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

[e-zero]

I have a question which states that a car's speed is 110km/h. Should I assume this speed to be 3 significant figures or 4 significant figures? why??

 

[technician]

the safest thing to do is stick to the number of figures given in the question, which is 3. If you feel that you must use 4 because calculations comu up like that then work with that but give 3 in your final answer.
Whatever you do...do not go beyond 4 !
The figures you do not know are, by definition, NOT SIGNIFICANT

 

[e-zero]

Ok, I figured it would be 3 in this case, but I was taught that the last non-zero digit is considered the first uncertain digit, which would mean that 110km would have 2 significant figures not 3.

This is where I'm confused. Could you explain?

 

[technician]

You have hit on a common problem !
It is difficult to know wheter the zero is significant or not.
If the speed was 115km/hr then there is no problem
I would say that in practice it is safe totake it to be 3 sig figs.
The problem is that these discussions about significant figures is not an exact science since they are used to convey accuracy of quantities. If greater accuracy is required then more significant figures are needed.
I am fairly certain you will get some other replies giving a variety of interpretations.

Hope this helps.
ps
if you think about it, giving a number to 3 sig figs implies that you know the value to better than 1%
when you write 115 you are saying that it is NOT 114 or 116, so you know the number to within +/- 1...about 1%
This is what makes it tricky deciding whether 110 is 3 or 2 sig figs. If it is only 2 then it means the value could be between 100 and 120 which is probably not intended ! because it implies an accuracy of only +/-10%

 

[mikeph]

I'd say that if the speed is 110 +/- 5 then it should be written 1.1x10^2 km/h, then it's far simpler to derive implicit errors from stated significant figures.

 

[Borek]

I was always under impression that in most physics courses nobody cares about sig figs, they are treated seriously only by chemists (and even then not by all).

 

[technician]

I was always under impression that in most physics courses nobody cares about sig figs, they are treated seriously only by chemists (and even then not by all).

They are treated as a means of assessing and conveying accuracy but it would be laborious if every example had to have this error analysis applied.
They are very important in practical situations and in practical exams students are penalised if they cannot show an awareness of the significance of significant figures.
This usually shows up when pupils write down the number from a calculator without any regard as to how many figures should be shown.

 

[Borek]

They are treated as a means of assessing and conveying accuracy but it would be laborious if every example had to have this error analysis applied.

I know what their application is, but I have seen on many occasions people being told to report the result with a "reasonable number of digits" - so obviously 1.23678923467854 doesn't make sense, but 1.24 or 1.237 is acceptable. Sig figs are about as wrong as easy to use, and treating them too religiously is a bad idea, as it gives students an unjustified feeling that they are doing something important.

 

[e-zero]

Let me use an example to see opinion: ½(23.1 cm3 − 20.32 cm3 + 19.0 cm3)

Would you equate that to 3 significant digits or 1? I am suggesting 1 because the numerator & denominator of the fraction 1/2 are both composed of 1 significant digit.

 

[mikeph]

I'd say common sense needs to be used, and the person doing the analysis needs to be aware of what terms are experimental data and what terms are exact. Without any further information (which we'd have if we did the experiment), I'd guess that 1/2 is an exact term and the other terms are to 3/4 s.f., so I'd quote the result to 3.

I agree there is some vagueness on what is acceptable, but there should definitely be clear pointers on what is unacceptable. I see all the time students measure results, round them, then average them (re-introducing false significant figures) and quoting the average to the same precision as the data. Unless you go through all the maths you'd never realize their result is ~10 standard deviations from the true mean. IMO that's very very bad practice. Rounding mid-calculation is obviously wrong, but a second error is quoting too many significant figures.

 

[Borek]

Would you equate that to 3 significant digits or 1? I am suggesting 1 because the numerator & denominator of the fraction 1/2 are both composed of 1 significant digit.

Are you suggesting that kinetic energy has never more that 1 sigfig, as in \frac{mv^2}2 2 has only one sigfig?

 

[e-zero]

No, I think it should be more, but just wanted to verify the ambiguity of sig figs.

 

[e-zero]

How about this example: 130km / 95km/h

Would your answer be 1.37h or 1.4h?

 

[jbriggs444]

How about this example: 130km / 95km/h

Would your answer be 1.37h or 1.4h?

1.4h pretty clearly.

The 130 km has either two or three significant figures depending on your guess as to whether the trailing zero is significant. The 95 has two significant figures. A quotient has as many significant figures as the less accurate of the dividend and the divisor. That's two either way.

However... That's a leading one on the 137 in the result. And that's a leading nine on the 95 in the input. That means that the relative error bounds on the input are tighter than "two significant figures" would normally suggest. And the relative error bounds on an output of 1.4 would be looser than "two significant figures" would normally suggest. So I would be quite willing to report 1.37 h if I were sure that the 130 km were accurate to +/- 0.5 km and that the 95 km/h were accurate to +/- 0.5 km/h. That way I wouldn't be throwing away accuracy in the name of significant figures.

 

[e-zero]

When you say there is a leading 1 and a leading 9, what significance does that have on deciding on the number of significant figures??

 

[jtbell]

To take a somewhat contrived example, consider two numbers, each with two significant figures:

(a) 11 - changing the last figure by ±1 is a change of about ±9%.

(b) 99 - changing the last figure by ±1 is a change of about ±1%.

Clearly and (a) and (b) have significantly different levels of precision, even though both have two sig figs.

Now consider the following number with three sig figs:

(c) 101 - changing the last figure by ±1 is a change of about ±1%.

Clearly (b) and (c) have about the same level of precision, even though one has two sig figs and the other has three.

 

[AlephZero]

If you want to estimate the error in the answer, then do it properly.

"Counting significant figures" is such a crude method as to be no real practical use. If the answer happens to be 99, then to 2 significant figures you have an error of about 1%. If it happens to be 101, to 2 s.f. you have an error of about 10%. If you think that makes any sense, then carry on counting significant figures...

Edit: I started typing this before jtbell's post appeared.

 

[e-zero]

I'm still a little confused. How about 65 * 1.96

What would you state that answer as?

 

[technician]

If you want to estimate the error in the answer, then do it properly.

"Counting significant figures" is such a crude method as to be no real practical use. If the answer happens to be 99, then to 2 significant figures you have an error of about 1%. If it happens to be 101, to 2 s.f. you have an error of about 10%. If you think that makes any sense, then carry on counting significant figures...

Edit: I started typing this before jtbell's post appeared.

101 is not given to 2 significant figures, it is 3. 101 means you know it is not 100 and it is not 102.
about 1% variation.
The problem arises when a value of 100 is quoted I would say

 

[technician]

I'm still a little confused. How about 65 * 1.96

What would you state that answer as?

You should give this to 2 significant figures, 130.
You should not use more significant figures than in the least precise number.

 

[AlephZero]

101 is not given to 2 significant figures, it is 3. 101 means you know it is not 100 and it is not 102.
about 1% variation.
The problem arises when a value of 100 is quoted I would say

Maybe the words were not quite clear, so I''l try again.

Suppose you measure an acceleration of 9.9m/s^2. The value is given to to 2 s.f, and the error is about 1%. Agreed?

Now suppose we convert the value into g. The standard value of Earth gravity is 9.80665 m/s2, exactly, by definition.

My calculator says 9.9 m/s^2 = 1.0095190508481489601443918157577g. Too much information!

But to 2 sf, that number is 1.0g, which would be interpreted as having an error of about 10%.

So if you believe significant figures mean something, just changing the units for a quantity can make it 10 times less accurate!

Of course the common sense thing to do is call it 1.01g, and not worry that the number of significant figures is "wrong".

 

[technician]

Suppose you measure an acceleration of 9.9m/s^2. The value is given to to 2 s.f, and the error is about 1%. Agreed?

Now suppose we convert the value into g. The standard value of Earth gravity is 9.80665 m/s2, exactly, by definition.

The DIFFERENCE between these values is 0.1...1% The nice bit of slight of hand does not change this.
If 'g' is defined then you can just leave it as a symbol (like c for speed of light) it is not a measured quantity. The 'measured' g of 9.9 implies it is between 9.8 and 10.0 Use these values and you will get the measured value = 9.9/9.8 = 1.0g or 10/9.8 = 1.0g...the .0 is significant.

The whole idea of significant figures is to do with reporting measurements which must have some degree of uncertainty about them. If you make measurements in an experiment and report values it is a big mistake to quote answers that can be shown to be 'too accurate' so you are obliged to use the number with the least number of sig figs as a 'worst case' example of the calculated value.
We can all dream up examples around 99, 100, 101 to show how easy it is to get confused and lost in the detail.

If we use e-zeros example of 65 x 1.96 this could represent the length of a steel bar and the width of the steel bar. If you are required to calculate the area then only 2 figures should be used because the 65 could be a value anywhere between 64 and 66 whereas the 1.96 could be any value between 1.95 and 1.97.
If these were important measurements this tells you that you need a better way to measure the length (or a less precise way to measure the width)

 

[jtbell]

65 x 1.96 this could represent the length of a steel bar and the width of the steel bar. If you are required to calculate the area then only 2 figures should be used because the 65 could be a value anywhere between 64 and 66 whereas the 1.96 could be any value between 1.95 and 1.97.

Let's check this using a more accurate method of propagating the uncertainties:

(65 ± 1)(1.96 ± 0.01)
= (65 ± 1.54%)(1.96 ± 0.51%)
= 127.4 ± √(1.54%² + 0.51%²)
= 127.4 ± 1.62%
= 127 ± 2 (after rounding off)

So the answer in this case is more precise than 130 ± 10, and not as precise as 127 ± 1. Nevertheless, I would say the third figure is still significant.

 

[Borek]

The whole idea of significant figures is to do with reporting measurements which must have some degree of uncertainty about them.

Sorry, but that's not the case. Significant figs are a poor mans way of dealing with uncertainties. Problem is, many people do believe it is THE way of dealing with them, while they are not. Sometimes they are better than nothing, but people that treat their results seriously don't use significant figs, but statistical approach and error propagation.

According to NIST electron mass is 9.10938291x10^-31 kg with a standard uncertainty of 0.00000040x10^-31 kg (which is sometimes written as 9.10938291(40)x10^-31). I wonder how you are going to tell that with significant figures.

 

[technician]

'Sometimes they are better than nothing'...exactly...that is the case...they do have some value and poor men (and women) making a start in physics with the hope of developing into 'people that will be able to treat their results seriously' need to be guided in the development of techniques.
Do you have any advice about sig figs that is constructive?
Do you know how the use of sig figs is taught in schools?

 

[Borek]

Do you have any advice about sig figs that is constructive?

Yes. Don't use them.

Do you know how the use of sig figs is taught in schools?

Unfortunately I do. 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.

 

[tiny-tim]

Let me use an example to see opinion: ½(23.1 cm3 − 20.32 cm3 + 19.0 cm3)

Would you equate that to 3 significant digits or 1? I am suggesting 1 because the numerator & denominator of the fraction 1/2 are both composed of 1 significant digit.

that's addition (and subtraction), so you don't use significant figures anyway, you use decimal places …

you use the highest decimal place of the given numbers, in this case one decimal place after the decimal point

then you divide by 2 which is exact (ie a million billion trillion sig figs)

 

[e-zero]

that's addition (and subtraction), so you don't use significant figures anyway, you use decimal places …

you use the highest decimal place of the given numbers, in this case one decimal place after the decimal point

then you divide by 2 which is exact (ie a million billion trillion sig figs)

So let me see if I understand this. If the original equation looked like this:

½(23.1 cm3 − 20.32 cm3 + 19.0 cm3) * 6

you would multiply by 6 which is exact (ie a million billion trillion sig figs)

but, if the original equation looked like this:

½(23.1 cm3 − 20.32 cm3 + 19.0 cm3) * 6cm3

then you would have to round to 1 significant figure.

correct?

 

[tiny-tim]

½(23.1 cm3 − 20.32 cm3 + 19.0 cm3) * 6cm3

then you would have to round to 1 significant figure.

correct?

correct

(unless you have some reason to believe the 6 cm³ is more accurately measured than is suggested)

 

[e-zero]

Pheww...I must say, this is more ambiguous than I thought.

 

[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.