Brian West , Jan 02, 2001; 06:26 p.m.
I am hoping some of you with better math skills than I can verify
my work. My teacher bought a lux-meter at a low price with the
interest of a discovering a cheap way to meter effectivly (he runs an
extremly low budget class at my high school). He and I have been
working to devise a simple guidline(aside from the sunny-16 rule) for
using with this meter, perhaps something that only requires a simple
calculator. Mixing what I found searching this forum and what I could
cull from The Negative I came up with this simple formula: f/stop=
square root of ISO (key stop) and shutter speed= 1/(lux*.0929). So
with 100 ISO and and lux reading of 6000 (I don't have it on me so
this could be a high reading, I'm not sure) I would shoot at f/11 and
1/60. This should give a rough exposure, I believe, but I am
questioning my math so I would like to verify it. Thank you for any
help you can give.
Michael Goode , Jan 02, 2001; 11:39 p.m.
well, I can't completely help you, but my Minolta Auto Meter IV manual
has a formula for calculating lux from Exposure Value, and the formula
is lux = 2.5 x 2 ^(EV). I just measured the light in this room, and I
got, at ISO 100, F2.8 at 1/30 sec, at EV8.0. So the lux value would be
640. Using your formula, I would get shutter speed = 1/(640*.0929). So
shutter speed = 1/10 second using your formula, whereas the actual
meter reading, if I had used an f-stop of root(ISO), would be f11 at
1/2 second. In which case, your rule would underexpose the negative by
5 stops, which is, um . . . bad.
Looks like your formula needs some work. I am too tired to do some
serious mathematical derivations now, and don't think I could manage
it anyway, so I'll leave it up to you to improve the formula.
M F , Jan 02, 2001; 11:58 p.m.
Curiously, both Mr. West's question and Mr. Goode's response contain arithmetic errors ...
Anyways, a few seconds at www.google.com will solve this problem. Hint: try 'lux "exposure value" formula' (and variations) as a search string.
Francisco Bernal Rosso , Jan 04, 2001; 06:09 a.m.
As I said yestarday, but my answer did not rise the forum:
I have been studying this problem and my conclusion is that the relation beetween lux end f is:
E=(269*f^2)/(s*t)
Where E is the illumination in lux.
f the number f.
s the exposure index (ASA)
t the time of aperture.
It is in relation with:
E=(269/s) * 2 ^ev
Where ev is the exposure value.
I have based my study in:
Several notes on different books.
A development of the formula from the basis of lightning.
The table of E, ev which come with Gossens photometeers (at least with my Profisix).
ALL the tables which comes with the stock film for movies at the Kodak's cinematographers web site. This was the main source to determinate the constant 269. which is the mean of all the constant I can determine from the Kodak's stock films dates.
More about it in my web: http://fotoluz.pagina.de
(in spanich).
