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Homework Help: I need to find the possible error for the volume of a cuboid

  1. Sep 18, 2011 #1
    I actually posted this on Yahoo answers, but I have no idea what they were saying...

    Let's say I used a micrometer to measure the dimensions of a cuboid.
    Length: 2.8850 cm (Uncertainty = 0.00005 cm)
    Width: 2.8750 cm (Uncertainty = 0.00005 cm)
    Height: 2.8950 cm (Uncertainty = 0.00005 cm)

    The volume is 24.01221563... cm3, and I have to round it off to 5 significant figures, hence the final answer is 24.012 cm3.

    Now I have to determine the possible error for the volume.
    There are two methods, i was told, one was to find the difference between the upper and lower limit. The other was to calculate the percentage error.

    So the percentage error for the length is: 0.00005/2.8850*100 = 0.0017331%
    % error for the width is: 0.00005/2.8750*100 = 0.0017391%
    % error for the height is: 0.00005/2.8950*100 = 0.0017271%

    Then I added them up. Total percentage error: 0.0051993%
    Multiply that by the volume and I get the possible error for the volume:

    Now the question is, how should I round up this possible error? Should I round it off to 5 sig. fig. (same as the volume), or should I round it off to 4 decimal places? I'm wondering, because if I round it off to 5 sig. fig., I'll have a ridiculously accurate possible error.
  2. jcsd
  3. Sep 18, 2011 #2


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    The volume you are giving runs to the 3rd decimal place. I would be giving the error rounded up to 3 decimal places.
  4. Sep 18, 2011 #3

    Actually, before I answer your original question, I'll point out that you made a small error in your last step. It appears that you accidentally multiplied the volume by 0.0051993, rather than 0.0051993% (or 0.000051993) to obtain the absolute error in the volume.

    I.e. you should get: 0.0012485.

    Building on PeterO's answer, in my personal experience, the general convention for the rounding of the absolute error of a quantity is to maintain decimal precision. That is, quote the same number of decimal places in the absolute error as are present in the quantity (assuming the quantity in question is quoted to the correct number of significant figures).

    I.e. in your case, since the volume V = 24.012cm^3 (which you've correctly quoted with 5 significant figures), one would then quote delta_V = 0.001 cm^3

    Notice that the absolute error contains only one significant figure. Typically it is decimal precision, not the number of significant figures, that matters when correctly stating the value of a measured quantity with it's absolute error.
    Last edited: Sep 18, 2011
  5. Sep 19, 2011 #4
    Thanks for the answers. I talked to my teacher about this today. He told me to use the same number of sig. fig., though I'll never attempt to calculate the error with percentages. I'll just use upper limits and lower limits. It saves a lot of time. My teacher says that sometimes using percentages is easier, but I guess I'll stick to upper and lower limits to save me the confusion... :D

    Thanks again.

    Oh wait, one more question.
    Let's say I have the length and width of a rectangle. I need its perimeter.
    Length: 0.50cm +- 0.05cm
    Width: 0.30 cm +- 0.05cm
    If I want the uncertainty for perimeter, do I add the uncertainty up to get 0.1 cm2 or should I add the uncertainty up and multiply it by two (0.2 cm2)?

    Therefore uncertainty = 0.05+0.05


    Therefore uncertainty = 0.05+0.05+0.05+0.05


    This is very annoying. I can't believe I'm racking my brains for the sake of significant figures... :uhh:
  6. Sep 19, 2011 #5
    I'd certainly listen to your teacher (since it is he who marks you) in this instance, but I have to say that I have rarely (if ever) seen uncertainties quoted to the same number of significant figures as the measured quantity.

    For example, 24.012 cm^3 ± 0.0012485 cm^3 looks very strange to me. I think my discomfort lies in fact that the value 24.012 has an intrinsic ambiguity of ±0.0005, which is far greater than that of the absolute error. Mind you, I'm not sure if it's reason or merely habit that motivates my argument.

    As for your second question, uncertainties add (notice the italics, multiplication/division is a different story) in much the same way as measurements do. Observe:

    l ± Δl + l ± Δl + w ± Δw + w ± Δw
    = 2*(0.50 ± 0.05) + 2*(0.30 ± 0.05)
    = 1.0 ± 0.10 + 0.60 ± 0.10
    = 1.6 ± 0.20
    = p ± Δp
  7. Sep 20, 2011 #6
    Thanks for the response. I'll ask him again just to make sure.

    As for the second question, I really shouldn't have asked it at all. It was a really lame question. I was considering it before sleep and realized that I've messed up sig. figs and uncertainty. Sorry for looking like a fool. :smile:
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