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Suppose, I am taking measurement with a scale whose least deviation is 1mm. the length of something seems to lie between 44mm to 45mm. But my instinct says, it's 44.7mm. So, the question is I should report 44.7+-0.5mm or 44.5+-0.5mm or something else?

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    $\begingroup$ There is a correct answer to this question and an answer that might be marked "Correct" by a grader. These two things are not always exactly the same. Nor should they necessarily be, as the correct answer may depend on the person doing the reading having developed a good sense for these matter while the classroom answer can be given to students still stumbling around trying to develop that sense. $\endgroup$ – dmckee --- ex-moderator kitten Aug 12 '19 at 19:20
  • $\begingroup$ Your question title should be, "how to interpolate when reading a scale?" I would not call a scale an "analog device." We say "analog" when one continuously variable physical quantity (e.g., the current flowing in a wire) represents some other physical quantity (e.g., the depth of liquid in a storage tank.) When you use the markings on a physical scale to assign a number to the length of some physical object, that's a "digital" process. $\endgroup$ – Solomon Slow Aug 12 '19 at 20:13
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    $\begingroup$ @SolomonSlow For good or for ill the use of "analog" to distinguish to instruments that read with a scale or needle and dial or similar from those that present a reading in numerals (which are "digital") is well established. $\endgroup$ – dmckee --- ex-moderator kitten Aug 12 '19 at 20:18
  • $\begingroup$ Thanks. I changed the title $\endgroup$ – Sk Shafayat Aug 12 '19 at 22:19
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There are a few different concepts we need to cover here.

First - the difference between precision and accuracy. Precision tells me the number of digits I can write down: in your case, you might be able to estimate that the number is close to 44.7 and you might even believe your ability to interpolate is such that the answer is 44.7 ± 0.1 mm. Of course, when you are reading an analog scale, you usually need to worry about parallax: depending on where your eye is with respect to the alignment of the measurement with the scale, you may get an error. Old analog meters often addressed this by putting a mirror behind the scale: you had to line up your eye such that the reflection of the needle fell right behind the needle - this is how you would minimize the parallax error.

The second issue is accuracy: when the supplier of your analog device (a ruler, in this case) put markings on the ruler, they did so with a certain tolerance. In a rare case, this tolerance might be noted on the ruler - most of the time, you assume it is "as good as the divisions on the ruler". In other words, when a ruler is marked in mm, you can reasonably assume that a measurement of 44 mm is closer to 44 mm than it is to 45 mm or 43 mm. But the manufacturer usually doesn't claim that the line that is drawn at 44 mm is actually drawn at 44.0 mm

When you try to interpolate, you are in effect assuming that 44 = 44.0 ; while that may sometimes be true (it would have to be independently confirmed), it won't always be. And if you start quoting numbers like 44.7 you are in fact claiming that your measuring instrument has accuracy beyond its markings.

In summary - there are actually multiple sources of error that affect your measurement. There is error aligning the ruler (including parallax), error reading the ruler (especially interpolating between markings), and errors in the ruler itself. Proper error estimation accounts for all these things.

If you cannot fully analyze all these sources of error, it is a useful shortcut to round to the nearest whole marking. When you add digits without knowing the accuracy of your ruler, you may be falsely claiming a precision that is not there.

To clarify: I recommend in your case to quote 45 +- 0.5 mm

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  • $\begingroup$ 45+-0.5 says that there’s a 50% chance that the value is over 45, and a 16% chance of over 45.5. That’s a gross overestimate of the error in reading the scale. $\endgroup$ – Bob Jacobsen Aug 15 '19 at 16:01
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Let's consider the process of interpolation.

Here is a part of a scale where two scale divisions of finite width are shown and you find that your reading instrument indicates that the value is somewhere between $44$ and $45$.

enter image description here

So how do you visually subdivide the interval between the $44$ and $45$ scale divisions?

What you should do is subdivide the interval $b$ but most probably you will subdivide the interval $a$ which will give an erroneous reading.

So I think that the best you can do is decide visually the whether the reading indicated by the instrument is closer to $44(.0)$ (left division) or $44.5$ (midway between divisions) or $45(.0)$ (right division).

Often half a scale division is a good rule of thumb when taking readings.

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For simplicity here, let’s ignore flawed rulers, poor technique (I.e. parallax), etc. The other answers address those.

The Question asks about quoting a central value and error, so let's bypass any discussion of significant figures, whether 45 vs 45.0 signifies something, etc.

In that case, you should report your best estimate of the central value and your best estimate of the error. Only in the significant-figures world would you ever go from "I estimate 44.7" to "But I'll report 45 to signify that that last digit is uncertain".

If you were so uncertain of the 44.7 reading that all you could estimate is that the measurement was between 44 and 45, you should report $44.7 \pm 0.3$ to represent that. The 0.3 is the (approximately) $1/ \sqrt{12}$ which is the RMS error for “somewhere in this unit length”.

Those RMS errors tend to be smaller than you think, so you should calculate them.

If you're best estimate is “somewhere between 44.6 and 44.8” then report $44.7 \pm 0.06$ (yes, that’s mismatched significant figures: that’s OK when reporting errors). The 0.06 is the full interval width of 0.2 times $1/\sqrt{12}$. That requires reliably dividing the division into 10 subparts, so perhaps that's a stretch.

You might feel most confident with something in between: "somewhere between 44.5 and 45". That error is then 0.5 times $1/\sqrt{12}$ = 0.15 so you'd report $44.7 \pm 0.15$ or perhaps even, if you've estimated "it really looks half-way" $44.75 \pm 0.15$: Remember, significant figures do not matter here.

Some might recommend i.e. $45 \pm 0.5$ in this case, but that says "There's a 50% chance the value is over 45, i.e. on the other side of the division from what was observed" Does that sound like it represents the observation properly? Taken to 1-sigma, it says "There's a 16% change the proper value is over 45.5". That's way to big a standard error to represent the usual use of a ruler...

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  • $\begingroup$ While I agree that the standard deviation of a uniformly distributed random variate can be computed this way, I know of no authoritative textbook that advocates for this method in the context of the question being asked. I worry this will only confuse, not enlighten. $\endgroup$ – Floris Aug 15 '19 at 12:02
  • $\begingroup$ Neither of the other extant answers discuss what error to report; the seem to just rely on significant figures. 44.7+-.5, 44.5+-0.5 are just wrong. What would you prefer? $\endgroup$ – Bob Jacobsen Aug 15 '19 at 13:54
  • $\begingroup$ “what door”? Autocorrect?? $\endgroup$ – Floris Aug 15 '19 at 13:57
  • $\begingroup$ My answer was “round to the nearest division” which is 45, so I would suggest 45 +- 0.5 as a reasonable way to report the measurement. I have added that explicitly to my answer - thanks for the suggestion. $\endgroup$ – Floris Aug 15 '19 at 14:01
  • $\begingroup$ That specifies there’s a 50% chance that the value should have been on the other side of the division than what’s observed. (>45 instead of approx 44.7) Authority for that being ok? $\endgroup$ – Bob Jacobsen Aug 15 '19 at 14:20

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