2
$\begingroup$

We research the aerodynamics of the vehicle. Usually, there exists turbulence, which will cause the time-varying measurement of the force. The unsteady flow field determines we could not get a definite result. Often, we get a rough result. However, this time I calculate the force with CFD (computational fluid dynamics).

However, due to the numerical error, there are some apparent flaws in the results. For example:

  1. The force coefficient is $0.00076$. However, in the physical world, we usually have two digits after the decimal point. Should I change $0.00076$ to $0$?
  2. One of the net forces should be $0$ because two forces $F_1$ and $F_2$ of the same size but opposite direction will cancel each out. However, because of numerical error again, $F_1 = 0.28$ and $F_2 = -0.27$ (after I keep only two digits after decimal point), which results in a net force of $0.01$. Should I leave it as $0.01$, or change it into $0$?

The numerical method produces too precise result, and they are too good to be true. In practice, it is impossible to get such a precise result. I have compared one of the results with the experimental results and found that there is 10% error between them.

$\endgroup$
6
  • 6
    $\begingroup$ "in the physical world, we usually have two digits after the decimal point" Say what? $\endgroup$
    – Tobias Kildetoft
    Jul 31, 2018 at 9:27
  • 1
    $\begingroup$ No, we don’t - we measure forces in the range they are applied and if that requires 6dp, then we use 6dp... $\endgroup$
    – Solar Mike
    Jul 31, 2018 at 10:22
  • 6
    $\begingroup$ Why would you only keep two decimals? That is the sort of thing you do if that is the precision you know you have, not some sort of universal rule. $\endgroup$
    – Tobias Kildetoft
    Jul 31, 2018 at 10:22
  • 1
    $\begingroup$ The easy solution to keeping two digits after the decimal point would be to make it 0.76 x 10^-3. Why round off calculation errors instead of managing them as one would manage uncertainties in physical measurements? $\endgroup$
    – Mick
    Jul 31, 2018 at 10:34
  • 1
    $\begingroup$ I guess he just wants to say than in this kind of experiment, numbers after the second decimal are meaningless because to small compared to other forces. However the 0.00076 may very well be "real". What makes you think it is a numerical error ? Concerning the equal forces, if you now that by symmetry they cancel then forget about the 0.01. However, if you present/ publish those results I'd suggest to mention those issues, i.e. by saying that you have non-zero forces but that in one case it is too small to be pertinent, and in the other case that it is non-zero due to numerical error. $\endgroup$
    – EigenDavid
    Jul 31, 2018 at 13:49

1 Answer 1

3
$\begingroup$

If you have theoretical expectations for some observables of simulations, I see two general ways of dealing with them:

  1. You exploit them to get more accurate results, e.g., you make your algorithm use the fact that $F_1=F_2$. In this case, the value of your observables is not interesting since it they match your theoretical result by construction.

  2. You use them as a sanity check or measure of simulation accuracy. In this case, there are two options for a publication:

    • You decide that the sanity check went well and is not of sufficient interest to be reported explicitly, i.e., it’s just you doing your homework.

    • You decide that the sanity is worthy to be reported. In that case you have to report how large your deviation from zero is. Depending on your kind of simulation, units, etc., 0.00076 could be large or small. That’s for you to decide.

This is not so different from experimental results where you would theoretical knowledge (that you are not seeking to confirm) either to increase your accuracy, to obtain a measurement at all, as a sanity check, or to gauge your accuracy.

Also note that 0, without any further specifications, is never a good way to present an empirical result¹, be it from experiment or simulations. Something can be:

  • zero within the measurement accuracy,
  • numerically zero,
  • 0.00076 ± 0.01 or better 0.00 ± 0.01,
  • complying with the null hypothesis of an expected value of zero,
  • …,

but not just 0.


¹ that is a real number, i.e., not the result of counting

$\endgroup$
1
  • $\begingroup$ It's a great answer. I'm more confident with my aritcle. Thanks. $\endgroup$
    – PureLine
    Aug 2, 2018 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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