Rounding uncertainties and measurements

[eddywalrus]

Hello,

I have some questions on uncertainties in measurement:

Say you have a ruler with 1mm graduations, and you are trying to measure the length of a metal rod. One end of the metal rod is at the 120.0mm graduation, and the other is in between the 19 and 20mm graduation, and you approximate its value to be 19.5mm. The uncertainty in these two measurements is half the smallest scale division, and therefore +-0.5mm.

The length of the metal rod = 120.00mm - 19.5mm = 100.5mm
The uncertainty in the length = 0.5mm + 0.5mm = 1mm

However, since the uncertainty in the length is 1mm, would we have to round the length of the metal rod up to 101mm since stating the length of the metal rod as 100.5mm would create false precision? (My textbook states that the length would remain unrounded as 100.5mm +- 1.00mm but I fail to see the reasoning behind it.)

If it does create false precision, then my second question would be:
Would uncertainties such as +- 0.14m, or +-0.32m, or +-0.54m, be contradictory in nature, since they are creating false precision? For example, wouldn't an uncertainty of +-0.14m be contradictory in nature, since it implies a precision of 0.01m, when the measurement, as stated by the uncertainty, is only precise to +-0.14m? Shouldn't an uncertainty of +-0.14m always be rounded upwards to +-0.2m, and all uncertainties in general be rounded to 1sf to avoid false precision?

I know these questions are quite rudimentary but I can't seem to figure them out -- thank you for all your help!

 

[zoki85]

Probably you should take measurement uncertainity ± 0.5 mm at both ends of metal rod.
- one end of the rod is at 120 ± 0.5 mm (min 119.5 mm, max 120.5 mm)
- the other end of the rod is at 19.5 ± 0.5 mm (min 19 mm, max 20 mm)

Therefore, length of the rod is somewhere between 119.5-20 = 99.5 mm and 120.5-19= 101.5 mm.
IOW, length of the rod is 100.5±1 mm.

 

[moriheru]

Hello,

I have some questions on uncertainties in measurement:

Say you have a ruler with 1mm graduations, and you are trying to measure the length of a metal rod. One end of the metal rod is at the 120.0mm graduation, and the other is in between the 19 and 20mm graduation, and you approximate its value to be 19.5mm. The uncertainty in these two measurements is half the smallest scale division, and therefore +-0.5mm.

I don't quite understand what you mean with the one end is at 120.0mm and the other is between 19mm and 20mm. If you could clarify this maybe I could help more,but as far as I have understood it I believe you are saying that the metall rod has two uncertantys that sum to 1mm. I think the metall rod has a total uncertainty of 0.5mm and the sum would thus be 0.25+0.25...?!

 

[eddywalrus]

I don't quite understand what you mean with the one end is at 120.0mm and the other is between 19mm and 20mm. If you could clarify this maybe I could help more,but as far as I have understood it I believe you are saying that the metall rod has two uncertantys that sum to 1mm. I think the metall rod has a total uncertainty of 0.5mm and the sum would thus be 0.25+0.25...?!

Hi, I've attached an image here which will hopefully clarify what I mean by how one end of the metal rod is at the 120mm graduation and the other is between 19mm and 20mm:

In order to calculate the length of the metal rod, we have to approximate/estimate where the left end of the metal rod is positioned and where the right end of the metal rod is positioned. The left end of the rod is between 19mm and 20mm, so we approximate it to be 19.5mm, and the right end of the rod is on the 120mm graduation, so we approximate it to be 120mm. Since the smallest scale division is 1mm, the uncertainty is half that, and therefore 0.5mm.

When we subtract one measurement from another (i.e. when we subtract 19.5mm from 120mm) we also have to add the uncertainties, so we add 0.5mm to 0.5mm to get a 1mm uncertainty.

Does this clarify things a bit?

 

[eddywalrus]

Probably you should take measurement uncertainity ± 0.5 mm at both ends of metal rod.
- one end of the rod is at 120 ± 0.5 mm (min 119.5 mm, max 120.5 mm)
- the other end of the rod is at 19.5 ± 0.5 mm (min 19 mm, max 20 mm)

Therefore, length of the rod is somewhere between 119.5-20 = 99.5 mm and 120.5-19= 101.5 mm.
IOW, length of the rod is 100.5±1 mm.

Firstly, thank you for your reply! I just wanted to ask -- wouldn't we have to round the 100.5mm to 101mm, since the uncertainty implies that the measurement is only accurate/precise to the nearest mm -- wouldn't giving the measurement to 1 decimal place create false precision/ suggest that the measurement is precise to 1 decimal place when it is not?

 

[jtbell]

The uncertainty in these two measurements is half the smallest scale division, and therefore +-0.5mm.

With a decent quality analog measuring scale, I always try to estimate to the nearest tenth of a division, which is 0.1 mm in this case. So in your case, the two ends of the rod would be measured as 19.5 ± 0.1 mm, and 120.0 ± 0.1 mm. Taking the difference and using your simple "add the uncertainties" rule gives a length of 100.5 ± 0.2 mm.

The "add the uncertainties" rule actually overestimates the total uncertainty a bit. A more correct method is to add the squares, and then take the square root: $\sqrt{0.1^2 + 0.1^2}$ = 0.14 mm. This is because the two uncertainties each act randomly, and independently of each other, so they can either reinforce or (partially) cancel each other out. But in practice, it's usually better to overestimate the uncertainty than to underestimate it, so the "add the uncertainties" rule is usually OK for a quick-and-dirty estimate.

wouldn't giving the measurement to 1 decimal place create false precision/ suggest that the measurement is precise to 1 decimal place when it is not?

It's always understood that the last digit is uncertain by some amount. In careful scientific work we usually state the uncertainty explicitly. If it's not stated explicitly, we usually assume ±1 in the last digit, for an analog scale like this one. With a digital scale like the one on most modern electronic meters, or a micrometer with a digital readout, I use ±0.5 of the last digit.

 

[eddywalrus]

Jtbell, first of all, thank you for clarifying things up for me! But I'm not still quite sure what you mean by

It's always understood that the last digit is uncertain by some amount. In careful scientific work we usually state the uncertainty explicitly.

So in the case I mentioned in the original post, would we still state the measurement as 100.5mm instead of 101mm? But stating the uncertainty explicitly doesn't negate the fact that stating the value as 100.5mm creates false precision, right? Since it implies the uncertainty is +-0.1mm or +-0.5mm when it is really +-1mm?

Thank you for your help!

 

[jtbell]

First, note that different textbooks and different instructors do things differently, because these "uncertainty and rounding rules" that rely only on the number of decimal places and significant figures are rather crude. In actual scientific work, we use more precise methods.

Using your original uncertainties, 19.5 ± 0.5 mm and 120.0 ± 0.5 mm, and the simple uncertainty-addition rule, I would calculate the length first as 100.5 ± 1.0 mm. I would then say the uncertainty indicates that the "ones" digit is uncertain, and round it off as 100 ± 1 mm. I use 100 not 101, because the usual rule for rounding when the last digit is 5 is to round the preceding digit to the nearest even number.

So you had the right idea (from my point of view), except that you rounded in the wrong direction. In general, in a final result, I always round the number and its uncertainty to a matching decimal position.

Note that using my "better" rule for adding uncertainties, I get 100.5 ± 0.7 mm. Some instructors don't use the "better" rule in introductory courses, though. If yours doesn't, stick to the simpler rule for now.

Finally, just to confuse you further, I sometimes use two significant figures in the uncertainty, not one, and round the result accordingly. In that case your result would be 100.5 ± 1.0 mm. Note that I wrote the uncertainty as 1.0 (two sig figs), not 1 (one sig fig). This is useful if you expect to use the result in further calculations, to reduce roundoff error.

 

[eddywalrus]

First, note that different textbooks and different instructors do things differently, because these "uncertainty and rounding rules" that rely only on the number of decimal places and significant figures are rather crude. In actual scientific work, we use more precise methods.

Using your original uncertainties, 19.5 ± 0.5 mm and 120.0 ± 0.5 mm, and the simple uncertainty-addition rule, I would calculate the length first as 100.5 ± 1.0 mm. I would then say the uncertainty indicates that the "ones" digit is uncertain, and round it off as 100 ± 1 mm. I use 100 not 101, because the usual rule for rounding when the last digit is 5 is to round the preceding digit to the nearest even number.

So you had the right idea (from my point of view), except that you rounded in the wrong direction. In general, in a final result, I always round the number and its uncertainty to a matching decimal position.

Note that using my "better" rule for adding uncertainties, I get 100.5 ± 0.7 mm. Some instructors don't use the "better" rule in introductory courses, though. If yours doesn't, stick to the simpler rule for now.

Finally, just to confuse you further, I sometimes use two significant figures in the uncertainty, not one, and round the result accordingly. In that case your result would be 100.5 ± 1.0 mm. Note that I wrote the uncertainty as 1.0 (two sig figs), not 1 (one sig fig). This is useful if you expect to use the result in further calculations, to reduce roundoff error.

Thank you so much for your help jtbell -- this really cleared things up for me quite a bit! I really do appreciate it!