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Thread: Portra 160 and 400 Reciprocity Failure

  1. #21

    Join Date
    Jan 2012
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    Re: Portra 160 and 400 Reciprocity Failure

    Wow, I still don't understand why mine aren't working, I'm using the same thing.

    ln(4) = 1.3862943611198906188344642429164 on my calculator
    x 0.5167 = 0.71629829639064748275176767431488 - 0.2006 = 0.51569829639064748275176767431488

    ln(4) = 1.3862943611198906188344642429164 on my calculator, still
    - .2006 = 1.1856943611198906188344642429164 x 0.5167 = 0.61264827639064748275176767431488

    You can also type ln(4) into Google search = 1.38629436112

    I don't get a .3something, STILL confused...

  2. #22

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    Re: Portra 160 and 400 Reciprocity Failure

    Did you try the numbers from the PDF I posted?
    My flickr stream

  3. #23

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    Jun 2013
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    Norman, Oklahoma
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    21

    Re: Portra 160 and 400 Reciprocity Failure

    Quote Originally Posted by buggz View Post
    Wow, I still don't understand why mine aren't working, I'm using the same thing.

    ln(4) = 1.3862943611198906188344642429164 on my calculator
    x 0.5167 = 0.71629829639064748275176767431488 - 0.2006 = 0.51569829639064748275176767431488

    ln(4) = 1.3862943611198906188344642429164 on my calculator, still
    - .2006 = 1.1856943611198906188344642429164 x 0.5167 = 0.61264827639064748275176767431488

    You can also type ln(4) into Google search = 1.38629436112

    I don't get a .3something, STILL confused...
    Ok, I see it. The 0.3something is the measured (from what I gather) data. The fitted function is an estimate. Higher degree of error in the shorter exposures.

  4. #24
    Random Pixel Generator
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    Aug 2009
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    Weimar, TX
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    302

    Re: Portra 160 and 400 Reciprocity Failure

    Beat me to it. The equation is "happier" at longer exposure times

  5. #25

    Join Date
    Dec 1999
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    Forest Grove, Ore.
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    3,696

    Re: Portra 160 and 400 Reciprocity Failure

    I believe that the values on the chart are the raw data. The calculation that you're using calculates what a model would predict. One would never expect that the calculated values would equal the raw data. But given the effectiveness of the model, we would hope that the calculated value would be good enough for use in the field. (So, pack a calculator with your gear.)

    As to "the" model, we have a choice of three. The model I provided is a standard least-squares fit, which is typically what is used in practice. I've verified the results I originally obtained from a TI calculator in SYSTAT, which is a well known statistics package.

    I'm not sure what method was used by Excel or used to get the results in the original post. But looking at how accurately each predicts the raw data (numbers provided on the chart), any of the three is usable. The least-squares fit (mine) appers to do better at a metered reading of around 4 seconds and will do acceptably well at higher values. The other two models appear to do better for higher metered readings.

    Given that the differences between them are on the order of a tenth stop, they appear to all work acceptably well. Forced to select, I would probably use the least-squares fit, because of it's improvied performance at the low end of the data.


    Quote Originally Posted by buggz View Post
    Wow, I still don't understand why mine aren't working, I'm using the same thing.

    ln(4) = 1.3862943611198906188344642429164 on my calculator
    x 0.5167 = 0.71629829639064748275176767431488 - 0.2006 = 0.51569829639064748275176767431488

    ln(4) = 1.3862943611198906188344642429164 on my calculator, still
    - .2006 = 1.1856943611198906188344642429164 x 0.5167 = 0.61264827639064748275176767431488

    You can also type ln(4) into Google search = 1.38629436112

    I don't get a .3something, STILL confused...

  6. #26

    Join Date
    May 2013
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    Derbyshire, England
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    Re: Portra 160 and 400 Reciprocity Failure

    The more I read this thread the more I love ILFORD's simple charts...

    RR

  7. #27

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    Aug 2013
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    Re: Portra 160 and 400 Reciprocity Failure

    On the stops of increase...

    I get whole stop increases. It is a doubling, right? So, 1 second + 1 stop = 2 seconds. A 2 stop increase would be 4 seconds, right? Another doubling?

    Same applies for other starting times, like 6 seconds. 1 stop is 12, 2 stops is 24, 3 stops is 48 seconds. Is that right?

    I know this is pretty basic, but it has always confused me.
    My flickr stream

  8. #28

    Join Date
    Aug 2013
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    Re: Portra 160 and 400 Reciprocity Failure

    Assuming I did the math right, here is the time vs. adjusted time chart that RR is looking for.

    Click image for larger version. 

Name:	PortraRecipCorrectionChart_0-60s.pdf 
Views:	299 
Size:	34.7 KB 
ID:	121444

    If it is not correct, let me know what is wrong and I will...correct it.

    The data is based on the fit that I did, which has an R^2 of 0.9954106.

    EDIT: I think I calculated my R^2 wrong. Still not sure I calculated it right, but I think it is slightly lower at 0.9885. Given the coarseness of the graph, I doubt the difference is visible.
    My flickr stream

  9. #29

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    May 2013
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    Re: Portra 160 and 400 Reciprocity Failure

    Quote Originally Posted by rbultman View Post
    Assuming I did the math right, here is the time vs. adjusted time chart that RR is looking for.

    Click image for larger version. 

Name:	PortraRecipCorrectionChart_0-60s.pdf 
Views:	299 
Size:	34.7 KB 
ID:	121444

    If it is not correct, let me know what is wrong and I will...correct it.

    The data is based on the fit that I did, which has an R^2 of 0.9954106.
    Yippee! That is exactly what I hoped for. Thank you very much indeed.

    (Kodak should have done this. Film is too expensive to waste on testing they should be doing anyway)

    Thank you again

    RR

  10. #30

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    Portland, OR USA
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    740

    Re: Portra 160 and 400 Reciprocity Failure

    Well, gee, I hope the data was somewhere near accurate in the first place. I originally sent Isaac Sachs some untested data I found on the internet, and he plugged it into a graph and posted it here. It seems to work well enough. I certainly hope you math whizzes out there haven't spent too many hours working this out.

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