# Thread: Portra 160 and 400 Reciprocity Failure

1. ## Re: Portra 160 and 400 Reciprocity Failure

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!

2. ## Re: Portra 160 and 400 Reciprocity Failure

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.)

Originally Posted by 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...

3. ## Re: Portra 160 and 400 Reciprocity Failure

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.

4. ## Re: Portra 160 and 400 Reciprocity Failure

So, roughly speaking if the meter says 60 seconds then the adjusted exposure should be 2 stops or 4 minutes? Have I understood this right?

How do we take these values and make them into easily read tables like the ones for FOMAPAN or the charts that work in seconds (not f stops) like the ILFORD ones (that are great if you superimpose them onto graph paper for the smaller gaps of seconds)?

RR

5. ## Re: Portra 160 and 400 Reciprocity Failure

I think I can probably generate both a chart and a table. I'll need someone to check the initial numbers. I am ok with even-numbered stop changes but have difficulty with fractional stop changes.

6. ## Portra 160 and 400 Reciprocity Failure

Originally Posted by Regular Rod
So, roughly speaking if the meter says 60 seconds then the adjusted exposure should be 2 stops or 4 minutes? Have I understood this right?
How do we take these values and make them into easily read tables like the ones for FOMAPAN or the charts that work in seconds (not f stops) like the ILFORD ones (that are great if you superimpose them onto graph paper for the smaller gaps of seconds)?

RR
You can make two column in Excel...
-- A | B
1 | 1 | =0.5167 * ln(A1) - 0.2
2 | 2 | =0.5167 * ln(A2) - 0.2
3 | 4 | =0.5167 * ln(A3) - 0.2
4 | 8 | =0.5167 * ln(A4) - 0.2
.
.
.

Sorry for the crudeness. I'm typing on my phone.

7. ## Re: Portra 160 and 400 Reciprocity Failure

I just want to know the correct formula, as none of the numbers I have tried coincide with the original table numbers.
This way, one can input arbitrary seconds for exposure, and get a corrected value.
Thanks!

8. ## Re: Portra 160 and 400 Reciprocity Failure

I plugged arbitrary times on my phone's calculator, and got results consistent with the graph.

9. ## Re: Portra 160 and 400 Reciprocity Failure

Thanks, so I am doing something wrong in my previous posting.
I get numbers that are not close to the graph numbers.
>> Correction_In_Stops = 0.5305 (Metered_Reading) - 0.2581

Originally Posted by jcc
I plugged arbitrary times on my phone's calculator, and got results consistent with the graph.

10. ## Re: Portra 160 and 400 Reciprocity Failure

Originally Posted by buggz
Thanks, so I am doing something wrong in my previous posting.
I get numbers that are not close to the graph numbers.
>> Correction_In_Stops = 0.5305 (Metered_Reading) - 0.2581
I was using:
Correction = 0.5167 * ln(metered_exposure_in_seconds) - 0.2

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