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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?
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  • $\begingroup$ 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

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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:

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

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