://www.normankoren.com/Tutorials/MTF.html<br>

This isn't very clear to me so I'll have to look into it further.</p>

 

<p>Graham. In those links you gave the addition of reciprocals is being used to add resolution figures, not MTF contrasts. An MTF curve can be used to find the limit to resolution, in cycles/millimetre (or more crudely lp/mm) of a system or individually of a film, lens, sensor or scanner say, but resolution is only one axis of the MTF curve.</p>

<p>So if we define the limit to resolution of the film in Zeiss's graph as 40% contrast (a bit high, but the lens contrast is only shown to just under 40%), and the limit of the lens as 40% as well, we can see that the respective <em>resolution</em> figures are 50 and 140 lp/mm. Now we plug these resolution figures into your addition-of-reciprocals formula and we get 1/(1/140 + 1/50) = 36.84, which isn't too far away from the real resultant of 42 lp/mm at 40% contrast as shown on the graph.</p>

<p>So, in the absense of full MTF data, and if we only have resolution figures to work with, the addition of reciprocals gives a reasonable approximation to the system resolution. But it's still an approximation, and no matter how many times it's been repeated, and with what authority, it still isn't the proper way to go about things. And you can clearly check with Zeiss's graph that adding contrast figures using that method gives completely the wrong result.</p>

<p>If you think about it, the product of percentage contrast makes absolute sense. If one component of the system reduces the contrast to half (50%) and another component reduces the contrast to one-quarter (25%), then the resultant contrast must be half of one-quarter, which is one-eighth, or 12.5%. Using the addition of reciprocals we'd get 1/(1/50 + 1/25) = 16.67.</p>

<p>Sorry Graham, but more than almost any other field of study, photography has a store of perceived "wisdom" that turns out to be based on almost no evidence, and most of what we think we know is wrong! Such as film being an analogue medium for example, or 140 line-pairs per millimetre being achievable on 35mm film with off-the-shelf equipment.</p>

<p>In a nutshell, Graham used the right (approximate) formula, but applied it to the y-axis values of the MTF graph instead of the x-axis values. A very understandable slip, since the y and x values often lie in the same numerical range (e.g. 0-100%, 0-100 lp/mm).</p>

<p>Yes, thanks Joe for the correction!</p>

<p>Apologies for the long-windedness of the above. I was trying to give a full explanation and got a bit carried away. Note to self - KISS!</p>