Ruler measurements and knowing how many decimal places to include

[flintstones]

Homework Statement

TL:DR; Suppose you have a ruler with markings every 0.2 cm as follows: 0 cm, 0.2 cm, 0.4 cm, 0.6 cm, 0.8 cm, 1 cm, etc. If you were trying to measure a length with this ruler, how many decimal places would your measurement need to have?

Relevant Equations

None

I teach high school and am trying to put together a resource that teaches students how to measure lengths properly with a ruler or meter stick. They don't understand how many decimal places to include in their answer (i.e. they will often write 1.7 cm when it should be 1.70 cm), and they struggle with knowing to estimate the last digit of their measurement. I am NOT trying to teach them about including uncertainties; only about recording their measurement to the correct number of decimal places for now.

So far, I've identified two steps for students to take:

1) Identify the resolution of the measuring tool (so for example, on a typical meter stick where there's a tick mark every 0.1 cm, the resolution is 0.1 cm)

2) Divide the resolution by 10 (continuing my example from step 1, take the resolution of 0.1 cm and divide it by 10. This gives 0.01 cm, meaning measurements with this meter stick should have two decimal places. So a measurement of 7.51 cm would be acceptable while simply 7.5 cm would not be).

These rules seem easy enough to follow until I came across an example where the resolution of the ruler is 0.2 cm. If we divide 0.2 cm by 10, we get 0.02 cm.

• Does this mean only measurements that are multiples of 0.02 cm are acceptable? i.e. 7.02 cm, 7.04 cm, & 7.06 cm are fine, but not 7.05 cm. This is incorrect, right? If you were using this ruler surely you could get a value of 7.05 cm?
• I believe the better way to think about it is this: After dividing 0.2 cm by 10 and getting 0.02 cm, we see that 0.02 cm has two decimal places. This simply means our measurement also needs to have 2 decimal places. So something like 7.05 cm would be perfectly acceptable since it has two decimal places.

Can someone check my reasoning and make sure everything I've written here is correct? I don't want to be teaching students wrong things and setting them up for problems when they get to college. Thank you!

 

[Orodruin]

I would say claiming a claim of sub-millimeter precicion when using a regular meter stic is completely absurd. It is questionable if you even have millimeter precision. Claiming precision to tenths of a millimeter severely underestimates the typical error.

Of course, the best thing to do is to estimate your actual error and do proper error propagation, but that is likely too advanced for your target audience.

they will often write 1.7 cm when it should be 1.70 cm

To underscore things here, there is no way you can be certain of that last digit using a regular ruler. It simply should not be there. The proper answer would bd 1.7 cm as this is as many digits that you can be confident in.

This is also not likely to change much if the grading changes to every 0.2 cm as the grading is only partially what goes into making a measurement with a ruler. Most people will have little problem mentally judging the mid-point between the marks. A little less precise perhaps, but probably well enough to still quote the mm value with high significance.

 

[haruspex]

To add a little to @Orodruin's comments…
Various practicalities make the limits nonlinear. Because of parallax, the thickness of the ruler is relevant; there will be limits to the accuracy of the ruler markings; the markings have thickness; the human eye has resolution limits, and those are affected by lighting.
Consequently, it may be reasonable to estimate the fraction of the distance between two 10cm marks to the nearest tenth, but less so between two 1cm marks, and certainly not for anything less.

 

[DaveE]

The way I learned it: If you say a length $l=7.51cm$ that means $7.505cm \leq l < 7.514 \bar 9...cm$. The resolution of your instrument must support that level of precision.

 

[berkeman]

I teach high school and am trying to put together a resource that teaches students how to measure lengths properly with a ruler or meter stick.

So as part of that instruction, you will cover parallax (already mentioned), and you may want to discuss the use of a Vernier scale to improve the resolution of the measurement.

https://en.wikipedia.org/wiki/Vernier_scale

 

[berkeman]

BTW, one other trick that I learned a long time ago that you may want to discuss with your students is how a standard ruler can be rounded and worn a bit on the ends, so it may not be as accurate when you try to "butt" it up against the end/base of something you want to measure. A better alternative is to use the line for the 1cm (or 1 whatever) mark as the start of the measurement, and just subtract that 1cm from the final length measurement. That way you are measuring between two well defined lines (that again you look directly down on each for the 2 measurements in order to avoid parallax).

 

[Orodruin]

A better alternative is to use the line for the 1cm (or 1 whatever) mark as the start of the measurement, and just subtract that 1cm from the final length measurement.

Or use a ruler like this one:

 

[Orodruin]

The way I learned it: If you say a length $l=7.51cm$ that means $7.505cm \leq l < 7.514 \bar 9...cm$. The resolution of your instrument must support that level of precision.

While not actually wrong to write it like that, it is worth noting that $7.514\overline{9} = 7.515$. But that’s a separate discussion.

 

[DaveE]

While not actually wrong to write it like that, it is worth noting that $7.514\overline{9} = 7.515$. But that’s a separate discussion.

Yes. It's not a rigorously correct definition. But it is easier to type this way.
It also illuminates the arbitrary boundaries of the "bins".
For example, $7.515000000000$ should be reported as $7.52$ according to this rule.
But $7.514999999998$ should be reported as $7.51$.
Does that mean that you can't report 3 digit results correctly without $1:10^{12}$ precision?

Ultimately more words are required. Or statistics, error bars, etc.

 

[Tom.G]

The definitive standard for rounding here in the States would be from The National Institute of Standards and Technology, WWW.NIST.GOV

A Google search such as this one will narrow it down somewhat:
https://www.google.com/search?&hl=en&q=rounding+site:nist.gov

Although for your example of Estimating, "rounding" may not be the appropriate term.

Cheers,
Tom