View Full Version : stacking neutral density filters??

chris jordan
22-Jun-2001, 20:47
hello! here's an easy one (hopefully): when stacking neutral density filters, do you add or multiply the filter factors together? for example, if i stack a 1 0-stop ND filter and a 3-stop ND filter, is the total filter factor 13 stops, or 30?

(yeah, i know, i'm a nut-- this means REALLY long exposures even in sunlight...)

~chris jordan, Seattle

Georges Pelpel
22-Jun-2001, 23:11
You add them. If the original speed was 1/500s, adding the 10 stop filter will require a speed of 2s (10 stops lower than 1/500s), and adding the 3 stop filter will lower the speed to 16s (3 stops lower than 2s and 13 stops lower than 1/500s) to which you will have to compensate for reciprocity failure.

Jorge Garcia
23-Nov-2001, 11:00
You have to multiply the factors the one as 8X (3 stops), 64X (6 stops) the stops is the exponent for number two ( 2^3=8, 2^6=64)

The number of stops and the density must be added, the density is the log of the factor

Eg: B+W 110 ND has factor 1000, stops=10, density=3.0 Eg: B+W 103 ND has factor 8, stops=3, density=0.9

Really 2^10 is 1024 but they say just 1000.

Hope it helps (sorry for my english)

Johnny Laidlaw
12-Sep-2006, 00:53
I have been researching this myself and I believe that filter factor and f-stop reduction are two separate things although, of course, related. You have to multiply filter factor and add f-stops. Go to B+W's filter website (I don't have it handy right now) but they have good informamtion on ND filters.

12-Sep-2006, 02:23
what johnny said.

depends on how your filters are annotated:

if its in factors then you multiply them.

if its in stops then you add them.

if its in logarithms (0.1, 0.3 etc) then you add them. (0.3 is one stop. 0.1 is 1/3 stop.)