ND grad for ISO compensation

[LADP]

ok, let's see.

if we have Sunny 16, we'd have:

f/16, ISO100, 1/125 shutter. Let's consider this EV0 (should be 15, but I think it is enough for this exercise)

if we change f/16 to f/2, we'd have:

f/2, ISO100, 1/125 shutter, which would yeld: EV6

if we then change the shutter speed to let's say: 1/1000 (2000 is tops, but let's keep it below):

f/2, ISO100, 1/1000 shutter, which would yeld: EV3

now let's be gentle, and bump the ISO to 400:

f/2, ISO400, 1/1000 shutter, which would yeld: EV6


this would mean I'd need a 6 stop ND filter, correct?

My math (which admittedly can be questionable at times) says 5 stops attenuation, or an ND1.5 is what you want.

I just posted about this in another thread.... ND's are commonly referred to with their change in log density. So, ND's work like this:

ND .3 = 1 stop
ND .6 = 2 stops
ND .9 = 3 stops
ND 1.2 = 4 stops
ND 1.5 = 5 stops
ND 1.8 = 6 stops

etc.

Back to the math of your example...

Going from f16 to f2, is 6 stops, obviously (or EV6 as you put it).

Then, changing shutter speed to 1/1000 from 1/125 is 3 stops back the other way, making the net stop change so far 3 stops (EV3).

Finally, bumping the ISO from 100 to 400 is 2 stops, making the final net compensation needed equal 5 stops (EV5), not 6.

So in this example, you'd want an ND 1.5 filter to attenuate 5 stops of light (which has a filter factor of 32).

I guess some filter manufacturers might designate their ND's with the filter factor rahter than the log density change, but the ones we use for motion picture use (Tiffen, Schneider, B+W) all use log density.

Hope this helps (and I hope I didn't screw up the math).