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Homework Help: Error propagation (tape shorter than measured length)

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data

    The problem is about calculating the error on the area of a rectangular field.
    What is known is that the sides of the rectangle are 120 m and 180 m, and they have been measured with a 10 m measuring tape. The tape has a sensitivity of 2 cm.

    2. Relevant equations

    Since the area is the product of the sides, one has to calculate its relative error to have its absolute error.
    The absolute error is the product of the relative error times the area.
    The relative error on the area is the sum of the relative errors on the sides.

    3. The attempt at a solution

    The whole point is what is the relative error on the rectangle sides. I'd say that this is the ratio of the sensitivity and the range of the measuring tape.
    For instance, the first side has been measured by summing 12 measures, each of which has a sensitivity of 2 cm = 0.02 m. The absolute error is 12⋅0.02 m = 0.24 m, and the relative error is 12⋅0.02/120=0.02/10=0.002.
    Likewise for the second side.

    So the relative error on the area is 0.004 and the corresponding absolute error is 0.004⋅120⋅180=86.4 ≈ 90

    I'm puzzled because the solution to the exercise says that the relative error is 0.002.

    Maybe I'm not getting the part where one uses a short measuring instrument to measure a longer distance.

    PS Actually the problem is about the price of the field, but this is obtained by multiplying the area by the price of a square meter. Since this is an exact number it won't affect the relative error, right?

  2. jcsd
  3. Dec 4, 2016 #2


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    Not sure what that means. Does that mean an error range of 2cm (so +/-1cm) or +/-2cm?
  4. Dec 5, 2016 #3
    Yes, that is it. In Italian "sensitivity" is the smallest difference that can be measured with a measuring instrument.
    This means that a measurement would be like (5,02±0,02)m.
    I did a quick web search and I had the impression that the same term was used in English. But maybe I just found poorly translated documents.
    Sorry about that, but I see your point... the solution would be ok in case the measure was (5,02±0,01)m.

  5. Dec 5, 2016 #4
    Still me,

    I quickly replied to haruspex this morning, before rushing to work. I still have some doubts.

    Is my line of reasoning right? That is,
    • I have a 10 m measuring tape, with an error range of 4 cm (there is a tick mark every 2cm),
    • I am measuring the side of a field, and I get 120 m (i.e. "exactly" 12 tapes)
    Am I correct in saying that the absolute error in my measurement is 12⋅2 cm = 24 cm, and hence the corresponding relative error is 24 cm/120 m = 0,002?

    Since this applies to both sides of the field, the relative error in the area (and in the corresponding price, assuming that makes sense) would be 0.004 and not 0.002.
    Does this make sense?

    Thanks a lot
  6. Dec 5, 2016 #5


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    Yes, I agree with you.
  7. Dec 5, 2016 #6


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    Ciao Franz,

    In a measurement there are usually several sources of errors, that can be roughly split in two groups:
    The instrument is calibrated with a certain accuracy
    The instrument is used with a certain accuracy

    If your 10 m tape is in fact 9.98 m long you have a systematic error of 0.2 % and the deviation is the same for all measurements, so the errors add up linearly to 0.4 % in the product of length and width.

    If you measure a length, you in fact measure the difference in position of two points: begin point and end point. In your case even 24 or 36 points, because you repeatedly drag the tape from end point to new beginning point. If you do that without bias and with an accuracy of 1 cm per point, you still have to add 24 (and 36) of these errors in quadrature. Not much, but not nothing either. And 1 cm outdoors isn't all that easy (ditches, walls, trees, uneven ground).

    That your tape has markings every 2 cm does not mean you cannot estimate fractions of 1 division: I don't think surveyors (even in the old days) do it that way, but if you place a stake and mark the begin point on the stake, then place another stake and mark the end point of the tape on that, you should be able to reasonably claim a reading to within one fifth of a division or better. That's just for the reading, it doesn't take away from the inaccuracy described in the preceding paragraph.

    I interpret this as: the measurement results are 120 and 180 m (the true dimensions are unknown).

    This is a form of erroneous reporting: if a measurement is accurate to within 0.2%, this should be reflected in the way it is reported, so here that would be 120.00 ##\pm## 0.24 m . That .00 is always suspicious in my paranoid brain.

    The reverse form -- frequenly occurring -- looks like: measuring and reporting 2.012693 m for the length of a table, where some digital device produces a lot of meaningless digits
  8. Dec 5, 2016 #7

    @ Aruspex: thanks a lot for your help.

    @ BvU: thanks for your insight also.

    I'm not sure one is really authorized to "guess" the measurement with more accuracy than allowed by the instrument, though.
    Rigorously one should choose the tick mark that's closer to the measure, and come to terms with the uncertainty, right?
    I think what you are suggesting is that the guy is actually (mentally) building a more precise instrument. That looks natural if the division is as large as 2cm., and makes sense at the practical level, but I think it gives rise to confusion if one is dealing with error propagation. For sure it's hard to do when the division is 1mm or less. But maybe we're saying the same thing here.

    About the problem: this is an exercise for people learning error propagation. I think that the idea was having them think about the "error" in their measure of the side of a large rectangle, if this is carried out with a "short" tape. It seems to me that you agree that the error is a suitable (obvious) multiple of the tick spacing of the tape.
    I agree with you that the measure should be then (120.00 ± 0.24) m. I'm using parentheses because the unit "belongs" to both the measure and the error, but that's a matter of notation I guess. Our textbook also gives a "rule of thumb" that usually an absolute error is rounded to one significant figure, so that would be (120.0 ± 0.2) m, if one is interested in giving the measure of just one side. The relative error would still be 0.002 to avoid rounding errors in further calculations.

    Anyway, my question was prompted by the fact that the solution to this exercise says that the relative error of the area is 0.002. I'd say it is 0.004.
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