1
$\begingroup$

In a calculation I have received a number of values in the range $1e4-1e6$. Now I would like to display the values with 3 significant figures according to the rules of using significant figures. But this causes some issues when the values vary in magnitude.

For instance, I have the values $34745$ and $211360$. It seems inconsistent to me to render these with 3 significant figures independently of each other, as $34700$ and $211000$, respectively.

On the other hand, choosing either as a baseline, rendering $35000$ and $211000$ or $34700$ and $211400$ seems to break the rules as well.

What is the standard course of action in this situation?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ 34745 to 3 significant digits is 34700 not 34800. ;-) $\endgroup$ – Andrew Jan 13 '17 at 11:51
  • $\begingroup$ Yikes, should I even be an engineer? $\endgroup$ – Toivo Säwén Jan 13 '17 at 13:12
  • 2
    $\begingroup$ Depending on your time zone you could claim it was Friday afternoon. Silly mistakes are allowed on Friday afternoons and Monday mornings. $\endgroup$ – Andrew Jan 13 '17 at 14:03
  • $\begingroup$ why do you think 34700 , and 211000 is inconsistent? Would you think 3.47E4 and 2.11E5 ok ? $\endgroup$ – agentp Jan 13 '17 at 20:01
1
$\begingroup$

The fundamental problem here is that you've got a bunch of numeric values which are not in engineering/scientific notation, so you have to make up some arbitrary guesses as to how many sig figs really are there.
Let's look at one of your values: 34745 . Where did it come from? If it's, say the reading on a digital display for a widget whose operating manual specifies an accuracy of +/-1 in the last digit, then you can use it as $(3.4745 \pm 0.00005)E5 $. But if it's the result of a calculator giving you a ton of digits after dividing two values which were accurate to, say, 1%, then you can't assign an accuracy to the result better than that of the inputs. Skipping the exact error propagation formula, you'd roughly be limited to $(3.47 \pm 0.03)E5$ .

Ultimately, the "rule" of sig figs depends on how you combined the source data to get the calculated result.

Improve this answer
$\endgroup$
0
$\begingroup$

It depends on what these different values are. If these different values are for different entities, then I would have each to 3 significant figures based on itself.

i.e. For the equation d = FL/EA, working in kN and m:

  • F = 5412, I would report as 5410
  • L = 6.911, I would report as 6.91
  • E = 210265, I would report as 210000
  • A = 0.043211, I would report as 0.0432

However, if I have a lot of deflections at different positions, then the relative deflection is important, and I would report all values based on which values are significant for the biggest deflection:

  • If Nodes 1-6 have deflections (in mm) of 0.000114, 0.256, 1.54, 17.9, 9.5211 and 0.0187, I would report them as: 0.0, 0.3, 1.5, 17.9, 9.5, 0.0

In this case, the accuracy of 0.000114mm for node 1 is unimportant: when node 4 moved 17.9mm, node 1 effectively didn't move at all.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Effectively choosing the highest value as the baseline for significant figure rendering. This definitely makes sense to me. $\endgroup$ – Toivo Säwén Jan 13 '17 at 13:24
  • $\begingroup$ note 210000 appears to have only two sig digits. Formally you might put a bar under the first zero or something to indicate it is significant. $\endgroup$ – agentp Jan 13 '17 at 20:03
  • $\begingroup$ "Significant figures" are a very crude way to express error estimates. It doesn't usually make any physical sense to report values on a scale that distinguishes 9.97, 9.98, 9.99, 10.0, but then suddenly jumps to 10.1, 10.2, etc instead of 10.01, 10.02, etc. Using "best estimate $\pm$ error" is often a better way. $\endgroup$ – alephzero Jan 14 '17 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.