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From what I read, when adding two $x$ and $y$ measurements with uncertainties $\delta x$ and $\delta y$, the resulting uncertainty is determined by doing:

$$\delta z = \sqrt{(\delta x)^2 + (\delta y)^2}$$

I am writing a software library to handle measurements and it includes the concept of precision/uncertainty. However, I have no way of knowing the context of the measurements. Is therefor not more correct for me to "assume the worst" and do:

$$\delta z = \delta x + \delta y$$

I know, from what I read, that this overestimates the uncertainty of $z$, and that it is very unlikely for both uncertainties to be their maximum. But on the other hand, it is theoretically possible and it's not up to me to make assumptions.

At least, that's my thinking process for now, but I'm no physics expert. But I would like my software library to be as correct as possible.

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marked as duplicate by Qmechanic Jul 9 '18 at 19:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You can't say that one is more 'correct' than the other. You can be realistic or safe; the only thing that's important is mentioning what calculation technique you've used to the user/reader. $\endgroup$ – user191954 Jul 9 '18 at 11:45
  • $\begingroup$ If your $\delta x$ variables are rms errors then their sum is not (generally) a worst-case error. $\endgroup$ – The Photon Jul 9 '18 at 14:54
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It depends what you mean by 'uncertainty'. As this is the physics stack exchange it presumably means the Gaussian standard deviation, and uncertainties add in quadrature thanks to the Central Limit Theorem. Just adding them is an elementary blunder and leads to severe over-estimation of the errors which - apart from anything else - means your software library will give results that look uncompetitive and nobody will use it.

If this were the engineering stack exchange uncertainty could mean 'tolerance', and tolerances do add in a linear way. When an engineer wants to insert a $10 \pm 1 $ mm rod into a $12 \pm 1$ mm socket they will expect it to fit every time, not just 'with good probability'.

Decide who your market is, adopt the appropriate definition, and stick with it.

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