Kodak Ultramax 400 Reciprocity Failure

Ricoh

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First off I hope the is the right area are for such a question.

I plan to do pinhole using Ultramax 400, but cannot find a data sheet on reciprocity failure for this film. I've searched the net, but nothing - always dead end leads, even Kodak's website is of no help. Also the cardboard box the film came in is blank, so no help there.
Anyone know where this data is, or can be found?
Thanks.
 
In all fairness, I recall that Kodak does not have reciprocity data for at least the color range. A rather generous "test yourself" is written instead.


Was curious about Portra's reciprocity for night long exposures and some people modelled it after testing: http://www.largeformatphotography.i...106441-Portra-160-and-400-Reciprocity-Failure


Being Kodacolor anyways, it should be a rather useful starting point. IDK but Ultramax may have been slightly updated with some of Portra's technology back at the time.
 
In all fairness, I recall that Kodak does not have reciprocity data for at least the color range. A rather generous "test yourself" is written instead.


Was curious about Portra's reciprocity for night long exposures and some people modelled it after testing: http://www.largeformatphotography.i...106441-Portra-160-and-400-Reciprocity-Failure


Being Kodacolor anyways, it should be a rather useful starting point. IDK but Ultramax may have been slightly updated with some of Portra's technology back at the time.
Thanks Prest, it sort of rules out my inability to conduct a simple internet search. Also, thanks for the link, it's a good start for me, something to base exposure on rather than guesswork.
Cheers!!
 
A good place to start is to use an exponential of 1.3. Your new exposure in seconds would be x<sup>1.3</sup>. With color neg that should be sufficient so you don't underexpose.

Color neg will also have a color shift, but you should be able to correct for that afterwards unless you are printing it in the darkroom.

Hope that helps.
 
A good place to start is to use an exponential of 1.3. Your new exposure in seconds would be x<sup>1.3</sup>. With color neg that should be sufficient so you don't underexpose.

Color neg will also have a color shift, but you should be able to correct for that afterwards unless you are printing it in the darkroom.

Hope that helps.
Thanks PRJ
In the Large Photography thread referenced above by PREST 400, the equation given is of the form y = f (x); more specifically y = 0.5167*ln(x) - 0.2

Where
x = metered exposure time in seconds
y = stops of correction


In the equation you've presented above, New Exposure = X<sup>1.3</sup>
I'm not familiar with the term 'sup'
Do you mean Y = log (base 1.3) of X ?
Where
Y = New Exposure
X = Metered Exposure

Thanks,
Steve
 
Had to take another look. The equation at LFF is to get the stops of correction. Took a deeper read and there's a PDF with another chart giving a graph with time instead.


Whereas PRJ suggests to just use an Exponential by just elevating a measured exposure time by 1.3.


So:
10s = 10^1.3 = 19.95s
30s = 83.2s
60s = 204s


It's quite interesting, there is some difference. When doing long exposures I tend to err on the long side as "it's hard to overexpose" as a few seconds can be a tiny difference in stops.
 
Thanks Prest_400 for the clarification.

I also homed in on the table in LPF (page 4 I believe) and looking at the figures in the table compared to x^1.3 the results are quite similar up to about 60 secs. And as you say, it's probably better to over expose than to under expose. I have the table saved on my phone, but x^1.3 is also easy to remember.

Right, time for some trials to commence!
 
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