Switch to English Language Passer en langue française Omschakelen naar Nederlandse Taal Wechseln Sie zu deutschen Sprache Passa alla lingua italiana
Members: 71,551   Posts: 1,573,102   Online: 764
      
Page 3 of 4 FirstFirst 1234 LastLast
Results 21 to 30 of 35
  1. #21

    Join Date
    Apr 2003
    Posts
    134
    The statement of nworth that the exposure can be scaled as the square of the lens distance is correct. (To be really precise, the image distance).

    Some of the other statements are either imprecise or not accurate.

    If you want to figure out the exposure time compensation by measuring linear dimensions of the prints, i.e., by the two magnifications, the correct formula is (M2 + 1) ^ 2 / (M1 + 1)^2. That is, the square of the ratios of the factors (magnification plus one). It seems obvious that the inverse square law should use the just the magnifications, but this neglects refocusing the lens and isn't accurate. The refocusing changes the effective f-stop.

    For a derviation, see Lenses in Photography by Rudolf Kingslake or Applied Photographic Optics by Sidney Ray.

    For example, going from M1 = 2 (8x10 from a 4x5 neg) to M2 = 4 (16 x 20 from a 4x5 neg), the change in exposure time should be x2.78 longer, not x4 longer. If you are doing enlargements from a small negative, the "+1" tends not to matter, for example, an 8x11 from 35 mm is M=8 and M and M+1 are about the same, and simplifying the equation gives an answer that is close.

    Of course, reciprocity effects or artistic choices depending on print size might cause departures from this equation, which only accounts for light.

    It also works to square the ratio of image distances because d_image = focal length * (magnifcation + 1). Maybe this is another way to see that there should be a "+1" in the magnification version of the equation.

    For the numeric example above, the 8x10 print from 4x5 neg has d_image = 450 mm and the 16x20 has d_image = 750 mm. The ratio squared is 2.78.

    In this example, the two distances between head and the paper (i.e., the the negatives and paper) are 675 mm and 937.5 mm. It doesn't work to take the ratio these distances and square.

  2. #22
    Pastiche's Avatar
    Join Date
    Aug 2005
    Shooter
    Multi Format
    Posts
    319

    summing up... or is that... squaring up . ..

    OK -
    Bottom line...
    slip a square, double the height of the height,
    punch your lights out,
    Meter for the low life,
    movements and jiggle aren't dance steps, but good derivations,
    if you inverse the law of doubling the print you get twice as many prints,
    and pack that into a pint real tight and it just may,
    turn out O... K...

    ... ah... and don't forget to pull the filter, or your meter will come off a foot,
    and you'll have to wobble on home without leaving a print...

    are we all on the same f* . . . STOP!?
    when you get there, you'll know, you'll be exausted.
    get off the bus.
    In the morning.. all you'll remember are the highlights baby, the highlights...

    Thanks all for the info. I think I've been overexposed. Matt- how you dealing?

  3. #23

    Join Date
    Jul 2005
    Shooter
    35mm
    Posts
    31
    hmm, i think somone needs to create a software for this - i have a pc in my darkroom, it would be so cool to just punch in the numbers and have the new exposure out. =P

    ...and i dont mean the calculator either...

  4. #24

    Join Date
    Jun 2003
    Posts
    1,656
    Images
    5
    Good Morning, Morpheuse,

    I think a computer would be overkill. A calulator does the job in just a few seconds, and, with experience, you soon reach the point where a guesstimate is usually about right.

    Konical

  5. #25
    Troy Hamon's Avatar
    Join Date
    Nov 2004
    Location
    Alaska
    Shooter
    Medium Format
    Posts
    291
    As a mathematical biologist, I find that very often people choose the most difficult line between two points when they are trying to develop a mathematical description. I'd say the same thing applies here. With the exception of Michael Briggs, who has given us a new result that is different than all the others, all of these equations are functionally identical, though not all identical in application (enlarger height versus print dimension).

    In terms of Michael's equation, there is no reason that you should have to add one to the two terms. As far as the physics of the situation, there is no basis for it. If it helps to obtain better prints, then sure, go ahead, and since I've never tried it that way perhaps it works reasonably well. As Michael said, for larger prints, the difference between his equation and the others in terms of the results provided becomes smaller and smaller. For small prints, though, it could provide pretty different prints at the different sizes.

    The basic physics of the problem is, as others have said, that light falls off at an exponential rate with distance. This does not mean that the problem is particularly difficult, though. A simple way to look at it is to think of it solely in terms of the light put out by the enlarger. There is a certain quantity of light that is coming through the enlarger lens. That same amount is going to come out of the lens whether the image on the print is covering 4x5 or 8x10 or 16x20. So, by simply taking the ratio of the areas in the prints (or the ratio of the squares of the enlarger heights) you can accurately measure the ratio of exposure times. So...a 4x5 is 20 inches square, an 8x10 is 80 inches square, an 8x10 takes 4 times the exposure of a 4x5. The mathematical version of this is

    Tn = To*(An/Ao)

    where Tn = new exposure time, To = old exposure time, An = new print area, and Ao = old print area. To alter this equation for enlarger distance, it simply becomes

    Tn = To*(Dn^2/Do^2)

    where Dn = new enlarger distance and Do = old enlarger distance. Neither of these are new, others above have already provided these results in different forms, but this presentation makes more sense to me, so I hope it is helpful to somebody else.

    The missing link in the above description is the light scattered during transmission through air. The greater the distance to the print, the greater the potential light loss, affecting our assumption above that there is a uniform amount of light falling on the prints of various sizes. In practice, the only way this would be important is if you were printing over exceptional distances or the air in the darkroom was particularly dirty...

    The overall line of reasoning laid out above is related to the reason that teleconverters reduce the effective aperture of your lenses, and why the physical aperture of a lens is larger to achieve the same F-stop (which is a ratio) for a longer focal length...but if people are interested in that I'll post it separately...and otherwise I'll happily keep it to myself.

  6. #26
    Helen B's Avatar
    Join Date
    Jul 2004
    Location
    Hell's Kitchen, New York, USA
    Shooter
    Multi Format
    Posts
    1,557
    Images
    27
    "...and why the physical aperture of a lens is larger to achieve the same F-stop (which is a ratio) for a longer focal length..."

    Just to clarify Troy's excellent post, the size of the physical aperture doesn't appear in the f-stop calculation, but the diameter of the entrance pupil (the appearance of the aperture, as seen from in front of the lens) does. I'm sure that Troy knows that, and I hope that he doesn't mind me mentioning it.

    Best,
    Helen

  7. #27
    Troy Hamon's Avatar
    Join Date
    Nov 2004
    Location
    Alaska
    Shooter
    Medium Format
    Posts
    291
    Hi Helen,

    You are right, and I'm sorry to have been somewhat opaque in my presentation. Optically, the aperture size (from which F-stops are calculated) is the size of the light-gathering element. Meanwhile we simultaneously have a mechanical aperture that controls the amount of light entering the lens that is transmitted to the film plane. As photographers, we tend to think of the mechanical device as 'the' aperture, since that's where the aperture-based exposure control happens.

    I've probably made this even more confusing...short version...Helen is right.

    And now to wander way off topic, I've wondered a bit about the front element size of some of the newer generation long lenses. The latest Nikon 600 f4 doesn't look like it is as large in the front element as the older versions. In fact, according to the data sheets, it is 6.6 inches for the new ones and 6.9 inches for the older ones. Is this because they have made the actual front element smaller? In order to do so, they must either change the focal length or the maximum aperture... It used to be that Nikon (and Canon I assume, though I'm not as familiar with their products) produced their super telephotos very close to the published specs, while Tamron, for example, produced a 280 mm instead of a 300 mm. This put them within legal tolerances for their product labeling, but also reduced the size, and presumably expense, of their lenses. In my irrelevant musings, I've wondered whether Nikon has made a similar adjustment in their latest 600. Since I've never touched one of these lenses and am not likely to, I'm free to wonder unfettered by details like the actual size of the lenses, which may not have changed in any real fashion.

    Random thoughts from my troubled mind...

  8. #28
    johnnywalker's Avatar
    Join Date
    Sep 2002
    Location
    British Columbia, Canada
    Shooter
    Multi Format
    Posts
    2,256
    Images
    78
    Troy, thanks for the clarification. I asked this question way back in this thread.
    If I had been present at the creation, I would have given some useful hints for the better arrangement of the Universe.
    Alfonso the Wise, 1221-1284

  9. #29

    Join Date
    Apr 2003
    Posts
    134
    Quote Originally Posted by Troy Hamon
    ..........

    In terms of Michael's equation, there is no reason that you should have to add one to the two terms. As far as the physics of the situation, there is no basis for it.........

    ......... A simple way to look at it is to think of it solely in terms of the light put out by the enlarger. There is a certain quantity of light that is coming through the enlarger lens. That same amount is going to come out of the lens whether the image on the print is covering 4x5 or 8x10 or 16x20. So, by simply taking the ratio of the areas in the prints (or the ratio of the squares of the enlarger heights) you can accurately measure the ratio of exposure times. So...a 4x5 is 20 inches square, an 8x10 is 80 inches square, an 8x10 takes 4 times the exposure of a 4x5. The mathematical version of this is

    Tn = To*(An/Ao)

    where Tn = new exposure time, To = old exposure time, An = new print area, and Ao = old print area. To alter this equation for enlarger distance, it simply becomes

    Tn = To*(Dn^2/Do^2)

    where Dn = new enlarger distance and Do = old enlarger distance. Neither of these are new, others above have already provided these results in different forms, but this presentation makes more sense to me, so I hope it is helpful to somebody else.
    First of all, the equation that I gave is not really my equation. Second, the "+1" added to the magnification has a physical basis. In fact, it is necessary for the equation to be correct. I tried explaining it in my first answer and I gave references to more detailed derivations than I can give in an internet answer. Do you really think that your mathematical optics by intuition is correcting an equation published by Rudolf Kingslake, who maybe Kodak's best lens designer and is one of the best authors of optics books?

    If you had checked the references that I gave, you will find that both Kingslake and Ray use the same numerical example, going from an enlargement of 2x to 4x. Both calculate an exposure time change of x2.8. The paper dial calculator in the Kodak Darkroom Dataguide also gives an exposure change from 10 to 29 s.

    What your light spread over paper area argument ignores is refocusing the enlarger lens and the consequent change in the effective f-stop.

    Tn = To*(An/Ao) and Tn = To*(Dn^2/Do^2) implies a relation between "enlarger distance" (image distance, I assume -- the term wasn't precisely defined) and print area that simpy isn't correct. If a 4x5 negative is enlarged to 8x10 with a 150 mm lens, the image distance is 450 mm. Switching to a 16x20 print, the image distance becomes 750 mm. The ratio of print areas is 4. The ratio of image distances squared is 2.78. If by "enlarger distance" is meant the total distance from negative to print, the two distances are 675 mm and 937.5 mm. The square of the ratios is 1.92, which still isn't 4.

    The correct equations relating image and object distance, focal length and magnfication are given near the top of the lens tutorial: http://www.photo.net/learn/optics/lensTutorial

    This numerical examle (same as my previous example) shows the consistency between squaring the image distance and the m+1 squared equation, and the inconsistency with the magnification squared equation. (750 / 450 )^2 = 2.78. (4+1)^2 / (2+1)^2 = 2.78. But 4 ^2 / 2^2 = 4. This should cause worry to those who believe in both the image distance squared and magnification distance squared equations.

  10. #30
    Helen B's Avatar
    Join Date
    Jul 2004
    Location
    Hell's Kitchen, New York, USA
    Shooter
    Multi Format
    Posts
    1,557
    Images
    27
    Slightly off topic:

    Troy wrote
    "Is this because they have made the actual front element smaller? In order to do so, they must either change the focal length or the maximum aperture..."

    Troy,

    The diameter of the front element does not directly determine the maximum aperture - ie it isn't the number that goes into the f-stop ratio. The diameter of the entrance pupil, with the lens wide open, does. The front element is not the entrance pupil, of course.

    Different lens designs of the same focal length, same maximum aperture and same coverage can have different front element diameters, but they will all have the same entrance pupil diameter.

    Back on topic:
    Michael wrote
    "What your light spread over paper area argument ignores is refocusing the enlarger lens and the consequent change in the effective f-stop."

    I agree with Michael on this one, and apologies to him if I suggested otherwise. If you want the correct answer this must be taken into consideration.

    Best,
    Helen

Page 3 of 4 FirstFirst 1234 LastLast


 

APUG PARTNERS EQUALLY FUNDING OUR COMMUNITY:



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