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error uncertainty of equation

Posted on 2013-06-05
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Last Modified: 2013-06-09
What is the uncertainty dA of this equation:

A=D/(T.cosB).(C1/C2-1)

assuming each of the variables D, T, B, C1, C2 have a propagation error dD, dT, dB, dC1, dC2
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Question by:bestnagi
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10 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 39224202
Assuming dD, dT, dB, dC1, dC2  are infinitesimal, you can differentiate A with respect to each of  D, T, B, C1, C2, but we'd still need to know whether dD, dT, dB, dC1, dC2 are independently distributed.
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Author Comment

by:bestnagi
ID: 39224214
Yes they are independently distributed
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LVL 84

Expert Comment

by:ozo
ID: 39224329
in that case, the net uncertainty would be the square root of the sum of the squares of the individual uncertainties.
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Author Comment

by:bestnagi
ID: 39224365
No its usually derived by taking the log of each side and then the derivative
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LVL 31

Expert Comment

by:GwynforWeb
ID: 39224499
Is the term ( C1/C2-1 )  

( (C1/C2) -1 )   or  ( C1/(C2 -1) )

Also is the term in the denominator or numerator? The parentheses of the whole equation are ambiguous..

By taking logs of both sides, expanding out and then differentiating gives the result fairly quickly. The A term substitutes out and the log terms go with the differentiation. I won't do the differentiation until I am sure what the equation is, as it is lengthy.
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LVL 84

Expert Comment

by:ozo
ID: 39224612
By individual uncertainties, I was referring to the individual contributions to dA, which can be derived from the derivative process
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LVL 31

Expert Comment

by:GwynforWeb
ID: 39225751
I will illustrate the process for A=D/(T.cosB), as I am not sure of the parentheses for the rest of the equation. This clearly extends to the full equation.

A=D/(T.cosB)

ln A = ln( D/(T.cosB) )

ln A = ln (D) -ln(T) -ln(cosB)

Now perturbing D, T and B by dD, dT, dB; through differentiation we get

  dA/A = dD/D - dT/T + dB.tan(B)

      dA =A( dD/D - dT/T + dB.tan(B) )

This may be the form you need, however you can continue further and substitute A = D/(T.cosB) to get

  dA = dD/(T.cosB)  - dT. D/(T²cosB ) + dB.D.tan(B)/(T.cosB)
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Author Comment

by:bestnagi
ID: 39226435
Thanks. The full equation is
A=[D/(T.cosB)].[(C1/C2)-1]
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Accepted Solution

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GwynforWeb earned 300 total points
ID: 39227368
A=(D/(T.cosB))((C1/C2)-1)

   =(D/(T.cosB))((C1-C2)/C2)

ln A = ln( (D/(T.cosB))((C1-C2)/C2))

ln A = ln (D) - ln(T) - ln(cosB) + ln(C1-C2) - ln(C2)

Now perturbing D, T, B, C1 and C2 by dD, dT, dB, dC1, dC2; through differentiation we get

  dA/A = dD/D - dT/T + dB.tan(B) + dC1/(C1-C2) -  dC2/(C1-C2) -  dC2/C2

      dA =A( dD/D - dT/T + dB.tan(B) + dC1/(C1-C2) -  dC2/(C1-C2) -  dC2/C2 )

This may be the form you need, however you can continue further and substitute

  A = (D/(T.cosB))((C1-C2)/C2) .

I am sure you can do the rest if required
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LVL 31

Expert Comment

by:GwynforWeb
ID: 39230828
A 'B'? Which part of it did I do wrong?
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