Sparrow
Veteran
I always thought it was my job to decide on the exposure, all the meter ever does is measure the light.
If you simply follow it without applying any reason to the reading I'd not be surprised if the results were variable.
If you simply follow it without applying any reason to the reading I'd not be surprised if the results were variable.
Joakim Målare
Established
You should download the .pdf from the film you are using (TriX for example), and see how much exposure compensation you should use on that specific film then run your own test and see what you come up with.
Right, that makes sense. Thanks.
Joakim, your ZI meter seems more accurate. Factor 5 is 2 1/4 stops not 3.
Cool. Do you know the math? I understand that the factor is 2 ^ stops, but can't figure out how to do the reverse equation
EdwardKaraa
Well-known
Cool. Do you know the math? I understand that the factor is 2 ^ stops, but can't figure out how to do the reverse equation![]()
Pretty simple really.
1x, 2x, 4x, 8x, 16x... etc correspond to 0, 1, 2, 3, 4 f/stops.
I don't know the formula but it's easy to interpolate the in between values.
Joakim Målare
Established
Pretty simple really.
1x, 2x, 4x, 8x, 16x... etc correspond to 0, 1, 2, 3, 4 f/stops.
I don't know the formula but it's easy to interpolate the in between values.
That's only valid for linear series. Which is why a factor of five is more like 2.322-something stops (trial and error on the calculator
Who cares really, 2-2.5 stops works for me
Thanks Edward.
kossi008
Photon Counter
The math is pretty easy:
The number of stops n is the logarithm to the base of two of the filter factor N:
n = log2(N)
Most calculators don't have log to the base of 2, so you can use log to the base of 10 by:
log2(x) = log10(x)/log10(2) = 3.32 * log10(x)
So, a factor of 5 is log2(5) = 2.32 (or 2 1/3) stops.
But if you are in doubt of your meter readings, wouldn't it be smarter (faster) just to do some test exposures instead of doing math and searching for internet links?
The number of stops n is the logarithm to the base of two of the filter factor N:
n = log2(N)
Most calculators don't have log to the base of 2, so you can use log to the base of 10 by:
log2(x) = log10(x)/log10(2) = 3.32 * log10(x)
So, a factor of 5 is log2(5) = 2.32 (or 2 1/3) stops.
But if you are in doubt of your meter readings, wouldn't it be smarter (faster) just to do some test exposures instead of doing math and searching for internet links?
Joakim Målare
Established
But if you are in doubt of your meter readings, wouldn't it be smarter (faster) just to do some test exposures instead of doing math and searching for internet links?
Definitely yes. Sometimes it's just fun to dive into a little theory.
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