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# Thread: How much over exposure? or A-Max (a REALLY big apeture)

1. Originally Posted by ic-racer
Of course the film and lens are usually capturing reflected rays, rather than the ambient ones. Only a small percentage of the reflected rays are hitting the film without the lens. So, I think of it like a "light funnel." A reflected light ray passing a foot to the left of your open film surface won't hit the film, but a huge one meter diameter lens can capture that ray and bring it in. (f 0.05)

BTW I have that lens I posted. It is actually a f0.90 Switar.
I seem to be missing it in the classifieds. Is it for a Bolex H16?

2. Originally Posted by Ray Rogers
Thanks, O.G.
I was thinking about wanting to see that conversion chart for another reason.
(Wouldn't it be possible for people who can meter off the film plane to use that data?)
I'll try to put the numbers up later.

If a camera metering in the film plane/behind the lens produces a certain EV as a result, could that be translated into Lux at the film plane/behind the lens?
Could be, if it were a simple, 1 on 1 thing. But is it?

Things start to get complicated when you remember that the table is there to convert an incident light reading into Lux. That is, it says something about the light falling on to the scene.
The meter behind the lens however does not see that light, but the proportion of it that is reflected by the things in the scene.

So you would have to figure out a way to take the reflective properties of your scene out of the equation. Perhaps point your camera at something reflecting 100% (or near enough) of the incident light, and let that fill the meter's angle of view.
But then, you'd still have to make sure that the reflecting thingy does not disperse or concentrate the light it reflects, else it would not be a true measure for the light falling onto it.

Originally Posted by Ray Rogers
My idea that in some cases, increasing hole size (I can only think in physical terms) will have no effect above some point is based on my logic that the light source is itself limited and has "coverage"; I cannot contnually increase the effect of lighting on my subject in the studio simply by using a wider lens..
if the lens I am using, already lets in all of the useful (incident) lght.
Yes.
Given (!) that the lens already lets in all the light ...

But it never does.
That is, it does so only if the light source (when direct), or the subject lit by it (if indirect), directs all of its light into that hole.
Imagine that scenario. You put a light up in your otherwise completely dark studio, point it at the subject, while not being able to see anything except when you look through the hole, through the lens.

Originally Posted by Ray Rogers

How much over exposeure is being given when one exposes for the same duration but without an aperture?
Yes, if you find a way to relate the light falling onto the film when held in the light by itself to te amount of light falling onto the film when it is behind a lens with a given aperture size.

An answer was given, saying that at f/1, both amounts would be the same.
I don't know that that is correct. Could be.

If so, the rest is easy.

Originally Posted by Ray Rogers
Hummm
I wonder if the pinhole people could throw any light onto this?
If they have ever determined the correct exposure time for a hole of infinitely large size, they'd have the answer straight away.

3. Originally Posted by Q.G.
Yes, if you find a way to relate the light falling onto the film when held in the light by itself to te amount of light falling onto the film when it is behind a lens with a given aperture size.

An answer was given, saying that at f/1, both amounts would be the same.
I don't know that that is correct. Could be.

If so, the rest is easy.
That's good to hear!

"Hummm,
I wonder if the pinhole people could throw any light onto this?"

If they have ever determined the correct exposure time for a hole of infinitely large size, they'd have the answer straight away.

---------

Re conversion data:
When you get a chance I look forward to those numbers!
(Here, or by PM)
Thanks...

4. fc = foot candles, and is measured with diffuser, and at ASA 50

EV = Lux = fc (approx)
0 = 5.5 = 0.5
1 = 11 = 1
2 = 22 = 2
3 = 44 = 4
4 = 88 = 8
5 = 175 = 16
10 = 5500 = 500
15 = 175000 = 16000

So, if using the diffuser at ASA 50 one gets an EV of 14, then the amount of foot candles falling on the scene is 8000.

If I was using 400 ASA film, that would be a reading of an EV of 17 which is 1/500 sec at f16 or 1/128,000 sec at f1.

So how much light is actually falling on the film? Hell, I have no idea -- but in order to get the right amount onto the film based on its sensitivity, I need to close the opening to f16 and limit the exposure to a short 1/500 second...according to the meter.

Vaughn

5. I really need to forget about "stops" or "f stops" for this exercise to make sense in my mind. What I need to do is compare the actual amount of light falling on the in-camera film to the actual amount of light falling on the film in the contact frame. Using a given exposure time makes this easiest to imagine.

At the same shutter speed, if just as much light hits the film in the contact frame as hits the in-camera film using an imaginary lens with a 1.0 max. t stop AND a 1.0 max. f stop (assuming 100% transmission, which is not possible)...then the OP's example of f/16 at '125 sec. with ISO 100 film is equivalent to f/1.0 at '32,000 sec. Now that both films are at the same f stop, all you do is count the shutter speeds between '32,000 sec. and 180 sec., and you have your answer...ignoring reciprocity failure, of course. '32,000-'16,000-''8,000-'4,000-'2,000-'1,000-'500-'250-'125-'60-'30-'15-'8-'4-'2-1-2-4-8-16-32-64-128-256 = 23 stops. Therefore 180 sec = approx. 22-1/2 "stops".

Now, there were a lot of assumptions made in the preceding paragraph. BAD assumptions too. However, there is your theoretical perfect-world answer: 22-1/2 stops. (...I think...)

As I first said, the real-world answer is: 100% MAXIMUM DENSITY on the negative for the film in the contact frame.

6. Originally Posted by ic-racer
Of course the film and lens are usually capturing reflected rays, rather than the ambient ones. Only a small percentage of the reflected rays are hitting the film without the lens. So, I think of it like a "light funnel." A reflected light ray passing a foot to the left of your open film surface won't hit the film, but a huge one meter diameter lens can capture that ray and bring it in. (f 0.05)

BTW I have that lens I posted. It is actually a f0.90 Switar.
Pretty trippy. Thanks for the explanation.

7. OK, very interesting...

Your combined efforts yielded a value close to what I had calculated for a related question that actually stiimulated this one. Let me run through the logic here punch in the numbers and see if it really all makes sense to me, then I should be able to respond to 2f/2f's concern about the 100% MAXIMUM DENSITY issue.

But, Thanks for thinking this out for me!

8. The biggest, and all important assumption is that f/1 is equal to an infinitely large aperture, i.e. nothing there to reduce light levels.
Is it?

If not, the calculations making use of that assumption are of course not correct.

9. Originally Posted by Q.G.
The biggest, and all important assumption is that f/1 is equal to an infinitely large aperture, i.e. nothing there to reduce light levels.
Is it?

If not, the calculations making use of that assumption are of course not correct.
I does not make sense to me that the exposure at f/1 (t/1 actually) is equal to the exposure to "raw" light. Even assuming 100% transmittance, just because the aperture is the same diameter as the focal length does not mean that the lens is passing every bit of light. Since f stop is a measurement of relative sizes, I fail to see what it has to do with actual transmittance; just relative transmittance. However, the one fellow said it is, and if so, I wanted to see what the answer would be.

10. Here are some approximate numbers.

An in-camera film receiving middling (Zone V, for example) exposure absorbs a certain amount of light energy. This amount of this energy is expressed as the product you get when you multiply the INTENSITY in the units LUX by the time in SECONDS. Exposure is Lux.seconds.

Middling exposure for all films is conventionally 10 times the reciprocal of the ISO speed expressed in Lux.seconds. For example a100 ISO film receiving middling exposure (gray card in sunlight, 1/100 second @ f16) gets a 0.1 Lux.seconds dose of light . For 400 ISO film a quantity of 0.025 Lux.seconds is a "standard", average, Zone V, middling exposure. This reciprocal rule is the secret (?) behind how exposure meters are derived out of light meters.

Direct sunlight at midday is pretty exactly 120 000 Lux. Exposing film to this intensity for 180 seconds will result in 21 600 000 Lux.seconds of exposure. This is enormously greater than the 0.1 Lux.seconds a 100 ISO film gets in camera; more than 200 million times greater!

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