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I've been inspired by this to try using Welder's glass as an ND filter.

For welders' glass, they use values of "shade" from about 4 up to 11 or so. I know 11 is really dark. Does anybody know if there is some sort of a formula to use to calculate how many f-stops a shade 10 filter would be, for example?

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It is best if you just go out and experiment. I've been doing a lot recently with stacked "black glass" going for long, 4 minute plus, exposure. Don't worry about the numbers, worry about the composition. And have fun, I love my ND filters! –  Paul Cezanne Apr 26 '13 at 20:44

3 Answers 3

up vote 7 down vote accepted

I've been looking around for this information too! I finally found the answer (after stumbling across your question first) at this website.

According to that page, the formula is:
OD = -log T
SN = 1 + (7/3) OD

where T = transmission rate, OD = optical density, and SN = shade number.
For example, shade #10 gives SN = 10, OD = 27/7, and T = 0.000139, or nearly 13 stops! (calculating stops merely involves using base 2 instead of base 10 for the optical density. A quick and easy method is simply to divide the OD by log 2. Actually, that simplifies the formula you seek to: F = (1 / log 2) * (3/7) * (SN - 1) ...combining the constant terms gives the approximate (and much simpler) equation of:

F = 1.4 * (SN - 1)

where F = number of equivalent f-stops!
Thus, if you wanted, say, 10 f-stops, then F = 10, giving SN = 8.)

It was surprisingly difficult to find this information online! It seems that most welding companies would rather tell you what shade number you need for specific applications rather than the actual formula they used to determine that shade number...


To clarify the math, the reverse of the top equations is:

T = 10 ^ ( -(3/7)*(SN-1) )

and since 1 f-stop is half as much transmission:

F = log T / log 0.5

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This seems right, from what Ive observed. –  Octopus Jul 24 '13 at 11:01

An ND filter is specially made to be of neutral-density ("ND"), meaning that it reduces the light level evenly across the visible light spectrum, therefore retaining the original white balance of the light source. It is also of an optical grade of material and design, so as to reduce any aberrations or ghosting caused by having a pane of glass in front of your lens. All this, of course, means that it's going to be a rather expensive "pane of glass."

A welder's glass alternative is going to have a strong color cast (the one I used for a solar eclipse was very green), as well as possible image quality issues such as distortion or ghosting (where off-angle light gets reflected internally within the filter) - especially with wide-angle lenses where the corners of the frame are receiving light at an angle far from perpendicular to the filter plane.

So if you're going to use welder's glass, expect to do some considerable white balance correction (you do shoot in RAW, right?) - or convert to black & white - and accept whatever optical peculiarities you may get from the particular sample you get.

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1  
This is not really answering to the question at all. –  Esa Paulasto Jul 28 '13 at 4:38

Looking at the images in the flickr group that is mentioned in the link you posted, people are using shade 10 glass to produce 1 minute exposures at f/16, ISO100. This is the exposure time you would expect outdoors in overcast conditions if you were using a 10 stop filter, so I would presume that the shade numbers correspond exactly to number of stops of light loss.

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Thats interesting and simple. Where are these 'images in the group' that you mention? –  Octopus Apr 26 '13 at 19:45
    
Yeah I agree with this, I have welding glass that I used and it was just below the shade value in stops. –  dpollitt Apr 26 '13 at 19:48
    
I was referring to the flickr group that I thought you had linked to directly, but I've just realised that group was actually in the post you linked to! –  Matt Grum Apr 26 '13 at 19:48

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