Sig Figs: Adding Accurate Lengths with Fewer Decimals

[Excalibur1152]

One thing I have always questioned is this:

Teachers always tell us that a final answer should never have any more significant figures than the least number of sig figs on a measured value used. But something doesn't add up here...


Suppose you knew the length of a metal rod to be 1.000000 meters (you knew the value up to 6 decimal places; you know it really accurately.)

And you know the length of another metal rod to be 1 micro-meter (you know it extremely accurately too, but there is only one sig fig because a micrometer is much smaller than a meter)

Now, if you were to add the lengths, you would obtain 1.000001 meters, but because of the teachers' rule, you leave this as 1 meter.

How is it possible that you know the value of a rod's length so accurately, and add on ~1 micrometer (almost no difference), and now you only know the final length to only 1 decimal?

 

[truesearch]

you would need to have measured the length of the first bar to better than
1 x 10^-6m to be able to give the length as 1.000000m
These zeros are not really significant figures. If they are then you would need to give the length as 1.000000 +/-0.000001m this implies an accuracy in measurement of +/- 0.0001%
I don't know of any measurement techniques (readily available) that can determine a length of 1m to this accuracy
PS it is not a 'teachers rule'... it is a practicality that your teacher is trying to bring home to you

 

[Excalibur1152]

you would need to have measured the length of the first bar to better than
1 x 10^-6m to be able to give the length as 1.000000m
These zeros are not really significant figures. If they are then you would need to give the length as 1.000000 +/-0.000001m this implies an accuracy in measurement of +/- 0.0001%
I don't know of any measurement techniques (readily available) that can determine a length of 1m to this accuracy
PS it is not a 'teachers rule'... it is a practicality that your teacher is trying to bring home to you

My point is that if you measured the length of the bar really accurately, and then measured a second one that was a seemingly infinitesimal length compared to the first one such that when you sum them, you can still know the final length to one part in a million (or similar).

BUT, because you are measuring the smaller length in a different unit of length, there is only one sig fig, so apparently your final answer should only contain one sig fig, which doesn't make any sense.Here's another analogy. Say that you know the length or a stretch of road to be 123901 meters (a really long road, and you measured it to six sig figs), and then you add one angstrom of length to it. Are you saying that I should now report the length to be 1x10^5 meters long? As if there is now so much inaccuracy now that the length could be 199999 meters long?

 

[Dickfore]

Teachers always tell us that a final answer should never have any more significant figures than the least number of sig figs on a measured value used. But something doesn't add up here...

This is a rule for multiplication and division.

Suppose you knew the length of a metal rod to be 1.000000 meters (you knew the value up to 6 decimal places; you know it really accurately.)

And you know the length of another metal rod to be 1 micro-meter (you know it extremely accurately too, but there is only one sig fig because a micrometer is much smaller than a meter)

Now, if you were to add the lengths, you would obtain 1.000001 meters, but because of the teachers' rule, you leave this as 1 meter.

How is it possible that you know the value of a rod's length so accurately, and add on ~1 micrometer (almost no difference), and now you only know the final length to only 1 decimal?

There is another rule for the operations of addition and subtraction. Namely, you write out the two quantities in decimal form. Then, your result has that many decimal places, as the one with the least number of decimal places.
The number of significant figures does not matter.
In your case:
$$L_1 = 1.000000 \, \mathrm{m}$$
and
$$L_2 = 1 \, \mathrm{\mu m} = 1 \times 10^{-6} \, \mathrm{m} = 0.000001 \, \mathrm{m}$$
The two lengths are expressed with the same number of decimals, namely 6 decimal places. Thus, the result of their addition must be expressed with 6 decimal places:
$$L_1 + L_2 = 1.000001 \, \mathrm{m} = 1000001 \, \mathrm{\mu m}$$

 

[Excalibur1152]

This is a rule for multiplication and division.



There is another rule for the operations of addition and subtraction. Namely, you write out the two quantities in decimal form. Then, your result has that many decimal places, as the one with the least number of decimal places.
The number of significant figures does not matter.
In your case:
$$L_1 = 1.000000 \, \mathrm{m}$$
and
$$L_2 = 1 \, \mathrm{\mu m} = 1 \times 10^{-6} \, \mathrm{m} = 0.000001 \, \mathrm{m}$$
The two lengths are expressed with the same number of decimals, namely 6 decimal places. Thus, the result of their addition must be expressed with 6 decimal places:
$$L_1 + L_2 = 1.000001 \, \mathrm{m} = 1000001 \, \mathrm{\mu m}$$

Much thanks, I think I am starting to get it now...but I can't control my mind from trying to think of places where the sig figs system doesn't work. If I am able to articulate anything else, I will post it but I probably won't.

 

[Borek]

And you know the length of another metal rod to be 1 micro-meter (you know it extremely accurately too, but there is only one sig fig

You can't know length extremely accurately with 1 sig fig, it is a self contradiction. If you know it extremely accurately you know it with several sig figs, like 1.0000 μm (doesn't matter at the moment if it is practically possible).

Not that it changes Dickfore explanation.

Note that sig figs are a rather lousy way of expressing accuracy. Some teachers treat them way too seriously.

 

[Excalibur1152]

You can't know length extremely accurately with 1 sig fig, it is a self contradiction. If you know it extremely accurately you know it with several sig figs, like 1.0000 μm (doesn't matter at the moment if it is practically possible).

Not that it changes Dickfore explanation.

Note that sig figs are a rather lousy way of expressing accuracy. Some teachers treat them way too seriously.

So just to make this clear, if you measure a certain dimension of something (whether it be time or length and so on), the number of decimal places kept can change depending on what units you use?

So that 1 micrometer >>conversion>> .000001 meters
each has one sig fig, but different numbers of decimal places.

What if you are converting units like feet to mile?
Say that you measure something to be '1 mile' and wish to express in feet:
1 mile >>>conversion>>> 5280 feet
Should this say 5.286 x 10^3 feet, or 5280??

Sorry if this is something that should be really obvious, my books only regard sig figs to numbers without dimensions or units

Thanks for the response by the way

 

[truesearch]

The important thing about significant figures is that, not only do they tell you what a value is, they also tell you what it is not.
If you quote a length to be 1.23m (this is 3 sig figs) then you are saying it is not 1.22m or 1.24m so there is an implied 'accuracy' of +/- 0.01m
With these figures that is better than 1%... more than adequate in most of physics.
Teachers try to get this across to students who are too keen to write down calculator readouts as 'answers'... 9N/1.23m is not 7.317073171N/m
Significant figures should be given greater priority in physics teaching.

 

[the_emi_guy]

Take your mile example. You say 1 mile has one significant digit? This means the distance is really 1.x miles where x is unknown. If the actual milage is 1.4miles, but you rounded it to 1 significant digit and got 1 mile, you would agree that it would be inappropriate to report 5280feet right?

 

[Excalibur1152]

Take your mile example. You say 1 mile has one significant digit? This means the distance is really 1.x miles where x is unknown. If the actual milage is 1.4miles, but you rounded it to 1 significant digit and got 1 mile, you would agree that it would be inappropriate to report 5280feet right?

Right. The 'true' value could be anywhere from 5280 to 2*5280 right? (it could actually have been 1 exactly or 1.9999... miles(the case where x=9999 or x=0)) right?EDIT: If the value could be 1.9999, couldn't it also be .0000001 ?

So 5280 plus or minus 5280 feet?

Either way, the answer should be reported to be 5x10^3 feet. Isn't this too precise then if it could actually be 5280*1.99999 ~10000 feet?

 

[Borek]

So just to make this clear, if you measure a certain dimension of something (whether it be time or length and so on), the number of decimal places kept can change depending on what units you use?

Yes, but number of significant digits stays the same. 1.0000 μm is 0.0000010000 m. 4 decimal places against 10, but 5 significant digits in both cases.

 

[truesearch]

As a teacher faced with this subject day in day out, this is why you should give numbers in standard form.
1.23 has 3 significant figures with a precision/accuracy of +/-0.01
0.00123 has 3 significant figures with a precision/accuracy of +/- 0.00001
much better to write 1.23 x 10^-3 +/-0.01 x 10^-3
0.00000123 ...need I say more??

 

[Borek]

In general you are right, but it becomes inconvenient during addition and subtraction.

1.23 has 3 significant figures with a precision/accuracy of +/-0.01

Bear in mind that's why significant figures are faulty by design. If 1.23 means 1.23±0.01, 1 means 1±1 - so 100% error, while 9 means 9±1 - only 11% error. Assuming both 1 & 9 were measured using the same device, there is no reason for such a difference.

 

[truesearch]

Absolutely correct!
1 implies the smallest scale division is 1 ! (100%)
9 implies the smallest scale division is 1 (otherwise you could have written 8 or 10) 10%(OK 11%)
95 implies the smallest scale division is 1 (YET AGAIN) ! 1%