Imagine to calculate the period of a pendulum using a software which takes his data from a cronometer that has an inerent instrumental uncertainty of 0.0025 s. In another discussion on this topic I found that the total uncertainty should be the sum in quadrature of both the instrumental and statistical error, so I would be tempted to say that the total uncertainty on my best value is: $\sqrt{\sigma_{statistical}^2+\sigma_{instrumental}^2} = \sqrt{\sigma_{statistical}^2 + 0.0025^2}$ and in this case all of my measurements would be perfectly compatible with each other. Despite this our professor once told us that the instrumental uncertainty should be divided by $\sqrt{12}$ and in this case our formula becomes $\sqrt{\sigma_{statistical}^2 + 0.0007^2}$. If I do this the means of different sets of measurements of the same period become pretty incompatible. Should I consider the first or the second formula? The first is more compatible with data, instead the second one should be mathematically more correct.

  • $\begingroup$ I think your prof means quantisation error, which gives that sqrt(12) factor. There are many different types of errors and quantisation error is not like the rest. dont just call it instrumental uncertainty. $\endgroup$ Apr 16, 2023 at 23:51
  • $\begingroup$ The factor "1/12" is used, if the uncertainty is described by a uniform distribution, see the variance on Wiki: en.wikipedia.org/wiki/Continuous_uniform_distribution. Note that in the nominator we actually have to use the width $b-a$. $\endgroup$
    – Semoi
    Apr 18, 2023 at 21:49

1 Answer 1


It depends on how "inherent instrumental uncertainty" is defined. If the chronometer counts "ticks" of duration $r$, and has no other source of error but this counting process, the maximum error is $r/2$. But a common measure of statistical error isn't the maximum, but the the variance, $\sigma^2$. That's $r^2/12$ for a perfect digitizer. That's what you add to other estimated variances to estimate the total variance.


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