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Percentage errors

  1. Sep 17, 2012 #1
    How can you get positive and negative errors to be different in magnitude?

    for example -

    when calculating the error of the area of an annulus from two circular areas (ie subtracting one from the other, why is the positive error greater than the negative error

    Thanks
     
  2. jcsd
  3. Sep 17, 2012 #2
    because the denominators are different.....

    like, say, you have $100 invested and you lose $20...that's a $20 loss....
    Now you have $80...What percentage gain do you need to get your $20 back.....
    20/80 is 25%.

    figures don't lie, but liars figure!
     
  4. Sep 17, 2012 #3
    i understand the money analogy, but not when talking about the areas - sorry! :S we are only subtracting once
     
  5. Sep 17, 2012 #4

    mathman

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    Could you illustrate by an example what you are concerned about?
     
  6. Sep 17, 2012 #5

    mfb

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    Imagine a square where both sides are known with a precision of 10% - they might be 10% shorter or 10% longer, but not more. What is the maximal deviation?

    Larger area: Both sides 10% longer, total area 1.1^2 = 1.21 of the original area (21% more).
    Smaller area: Both sides 10% shorter, total area 0.9^2 = 0.81 of the original area (19% less).
    Do you see the difference?
     
  7. Sep 18, 2012 #6

    mathman

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    I see the difference, but why do you see this as a problem?
     
  8. Sep 18, 2012 #7
    The error depends on how the measurement was taken and what analysis was done. When you do math on the measurement, the errors will change shape. If you measure a circle's radius with a ruler with a symmetric error in the length, then the error in the area is not symmetric, because area goes as length squared.
     
  9. Sep 19, 2012 #8

    mfb

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    Where did I say that it is a problem?
    I just said that you can get asymmetric uncertainties in this way.
     
  10. Sep 21, 2012 #9
    ah great :) thanks :)
     
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