Reciprocity failure

[E. von Hoegh]

Who's arguing E.?

I asked an honest question(a couple actually), and was summarily insulted. And there were no answers(informative or otherwise) to them, go back and read the thread again.

I gave an example for Leigh, which I ask of you. I am sure you have printed a print with very close and middle of the road reciprocals, did those two prints look the same to you E.? Mine don't.

Why do all the reciprocal "corrections" with film increase continuously as the time gets longer? There are charts above in this thread that show that, I assume that they are correct "enough"; and other charts I have read show the same thing(off hand iirc Simmons has a chart in The View Camera). Does it make sense to you then that, reciprocity "failure" occurs only at either end, and it just clicks itself off in the middle?

Answer this too,E., in the context of long exposures for film(which is the topic of this thread), what is so offensive of my statement that The Law of Reciprocity is more of a suggestion than a Law? Was my suggestion to the OP to experiment(because reciprocal values no longer work), and figure out usable apertures and shutter times offensive?

Go back and read post 5 please. By all means quote what is so offensive about that post, E. or what is wrong with it.

If you had ever bothered to look, you would have found that what you speak of is properly called reciprocity law failure and it is a result of either very low light levels at one end, or very short exposure times at the other end. It is not a suggestion. All of the questions you ask above can be answered by a short study of sensitometry. Read up on it, like I did. http://www.amazon.com/Photographic-S.../dp/0871001802

[RichardSperry]

Leigh,

Except with long exposures and very short exposures.

And the "failure" increases exponentially as the time increases proportionately.

You haven't received an answer to that question because it makes absolutely no sense.

It doesn't make sense to me, when there is an obvious change of rate(of needed correction) as time increases that there is no "failure", no matter how small, in the middle. You yourself stated that Failure applies to all photographic media. Its a given that the failure of film, in the middle average values, is imperceptible, so I gave a model of photo paper instead. Print a sheet at f/8 for 10 seconds, print a second at f/11 for 20 seconds...they will not look the same Leigh. You know this is true.

[Leigh]

Except with long exposures and very short exposures.

"Except" what? Your sentence does not reference anything.

The same toe and shoulder curves will result from changes in aperture, although we normally cannot change
that value over the same wide range as shutter speeds due to physical limitations.

It doesn't make sense to me, when there is an obvious change of rate(of needed correction) as time increases that there is no "failure", no matter how small, in the middle.

Then you don't understand the mathematical meaning of the term 'product', which is the result of multiplying two numbers.

Print a sheet at f/8 for 10 seconds, print a second at f/11 for 20 seconds...they will not look the same Leigh.

They look the same on my equipment, provided the times are within the straight line section of the paper's exposure curve.

If they don't on yours, I suggest you get your equipment repaired.

- Leigh

[Lenny Eiger]

Does it make sense to you then that, reciprocity "failure" occurs only at either end, and it just clicks itself off in the middle?

It does. I once had it explained to me by a physicist. (I am not a physicist, so this is for explanation only, I can't say I did this test myself.)

Here's what he said: It has to do with the difference between what happens to one of these grains of silver salts from being hit by photons. If they are hit by enough photons, they move from the "undevelopable" to a "developable" state. We know what happens in the developer....

However, at some levels of energy (less photon strikes) the grain moves to a third state, which is "I don't know". It may like the bowling pin that gets hit by the ball but not with enough force and so tips back and forth, and will either fall over, or right itself.

In the case of "not enough light" this third state rules. Some of the grains move back from the precipice of being developable and become undevelopable again.

The less the amount of light, the more of them are in the state where they might change back, and so there is a predictable rate of reciprocity failure. I don't care what AA said, it's a failure of a grain to turn to the developable state by being hit with photons. Failure is a fine word, IMO. I think the distinction is fairly meaningless.

At any rate, I hope this helps you understand what happens at the toe of the curve, at least. What happens at the top is a different matter.

Regards,

Lenny

[E. von Hoegh]

It does. I once had it explained to me by a physicist. (I am not a physicist, so this is for explanation only, I can't say I did this test myself.)

Here's what he said: It has to do with the difference between what happens to one of these grains of silver salts from being hit by photons. If they are hit by enough photons, they move from the "undevelopable" to a "developable" state. We know what happens in the developer....

However, at some levels of energy (less photon strikes) the grain moves to a third state, which is "I don't know". It may like the bowling pin that gets hit by the ball but not with enough force and so tips back and forth, and will either fall over, or right itself.

In the case of "not enough light" this third state rules. Some of the grains move back from the precipice of being developable and become undevelopable again.

The less the amount of light, the more of them are in the state where they might change back, and so there is a predictable rate of reciprocity failure. I don't care what AA said, it's a failure of a grain to turn to the developable state by being hit with photons. Failure is a fine word, IMO. I think the distinction is fairly meaningless.

At any rate, I hope this helps you understand what happens at the toe of the curve, at least. What happens at the top is a different matter.

Regards,

Lenny

Which is precisely how it is explained in the book I posted the link to above. Available used from $1.98, and it has a host of other great info.

At the top of the curve, for very short exposures, there isn't enough time for the photons to initiate the change in state in all silver grains, and so the compensation - preferably with the aperture.

[RichardSperry]

If you had ever bothered to look, you would have found that what you speak of is properly called reciprocity law failure and it is a result of either very low light levels at one end, or very short exposure times at the other end. It is not a suggestion. All of the questions you ask above can be answered by a short study of sensitometry. Read up on it, like I did. http://www.amazon.com/Photographic-S.../dp/0871001802

The book in your link is out of stock. I know what reciprocal law failure is, E. by the way; I have bothered to look.

Anyway,
Of course, the "Law" is not even a good suggestion at very long exposures(which is the topic of this thread). And most light meters don't work reliably there anyway(the Luna Pro SBC being a notable exception).

Add to that, differences between films, say when Acros 100 becomes faster that a 400 film. The law is useless for determining what is an equivalent time for say an f/11 8 minute shot on Acros compared to HP5+, for example. E., how long would you time that HP5 for, using the Law?

[E. von Hoegh]

The book in your link is out of stock. I know what reciprocal law failure is, E. by the way; I have bothered to look.

Anyway,
Of course, the "Law" is not even a good suggestion at very long exposures(which is the topic of this thread). And most light meters don't work reliably there anyway(the Luna Pro SBC being a notable exception). Why it's called "failure"

Add to that, differences between films, say when Acros 100 becomes faster that a 400 film. The law is useless for determining what is an equivalent time for say an f/11 8 minute shot on Acros compared to HP5+, for example. E., how long would you time that HP5 for, using the Law?

For HP5 I would use my LunaSix and compensate per the Ilford datasheet. Films respond differently, but you knew that right?
There are 12 used copies of the book available. It's out of print, so new ones are scarce.

[Leigh]

The film datasheets normally include correction curves for long exposures.

As usual, you would derive more benefit and understanding from reading than by posting nonsense here.

- Leigh

[RichardSperry]

If they don't on yours, I suggest you get your equipment repaired.

Really? I will test it again, on a different enlarger. Thanks for the suggestion.

"Except" what? Your sentence does not reference anything.

I was referring to your previous post, never mind it now.

Then you don't understand the mathematical meaning of the term 'product', which is the result of multiplying two numbers.

When I see an increasing rate of change, I think of differential calculus not multiplication.

[RichardSperry]

The film datasheets normally include correction curves for long exposures.

I have my times for my film memorized, Leigh.

As usual, you would derive more benefit and understanding from reading than by posting nonsense here.

- Leigh

I should go out a shoot some film, and not waste my time with you lot.