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Method to usde to find percentage error

  1. May 8, 2007 #1
    Simple question, im not sure of the method to use to find percentage errors

    Example :

    Values in time = 2s Error in time = +-0.05

    Percentage Error = (0.05/2)*100

    +-2.5%?

    Should i be using relative errors?
     
  2. jcsd
  3. May 8, 2007 #2

    berkeman

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    Staff: Mentor

    Your answer looks correct to me. +-0.025/1 is +- 2.5%.
     
  4. May 8, 2007 #3
    so the method is correct? :)

    and i shouldnt be using relative errors?
     
  5. May 8, 2007 #4

    berkeman

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    What do you mean by relative error? The error would normally be relative to the quantity being measured, I would think. Is there a different kind of error that you are studying also?
     
  6. May 8, 2007 #5
    dont worry :P the example i used is what i originally thought was correct, just wanted confirmation :)

    edit*

    one quick question though, if i am using the equation s = ut + 0.5at^2

    and the % error in ut is 1% and the % error in 0.5at^2 is 3% then what would the overall error in s be?
     
    Last edited: May 8, 2007
  7. May 8, 2007 #6

    berkeman

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    It's usually safest to compute what the answer is without error, and then what it is with all the errors added in, and then take the ratio to see what your final errors are.

    So what is the nominal anwswer in your equation with no error?

    And what is the largest you can make the answer with errors included?

    And what is the smallest you can make the answer with errors included?

    Then your +- errors would be

    (1 - biggest/nominal) * 100%

    (1 - smallest/nominal) * 100%

    So in your question above, you need real numbers to figure out what the % errors are. You can't just add the percentages, because they can act on very different size numbers. Does that make sense?
     
  8. May 8, 2007 #7

    berkeman

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    BTW, what I said above applies when you are adding terms.

    Quiz Question -- Why is it different when you are multiplying terms? What is the total error for this:

    A = B * C

    when the error in B is +-2% and the error in C is +-3% ?
     
  9. May 8, 2007 #8
    i think its +-5% from what ive done before

    if im correct when u multiply approximations you add the % errors

    also from what u said before i assume my final error will lie in a range of the answer , s

    such as if s=30 then x<30<y
     
  10. May 8, 2007 #9

    berkeman

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    1 - (1.02) * (1.03) = ?

    1 - (0.98) * (0.97) = ?

    So it's close to adding, but not exact.

    Sorry, I'm not tracking what you are saying.
     
  11. May 8, 2007 #10

    so i can say that adding the % errors when multiply approximations is ok for small % errors?

    forget what i was saying before, was babbling on a bit :)
     
  12. May 9, 2007 #11
    *edit soz was being stupid
     
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