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.