Switch to English Language Passer en langue française Omschakelen naar Nederlandse Taal Wechseln Sie zu deutschen Sprache Passa alla lingua italiana
Members: 70,536   Posts: 1,544,245   Online: 729
      
Page 2 of 3 FirstFirst 123 LastLast
Results 11 to 20 of 23
  1. #11
    2F/2F's Avatar
    Join Date
    Apr 2008
    Location
    Los Angeles, CA
    Shooter
    Multi Format
    Posts
    8,008
    Images
    4
    Quote Originally Posted by ic-racer View Post
    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?
    2F/2F

    "Truth and love are my law and worship. Form and conscience are my manifestation and guide. Nature and peace are my shelter and companions. Order is my attitude. Beauty and perfection are my attack."

    - Rob Tyner (1944 - 1991)

  2. #12

    Join Date
    Jul 2007
    Location
    Netherlands
    Shooter
    Medium Format
    Posts
    5,686
    Quote Originally Posted by Ray Rogers View Post
    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.

    Quote Originally Posted by Ray Rogers View Post
    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.


    Quote Originally Posted by Ray Rogers View Post
    Is my orignal question answerable?

    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.

    Quote Originally Posted by Ray Rogers View Post
    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. #13

    Join Date
    Aug 2005
    Location
    Earth
    Shooter
    Multi Format
    Posts
    1,560
    Quote Originally Posted by Q.G. View Post
    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. #14
    Vaughn's Avatar
    Join Date
    Dec 2006
    Location
    Humboldt Co.
    Shooter
    8x10 Format
    Posts
    4,649
    Images
    40
    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
    At least with LF landscape, a bad day of photography can still be a good day of exercise.

  5. #15
    2F/2F's Avatar
    Join Date
    Apr 2008
    Location
    Los Angeles, CA
    Shooter
    Multi Format
    Posts
    8,008
    Images
    4
    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 2F/2F; 05-07-2009 at 04:43 AM. Click to view previous post history.
    2F/2F

    "Truth and love are my law and worship. Form and conscience are my manifestation and guide. Nature and peace are my shelter and companions. Order is my attitude. Beauty and perfection are my attack."

    - Rob Tyner (1944 - 1991)

  6. #16
    2F/2F's Avatar
    Join Date
    Apr 2008
    Location
    Los Angeles, CA
    Shooter
    Multi Format
    Posts
    8,008
    Images
    4
    Quote Originally Posted by ic-racer View Post
    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.
    2F/2F

    "Truth and love are my law and worship. Form and conscience are my manifestation and guide. Nature and peace are my shelter and companions. Order is my attitude. Beauty and perfection are my attack."

    - Rob Tyner (1944 - 1991)

  7. #17

    Join Date
    Aug 2005
    Location
    Earth
    Shooter
    Multi Format
    Posts
    1,560
    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. #18

    Join Date
    Jul 2007
    Location
    Netherlands
    Shooter
    Medium Format
    Posts
    5,686
    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. #19
    2F/2F's Avatar
    Join Date
    Apr 2008
    Location
    Los Angeles, CA
    Shooter
    Multi Format
    Posts
    8,008
    Images
    4
    Quote Originally Posted by Q.G. View Post
    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.
    Last edited by 2F/2F; 05-07-2009 at 03:47 PM. Click to view previous post history.
    2F/2F

    "Truth and love are my law and worship. Form and conscience are my manifestation and guide. Nature and peace are my shelter and companions. Order is my attitude. Beauty and perfection are my attack."

    - Rob Tyner (1944 - 1991)

  10. #20
    Maris's Avatar
    Join Date
    Jan 2006
    Location
    Noosa, Queensland, Australia.
    Shooter
    Multi Format
    Posts
    729
    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!
    Photography, the word itself, invented and defined by its author Sir John.F.W.Herschel, 14 March 1839 at the Royal Society, Somerset House, London. Quote "...Photography or the application of the Chemical rays of light to the purpose of pictorial representation,..". unquote.

Page 2 of 3 FirstFirst 123 LastLast


 

APUG PARTNERS EQUALLY FUNDING OUR COMMUNITY:



Contact Us  |  Support Us!  |  Advertise  |  Site Terms  |  Archive  —   Search  |  Mobile Device Access  |  RSS  |  Facebook  |  Linkedin