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

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

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?

2. 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. 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.

4. 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.

5. Originally Posted by Q.G.
That would be the original question...To monitor that, you will need to know how many stops* difference there is.
Which is the original question again.
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. Originally Posted by Ray Rogers
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?
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.

Originally Posted by Ray Rogers
I think there must be a diameter that after reaching, no longer allows enough more light in to change anything...
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.

Originally Posted by Ray Rogers
What is the definition of "F-stop" anyway?
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. 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. 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.

9. Originally Posted by 2F/2F
I thought that theoretically, f/1.0 should give you 100% of the light at the scene onto your film, and he 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.
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. Originally Posted by Q.G.
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.
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)
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.
Hummm
I wonder if the pinhole people could throw any light onto this?

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