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  1. #1

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    Stacking Neutral Density Filters

    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

  2. #2
    Leon's Avatar
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    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?

  3. #3
    reellis67's Avatar
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    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

  4. #4

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    Quote Originally Posted by matherto
    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.

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    reellis67's Avatar
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    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

  6. #6
    reellis67's Avatar
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    Here's a good site for reference...

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

    - Randy

  7. #7
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    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

  8. #8

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    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

  9. #9
    Loose Gravel's Avatar
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    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.
    Watch for Loose Gravel

  10. #10
    Leon's Avatar
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    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.

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