Just forgot to add the picture...
Sorry, Paul Attached files
Just forgot to add the picture...
Sorry, Paul Attached files
Originally Posted by 1563
Sorry Paul, that is a phrase used to suggest "I cannot accept what you are proposing".If - then - you want to "buy" the results, how much are you ready to pay
Whilst I find your arguments & formulas very compelling, for me they just do not relate to real world results. I would very much like to understanding why that is.
Hi Tony,
When in the Sciences world (with capital S !) some theory is proposed, there are two things to do :
1/ check the computations, together with the assumptions that are made (are they complete ? relevant ?)
after that :
2/ design experiments to confirm or infirm (the facts already explained, new predictions)
The truth lies in experiments, not sentiments.
of course "buy" was used as a joke (you should know me by now)
Paul
Happy birthday, Peter !
Paul
Paul and Peter:
What are the units along the x-axis in the light falloff charts? If it is degrees, then shouldn't the line become discontinuous at 90? Theoretically, the pinhole has ceased to exist.
Until the numbers are tested with densitometer readings, I don't think that we have a means of checking the theory experimentally. It is difficult to translate light falloff in the camera to density falloff on the negative, especially when we are looking at scanned negatives where the negative densities have been re-scaled and 'normalized' in the scanning process. By this I mean that we tend to take the density range of the negative and spread it out to the full 256 value range of the digital white/black value.
Happy birthday, Peter.
Earl, the x-axis is millimetres on the film plane from the centre of the frame, not degrees.
I have found an exposure I made of a flat uniform, white area.
This exposure was intended for use as a centre spot neutral density filter, whereby the filter (the negative) would exactly reverse the exposure fall off of the pinhole.
The exposure was shot in a camera with 0.010" pinhole in 0.002" brass plate. The focal length was 18mm.
This exposure has been scanned. As Earl has suggested I have tried to hold all black/white values of the scan by not clipping the output & by 'pegging' the black & white points of the histogram so that all values are included & not clipped.
The histogram can explain it better than I can.
Once this scan is in Photoshop, I take a duplicate image. On the duplicate, I apply a one stop under exposure & snip a sample from the centre of the frame.
This is then pasted into the original frame & moved until it 'vanishes'. This suggests that the one stop under sample is at the same exposure value on the original frame.
This process is then repeated for further one stop under exposure steps.
The attached pictures should explain further. To do a test like this with different negs, I think you would need to conduct the measurements in the darkroom. If you tried to do it by scanning, in my experience even if you set the scanner exposure control to manual, it always varies from scan to scan.
Therefore, I think the most reliable way to work on the computer is by using a single scan.
Please note, my scanner is not intended for 120 film so I cannot get a full frame scan (hence this 6x6cm frame not being quite square).
I shall allow you fellows to draw conclusions from this, if you wish, and if you think it offers anything meaningful. Attached files
Tony,
It will be interesting to see what the results are from your experiment. It would be nice to see what the difference is to using Paul's approach where the exposure is made on film and the non-linearities of the shadow exposures are taken into account.
I know that on my folding 4x5 pinhole (0.37mm on 0.01mm stock), at 28mm and when I shoot on photo-paper and develop it fully, the centre can be blown and the corners still blocked up black suggesting that the exposure range was greater than the 4 stop dynamic range of the paper (at both ends the transition region is very short on the paper and it clips rapidly). Paul's equation says I should be seeing about 6.5 stops loss at the corners. I have an image which blows at both ends, even with a scene that has 2 stop dynamic range (one I spot-metered with my DSLR), suggesting that the 6.5 stop range may be right.
At a 45mm extension, for a flat scene with about a stop dynamic range, I can just get a bit of highlight detail before the shadows have blocked up too badly. The experience seems to fit the software's prediction of 4 stop loss at the corners.
Even with decent film, the roll-off in the highlights and shadows is far from linear. Could it be possible that the non-linear response is giving the impression that the light fall off is less than it really is? A test would be to use a DSLR with a body cap to capture an image as the results should be linear (if captured RAW).
Best regards,
Evan
There could be another reason : my model of light falloff is strictly geometric, it does not take in account diffraction ; which of course may become very important at grazing incidences, reducing contrast and also letting maybe some light coming from elsewhere "flare" the image in the corners ; thus the test image that I propose for checking the model while reducing this effect : with a white central square, four white squares in the corners and pure black in between.Could it be possible that the non-linear response is giving the impression that the light fall off is less than it really is?
Diffraction comes only into play in my models to compare the geometric "confusion images" of the pinhole (as seen by the film, thus not circles except on the axis) with the (approximate) dimensions of the central part of the diffraction blurring : not at all a rigorous integration, I confess : the formulas are only "heuristically suitable" (otherwise, they would not have been analytically expressible for simulation tools like Peter's).
Paul
A scan of my paperwork (relevant to this discussion)
Paul Attached files
Paul:
I like the word 'vignettage' - I think that I will have to use it as a word in English. It has a nice ring to it. To complicate your observations about diffraction, all the diffraction calculations that I have seen simplify the light frequency to a single value. We are working with the whole visible (and beyond) spectrum here.
However, your work is much appreciated. It is the most rigorous geometrical analysis that most of us non-science/math guys have seen around here. It gives a predictive model that is of great value in making camera design decisions.
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