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F-Stop Printing Calculators. now I am confused

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lhalcong

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Up until now it was my clear understanding that if the normal print time (hypothetically speaking) for a specific print was 8 sec, then 1/2 stop would be 4 sec. and one extra stop would be 16 sec..... This is the way we do it on camera right. ! However, I recently downloaded (for the iphone) a couple of F-Stop print calculators. To my surprise 1/2 stop of 8 sec is reported as 5.7 in both. What tha ? I am confused now. I have been printing for long time based on my prior understanding . Can someone clarify this and why ?
 
lhalcong said:
Up until now it was my clear understanding that if the normal print time (hypothetically speaking) for a specific print was 8 sec, then 1/2 stop would be 4 sec. and one extra stop would be 16 sec.....
Click to expand...

Nope.

Going from 8 seconds to 4 seconds is one full stop less exposure (1/2 the actual exposure). Going from 8 seconds to 16 seconds is in fact, 1 full stop more exposure, so you are half tight.
 
Sorry. I mis-spoke.(or typed) I meant . half an exposure would be 6 sec. (right between 4 and 8) why does the calculator reports 5.7sec. I guess that's not much difference but I am curious.
 
It seems two writers of programs simply do not understand...or you are misunderstanding '1/2 stop'. The latter seems true.

From f/8, -1/2EV smaller is f/9.5 or +1/2EV larger is f/6.7

Referring to a Kodak Darkroom DataGuide (c1965) if print time is 8 sec, then -1EV (smaller) in aperture yields 17 sec. (yes, that's right!) and +1EV (larger) in aperture yields 4 sec. -2EV (smaller) aperture yields 37 sec. according to Kodak.
 
wiltw said:
Referring to a Kodak Darkroom DataGuide (c1965) if print time is 8 sec, then -1EV (smaller) in aperture yields 17 sec. (yes, that's right!) and +1EV (larger) in aperture yields 4 sec. -2EV (smaller) aperture yields 37 sec. according to Kodak.
Click to expand...

I saw that too... then I looked closer. You may have overlooked the phrase "(Modified to compensate for reciprocity effect)"
 
lhalcong said:
Sorry. I mis-spoke.(or typed) I meant . half an exposure would be 6 sec. (right between 4 and 8) why does the calculator reports 5.7sec. I guess that's not much difference but I am curious.
Click to expand...

Going from 8 seconds to 4 seconds is 1 STOP less exposure. Going from 8 seconds to 5.7 seconds would be 1/2 STOP less exposure.
 
Last edited by a moderator:
While "a half-stop" may not be very well defined, remember a "stop" is defined (on camera) as ratio of lens aperture size to focal length and one stop is factor of 2 exposure; exposure is then proportional to f^2 (or better, relative amount of light is proportional to relative f's^2), it is a geometric (or "logrithmic") function. So a "half-stop" would more naturally be a sqrt(2) relative change in exposure. Thus, 8 sec goes to 8/sqrt(2) = 5.7 sec, etc.
 
A "stop" is a factor of two.

For exposure times, and full stops, it is quite simple: twice the time means one more stop exposure, and half the time means one less stop exposure.

But what about partial stops.

Exposure is, as mentioned, essentially logarithmic. In a logarithmic world, the "half way" point between 2 and 4 is about 2.8, the half way point between 4 and 8 is about 5.6, the half way point between 8 and 16 is about 11 ...

hmm, does 2, 2.8, 4, 5.6, 8, 11, 16 look familiar?

Each entry in that progression is equal to the square root of 2 times the previous entry.

When you are measuring the area of something circular, like a lens aperture, the radius of the aperture is what you measure, and the area increases as that radius increases. The rate of increase is related to the square of the radius, and one common version of that progression is, again:

2, 2.8, 4, 5.6, 8, 11, 16

If you want to do test strips that have half stop progressions, it is very convenient to use these times: 2, 2.8, 4, 5.6, 8, 11, 16

Hope this helps!
 
1EV of time difference is doubling (or halving) the time

1EV of Aperture difference is a full f/stop, or the progressions 2, 2.8, 4, 5.6, 8, 11, 16...arithmetically the Sqrt(2) or factor of 1.414 when going smaller aperture (larger f/stop number) (or 0.707 if going larger aperture/smaller f/stop number)

Each provides twice as much light (longer time doubling or larger aperture
 
Prof_Pixel said:
Going from 6 seconds to 4 seconds is 1 STOP less exposure. Going from 8 seconds to 5.7 seconds would be 1/2 STOP less exposure.
Click to expand...

Prof_Pixel, was that a typo 6 for 8? Didn't you mean to write the following?

Going from 8 seconds to 4 seconds is 1 STOP less exposure. Going from 8 seconds to 5.7 seconds would be 1/2 STOP less exposure.
 
Bill Burk said:
Prof_Pixel, was that a typo 6 for 8? Didn't you mean to write the following?

Going from 8 seconds to 4 seconds is 1 STOP less exposure. Going from 8 seconds to 5.7 seconds would be 1/2 STOP less exposure.
Click to expand...

Yup, it should have been an 8. Sorry. I've corrected the posting.
 
A stop either way is a factor of 2 different.

Half a stop either way is a factor of the square root of 2 or 1.414

A third of a stop either way will be the cube root of two or 1.26

To check the maths, two half stops = 1.414 x 1.414 = 2, or 1 stop.

and three half stops = 1.26 x 1.26 x 1.26 = 2, or 1 stop again.


Steve.
 
Yes, but the those numbers when used as aperture values are in whole stop steps. When used as time they are half stops.


Steve.
 
I think this is why I never got into Nocon's concept of f-stop printing.

I have always been more comfortable with "the test strip looks about 20% too dark/light" and then corrected from there.

Of course, as I always print to the same size, I have a relatively consistent base exposure.

Bests,

David.
www.dsallen.de
 
My head hurts. Thank goodness I went to a metronome for timing.
 
lhalcong said:
Up until now it was my clear understanding that if the normal print time (hypothetically speaking) for a specific print was 8 sec, then 1/2 stop would be 4 sec. and one extra stop would be 16 sec..... This is the way we do it on camera right. ! However, I recently downloaded (for the iphone) a couple of F-Stop print calculators. To my surprise 1/2 stop of 8 sec is reported as 5.7 in both. What tha ? I am confused now. I have been printing for long time based on my prior understanding . Can someone clarify this and why ?
Click to expand...

it works logaritmically not on a linear time line:laugh:.rather than usingthese calculators you may be better of using an f/stop timing table, printed large enough so you can see it in the darkroom:smile:
 

Attachments

  • TimingExposureEd1a.pdf
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Why do I like to use the logarithmic (i.e. in stops)?

Because it mirrors the response of the paper.

The different segments of a test strip printed that way end up looking much more evenly spaced than one that is printed with a sequence like 2, 4, 6, 8, 10, 12, 14, 16.

And once you get to the point of recognizing what a "half stop too dark" looks like, the process for making the correction is essentially the same whether your base exposure is 11 seconds or 32 seconds.
 
Why not simply buy a simple calculator with natural logarithmic function on the keyboard? With this you can compute any time change wanted in terms of f-stops of difference. It’s quite simple and you don’t need to know any math, just press a few keys to find the result you want.


If you made a print at time t and you want to increase the exposure by ∆ stops, then the new time T is found by


T = t*2^∆, where ∆ is positive for an exposure increase, and ∆ is negative for an exposure decrease.




Example 1: You made a print at 8 seconds and you want a 1-stop increase on the next one. ∆ = 1.

T = 8 seconds*2^1 = 16 seconds.



Example 2: You made the print at 16 seconds but find that the estimate of a 1-stop increase is too much by about 1/3 stop. So, you want the next exposure 1\3 stop less than 16 seconds. ∆ = -1/3.

T = 16 seconds*2^(-1/3) = 12.7 seconds.


Once you get used to doing this the calculation gets easy and you can do so quickly.

For the examples you cited in post #1 based on an 8-second exposure, if you want 1/2 stop exposure increase

T = 8 seconds*2^(1/2) = 11.3 seconds



If you wanted 1/2 stop LESS exposure

T = 8 seconds*2^(-1/2) = 5.7 seconds.
 
MattKing said:
Why do I like to use the logarithmic (i.e. in stops)?

Because it mirrors the response of the paper.
Click to expand...


We do the same in live sound with the decibel scales because they mirror the response of the human ear.

Also 6dB is a doubling or halving of signal voltage and 3dB is a doubling or halving of power.

So adding 3dB to a 10db signal gives the same perceived effect as adding 3dB to a 38dB signal... just as adding a stop doubles the light whatever its level was to start with.



Steve.
 
Toffle said:
I've got the chart (and the book). The chart is pinned above my Gralab in my darkroom. It's one of my most used darkroom resources. Highly recommended.

Cheers,
Tom
Click to expand...

Pinning the book above your Gralab is however NOT recommended .... :whistling:
 
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