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% error over many, many measurments

  1. Feb 2, 2012 #1
    Wasn't exactly sure where I should put this question. It is from my phys class but is just about figuring error. This is for a lab that used a spark timer, tape and toy car to measure velocity/accel/distance.

    1. The problem statement, all variables and given/known data
    I have a set of calipers with +-0.025mm error. I used these calipers to measure about 180 positions on the spark tape ranging from 0.05mm to 18mm and a total of 1883.8mm. The question is how should I do the overall error? I realize that from one point to the next its the 0.025 but what about overall length?

    2. Relevant equations



    3. The attempt at a solution
    Simply adding the max error I find a +-5mm but this seems excessive, especially if I use a ruler to check the overall length.

    Thanks for help/information
     
  2. jcsd
  3. Feb 3, 2012 #2

    vela

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    Can you be a bit more specific about what quantities you measured and what quantity you're trying to calculate an error for? Equations would be good.
     
  4. Feb 3, 2012 #3
    sure, so the car takes off with the tape stuck to the top of it and as the tape is pulled a spark timer marks the tape at 60hz or 0.0167seconds. it looks like this:

    .. . . . . . . . . . . . . . . . . . . . . . . .. ..

    I need to measure the distance of each point from one another so I can find the velocity, acceleration and distance traveled.

    The issue is I measured from point to point with calipers that have an error of ±0.025mm. Given this my only prior experience tells me that I should measure each point and sum each to get my overall distance. Then add 0.025mm to each point and sum again. The difference between the (measured distance) and the (measured distance + error) would be my total ±error. That ends up being several millimeters off my measured value.

    P1+P2+Pn=Ptotal
    (P1+0.025)+(P2+0.025)+(Pn+0.025)=Ptotal+error
    Ptotal+error-Ptotal=±error

    Would this be the correct way?

    Thanks
     
  5. Feb 3, 2012 #4

    vela

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    Nope. For each measurement, sometimes the error is positive, and sometimes it is negative. The error for one measurement may partially offset the error for another. Your assumption that you overestimated on every single measurement is a highly unlikely outcome, so if you use that to calculate your error estimate, you'll be overestimating your error bounds.

    To calculate the error of a sum, what you want to do is square each error, sum them, and then take the square root of the total.
     
  6. Feb 3, 2012 #5
    Not 100% on what you're meaning.

    Are you saying that I should sum my 0.025mm then take the square root of it? Or are you saying squaring the (P1+0.025)+(P2+0.025)+(Pn+0.025)=Ptotal+error and taking the square root of that?

    The first way I get 0.34mm and the other is way high, 156mm. The first seems closer to reality but maybe low?
     
  7. Feb 3, 2012 #6

    vela

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    Here's an example with two variables:

    Suppose you have two measurements x1 = 3.0 ± 0.3 and x2 = 2.3 ± 0.4. Let y = x1+x2. You would calculate the error as follows:
    $$\Delta y = \sqrt{(0.3)^2 + (0.4)^2} = 0.5$$ so y = 5.3 ± 0.5.
     
  8. Feb 3, 2012 #7
    Okay, so the lower number is correct. My calculation went like:

    [itex]\sqrt{(0.025mm)(186)}[/itex] (186 is the number of measurements, 0.025 being the error from the calipers).
     
  9. Feb 3, 2012 #8

    vela

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    No, the units don't work out the way you did it.
     
  10. Feb 3, 2012 #9
    sorry, meant to square the 0.025 in there.

    so [itex]\sqrt{(0.025)^{2}(186)}[/itex]

    In excel I just summed the square of the error and took the square root. here its just simpler to show a multiplication.
     
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