Posted 31 January 2003 - 05:43 AM
Posted 31 January 2003 - 06:24 AM
Remeber, it is half the area, not necessarily half the number. Above link is a nice article on your question.
f/1 f/1.4 f/2 f/2.8 f/4 f/5.6 f/8 f/11 f/16 f/22 f/32 and on to infinity and beyond. For those who don't want to look up the site.
Posted 31 January 2003 - 07:06 AM
Posted 08 February 2003 - 09:12 AM
Someone said above, "remember it's half the area, not necessarily half the number." Certainly true, but that doesn't directly answer the question, "How can I calculate?"
The first little problem is that as the f/stop number gets bigger, the amount of light getting through is less. A help here is to think of the f/stop number as the denominator of a fraction that relates to the diameter of the diaphragm, like f/2 is a diameter of 1/2, f/8 is a diameter of 1/8, etc. Smaller diameter, less light.
Now, the amount of light getting through is proportional to the area of the diaphragm opening. To double the area and light (to multiply it by 2), you have to multiply the diameter by the square root of 2 (or 1.414-ish).
So each full f/stop is derived from the previous one by multiplying by 1.414-ish and rounding a bit. Just remember that multiplying gives you a bigger number so you're really reducing the amount of light by 2x, not increasing.
f/4 x 1.414 = f/5.656 (mfr round to 5.6) = half the exposure of f/4
f/8 x 1.414 = f/11.312 (mfr round to 11) = half the exposure of f/8
Posted 08 February 2003 - 03:36 PM
The math works, are we correct in all this?
Posted 09 February 2003 - 08:19 PM
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