Lets say I have built a rotating device. Now I want to measure how accurate the rotation is. For this I use a measuring instrument with a resolution of $0.1^\circ$ with an uncertainty of $\pm0.2^\circ$ . I first rotate the device to a starting point ($-20^\circ$) and measure the actual value (val_start). I now let the device rotate, lets say about $35.0^\circ$ to $+15^\circ$ and take a measurement after the rotation (val_end) with my instrument. I repeat this N times. Now I want to estimate the how accurate my device is and how strong it deviates. Here is some exemplary data (N=4):

ID val_start / degree abs_start_error / degree val_end / degree abs_end_error / degree
1 -20.8 -0.8 14.7 -0.3
2 -19.3 0.7 15.1 0.1
3 -20.1 -0.1 15.3 0.3
4 -18.9 1.1 14.9 -0.1

From both absolute errors I can calculate a standard deviation:

$$ \sigma_{start} = \frac{1}{4}\sum x_i^2 = 0.5875^\circ $$ $$ \sigma_{end} = \frac{1}{4}\sum y_i^2 = 0.05^\circ $$

As you can see, $\sigma_{end}$ is smaller than the uncertainty of the measuring device. And now to my questions:

  1. How to proceed correctly? Is the total uncertainty just $\sigma_{start}^{tot} = \sqrt{0.2^2 + 0.5875^2}$ ?
  2. Am I even allowed to use that many decimals?
  • $\begingroup$ what are $x_i$ and $y_i$? $\endgroup$
    – JEB
    Nov 25, 2022 at 20:21
  • $\begingroup$ corresponding abs_start_error and abs_end_error $\endgroup$
    – Andi
    Nov 25, 2022 at 22:13
  • $\begingroup$ Note that $ +15^{\circ} - (-20^{\circ}) = 35^{\circ} \ne 45^{\circ}$ $\endgroup$
    – JEB
    Nov 25, 2022 at 23:12
  • $\begingroup$ Sorry, stupid mistake. Fixed it. $\endgroup$
    – Andi
    Nov 26, 2022 at 16:21

1 Answer 1


The question is confusing because your errors aren't errors. I mean they are, but the language is confusing. One error is a bias, and another error is an uncertainty.

The first error ("abs_start_error") is not a measurement uncertainty, it is in fact a measurement of your instrument's bias. As stated in the problem, the uncertainty of those measurements is $\sigma = 0.2^{\circ}$.

Skipping the error analysis of the start/stop accuracy, we can go straight to the 35 degree rotation, defining the angle $a$ to be "val_end - val_start":

$$ a_i = [35.5^{\circ}, 34.4^{\circ}, 35.4^{\circ}, 33.8^{\circ}]$$

Those have a mean value:

$$ \bar a = 34.8^{\circ} $$

Each end point has a $\sigma = 0.2^{\circ}$ uncertainty, so any rotation measurement has an uncertainty:

$$ \sigma_R = \sqrt{\sigma_{start}^2 + \sigma_{end}^2} = \sqrt 2 \sigma = 0.3^{\circ} $$

Meanwhile, the standard deviation of the $N$ measurements: $a_i$ is:

$$ \sigma_a = 0.71^{\circ}$$

You can assume the rotation angle variance is the sum of your instrument's variance and the measurement variance, that is:

$$ \sigma^2_a = \sigma^2_I + \sigma^2_R $$

so that:

$$ \sigma_I = \sqrt{\sigma_a^2 - \sigma_R^2} = 0.65^{\circ} $$

So that is the precision of your rotation. You asked for accuracy. That is determined by the deviation of $\bar a$ from 35 degrees:

$$ \delta_a = 35^{\circ} - \bar a = -0.22^{\circ} $$

The uncertainty of that measurement is the standard error of the mean:

$$ \sigma_{\bar a} = \frac{\sigma_a}{\sqrt{N}} = 0.50^{\circ} $$

That is, the accuracy is:

$$ \delta_a = -0.22^{\circ} \pm 0.50^{\circ} $$

For completeness, you can also get an uncertainty on the precision using the standard error on the variance:

$$ \sigma_{VAR} = (N-1)[(N-1)\mu_4 - (N-3)\mu_2^2]/N^3\frac{N}{N-1} $$

where the $\mu_n$ are the $n^{\rm th}$ central moments of the distribution. That gives:

$$ \sigma_I = 0.65^{\circ} \pm 0.15^{\circ} $$


Your Answer

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

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