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Thread: Exposure mapping

  1. #11

    Exposure mapping

    My models have both a pinhole diameter and pinhole thickness parameter (edit : after considering, I think that perfect circular shape is in fact not required for accuracy by the light falloff model - approximate circular shape will do).

    And when inputting a non negligible pinhole thickness, they give a "tunnel effect" : grazing over some angle, the light gets completely blocked. Look below the post I spotted in the same thread

    To Tony, my curves give the EV decrease as a function of distance to the center of the image. You simply have to point on the graph the distances for -1EV, -2EV, etc. and plot the circles with this radius.

    I have also an idea to check these results experimentally : with a wide angle flat plane camera do 3 or 4 pictures (or more) on the same film of some uniform surface, with decrease of exposure times by factor 2 ( T for correct exposure at the center, T/2, T/4, T/8, etc.), then after developping it : with a densitometer (or a digital scanner) check at what distance from center the density is the same on the picture exposed at T, as the density at the center on the pictures taken at T/2, T/4, T/8, etc. (edit : this will indeed work well only with films with lesser sensitivity to reciprocity failure)

    Paul

  2. #12

    Exposure mapping

    A perfectly thin, two-dimensional pinhole - a circle - (a mental construct only) will be elliptical when viewed at any angle other than dead-on. So these off-axis exposure calculations need to at least take this into account.

    A real-world pinhole, when viewed off-axis, will also be elliptical, but with some portion of the area reduced by the thickness of the material. So a real-world pinhole, when viewed off-axis, will have less aperture area than the ideal two-dimensional pinhole.

    You can picture this by imagining the pinhole as two concentric circles, one behind the other (the front and back edges of the pinhole), separated by the thickness of the material. Now imagine viewing these two circles off-axis; each circle becomes an ellipse, but the open aperture area becomes reduced by the "side wall" obstructing the front circle. This side-wall obstruction takes the shape of a crescent moon; you can imagine, with a very thick material, the clear off-axis aperture would be very small, and the side-wall obstruction would take on more than a half-moon shape.

    The off-axis exposure formula should take this true off-axis aperture area into account. There's probably a way of calculating this true off-axis area; I'll leave that to you math wizards.

    ~Joe

  3. #13

    Exposure mapping

    Quote Originally Posted by 1351
    The off-axis exposure formula should take this true off-axis aperture area into account. There's probably a way of calculating this true off-axis area; I'll leave that to you math wizards.
    Done !

    Suppose you have a 6x9 pinhole camera (half-diagonal : 55mm) with a focal length of 40mm

    Pinhole designer (fitted with "my" user defined constant 1.56 instead of Raleigh's value of 1.9) gives an optimal pinhole diameter of 0.242mm. Suppose your pinhole plate is 0.2mm thick (not negligible versus pihole diameter)

    Paste following formula into Notepad :

    lb(0.637*(acos(e*x/f/d)-e*x/f/d*sqrt(1-(e*x/f/d)^2))/(1+(x/f)^2)^2)

    And from within its edit menu "replace everywhere" :

    f by 40[mm]
    d by 0.242[mm]
    e by 0.0[mm] (red curve)
    e by 0.2[mm] (green curve)

    you get :

    lb(0.637*(acos(0.0*x/40/0.242)-0.0*x/40/0.242*sqrt(1-(0.0*x/40/0.242)^2))/(1+(x/40)^2)^2)
    lb(0.637*(acos(0.2*x/40/0.242)-0.2*x/40/0.242*sqrt(1-(0.2*x/40/0.242)^2))/(1+(x/40)^2)^2)

    when you plot the curves you see attached figure : x axis, distance from center of film plane in mm, y axis light falloff relative to center in EV.


    Paul


    Attached files

  4. #14

    Exposure mapping

    Polka,

    That's very interesting!

    I was wondering, what is the 0.637 value?

    I tried to make a small javascript thingy to plot it: http://www.durian.se/pinhole/evdrop.html (You can drag the sliders to change the values - no idea if it works on other than Firefox. You can drag the red dot over the curve.).

    -peter

  5. #15

    Exposure mapping

    Hi Peter, congratulations !

    This is just the tool needed, that I was not smart enough programmer to code. And it seems to work (on my µsoft IE too) and give sensible results.

    The 0.637 is a reasonable approximation for constant 2/PI that is needed for scaling the trig functions in their math units (right angle : PI/2) which were used in pinhole ovalisation formula. I have to find and check with my paperwork to remember and give you the exact demonstration.

    I would be very interested if you continued your programming with the other formulas (diffraction blurring and anamorph cams).

    Paul

  6. #16

    Exposure mapping

    Paul & Peter, perhaps you can help me with this.....

    Firstly, well done Peter, that is a super piece of JS programming (why doesn't blasted JS work in Opera by the way??? ).
    Having entered all the relevant figures for my super short FL design (19mmFL on 6x6cm) to test the figures, I end up with an exposure map as shown.
    As you can see, this tells me that the corners of the frame will lose 5 stops. I just can't buy it to be honest, look at the examples I have listed in reply 2 above & tell me those pictures have a 5 stop loss of light in the corners (full frame pictures).......
    Maybe it is true, can you convince me it is Paul/Peter?
    And why does my map look very linear? I would expect to see a more 'logarithmic' effect.
    Thanks. Attached files

  7. #17

    Exposure mapping

    I think that in the corners of your 4 examples taken with focal 19mm you have way over 2 stops darkening (as your model predicts).

    In post #9, I suggested a way to test experimentally the models ; I cannot do it myself, because for the moment I have only anamorph cams.

    To be sure to get pictures minimizing internal flare, I would suggest to take full frame pictures (with your 19mm 6x6) of a test image like this one*, and apply the protocol : correct exposure at the center => T, then other shots at T/2, T/4, T/8, etc. and compare the densitiy of the white squares in the corners of the correctly exposed neg with the density of the center square of the underexposed negs.

    If - then - you want to "buy" the results, how much are you ready to pay

    Paul

    * with wide enough black baffle all around ; but to see also if internal flare may occur and how it may perturb the experiment, you could reapeat the test with a white uniform surface. Attached files

  8. #18

    Exposure mapping

    Quote Originally Posted by 1319
    Hi Peter, congratulations !

    This is just the tool needed, that I was not smart enough programmer to code. And it seems to work (on my µsoft IE too) and give sensible results.

    The 0.637 is a reasonable approximation for constant 2/PI that is needed for scaling the trig functions in their math units (right angle : PI/2) which were used in pinhole ovalisation formula. I have to find and check with my paperwork to remember and give you the exact demonstration.

    I would be very interested if you continued your programming with the other formulas (diffraction blurring and anamorph cams).

    Paul
    Thanks!

    Ok, 2/PI it is. (The javascript uses arguments in radians for the trig. functions, is that relevant here too?)

    I have been working on the blurring, but I'm not totally satisfied yet :-)

    O, and I was also wondering, it shouldn't be too hard to get the x (or the distance from the centre) as a function of the EV (to be able to compile a table for distances at EV(drop-off) 1, 2 etc), should it?

    -peter

  9. #19

    Exposure mapping

    Quote Originally Posted by 1563
    Paul & Peter, perhaps you can help me with this.....

    Firstly, well done Peter, that is a super piece of JS programming (why doesn't blasted JS work in Opera by the way??? ).
    Having entered all the relevant figures for my super short FL design (19mmFL on 6x6cm) to test the figures, I end up with an exposure map as shown.
    As you can see, this tells me that the corners of the frame will lose 5 stops. I just can't buy it to be honest, look at the examples I have listed in reply 2 above & tell me those pictures have a 5 stop loss of light in the corners (full frame pictures).......
    Maybe it is true, can you convince me it is Paul/Peter?
    And why does my map look very linear? I would expect to see a more 'logarithmic' effect.
    Thanks.
    Thanks!

    I'll test Opera later today - the graph drawing library I us should work with it as far as I know.

    As for the results - I don't know. With 19 mm on 6x6 one would expect quite some drop-off I guess. But it is possible there is an error in my Javascript version of the formula. More checking and "common sense"/real-world comparison is always a good thing to do!

    -peter

  10. #20

    Exposure mapping

    By the way, my ovalisation model supposes pinhole section 1 ; if you believe that 2 is more suitable, then pose e=0.

    But if e < d/10 (say), there is not much difference (only with very very short focal length, do the curves diverge)

    Paul

    P.S. I retrieved my paperwork and either I will scan it and post it here as jpegs, or I will try to convert it to text (or both)

    To Peter : I ran your program with posted results of my model, they correspond, so your coding seems perfect ; if something is wrong, it has to be my model... but I don't think so because it is agreed elsewhere that for 45 degrees incidence you loose 2 stops and this is what my model says, so...

    To answer your other question, I will look if I can invert the formulas (get x(EV) instead of EV(x) ?) it might not be trivial, I will do it first posing e=0 (simplified model).

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