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# Calculate uncertainty

Hello Experts:

I need some assistance with applying a formula/equation (based on calculating "uncertainty".

As the saying goes "A picture is worth a thousand words", I've attached a spreadsheet that contains the following:

1. Calculation of sound in sea water (this is complete)
2. Reference to website that includes equations for "uncertainty"

Based on step #2, I need to calculate the value for uncertainty based on a) multiple inputs and b) single input.

Given the listed equations (on NIST's website), how to I calculate a) and b) in MS-Excel?

Thank you in advance,
EEH
Calculate-uncertainty.xlsx
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ExpExchHelp
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1 Solution

Commented:
Hi EEH

Not really my field, but it seems clear to me that u(xN) is the uncertainty in the measurement of xN, not the measurement itself.

For example, if your temperature estimate is (2 ± 0.5) °C, then u(T) is 0.5.

It seems that you multiply these by the squares of the constants in the original formula, so the first term a1^2*u(x1)^2 is:

4.591^2 * 0.5^2

I'm not sure what happens when the estimated value is raised to a power, as in the terms like (5.304*10^-2)*(T^2)

It's even trickier when they are exponentiated and multiplied:
(7.139*10^-13)*T*(D^3)

All the best,
Graham
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Author Commented:
Graham:

How did you arrive at 4.591?

=2.05^2 = 4.2025... multiplying it by 0.5^2 doesn't give me the value either.

Would you be ok to send me your version of the spreadsheet w/ these calculations?

Thanks,
EEH
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Commented:
Hi EEH (what's your real name, by the way?  :)

Thanks for awarding the points.  I'm puzzled because I posted a followup answer to your question 24 hours ago and it seems to have disappeared.

The 4.591 is the coefficient of the first non-constant term in the Mackenzie formula:

1448.96 + 4.591T - ...

As I understand the article on combined uncertainty, it is therefore the a1 in the term:
a1^2*u(x1)^2
and u(x1) is the uncertainty of T.

I don't have my own version of the spreadsheet, sorry.  I just took a peek at your copy.

All the best,
Graham
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Author Commented:
Graham:

Thanks... appreciate the additional feedback.    ;)

Cheers,
Tom
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