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

[e-zero]

I have another example, which includes a measurement of uncertainty.

If I was going to convert minutes to hours I would use th = 1/60 * tm, where th is hours and tm is minutes.

If I had a set of values for tm with an uncertainty of +/- 0.2min, then I could determine my uncertainty of th to be:

(delta)th = 1/60 * (delta)tm
(delta)th = 1/60 * 0.2
(delta)th = 0.003

If I calculate th when tm = 20.0, then I would get th = 0.333. I keep th at 3 significant figures since tm is at 3 significant figures during calculation (i.e th = 1/60 * 20.0).

My question is this: If I calculate th when tm = 80.0, then I would get th = 1.33. How can (delta)th = 0.003 if, in this case, my value for th = 1.33?? That doesn't make sense to me.

 

[jtbell]

You should take that as an indication that you haven't allowed enough significant figures in the answer, and that it really should be 1.333 ± 0.003 hr. Let your calculated uncertainty determine the number of significant figures.

 

[technician]

Let your calculated uncertainty determine the number of significant figures.

This is excellent advice. If you are a physics student be sure to see the link between uncertainty in measurement and significant figure.

Yes. Don't use them

Ignore this advice !

Which is one of the reasons why I am against them and I always advice people to not treat them too seriously. You were given several examples of why they are misleading and confusing.

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.

 

[e-zero]

Thanks, I understand the importance of sig. figs. better now, and I'm also more aware of how to deal with them properly. The textbook I'm using has pretty specific language it specifies, for instance when using the word 'about' to describe a measure. I guess sig. figs. are good up to a point where there has to be some sort of common acceptance, i.e. between a teacher and a student.

 

[e-zero]

I do, however, have another example that involves sig. figs...yes another one :)

If I'm given length, width, and height:

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

 

[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