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Sound attenuation barrier height (trigonometry)

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img543.imageshack.us/img543/4598/photosfr.jpg [Broken]

    Find dss, dsr and thus hs.

    2. Relevant equations

    trig

    3. The attempt at a solution
    dss = hs/(sin[θ])
    dsr = hs/(sin[∅])

    dss = 3.059/(cos[θ])
    dsr = 47.191/(cos[∅])

    θ = arctan(hs/3.059)
    ∅ = arctan(hs/47.191)

    I have tried many times to solve this; usually by substituting the last two equations into the first 4 and trying to solve but without success.

    The problem is to do with ISO9613 where Z (path difference between dss+dsr and d) = 7.81 m and d = 50.25 m.

    http://www.cevreselgurultu.cevreorm...ps/assessment_methods/industry_ISO_9613-2.pdf (page 21-25)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 9, 2012 #2
    http://img543.imageshack.us/img543/4598/photosfr.jpg [Broken]

    OP's image.
     
    Last edited by a moderator: May 6, 2017
  4. Nov 9, 2012 #3

    haruspex

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    Let d comprise d1 (dss side) and d2. Write D = dss+dsr.
    D = √(d12+h2) + √(d22+h2)
    D2 = d12+2h2+d22+2√((d12+h2)(d22+h2))
    D2 - d12-2h2-d22=2√((d12+h2)(d22+h2))
    Squaring again leaves a quadratic in h2.
     
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