Switch to English Language Passer en langue française Omschakelen naar Nederlandse Taal Wechseln Sie zu deutschen Sprache Passa alla lingua italiana
Members: 55,696   Posts: 1,143,524   Online: 565
      
+ Reply to Thread
Results 1 to 9 of 9
  1. #1
    Photo Engineer's Avatar
    Join Date
    Apr 2005
    Location
    Rochester, NY
    Shooter
    Multi Format
    Posts
    19,238
    Images
    65

    Sensitometry pictorially

    There have been a lot of questions on the latitude of this and that film and this and that paper.

    I have taken some curves that I know are realistic and put them on paper for you to see.

    In the attached file, you will see 3 lines (curves). The horizontal (X) steps go from 1 - 21 and represent 0.15 Log E of exposure increments for each step. The vertical (Y) steps go from 0 to 3.6 and represent density units in 0.2 density units per vertical step.

    The three curves are, in order of your normal use:

    1. A normal professional film with a contrast of 0.60 taken from an actual coating.

    2. A contrast grade 2.0 paper which has a mid scale contrast of 2.5. The four points marked on it are the speed (at D=1.0) and the ideal toe (2 points measured) and shoulder inflection point. Again taken from an actual coating.

    3. A normal print from this combination of paper and film. It has a contrast of 1.50 which is 0.60 x 2.50 in the mid scale.

    Curves #1 and #2 are to be found (or were last I looked) on the Kodak web site.

    All three curves are defined mathematically by cubic splines.

    The usable latitude at the ISO speed is the speed between the two vertical lines which are about 5 stops apart giving you a latitude of that much. In terms of film density, this is from 0.4 to 1.40. Of course the paper is fully used from Dmin to Dmax.

    In the area of data loss, or compression, the actual 'reproduction rate' is the gamma of the film * gamma of the paper, but the film is constant and the paper is decreasing in the toe and shoulder areas, so the actual decrease in gamma of the paper imposed on the 0.6 gamma of the film results in a rapid data loss. I have drawn a sharp toe and shoulder in the print example I used, but in actuality, prints have much softer toe and shoulder curvature than what I have drawn.

    I hope this explains some of the facts about films, papers and how they interact.

    PE
    Attached Thumbnails Attached Thumbnails sensitometry.jpg‎  

  2. #2
    hal9000's Avatar
    Join Date
    Sep 2005
    Location
    Berlin, Germany
    Shooter
    Medium Format
    Posts
    228
    Images
    8
    Hi Photo Engineer,
    Thanks very much for this post. I had been trying to figure out what mathematical function to use to smooth density curves, your post is the first I've seen describing them as cubic splines. I'll give it a try in my plotting program!

  3. #3
    eddym's Avatar
    Join Date
    Jan 2006
    Location
    Puerto Rico
    Shooter
    Multi Format
    Posts
    1,918
    Images
    26
    Quote Originally Posted by Photo Engineer View Post
    1. A normal professional film with a contrast of 0.60 taken from an actual coating.
    Hmm... short toe, the rest is all straight line, no shoulder. Super XX pan?

    Also, aren't you ignoring flare loss in taking and enlarging lenses?
    Eddy McDonald
    www.fotoartes.com
    Eschew defenestration!

  4. #4
    Photo Engineer's Avatar
    Join Date
    Apr 2005
    Location
    Rochester, NY
    Shooter
    Multi Format
    Posts
    19,238
    Images
    65
    Of course I am ignoring lens flare Eddy. In fact, amateur films have a contrast of 0.63 or higher to compensate for lens flare, and that is one of the reasons why Gold is higher in contrast than Portra.

    The film shoulder is off the graph to the right.

    PE
    Last edited by Photo Engineer; 03-12-2007 at 09:28 PM.

  5. #5

    Join Date
    Apr 2006
    Shooter
    Medium Format
    Posts
    169
    So what happens in the case where you correct for underexposure or overexposure in printing? I think I know what the equivelent chart would look like, but I could be wrong.

  6. #6
    Photo Engineer's Avatar
    Join Date
    Apr 2005
    Location
    Rochester, NY
    Shooter
    Multi Format
    Posts
    19,238
    Images
    65
    If under or over exposure is severe, you are printing on the toe or shoulder of the film and lose detail. If not, then a portion of the image comes from the straight line and only the shadows or the highlights suffer. In all cases, you suffer from data loss (to use a digital term).

    PE

  7. #7

    Join Date
    Apr 2006
    Shooter
    Medium Format
    Posts
    169
    And, numerically, you have one stop of underexposure before you are printing on the toe and about 2 stops before you are printing on the shoulder?

  8. #8
    Photo Engineer's Avatar
    Join Date
    Apr 2005
    Location
    Rochester, NY
    Shooter
    Multi Format
    Posts
    19,238
    Images
    65
    Wire;

    That is simply due to the way I drew the curve. Actually, the example I used would probably be a stop underexposed, but that is just a guess. I would have to do some figuring to prove the point. I did this due to the film curve running off the right of the graph. In actual practice, we plot the film curves on horizontal graph paper and paper and prints on vertical graph paper due to film latitude. I just used a sheet of what I had.

    Like the bunny, the film curve keeps on going.

    PE

  9. #9

    Join Date
    Apr 2006
    Shooter
    Medium Format
    Posts
    169
    Okay, I was figuring that the curve is an abstraction of what things really are, but wanted to make sure that I actually understood what I was thinking to be true.



 

APUG PARTNERS EQUALLY FUNDING OUR INFRASTRUCTURE:


 
                     

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