1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Estimating the maximum possible percentage error

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi all! Im new to the forum, came across it from googling the problem which I had and someone was having difficuilty with a question like this on here! Here's the question I have...

    The coefficient of rigidity, n, of a wire of length L and uniform diameter d is given by


    If A is a constant and the parameters L and d are known to within ± 2% of their correct values, estimate the maximum possible percentage error in the calculated value of n.

    2. Relevant equations


    3. The attempt at a solution

    dn = dn/dL x dL + dn/dA x dA

    dn = A/d^4 dL + L/d^4 dA

    n = A/d^4 x ΔL + L/d^4 x ΔA

    n = ΔL/L + ΔA/A

    Where do I go from here? Any help will be greatly appreciated :)
  2. jcsd
  3. Aug 10, 2010 #2


    Staff: Mentor

    Welcome, benji. You have come to a good place for help.
    A is a constant, so dn/dA doesn't make any sense. Instead, you want dn/dd. For the record, dn/dL and dn/dd are really the partial derivatives of n. I.e.,
    [tex]\frac{\partial n}{\partial L} \text{and} \frac{\partial n}{\partial d}[/tex]
    For the percent change in n, you want dn/n.
  4. Aug 10, 2010 #3
    Thanks for the reply Mark :) and correcting me with the dn/dA and dn/dd situation!

    I was wondering, how do I get the estimation when I have no values to work from for L and d?
  5. Aug 10, 2010 #4


    Staff: Mentor

    You don't have L and d, but you have their relative errors: ΔL/L and Δd/d.
  6. Aug 10, 2010 #5
    What do I do with the relative error? As "Δd" will be 2% what will I use for "d" ?
  7. Aug 10, 2010 #6


    Staff: Mentor

    No, Δd is not a percent value; it is the error in determining d. The relative error, which is what you are given, is Δd/d, and that is .02. Actually, what is given is that |Δd|/d <= .02. Same with L and ΔL.
  8. Aug 11, 2010 #7
    So would it be...

    % change in n = % change in L - (4 times % change in d)
    % change in n = -2 - (4 times -2)
    % change in n = -2 +8
    % change in n = + 6%

    Therefore, ±2% change in L and d = ±6% change in n?
  9. Aug 11, 2010 #8


    Staff: Mentor

    No. Where are you getting the -2 numbers?

    You started off in the right direction, but didn't fix the errors I pointed out earlier, so let's start from the beginning.

    n = AL/d4
    [tex]dn = A(\frac{\partial n}{\partial L}~dL + \frac{\partial n}{\partial d}~dd)[/tex]

    The percent change in n is Δn/n. Fill in the partial derivatives in the equation above and divide both sides by n.
  10. Aug 11, 2010 #9
    There is a more elementary way.

    n \propto L

    means that n increases with increasing L, while

    n \propto d^{-4}

    means that n decreases with increasing d.

    Therefore, the smallest value for n is obtained by taking the smallest value for L and the largest value for d:

    n_{\mathrm{min}} = A \, \frac{L_{\mathrm{min}}}{d^{4}_{\mathrm{max}}}

    The largest value for n, on the other hand, is obtained by taking the largest value for L and the smallest value for d:

    n_{\mathrm{max}} = A \, \frac{L_{\mathrm{max}}}{d^{4}_{\mathrm{min}}}

    In this way, you obtain an interval for the possible values of n:

    n \in [n_{\mathrm{min}}, n_{\mathrm{max}}]

    Instead of the interval notation, one usually uses the "techincal notation":

    n = \bar{n} \pm \Delta n

    which actually means:

    n_{\mathrm{min}} = \bar{n} - \Delta n \\

    n_{\mathrm{max}} = \bar{n} + \Delta n
    \end{array}\right. \Leftrightarrow \left\{\begin{array}{l}
    \bar{n} = \frac{1}{2} \, (n_{\mathrm{min}} + n_{\mathrm{max}}) \\

    \bar{n} = \frac{1}{2} \, (n_{\mathrm{max}} - n_{\mathrm{min}}) \\

    Then, of course, the relative uncertainty, expressed in percent, is defined as:

    \delta_{n} \equiv \frac{\Delta n}{\bar{n}} \cdot 100\%

    It is up to you to:
    1. Find Lmin and Lmax by knowing [itex]\bar{L}[/itex] (the nominal value) and [itex]\Delta L = \delta_{L}/{100 \%} \, \bar{L}[/itex] (the absolute uncertainty);

    2. Do the same for dmin and dmax;

    3. Find nmin and nmax according to the above formulas;

    4. Find [itex]\bar{n}[/itex] (nominal value) and [itex]\Delta n[/itex] (absolute uncertainty) according to the above formulas;

    5. Find the relative uncertainty [itex]\delta_{n}[/itex].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook