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Kevin Crisp
1-Nov-2003, 16:57
I'm putting a lens in shutter and was wondering what the numeric f stops are if I go by thirds between f:16 to f:45. What is the numeric f stop if I stop down from f:16 by 1/3rd stop, in other words, and so on. Or, to make it easy, how do I compute this? Thanks.

Paul Butzi
1-Nov-2003, 17:52
The sequence of f stop values (in 1/3 stop increments) are: 1.00 1.12 1.26 1.41 1.59 1.78 2.00 2.24 2.52 2.83 3.17 3.56 4.00 4.49 5.04 5.66 6.35 7.13 8.00 8.98 10.08 11.31 12.70 14.25 16.00 17.96 20.16 22.63 25.40 28.51 32.00 35.92 40.32 45.25 50.80 57.02 64.00 71.84 80.63 90.51 101.59 114.04 128.00 143.68 161.27

Note that some of these are rounded weirdly by convention.

Kevin Crisp
1-Nov-2003, 17:54
Thanks, that was quick and helpful!

Kevin

Steve Baggett
1-Nov-2003, 18:05
For whole stops, start at 1 and multiply by 1.414 (square root of 2). Each stop is "rounded" to the nearest whole number but the product should be maintained to 2 or 3 decimal places for the next term in the series. For 1/2 stops, do the same but use the 4-th root of 2 (1.189). For 1/3 stops, do the same but use the 6-th root of 2 (1.122). Here's the series for 1/2 stops with the rounded stop in brackets [] using 3 decimal places :

1[1] 1.189[1.2] 1.414[1.4] 1.681[1.7] 1.999[2] 2.376[2.4] 2.825[2.8] 3.359[3.4] 3.994[4.0] 4.749[4.8] ..... 15.995[16] 18.971[19] 22.557[22] 26.820[27] 31.889[32] 37.916[38] 45.082[45] 53.602[54] 63.733 [64]

Note that f/22 should really be f/22.5 or f/23.

Kevin Crisp
1-Nov-2003, 18:50
The sixth root of two. Of course! What was I thinking? (thanks.)