Does it ever make sense to have more precision than accuracy? E.g. if you had a variable resistor which you could control with a resolution of 0.01 ohms but your multimeter only had 1% accuracy, then that extra resolution would be pretty useless wouldn't it?

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    \$\begingroup\$ "make sense" and "useless" are not very accurate or precise terms for defining a question. ;) \$\endgroup\$ Dec 16, 2013 at 2:01
  • \$\begingroup\$ Haha, fair enough, will take note for any future questions! \$\endgroup\$
    – jrandj
    Dec 16, 2013 at 2:25

3 Answers 3


Absolutely. There are many applications in which you're much more interested in the relative changes in a signal than it's absolute magnitude.

Take digital audio, for example. It is the precision of the converters that gives you the high signal-to-noise ratio you're interested in. The absolute accuracy (i.e., whether full-scale is 1.0 or 1.1 V) is not generally all that important — it just means the signal overall sounds a little softer or louder, which you can compensate for by adjusting the volume control.

  • \$\begingroup\$ measuring parameter variance of high-volume prodution hardware also comes to mind \$\endgroup\$
    – scld
    Dec 16, 2013 at 13:47
  • \$\begingroup\$ There is also a distinction between measurement precision and value precision. A 12-bit ADC provides 12 bits of value precision by definition but might provide less measurement precision. Deviation of a value from reality may be more or less systematic (with constant offset being a simple case) with the system being more or less easily modeled (randomness and limited measurement precision related to quantum mechanics being an extreme). \$\endgroup\$
    – user15426
    Dec 17, 2013 at 23:07
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    \$\begingroup\$ @PaulA.Clayton: I think what you're calling "value precision", I would simply call "resolution". But we're really getting into hair-splitting territory here. In terms of the actual performance of the application, it's the "measurement precision" that's relevant. \$\endgroup\$
    – Dave Tweed
    Dec 17, 2013 at 23:20

Yes it does, because when calibrating it allows you to measure the error with accuracy equal to the precision.

  • \$\begingroup\$ Exactly. Accuracy is typically just how a component is specified... with characterization and calibration you can do sometimes much better. \$\endgroup\$
    – darron
    Dec 17, 2013 at 14:42

Errors add up. If I measure a circuit in production at 1%, and it tells me that I need 1R, I'd like to be able to set 1R, and be within 1% of correct value. If I set 1R +/- 1%, I'm within 2% of correct value.


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