How do welder’s glass shade numbers compare to ND filter stops?
Asked 4/26/2013
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2 answers
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I’m considering using welder’s glass as a neutral density filter for long exposures. Welder’s glass is labeled by shade number (for example shade 10 or shade 11), but ND filters are usually described in stops. Is there a reliable way to convert a welder’s glass shade number into equivalent f-stops of light reduction?
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
13y ago
2 Answers
16
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
Originally by user21249. Source · Licensed CC BY-SA 4.0
user21249
13y ago
0
Generated from our catalog & community — verify before relying on it.
Yes. One community answer provided the standard relationship between welder’s shade number and optical density:
- OD = -log10(T)
- shade number (SN) = 1 + (7/3) × OD
From that, the equivalent light reduction in stops is approximately:
stops ≈ 1.4 × (SN - 1)
So for shade 10:
- stops ≈ 1.4 × (10 - 1)
- stops ≈ 12.6 stops
That’s commonly rounded to about 13 stops.
A second answer compared real-world exposures and estimated shade 10 at around 10 stops, but the formula-based answer is the more precise conversion given in the discussion.
In practice, exposure may vary a bit depending on the actual glass, so it’s wise to test your specific piece. Also note that welder’s glass often has a strong color cast, so while it can work for long exposures, it may not behave like a true neutral ND filter.
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