Tom Diaz
Well-known
Jan,
I looked at the vignetting data for the 24mm f/1.4. (By the way, do you actually own that gorgeous lens?) The Y axis shows light falloff as you move away from the center, where "100%" means "no vignetting." Doesn't that mean that 0% equals 1 f/stop of falloff?
Anyway, the X value of interest would be 10-12 mm, right? The edge of an APS-C sensor is about that far from the center. Anyway, difference between f/5.6 and f/1.4 is about 25%. I'm not sure how may stops that equals--1/4?
If my reasoning is valid, which it may not be, it seems like a vignetting correction for wide open would be reasonably good for a couple of stops down, too.
Tom
jankap
Established
No, no, this is just an example.
The diagonal of full format is 42mm (43,2mm to be exact). This gives an X value of 14mm (radius and crop factor).
The Y value of 5.6 at 14mm is 60%, that is almost one f-stop from the ideal value (100%).
At 1.4 it is 30%, that is another full f-stop less.
Jan
The diagonal of full format is 42mm (43,2mm to be exact). This gives an X value of 14mm (radius and crop factor).
The Y value of 5.6 at 14mm is 60%, that is almost one f-stop from the ideal value (100%).
At 1.4 it is 30%, that is another full f-stop less.
Jan
Tom Diaz
Well-known
1. OK. In that case I can see the reason to aim the vignetting correction at whatever aperture you use most frequently. With something like the 24mm f/1.4, I'd probably be shooting wide open a lot, because that's the reason for owning the lens!
2. Just checking my math: is the factor of light falloff the antilog of the percentage of falloff? If so, then "60%" means a falloff of 2.5x (antilog .4), even more than one stop. Please set me straight if this is not the right formula.
Tom
jankap
Established
Isn't it like exposure times, f-stops and ISOs? Double or half.
One f-stop is 50% of the light (-intensity), two f-stops are 25%, etc.
At the top: 100% means no f-stop compensation necessary. And 0% means absolutely no light in this area, say an infinity number of f-stops
.
I propose the following formula: y = 1/log(2.) * log(1/x)
where
x is the light fall-off in parts (in stead of 25% -> 0.25).
y the vignetting in f-stops
log function at base 10
constant log(2.)=0.30103
These compensations (M9 and Ricoh) stay emergency measures. In these areas the electric current is amplified simply, giving the same effect as in Photoshop, namely more noise. The light (the information) is dimmed in these areas.
Jan
PS
A light fall-off of 60% gives 0.74. So between 2/3 and 1 f-stop.
One f-stop is 50% of the light (-intensity), two f-stops are 25%, etc.
At the top: 100% means no f-stop compensation necessary. And 0% means absolutely no light in this area, say an infinity number of f-stops
.I propose the following formula: y = 1/log(2.) * log(1/x)
where
x is the light fall-off in parts (in stead of 25% -> 0.25).
y the vignetting in f-stops
log function at base 10
constant log(2.)=0.30103
These compensations (M9 and Ricoh) stay emergency measures. In these areas the electric current is amplified simply, giving the same effect as in Photoshop, namely more noise. The light (the information) is dimmed in these areas.
Jan
PS
A light fall-off of 60% gives 0.74. So between 2/3 and 1 f-stop.
Pete B
Well-known
Still no m9/GXR noise comparison?
Pete
Pete