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Re: [OM] E-1 4/3 vs. full sensor (stop steps)

Subject: Re: [OM] E-1 4/3 vs. full sensor (stop steps)
From: Joe Gwinn <joegwinn@xxxxxxxxxxx>
Date: Fri, 5 Dec 2003 20:08:45 -0500
At 1:57 PM +0000 12/5/03, olympus-digest wrote:
>Date: Fri, 05 Dec 2003 00:57:11 -0800
>From: Jim Brokaw <jbrokaw@xxxxxxxxxxx>
>Subject: Re: [OM] E-1 4/3 vs. full sensor
>
>I'm not sure of the math, but I thought a 1.2 was one-half stop faster than
>a 1.4 lens...?  The numbers are not linear, but that's as far as my
>twice-flunked Intro to Calculus will take me... if I could of learned math,
>I could have been a engineer instead of a writer <g>

Only need arithmetic.  Stops are chosen so the *area* of the aperture doubles 
or halves, so it's the square of the f/ number that doubles or halves as one 
goes down or up by a stop.

(1.4/1.2)^2= 1.36; log2(1.36)= 0.444= 1/2 stop, or a little less, so your 
understanding is correct


f/8 is the center of the system.  The f/ of the stops will each be (Sqrt[2])^N 
larger or smaller than f/8, where N is the number of stops away from f/8.  The 
resulting values are rounded to get the traditional values:

4/Sqrt[2]= 2.83= 2.8

5.567/Sqrt[2]= 4

8/Sqrt[2]= 5.567= 5.6

8*Sqrt[2]= 11.

11.3137*Sqrt[2]=16

16*Sqrt[2]= 22.6= 22  (should have been 23, but tradition ruled)

22.6*Sqrt[2]= 32


Joe Gwinn


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