Why no long telephoto zoom lenses specifically for cropped frame DSLR's

[joseph_wisniewski]

<blockquote>

<p>Joseph's argument is that a larger film or sensor has more silver or photo cells than a smaller film of sensor so it is more sensitive to light. This is convoluted logic.</p>

</blockquote>

<p>No. That is a convoluted interpretation.</p>

<p>Joseph's argument is that, in a particular scene, the number of photons on a trajectory to make a particular picture (same field of view, same depth of field, same exposure time, same dynamic range) is fixed.</p>

<p>It's hard to explain grad school level photonics without talking in terms of things like photon flux or quantum efficiency.</p>

<p>A scene emits a particular number of photons every second. That's the photonic flux Φ_p. Some of those photons are on the right course to go through the aperture of your lens and end up as part of the picture. When you multiply that flux by the exposure time, you get the photonic energy of your particular picture, N_p. That's what makes a particular picture, a certain number of photons. Not "f-stops" and not "ISO sensitivity". Photons. Light. The "essence" of photographs.</p>

<p>If you shoot at 1/100 sec with a 35mm camera with a 50mm lens at f2 at , and with an 8x10 with a 375mm at f15 (sorry for the ugly numbers, the ratio of diagonals is 7.5:1) you've still got the same number of photons, on the same trajectories, passing through a 25mm hole in space. It will, obviously, take film, or sensors, of differing ISOs to deal with that particular number of photons. Too fast a film, and the photons outnumber the available grains, and you get what is called "overexposure". Same thing if you turn the gain up too high on a digital sensor. To slow, and there's too much space between flipped grains, and you can't print it easily. The photons make the picture, the film or sensor has to be adjusted to match the number of photons. And the number of photons has to be sufficient for the picture. If you only have one photon, you have a "dot". Print that dot at any size you want, and it's still a dot. Add a few more photons, and the picture starts to take shape. What matters is the photon energy, N_p the total number of photons. It's actually possible to calculate the N_p, the number of photons you need to capture to make a print or screen image of a particular size with a particular resolution and a particular dynamic range. I'll repeat it yet again: photons are pictures. "ISO sensitivity" isn't the picture, it's a parameter, related to photonic irradiance, E_p, the number of photons per unit area of the film or sensor, as illuminated by the lens. ISO sensitivity is tied to film chemistry (and not particularly valid, as any zone system adherent could tell you) that the ISO never managed to successfully translate to digital (the ISO standard for digital sensitivity has 6 different computational methods and boils down to "do what you want"). That's why we have people looking at histograms and chanting mantras like "expose to the right": they're exposing by sensor saturation, dealing directly with the photons.</p>

<p>If you make the very common mistake of holding the ISO constant (I think I shot below 100 once on my 4x5, never shot anything below 320 on an 8x10, and shot all the way down to 3 on a 35) you end up taking very different pictures. I'm continually amazed by photographers with lots of film experience who somehow end up blind to that simple truth.</p>

<p>But overall, it's a simple concept. <strong>How many photons go into the final, deliverable image: N_p</strong> (say "N sub P"). The example that started this whole thing was that the Olympus 300mm f2.8, used with a 21.8mm diagonal sensor, was somehow "equivalent" to a 600mm f2.8 on a 43.3mm diagonal sensor. It isn't. Wide open, it delivers, to its sensor, 1/4 the photons that a 600mm f2.8 does to a FF sensor. It's equivalent, in number of photons (which is the final arbiter of "sensitivity", at the print) to a 600mm f5.6. With the same level of sensor technology, you get the same picture. Obviously, you need to set the FF camera to a "sensitivity" 2 stops higher to get that particular number of photons into the range of the sensor (same clipping, same shadows). And, although the current crop of four thirds sensors are made by different companies than the current FF sensors, they're at sufficiently similar levels of technology to see this relationship in actual photos. Shoot a four thirds at ISO 800, it looks a lot like an FF at ISO 3200. Both sensors received the same number of photons.</p>

[joseph_wisniewski]

<blockquote>

<p>And please, leave the sarcasm at your house.</p>

</blockquote>

<p>Mr. Pot, you dragged out the sarcasm super-soaker and wet down everyone else in this thread with gallons of the stuff. No one else here has, in any way, shape, or form, approached your own level of conduct.</p>

<blockquote>

<p>If you look at it like that there is no image quality advantage to using a larger format, myself, and everybody I know who uses medium format and larger, is doing so for the sole reason of increased image quality,</p>

</blockquote>

<p>It appears that I either know a lot more people than you; or that the people I know are more varied in their interests and abilities; or that I am simply sufficiently open minded to actually see what the people I know are really doing, instead of filtering it through my own preconceptions.</p>

<ul>

<li>I know Michael Pavlov, editor of "Light Leaks" magazine, who owns an awful lot of Holgas, Dianas, etc. because their image quality is so flipping bad. There's an amazingly large number of people who go with MF specifically for bad images.</li>

<li>Personally? My 4x5 has a standard 11 movements, the best I can do on my 35mm with a tilt/shift lens, is 4.</li>

<li>I've seen medical work, pre-digital, where they used a 4x5 because you could toss the film into the processor and have a viewable image on the light panel in 3 minutes flat.</li>

<li>I know someone who uses MF for no other reason than it has a leaf shutter and can sync flash at a higher speed, which gives her a better flash to ambient ratio on location.</li>

<li>I know someone who does a lot of 8x10 and a bit of 11x14 field camera work because it gives negatives that can be contact printed for platinum, instead of having to screw around with enlarging to inter-negatives.</li>

<li>I know way too many people who shoot MF or LF because they say it "forces them to slow down". That's not a difference in formats, that's a failure in a particular photographer's self discipline.</li>

<li>I know someone who shoots LF so she can scratch up the negatives, draw on them in ink, then contact print them.</li>

<li>I know someone whose only MF or LF camera is a 4x5 box pinhole camera.</li>

</ul><div></div>

[joseph_wisniewski]

<p>Oh, and I do know many photographers who shoot MF and LF for an image quality advantage, real or imagined.</p>

[James G. Dainis]

<I>"If you shoot at 1/100 sec with a 35mm camera with a 50mm lens at f2 at , and with an 8x10 with a 375mm at f15 (sorry for the ugly numbers, the ratio of diagonals is 7.5:1) you've still got the same number of photons, on the same trajectories, passing through a 25mm hole in space."</I><P>

 

You forgot about the inverse light law. Light intensity varies inversely as the square of the linear measure. Since the light has to travel further with the 375 mm lens, you need that 25 mm hole just to get f/16 with the 375 lens. <P>

 

Try to visualize what is happening in the camera. With the same lens and aperture opening just as many photons per sq mm are falling on the smaller sensor as on the larger sensor which has just as many photo sensors per sq mm as the smaller sensor.

<P><center><img src="http://jdainis.com/film_photons2.gif"></center>

[joseph_wisniewski]

<blockquote>

<p>Try to visualize what is happening in the camera.</p>

</blockquote>

<p>That is good advice. Since I understand, apparently better than anyone else participating in this discussion, what is actually happening in the camera, let's see if I can help you visualize it.</p>

<p>N_p photons enter the camera and are incident on N_sensor pixels of the sensor, an average of N_p/N_sensor photons per pixel. Those N_sensor pixels are then scaled until they are N_printer pixels (an awful lot more pixels, because printers print with an enormous number of pixels, often as many as 5,760 per linear inch, or 33 million per square inch), a factor of N_sensor/N_printer reduction in the number of photons per pixel. And the print ends up with</p>

<ul>

<li>N_p/N_sensor x N_sensor/N_printer = N_p/N_printer photons per pixel</li>

</ul>

<p>of information content. The sensor terms cancel out. That's not magic, it's science. It works for film, where the "pixels" are grains, and they only have 2 states, quiescent and flipped, but there's as many as 10 million of them per mm², and it works for digital, where a pixel is a "count" of incident photons (times quantum efficiency, but that's a bit beyond the scope of this) with as many as 65,536 counter states, but something more like 40,000 per mm². It's all about photon energy, which is another way of looking at information density. Your "inverse light law" doesn't alter the fact that there's exactly N_p photons of information in that photograph.</p>

 

[scott_ferris]

Thank you so much for being so magnanimous Joseph, if I wasn't so interested in the answer I can assure you I would

treat you with the same distain, oh, no I wouldn't.

 

However, I am interested and I see a contradiction. I fully understand your finite number of photons per exposure, but

how do I reconcile that with the fact that if I cut an 8x10 100 iso sheet film into four and use it in a 4x5 and expose for

100 iso I don't end up with a seriously overexposed image?

 

In the meantime I am happy to acknowledge you know many more people than me, in fact when I grow up I really

hope I am just like you, oh, no I don't.

[James G. Dainis]

"Your "inverse light law" doesn't alter the fact that there's exactly Np photons of information in that photograph"

 

No it doesn't but there are the exact number of Np photons of information in that photograph because the correct exposure was used. With the film 375 mm behind the read nodal plane the 25mm aperture opening had to be used because of the light fall off at that distance. This seems like teacher confusing the issue so he won't have to admit he made a mistake or left an important part out. Now you are implying that the distance of the film from the light source (the lens) has no bearing on the intensity of light reaching the film.

 

Scott,

 

Joseph's contention is the 4 x 5 film will have 1/4 as much silver because it is four times smaller in area than the 8 x 10 film. Less sensitivity means underexposed using the same exposure settings. That is what I meant by convoluted logic. There is just as much silver per sq mm on the 4 x 5 film as on the 8 x 10 film. None of the silver halide crystals fell off when you were cutting the film. I have mentioned that I have used 4 x 5 film on a reducing back along with 8 x 10 film under the same lighting, with the same lens and using the same exposures. After developing in the same developers for the same time both negatives were of equal density. If I had opened up two stops with the 4 x 5 film, the film would have been two stops overexposed.

 

Joseph,

 

Take an 8 x 10 film holder and load one side with a sheet of Tri-X 320 ISO film. Turn the film holder over. Cut out about one frame from a 35mm roll of Tri-X 400 ISO Tape that on the center of the film holder. Take the camera and use whatever method you want to get a good exposure on the 8x10 film. Turn the film holder around and recock the shutter. Change nothing and expose the 35mm film frame using the same settings as before. Develop normally. Do you really think that the 35mm film frame will be 8 stops underexpose? If you do you are wrong. I suggest that you actually do it to see. Or if you have them available try using the same lens and exposure setting on a full frame and 4/3 digital camera.

 

 

I would remind you that this forum is a public record. Anyone may do a Google search on your name and find it. Your original post:

 

"Maury, the Olympus 300mm f2.8 is effectively a 600mm f5.6 given the 4/3 crop factor. It applies just as much to aperture as to focal length."

 

is wrong, dead wrong. If I make a mistake I admit it. I think people respect that more than someone who continuously argues an untenable position.

 

What others say:

 

http://www.bobatkins.com/photography/tutorials/crop_sensor_cameras_and_lenses.html

 

http://forums.hardwarezone.com.sg/showthread.php?t=2361438

 

http://photography-on-the.net/forum/showthread.php?t=557405

 

http://www.bobatkins.com/smf/index.php?topic=582.0

[scott_ferris]

<p>James,</p>

<p>If Joseph's position is as baseless as your suggestion I am very disappointed, I had always held his replies, particularly regarding macro imaging, in high regard.</p>

<p>The only reference I can find regarding increased sensitivity as format size increases is when talking about the scenario I laid out previously, to get comparable prints, with the same characteristics, from different formats, what I have now learnt is called <strong><a href="http://www.josephjamesphotography.com/equivalence/">equivalence</a>. </strong></p>

<p>That linked article does state<br>

<em><br /></em></p>

<blockquote>

<p><em>"• 5DII at 50mm, f/5.6, 1/200, ISO 400</em><br /><em>• D300 at 33mm, f/3.5, 1/200, ISO 160</em><br /><em>• 7D at 31mm, f/3.5, 1/200, ISO 160</em><br /><em>• E30 at 25mm, f/2.8, 1/200, ISO 100</em><br>

<em>If the same scene is shot from the same position, all four systems will have the same <a href="http://www.josephjamesphotography.com/equivalence/#perspective">perspective</a> (subject-camera distance) and AOV. If the resulting images have the same <a href="http://www.josephjamesphotography.com/equivalence/#outputsize">display dimensions</a>, they will also have the same <a href="http://www.josephjamesphotography.com/equivalence/#DOF">DOF</a>. In addition, because the shutter speeds are also the same, they will also have the same <a href="http://www.josephjamesphotography.com/equivalence/#exposure">apparent exposure</a>. The level of <a href="http://www.josephjamesphotography.com/equivalence/#noise">apparent noise</a> will depend on the efficiency of the sensor, although, typically, for a given generation of camera and at the same level of detail, the apparent noise levels will generally be very close for equivalent settings. However, the level of <a href="http://www.josephjamesphotography.com/equivalence/#4">detail</a> will depend on both the pixel count of the sensor and the sharpness of the lenses used. In this scenario of fully equivalent images, the differences in <a href="http://www.josephjamesphotography.com/equivalence/#iq">IQ</a> between the systems will be at a minimum. As always, which system has the "IQ advantage" will be a subjective measure, but most likely will go to the system that is able to render the greatest amount of detail, which will often be the system that has the largest native pixel count and/or best performing lenses at the f-ratios used to capture the images."</em></p>

</blockquote>

<p>If that is what Joseph meant, and I laid out earlier, then he is 100% wrong to say that a larger piece of 100iso film is more sensitive than a smaller piece of 100iso film.</p>

<p>Another quote from the earlier linked article,</p>

<blockquote>

<p>"The <em>exposure</em> tells us the <em>density</em> of the light that falls on the sensor, where density means the number of <a href="http://en.wikipedia.org/wiki/Photon" target="_blank">photons</a> per unit area. For a given scene illumination, the exposure is determined <em>solely</em> by the f-ratio and shutter speed, <em>and does not depend on the sensor size or the ISO</em>. For example, two pics of the same scene, one at f/2.8 1/200 ISO 100 and another at f/2.8 1/200 ISO 400 will both have the same exposure, <em>regardless of the format</em>, because the same number of photons per unit area will fall on the sensor regardless of what the ISO is set at (the ISO simply applies a <em>gain</em> to the actual exposure) or how large an area the light falls on."</p>

 

</blockquote>

<p>Joseph, time to stop tilting at windmills, you are wrong.</p>

[ShunCheung]

<p>Scott, Joseph, and James: I think the right thing to do is to give this thread a rest.</p>

[richardsperry]

<p>Joseph,<br /> "let's see if I can help you visualize it."</p>

<p>Nope, I can't see it yet.</p>

<p>I still can't see how if James, with all other factors being equal(lens, focal length, aperture, shutter speed), that if he exposes a 4x5 film in his 8x10 camera, that he must use a different aperture or shutter speed setting if he uses 8x10 film instead.</p>

<p>What you are seeming to say is that light meters are completely inaccurate, to the point of being useless. A light meter reading would be relative to 35mm, or 120, or large format film, at the least.</p>

<p>"because printers print with an enormous number of pixels, often as many as 5,760 per linear inch"<br /> <br /> Really? 5760 dots per inch or pixels per inch is pretty intensive. Can the average human eye discern the difference between 300 compared to 600 PPI printed? What printers are you going to, to get almost 6000 PPI?</p>

[joseph_wisniewski]

<p>Shun, you're probably right. But, since Richard brought up some new points, and he did it without ad-homs, I hope you won't mind if I address his comments...</p>

<blockquote>

<p>Really? 5760 dots per inch or pixels per inch is pretty intensive. Can the average human eye discern the difference between 300 compared to 600 PPI printed?</p>

</blockquote>

<p>Seriously, printers really do print at that level. A printer lays down dots of ink with very little ability to control density. A modern inkjet that lays down 4,800 or 5,760 DPI can only do that with 3 different size drops of ink (measured in picoliters, millionths of milliliters, they're that small). That's not much tonal range, a typical image has a lot more than 3 tones. Typically, a critical observer can discern over 500 tones on a B&W image, different densities of black. So, they trade "spatial resolution" for dynamic range, through a process known as "dithering" or "oversampling". The eye can't resolve 5,760dpi, so the individual printer dots are, essentially, invisible. The eye can, under decent conditions, resolve 300-600, so it "blurs" about 100-400 of those "invisible"5,760dpi dots with 3 tones into 1 dot with 300-1200 tones. That's "information content" again. Photons per picture.</p>

<p>And that's what the eye actually resolves. The ISO standard is a 25cm viewing distance, and 6 line pairs/mm. That will satisfy about 50% of observers, it's what they chose as an "average" viewer. 6 line pairs is 12 pixels or dots, and 12 dots/mm is 300dpi. 8 lpmm, 400dpi satisfies 90% of observers. So, yes, the average observer can tell the difference between 300dpi and 400 dpi (see note 1). And, if you give that observer the ability to move the print closer than</p>

<p>This is how film photography works, too. At the microscopic level, film has no tonality, at all. A grain of silver is either flipped (dark) or unexposed (dissolved away during developing). There's an insane number of them. Photographic papers run about half a billion grains per square inch, or 22,000 dpi, linearly. Each of those dots can either be on or off. If you corral a 300dpi "chunk" of 22,000 dpi grains, you get 5,500 grains in a 300dpi area. The eye can't see the individual paper grains, so it blurs 22,000 dpi with 2 tones into 300dpi with 5,500 tones.</p>

<p>Slick, huh? Film photographers have been using a digital binary "on/off" medium and the sophisticated technique of "oversampling" for almost 2 centuries, without noticing it.</p>

<blockquote>

<p>What printers are you going to, to get almost 6000 PPI?</p>

</blockquote>

<p>Epson. Even my ancient Epson 2200, over 10 years old, runs up to 2,880 dpi. Canons are based on multiples of 300 dots, they typically only run to 4,800 dpi.</p>

<blockquote>

<p>I still can't see how if James, with all other factors being equal(lens, focal length, aperture, shutter speed), that if he exposes a 4x5 film in his 8x10 camera, that he must use a different aperture or shutter speed setting if he uses 8x10 film instead.</p>

</blockquote>

<p>He doesn't, if his desired result is nothing more than "a properly exposed piece of film". But one doesn't normally utilize randomly sized pieces of film, the final goal being constant sized prints. If you move to a smaller format, you typically move to a slower film, to get the same result. You don't keep aperture or focal length equal if you're trying to get the same image. That's pretty obvious. That's why people talk of "equivalent" lenses. Even team ad-hom admits that. What I just don't get is the insistence that you have to keep ISO constant. We don't do that in real photography. I never shot Tech Pan on a 4x5. I don't know if they even made it in that size. It resolved over 200 lpmm at MTF 30, and no 4x5 or 8x10 format lens resolves even half that. It's a film only for 30mm or MF, and then, only with exceptional lenses and technique. At the other end of the scale, I shoot portraits on the Nikon D3 at ISO 1600. I've shot enough four thirds cameras to know what their tonality is like at ISO 1600, and it's not adequate.</p>

<p>So, an example where you hold ISO constant and just keep cutting the film into smaller and smaller pieces is complete BS. You can extend it to its logical conclusion, half of half of half of half until you've got a microscopic piece of film with just one grain on it. It's not a "picture", it's not got any useful information content.</p>

<blockquote>

<p>What you are seeming to say is that light meters are completely inaccurate, to the point of being useless. A light meter reading would be relative to 35mm, or 120, or large format film, at the least.</p>

</blockquote>

<p>It is. It has provision for setting film speed. You use a different range of speeds for 35mm and large format. The light meter doesn't judge. If you do something silly like loading a 4x5 with ISO 3 holographic film or ISO 25 Tech pan, and you set the meter accordingly, you'll get correctly exposed, but technically useless results. The meter won't care that setting a four thirds camera to ISO 3200 doesn't make any sense: it will happily help you create trash.</p>

<p>note 1) I'd like to say that 400 dpi being a "critical" observer vs. 300dpi being an "average" one is why Epson chose 360dpi as the "base" resolution for their machines while Canon, HP, OKI, Centronix, etc. chose 300, but I think Epson actually picked a number that was easier to express in a mix of printer's units like "points" (1/72 of an inch). In any case, they "got lucky" because that extra bit of resolution gave them an edge in the art printer market.</p>

[studio460]

<p>Joseph's printer post makes wicked good-sense, and contains an excellent examination of "dithering" as it applies to modern inkjet printers. And I'm inclined to believe his "sensor" argument is also correct. Scott's arguments, too. I think the disconnect here is that the assumed conclusions from Joseph's arguments are being misinterpreted. In other words, you're both right; it's simply the precise meaning of the disputed conclusion(s) which remains in question.</p>

<p>I still don't understand it quite well enough to re-state it, but I believe Joseph is using as his "givens," equivalent noise levels of various sized sensors, or equivalent degrees of film grain of various sizes of sheet film, via-a-vis, the final output. I think the discussion here (on both sides) contains valuable information, and is worth fully understanding.</p>

[richardsperry]

So the two stop comment is in relation to printing to paper differences(with enlarger) between the 4x5 compared to the

8x10?

 

Not the film itself, but what one must do with the enlarger to get the same size paper print?

 

If that is what you mean, I can easily see that. I need 2 stops more light using the 4x5 to make the enlarged prints the

same size.

 

But it does not hold for a contact print or scanning, correct?

[richardsperry]

"It is. It has provision for setting film speed"

 

With the same film?

That does not make sense.

 

"the final goal being constant sized prints"

 

Why is that? One may print to any size he or she can.

[joseph_wisniewski]

<p>Ralph, thank you. Glad you found the dithering explanation useful. And you're right on about varying sensitivity.</p>

<p>Richard, it sounds like we're converging. But you went back to</p>

<blockquote>

<p>With the same film? That does not make sense.</p>

</blockquote>

<p>I have said, over and over and over and over again, with different film. The only thing that "does not make sense" is the insistence on using the same film with different size cameras. I know that from decades of experience with everything from half frame Oly Pen cameras to 8x10. I'm pretty sure you know it, too, and are, for reasons unknown, just holding the "same film" stand to be stubborn.</p>

<blockquote>

<p>"the final goal being constant sized prints"<br>

Why is that? One may print to any size he or she can.</p>

</blockquote>

<p>Why is that? Because Maury Cohen, the person who brought up the whole "this lens on one format is equivalent to that lens on a different format" made that a condition of the argument, back on page 2 of this meandering discussion.</p>

 

<blockquote>

<p>Shun... The Olympus 300 2.8 is effectively a 600mm f/2.8 lens given the 4/3 crop factor.</p>

</blockquote>

<p>If you don't hold the print size constant, nothing is "effectively" anything else, because you've now changed the parameters of the task: you're taking two entirely different pictures. Camera companies and photographers both use that term "equivalent" constantly. "Equi valent", having the same value. If you set out to do two entirely different tasks, the whole idea of what is "equivalent" goes right down the tubes. Sometimes, we do that when changing formats. There are tasks that certain formats simply aren't suited to. When was the last time you saw someone on the sidelines at a football game with an 8x10? And there's a reason Ansel Adams wasn't at Half Dome with a Leica M, while HCB though it a great camera for "the decisive moment".</p>

 

[richardsperry]

Not stubbornness.

When making comparisons, it is better to remove all the variables possible. Then the remaining single variable, with everything else being constant, is what is being compared.

 

If, in our 4x5 in an 8x10 model, one changes the film there is no constant for comparison in stating that the film is

more sensitive or not; when the film is different. And why would one? Any film available in 8x10 sheets would be

available in 4x5 sheets. In the given model, the film size is the variable, everything else needs to be constant to make

a comparison. If you introduce other variables, your model is no longer controlled. And the comparison is no longer

meaningful.

 

If what you are comparing is one of many variables in the experiment, then there is no comparison to be made.

We already know that an ISO 100 film is different than an ISO 400 film. Stating one is more sensitive than the other is

already known.

 

One needs 2 stops more light at the enlarger to make the print of the 4x5 the same size as the print from the 8x10 film. That's a given. Because my lens is now twice as far from the paper, I need 4 times the light for exposure of the paper. But stating this as that the 8x10 film is 2 stops faster is obfuscating the discussion. I assume it's theoretical, if one does have an 8x10 film exposed at 2 stops lower, the two prints with the same enlarger f stop are not going to look the same anyway. There are now introduced variables with the paper exposure.

 

For example with my enlarger experiments, a 30 second exposure of paper at say f/11 is NOT the same as a 60 second exposed print at f/16. A stop difference in time is NOT equivalent to a stop in aperture on paper(with my enlarger). They are close, but different enough to see a difference in density with the human eye. There is a flaw in the reciprocals. Most likely in the real world exposure curve of the paper, or enough defect in my lens to skew the exposures.

 

And thirdly, if the two films are contact printed or scanned, there is going to be no difference in the "ISO" of the two different sized films, everything else being constant.

[joseph_wisniewski]

<blockquote>

<p>When making comparisons, it is better to remove all the variables possible. Then the remaining single variable, with everything else being constant, is what is being compared.</p>

</blockquote>

<p>But in holding the film formula constant, you are changing one of the most important variables, the number of photons required to produce a picture. If you use the same film formula at two different sizes, you've changed the number of silver grains (and probably their shape and sensitizing chemistry, because "technology" isn't equal across emulsions). It's a lot closer to equal on digital, which can do so many things. Film can't change speed without also changing its contrast characteristics, it's non-linear, and annoying. Digital is linear, it does a damn good job of tracking photons, and this is a digital discussion. The OP brought up cropped DSLRs. The annoying equivalence guy brought up four thirds specifically. Cameras "gain up" consistently over a 5 stop range, that's the entire gamut of sheet film, from ISO 25 (if you can even get that in sheet) to 800 (the "native" speed of TMAX 3200 without pushing).</p>

<p>So, you're creating the very same multivariate condition that I'm seeking to eliminate.</p>

<blockquote>

<p>But stating this as that the 8x10 film is 2 stops faster is obfuscating the discussion</p>

</blockquote>

<p>I quite agree. For God's sake, stop saying it, and stop saying that I said it. You keep shoving those words in my mouth, over and over again, and it makes it incredibly difficult to have a rational conversation with you.</p>

<p>Seriously, how would you like it if I made up some off the wall interpretation of something you said, and then, no matter what else you said, I just kept repeating that twisted distortion over and over again. You'd probably find that pretty maddening. I could do it. You just made a statement about a specific printing condition where reciprocity is not holding. How would you like it if I generalized that, and repeatedly maintained that you stated, categorically, that aperture and shutter speed do not correlate in any meaningful way, ever.</p>

<p>That would drive you up the wall, wouldn't it. I believe you're doing that to me deliberately, and you're quite aware of what you're doing.</p>

<p>Oh, and there's a very simple explanation for this.</p>

<blockquote>

<p>For example with my enlarger experiments, a 30 second exposure of paper at say f/11 is NOT the same as a 60 second exposed print at f/16. A stop difference in time is NOT equivalent to a stop in aperture on paper(with my enlarger).</p>

</blockquote>

<p>It's probably not the standard "reciprocity failure" we're used to seeing with film. Printing paper is formulated to avoid that, at a tremendous cost in speed. You're dealing with an angular illumination problem. We see that effect in all transmissive focal optical systems. Your condenser is too slow, or your diffuser is not being properly illuminated. It's also a big problem in microscopy. The resolution is decreased by a condenser with a numeric aperture (yes, other fields of optics don't use flippin' f-stops, they're totally arbitrary) that is smaller than the NA of the objectives.</p>

[richardsperry]

"James Dainis - Sensitivity doesn't vary with sensor (or film) size. I can make an exposure at f/32 with an 8x10 camera

then slap a 4x5 reducing back on and make another exposure at f/32 on the 4x5 film. A 4x5 inch area of the 8x10 inch

film will be the same as that of the 4x5 film."

 

"Of course it does. One of your examples has 4x as many silver grains as the other. If you want to have anything

resembling the same picture at the same print size, you need to use a slower film on the smaller format"

 

I am rebutting these statements, Joseph. I'm not putting words in your mouth. His model is an 4x5 film in an 8x10 camera. It could be the same 8x10 sheet of film with a mask. It's not a different camera or format.

 

Honestly Joseph, I am not going to change my film, film exposure, or film speed depending on if I'm going to enlarge

to 8x8" or 16x16". I would change the aperture or exposure on the enlarger first. Doing it your way would be unnecessarily cumbersome and tedious for me. I don't know what size I am going to enlarge to anyway before or when I expose the frame. I'll know that after I've developed the film and I'll pick the best shots to enlarge. I have rolls of developed film that I will never enlarge, the sensitivity of the film of those exposures did not change because they never hit paper.

 

James' model was only used to show an analogy to FF to DX digital sensors. I am not the one who side tracked the

thread to film, originally. My reciprocal failure may be what you say, a flaw with my enlarger. But it is easily

compensated for. If I enlarge twice the size(of 6x6cm) I get the same relative exposure on the paper(say 8x8" to 16x16") if I open two

stops, remaining with a constant time. The two prints look the same, save size.

 

Printing to paper, which was your sidetrack, may never enter into most digital captures, ever. I suspect that the vast majority of digital images exposed, by everyone, likewise will never go to paper, and remain as digital files indefinitely. Which makes the whole set of stepped disconjugate analogies moot in the end; even if they were not disconjugate.

 

The leap to paper was yours, if you want to argue that paper size changes film(and sensor) sensitivity, at least try and be clear about it. I won't buy it, of course.

[richardsperry]

"But in holding the film formula constant, you are changing one of the most important variables, the number of photons

required to produce a picture. "

 

You are assuming that a "picture" is enlarged and printed to paper. And you are not stating that.

 

That assumption is a false premise.

 

And I got the 2 stops thing from this...

"Les, that's only true if you hold the ISO constant. Given equal sensitivity, a FF is 2 stops more sensitive than 4/3"

 

I'm still trying to figure out what you mean by this.

[joseph_wisniewski]

<blockquote>

<p>You are assuming that a "picture" is enlarged and printed to paper. And you are not stating that.</p>

</blockquote>

<p>No, I am assuming no such thing. I specifically spelled out printed, projected, or viewed on a monitor.</p>

<p>With that, I am done with you. Score yourself a victory in your little game of twisting other people's words around. Dance your victory dance.</p>

[jeff_livacich]

<p>The fact is, Joseph W., your initial statement that the 300/2.8 is equivalent to a 600/5.6 on a 4/3 camera is wrong. It is equivalent to a 600mm/2.8.<br>

<strong><em>The f/number of a lens remains constant, regardless of format.</em></strong></p>

<p><strong><em><br /></em></strong><br>

Compared to FF, 4/3 format receives half the total number of photons, but is also half the physical size. It receives the same number of photons as FF at any point on the frame. With any FF lens, the 4/3 is just a masked FF image. Exposure does not change.<br>

Switching between panoramic and 24x36 on an X-Pan will not change exposure, nor will changing between 6X8 and 6X4.5 in a Fuji GX680.<br>

Using the adapter Pentax made, put a 200mm/4 lens from a Pentax 67 on a Pentax LX (which has about 1/4 the image area as the 67), and it's still a 200mm/4, and will require the same exposure as a K-mount 200mm/4.</p>

<p> Compared to a 100mm lens, a 200mm lens magnifies the subject by a factor of 2. Images of objects within the frame will be physically twice the linear size, and have 4X the area.<br>

The 200mm lens produces an image of an object with 2X the linear size and 4X the area, so, it must transmit to the sensor 4X the photons from that object to produce the same exposure. The physical aperture must be 2X the diameter, therefore 4X larger in area, to transmit 4X the photons, and if it is, it will have the same f/number as the 100mm lens.</p>

<p>The smaller physical aperture of (for example) a 100mm/2.8 compared to a 200mm/2.8, and corresponding smaller number of photons, is compensated exactly by the fact that those photons cover a smaller area on the sensor due to less magnification. The image of an object is physically reduced to the same degree that the number of photons coming through the lens is reduced.<br>

1/4 the photons, in 1/4 the area=same exposure.</p>

 

[jeff_livacich]

<p>Addendum:<br /> RE: Example of the Pentax 200mm/4 lens: the 200mm on the Pentax LX will be equivalent to having a 400mm/4 on the 67, just as a 300mm/2.8 on 4/3 is equivalent to having a 600mm/2.8 on FF. It does not become equivalent to a 200mm/8 on the LX.<br>

Mounting either 200mm/4 on the LX, i.e., the 200mm/4 SMC Pentax in 6X7 mount, or the 200/4 SMC Pentax in K-mount, will result in the same exposure requirement. The only change is field of view.</p>

[jeff_livacich]

<p>Edit timed out.<br /> Misstated last sentence. The only change using the 67 lens on the LX is field of view between the two formats.</p>

[joseph_wisniewski]

<blockquote>

<p>The fact is, Joseph W., your initial statement that the 300/2.8 is equivalent to a 600/5.6 on a 4/3 camera is wrong. It is equivalent to a 600mm/2.8.</p>

</blockquote>

<p>The fact is, Jeff L., your current statement is the one that is wrong.</p>

<p>Now, this may come as quite a shock, but "equivalent" is a <a href="http://en.wiktionary.org/wiki/real">real</a> <a href="http://en.wiktionary.org/wiki/word">word</a>, having a <a href="http://en.wiktionary.org/wiki/meaning">meaning</a> that you can look up in a dictionary.</p>

<p>http://en.wiktionary.org/wiki/equivalent</p>

<blockquote>

<p><strong>Etymology</strong><br>

From Latin <em>aequivalentem</em>, accusative singular of <em>aequivalēns</em>, present active participle of <em>aequivaleō </em>(“I am equivalent, have equal power”).<br>

<strong>Adjective</strong></p>

<ol>

<li>equivalent (comparative more equivalent, superlative most equivalent)</li>

<li>similar or identical in value, meaning or effect; virtually equal</li>

<li>(mathematics) of two sets, having a one-to-one relationship</li>

<li>(mathematics) relating to the corresponding elements of an equivalence relation</li>

<li>(chemistry) having the equal ability to combine</li>

<li>(cartography) of a map, equal-area</li>

</ol></blockquote>

<p>Now, it is possible that you may never have actually shot a 300mm f2.8 on four thirds (as I have) or a 600mm f4 on FF (as I also have) or some of the other commonly claimed "equivalence miracles" of the four thirds adherents, and if so, you should give it a try, because you'd realize that the lenses are not "similar or identical in value, meaning or effect" or "virtually equal".</p>

<p>The 300mm f2.8 on four thirds cannot achieve the degree of background blurring that an actual 600mm f4 can on FF, let alone the hypothetical 600mm f2.8 that the four thirds lens is supposed to be of equal value to. The best that it can do is matching the DOF of a 600mm f5.6, and that is one reason why that is what it is equivalent to. This is not a hypothetical case, shallow DOF is often used in sports and wildlife (and sometimes, even in fashion) to isolate the subject from the background. Those billboards in sports arenas are distracting as all get-out. If a 600mm f5.6 would do the job, you don't think people would be lugging a 600mm f4 that costs 4x as much and weighs twice a much, do you? There's a reason that four thirds didn't sweep those fields. The 600mm f4 has an "effect" that the 300mm f2.8 simply cannot match. They are not equivalent, and the 300mm f2.8 is certainly not equivalent to a hypothetical lens that actually exceeds the "value" and "effect" of the 600mm f4.0.</p>

<p>The 300mm f2.8 on four thirds cannot achieve the degree of low light shooting ability that a real 600mm f4 or hypothetical 600mm 2.8 can, either. And that's important, when you're a sports shooter competing with other sports shooters, and you need to match the noise and shutter speeds of your competitors (there's that whole "equal power" or "equal value" concept again).</p>

<p>Oly users brag a lot about the 14-35mm f2.0 and the 35-100mm f2.0. Even the ones that haven't actually shot the lenses brag about them. I can go out with a 70-200mm f4 on FF and come home with exactly the same pictures in the same situations as the 35-100mm f2.0 on four thirds can do. Same DOF, pretty much the same noise at the same shutter speeds. Yes, I'm shooting the FF camera at a 2 stop higher ISO. But remember what the dictionary said about "equivalent", that it's "similar or identical in value", and "virtually equal". If you shoot a four thirds and a FF camera at the same ISO, you get pictures that are radically different. Radically different is not "similar or identical". "Similar" only happens when you shoot the FF camera at 2 stops higher ISO.</p>

<p>I've heard four thirds proponents say that the 25mm f1.4 Panasonic is (was?) equivalent to a 50mm f1.4 on FF. I can shoot a single candle light portrait on FF at 50mm f1.4 1/15 second, ISO 6400 and get a sellable picture from most subjects. If I try an ISO 6400 shot on a four thirds, I get something at home in the waste bin, not in a picture frame.</p>

<p>Oh, and one minor detail...</p>

<blockquote>

<p>Compared to FF, 4/3 format receives half the total number of photons, but is also half the physical size.</p>

</blockquote>

<p>It receive one quarter the total number of photons, not half. That is why it needs to be set to an ISO 2 full stops faster to deliver an equivalent ("similar or identical in value, meaning or effect") quality print at the same print size.</p>

<blockquote>

<p>It receives the same number of photons as FF at any point on the frame.</p>

</blockquote>

<p>So does the sensor on a point and shoot or a camera phone. Are you going to argue that a point and shoot with a 24-580mm "equivalent" focal length is equivalent to a 300mm f2.8 on four thirds or a 600mm f4 or f2.8 on FF? It may have an "equivalent" focal length, but as soon as you specify other parameters, spelling out both focal length and aperture, for example, you have to make each of those parameters equivalent to make the two lenses equivalent.</p>

[richardsperry]

We should hold a shoot off. Then post the results image results.

 

I think that's a wonderfull idea. An assignment with presentation.

 

Who's interested?