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Measurement uncertainties

  1. Sep 12, 2008 #1
    Hello, first of all sorry for my bad english it is not my mother tongue

    I am new to the uncertainty and significant figures concept and I have measured some things, I want to write the results using them.

    The length of my table is 273,3 cm. The instrumental uncertainty is ± 0,3 cm, and I have approximated the non-instrumental uncertainty to ± 1cm. Therefore, the total uncertainty is equal to ± 1,3 cm.

    Question: do I write 273,3 ± 1,3 cm? it looks to me that writing it this way does not make sense because the 3 in the units position in 273,3 is not 100% sure, so the 3 in the decimal part doesn't have any value. am I right? what would you write down?

    thank you
     
  2. jcsd
  3. Sep 12, 2008 #2

    Chi Meson

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    By writing " 273,3 ±1,3 cm" You are saying that the proper measurement is somewhere between 272,0 and 274,6 cm. There is nothing wrong with that. Divide the uncertainty by the measurement, and you find that this is a relative uncertainty of 0,5%. This is a valid statement.

    If you dropped that ",3" then you would be saying that the proper measurement is between 272,3 and 274,6. Now it looks like your uncertainty has been "improved" to 0,4%. The original analysis was better.

    In general (but not always the case) your uncertainty should have the same precision as the measurement. That is, the last digit of the uncertainty should be in the same decimal place as the last digit of the measurement.
     
  4. Sep 12, 2008 #3

    Redbelly98

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    Just to add to Chi Meson's excellent answer: it's customary to quote an uncertainty to either 1 or 2 significant figures, but never 3 or more. The ± 1,3 cm figure is consistent with this.
     
  5. Sep 13, 2008 #4
    ok, thank you for the clear answers. I have 2 other questions:

    273.3 * 76.34 = 20880.12
    "The LEAST number of significant figures in any number of the problem determines the number of significant figures in the answer."
    My answer must have 3 significant figures. How do I write 20880.12 using 3 significant figures? I really don't see how..


    2nd question
    perimeter of a table = (273.3±1.3)+(273.3±1.3)+(76.4±0.3)+(76.4±0.3) = 699.4 ± 3.2
    this is correct isn't it? no problems with the significant figures right?

    thank you
     
    Last edited: Sep 13, 2008
  6. Sep 13, 2008 #5

    Redbelly98

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    For starters, 273.3 and 76.34 both have 4 significant figures. Do you see why?
    So 4 is the least number of sig. fig's, and that should be the number of sig. fig's in the answer.

    Next, take the answer 20880.12. Start with the most significant figure 1st, then the 2nd-most sig. fig., etc., until you have 4 figures:

    1st: 2
    2nd: 0
    etc.

    Stop after the 4th figure, but add any zeroes as necessary until you've reaches the "1's" digit.

    Yes, looks good.
     
  7. Sep 13, 2008 #6
    ok, thank you very much

    I have another question:shy: sorry hopefully it's the last one

    let's say the diameter of a circle is 12.7 ± 0.3 cm
    I want to find the radius
    I divide 12.7 by 2 = 6.35 => 6.4 (because 3 is odd)
    I don't touch the uncertainty (because 2 is not a measured number, no uncertainty)
    so the final answer is 6.4 ± 0.3 cm
    correct?
     
  8. Sep 13, 2008 #7

    Redbelly98

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    No, the error for the radius will be different than the diameter error.

    Let's try a different example, one with no round-off issues, to see what's going on.

    Let's say a circle's diameter is 10.0 ± 0.4 cm.
    That means the diameter is between 9.6 and 10.4 cm.
    Taking 1/2 of those numbers to get the radius, we find the radius could be between
    9.6/2 and 10.4/2 cm, or 4.8 and 5.2 cm

    This range, 4.8 to 5.2, is expressed as 5.0 ± ???
    Notice the relation between the ??? radius error and the 0.4 cm diameter error.
     
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