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error uncertainty of equation

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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bestnagi
Asked:
bestnagi
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1 Solution
 
ozoCommented:
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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bestnagiAuthor Commented:
Yes they are independently distributed
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ozoCommented:
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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bestnagiAuthor Commented:
No its usually derived by taking the log of each side and then the derivative
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GwynforWebCommented:
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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ozoCommented:
By individual uncertainties, I was referring to the individual contributions to dA, which can be derived from the derivative process
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GwynforWebCommented:
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/(TĀ²cosB ) + dB.D.tan(B)/(T.cosB)
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bestnagiAuthor Commented:
Thanks. The full equation is
A=[D/(T.cosB)].[(C1/C2)-1]
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GwynforWebCommented:
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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GwynforWebCommented:
A 'B'? Which part of it did I do wrong?
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