WHat is the uncertainty in a metre rule?

[Dickfore]

Ultimately, the proper ± figure for scale-reading uncertainty depends on the person who is making the measurement...

No. Wrong experimental procedure leads to systematic errors that are not a measure of the uncertainty.

...and the instrument that he is using,...

Yes. But:

I feel confident in assigning ±0.1 mm when using a metal scale with finely-engraved lines, in a way that eliminates or minimizes parallax error due to the thickness of the scale.

This is definitely wrong if the "finely-engraved lines" are a distance 1 mm apart, as given in the OP.

 

[K^2]

No. Wrong experimental procedure leads to systematic errors that are not a measure of the uncertainty.

They can lead to either or both. Eye-balling a value, for example, vs using a precise measurement is a technique flaw that introduces a random error, rather than a systematic one.

 

[Dickfore]

Not at all to do with statistics.

But everything to do with technique which contains inherent sources of error v technique which avoids these.

When you place your test piece and ruler against the stop end you have a guaranteed square and reproducible 'zero'.

When you estimate the alignment of two lines along a sight line that may or may not be square and hold the ruler and testpiece at some random (albeit small) angle to each other you have a recipe for variability of measurement. Notice I said 'sight line'. Two operators will align the pieces slightly differently by sight. They cannot do this with a stop end.

Again, if you consider the error due to alignment of zero to give an uncertainty of 0.5 mm (on a ruler with a division of 1 mm), you are overstating the error. This is also a mistake.

Consider the error propagation formula:
<br /> \sigma_{L} = \sqrt{\sigma^{2}_{\mathrm{left}} + \sigma^{2}_{\mathrm{right}}}<br />
Now, by the nature of the measurement, we must have \sigma_{\mathrm{left}} \ll \sigma_{\mathrm{right}}. Then, we may expand:
<br /> \sigma_{L} = \sigma_{\mathrm{right}} \, \left( 1 + \left( \frac{\sigma_{\mathrm{left}}}{\sigma_{\mathrm{right}}} \right)^{2} \right)^{\frac{1}{2}} \approx \sigma_{\mathrm{right}} \, \left[ 1 + \frac{1}{2} \, \left( \frac{\sigma_{\mathrm{left}}}{\sigma_{\mathrm{right}}} \right)^{2} \right]<br />
This is much smaller than \sqrt{2} \, \sigma_{\mathrm{right}} if:
<br /> \frac{\sigma_{\mathrm{left}}}{\sigma_{\mathrm{right}}} \ll \sqrt{2 (\sqrt{2} - 1)} = 0.92<br />
which is certainly the case.

 

[K^2]

I think, I misread the OP. I was reading it as length of object requiring two measurements, because ruler isn't long enough. But for matching zero, of course, the precision is much better than 0.5mm. It's on the order of the width of the tick, rather than order of distance between ticks.

 

[Dickfore]

They can lead to either or both. Eye-balling a value, for example, vs using a precise measurement is a technique flaw that introduces a random error, rather than a systematic one.

Not if you eyeball from the same direction consistently, it isn't.

 

[K^2]

If the actual values are different, you'll always have a different error in eye-balling it.

You can go ahead and do your own experiment on that. You will note a random error. There might also be a systematic one, but random error will dominate.

 

[Dickfore]

Actually, parallax error may be negligible if the scale is next to the measured object.

 

[K^2]

We are talking about different things. Try eye-balling distances/sizes without a scale at all, so that you can use the scale measurement to compare it to. You'll find the error to be mostly random. Though, a bias may be present as well. This is extreme case, of course, but eye-balling distances between ticks works the same way. It's just harder to check yourself.

 

[Dickfore]

Try eye-balling distances/sizes without a scale at all, so that you can use the scale measurement to compare it to.

I'm afraid I don't understand what you're trying to describe.

 

[haruspex]

I understand the temptation to leap into discussing standard deviations, but the OP asked what the uncertainty is. If the two input errors are +/- A and +/- B then the uncertainty in the difference (or sum) is +/-(A+B). Whether it is more appropriate to use that value or one based on s.d. depends on the purpose to which the answer will be put. If my life depends on staying within a specific bound, I'll take the conservative approach. Note that using s.d. then being conservative by demanding 3 s.d.s of tolerance actually produces a larger safety margin than necessary.
Where s.d. is considered appropriate, it's worth thinking about the distribution of the input errors. In this case, they'll follow a uniform distribution over the stated range (perhaps with a little rounding at the edges). The error in the difference will therefore be distributed as a symmetric trapezium ('trapezoid' in US). In the special case of the ranges being equal, like the sum of two dice, this simplifies to an isosceles triangle. The s.d. will be √((A²+B²)/3).