Need a Math Fiend to Help Me!

[HenryChavez]

I am working on a program to use the zone system. Basically, the program documents the aperture, shutter speed and gives the EV number. You choose what zone you want to put the reading at and the program gives you the new aperture and shutter speed based on what zone you selected. Here is my question; does anyone know what the calculation/formula is to take an existing shutter speed or aperture and obtain the difference based on the number of stops? For example, f/8 @ 1/8 and I want to increase exposure by 2 stops. So I want, in this case to increase the aperture by 2 stops. I know the answer is f/2, but I want to know the equation/calculation to get to this answer. Anyone have any ideas? I did a search and could not find anything.

Kindly,

Henry

[JoeVanCleave]

I'm no math fiend, but I believe the f-stop sequence is based on SQRT2, which is about 1.414. So one stop below f/8 is 8/1.414=f/5.6. And the next stop below f/5.6 is 5.6/1.414=f/4, etc.

~Joe

[HenryChavez]

I'm no math fiend, but I believe the f-stop sequence is based on SQRT2, which is about 1.414. So one stop below f/8 is 8/1.414=f/5.6. And the next stop below f/5.6 is 5.6/1.414=f/4, etc.

~Joe

Thanks Joe, I'd seen the number, but didn't know it was the square root of 2. This is the missing piece. Thanks!

Best,
HC

[DaveBell]

Joe of course is correct, Henry. The reasoning might help with grasping the whole concept:
f ratio is equal to the focal length divided by the diameter of the aperture.
But the amount of light that gets through is proportional to the *area* of the aperture, which is of course proportional to the square of the diameter.
So, doubling the diameter increases the light by four times (or decreases the exposure time by four.)
To get exactly double the light, you need 2 times the area, so sqrt(2) times the diameter, which reduces the f ratio by sqrt(2).

Pedantically,
Dave

[HenryChavez]

Joe of course is correct, Henry. The reasoning might help with grasping the whole concept:
f ratio is equal to the focal length divided by the diameter of the aperture.
But the amount of light that gets through is proportional to the *area* of the aperture, which is of course proportional to the square of the diameter.
So, doubling the diameter increases the light by four times (or decreases the exposure time by four.)
To get exactly double the light, you need 2 times the area, so sqrt(2) times the diameter, which reduces the f ratio by sqrt(2).

Pedantically,
Dave

Thanks Dave. Very cool to know.

HC

[earlj]

it's easy to generate an f number table in Excel: first column - integer numbers from 1 to whatever (1,2,3, ...) 2nd column: =(SQRT(2))^A1, =(SQRT(2))^B1, =(SQRT(2))^C1, ...
Each f number is the square root of two raised to the power of an integer number. This relationship also gives you an easy calculation for the number of stops between two known f numbers, using the logarithm with a base of the square root of two. So, let's say your pinhole camera is f295, and your light meter reads to f22. To calculate the number of stops between those two f numbers, subtract the log of each using a base of the square root of two: =LOG(295,SQRT(2))-LOG(22,SQRT(2)) The result is 7.49, so there are 7 and 1/2 stops between f22 and f295.

[HenryChavez]

it's easy to generate an f number table in Excel: first column - integer numbers from 1 to whatever (1,2,3, ...) 2nd column: =(SQRT(2))^A1, =(SQRT(2))^B1, =(SQRT(2))^C1, ...
Each f number is the square root of two raised to the power of an integer number. This relationship also gives you an easy calculation for the number of stops between two known f numbers, using the logarithm with a base of the square root of two. So, let's say your pinhole camera is f295, and your light meter reads to f22. To calculate the number of stops between those two f numbers, subtract the log of each using a base of the square root of two: =LOG(295,SQRT(2))-LOG(22,SQRT(2)) The result is 7.49, so there are 7 and 1/2 stops between f22 and f295.

Thanks Earl, that is what I am using. This tells me I am going in the right direction.

Best,
Henry

[Ned.Lewis]

HC: Here's some more, if it seems like too much then just ignore.... you already have good answers!

Joe's answer is completely practical.

Dave's answer explains why the square root of 2 comes into play.

Earl's answer uses logarithms to further simplify the problem. Logarithms are used to turn problems with powers and multiplication into arithmetic problems, and the f-stop scale is naturally a logarithmic one. Earl's answer simplifies the calculation of a difference of f-stops, and it simplifies the understanding of that difference if you are comfortable and intuitive with logarithms.

If you only have a calculator and not a program that can perform logarithms in any base, then you can change the base of a logarithm by dividing as follows ( lower case "log" is the log base 10 function on your calculator )

LOG( x, sqrt(2) ) = log( x )/log(sqrt(2))

This comes in handy when doing quick calculations designing pinhole or lensed cameras! ( Like estimating how many "stops" falloff will there be at the corner of your film in a pinhole camera... )

I don't know that we do such a great job teaching mathematics. The story of Napier and Briggs and the invention of logarithms is fascinating and it was a revolution in science. Before Napier, people solved difficult multiplicative problems with tables of sines and other tricks, but logarithms freed scientists from time-consuming tedious calculations. More importantly maybe, it led to thinking and intuition about powers arithmetically, which is often very natural just like it is with f-stops. There are some hints that Kepler got to his 3rd law by "logarithmic thinking" he certainly knew the work of his contemporaries Napier and Briggs. Today's scientists use that thinking unconsciously having been steeped in it through school. It was a major stepping stone along the way to modern science, with amazing consequences, yet the story is hardly known.

[HenryChavez]

HC: Here's some more, if it seems like too much then just ignore.... you already have good answers!

Joe's answer is completely practical.

Dave's answer explains why the square root of 2 comes into play.

Earl's answer uses logarithms to further simplify the problem. Logarithms are used to turn problems with powers and multiplication into arithmetic problems, and the f-stop scale is naturally a logarithmic one. Earl's answer simplifies the calculation of a difference of f-stops, and it simplifies the understanding of that difference if you are comfortable and intuitive with logarithms.

If you only have a calculator and not a program that can perform logarithms in any base, then you can change the base of a logarithm by dividing as follows ( lower case "log" is the log base 10 function on your calculator )

LOG( x, sqrt(2) ) = log( x )/log(sqrt(2))

This comes in handy when doing quick calculations designing pinhole or lensed cameras! ( Like estimating how many "stops" falloff will there be at the corner of your film in a pinhole camera... )

I don't know that we do such a great job teaching mathematics. The story of Napier and Briggs and the invention of logarithms is fascinating and it was a revolution in science. Before Napier, people solved difficult multiplicative problems with tables of sines and other tricks, but logarithms freed scientists from time-consuming tedious calculations. More importantly maybe, it led to thinking and intuition about powers arithmetically, which is often very natural just like it is with f-stops. There are some hints that Kepler got to his 3rd law by "logarithmic thinking" he certainly knew the work of his contemporaries Napier and Briggs. Today's scientists use that thinking unconsciously having been steeped in it through school. It was a major stepping stone along the way to modern science, with amazing consequences, yet the story is hardly known.

Ned,
Seriously fascinating. I was a poor student with math as a youngster, only as a result of my arrogance. As an adult I have found a great need for it and have had to teach myself as a necessity. The necessity coming from my hobbies. This is a prime example of needing math. I do so enjoy learning when I play. I had to research and understand log, albeit rudimentary, to work on this program. Your insight is greatly appreciated and enjoyed.

Best,
Henry

[HenryChavez]

Well, I have a working copy up and running of the program I was writing. I'm still debugging it, but it seems to work fairly well. I'm thinking of adding ISO, but I haven't decided yet. I'll see if I find it useful before I release it to the library. I had fun working on it and I really appreciate all the help I got here. It was ALL VERY useful.