How much faster is f1?

[triplefinger]

I'm fine calculating the normal f stops, but once i get into very low numbers, i get confused...

starting at f2
is 1.4 or 1.5 2 times faster?
how much faster is 1.2?
how much faster is 1.0?

this is all leading to the question...noctilux? summilux? or wait for the re-issue sonnar 1.5?

thanks, Mike

 

[Stephan]

One stop slower is multiplying the number by 1,41 (sqrt of 2). So in full stops the scale goes: 1 - 1.4 - 2 -2.8 - 4 - 5,6 - 8 -11 -16 etc...

 

[Stephan]

And a full stop is always 2 times slower (going up the scale).


As for which lens to choose, I'll leave you in the capable hands of the bunch of lens geeks that hang out here 😀 You'll get as many opinions as people you ask, it's very much a matter of taste 😉

 

[SDK]

starting at f2
is 1.4 or 1.5 2 times faster?
how much faster is 1.2?
how much faster is 1.0?
thanks, Mike

If an F number is outside the standard stops, like those Stephan described, you can use the equation v=2 log(N/N0)/log 2 to figure out the proportion of a stop difference between them, where N is the first stop in question and N0 is the other one.

Examples: 2x log (1.5/2.0)/log 2 = -.83, so f/1.5 is 8.3 tenths of a stop faster than f/2, almost a whole stop faster (you get a negative number if N<N0).

2x log (1.2/1.4)/log 2 = -0.44, so f/1.2 is 44% of one stop faster than f/1.4, not quite a half stop faster.

For the Leica 135mm/4 compared to the newer 135mmf/3.4, you get 0.47 which is very close to 1/2 a stop slower (you get a positive number if N>N0).

You can use either ln (loge) or log10, and the equation works either way giving the exact same results (but don't ask me why!).

Equation from "Some Thoughts on View Camera Calculations" by
Leonard Evens, Department of Mathematics, Northwestern University, 2003, a PDF which I found in the Large format forum Photo.net.

 

[ClaremontPhoto]

I've always suspected those f1.8's and f1.2's were designed at the behest of the marketing director.

I mean, an f1.8 must be much better than an f2 mustn't it? Way better.

 

[triplefinger]

very helpful math. thank mucho.

now the math is in the wallet!

 

[gns]

Why f numbers increment by a factor of 1.4...
If you move a light source closer to or farther from a subject by a distance of 1.4x, you double or halve its intensity. I think that's right.

 

[Mazurka]

If you have the Noctilux in mind: there is almost 3 stops of light fall-off at the extreme corners when used wide-open, so only the very centre has transmission equivalent to f/1.

f/1.2 (or 1.189) is half a stop slower. At such speed, you can expect at least 1 stop (usually more) of light fall-off at the corners.

f/1.7 is half-way between f/1.4 and f/2. 1/3 stop sequence is f/1.4, 1.6 (or rather 1.59), 1.8 and f/2.

 

[dmr]

Why f numbers increment by a factor of 1.4...
If you move a light source closer to or farther from a subject by a distance of 1.4x, you double or halve its intensity. I think that's right.

It's the inverse square law. 1.414... is the square root of 2, and yes, you are correct. 🙂

 

[dmr]

You can use either ln (loge) or log10, and the equation works either way giving the exact same results (but don't ask me why!).

It's because you're actually converting to a base 2 logarithm in both cases. 🙂

You can very easily convert logarithms from one base to another by dividing by a constant, that being the log (in the current base) of the base you are converting to, in this case, log 2. 🙂

Don't ask me why I know this stuff, or why I actually remember it. 🙂

 

[Mazurka]

No need to mess with all that logarithm stuff (at least not directly) for such calculations.

f/stops are all in powers of SQRT 2.

e.g. from f/2, mulitply by SQRT 2 to get the next smaller stop, i.e. f/2.8

1/2 stop smaller than f/2: 2 x (SQRT 2 ^ 0.5) = f/2.4

1/3 stop smaller than f/2: 2 x (SQRT 2 ^ 0.333) = f/2.25

2/3 stop smaller than f/2: 2 x (SQRT 2 ^ 0.6667) = f/2.5, etc.

 

[Kyle]

This thread is making my head hurt.

 

[wyk_penguin]

Looks like it is time we get the slide rules out.

 

[jdos2]

Looks like it is time we get the slide rules out.

My linear slide rule sits next to the Mac- just to remind it I did without for years, and can do so again.
My circular one is in the closet, in the "aviation" pile (sectionals, logs, &c.)

🙂

 

[ClaremontPhoto]

If you find yourself without your log tables and/or slide rule remember...

Alternate f stops double numerically: 1 2 4 8 16 32

interspersed with: 1.4 2.8 5.6 11 22

 

[Ergo]

...

...

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If an F number is outside the standard stops, like those Stephan described, you can use the equation v=2 log(N/N0)/log 2 to figure out the proportion of a stop difference between them, where N is the first stop in question and N0 is the other one.

Examples: 2x log (1.5/2.0)/log 2 = -.83, so f/1.5 is 8.3 tenths of a stop faster than f/2, almost a whole stop faster (you get a negative number if N<N0).

2x log (1.2/1.4)/log 2 = -0.44, so f/1.2 is 44% of one stop faster than f/1.4, not quite a half stop faster.

For the Leica 135mm/4 compared to the newer 135mmf/3.4, you get 0.47 which is very close to 1/2 a stop slower (you get a positive number if N>N0).

You can use either ln (loge) or log10, and the equation works either way giving the exact same results (but don't ask me why!).

Equation from "Some Thoughts on View Camera Calculations" by
Leonard Evens, Department of Mathematics, Northwestern University, 2003, a PDF which I found in the Large format forum Photo.net.

 

[jano]

you can use the equation v=2 log(N/N0)/log 2

This thread is making my head hurt.

Kyle, sing with me! 😀

"When the log rolls over we will all be dead!" 😛 😀

 

[ClaremontPhoto]

Who wants a Leica/Voigtlander etc with program mode after all this?

 

[Kyle]

Kyle, sing with me! 😀

"When the log rolls over we will all be dead!" 😛 😀

Hahaha... now thats funny!

 

[triplefinger]

Wow!!

Wow!!

I am both thrilled and nauseated(spelling?) by the thread I started. Thanks everyone for the excellent input. Now, I will have 3 advil.

Mike