I am an environmental scientist and it's been too long since I've taken a math class. In my research I've had to use this equation for gas loss constants, but I am going back to assess my uncertainties. I did all the easy uncertainties in my calculations, but this formula has me stumped. I am not sure how to apply error propagation to a variable raised to a variable. It comes up in this equation:

$$k= 0.45W^{1.6\left(\frac S{600}\right)^{-0.5}},$$

(I gave actual numbers from one row of my data for ease of use) where W is $5.376 \pm 17.383$ and S is $796.825\pm 258.418$

K is a gas loss constant, S is the Schmidt number and W is the wind velocity in m/s.

I have run into the formulas (simplified ones) for a variable raised to a power as in: for $a^n=Q$, where $\text{d}a$ and $\text{d}Q$ represent the error of $a$ and the error of $Q$,

$$dQ= |Q||n|\bigg(\frac{\text{d}a}{a}\bigg)$$

and for a constant raised to a variable as in: for $e^a=Q$, where e is a constant, a is a variable with an known error (d$a$) and $Q$ is the resultant number with its own resultant error (d$Q$),


I don't think either of those apply in my case as the base nor the exponent are constants. It has been way to long since I took Calculus and I need some help.

Please if I have my formula's for error propagation wrong let me know and more importantly if you know how to propagate these errors please show me how.

Thank you anyone who takes the time to read this.


1 Answer 1


We have $$k=0.45W^{\frac {16\sqrt 6}{\sqrt S}}$$ Taking logarithm, we have: $$\ln k=\ln(0.45)+\frac {16\sqrt 6}{\sqrt S} \ln W$$ Taking differentials, $$\frac {dk}{k}={\frac {16\sqrt 6}{\sqrt S}}\frac {dW}{W}+\ln W{\frac {8\sqrt 6}{S^{\frac 32}}}dS$$ Now you simply have to plug in values. Notice here that the second differential would have had been negative, but it was still added because we are always interested in finding out the maximum possible error.

Note: Your measurements are not consistent. You have evaluated, for example, wind velocity upto $4$ significant figures, so your least count must have been $10^{-3} \frac ms$, so your error could not have been more than this. This is similar to measuring the length of a rod with a meter scale with a least count of $1 mm$, in such a case you can't have an error more than $1 mm$.

  • $\begingroup$ Probably the two terms should add in quadrature. $\endgroup$
    – rob
    Commented May 17, 2021 at 20:53
  • $\begingroup$ Im not so sure about that...thus is the way Ive seen it being done in books and lectures. $\endgroup$ Commented May 17, 2021 at 20:58
  • $\begingroup$ Besides, I think this makes more sense, because this follows from the product rule of differentiation, hence for small errors it's an accurate estimation of propagation. $\endgroup$ Commented May 17, 2021 at 21:22
  • $\begingroup$ For independent uncertainties in $W$ and in $S$, the errors have some probability of going in opposite directions, so the linear sum will in general overestimate the error. You have already noted that a linear difference will underestimate the error and changed the sign of the second term. This derivation of the sample variance shows how the sum-in-quadrature arises when you’re computing the sample variance, including an additional term that appears if errors in $W$ and $S$ are correlated with each other. $\endgroup$
    – rob
    Commented May 17, 2021 at 23:17

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