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Total error in a measurement.

by Topher925
Tags: error, measurement
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Sep8-08, 11:40 AM
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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)

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?
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Sep8-08, 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.
Sep8-08, 11:54 AM
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Quote Quote by gmax137 View Post
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.

Sep8-08, 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.
Andy Resnick
Sep8-08, 12:17 PM
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Quote Quote by Topher925 View Post
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:

[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.
Sep8-08, 12:50 PM
P: 856
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...
Sep8-08, 01:58 PM
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PF Gold
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P: 2,252
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.
Sep8-08, 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...
Sep9-08, 01:32 PM
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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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