Date: Mon, 15 Feb 1999 18:08:35 +0000 From: slre@LGC.CO.UK To: iso25@fasor.com, iso25@quality.org Subject: Re14: Uncertainty in Cal lab and Testing labs? (fwd) -Reply Jim: thanks for the addition. I agree; you have to be a BIT careful! Specifically, non-linearity CAN be a problem. But I think your example overstates the case, possibly because I wasn't completely clear in what I'd do. The main point is, given individual identifiable effects, treat the parameters separately (ie not taking the combined effect of all the terms, but looking at one at a time); also, few metrologists now use linear summation for independent effects, 'cos the ISO guide says otherwise. With those two caveats, my interpretation of your example becomes: I=V/R V= 20 volts, +/-5%, (ie V= 19 to 21 V) R = 10 ohm +/-1%. (ie V=9.9 to 10.1 ohm) For the voltage uncertainty, taking -5% gives I=1.9A; +5% gives I=2.1A. exactly 0.1A in both cases. That's the Current uncertainty arising from the voltage uncertainty contribution. For Resistance, -1% gives 2.0202A; +1% gives 1.9802A. Either 0.0202A or 0.0198A difference as the Current uncertainty contribution arising from uncertainty in R. Notice that these latter two uncertainty contributions differ by about 1% of the uncertainty. That's because of non-linearity, and is the reason I'd agree with a need to check both sides. But in this (as most) cases, 1% in an uncertainty estimate is a tiny approximation error, and can nearly always be neglected (anybody who expresses uncertainty to more than two figures is almost certainly deluding someone ;-)) . More, on combination by root sum of squares with the much bigger Voltage term, that 1% difference in a minor uncertainty contribution vanishes off into the mists; we get uncertainty estimates of differing by less than 0.1% of one another. (5.097% and 5.101% of the current value, respectively). I doubt that any assessor would quibble over that difference! In practice, when I'm being fussy (often!) I would estimate these uncertainties using (y(x+dx)-y(x-dx))/2, which is (a) just numerical differentiation, and (b) gives a fair average. But it is still sensitive to non-linearity; you simply cannot represent an asymmetric uncertainty interval with only one number. You have to interpret the ISO Guide's 'parameter' rather broadly (eg as a multivariate parameter or an array of values!) Steve Ellison.