Is a Car's Speed of 110km/h Accurate to 3 or 4 Significant Figures?

[Borek]

They are part of the teaching and understanding physics.

You are half right - they are part of the teaching. But even that doesn't make them right.

 

[technician]

You are half right - they are part of the teaching. But even that doesn't make them right.

A statement like this makes no sense at all.
What are significant figures? Can you explain a reason for their existence?

 

[Borek]

A statement like this makes no sense at all.

Yes it does. They are taught - and that was the half of your statement that was correct. They are not necessary to understand physics, that was the incorrect half. And the fact that they are taught doesn't make them correct - they are still the wrong way of dealing with uncertainties.

What are significant figures? Can you explain a reason for their existence?

Significant figures are an approximation that pretends to be an efficient way of dealing with uncertainties. You were shown several times in this thread that they are not. I don't know any reason FOR their existence, I know a reason WHY they still exist - they exist because they are still taught. There is really no need for that, as we can talk about uncertainties without significant figures.

Consider e-zero's example:

L=26.2 +/-0.1cm, W=20.6 +/-0.1cm, T=3.9 +/-0.1cm

We are interested in the volume. From the data given L is at least 26.1 and at most 26.3 cm, W is at least 20.5 and at most 20.7 cm, T is at least 3.8 and at most 4.0 cm. That means the real volume is somewhere between 26.1*20.5*3.8 (2025.4) and 26.3*20.7*4.0 (2177.64) - you don't need significant figures to see that the reasonable way to report the volume is 2101±76. This is much better result than 2.1*10³ which incorrectly suggests something between 2050 and 2150.

 

[e-zero]

Could someone explain why the measurement of uncertainty of volume needs to be rounded to 100??

 

[jtbell]

L=26.2 +/-0.1cm, W=20.6 +/-0.1cm, T=3.9 +/-0.1cm, then

V = L * W * T = 2104.908cm^3 = 2100cm^3 (2 sig. figs. b/c of T)

and measure of uncertainty for V is:

(delta)V = |V| ((delta)L / L + (delta)W / W + (delta)T / T)
(delta)V = 72.224cm^3

I'm told

By whom?

that you should round up (delta)V to 100cm^3. Could someone explain why??

Using your numbers, I would write the volume as 2100 ± 70 cm³. In this case the first zero in 2100 would be significant, and the second zero not significant. The first zero is there because it's in your original raw calculated result of 2104.908 and it corresponds to the first digit of the uncertainty (72.224 which gets rounded to 70); not because of a sig-fig rule.

If you're not going to state the uncertainty in V explicitly, then I suppose one could argue that it's better to state V to only two sig figs (which effectively rounds the uncertainty up to 100) than to three sig figs (here of course that figure is 0, but suppose for the moment it's something else, say 2140 for the sake of argument) because that third digit (4) is not very well known at all.

However, if you're going to state the uncertainty explicitly, then I don't see any need to round it up like that.

 

[cjl]

Look at the advice supporting the use of sig figs.
You will lose marks in a physics exam if you show no appreciation of sig figs.
They are part of the teaching and understanding physics.

Just because the knowledge of something is required to pass certain exams does not make the knowledge useful or correct (other than the obvious use of scoring well on exams). Sig figs are a potentially misleading and largely incorrect way of dealing with uncertainty, and their religious use by many a high-school physics teacher has probably caused more confusion than knowledge among students. I'm firmly in agreement with Borek on this one - they are not necessary to understand physics, and in situations where uncertainty matters, proper propagation of uncertainty should be taught, rather than sig figs (which are a poor substitute).

 

[e-zero]

I have a simpler, but similar example to last:

If I want to calculate the area of a circle whose radius is 7.3 +/-0.2cm, then I would get A=167.415cm^2 which would be rounded to 2 significant figures and end up being A=170cm^2.

Now if I calculate the uncertainty of A, then I would do the following:

(delta)A = PI * 2r * (delta)r
(delta)A = PI * 2(7.3) * (0.2)
(delta)A = 9.173cm^2

I would now round my answer for (delta)A to 10cm^2 since the area of the circle, A=170cm^2, was rounded to 2 significant figures. Correct?

 

[technician]

That means the real volume is somewhere between 26.1*20.5*3.8 (2025.4) and 26.3*20.7*4.0 (2177.64) - you

There is a calculator mistake here. 26.1 x 20.5 x 3.8 = 2033.19 not 2025.4

Teachers try to ensure that students check what they are doing and check their working.
Incorrect answers do not give rise to confidence in any conclusions that are drawn.
It is a disgrace if one needs to criticize education (teachers, schools, textbooks, exams etc) to put forward 'physics' arguments.

 

[cjl]

I have a simpler, but similar example to last:

If I want to calculate the area of a circle whose radius is 7.3 +/-0.2cm, then I would get A=167.415cm^2 which would be rounded to 2 significant figures and end up being A=170cm^2.

Now if I calculate the uncertainty of A, then I would do the following:

(delta)A = PI * 2r * (delta)r
(delta)A = PI * 2(7.3) * (0.2)
(delta)A = 9.173cm^2

I would now round my answer for (delta)A to 10cm^2 since the area of the circle, A=170cm^2, was rounded to 2 significant figures. Correct?

Assuming your calculations are correct, I would report that as 167 +/- 9. You don't necessarily need to follow significant figures, especially if you are going to properly calculate the uncertainty.

 

[jtbell]

I agree. As I wrote before, let the uncertainty decide the number of sig figs, not the other way around.

 

[e-zero]

Ok, so you all agree that in that second last example V=2100 and uncertainty is +/- 70

and in that last example A=167 and uncertainty is +/- 9

This is contradicting what I am being told by my tutor. Is there a right and wrong here? Or is it just the way that my tutor wishes to complete his/her answer.

 

[technician]

Ok, so you all agree that in that second last example V=2100 and uncertainty is +/- 70

I could go along with this, it looks like you have an appreciation of the meaning of significant figures.
If we debated this any further we would be talking about the INSIGNIFICANT figures and I am sure you know what INSIGNIFICANT means.

Do you realize that 90% of the numerical information here is only available because we all have electronic calculators...the information is an artefact of calculators not of physical measurements.
20, 30 years ago the only common calculator available was a slide rule which could only give answers to 3 or maybe 4 sig figs.
Dealing with uncertainty, errors and sig figs was no problem then !
It was well understood and not subject to confusion by the production of 10 digit numbers.

 

[technician]

Just because the knowledge of something is required to pass certain exams does not make the knowledge useful or correct (other than the obvious use of scoring well on exams). Sig figs are a potentially misleading and largely incorrect way of dealing with uncertainty, and their religious use by many a high-school physics teacher has probably caused more confusion than knowledge among students.

This is a ridiculous statement in a forum designed to promote education in physics.
What do you suggest,...get rid of teachers, burn the textbooks don't bother with exams ?

 

[mikeph]

Just because the knowledge of something is required to pass certain exams does not make the knowledge useful or correct (other than the obvious use of scoring well on exams). Sig figs are a potentially misleading and largely incorrect way of dealing with uncertainty, and their religious use by many a high-school physics teacher has probably caused more confusion than knowledge among students.

This is a ridiculous statement in a forum designed to promote education in physics.
What do you suggest,...get rid of teachers, burn the textbooks don't bother with exams ?

I agree with the red text too. I think most people would.

For one thing, they don't give you any concept of the distribution of the error, whereas a standard error figure tells everyone you're using a Gaussian. Reporting your error using only significant figures arbitrarily biases your error estimates to the decimal system, i.e. 15 would be read as 15 +/- 0.5. What if the actual error is +/- 0.3? Sig.figs cannot account for this. And finally, they are also problematic when you want to quote more than one digit on your error- e.g. 15.54 +/- 0.12 is a perfectly reasonable distribution but there is no way of writing this using significant figures only. You aren't allowed to round the data for no reason, so you have to just write 15.54 and the reader will assume your error is 15.54 +/- 0.005, which is far more precision than you have.

In my opinion it's another case of teaching students one thing because they haven't got the prerequisites of multivariate calculus and statistics requited for proper error analysis.

 

[cjl]

Just because the knowledge of something is required to pass certain exams does not make the knowledge useful or correct (other than the obvious use of scoring well on exams). Sig figs are a potentially misleading and largely incorrect way of dealing with uncertainty, and their religious use by many a high-school physics teacher has probably caused more confusion than knowledge among students.

This is a ridiculous statement in a forum designed to promote education in physics.
What do you suggest,...get rid of teachers, burn the textbooks don't bother with exams ?

Not at all. I just wish teachers would perhaps change the curriculum to emphasize the correct way to calculate uncertainty (and I wish the textbooks and exams would be updated to reflect this). I can disagree with a portion of the curriculum without wanting to burn all the textbooks.

 

[technician]

And finally, they are also problematic when you want to quote more than one digit on your error- e.g. 15.54 +/- 0.12 is a perfectly reasonable distribution but there is no way of writing this using significant figures only. You aren't allowed to round the data for no reason,

I am curious to know what measurements in what experiment resulted in a number to be quoted like this.
To show what this number 'looks like' I will rewrite the number as 1554 +/-12 (to avoid messing about with the decimal point)
I have taken this to represent 1554mm +/-12mm and cut a wooden slat to show the number on a line.
I wonder what application in practical physics would find this number, represented as it is, to be useful. I cannot see that the +/-12 (mm) has any significance.
The only thing I could think of is perhaps the height of a stack of 12mm blocks was measured, found to be 1554mm and the number of blocks in the stack was calculated. If the count was thought to be out by 1 block that could be expressed as +/-12mm.
Or perhaps 12mm represents a wavelength and a different sort of counting (of wavelengths) was involved.
For me the moral is...LOOK at the number, each digit to the right is 10x smaller than the one before...how many moves to the right do you need to make for the number you meet can be considered to be 'insignificant'...for most everyday physics I would say 3 or 4 at the most.

 

[jtbell]

It's not unheard of to quote uncertainties to two significant figures. See for example this table of physical constants from the Particle Data Group:

http://pdg.lbl.gov/2012/reviews/rpp2012-rev-phys-constants.pdf

 

[milesyoung]

I am curious to know what measurements in what experiment resulted in a number to be quoted like this.
To show what this number 'looks like' I will rewrite the number as 1554 +/-12 (to avoid messing about with the decimal point)
I have taken this to represent 1554mm +/-12mm and cut a wooden slat to show the number on a line.
I wonder what application in practical physics would find this number, represented as it is, to be useful. I cannot see that the +/-12 (mm) has any significance.

How do you know that you've cut the wood to 1554 mm? Would it be fair to say that you're not entirely sure that it's 1554 mm?

 

[technician]

How do you know that you've cut the wood to 1554 mm? Would it be fair to say that you're not entirely sure that it's 1554 mm?

You have hit the nail square on the head ! The numbers in the post I quoted (#49) were 15.54+/-0.12 and I would say the same question should be asked here.
My piece of wood was cut with care but it was to illustrate the idea, I could equally easily have just done a sketch.
You see the +/-12mm block at the end, it 'swamps' the last figure (4).

You have highlighted the need to be aware of the significance of significant figure !

 

[milesyoung]

I was responding to this:

I wonder what application in practical physics would find this number, represented as it is, to be useful. I cannot see that the +/-12 (mm) has any significance.

If for the sake of argument we assume your ruler, or whatever instrument you used to place the 1554 mm mark for the cut, is exact.

You might be a good craftsman with a saw, but you're probably still going to cut off a bit too much or a bit too little here and there. If you had a production of these wood pieces, you might take a large random sample of your work and conclude on the basis of this sample that you're able produce lengths between 1542 and 1566 mm with a mean of 1554 mm.

Wouldn't you find it significant to include a measure of this accuracy for your customers?

I can't follow the thought process that brought you to this:

You have highlighted the need to be aware of the significance of significant figure !

 

[technician]

It's not unheard of to quote uncertainties to two significant figures. See for example this table of physical constants from the Particle Data Group:

http://pdg.lbl.gov/2012/reviews/rpp2012-rev-phys-constants.pdf

It is not unheard of but it is not common. The values of physical constants is a very specialist branch of physics, the vast majority of constants in that table have 9, 10 sig figs which means they are already known to an accuracy equivalent to measuring the length of a 1m bar to within +/- 1atom. This is hardly typical laboratory level. The heading of the paper also defines how the uncertainties are calculated.
The atomic masses of atoms is quoted to about 5 or 6 sig figures and these are all quoted to +/-1 in the last figure...more typical of use of significant figures.
This shows that there is a need to understand the significance of significant figures, it is an essential aspect of treating measurements in physics.

 

[technician]

I was responding to this:


If for the sake of argument we assume your ruler, or whatever instrument you used to place the 1554 mm mark for the cut, is exact.

You might be a good craftsman with a saw, but you're probably still going to cut off a bit too much or a bit too little here and there. If you had a production of these wood pieces, you might take a large random sample of your work and conclude on the basis of this sample that you're able produce lengths between 1542 and 1566 mm with a mean of 1554 mm.

Wouldn't you find it significant to include a measure of this accuracy for your
customers?

I can't follow the thought process that brought you to this:

I wasn't making anything for customers ! I was making something to illustrate an idea.
Would a sketch have been better ??
You are getting away from the point!

 

[cjl]

I wasn't making anything for customers ! I was making something to illustrate an idea.

That's not the point. The customer question was hypothetical: if you had customers who wanted a piece of wood, and you made a hundred pieces of wood (for whatever reason), and they measured anywhere from 1542 to 1566 mm, with a mean of 1554, how would you report their length? 1554 +/- 12 seems to be a good way to me, as it accurately represents both the mean and the variance of the length of the wood.

 

[Borek]

The atomic masses of atoms is quoted to about 5 or 6 sig figures and these are all quoted to +/-1 in the last figure...

Not true. Check this list:

Atomic Weights and Isotopic Compositions for All Elements

First on the list - hydrogen - is given as 1.00794(7). Less than 20 are quoted as ±1.

.more typical of use of significant figures.
This shows that there is a need to understand the significance of significant figures, it is an essential aspect of treating measurements in physics.

You are twisting facts to support your opinion. It won't work.

 

[technician]

Not true. Check this list:

Atomic Weights and Isotopic Compositions for All Elements

First on the list - hydrogen - is given as 1.00794(7). Less than 20 are quoted as ±1.
You are twisting facts to support your opinion. It won't work.

You are correct ! the uncertainty is IN THE LAST FIGURE is what I meant...unintended twisting of a fact.
I have not looked in any great detail...are there any uncertainties that encompass the last 2 figures??
At least I use facts...I have nothing to gain whether it works or not... I enjoy the debate...it has given some people something to think about. I will contribute as long as I find it interesting.

extra: I have checked the table of atomic masses, all of the standard atomic masses have uncertainty only in the last figure.
The relative atomic masses are quoted to 9, 10, 11 figures representing an uncertainty equivalent to quoting the length of a 1m rod to +/- 1atom...I am not surprised to find doubts about uncertainty at this level. I certainly would not quote these in any argument about the role of significant figures ! references are supplied in the table for anyone interested in detail

 

[technician]

That's not the point. The customer question was hypothetical: if you had customers who wanted a piece of wood, and you made a hundred pieces of wood (for whatever reason), and they measured anywhere from 1542 to 1566 mm, with a mean of 1554, how would you report their length? 1554 +/- 12 seems to be a good way to me, as it accurately represents both the mean and the variance of the length of the wood.

I agree with you but this is not the uncertainty in the measured length of a plank.
I would quote all of the lengths to +/-1mm, (some would argue that this should be +/-0.5mm) I would use a steel tape measure and the smallest divisions are 1mm.
The fact that all of the planks could not be cut to within +/-1mm of each other is a different thing.

 

[milesyoung]

So you now know why measurements are presented in such a way?

I am curious to know what measurements in what experiment resulted in a number to be quoted like this.

 

[technician]

So you now know why measurements are presented in such a way?

In that kind of example...of course, there is no problem.
It is like reporting the heights of people in a population (I would even say sig figs are meaningless !)

The original figures related to reporting a measurement, ie the height of a person, (sig figs are not meaningless !) not representing the range of measurements in a sample of 100.

Extra
I have done a quick Google search for 'significant figures'...there is a wealth of information relating the importance of sig figs and the links with accuracy.
Nowhere will you be told to ignore them, nowhere will criticism of education systems be used to discredit sig figs.
I think that I will rest my case in those references.

 

[milesyoung]

It is like reporting the heights of people in a population (I would even say sig figs are meaningless !)

In this case you'd probably give the data in the form of a distribution instead. The mean and range of heights isn't very informative unless these parameters happen to uniquely describe the distribution.

The original figures related to reporting a measurement, ie the height of a person, (sig figs are not meaningless !) not representing the range of measurements in a sample of 100.

Be it a measurement of height or the length of one of your wood pieces, we're still just specifying a number with some uncertainty.

If you care about your data, you'd apply some method of error analysis when you do calculations with it. You'd most definitely not rely on counting significant figures.

I have done a quick Google search for 'significant figures'...there is a wealth of information relating the importance of sig figs and the links with accuracy.

I'd be suprised if "significant figures" and "accuracy" didn't generate a lot of hits.

Nowhere will you be told to ignore them, nowhere will criticism of education systems be used to discredit sig figs.

I have the following resource for you from the PHYS-L mailing list of the Buffalo State University in New York. Some background:

PHYS-L is a list dedicated to physics and the teaching of physics with about 700 members from over 35 countries, the majority of whom are physics educators. Traffic varies from zero to sixty messages/day with an average of about ten per day. All postings are archived. Noninflammatory, professional and courteous postings intended to inform members on how to better understand, teach and learn physics are always welcome.

PHYS-L is officially supported by the SUNY-Buffalo State College Department of Physics, SUNY-University at Buffalo Department of Physics, and unofficially by the American Association of Physics Teachers (AAPT).

The resource was compiled by John S. Denker, formerly of Bell Labs, from the PHYS-L archives. Here's a quote from it:

Executive summary: No matter what you are trying to do, significant figures are the wrong way to do it.

Measurements and Uncertainties versus Significant Digits or Significant Figures

You can find plenty more in the public domain on error analysis.

I think that I will rest my case in those references.

You didn't provide any.

 

[technician]

I could not resist looking at this reference. 99.9%(+/-0.1%) of this is concerned with analysing data extracted from a distribution NOT with uncertainty in making a measurement. Like my planks, the spread in 100 planks may be +/-12mm but the uncertainty in anyone plank is +/-1mm...different things.
It is full of contradiction and confusion apparently 1.8 is the same as 1.80...it is not in the world of measurement ! 'hogwash' to use the author's vernacular.

Below is one of his lists of evidence, I have highlighted in red evidence that would be called 'anecdotal', ie no evidence of any worth.
I have highlighted in blue statements that make sense.

The disadvantages of sig figs include:

Given something that is properly expressed in the form A±B, such as 1.234±0.055, converting it to sig figs gives you an excessively crude and erratic representation of the uncertainty, B. See section 7.5.3 and especially section 15.5.
Sig figs also cause excessive roundoff error in the nominal value, A. This is a big problem. See section 6.8 for a concrete example.
Sig figs cause people to misunderstand the distinction between roundoff error and uncertainty. See section 6.8 and section 5.4.
Sig figs cause people to misunderstand the distinction between uncertainty and significance. See section 13, especially section 13.3.
Sig figs cause people to misunderstand the distinction between the indicated value and the corresponding range of true values. See section 4.6.
Sig figs cause people to misunderstand the distinction between distributions and numbers. Distributions have width, whereas numbers don’t.Uncertainty is necessarily associated with some distribution, not with any particular point that might have been drawn from the distribution. See section 1.2, section 5.4, and reference 1.
As a consequence, this makes people hesitate to write down numbers. They think they need to know the amount of supposedly “associated” uncertainty before they can write the number, when in fact they don’t. Very commonly, there is not any “associated” uncertainty anyway.
Sig figs weaken people’s understanding of the axioms of the decimal numeral system. See section 15.5.7.
Sig figs give people the idea that N nominal values should be associated with N uncertainties, which is just crazy. In fact the number of uncertainties scales like (N2 + N)/2, as discussed in section 8.1.
The sig figs approach cannot possibly apply to algebraic variables such as A±B, so you are going to have to learn the A±B representation anyway. Having learned it, you might as well use it for decimal numerals such as 1.234±0.055. See section 15.5.5.
Et cetera

There are 89 pages in this reference...full of contradiction and condescending, snappy quotes
Thats all for now

 

[milesyoung]

I could not resist looking at this reference. 99.9%(+/-0.1%) of this is concerned with analysing data extracted from a distribution NOT with uncertainty in making a measurement.

I think you have misunderstood what you've read. Yes, the resource is in large part based on statistical methods, but that's not really suprising, given the subject.

Like my planks, the spread in 100 planks may be +/-12mm but the uncertainty in anyone plank is +/-1mm...different things.

If the uncertainty was +/- 1 mm, how could you ever produce a piece with a length of 1542 mm?

Below is one of his lists of evidence, I have highlighted in red evidence that would be called 'anecdotal', ie no evidence of any worth.

You've highlighted portions of a summary, the "evidence" is given in the sections that are linked to.


I'd be inclined to ask you for examples of what you disagree with, but your responses in this thread, and others where we have posted, have left me flabbergasted. I have tried to argue what I believe to be true to the best of my ability, but somehow it seems that, regardless of the argument, you discard it just the same, and that's not how I wish to spend my time on this forum.

 

[technician]

If the uncertainty was +/- 1 mm, how could you ever produce a piece with a length of 1542 mm?
I couldn't !
BUT I could produce a plank that was measured to be 1542 +/-1mm (the 2 at the end let's you know that the measurement was made with a mm scale)

I'd be inclined to ask you for examples of what you disagree with, but your responses in this thread, and others where we have posted, have left me flabbergasted. I have tried to argue what I believe to be true to the best of my ability, but somehow it seems that, regardless of the argument, you discard it just the same, and that's not how I wish to spend my time on this forum.

couldn't have said it better myself.
I take it this will be the last in this post.

Do you think that 1.8 is the same as 1.80 is the same as 1.800?

 

[cjl]

Do you think that 1.8 is the same as 1.80 is the same as 1.800?

They are mathematically equivalent, and they all represent precisely the same number.

 

[technician]

In terms of MEASUREMENT these numbers are not the same. This is the importance of knowing what is meant by significant figures.
!.8 means that the measured quantity lies between 1.7 and 1.9
1.80 means the quantity is between 1.79 and 1.81
and so on
I include a sketch of scales and scale divisions that lead to this logic.
For the number 1.80 the last zero is significant

 

[Borek]

This is getting boring. How do you represent 1.80±0.02?