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

Discussion in 'Miscellaneous Equipment' started by Ray Rogers, May 5, 2009.

1. ### Ray RogersMember

Messages:
1,555
Joined:
Aug 27, 2005
Location:
Earth
Shooter:
Multi Format
And now for something a bit different from your typical question!

I want to know
if the sky was a lens,
what would it's aperture value be?

That is, (to put the question differently)
How many stops difference is there between:

When film recieves an in-camera exposure of lets say
F/16 1/125th sec.
being properly exposed with 100 ISO film,
metering 18%gray at ISO 100.............

and

When the light hits that same speed film
having been placed in a printing frame...
and exposed out-of-camera for 180 seconds
in strong open sunlight.

.............

How much over exposure is the film in the second situation getting?
:confused:

2. ### VaughnSubscriber

Messages:
5,614
Joined:
Dec 13, 2006
Location:
Humboldt Co.
Shooter:
8x10 Format
I think you will have to back away from the aperture/shutter speed model and consider the amount of light filter through the sky in foot-candles or other similar unit of measurement. Luna Pros had a conversion table on them to go from EV to lux and/or foot-candles.

Vaughn

3. ### 2F/2FMember

Messages:
8,003
Joined:
Apr 29, 2008
Location:
Los Angeles,
Shooter:
Multi Format
First of all, practically speaking, a 100 film in a contact printing frame for 180 seconds in sunny 16 light ends up black when developed, and you don't need to know by how many stops, because it will be by a very healthy amount.

You can't apply an f stop in this case, because you need a focal length and a hole for an f stop to exist. When you are exposing something to raw light with nothing in place to reduce that light (like a lens or aperture), the concepts of aperture or f stop cease to apply, since the aperture is effectively infinite; as large as it can get, at any rate.

So, what you are really asking is how many times more light. If you prefer to think of that in terms of "stops", that is fine (almost all of us do it, I think), but realize that it is simply a choice of terminology that refers to doublings and halvings of amounts of light, and not a true measurement of f stops. First, you have to figure out how many times more light, then you can figure out how many "stops" to call it.

Regardless of whether or not you are taking a photograph, or what lens, shutter, aperture, ISO, etc. that you are using, a certain amount of light falls in a certain lighting condition. This light exists in the world, regardless of photography, and can be referred to using various systems of units. This amount is usually given to you by a light meter as an exposure value (EV); a term that takes a measurement of light that is falling, and puts it into photographic terms. Each EV represents a shutter speed/aperture/EI combination (and equivalent exposures) that theoretically will make the metered surface a mid tone grey.

So, to know how much light hits a piece of film in a contact frame versus how much hits a piece of film in a camera, you basically have to figure out just how much light the lens and aperture are cutting from the amount of light that exists at the scene; what percentage of the light that actually exists at the scene actually strikes the film when you take a picture with a camera. Once you know how much light is cut from the light at the scene just by using a lens or aperture, you can compare it to the amount that falls on the film without a lens or aperture (100% of the light that exists), then figure out how many "stops" difference there is at a given exposure time.

So, what you really want to do to compare is to pick a constant shutter speed for both the in-camera film and the film in the contact frame. If your in-camera exposure is '125 at f/16, and this causes to the film to achieve a certain amount of density, then how much less light does the film in the contact frame need in order to get the film to that same density with a '125 exposure?

It's impossible to test without either a giant shutter for the contact frame, or much slower exposure times (ND filters for the camera and ND gel sheets for the contact frame). Try to get a film that does not suffer from bad reciprocity failure. I might choose Fuji T64. It is nice and slow, and it is listed as being good out to four minutes with no exposure adjustments (though IMO it can go much longer). It has to be filtered with orange (Wratten #85) to use in daylight, which reduces its effective EI by 2/3 stop to EI 40. So, in sunny 16 conditions, you get '30 at f/16-1/3. Say you want a four second exposure time. To get this, you need to add 7 stops (see!) of ND; both to the camera and to the printing frame. If you want longer exposures than four seconds, you can add more ND. I would sandwich the film with a homemade film strip that is a series of bracketed grey card exposures of known density, or a premade step wedge.

Last edited by a moderator: May 5, 2009
4. ### Q.G.Inactive

Messages:
5,682
Joined:
Jul 23, 2007
Location:
Netherlands
Shooter:
Medium Format
That would be the original question, yes: how does the illuminance at film level relate to the illuminance in the scene, an image of which is projected on the film.
Lux per lux, 'mediated' by f-stops. Something like:
Illuminance_film = Illuminance_scene/f-stop * Constant.

(As Vaughn wrote, some meters have/had a table converting EVs to Lux.
An rough (value of constants is an approximation) formulae to do that is:

Illuminance in the scene (in Lux) = 2^(EV -(log(ISO * 0.32)/log(2)) * 85

But that doesn't tell us how many Lux the film is receiving.)

The question however should not be "how much less light does the film in the contact frame need in order to get the film to that same density with a '125 exposure?"
Time is the only variable, so we cannot set it to one fixed value for both situations. Unless, of course, you let light levels vary as well, and wait for the light level drop to the same level it is at the film level.
To monitor that, you will need to know how many stops* difference there is.
Which is the original question again.

*stops as a measure, indicating a halving or doubling of exposure. Not f-stops, as in the physical diaphragm and what state that is in.

Last edited by a moderator: May 5, 2009
5. ### Ray RogersMember

Messages:
1,555
Joined:
Aug 27, 2005
Location:
Earth
Shooter:
Multi Format
Well...
I was sitting on the edge of my chair for a moment there, but I am just as confused as when I started!

Time is intuitive...
but how much incident light do apertures reduce the light by?

Isn't it be proportionial across all lighting levels?

If so, then can't we find a % value that the light is reduced by?

I think there must be someone here who can do this sort of figurin', but it
is pretty much a brain teaser for me!

Still, I can't help thinkng the solution is going to be ridicously simple....
---
Vaughn mentioned Luna Pros had a conversion table on them to go from EV to lux and/or foot-candles... does anyone have that conversion table?
---
2f/2f:
I think there must be a diameter that after reaching, no longer allows enough more light in to change anything...

Imagine a shutter in the sky... closeing it at first has little effect overall, but eventually, the "apetuer" gets so small that the light reaching the printing frame begins to fall off... so there is probably some fixed diameter which is less than infinite, above which greater diameters do not produce greater effective exposure.
---
What is the definition of "F-stop" anyway?

6. ### Q.G.Inactive

Messages:
5,682
Joined:
Jul 23, 2007
Location:
Netherlands
Shooter:
Medium Format
Yes.
But is is of no use. All it does is tell you how many Lux the light illuminating the meter is.
You still do not know how that relates to the illuminance at the film plane.

You're not just dealing with a hole. There is glass in front and behind the thingy.
That glass can indeed concentrate the light into a point, or an area about the size of your film frame.
And given that, a bigger hole gives more light for the glass to concentrate and project in your image frame.
So theoretically, there is no hole big enough, as long as the glass plays along.

There is no "fixed diameter which is less than infinite, above which greater diameters do not produce greater effective exposure."
(There is, if you still want the lens to produce a useable image. But that (loss of image at f-stops larger than f/0.5) can be counteracted by making the lens longer; bigger hole, yet not so impressive f-stop.)

Conversely, starting with a huge hole, closing it down, every time you reduce the size of the hole the light available to project onto the film gets less.
From the very first moment you begin making the hole smaller.

It's the ratio between hole size (optical, not physical. That glass again ...) and focal length.
Invented so you know that the illuminance at film level is always the same as long as the f-stop is the same. No matter what lens you are using.
Very useful, that.

7. ### ic-racerMember

Messages:
7,808
Joined:
Feb 25, 2007
Location:
Midwest USA
Shooter:
Multi Format
Simple answer; when you take the lens off you are at F1.0.

This also means that when you put that fancy F0.9 lens on the camera you actually are getting more light to the film than without the lens

8. ### 2F/2FMember

Messages:
8,003
Joined:
Apr 29, 2008
Location:
Los Angeles,
Shooter:
Multi Format
I thought that theoretically, f/1.0 should give you 100% of the light at the scene onto your film, and the same as no lens at all...but then I remembered the Canon f/0.95 lens. How is it physically possible for a lens to increase the ambient light? I can't wrap my head around it.

Last edited by a moderator: May 6, 2009
9. ### ic-racerMember

Messages:
7,808
Joined:
Feb 25, 2007
Location:
Midwest USA
Shooter:
Multi Format
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.

10. ### Ray RogersMember

Messages:
1,555
Joined:
Aug 27, 2005
Location:
Earth
Shooter:
Multi Format
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?)

Your point that the aperture is more than just a hole is noted.

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.

Does that make sense?

Now that stuff about bringng in non-incident light to make it incident...
that is not what I had in mind.

I am clearly in over my head here, but the bottom line is,

How much over exposeure is being given when one exposes for the same duration but without an aperture?

That is, (to put the question the way I did initally)
Hummm
I wonder if the pinhole people could throw any light onto this?

Last edited by a moderator: May 6, 2009
11. ### 2F/2FMember

Messages:
8,003
Joined:
Apr 29, 2008
Location:
Los Angeles,
Shooter:
Multi Format
I seem to be missing it in the classifieds. Is it for a Bolex H16?

12. ### Q.G.Inactive

Messages:
5,682
Joined:
Jul 23, 2007
Location:
Netherlands
Shooter:
Medium Format
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.

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.

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.

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

13. ### Ray RogersMember

Messages:
1,555
Joined:
Aug 27, 2005
Location:
Earth
Shooter:
Multi Format
That's good to hear!

---------

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

14. ### VaughnSubscriber

Messages:
5,614
Joined:
Dec 13, 2006
Location:
Humboldt Co.
Shooter:
8x10 Format
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

15. ### 2F/2FMember

Messages:
8,003
Joined:
Apr 29, 2008
Location:
Los Angeles,
Shooter:
Multi Format
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.

Last edited by a moderator: May 7, 2009
16. ### 2F/2FMember

Messages:
8,003
Joined:
Apr 29, 2008
Location:
Los Angeles,
Shooter:
Multi Format
Pretty trippy. Thanks for the explanation.

17. ### Ray RogersMember

Messages:
1,555
Joined:
Aug 27, 2005
Location:
Earth
Shooter:
Multi Format
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!

18. ### Q.G.Inactive

Messages:
5,682
Joined:
Jul 23, 2007
Location:
Netherlands
Shooter:
Medium Format
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.

19. ### 2F/2FMember

Messages:
8,003
Joined:
Apr 29, 2008
Location:
Los Angeles,
Shooter:
Multi Format
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.

Last edited by a moderator: May 7, 2009
20. ### MarisMember

Messages:
947
Joined:
Jan 17, 2006
Location:
Noosa, Australia
Shooter:
Multi Format
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!

21. ### Ray RogersMember

Messages:
1,555
Joined:
Aug 27, 2005
Location:
Earth
Shooter:
Multi Format
ca. 27.7 stops more ?

22. ### Q.G.Inactive

Messages:
5,682
Joined:
Jul 23, 2007
Location:
Netherlands
Shooter:
Medium Format
You're still ignoring the bit mentioned earlier, that the Lux numbers are a measure for how much light is falling on the scene. It is a measure, not for what a lens (and the film behind) it sees, but of the maximum amount it could see.

A lens always projects light that is reflected off something in that scene (unless you point the lens at the light source itself, and have it fill the lens' entire field of view).
And unless that bit the lens sees reflects all the light that is falling on it, and does so - as far as the field of view of the lens (and the film behind it) is concerned - without concentrating or dispersing the light, it is not the same amount.

The relation between incident and relected light (the latter being what the lens, and the film, get to see) depends on unknowns. You can't derive a ' Lux level' at film plane, just taking the 'Lux level' in the scene as measure.

And that is still ignoring what a lens does, assuming instead that all it does is relay all of the light to the film.

23. ### Ray RogersMember

Messages:
1,555
Joined:
Aug 27, 2005
Location:
Earth
Shooter:
Multi Format
Thank you for the signposts...
I will do a reality check on it once I get the numbers ostensibly "working".

Ray