Error propagation question

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?
• what are $x_i$ and $y_i$?
– JEB
Commented Nov 25, 2022 at 20:21
• corresponding abs_start_error and abs_end_error
– Andi
Commented Nov 25, 2022 at 22:13
• Note that $+15^{\circ} - (-20^{\circ}) = 35^{\circ} \ne 45^{\circ}$
– JEB
Commented Nov 25, 2022 at 23:12
• Sorry, stupid mistake. Fixed it.
– Andi
Commented Nov 26, 2022 at 16:21

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:

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