# Total error in a measurement.

by Topher925
Tags: error, measurement
 P: 1,672 I have a question that has been bugging me lately. How is it that you determine the total error of a measurement? For example, if we are trying to measure the flow rate of water coming out of a hose. We let the water flowing through the hose fill a graduated cylinder and measure the time it takes to do it. So we would have: Flow rate = Volume / Time However lets say that we need to know the error of this measurement. Would we say that the total error is: Error = Ev*dQ/dv + Et*dQ/dt (d's are partial derivatives) Where, Q = function for flow rate Ev = max error from volume measurement Et = max error of time measurement Error = total error So the formula would ultimately be: Error = Et*-V/t^2 + Ev*1/t Would this be correct? Haven't done this in a while and its just not making sense to me?
 P: 812 I think you take the *partial* derivatives of the function wrt each variable, and multiply each pd by the error, then sum them up. So in your example where F = V/t you have err = ev d(V/t)/dV + et d(V/t)/dt (where these d's are partials) You have to be careful to keep track of the units, sometimes errors are given as percent of scale and sometimes in absolute units.
P: 1,672
 Quote by gmax137 I think you take the *partial* derivatives of the function wrt each variable, and multiply each pd by the error, then sum them up.
Thanks for the reply. Thats what I meant to say, I should probably correct that. I always get confused about this formula for some reason it just doesn't make much "physical" sense to me.

P: 812

## Total error in a measurement.

The physical meaning is this - the partial derivative of f(x,y) wrt x is how much f changes for a given change in x. Now consider ex, the error in x, as that "given change in x". Then the change in f for the change in x is ex times partial of f wrt x.
P: 5,444
 Quote by Topher925 I have a question that has been bugging me lately. How is it that you determine the total error of a measurement?
John Taylor has written an excellent book "An Introduction to Error Analysis", and it's worth reading.

Under most conditions- errors in one quantity (say cylinder radius) are independent of errors in another quantity (say time to fill)- then the different errors add in quadrature. Another importnat assumption is that repeated measurements form a Gaussian distribution. Then, given a function of several variables q=F(x,y,v,...t) then:

$$\delta q = \sqrt{(\frac{\partial q}{\partial x} \delta x)^{2}+(\frac{\partial q}{\partial y} \delta y)^{2}+(\frac{\partial q}{\partial v} \delta v)^{2}+...(\frac{\partial q}{\partial t} \delta t)^{2}}$$

or variants thereof. The formula your wrote is also true, but it's an extremum value.
 P: 812 sorry for any confusion - you are quite right about summing inquadrature (a.k.a, root-sum-square or RSS). This is probably where the OP has trouble with the "physical meaning." I remember when I was trying to get to understand the meaning of the RSS formula, I found Taylor's book and it helped me out. Now that was a few years ago and I can't remember how he explained it...And my copy of the book is not at hand...
 Sci Advisor PF Gold P: 2,175 Error estimation is very tricky and there are often no single "correct" estimate. Nowadays Monte Carlo calculations are used in more advanced applications but even then there are problems; mainly related to how we interpret probability (i.e. what does the limits actually mean). From what I understand even GUM* is a bit messy in that regard, mixing Baysian and functionalistic approaches. *GUM=Guide for measurement uncertainty, a publication from ISO, IEC and a few other standardization organizations.
 P: 550 I believe your equation for the error is correct, apart from absolute values! I think it's supposed to be: $$dQ = \left| \frac{\partial Q}{\partial V} \right| \, dV + \left| \frac{\partial Q}{\partial t} \right| \, dt = \frac{1}{t} \, dV + \frac{V}{t^2} \, dt$$ (Assuming t and V are both always positive as would be the obvious case with time and volume) Note the + V/t^2..! It doesn't make sense to have your error become less (or even negative!) if some partial derivative happens to be negative...
 P: 1,672 Thanks for all the replies. I am now a little more knowledgeable about error analysis. I'll try to get to that book someday but right now I just don't have the time due to other books that I have to read.

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