I need to find the total margin of error for calculating velocity, while I have margins of error for time and distance. Actually the margins are the same (as both measurements were based on GPS - but this is not important here), and are given as 1/365000 (20 cm for 730 km).

So, I spent quite some time studying various information, most of which was about standard deviation (which I understand vaguely) and found here (but not only) that the formula I should use is:

if $S=A×B$ or $A/B$ then $σ_S/S=\sqrt{(σ_A/A)^2+(σ_B/B)^2}$

Well, let's take an easy example. Say I have two measurements: distance of 40 m and time of 4 sec; both margins are the same and equal 0.2 (huuuuge, I know). The above formula (if I am interpreting it correctly) would give me the margin of $σ_S/S=\sqrt{(0.2)^2+(0.2)^2}=\sqrt{0.04+0.04}= 0.2828$ (for the calculated velocity of 10 m/s). Therefore the lowest actual velocity can be $v=7,172 m/s$.

Now, let's try to calculate the maximum error from actual numbers for distance and time. For the distance I can have measured up so the actual distance might even be as low as 33,33333 m, while for the time I can have measured down, which gives me the maximum possible time of 5 s. The real velocity would have been then $v=33,3333m/5s=6,66667 m/s$, which means I was wrong by 0,3333.

Obviously, the above calculation shows the theoretical equation underestimated the margin of error, as it said the error cannot exceed 0.2828.

On the other hand, I found elsewhere, yet without any explanation (but from a credible source) that in such case I should have calculated the total margin of error as simply a square root of 0,2 (or 1/365000 in my original problem). In such case the total margin of error equals 0.4472, which - although much higher than what I calculated in my example - is not underestimated at least.

What do I do wrong, and - if the error of margin in my simple example really is 0.4472 (i.e. square root of a margin of error for the distance or time) than - why do I calculate it this way?

  • $\begingroup$ The formula you cite is derived on the assumption that the error is very small in comparison to the value. No wonder your large errors give a bad approximation $\endgroup$
    – vonbrand
    May 10, 2014 at 12:36
  • $\begingroup$ OK, thank you. Now, is there any broader formula? One that is not confined to relatively small errors? And how about the square root of the margin error (identical for both variables) formula? Any idea where this comes from and how it can be explained? $\endgroup$ May 10, 2014 at 13:22
  • $\begingroup$ Sadly, any formula that doesn't rely on using a linear approximation (i.e., very small errors) will be horribly complex. Easier to just do as you did, get a interval where the result lies by computing lowest/hghest value of the expression. $\endgroup$
    – vonbrand
    May 10, 2014 at 13:29

1 Answer 1


What you're doing wrong is assuming too much about the error distribution.

There are two common ways to look at the error in a measurement: normally distributed and absolute bounds. Normal (or Gaussian) distributions have a bell shape, with a 66% chance of the actual value being within one standard deviation of your measured value and a 95% chance of the actual value being within two standard deviations. With absolute bounds, you assume a 100% chance of the actual value being between the measurement bounds, but little more.

Your first formula gives a relative standard deviation of 0.2828, assuming normally distributed data. There's a 34% chance of the actual value being off more than that, under that assumption.

Your second formula establishes a relative maximum error of 0.3333, with a 0% chance of the error being more.

These two statements aren't in direct conflict yet - there could be a 34% chance the error is between 0.2828 and 0.3333. But that's not how it works. The normal distribution also has a 2.5% chance of the speed being lower by at least two standard deviations, 0.5656 (i.e. v=4.344 m/s), but the absolute bounds assumption has the chance of that at 0%.


Note that your example is quite misleading. You first quote a standard deviation of 20% for the speed (so measuring a speed of less than 32 m/s happens in 2.5% of cases), and then state that the minimum you can measure is 33.3 m/s. Those are quite different distributions. If 33.3 m/s is a truly unexpected outcome, you're talking about 5 or 6 standard deviations, not 1.7

[Edit 2]

So, the 0,2 error isn't a standard deviation, and the first formula used is therefore wrong. The result bounds are then :[10 * (1-0,2/1+0.2) , 10 * (1+0,2/1-0,2)] = [6.667, 15]. Note that the bounds are asymmetric.

  • $\begingroup$ I understand a little more now (although I would be obliged if you could give me a link to some recommended source on this type of error calculation). But, perhaps more importantly, do you have any idea, where this square root equation I mentioned at the end of my question comes from? In your physics answer to the quantum-related question you used something similar (at first glance at least) and therefore I though you might be able to unwind it (I will stress it again - it comes from a person I have quite a trust in, but I am not able to consult him now). $\endgroup$ May 27, 2014 at 9:01
  • $\begingroup$ The standard deviation is the square root of the variance, it's that simple. For independent errors, variances add up. As for the formula at the end of your question, I have no idea. It doesn't make sense. To see it's nonsense, consider the case where you have three independent variables as input, not two. E.g. S=A*B/C. $\endgroup$
    – MSalters
    May 27, 2014 at 10:34
  • $\begingroup$ Sure (but perhaps it would be cube root then?), but on the other hand, 2 standard deviations producing 0,5656 seems also odd, if maximum possible error is 0,3333 as I calculated. $\endgroup$ May 27, 2014 at 10:50
  • $\begingroup$ @brightmagus: That's because the 2 SD's are under the assumption of normally distributed errors, and your calculated maximum assumes non-normal hard bounds. $\endgroup$
    – MSalters
    May 27, 2014 at 11:56
  • $\begingroup$ OK, I guess I need to read more about it to understand why "normal" error is different than possible but "non-normal" error. I just don't get it why accepted calculation provides either too low or too large margins of error (i.e. incorrect ones), while I simply want to calculate what maximum error is possible (without considering the probability of extreme values). So is the only way to calculate it the way I did when I arrived at the 0,3333 figure? $\endgroup$ May 27, 2014 at 12:06

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.