Calculating Uncertainty in multiplication/division (Volume&Ration)

  1. Jan 21, 2009 #1
    [SOLVED]Calculating Uncertainty in multiplication/division (Volume&Ration)

    Hello all.

    1. The problem statement, all variables and given/known data

    As you eat your way through a bag of chocolate chip cookies, you observe that each cookie is a circular disk with a diameter of 8.50 +/- .002cm and a thickness of (7.0×10^−2) +/-0.005.

    1. Find the average volume of a cookie. > Express your answer using two significant figures
    2. Find the uncertainty in the volume of a cookie. > Express your answer using one significant figure.
    3. Find the ratio of the diameter to the thickness. > Express your answer using two significant figures.
    4. Find the uncertainty of the ratio. > Express your answer using one significant figure.

    2. Relevant equations

    I already got the average volume and ratio, but I'm having a hard time solving for the uncertainty. BTW, this is a MasteringPhysics problem, so it's online and we get feedback, but every incorrect attempt lowers your possible points.

    1. Volume:
    V=A*thickness = (pi*4.25^2)*.07 =^3 = 4.0cm^3 (2 sig. figs)

    3. Ratio = diameter/thickness = 8.50/.07 = 121.4285714 = 120 (2 sig figs)

    3. The attempt at a solution

    I read the physics textbook, but I'm still confused. It talks of implicit uncertainty (sig. figs). The example it gives is:

    "We give the thickness of the cover of this book as 2.91mm, which has 3 sig.figs. By this we mean that the first two digits are known to be correct, while the third digit is uncertain. The last digit is in the hundredths place, so the uncertainty is about .01mm"

    So I tried using the volume (4.0 cm) - it has 2 sig. figs. According to the book, the last digit is the 0, and it's in the tenths place, so the uncertainty would be = 0cm?! That's obviously wrong. Same thing when you try to repeat the process for the ratio (120) - there are two sig figs, the first digit is known to be correct, and the second is uncertain 2 is in the tens spot, so the uncertainty would have to be 20 - which again is wrong - I plugged it in and it was wrong.

    I was trying to find help online, and I came to this site:

    If you scroll down, it comes to uncertainty in multiplication or division, and they give you the equation:

    uncertainity = sqrt [(uncertainty of A/A)^2 + (uncertainty of B/B)^2]

    I plugged in:

    sqrt [(.02/8.5)^2 + (.005/.07)^2] = .07146... = 7*10^-2 (one sig fig)
    ...and it was wrong.

    Essentially, I'm stuck and not sure how exactly you *find* uncertainty. I read a lot of the uncertainity posts on here, but it still hasn't helped, the same formula as the dartmouth site, and I still get that the answer is wrong.

    I wish I could show more work, but all of my work is wrong, and after that, I was just trying to guess using random combinations of the given uncertainties (.02cm and .005cm), and I doubt it'd be much use. Other than going in to ask the professor himself, I've exhausted all the other options, none of what I read online or in the textbook clicked to show me how to go about answering the problem.

    Thank you very much for any help!
    Last edited: Jan 21, 2009
  2. jcsd
  3. Jan 21, 2009 #2


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    Welcome to PF.

    The formula from the Dartmouth site which uses the RSS of the relative errors should be OK. To apply it properly you need to account for the area uncertainty as the square of the dimension that you would have used. If you would have calculated the area as πd²/4, then you would want to add d's contribution to uncertainty twice.

    Hence ((Δd/d)² + (Δd/d)² + (Δh/h)²)1/2
  4. Jan 21, 2009 #3


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    On the other hand your course may also be treating uncertainty in a more conservative manner and the multiplication rule your teacher is using may only be based on the simple sum of the relative uncertainties.

    In which case you may be expected to use (Δd/d) + (Δd/d) + (Δh/h)
  5. Jan 21, 2009 #4
    Thank you! I just realized that to find the volume of the cylinder, I used the radius (1/2 diamter > 8.5*.5 = 4.25), not the diameter.

    So sqrt[(.02/4.25)^2 + (.005/.07)^2] = .0715834209 - which would not make a difference in the final result since we need to round to one sig. figure, and it still works out to be 7*10^-2.


    If I use ((Δd/d)² + (Δd/d)² + (Δh/h)²)1/2
    = sqrt[(.02/4.25)² + (.02/4.25)² + (.005/.07)²]
    = .0717379361 > which still gives the same answer with 1 sig. fig

    If I use the diameter, and not the radius:
    = sqrt[(.02/8.5)² + (.02/8.5)² + (.005/.07)²]
    =.0715060381 > again, works out to the same number.


    another thing, if I halve the diameter, do I also need to halve the uncertainty? (.02 > .01cm?) It seems logical in the sense that had you not halved the uncertainty, it would grow each time you reduce the actual measurment, to the point where the uncertainty will be greater than the measurement itself, and that's impossible, right?

    And by the same logic, the uncertainty of the thickness would not change becuase it hasn't been manipulated:

    If so, then I could use the formula again and:
    =sqrt[(.01/4.25)² + (.01/4.25)² + (.005/.07)²]
    =.0715060381 > Again. I guess this makes no difference.

    with the diameter instead of radius measurements:
    = sqrt[(.01/8.5)² + (.01/8.5)² + (.005/.07)²]
    =.071447946 > Yet again.

    Also, I'm not sure what RSS means? Thank you again though, and I appreciate your reply.

    [EDITED TO ADD:] I tried adding the uncertainties (.005+.02) for both the ratio and volume, and it was wrong.
  6. Jan 21, 2009 #5


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    This was why I suggested you use the πd²/4 for area, as it should clarify how you would take the relative uncertainty.

    (Δd/d)² ---> (.002/8.5)²

    Yes. Add it twice.

    Note your error ±.002 over 8.5 cm as per the original statement
  7. Jan 21, 2009 #6


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    RSS = Root Sum of the Squares
  8. Jan 21, 2009 #7
    Alright. I don't think I see what you're saying, to me it looks like you're just repeating to do what I've already done, which is:


    Are we speaking past each other? Becuase as I said, the .07....number I keep coming up with is incorrect. MasteringPhysics automatically checks answers, it's deemed that one wrong...
  9. Jan 21, 2009 #8


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    Let me say it again:

    Note your error ±.002 over 8.5 cm as per the original statement
  10. Jan 21, 2009 #9
    Aagh - I'm so sorry, I copied the original problem wrong, it's not .002, but .02cm (and still .005cm for the other). Sorry for the confusion. :/
  11. Jan 21, 2009 #10


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    Ok. That's fine. Then are you sure your instructor is expecting you to take the RSS to determine uncertainty propagation?

    Because regardless, the error of the diameter is so much smaller than the error of the thickness, that the RSS will be dominated by the thickness error.

    Might it simply be the sum of the relative errors in your course for taking multiplication/division uncertainties?

    In which case that sums to .076.

    But remember what you get from either method is a relative uncertainty result that would be expressed as a %. If they want the error expressed as an absolute volume then you still need to multiply that % by the nominal calculation.
  12. Jan 21, 2009 #11
    No, I have no idea what he expects. Of the entire problem set for the section, this is the only problem with uncertainties. This is the first homework set though, but/and it's 'physics for engineering majors'. We haven't covered before, so maybe you're right and he's looking for a simpler solution.

    What you're saying, is that I take the 7*10^-2, and multiply it by the nominal calculation, so .076*4.0(volume) and .076*120(ratio).

    YEEEEEEEEESSSSSSS!!!!!! Thank you! That works, and I also get it.
  13. Jan 21, 2009 #12


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    If the .076 is correct then the instructor is not using the RSS, he is using the simple sum of the relative errors as his multiplication/division rule, with the relative error of the diameter taken twice because of its effect on area.

    Good luck.
  14. Jan 21, 2009 #13
    Yeah, .076 multiplied by the actual result (ratio, volume) was right. Thanks again!
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