Wow - read this re: Film versus Digital debate!

[vijay_nebhrajani]

Bravo, Hugh. This resonates well with my experience too. By the way, I am a Kodachrome fan.

[mark_l3]

you can argue these points forever, which from the length of this discussion it already has. Regardless of what you use a quality image will shine through.

 

I DON'T CARE WHAT YOU USE, SHOW ME WHAT YOU DO WITH IT!!!

[vijay_nebhrajani]

Why does this discussion bother you that much, mark L?

[rich_evans]

Rishi/Mark - I've been enjoying this discussion and buy alot of your input (both sides) but let me add a few comments/corrections:

 

1 - film format has nothing to do with resolving power - film is film whether its 35mm or 110 or 2 1/4 x 2 1/4. Resolving power is a function of the structure of the film and emulsion. However if you wish to compare a given area of any film - say 1 square millimeter - with an equal area on a digital capture sensor, then you've got the basis for valid comparisons.

 

2 - the 'atomic level' at which this conversation seems to be based is totally out of scale with what is translated into a visible image. Density is the final arbiter of grey levels - density depends upon the multiplicity of crystals which are 'piled up' upon each other. You forget that the crystals at this level are NOT opaque. A single silver crystal as shown in the referenced micrograph is probably transparent or translucent.

 

3 - the electron micrograph used as a basis for this discussion was obtained from a transmission electron microscope (TEM) - and is a two-dimensional representation of the structures. To get a true idea of the 'transparency' of the crystals in question, one must use a scanning electron microscope (SEM) to obtain a more accurate topographic image which illustrated the true dimensionality of the features in question.

 

I'm not a film scientist or a chemist, however I am a research microscopist who uses SEMs every day, and it appears to me that you may be relying to much on the TEM images in these conversations to make assumptions about how images are formed at the molecular level.

 

--Rich Evans, (Ph.D. - Optics)

[rishij]

Ron: you call us 'you digital people'. I shoot film (Velvia 50) exclusively. The digital image I posted above was a test with a friend's camera. So my argument in this debate is purely for the sake of knowledge, not because I have a thing for digital because, uh, I still prefer film.

 

Since you do seem to know a lot about this, are you basically saying that there are many 'sensitivity centers' per silver crystal/grain, and that the number of such centers, on each grain, that contain at least 3 reduced silver metal atoms (b/c you need at least that many to support electron transport from the developer), as well as the number of reduced silver metal atoms at such sensitivity centers, determine the rate at which the reduction reaction that tries to turn all the silver ions (associated with halides) in the crystal to black metallic silver? Meaning the more sensitivity centers that support electron transport (b/c they have 3 or more metal silver atoms that have migrated to those sites), the more Ag ions that get reduced to metallic silver, and so the more dense the final resulting silver crystal is?

 

As you can see, I'm not going for a 'gut argument' here... I'm trying to come up with a sound chemical rationale that supports your/Mark Smith's argument.

 

Vijay: 10x10 microns? What kind of grain are you talking about? Most silver grains are 2 microns (largest) to sub-micron size.

 

You say: "Gazillions of photons are hitting each grain simultaneously in my example. Yet one R-grain magically decides to be unaffected by these photons,"

 

Not sure what you're arguing here. You realize there ARE areas of the film where not enough photons productively fall to even reduce 3, yes just 3, silver ions to silver metal, rendering those grains 'un-developable', which are then later dissolved away to give clear film base. So, yes, sub-micron (not 10 micron) size 'R-grains' do, from time to time, not get enough productive collisions with photons to generate a conductive site of 3-4 (at least) metallic silver atoms and so hence do not efficiently support transport from the developer during the development process.

 

So... how am I going off tangent?

[rishij]

Hugh,

 

Very interesting. I have to agree with you conclusions there.

 

I was guessing that my 35mm Velvia contained ~12 megapixels worth of info, but, maybe with an even better scanner or a different test (or a different subject matter, since Vijay brought up the possibility of bridge resonance introducing softness in my example), I may also see closer to 20 megapixels worth resolved.

 

Heck, maybe in my example above I underestimated how much the film resolved. I did argue earlier that it resolved more than the 10MP digital camera; but lines appeared 'sharper' in the digital image. Did you take a look at my side-by-side comparison of the shot of I-5?

 

Cheers,

Rishi

[rishij]

Rich,

 

Thanks for your input :) The only reason I brought up the film format was because I was using a bunch of examples where I was saying '10 billion crystals on one frame'... so I wanted to clarify that I meant 10 billion crystals on a 35mm film frame, because a large format film frame certainly has more than 10 billion crystals (since the density of crystals shouldn't vary from format to format for the same film). So, yes, I understand your point, and that's the only reason I brought that up

 

Rishi

[rishij]

Hugh, forgot to ask -- was any sharpening applied to your image? That could affect your guess of resolving power.

[Landrum Kelly]

Ken Rockwell? Trust but verify. He is usually worth reading, but check his opinion against those of others.

 

As for film, why not? I have some Bronica gear that I bought some time back that I have been wanting to try.

 

Right now, what deters me about seriously going back to film is indeed the scanning. It is simply a lot of trouble--but a lot of people will go to the trouble. Film is expensive, too, and using film is not necessarily going to get one away from the computer if one digitizes everything that one shoots.

 

Maybe we should combine all "film v. digital" threads into one mega-thread and just put it in one special place where it can get bigger and bigger, while people continue to cover the same territory over and over--myself included.

 

--Lannie

[vijay_nebhrajani]

Rishi, I don't know how to make this clearer - 10 microns, 2 microns - they are all at the macro scale, and the

phenomena are identical.

 

All I'm asking is that if film grains can only have two possible states - opaque, by the presence of the grain,

and transparent, by its absence - how is it possible to create a middle gray - any middle gray - under constant

illuminance?

 

The point is, the grains are identical, the illumination is completely uniform, so all the grains will behave

identically - either all will go opaque, or all will go transparent. That is the nature of a binary system. If

you replace binary grains with phototransistors, all the (binary) phototransistors will behave identically too -

either switching on, or remaining off. This is the nature, once again, of binary systems.

 

This is exactly why the phototransistors that make up a pixel are biased in the active region - i.e., as analog

(not switching) devices, and an analog to digital converter is used to quantize the response of the phototransistor.

 

Once again, if grain were binary, creating grayscale images would be impossible. If the phototransistor in a

digital pixel were made binary, creating grayscale images electronically would also be impossible.

 

The response of a single unit (grain or phototransistor) has to vary (linearly, logarithmically, exponentially,

who cares - all curves can be linearized) with illuminance; you can't achieve the same effect by having multiple

units, because all the units will sample the same thing.

 

Clear enough?

 

Now, postulating that all the grains or phototransistors don't behave identically is tantamount to imparting

intelligence or sentience to them, which is tantamount to absurdity.

 

My example was supposed to highlight this absurdity, but it evidently did not have the desired effect.

[rowland_mowrey]

Rishi;

 

I spent 32 years doing R&D at Kodak and many years + doing professional work before and after working for Kodak. There can be many centers on one crystal. That helps it be analog. :D

 

Ron Mowrey

[rishij]

Ron,

 

OK, sure, I'll buy that there can be many centers on one crystal. Whether or not that helps it be analog -- I'm not convinced yet. Because you have to show that:

 

A crystal with 4 metallic silver atoms (reduced) at one sensitivity site (enough to make this crystal 'develop-able', according to texts) will, upon development, result in a less dense overall silver crystal/grain than a crystal with, say, 4 metallic silver atoms (reduced) at 5 sensitivity sites.

 

Is that what you are saying and, if so, can you please find me like ONE reference paper that actually says/shows this?

 

Also, as for the spectral sensitizers that are first excited and then excite the halide to give up an electron: doesn't there have to be one spectral sensitizer molecule PER possible reduction event? Isn't that a lot of spectral sensitizer molecules per silver grain/crystal? Or can these spectral sensitizers recycle electrons from somewhere else?

 

Thanks,

Rishi

[rishij]

Vijay: "All I'm asking is that if film grains can only have two possible states - opaque, by the presence of the grain, and transparent, by its absence - how is it possible to create a middle gray - any middle gray - under constant illuminance?"

<p>

<img src="http://upload.wikimedia.org/wikipedia/commons/9/97/Halftoning_introduction.png">

<p>

Left: Halftone dots. Right: How the human eye would see this sort of arrangement from a sufficient distance.

<p>

For further reading, go here:<br>

http://en.wikipedia.org/wiki/Halftone

<p>

And let me quote yet another line from this entry: "At a microscopic level, developed black and white photographic film also consists of only two colors, and not an infinite range of continuous tones. For details, see film grain.

<p>

I mean, so far, every article I look up says this. It is you couple of guys who are arguing otherwise. If what you are saying is well established *please* point me to a reference that proves your point. If one doesn't exist, don't you think it's about time you published your idea in 'Science' or 'Nature' as a "this contradicts traditional thinking" article??

<p>

-Rishi

[bernardwest]

Rowland... Just for clarity sake, your answer doesn't fully answer Rishi's question (which was: "Meaning the more sensitivity centers that support electron transport (b/c they have 3 or more metal silver atoms that have migrated to those sites), the more Ag ions that get reduced to metallic silver, and so the more dense the final resulting silver crystal is? ") Is this the case?

[rishij]

Ron/Mark: All you'd have to do to prove your point would be just show an image with a few grains showing variable levels of black filamentous density (EM images).

 

I don't really understand the initial image... I see two grain crystals, and then some other filamentous black masses... where are the latter coming from?

[rishij]

And, Vijay, your argument makes no sense. Of course there can be some grains that remain unexposed even when a lot of light shines on a given area of the film *for a given exposure time* because grains vary in their sensitivity based on size, position, orientation, etc.

 

All of them will be exposed if you expose for sufficiently long enough.

 

Read about halftone processing as another way to achieve tones.

 

You have to realize that BOTH what Ron/Mark are saying AND what has been in established literature for years ARE possible methods of attaining photographic images. You trying to prove that you JUST CAN'T get grays from the method that has been established in the literature for years and is proven to work as it does for the halftone process is just going to make you look foolish.

 

I'm trying to come up with an argument that'll show that Ron/Mark's theory IS plausible. Because even if it is, the clumping of grains to give varying densities WILL STILL HOLD TRUE. So stop trying to disprove it.

 

Rishi

[vijay_nebhrajani]

Nice image Rishi, but no workey. For this to work, either

 

a) The smallest grain in the emulsion must respond first to the incident illumination and as light continues to

impinge, larger grains must start responding later, i.e., the response time of a grain must vary with light

intensity. Further since the grain is binary, it must "store" photons as they impinge, and when some correct

threshold is met, the entire grain must suddenly go from clear to black. This is the nature of binary systems.

The threshold has to vary with grain size, of course, so that smaller grains switch earlier in time.

 

or

 

b) A grain must grow in size (increase mass while isolated in an emulsion) proportional to the illuminance.

 

No other explanation is possible because the grain is binary - i.e., must have all AgBr or all Ag in it.

 

So which of the above is it?

[rishij]

Vijay,

 

I'm beginning to think perhaps you're a troll. Trying to waste my time. Either that, or you really need to read a paper on the chemistry of photographic development. Like the one I linked to.

 

Not that I even understand WHY you make this statement: "Further since the grain is binary, it must "store" photons as they impinge, and when some correct threshold is met, the entire grain must suddenly go from clear to black. This is the nature of binary systems."

 

But...

 

Yeah, that's exactly what is said to happen. Photons knock electrons off of halides, these electrons are trapped by an electron trap "gold or silver sulfide", some Ag+1 ions then eventually migrate to this electron trap site on the surface of the crystal, then the electron from the trap reduces the Ag+1 ion to metallic silver (Ag charged 0). At least 3 or 4 of these events must occur to render the crystal 'develop-able'.

 

I can't believe I just wasted my time even repeating that, which I've already stated in some posts above.

 

If anyone else has anything to chime in, please do. Otherwise, Vijay, I'm sorry, I just can't.

 

Rishi

[vijay_nebhrajani]

<i>Ron/Mark: All you'd have to do to prove your point would be just show an image with a few grains showing

variable levels of black filamentous density (EM images). </i><p>

 

From Mark's blog - a photomicrograph of Ilford Delta grain - you see lighter and darker grains; you also see the

same grain (a three dimensional crystal) lighter on one side and darker on the other.

<p>

<i>

I don't really understand the initial image... I see two grain crystals, and then some other filamentous black

masses... where are the latter coming from?

</i><p>

There are five partial grains in the image. The top left hand corner is a black, fully exposed grain. Below that

is a partially exposed grain, below that another fully exposed grain. On the top right is a fully exposed grain

and below it a partially exposed one. The TEM creates "shadows" of a three dimensional structure, so what you see

is a 2D representation of the grain. Black specs "outside" the tablet boundary are the ears. Mark, Ron - please

correct me if I am wrong.<div></div>

[rishij]

Vijay,

 

Thanks for pointing that out and linking the image. OK I admit... I got so caught up in the discussion I didn't check this picture out on Mark's blog (although I read his blog, so I must have spaced out and not seen the image).

 

Now I'm beginning to be convinced. As long as the darker areas in the image above aren't imaging artifacts.

 

And someone please confirm this... and that those ears arranged around the tablet boundaries are in fact sensitivity specks?

 

I'm leaning now towards: it's a combination of both camps... varying intensities of crystals themselves, as well as a greater dynamic range achieved via the concentration of exposed crystals controlling the perceived density of a section of film.

 

Rishi

[vijay_nebhrajani]

Rishi, please keep the verbal abuse to yourself. You don't seem to understand what I am saying and I have no

problem with that - I can explain the same thing again for your benefit, but you seem to be hell bent on seeing

only what supports your viewpoint.

<p>

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 <b>not</b> met, the grain must remain

<b>transparent</b>, despite the fact that it has lost its electron and blah blah. Is this said to happen in the

literature? That once the AgBr trap blah blah occurs, the grain still remains perfectly transparent? Once again,

it <b>must</b> remain perfectly transparent, since the only other state is perfectly <b>opaque</b>. If the grain

reaches any

intermediate state between transparent and opaque it is <b>not binary</b>.

<p>

Get it? It is not binary.

<p>

If you feel that what I am saying is absurd, then you are right - it is; because it

stems from the assumption that grain is binary. Go look up <a

href="http://en.wikipedia.org/wiki/Reductio_ad_absurdum">reductio ad absurdum.</a> It's a way of proving that the

initial assumption was false.

[keith_anderson7]

Wow, it's like you've unlocked the greatest mystery of the universe. Can we go take some pictures now?

[vijay_nebhrajani]

<i>Wow, it's like you've unlocked the greatest mystery of the universe. Can we go take some pictures now?</i>

<p>

If you were waiting for my permission, you have it. Go take pictures, now.

[vijay_nebhrajani]

<i>As long as the darker areas in the image above aren't imaging artifacts. </i><p>

 

Well, you also see semi-transparent grains: the grain behind that dark grain near the top center is visible through the grain in front.

[bernardwest]

<i>Now I'm beginning to be convinced</i><p>

 

Well, I'm not. I won't pretend that I understand all the minutiae of this debate, and I really don't have a clue which

side of the debate is right (although I would love to know!). But I see a couple of holes in the "non-binary" camp. 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. How do you explain this and maintain that grains are continously toned?

This example fits the halftone argument perfectly.<p>

 

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.