Simple question: time and "stops"

JeffD · Feb 27, 2005

Probably and easy question:

How do I figure out how to add a third of a stop to a particular amount of time (keeping fixed aperature)?

For instance, if I want to add 2/3 a stop to 19 seconds of time, how do I determine that?

I know adding a stop doubles the time, it is the 1/3 and 2/3 increases that I'm not sure how to figure out properly..

JeffD · Feb 27, 2005

Michael Scarpitti said:
1/3 of a stop is 1/3 of 2x.

19 secs + 19 secs = 38 secs would be one stop

1/3 of 19 secs is about 7 secs.

19 + 7 = 26 secs.

That sounds about right. But, doesn't exposure increase at a non arithmetic fashion? In otherwords, isn't 2/3 a stop actually more than 2 * 7 seconds?

Or, maybe I'm confusing something I have read.

Short times it probably doesn't matter much at all, but I am trying to think through a reciprocity problem.

Donald Miller · Feb 27, 2005

Using your example. One stop would be 38 seconds total time. One third stop additional would be 25.3 seconds total time. Two thirds stop addtional would be 31.6 seconds.

JeffD said:
Probably and easy question:

How do I figure out how to add a third of a stop to a particular amount of time (keeping fixed aperature)?

For instance, if I want to add 2/3 a stop to 19 seconds of time, how do I determine that?

I know adding a stop doubles the time, it is the 1/3 and 2/3 increases that I'm not sure how to figure out properly..

Flotsam · Feb 27, 2005

There are a couple of good F stop printing charts available on the internet too. I keep one on my darkroom wall.

MurrayMinchin · Feb 27, 2005

Hi Jeff, are there any simple questions?

My method is to use percentage increases/decreases. A one stop increase of exposure for 19 seconds is a 100% increase; 38 seconds. A 1/3 stop increase would be 19 seconds + 33%, 0r 25.27 seconds...I'd round it up to 26 seconds.

This way, if at a later date you're making a larger print of the same image and want to burn the sky the same way, you just apply the original percentage increase to your new exposure for the larger print.

Murray

rjr · Feb 27, 2005

One chart is available at

Dead Link Removed

It´s the chart published by Lambrecht and Woodhouse in their book "Way beyond monochrome".

Bruce Osgood · Feb 27, 2005

From Ed Buffaloes' page on test exposure for printing, I think principle is the same for camera exposure increases:

BASE 19.0
PLUS------Net change---new exp
1/4 STOP----0.19--------22.6
1/2 STOP ----0.41--------26.8
3/4 STOP----0.68--------31.9
1 STOP--- --1.00--------38.0

http://unblinkingeye.com/Articles/TestExpo/testexpo.html

The queston was about 1/3 stop increase and my answer is about 1/4 stop increases but for greater detail on the reasoning please see Ed Buffaloes' site, he explains it all much better than I can.

Bruce Osgood · Feb 27, 2005

OOPS,

Bruce (Camclicker) said:
From Ed Buffaloes' page on test exposure for printing, I think principle is the same for camera exposure increases:

BASE 19.0
PLUS------Net change---new exp
1/4 STOP----0.19--------22.6
1/2 STOP ----0.41--------26.8
3/4 STOP----0.68--------31.9
1 STOP--- --1.00--------38.0

http://unblinkingeye.com/Articles/TestExpo/testexpo.html

The queston was about 1/3 stop increase and my answer is about 1/4 stop increases but for greater detail on the reasoning please see Ed Buffaloes' site, he explains it all much better than I can.

My colulmn titled Net change should be titled MULTIPLIER and the numbers should be expressed as 1.19; 1.41; 1.68 and 2.00.

I was once told by a boss that you can make all the mistakes you want as long as you keep them on your desk. Sorry this got off.

oriecat · Feb 27, 2005

I was reading in my Way Beyond Monochrome book about f/stop printing, but it went way over my head at the time. Does anyone have a good basic link that I could start with to try and get into this?

Kevin Caulfield · Feb 27, 2005

As I understand it, to calculate new times, you need to raise the number 2 to the power of the number of stops.

Therefore, for a 19 second exposure, one stop increase would be 2 to the power of 1, which is 2, and 2 x 19 is 38s.

For a 1/3 stop increase, 2 to the 1/3 = 1.2599
1.2599 x 19 = 23.94s

For 2/3 stop, 2 to the 2/3 = 1.5874
1.5874 x 19 = 30.16s

KenM · Feb 27, 2005

Kevin Caulfield said:
As I understand it, to calculate new times, you need to raise the number 2 to the power of the number of stops.

Therefore, for a 19 second exposure, one stop increase would be 2 to the power of 1, which is 2, and 2 x 19 is 38s.

For a 1/3 stop increase, 2 to the 1/3 = 1.2599
1.2599 x 19 = 23.94s

For 2/3 stop, 2 to the 2/3 = 1.5874
1.5874 x 19 = 30.16s

If your calculator doesn't handle raising numbers to fractional exponents, you can use the log function to caculate the same thing:

You want to find z = 2 ^ (2/3). Tough to do, but easy with the log function:

log( z ) = (2/3) * log( 2 )
log( z ) = 0.20068...

Then, take the inverse log of 0.20068, and you get:

z = 1.58740...

Then, 19 * z gives you 30.16s. You can use this method to determine any fractional stop exposure change. I used the log function which determines logarithms based on '10' - the ln function determines logs based on 'e'. You can use either log function for this task, as long as you're consistent.

Bob F. · Feb 27, 2005

Kevin & Ken above give the real mathematical deal - if you have a calculator/computer available, use their equations.

Easier to do sans calculator is to add 25% for +1/3 stop & 60% for +2/3 stop (or subtract 20% for -1/3 stop & subtract 40% for -2/3 stop).

So, 19 secs + 1/3 stop = 19+5(ish)=24 secs. 19 secs - 2/3 stop = 19-8(ish)=11 secs. Not exact, but close enough for most practical purposes.

Cheers, Bob.

JeffD · Feb 27, 2005

Bruce (Camclicker) said:
My colulmn titled Net change should be titled MULTIPLIER and the numbers should be expressed as 1.19; 1.41; 1.68 and 2.00.

That is good info. I wonder what the multiplier is for 1/3 stops?

I am realizing that and additional 2/3 stops worth of time is different than 2 * (1/3 stops worth of additional time).

Maybe I am nitpicking.

Thanks for all the answers!

Kevin Caulfield · Feb 27, 2005

It's what I posted above. The multiplier for 1/3 stop is 1.26, and for 2/3 stop it is 1.59.

So, Bob's advise above to use 25% for one stop and 60% for two stops is spot on (near enough!!).

Kevin Caulfield · Feb 27, 2005

Oops, I meant "advice" not "advise".

smieglitz · Feb 28, 2005

A very quick and dirty way to do this and get in the ballpark is to reference the standard ISO progression of films. As an example using 20 seconds (ISO 20) one-third up would be 25 seconds (ISO 25) and + 2/3 stop would get you to 32 seconds (ISO 32). If you had a 12.5 second print (think ISO 125) 1/3 up would be 16 seconds (think ISO 160) 2/3 up would be 20 seconds (think ISO 200).

So, if you memorize the ISO sequence of film speeds, you can relate it to exposure times as well. Since every third number in the sequence doubles, the entire sequence can be derived from just remembering 3 of the ISO speeds. For example, if you can remember ISO 100 (TMX), ISO 125 (Plus-X) and ISO 160 (Portra) you can generate:

...25, 32 , 40 , 50, 64 , 80 , 100, 125 , 160 , 200, 250 , 320 , 400 ...

The jump from 25 to 32 is 1/3 stop just as the jump from 2.5 to 3.2 would be 1/3 stop.

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Simple question: time and "stops"

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