Hi, CH and all,
From: "C.H.Ling" <ch_photo@xxxxxxxxxxxxx>
>Hi Dr. Flash, F11.8 is equal to F11 +0.8 stop? So F12 is equal to F11 +1
>stop? then F12=F16??? The formula should be more complex than you think :-)
Multiply by 1,03526492384 for each tenth of a stop... thus F11.8 is really
little more than 0.2 stops over F11.
For thirds of a stop, multiply (or divide) the aperture by 1,12246204831
each. Typical secuence (rounded):
[1.4] - 1.6 - 1.8 - [2] - 2.2 - 2.5 - [2.8] - 3.2 - 3.5* - [4] - 4.5 - 5 -
[5.6] - 6.3 - 7.1 - [8] - 9 - 10 - [11**] - 12.7** - 14.2** - [16] - 18 - 20
- [22]
*) It's closer to 3.6, but usually rounded to this value.
**) F11 should be really about 11.2, the intermediate values are calculated
from this.
Cheers,
--
Carlos J. Santisteban Salinas
IES Turaniana (Roquetas de Mar, Almeria)
<http://cjss.sytes.net/>
--
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