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Borek
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technician said: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 said:They are part of the teaching and understanding physics.
Borek said:You are half right - they are part of the teaching. But even that doesn't make them right.
technician said:A statement like this makes no sense at all.
What are significant figures? Can you explain a reason for their existence?
L=26.2 +/-0.1cm, W=20.6 +/-0.1cm, T=3.9 +/-0.1cm
e-zero said: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??
technician said: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 said: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 said: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 ?
technician said: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 ?
technician said: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.
milesyoung said: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 said: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.
technician said:You have highlighted the need to be aware of the significance of significant figure !
jtbell said: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 said: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:
technician said:I wasn't making anything for customers ! I was making something to illustrate an idea.
technician said: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.
Borek said: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.
cjl said: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.
technician said:I am curious to know what measurements in what experiment resulted in a number to be quoted like this.
In that kind of example...of course, there is no problem.milesyoung said:So you now know why measurements are presented in such a way?
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.technician said:It is like reporting the heights of people in a population (I would even say sig figs are meaningless !)
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.technician said: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.
I'd be suprised if "significant figures" and "accuracy" didn't generate a lot of hits.technician said: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 have the following resource for you from the PHYS-L mailing list of the Buffalo State University in New York. Some background:technician said:Nowhere will you be told to ignore them, nowhere will criticism of education systems be used to discredit sig figs.
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).
Executive summary: No matter what you are trying to do, significant figures are the wrong way to do it.
You didn't provide any.technician said:I think that I will rest my case in those references.
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.technician said: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.
If the uncertainty was +/- 1 mm, how could you ever produce a piece with a length of 1542 mm?technician said:Like my planks, the spread in 100 planks may be +/-12mm but the uncertainty in anyone plank is +/-1mm...different things.
You've highlighted portions of a summary, the "evidence" is given in the sections that are linked to.technician said: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.
technician said:Do you think that 1.8 is the same as 1.80 is the same as 1.800?