What is the formula to calculate absolute uncertainty?

Most numerical values in science are measurements, and every measurement has an uncertainty. Scientists communicate the uncertainty of their measurements to show their level of confidence in their measurements. If the length of a ramp had been measured to be 80 cm, the absolute uncertainty of the measurement would likely be ± 1 cm. For a time measurement taken by a person with a stopwatch, the absolute uncertainty is normally ± 0.2 s. These are examples of absolute uncertainty.

If a series of measurements is being added together or subtracted, the absolute uncertainty of the calculated value is found from the sum of the individual uncertainties on the measurements. For example, consider the addition of the following lengths and their uncertainties.

$\hspace{2em}$(2.0 ± 0.1) cm + (8.2 ± 0.2) cm + (2.5 ± 0.5) cm

The result will be the sum of the lengths, and the absolute uncertainty will be the sum of the individual uncertainties:

$\hspace{2em}$Length = 2.0 + 8.2 + 2.5 = 12.7 cm

$\hspace{2em}$Absolute uncertainty = 0.1 + 0.2 + 0.5 = 0.8 cm

$\hspace{2em}$The final answer is expressed as 12.7 cm ± 0.8 cm.

If two measurements are being multiplied or divided, then the calculation of the absolute uncertainty on the result is more complex. During multiplication or division operations, the relative or fractional uncertainties need to be added together. To do this, the absolute uncertainties on the original measurements must first be converted into fractional uncertainties, and then the fractional uncertainty of the result must be converted back into an absolute uncertainty.

Consider the following calculation of the density of an object. The mass m and volume V of the object are measured to be

$\hspace{2em}m$ = 4.5 g ± 0.2 g

$\hspace{2em}V$ = 2.5 cm$^3$ ± 0.5 cm$^3$

The density can be found from the formula

$\hspace{2em}\rho=\dfrac{m}{V}$

Giving

$\hspace{2em}\rho=\dfrac{4.5\ \ce{g}}{2.5\ \ce{cm}^{3} }=1.8$ g cm$^{-3}$

Now we need to find the absolute uncertainty on the result. First we find the fractional uncertainties on the mass and the volume:

$\hspace{2em}$ $\dfrac{\Delta{m}}{m}=\dfrac{0.2}{4.5} \hspace{3em} \ce{and} \hspace{3em} \dfrac{\Delta V}{V} =\dfrac{0.5}{2.5}$

Because we are dividing the terms, we add the fractional uncertainties

$\hspace{2em}$ $\dfrac{\Delta{m}}{m} + \dfrac{\Delta V}{V} = \dfrac{0.2}{4.5} + \dfrac{0.5}{2.5} = 0.24$

Knowing that

$\hspace{2em}$ $\dfrac{\Delta \rho}{\rho}=0.24$ (this can also be expressed as 24%)

We can now solve for ∆p:

$\hspace{2em}$ $\Delta \rho = (0.24)\rho =(0.24)(1.8)=0.4\ \ce{g cm}^{-3}$

Our final value for density, with its absolute uncertainty, becomes

$\hspace{2em}$ $\rho =1.8\ \ce{g cm}^{-3} ± 0.4\ \ce{g cm}^{-3}$