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Total error in a measurement. 
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#1
Sep808, 11:40 AM

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? 


#2
Sep808, 11:50 AM

P: 856

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. 


#3
Sep808, 11:54 AM

P: 1,672




#4
Sep808, 12:03 PM

P: 856

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.



#5
Sep808, 12:17 PM

Sci Advisor
P: 5,523

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: [tex]\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}}[/tex] or variants thereof. The formula your wrote is also true, but it's an extremum value. 


#6
Sep808, 12:50 PM

P: 856

sorry for any confusion  you are quite right about summing inquadrature (a.k.a, rootsumsquare 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...



#7
Sep808, 01:58 PM

Sci Advisor
PF Gold
P: 2,250

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. 


#8
Sep808, 02:22 PM

P: 550

I believe your equation for the error is correct, apart from absolute values!
I think it's supposed to be: [tex]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[/tex] (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... 


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