I'm trying to compute uncertainty for the density of the ball.

I measured its radius 6 times, so I was able to compute the stastistical uncertainty (we call it uncertainty type A, I don't know, if that's common used designation) and I knew accuracy of the micrometer, so I got standatd uncertainty (we call it Type B uncertainty) and made combined uncertainty, which is $4.3 \cdot 10^{-6}~m$.

Then I measured the weight of the ball, but only once and computed standard uncertainty (type B) as $0,29 \cdot 10^{-4}~g$.

Now I have this formula for the density: $\rho = \frac {m}{V} = \frac {m}{\frac 43 \pi r^3}$ and I need to find a way how to express uncertainty of this quantity.

Does anybody know how to do that?


1 Answer 1


Take its differential form :

$$\mathrm{d} \rho = \frac{1}{4/3 \pi r^3} \mathrm{d}m - \frac{m}{4\pi r^4}\mathrm{d}r$$

The greatest variation in $\rho$ will be achieved when all the terms add positively

$$\delta \rho =\frac{1}{4/3 \pi r^3} \delta m + \frac{m}{4\pi r^4}\delta r$$

Factor of $-3$ appears as matter of integration. Which you can recast into

$$\frac{\delta \rho}{\rho} = \frac{\delta m }{m} + 3\cdot\frac{\delta r}{r}$$

This will be accurate if the uncertainties are small enough compared to the value.

Otherwise, you can use the more general formula. For a function $f(\{x_i\})$, where $x_i$ is a quantity you measured with an uncertainty $\Delta x_i$, then

$$\Delta f = \sqrt{\sum \limits_{i=1}^{n} \left(\frac{\partial f}{\partial x_i}\right)^2 (\Delta x_i)^2}$$

  • $\begingroup$ +1 for both methods: Your relative error addition is corect if you multiply a factor of three to the radius term. It's from integration constant of $r^{-3}$ :) Brilliant error calculus. $\endgroup$ Commented May 17, 2013 at 14:30

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