Filtration for B&W

[fastfirstshooter]

Has anyone tried combining filters for b&W. Say a red #25 with a

yellow #11 or green #58? What results?

 

Jim

[peter_dendrinos1]

Not to be an SA, but what would be the point

[jimvanson]

SWAG here -- you're trying to really darken the blue sky? (Red's opposite is blue, yellows opposite is kinda blue?).<P>I'm not a filter expert -- I'd think that depending on the angle of the sun a red & polarized might be a better way.<P>One thing for sure though the red, yellow will act as one hell of a density filter.

[__wally_klinger]

result? black.

[dave_s]

You can answer this by just superimposing the absorbance spectra. You can get these from Kodak literature, or off the Web at http://www.geocities.com/thombell/curves.html

 

Adding a yellow-green #11 to a red #25 would give you an effect similar to the #25 itself, which eats everything shorter than about 600nm. The #11 also absorbs a second hump in the reds, so you would indeed get a very dark density filter, but I bet you wouldn't be able to detect a difference from the #25 under normal conditions.

 

The red #25 and green #58 will absorb everything shorter than about 700nm. The combination would be close to visually opaque, but would work with infrared film.

[rowland_mowrey]

Just never combine filters that are opposites.

 

Red, Green and Blue are additive filters.

 

Cyan, Magenta, and Yellow are subtractive filters.

 

Never combine blue and yellow for example. The filters are listed above in order of opposites.

 

Never combine Red and Cyan.

 

Filters can be less than additive or more than additive in filter factor, so be careful.

 

Ron Mowrey

[ronald_moravec1]

For really dark sky, put a poloroiser over a red. Careful of vignetting. It will cost 5 stops exposure.

[andy_ly1]

The middle filter would be of no effect, the first filter converts the scene to one color where the second filter will only act as a neutral density since no color is further filtered out.

[dave_s]

That's right, folks. The photons know the order you put the filters on your lens. Sneaky devils. (And fast!)

[daniel_janssen]

I am no expert but AA was. Combining red, yellows and green would give the same effect as using the strongest filter: red. You will hvae to compensate your exposures for all filters though. At least, that is what I read in The Negative.... So there is no use in doing this unless you need very slow shutter speeds...

 

Daniel

[dave_s]

<Combining red, yellows and green would give the same effect as using the strongest filter: red.>

 

Maybe you can tell us where in "The Negative" that Adams says that, because it's wrong. And even if he did say it, it's still wrong.

 

Combining a red 25 and a yellow 8 gives exactly the same result as a red 25. The yellow 8 is transparent in the wavelengths passed by the 25.

 

Combining a red 25 and a yellow-green 11 will give a very slightly darker red, because of the slight green absorbance peak of the 11. Trust me, I just did this, and this is what happens.

 

Combining a red 25 and a serious green filter, like a 58, absorbs nearly everything in the visible spectrum. It transmits everything longer than 700nm, so the combination is effectively an IR filter. It's not the same as "the strongest filter: red" whatever the "strongest filter" means.

[robert_davis2]

If you stay within the family [Yellow-to orange to red] then you can go by the higher filter filtering more light out. The problem is when you jump outside the family.

 

The Kodak Wratten are covered here.

 

http://www.geocities.com/thombell/curves.html

 

I guess people could just print them out and over lay them.

[fastfirstshooter]

Peter D., curiosity is the lubricant and learning, the point.

 

Dave S, what a great way of looking at it. Just as soon as you suggested it, I immediately recognized absorbance spectra were the way to go. So simple, so obvious ? as soon as you told me, I knew immediately.

 

And BTW, I had been wondering just how quickly those sneaky little photons would figure out my filter order. Cunning little buggers, those photons.

 

Seriously, I had been thinking in terms of color. If one places filter X in front of filter Y, what color would be left for filter Y to filter, after filter X had had its way with her ? I mean it. But, absorbance spectra make way more sense. So now, I know: just add or subtract nm. The graphs at ?curves.html? explained it very clearly.

 

A great deal of my interest is in SFX. For example, at White Sands National Monument, I shot color transparencies of brilliant white sand against august azure skies ? with a #25. You can FEEL the heat. I also shoot amusement parks with color filters on transparency films just to see what I shall see. I wondered what the effect with B&W would be. Now, I know.

 

Thanks everyone, I much appreciate your responses. Blessings.

 

JPM

[peter_dendrinos1]

I certainly did not mean to offend, I agree with you, god knows I have learned most of what I know from experimentation. And I teach young people, some craving to know. I guess I answered in a moment of frustration. I guess I am at a point where I prefer the comfort of experience over experimentation.

 

Pete

[andy_ly1]

Say you slap a green filter on, then you slap a red filter on top of that. The objective of the red filter has already done it's job, (darken blues, lighten whites, etc). The green filter sees everything as red, which is linear in all amounts, therefore the green filter simply acts as a neutral density filter.

[dave_s]

Nope. Wrong. Sorry.

[daniel_janssen]

Sorry, I wasn't accurate enough. My statement (and AA's) only holds for filters that work in the same spectrum (of course). I did write green, but I meant to write Yellow/green.

 

Daniel

[dave_s]

(Red 25 + Yellow 8) = Red 25.

 

(Red 25 + Yellow-Green 11) = (Red 25 + 0.3ND) (roughly)

 

(Red 25 + Green 58) = Infrared 89B (roughly)