Is the Calculation of Uncertainty for f Correct?

AI Thread Summary
The discussion revolves around calculating the value of f and its uncertainty based on given variables a, b, c, and d. The initial formula for f was incorrectly stated, leading to confusion, but it was clarified that f-1 should be expressed as ((c/d)-1)/((a/b)-1). The method of adding percentage uncertainties in quadrature was confirmed as appropriate, yielding f=1.0038±0.0012. Additionally, a separate inquiry about estimating the uncertainty in the half-value layer of X-rays through aluminum was addressed, affirming that independent sources of error can be added in quadrature for an overall uncertainty estimate. The discussion highlights the importance of clarity in formulas and the correct application of uncertainty calculations.
lavster
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Homework Statement



let a=360.9±0.1, let b=180.4±0.1, let c=212.5±0.13 and let d=211.7±0.16

whats f (including uncertainty) where

f-1 = ((a/b)-1)/((c/d)-1)

Homework Equations



add the percentage uncertainties in quadrature if dividing two variables
add the absolute uncertainties in quadrature of adding two variables?


The Attempt at a Solution



from plugging in the numbers f=1.0038

the percentage uncertainty for a is 0.03%
the percentage uncertainty of b is 0.06%
adding this in quadrature gives 0.07% (call this y)

the percentage uncertainty for c is 0.063136037%
the percentage uncertainty of d is 0.077617746%
adding this in quadrature gives 0.1% (call this z)

now find the uncertainty of the quotient of these is adding y and z in quadrature giving 0.12%

therefore f=1.0038±0.0012


is this correct?!

thanks
 
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lavster said:

Homework Statement



let a=360.9±0.1, let b=180.4±0.1, let c=212.5±0.13 and let d=211.7±0.16

whats f (including uncertainty) where

f-1 = ((a/b)-1)/((c/d)-1)
Why f-1? Why isn't the relationship specified as f = ...
Or is the -1 supposed to be an exponent? Are the other -1's also exponents?

Homework Equations



add the percentage uncertainties in quadrature if dividing two variables
add the absolute uncertainties in quadrature of adding two variables?


The Attempt at a Solution



from plugging in the numbers f=1.0038
I'm not seeing how you arrive at this number from the initial values and relationship. Can you elaborate, perhaps break down and present the calculation in smaller steps?
 


I stupidly typed in the wrong equation its meant to be:

f-1 = ((c/d)-1)/((a/b)-1)

It just gives it as f-1 =... it could equally be:

f=1+(((c/d)-1)/((a/b)-1)). You don't know where the uncertainties come from either? i just added it in quadrature eg

does that make more sense? Do you understand where the uncertainties comes from?

thanks
 


Okay, with your correction I can now see what's what.

Your method looks okay to me, and the result appears to be reasonable. You might want to keep a couple more decimal places in intermediate results and round the final value only at the end.

As a check I calculated the uncertainty using another method (partial differentials) and arrived at a value very close to yours.
 


great, thanks! I have a difficulty with uncertainties. I have another general question, if you don't mind - I am trying to give an estimate of the half value layer of xrays going through aluminium. From a graph I have determined the half value layer to be 3.03mm

Im now trying to give an estimate of the error. Here are the sources of uncertainty i have come up with:

Repeatability of output - 1%
Influence of scatter - 1%
electrometer - response 0.5\%
scale reading 0.01
standard deviation - calculated as usual

effect on environment - temperature 1.6\%
- pressure neglible
Aluminium thickness - 0.05mm (1.7\%)

can I just add the percentage errors in quadrature and so get an overal uncertainty of 3.17% and thus get an answer of (3.03±0.1)mm??

or do i do it another way. I have spent hours searching the internet but I am just getting really confused!

thanks
 


If the sources of error all contribute to a single measurement and they are truly independent of each other, then yes, add them in quadrature. If your plotted values are the result of calculations on several measured values that are affected individually by these things, then apply the rules for combining uncertainties according to the equation(s) used.
 


so its not the absote uncertanties you add?

Ive just thought -

if I am just plottng the graph with all the error bars and I want to quote the result from the graph - not the equaton i.e. i just look to see where ts half the ntensty and quote the correspondng depth. what's the uncertainty there. a combinaton of the uncertanties described previously or smply the scale reading error on the graph?

because the expermental errors is to do wth measurng the indvidual points, not how accurately i can measure my graph...

thanks
 


Usually the graph points are plotted with error bars and a least-squares type of curve fit is done. Presumably there's curve plotting software that will take the individual point uncertainties into account, but I'm not familiar enough with what's available to make a recommendation.

If you draw the curve by hand then you're 'connecting the dots' by eye, making a smooth curve that passes within the error bars from all the points. This is a type of averaging/estimating that's hard to characterize precisely. I suppose for the uncertainty you could then make an estimate from the size of the error bars in the neighborhood of the point in question.
 
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