Portra 160 and 400 Reciprocity Failure [Archive] - Large Format Photography Forum

View Full Version : Portra 160 and 400 Reciprocity Failure

RHITMrB

23-Aug-2013, 10:08

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.

http://i.imgur.com/hNxJiaY.png

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

23-Aug-2013, 15:42

That's great, thank you!

JeRuFo

24-Aug-2013, 03:39

That is definetely good info, thanks!

Did you find anything about the increase in contrast?

Eric Bryan

4-Sep-2013, 07:27

Great chart! I look forward to this data being added to the Reciprocity app. Until now, I have simply used the Ektar numbers in the app, cross references with a chart that I found a while back and that has yielded good results. Probably not absolutely perfect but since Portra is so forgiving, I have not had any issues. I'm not shooting extremely long exposures though....no more than 1 or 2 minutes.

I do have a question, as I am mathematically challenged. I understand the chart and how to calculate reciprocity based on the numbers you provide (eg. 4, .321928095), but in the formula what is ln?

I have tried plugging in various numbers for ln but haven't gotten anything correct. If you could briefly explain it I would greatly appreciate it, thanks!

Dave Langendonk

4-Sep-2013, 07:54

I do have a question, as I am mathematically challenged. I understand the chart and how to calculate reciprocity based on the numbers you provide (eg. 4, .321928095), but in the formula what is ln?

I have tried plugging in various numbers for ln but haven't gotten anything correct. If you could briefly explain it I would greatly appreciate it, thanks!

ln is not a variable. It stands for natural logarithm. So, y=0.5167 times the natural log of x - 0.2.

See the definition of natural logs for more... http://en.wikipedia.org/wiki/Natural_logarithm

JeRuFo

4-Sep-2013, 07:54

ln stands for natural logarithm. I won't bore you with the details, but it is basically the logarithm of x to the base e (a mathematical constant).
If you are a windows user you can just calculate the number for ln(x) by putting in x into your calculator and then hitting the 'ln' button in expert(?) mode (see attachment). Probably easiest to do that up front and just store that number in the memory and then put in the whole sum to get y. Other calculators have probably similar buttons.

Sorry, I can't explain it any clearer, I took most of my basic algebra in Dutch.

101380

Drew Wiley

4-Sep-2013, 08:38

Any ten buck math calculator will have a log function.

gregmo

4-Sep-2013, 10:35

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

Eric Bryan

5-Sep-2013, 07:07

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

4-Sep-2014, 14:15

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

5-Sep-2014, 13:56

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!

neil poulsen

5-Sep-2014, 23:14

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...

rbultman

7-Sep-2014, 07:35

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.

121416

Below is the chart showing the original and fitted curves.

121415

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.

Regular Rod

7-Sep-2014, 07:50

rbultman

7-Sep-2014, 08:14

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.

jcc

7-Sep-2014, 08:23

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.

buggz

7-Sep-2014, 09:08

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!

jcc

7-Sep-2014, 09:11

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

buggz

7-Sep-2014, 09:26

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

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

jcc

7-Sep-2014, 09:31

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

I was using:
Correction = 0.5167 * ln(metered_exposure_in_seconds) - 0.2

buggz

7-Sep-2014, 09:53

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...

rbultman

7-Sep-2014, 09:55

Did you try the numbers from the PDF I posted?

jcc

7-Sep-2014, 10:00

Michael Lloyd

7-Sep-2014, 10:05

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

neil poulsen

7-Sep-2014, 12:16

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.

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...

Regular Rod

7-Sep-2014, 13:42

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

rbultman

7-Sep-2014, 13:58

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.

rbultman

7-Sep-2014, 14:36

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

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.

Regular Rod

7-Sep-2014, 14:48

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

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

Peter Gomena

7-Sep-2014, 23:39

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.

neil poulsen

8-Sep-2014, 06:42

I have a question: In using this chart, are you adjusting the aperture by the given number of stops?

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.

http://i.imgur.com/hNxJiaY.png

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!

rcmartins

2-Nov-2014, 13:45

Looking for some reciprocity figures for Fuji 160NS I stumbled upon this fairly recent thread for Portra 160 and Portra 400. Apparently I missed this thread when it appeared. The original graphic is very useful, specially considering the lack of information that there is regarding negative color film. I would thus like to thank the OP for doing it.
The OP decided to make the graph expressed in EV increments. These increments can be carried out in time compensation or aperture. I would be careful though because the reciprocity failure has to do with the amount of light that falls on the film and thus should imply manipulation of time for a give chosen aperture. I don't think one should change the aperture because it changes the amount of light that hits the film and thus requires a different time altogether.
To compensate for time one just have to do Ts = Tm*2^(Compensation EV). Given the expression the OP showed this means that:

Ts = Tm*2^(0.5167 * ln(Tm) - 0.2)

with Ts the new corrected shutter open time and Tm the measured time for the scene.
Hope this helps clarify it's use.
raul

PS: I have not enough experience, but is it a matter of fact that the two Portras fail in the same way?

jbenedict

3-Nov-2014, 03:50

I have a question: In using this chart, are you adjusting the aperture by the given number of stops?

For the purposes of exposure, "a stop is a stop" whether adjusted via aperture or shutter.

polyglot

3-Nov-2014, 04:05

For the purposes of exposure, "a stop is a stop" whether adjusted via aperture or shutter.

Not when reciprocity is failing! That's the whole thing, it's reciprocity failure, i.e. the film fails to follow the whole a-stop-is-a-stop reciprocity rule wherein film-plane illuminance (~= scene brightness * effective aperture area) and time can be traded off as equivalent.

Reciprocity corrections are time corrections. Correcting the aperture increases the illumination on the film and therefore reduces the amount of failure.

BM PHOTOS

15-Jun-2016, 20:02

Ts = Tm*2^(0.5167 * ln(Tm) - 0.2)
Ts the new corrected shutter open time and Tm the measured time for the scene.
Hope this helps clarify it's use.

Thanks a lot Raul! This helps tremendously

Here are a graph and a Table I made using your formula and rounding the numbers.
Hope it's useful for people in the field.

Best,

Nico

151827

151826

StoneNYC

16-Jun-2016, 03:51

There's a member here who did a lot of testing with seemingly accurate results and those numbers were added to the app "Reciprocity Timer" for iPhone I highly recommend it, works great every time.

Christopher Barrett

16-Jun-2016, 05:18

There's a member here who did a lot of testing with seemingly accurate results and those numbers were added to the app "Reciprocity Timer" for iPhone I highly recommend it, works great every time.

Yeah, the app has worked awesomely for my lakeshore images shot on Portra 160, usually 3-5 minutes. It also handles bellows factors really nicely.

Serge S

16-Jun-2016, 14:37

Hi Stone,

I can't find that app...does it go by another name?

There's a member here who did a lot of testing with seemingly accurate results and those numbers were added to the app "Reciprocity Timer" for iPhone I highly recommend it, works great every time.

StoneNYC

16-Jun-2016, 15:58

Hi Stone,

I can't find that app...does it go by another name?

Link...

Reciprocity Timer by Pump Interactive
https://appsto.re/us/-tLzB.i

App looks like this...

http://uploads.tapatalk-cdn.com/20160616/52cae03b921a15150d474e01ef81e529.jpg

Serge S

16-Jun-2016, 16:52

Link...

Reciprocity Timer by Pump Interactive
https://appsto.re/us/-tLzB.i

App looks like this...

http://uploads.tapatalk-cdn.com/20160616/52cae03b921a15150d474e01ef81e529.jpg

Great thanks a bunch. Got it. Not sure why the search function was not pulling it up.

BM PHOTOS

16-Jun-2016, 18:31

Thanks a lot for this app! Testing it now.

Ari

16-Jun-2016, 19:23

The only photo app I use, and it's a damn useful one in all kinds of situations.
Highly recommended, worth the price.

neil poulsen

16-Jun-2016, 21:06

I would use aperture. Using shutter speed only puts one further down the curve.

For the purposes of exposure, "a stop is a stop" whether adjusted via aperture or shutter.

Christopher Barrett

17-Jun-2016, 04:42

I would use aperture. Using shutter speed only puts one further down the curve.

Unless you're using aperture for a very specific depth of field (for esthetic reasons), which is why I always bracket via shutter speed.

BM PHOTOS

17-Jun-2016, 22:00

senderoaburrido

14-Jan-2017, 14:37

Well, I'm not too much of an app person, I prefer a little chart to have in my pocket. Here's one in 1/3 stops based on time as I prefer to shoot with very small apertures. Much better than my previous one :-)

151961

Hope this can help :-)

Sorry BM, I don't understand your chart. What do you mean by 1/3 stops? Is it in 1/3 of a minute? I don't normally have my light meter output in stops. Didn't even know that was a thing.

mitrajoon

15-Jan-2017, 18:32

arkadi

30-Apr-2019, 10:27

Guys I guess you all love complications, formulas and such. Why not make it the simplest way there? 1 sec -> x stops, 2 sec -> y..

kylejosephbuckley

8-Sep-2019, 11:29

Thanks a lot Raul! This helps tremendously

Here are a graph and a Table I made using your formula and rounding the numbers.
Hope it's useful for people in the field.

Best,

Nico

151827

151826

Thank you for making this! I'm coming in 3 years later haha with what I just want to clarify. "Tm" stands for the time measured by whatever device you're using to meter the scene and "Ts" stands for what your actual shutter speed should be? So for example in your chart, it says "Tm 1" and then next to it it says "Ts 2"; that means if the light meter is telling me I should expose the scene for 1 second, I should apply your chart and expose it for 2 seconds?

Thanks again!

Steven Ruttenberg

11-Sep-2019, 08:37

I love the app. Use it for it for Portra, etc. Just used it this weekend for a sunrise/sunset session. I also use it for bw film. I like the timer and being able to account for bellows extension and stops compensation (like using a red25A filter, etc)

It will be interesting to see how long an exposure you can make with Portra before there is a color shift if any. And the app says Portra 100/400, should say 160/400 I believe.