Wow - read this re: Film versus Digital debate!

<i>The most glaring (i think) is the images DAniel posted above (11/11 at 3.30pm) and the observational FACT that

at magnifications greater than what would normally be involved in the printing of a 35mm negative, the negative

consists of ONLY black clumps and clear sections.</i><p>

Because you are looking at filamentary silver <b>within</b> a grain - a subgranular feature, if you will, that

<b>nobody</b> denies can either be present or absent (binary). The more such subgranular "dots", the blacker the

grain, the lesser these "dots", the lighter the grain.

<p>

Think of this as "shading" a <i>grain</i> via a sort of halftone representation, the density of the filamentary

"shading" varying with illuminance etc.

<p>

<i>The other hole I think I see is that the "non-binary" camp are relying on EM images. Correct me if I am wrong,

but relating these images to how the crystals interact with light is somewhat of an oxymoron. EM's work at

wavelengths shorter than light.</i>

<p>

Yes, but what they show is the distribution of silver; visible light may be of much longer wavelengths, but that

doesn't mean that it can't pass through interfilamentary space in a grain. Besides, atoms are neither opaque or

transparent - opacity is a resultant property as we go to macro sizes. After all carbon is black and opaque but

sugar is transparent.

Vijay: "You clearly do not understand binary, or switching systems - you don't understand the concept that "storing photons till a threshold is met" is an absurdity because if the threshold is not met, the grain must remain transparent, despite the fact that it has lost its electron and blah blah. Is this said to happen in the literature?"

<p>

I don't understand binary? Dude, a kid could understand binary.

<p>

'Storing photons till a threshold is met' IS an absurdity, but the SAME EFFECT can be reached by the CHEMICAL DESIGN of film. Instead of sitting here telling me I don't understand what binary, <b>why don't you read the paper I linked to</b>?

<p>

"the grain must remain transparent, despite the fact that it has lost its electron"? <b>It</b> has lots <b>its</b> electron?? What, the grain lost its electron? You are all over the place. Maybe if I say it <b>one</b> more time, you'll <b>actually read</b> what I've already written 10 times before:

<p>

When a productive photon collision occurs, an electron is knocked off of ONE OF BILLIONS of halides within ONE silver crystal/grain. This electron is trapped by the gold sulfide electron trap (e.g. the 'ears'). The purpose of the electron trap is to hold the electron until a silver ion that has migrated near the electron trap can pick it up and be reduced to uncharged silver (metallic). At least 3 silver ions must migrate to one of these 'ears', each picking up an electron... so that means, all together, 3 electrons have to come from 3 halides; these 3 electrons are typically held at the electron trap and then given to silver ions that have migrated near the trap. Once there are 3 reduced silver metal atoms clustered at this 'ear', which is near the surface (which is important b/c developer can easily access the surface), developer can easy start giving electrons to other silver ions within the grain, during development, because electrons can be transported easily through this cluster of CONDUCTING SILVER METAL ATOMS. If the threshold of 3 reduced silver atoms is not met at any ear on a given grain, then, in the time span of development, the grain doesn't get any electrons. It stays as a grain of silver ions (silver halides, to be exact). These are then dissolved away in the next step of development, and then washed away, leaving clear film behind.

<p>

That is how the <b>threshold of 3</b> is established in film!

<p>

Please read some basic chemistry of photography. I'm really not confident that you know anything about it if you were touting earlier that <i>prolonged development wouldn't reduce unexposed grains</i> because if it did you would never have gotten back pretty portraits from the lab. I mean, c'mon, that's one of the first things you learn when you read anything about the chemistry of photography.

<p>

Rishi

Look, Vijay, I'm not trying to be mean. I think that our discussion/argument would proceed much more smoothly, however, if you read up on the basic chemistry of film.

 

My paper I linked to does cover it, but it might be a bit too technical. I'll try and find a better, concise reference, and post it back here.

 

I apologize if I sound offensive. I really do want you, though, to thoroughly read my paragraph above on the chemical rationale of exposure/development. I myself am getting pretty frustrated that this debate can't be put to bed!

Rishi

<i>Dude, a kid could understand binary. </i><p>

 

And yet evidently you don't. Listen, a binary system may be 1's and 0's for you logically, but it is far more difficult to achieve in practice. You have to create a system with two stable states - and you have to have a threshold mechanism for some phenomenon - when that threshold is tripped, the system rapidly toggles state. It must do this in vanishingly small time. In electronics, this is made possible by a rather contrived circuit called a bistable multivibrator or a simplified version called a flip-flop. A flip-flop is the basic element that is called a "bit", and has the ability to store state (i.e., has memory). Such a system is called a non-linear or switching system. A characteristic of a binary switching system is that it has only two stable states, and deliberately putting the system in any other state, makes the system rapidly go to one of its two stable states.

<p>

Postulating that grain is binary (i.e., can exist only in one of two states) must imply that there is a threshold of something - light - whatever - that causes the grain to trip from being transparent to being opaque. It must imply that transparent and opaque are the only two stable states for the grain, and intermediate states must rapidly resolve to one of the two stable states. Having a threshold requires a memory and a comparator (has that thing - light - crossed (comparison) the threshold that is stored (memory)). Don't try to invent theories to bypass this - all this is absolutely mandatory for a binary system.

If even one of these criteria is not possible within a grain, then grain is not a binary system. Period.

<p>

Oh and that losing an electron thing was said to show you that it didn't matter - didn't you notice the blah blah afterwards?

<p>

For the umpteenth time, postulating that a grain is binary shows no understanding of binary systems; I couldn't care less about the chemistry involved. If the math doesn't work, the chemistry is irrelevant.

<p>

I already pointed out that if you had a digital sensor where every pixel were binary (if some value of light energy touched it, it went to a logic 1 state, else remained logic 0), you couldn't create grayscale images, period. This is a property of binary systems in general, and has nothing to do with electronics or film or chemistry. Disprove this, and I'll agree with you that grain is binary.

<i>Because you are looking at filamentary silver <b>within</b> a grain - a subgranular feature</i><p>

 

Vijay, I need some clarification with this statement. <i>Is</i> what we are looking at in image (d) of Daniel's post <i>within</i> a grain, or a <i>collection</i> of grains? I read the caption as suggesting that it is a <i>collection</i> of grains. Further, if you do the maths, something doesn't add up in my book. 400x magnification of a 2 micron grain (if 2 microns is more or less correct), adds up to 0.8 of a millimeter. This image is clearly much larger than 0.8mm. What's missing here?

Vijay, I think Rishi is saying that the 3 electron trap is the threshold you talk about. Short of 3 electrons - no black.

 

Regards the greyscale images from binary pixels... I suggest that you can achieve this as shown with the example of halftones.

Bernie, you're right - that needs clarification. I should have looked at Daniel's images more carefully. There is a fundamental problem with those images - they progressively enlarge a binary section of the image - the catchlights in the model's eyes. Those catchlights are completely black, the pupil is clear. The source information is "binary", so sorry, that can't be used to "prove" that grain is binary.

Hmm... so Vijay, you understand binary systems so well that Mr. Bernie West here had to school you when he said: "Vijay, I think Rishi is saying that the 3 electron trap is the threshold you talk about. Short of 3 electrons - no black."

 

Funny how Bernie & I can understand how this system *could* be binary. I'm not saying it *is* for sure, because for me to say that it *is* for sure, someone would have to corroborate the statement "develop-able grains are fully reduced to black metallic silver". Oh, wait, that's what all the literature says! And here I was being nice giving you, Mark, and Ron the benefit of the doubt by doubting the literature and allowing you to try and show how grain is NOT binary.

 

"Postulating that grain is binary (i.e., can exist only in one of two states) must imply that there is a threshold of something - light - whatever - that causes the grain to trip from being transparent to being opaque." -- Yup, and Bernie referred you to my previous statement that explained to you the chemical rationale for this 'threshold'. Funny how he picked it out while, clearly, you missed it the first 10 times 'round. Maybe you'll read it now that Bernie, a third party, pointed it out? But, anyway, don't let me take the credit for the basis of this 'threshold'... It's straight out of page 86 of that Chem. Review paper I linked to.

 

"I couldn't care less about the chemistry involved" -- Yup, that's apparent.

 

Rishi

<i>Regards the greyscale images from binary pixels... I suggest that you can achieve this as shown with the

example of halftones.</i><p>

That is exactly the point. You can't do that. You really can't. Suppose you were photographing a gray wall with

that sensor. If the light falling on the sensor (from the lens) were enough to trip the threshold, you'd get a

white wall. If it were below the threshold, you'd get a black wall. Either all pixels would trip, or none would.

<p>

Halftones would require that two identical pixels, illuminated by the same amount of light (same uniform grey

wall) <i>do two different things</i> - (and proportional to the grayness of that wall, no less) - that's

impossible in the universe we inhabit.

<p>

Film can't magically do something that is a <i>mathematical</i> impossibility. You don't have to know chemistry

to understand that.

<i>Vijay, I think Rishi is saying that the 3 electron trap is the threshold you talk about. Short of 3 electrons - no black. </i><p>

You'll have proved its binary if you prove that 3+ electrons = ALL black, no gray. Saying less than 3 electrons = clear is not the same thing.

Phil,

Film is definitely alive! I shoot Kodachrome when shooting with my Nikon FE2 AND Velvia and TMax 100 when shooting with MY PENTAX

6x7. I get great results and the guys at my pro lab and pro camera store are always amazed at my 11x14 prints, either b&w or color prints

made from slides. The TOP wedding photographer in my city shoots film and he gets anywhere from $5,000 - $9,000 for a wedding. Just my 2

¢ worth. You can buy anything you want at those two main stores in NY.

Chris

"I already pointed out that if you had a digital sensor where every pixel were binary (if some value of light energy touched it, it went to a logic 1 state, else remained logic 0), you couldn't create grayscale images, period. This is a property of binary systems in general"

 

No... it can't be a property of binary systems in general if the halftone process works. For example, if you take this digital 'sensor' and make it a digital 'display' instead, then, if every pixel were binary, it could still generate grayscale images *if the viewer views it from a distance*. You do agree with that, yes?

 

I think what you're trying to argue here is whether or not a halftone-like pattern representing gray could be recorded by such a digital sensor... that I need to think about a bit longer.

I admit, I am a bit stumped on the gray wall example. I guess it comes back to your statement about luminance being uniform. I'm not familiar with this phenomenon (not saying it doesn't exist, just never heard about it). I would have thought that the grey wall itself isn't a uniform grey, but is made up of many darker and lighter points, which would be reflected (bad choice of word) in the image projected onto the sensor.

<i>No... it can't be a property of binary systems in general if the halftone process works. </i><p>

The halftone process is not binary: the dots can be any diameter from zero to max in a continuous range of

values.<p>

<i>I think what you're trying to argue here is whether or not a halftone-like pattern representing gray could be

recorded by such a digital sensor... that I need to think about a bit longer.</i><p>

I am making a broader argument - that any gray tone could not be recorded by

<i>any</i>

mechanism where the individual "pixels" could have only two states.

"You'll have proved its binary if you prove that 3+ electrons = ALL black, no gray. Saying less than 3 electrons = clear is not the same thing."

<p>

<i>My word</i>, Vijay, that's what I've been saying ALL ALONG & I even had two posts above where I cited literature that said EXACTLY that -- that exposed grains turn ALL black.

<p>

For those disinclined to read my posts, Vijay, I'll just copy & paste here, though I dunno if that'll treat your symptom of glossing over & not actually reading what I write:

<p>

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~<br>

Under 'Silver Halide' in Wikipedia, we find:<br>

"When a silver halide crystal is exposed to light, a sensitivity speck on the surface of the crystal is turned into a small speck of metallic silver (these comprise the invisible or latent image). <b>If the speck of silver contains approximately four or more atoms</b>, it is rendered developable - meaning that <b>it can undergo development which turns the entire crystal into metallic silver.</b>"

<p>

Now, let's turn to this website: http://www.cheresources.com/photochem.shtml

<p>

It says: "In general, as the grain size in the emulsion increases, the effective light sensitivity of the film increases - up to a point. An optimum value of grain size for a given sensitivity is found to exist because <b> the same number of silver atoms are needed to initiate reduction of the entire grain by the developer despite the grain size</b>, so that producing larger grains reaches a point of diminishing returns and no further benefit is obtained."

<p>

Look also at the image there, it shows the entire crystal turning into metallic silver.

<p>

Now, let's turn to this article: http://findarticles.com/p/articles/mi_m1511/is_8_21/ai_63583781/print?tag=artBody;col1

<p>

"<i>If a silver halide crystal already has a light-blackened duster containing a critical minimum of silver atoms--four, typically</i>--that cluster's electric field draws electrons to the silver ions around it, and <b>the whole billion-ion crystal is quickly blackened</b>. All the other crystals, unexposed or barely exposed, the developer leaves alone; the crystals react individually because the gelatin isolates them from one another."

<i>I guess it comes back to your statement about luminance being uniform.</i><p>

No, no - what I said meant creating a condition where exactly the same amount of light falls on every point of the film or sensor. Luminance can be uniform or not - it is just a function of the light source; think of luminance as sort of the brightness per unit area. Such a condition can be created by trying to photograph a smooth white wall, or by removing the lens from an SLR camera and pointing it to the sun, etc.

<p>

Also forget trying to focus on the wall, or worrying that it could be made of specks of black and white - all we want is that the light that falls on one pixel of the sensor be the same as falling on every other pixel.

"The halftone process is not binary: the dots can be any diameter from zero to max in a continuous range of values."

 

Yes but it CAN be. The dots CAN be all the same size, and be only white or black, and you CAN still create gray scale values if viewing from a distance.

 

Are you arguing just for the sake of arguing? Or do you like to throw in a red herring from time to time?

"Also forget trying to focus on the wall, or worrying that it could be made of specks of black and white - all we want is that the light that falls on one pixel of the sensor be the same as falling on every other pixel."

 

I see what you're saying but your argument hinges upon every single pixel receiving the same number of photon hits per unit time. And I'm not entirely sure that is correct for vanishingly small instances of time.

 

This is a bit of a thought experiment. I need to ponder longer.

A grain that "can undergo development which turns the entire crystal into metallic silver" doesn't necessarily do that all at one go. If it did, you'd have some grains that were developable, and went all black immediately, and some that didn't cross the threshold of 4 atoms that are undevelopable, i.e., would never develop, no matter how long you kept the film in the developer. Which means that there would be no relation between density and development time.

 

Rather as you keep the film in the developer the grain goes from four atoms of metallic silver to all atoms of metallic silver gradually, going through all possible tones from clear to black. This is what allows us to push or pull film.

 

"... the same number of silver atoms are needed to initiate reduction of the entire grain by the developer despite the grain size." Of course, but that reduction can't happen from clear to black instantly as soon as you put it in the developer - the grain has to get darker with time in the developer. Linear process - not binary.

 

In the images, you see a small Ag speck become a bigger Ag speck within the grain. Does this happen instantaneously with contact with the developer? Will that Ag speck be smaller if I take the film out of the developer earlier?

 

Think, man think - everything you show me points out a linear (not binary) system and process.

<i>Yes but it CAN be. The dots CAN be all the same size, and be only white or black, and you CAN still create gray scale values if viewing from a distance.</i>

<p>

Yes, and you can create sort of halftone images with that, but the process does not work in reverse. This is because of causality - if we know that an area is light we put less dots there, if we know another area is dark we put more dots there. In reverse, the light hitting the two binary sensor pixels is the same, they won't do different things. Basic law of cause and effect; if the inputs to two identical systems are identical, the outputs have to be identical.

<p>

<i>Are you arguing just for the sake of arguing? Or do you like to throw in a red herring from time to time?</i>

<p>

I do like to throw in red herrings from time to time, but not this time. I'm not arguing for arguments sake at all, I'm honestly trying to educate and learn.

<i>I see what you're saying but your argument hinges upon every single pixel receiving the same number of photon hits per unit time. And I'm not entirely sure that is correct for vanishingly small instances of time. </i><p>

If you were to photograph a sunlit wall, your exposure would be determined by the sunny 16 rule - for 100 ASA film, you'd expose for 1/125 sec, or 8 milliseconds. That is not vanishingly small at all.

<i>I see what you're saying but your argument hinges upon every single pixel receiving the same number of photon hits per unit time. And I'm not entirely sure that is correct for vanishingly small instances of time.</i><p>

 

And here, Rishi, you are correct. The amount of light hitting a surface will never be uniform due to quantam shot noise which is a poisson distribution. Poisson approaches a normal distribution for large numbers, hence an equal amount of photons will 'undershoot' the average, and an equal amount will 'overshoot' the average. So it follows that half the pixels will be black, and half white, presumably randomly distributed, and therefore a grey image when viewed at distance.

"Think, man think - everything you show me points out a linear (not binary) system and process."

 

No, not everything, as I quoted Ansel Adams as saying: ""Examination of a photographic negative with a magnifier reveals that it is not made up of a continous range of white-to-black values, but that such values are simulated using a controlled deposit of individual black specks. These specks are the grain of the emulsion, the reduced metallic silver deposited when a halide crystal responded to light and 'developed'."

 

That being said, as I said about 10 posts ago, I'm starting to lean towards the analog camp, though I don't doubt that the number of exposed grains also has a lot to do with perceived density of any particular feature being imaged... since any such feature, even the tiniest one, will still be represented by hundreds if not thousands of silver *grains* (and not just one grain, especially seeing as how a lens couldn't even resolve down to that level of detail... that is, if a lens is focused on some feature, and if you take the smallest element of that feature capable of being resolved by this lens/media system, and you take all light paths from that element that are also running through the lens, the lens wouldn't be capable of focusing all these light paths onto a point on the film as small as one silver grain... it just doesn't have that kind of resolving power. Some of these light paths would fall in and around said silver grain... therefore, that particular element will be represented by many silver grains. Thus the number of silver grains exposed within this area representing this element will have an effect on the resulting tone. This argument about the resolving power of the lens is exactly what I was invoking earlier to argue that perhaps some of the light paths *won't* fall on certain silver grains within this element-resolving area on the film... or they will but not enough if said silver grain is more insensitive than some of its neighbors... in which case they won't be exposed... and this itself could give rise to a tone that is less-than-black.).

 

"Yes, and you can create sort of halftone images with that, but the process does not work in reverse. This is because of causality"

 

I have to say, I like that argument. I still have to ponder on it longer though.

Bernie, don't invoke quantum mechanics when simple Newtonian physics will suffice. We aren't underexposing so badly that we get into shot noise. Assume enough exposure that the signal hugely dominates the noise.

 

Also try the thought experiment as a simple logical experiment - of course when you create a real binary sensor, and hugely underexpose, getting to very low S/N ratios quantum shot noise will dominate etc. That can't be the basis for predictably creating grayscale images from binary pixels though - by hugely underexposing. When the S/N ratio is that low, you can't form any kind of image, period.

Oh, thanks Bernie for that explanation! Would you mind then reading my really long paragraph in my last post above & tell me if that makes sense? The whole thing about the lens, resolving power, smallest element being imaged, etc., and tell me if that makes sense? I have to say I'm really going into unfamiliar territory at this point, and just trying to reason at this point.