Stacking Neutral Density Filters

[sqphoto]

I have several neutral density filters that require various lengths of additional exposure. For example one filter requires 2 stops of additional exposure and another filter requires 6 stops of additional exposure.
I want to try and stack these filters when exposing black and white film so that I can get long exposures - like 1 minute real time - which is a lot longer when considering reciprocity effect - but if I have an accurate combined filter factor then the experiments start. Has anyone got any experience? Do I add or multiply filter factors? Other sources of information? Thanks. Michael

 

[Leon]

I'm not sure what you are asking for here ... but when I stack filters (usually a 3 stop nd, a 2 stop nd and a 2 stop orange, I multiply the stop values together : 3 x 2 x 2 = 12 so I give + 12 stops then apply reciprocity compensation as necessary - hope that helps?

 

[reellis67]

My understanding is that the factors are multiplied. I have not used this myself so I cannot say this works from personal experience though...

- Randy

 

[Donald Miller]

I have several neutral density filters that require various lengths of additional exposure. For example one filter requires 2 stops of additional exposure and another filter requires 6 stops of additional exposure.
I want to try and stack these filters when exposing black and white film so that I can get long exposures - like 1 minute real time - which is a lot longer when considering reciprocity effect - but if I have an accurate combined filter factor then the experiments start. Has anyone got any experience? Do I add or multiply filter factors? Other sources of information? Thanks. Michael

Filter factors are added. In other words if you used the combination of your 2 stop ND and your 6 stop ND the combined effect would be 8 stops of ND and not 12 stops of ND.

 

[reellis67]

I thought you added the change in f/stop or multiplied the filter factor.

for example:

filter A is 2x factor, which is a +1 stop compensation.
filter B is 4x factor, which is a +2 stop compensation.

filter A + filter B = +3 stops combined
-or-
filter A + filter B = 8x factor (which is a +3 stop adjustment)

- Randy

 

[reellis67]

Here's a good site for reference...

http://www.fineart-photography.com/bwfilter.html

- Randy

 

[simulatordan]

Hello matherto,

I hope this doesn't add to you confusion but I just thought I'd add my two penneth.

A 1stop Neutral Density filter is often designated as, for example ND2, meaning that it doubles the length of the exposure, A 2stop as ND4, 3stops as ND8 and so on: they can also be shown as 0.30, 0.60, and 0.90 as well.
If you use a 2stop and a 3stop together the resulting increase is 5stops ( 2 + 3 ), if you take the factor e.g. nd 4 + nd 8 then if you multiply the factors ( 4 * 8 ) = 32 the number you end up with is the multiplier for the exposure without filtration.

For example: say you calculate the exposure as 1 sec at f8 without filters

1 sec + 2 stops = 1 -> 2 - >4 secs ( nd 4 )
4 secs + 3 stops = 4 -> 8 -> 16 -> 32 secs ( nd 8 )

or to put it another way : 1 sec at f8 multiplied by ( 4 * 8 ) = 32 secs = 5stops

Does this make sense to you ? hope it helps Daniel

 

[rrankin]

I think this confusion is common among new filter users or people who don't let the TTL meter handle it all (a hope and a prayer on that one...).

Sometimes the effect of a filter is stated as "Increase exposure by X times" e.g. 2 times, and another place something might say "open lens 1 f-stop". When the exposure effects are given the first way, they are multiplied for multiple filters. When given as f-stops, they are added for multiple filters.

So, for example, let's say you have 2 filters together. If they say "increase exposure by 2 times and 4 times', respectively, you would MULTIPLY your indicated exposure by 2 * 4 = 8. If they were given as 'open lens 1 stop and 2 stops', respectively, they would be ADDED and you'd get 1+2 stops = 3 stops to open up.

Multiplying your indicated exposure by 8 or adjusting by 3 stops is the same thing. It's not really appropriate here to go into WHY that is true, but the f-stop system could be called an alternate mathematical system to get similar results more easily.

Hope that helps.

Cheers,
Richard

 

[Loose Gravel]

Randy,

NDs have numbers that are given as the log of the attenuation, so a 1 stop (2x) adjustment has a filter factor of 0.3. 2 stops is 0.6, 3 stops is 0.9, etc. A 3.0 ND is 10 stops. It is the logs that you must add. Also, as you pointed out, add the compensation.

 

[Leon]

so, in multiplying my f stops, I am over exposing? I 've not really noticed that I am - they always seem to be pretty well exposed to me.

 

[sqphoto]

That answered my question. Thanks for the info. I asked the same question to another photographer and he said add them. But, your information agrees with the data in one of Ansel Adams' books.What prompted me to ask the question in the first place was I could not figure out what would happen if you had for example a K2 yellow filter with a 1 stop density and then a nd filter say 3 stops? If you multiply that you get 3 stops. That doesn't seem to work. Any more thoughts? Michael

 

[pentaxuser]

That answered my question. Thanks for the info. I asked the same question to another photographer and he said add them. But, your information agrees with the data in one of Ansel Adams' books.What prompted me to ask the question in the first place was I could not figure out what would happen if you had for example a K2 yellow filter with a 1 stop density and then a nd filter say 3 stops? If you multiply that you get 3 stops. That doesn't seem to work. Any more thoughts? Michael

Yes it does work. See reelis67 post again. If you add the stops you get 4. If you multipy the factors you get Yellow a factor of x2 and an ND a factor of x8 which is a total of 16 if multiplied. A factor total of 16 is 4 stops. It progresses 2,4,8,16. Each stop is double the value of its predecessor.

Pentaxuser