Hi,
I was going thru the photometrics specification given in the following link.
http://emea.rosco.co...roduct/silk210
Is there any way to convert the lumen/lux/ fc output given in the light specification into T stops?
Mathew Collins.
lumen/ lux/ fc output to Tstops
Started By Mathew Collins, May 23 2017 09:56 PM
2 replies to this topic
#1Posted 23 May 2017  09:56 PM Hi,
I was going thru the photometrics specification given in the following link.
http://emea.rosco.co...roduct/silk210
Is there any way to convert the lumen/lux/ fc output given in the light specification into T stops?
Mathew Collins. Support Cinematography.com and buy gear using our Amazon links! PANASONIC LUMIX GH5 Body 4K Mirrorless Camera, 20.3 Megapixels, Dual I.S. 2.0, 4K 422 10bit, Full Size HDMI Out, 3 Inch Touch LCD, DCGH5KBODY (USA Black) #2Posted 23 May 2017  10:37 PM The simple rule to remember is that 100 footcandles on 100 ASA film gets you a T/2.8 exposure (at 24 fps with a standard 180 degree shutter).
From those ratios you can figure out what happens when you change any one of those three, for example, 50 fc at 200 ASA also gets you T/2.8, so does 25 fc at 400 ASA, 12 fc at 800 ASA, etc.
There is an actual conversion formula that someone can dig up... #3Posted 24 May 2017  03:22 AM There is no completely unequivocal conversion, because the calculations rely on a "meter calibration constant" which represents the average observer's idea of a properlyexposed image. For reasons too longwinded to explain, 330 is a common value.
With that in mind, the required exposure value for a scene is equal to the binary logarithm of the light level in lux multiplied by ISO over 330.
This is complicated by the fact that most calculators don't have a direct way of getting the binary logarithm of a value, but you can get the same result by dividing the natural logarithm of that value by the natural logarithm of 2.
Assume 1000 lux and ISO 100:
log_{2}(1000 * 100 / 330) = 8.24
Similarly, the effective exposure value of a particular camera configuration is equal to the binary logarithm of the aperture squared over the shutter time in seconds:
log_{2}(2.8^{2}/(1/48)) = 8.56
Thus 1/48s at f/2.8 is about a third of a stop overexposed in an incident light of 1000 lux at ISO 100.
P 
