Portra 160 and 400 Reciprocity Failure

[RHITMrB]

I was talking to Peter Gomena last month about his experiences with Portra sheet film in pinhole cameras and he said he'd found some good data on Portra's reciprocity failure but that he hadn't had much chance to test it himself. Surprisingly, the reciprocity failure for 160 and 400 seem to follow the exact same curve. I put the data into Excel and cleaned it up a bit (for example, 8 seconds metered exposure time certainly needs more correction than 4 seconds, contrary to the data I got from Peter) and found a best-fit curve. Myself and dukeku have been using this as a guide for nighttime exposures over the last month and so far it seems to be pretty much dead on.



The important part is this:

y = 0.5167 * ln(x) - 0.2
x = metered exposure time in seconds
y = stops of correction

Now, this doesn't mean you have to carry a calculator with you, since you could easily just carry a table with the data points on it. But if you're looking to do extremely long exposures - whether at night or with pinholes or both - that equation should work for arbitrarily long exposure times, even out past the end of the chart above.

It seems really hard to find solid data on Portra reciprocity failure out there on the internet, so I hope this helps many of you!

[mijosc]

That's great, thank you!

[JeRuFo]

That is definetely good info, thanks!

Did you find anything about the increase in contrast?

[Drew Wiley]

Any ten buck math calculator will have a log function.

[gregmo]

Thank for the time & effort to create the chart. This is very useful for the type of shooting I do.

[Eric Bryan]

Thank you everyone for your explanation! I understand the formula now

I have been using Portra more and more for my landscape work, so this proves to be VERY handy.

[buggz]

Okay, I could use this, thanks!
However, I do not get the same numbers as is shown in the chart.
Yes, I am mathematically challenged.

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, confused...

[neil poulsen]

So, I'm trying to figure out what's going on.

As to the first question, are the "Y" numbers on the graph calculated from a model, or do they represent the data used to calculate the model? Not knowing the method that was used to obtain the data, I suspect that they represent the raw data. As you show, they certainly aren't calculated from the model that's given below the graph. (Under any circumstances, one would not expect the calculated values in this kind of modeling to be necessarily the same as the raw data.)

Assuming that the numbers provided on the graph represent the raw data, I get a different model than that provided below the graph. (Slope: 0.5305 Intercept: -0.2581 R-Square: .995) Note that, I entered all the significant digits given on the graph. Plus, had the numbers on the graph been calculated from an actual model, we would have expected R-Square=1.0000000. That wasn't the case. Again, they're probably the raw data. (The only other possibility is if the "2" for 60 seconds was incorrectly recorded, and this made the R-Square different from 1.00000.)

If in fact the numbers on the graph are the raw data, to use this study, I would use the following calculation to determine the exposure correction:

Correction_In_Stops = 0.5305 (Metered_Reading) - 0.2581

Except at a metered reading of 8 sec., the difference between this calculated value and the data is less than a tenth stop. And even at 8 sec., the calculated value is only 1/3rd stop above the raw data. Given the close agreement between the calculated value and the raw data for all other metered readings, I would wonder if the data collected at 8 sec. might be a little off. I would use the calculated value for a metered reading of 8 sec. as well.

Since I don't really understand methods used to determine long exposure corrections, I'd be curious to learn the experimental procedure that was used to determine the exposure correction for each of the metered readings on the graph? (e.g. 4, 8, 15, 30, 60, etc.)

Okay, I could use this, thanks!
However, I do not get the same numbers as is shown in the chart.
Yes, I am mathematically challenged.

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, confused...

[buggz]

No one?
Please understand, I am not trying to be condescending, I just really would like to understand this.
I use Portra 160, and would love to start shooting zoneplates/pinholes.
Thanks!

[rbultman]

I think the issue with the values printed on the original chart are that the line fit facility does not print enough significant digits on the chart. I took the original data and used Solver in Excel to determine the coefficients for the equation. Here is a PDF of the spreadsheet with the data. The value below the total error value is the R^2 value for the curve fit.



Below is the chart showing the original and fitted curves.



Note that the worst error is at the 4 second time but is a 1/4 of a stop. The rest of the error relative to the original values is less than 1/10 of a stop.

Hopefully these coefficients can be used in the reciprocity app.