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Error calculation

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Two Sides of a triangular plate are measured as 125mm and 160mm, each to the nearest millimetre. The included angle is quoted as 60+-1. Calculate the length of the remaining side and the maximum possible error in this result

    My first question is to the nearest millimetre? What does this mean, i think +-0.5 but when i think about this if it was 0.5 biggest it would be rounded up..

    2. Relevant equations
    δu = Fxδx + Fyδy + Fzδz


    3. The attempt at a solution
    So firstly to find the length of the other side i used the cosine rule
    a^2 = b^2 + c^2 - 2bccos(A)
    Where b = 125
    c = 160
    A = pi/3
    Which comes out as 145.7 which is correct.
    The next stage i guess is to take partial derivatives.
    My equation for a is a^2 so i assume i have to square root it so i can find δa im not sure but i guess i have to??
    δa = partial with respect to b * δb + partial with respect to c *δc + partial with respect to A δA

    Therefore a = (b^2 + c^2 - 2bcCos(A))^1/2
    Okay this is where it gets messy.
    Partial with respect to b:
    [1/2 * 2b-2cCosA/(b^2+c^2-2bcCos(A))^1/2] *δb

    Partial with respect to C
    [1/2 * 2c - 2bcos(A)/((b^2+c^2-2bcCos(A))^1/2) ]*δc

    Partial with respect to A
    [2bcsinA/((b^2+c^2-2bcCos(A))^1/2)] *δA

    When i input values of δb and δc = +-0.5 and δA = +-1 i don't seem to get the right answer which is +- 2.6, i really don't see where I'm going wrong, as i did the similar method with a previous exercise.
     
  2. jcsd
  3. Sep 27, 2013 #2

    mfb

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    Right.
    True, but 0.4999999 (add as many finite 9 as you like) would be rounded down, the smallest upper bound is 0.5.

    You can calculate δa^2 as intermediate result if you like.

    I think there are some brackets missing in your derivatives.
    Which answer do you get?
     
    Last edited: Sep 28, 2013
  4. Sep 27, 2013 #3
    My answers are
    Partial b = [((2b-2cCosA)/2*(b^2+c^2-2bcCos(A))^1/2] *δb
    = [((250 - 160)/2*((125^2 + 160^2 - 2*160*125*0.5)^0.5))] * 0.5
    = (90/291.4) * 0.5 = +-0.154
    partial c has the same denominator
    [((2c - 2bcos(A)) / 291.4) ]*0.5
    = 195/291.4 * 0.5 = +-0.335
    Partial A has the same denominator aswell
    = (125*250*2*sin(pi/3)) / 291.4
    Which is a really large number, not sure whats going wrong there..

    The correct answer is 2.6, i don't seem to be close to that at present, i guess there must be a simple mistake i'm making.
     
  5. Sep 27, 2013 #4

    vela

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    Did you convert ##\delta A## to radians as well?
     
  6. Sep 27, 2013 #5
    Nope completely forgot about that, thanks
    Well that leaves me with
    [(125*250*2*sin(pi/3) / 291.4 ] *pi/180 = 3.24 which is still off, i cant see where i've made a mistake, i've done problems after this one and been fine, i just can't seem to get this one.
     
  7. Sep 27, 2013 #6

    vela

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    The 250 in the numerator should be 160, no?

    Also, I get a slightly different expression for the partial derivative wrt A than you do.
     
  8. Sep 28, 2013 #7
    δa = [itex]\frac{\partial a}{\partial b}[/itex] δb + [itex]\frac{\partial a}{\partial c}[/itex] δc + [itex]\frac{\partial a}{\partial A}[/itex] *δA

    δb = 0.5 δc = 0.5 δA = pi/180

    [itex]\frac{\partial a}{\partial b}[/itex] = [itex]\frac{b-cCos(A)}{\sqrt{b^{2}+c^{2} - 2bccos(A)}}[/itex]
    [itex]\frac{\partial a}{\partial b}[/itex] = [itex]\frac{125-160*Cos(\frac{pi}{3})}{\sqrt{125^{2}+160^{2} - 2bccos(\frac{pi}{3})}}[/itex] = [itex]\frac{45}{145.688}[/itex]
    [itex]\frac{\partial a}{\partial b}[/itex] = 0.309
    [itex]\frac{\partial a}{\partial b}[/itex] * δb = 0.1545

    [itex]\frac{\partial a}{\partial c}[/itex] = [itex]\frac{c-bCos(A)}{\sqrt{b^{2}+c^{2} - 2bccos(A)}}[/itex]
    [itex]\frac{\partial a}{\partial c}[/itex] = [itex]\frac{160-125*Cos(\frac{pi}{3})}{\sqrt{125^{2}+160^{2} - 2bccos(\frac{pi}{3})}}[/itex] = [itex]\frac{97.5}{145.688}[/itex]
    [itex]\frac{\partial a}{\partial c}[/itex] = 0.669
    [itex]\frac{\partial a}{\partial c}[/itex] * δc = 0.335

    [itex]\frac{\partial a}{\partial A}[/itex] = [itex]\frac{cbsin(A)}{\sqrt{b^{2}+c^{2} - 2bccos(A)}}[/itex]
    [itex]\frac{\partial a}{\partial A}[/itex] = [itex]\frac{160*125*sin(\frac{pi}{3})}{\sqrt{125^{2}+160^{2} - 2bccos(\frac{pi}{3})}}[/itex] = [itex]\frac{365.52}{145.688}[/itex]
    [itex]\frac{\partial a}{\partial A}[/itex] = 2.509
    [itex]\frac{\partial a}{\partial A}[/itex] * δA = 0.044

    I know for a fact these won't add up to 2.6, so i very confused at which stage im going wrong.
     
  9. Sep 28, 2013 #8

    mfb

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    160*125*sin(pi/3) is not 365.52. Make sure that you use radians for the sine, not degrees.
     
  10. Sep 28, 2013 #9
    oo what a silly mistake
    [itex]\frac{\partial a}{\partial A}[/itex] = [itex]\frac{160*125*sin(\frac{pi}{3})}{\sqrt{125^{2}+160^{2} - 2bccos(\frac{pi}{3})}}[/itex] = [itex]\frac{17320.51}{145.688}[/itex]
    [itex]\frac{\partial a}{\partial A}[/itex] = 118.888
    [itex]\frac{\partial a}{\partial A}[/itex] * δA = 2.075

    2.075 + 0.1545 + 0.335 = δa
    δa = 2.56 → 2.6 which is the correct answer yey, thanks very much.
     
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