Suppose the volume of a cylinder with diameter $d=$11.92 ± 0.01 mm and height $h=$38.06 ± 0.02 mm. Calculating $\frac{\pi d^2h}{4}$, the volume is 4247.282773 mm^3, without rounding off. Now, I read that "The rule in multiplication and division is that the final answer should have the same number of significant figures as there are in the number with the fewest significant figures." As I understand, the volume should be reported as 4247 mm^3.

However, one colleague insists that the result should be written as 4247.28 mm^3 because otherwise we are losing information from the measurement instruments. Is this another thing to take into account when rounding values from indirect measurements?

The quote is from: https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.04%3A_Significant_Figures_in_Calculations

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    $\begingroup$ More on significant figures. $\endgroup$
    – Qmechanic
    Apr 12 at 6:28
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    $\begingroup$ Please state references when quoting. Where did you read that from? Is it an reliable source or some article floating around on the web? $\endgroup$
    – ananta
    Apr 12 at 6:40
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    $\begingroup$ Significant figures is a crude, non-rigourous method taught to students because it is simple. It's a "better than nothing" method. When it actually matters you should be using error/uncertainty propagation which depends on the operations and equations being used: geol.lsu.edu/jlorenzo/geophysics/uncertainties/… $\endgroup$
    – DKNguyen
    Apr 12 at 14:00

5 Answers 5


Perhaps a way forward is to estimate the error in the volume?

The fractional error in $d$ is $1/1192$ and in $d^2$ is $2/1192$, and the fractional error in $h$ is $2/3806$.

An estimate of the fractional error in the volume is $\sqrt{(2/1192)^2+(2/3806)^2}\approx 0.00176$.

Thus an estimate of the error in the volume is $0.00176 \times 4247.48 \approx 7.47$

So the volume is $\bf 4247\pm 7 \,\rm mm^3$.

  • $\begingroup$ Upvoted because 7.47 is the correct error figure, but the last number should read 4247 rather than 7247. $\endgroup$
    – CR Drost
    Apr 12 at 8:29
  • $\begingroup$ @CRDrost Thanks - correction made. $\endgroup$
    – Farcher
    Apr 12 at 9:44
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    $\begingroup$ In other words, significant figures aren't nearly as accurate or useful as proper error analysis - keeping track of error bars on every number and adding errors in quadrature. $\endgroup$
    – AXensen
    Apr 12 at 13:37
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    $\begingroup$ True, but in this case it's worth noting the significant figures correctly identified the order of magnitude of the uncertainty (i.e. uncertainty in the "ones" decimal place), and that it would be meaningless to report numbers to the right of the decimal. $\endgroup$
    – RC_23
    Apr 12 at 15:59
  • $\begingroup$ I see that estimating the error is more useful, as it allows to state the order of magnitude of uncertainty as RC_23 has commented. Thanks! $\endgroup$ Apr 13 at 12:23

I think 4247 mm^3 is right. Even if you rounded to 4247.28277 and only took the last digit off, that would technically be losing a tiny bit of information on what the volume most likely is. However, the .28277 or just .28 (or .2) is not known to be exact, so it isn't reported. In some cases, like if you are trying to find the most likely exact value or something (idrk), you might not want to round, but it seems like in this case reporting would follow the rules for significant figures.

  1. The uncertainties in your example are given with one significant digit only. In some situations or communities, two or even three significant digits are preferred. But since here we have only one, there's nothing we can do about it. In fact, here we deal with the worst possible case, as "0.01" can represent a number from the range of 0.005 to 0.015, so the relative uncertainty of this uncertainty is very large.

  2. The relative uncertainty of the volume is 7.46... mm³. How many significant digits for the uncertainty are needed? As we started with one, anything beyond one is a distraction. You may want to calculate the uncertainty of the uncertainty of the volume, if you want to understand this reasoning. The rule is to report the value with the same number of significant digits as the uncertainty, and then indeed it's correct to the result as 4247±7 mm³. Maybe one could even argue that given high uncertainty of uncertainty (4.25±0.01)·10³ mm³ is more fair.

  3. But also indeed any rounding introduces some error, so in your internal calculations avoid rounding, if reasonably possible. These days, it's easy, as we use computers to do all the maths. Round only the final result you report.


According to measurement absolute errors propagation in any calculation : $$ \Delta_f = \sum_n \left| \frac{\partial f(x_1,x_2,\ldots, x_n)} {~\partial x_n} \right| \cdot \Delta {x_n} \tag 1 $$

You should compute such compound, because you have 2 measurement variables $d,h$ :

$$ \Delta_{volume} = \left| \frac{\partial}{\partial d} \left( \frac{\pi d^2h}{4} \right) \right| \cdot \Delta d + \left| \frac{\partial}{\partial h} \left( \frac{\pi d^2h}{4} \right) \right| \cdot \Delta h \tag 2$$

Evaluating (2) gives :

$$ \Delta_{volume} = \frac {\pi d h}{2}\cdot \Delta d + \frac {\pi d^2}{4}\cdot \Delta h \tag 3$$

If I correctly substituted your given measurement data into (3) :

$$ \begin {align} \Delta_{volume} = (0.5\cdot \pi \cdot 11.92mm \cdot 38.06mm \cdot 0.01mm)&+\\(0.25 \cdot \pi \cdot (11.92mm)^2 \cdot 0.02mm) \\ \approx \boxed {\boxed{ 9.4~\text{mm}^3}} \tag 4 \end {align} $$

So answer reported should be :

$$ V = 4247mm^3 \pm 9 mm^3 \tag 5$$

Or even better $\pm 10mm^3$, even when 0.4 fraction does not pass rounding rule requirement,- in that way you'll be more safe in guarantees.


You can figure this out yourself, even without knowing much about error propagation, in a naive way.

d = 11.92 ± 0.01 indicates that d can be as low as 11.91 or as high as 11.93. Likewise for h, it could be 38.04 or 38.06 or anything in between.

Then you calculate the minimal possible volume using the low figures (11.91, 38.04), and the maximal possible volume using the high figures. You will get 4237.93 and 4256.65 respectively. Our estimate is that the 'true' volume is in this 18 mm^3 wide range.

This result clearly shows that there is little use talking about digits after the decimal point.

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    $\begingroup$ Caveat lector: for uncorrelated uncertainties which are drawn from a normal distribution, this method overestimates the final uncertainty. $\endgroup$
    – rob
    Apr 12 at 13:42

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