Multiplying Uncertainties in Different Units

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How do you multiply quantities with their uncertainties when the units are different?
I could not find any clear explanation on multiplying quantities with different units while including their uncertainties. For example, how would you compute the following product with their uncertainties? 3.4 Newtons +/- .12 Newtons x 1.7 seconds +/- .23 seconds
 
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You treat it the same as any other uncertainty. You have the formula ##f(F,t)=F \ t## so according to the standard propagation of uncertainty $$\sigma^2_f=\left( \frac{\partial f}{\partial F} \right)^2 \sigma^2_F + \left( \frac{\partial f}{\partial t} \right)^2 \sigma^2_t + 2 \frac{\partial f}{\partial F} \frac{\partial f}{\partial t} \sigma_{Ft}$$

Notice that the units work out naturally.
 
Dale said:
You treat it the same as any other uncertainty. You have the formula ##f(F,t)=F \ t## so according to the standard propagation of uncertainty $$\sigma^2_f=\left( \frac{\partial f}{\partial F} \right)^2 \sigma^2_F + \left( \frac{\partial f}{\partial t} \right)^2 \sigma^2_t + 2 \frac{\partial f}{\partial F} \frac{\partial f}{\partial t} \sigma_{Ft}$$

Notice that the units work out naturally.
Sorry, I don't understand your reply. A little over my head. Could you give me a concrete example?
 
e2m2a said:
Sorry, I don't understand your reply. A little over my head. Could you give me a concrete example?
In your case, if the uncertainty in ##F## is un correlated with the uncertainty in ##t## then ##\sigma_{Ft}=0## so the last term drops out. Then $$\frac{\partial f}{\partial F}=t$$ and $$\frac{\partial f}{\partial t}=F$$ and you already know ##F = 3.4##, ##\sigma_F = 0.12##, ##t = 1.7##, and ##\sigma_t= 0.23##. So just plug in and calculate.
 
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e2m2a said:
Sorry, I don't understand your reply. A little over my head. Could you give me a concrete example?
You are trying to calculate a quantity called impulse, ##I##, which satisfies ##I=Ft##. The standard error in ##I## is what you are looking for, and is ##\sigma_I## (@Dale called this ##\sigma_f##). This relates to the standard errors in ##F## and ##t##, ##\sigma_F## and ##\sigma_t## respectively, and their covariance, ##\sigma_{Ft}##, through the formula Dale gave.

You gave us ##\sigma_F## and ##\sigma_t##. Do you know how to calculate the partial differentials? If not, it's really easy - ##\frac{\partial I}{\partial F}## is the derivative of ##I## with respect to ##F## when everything else is treated as a constant. Finally, do you think your measurement error in ##F## depends on your measurement error in ##t##? If yes, you need to measure the covariance. If not (which I would think is the case) then ##\sigma_{Ft}=0## and you can ignore the last term.

Plug in the numbers - you'll find that each term has units of ##(\mathrm{Ns})^2##, so the units work out.
 
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