Sound attenuation barrier height (trigonometry)

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SUMMARY

The discussion focuses on calculating the height of a sound attenuation barrier using trigonometric principles and the ISO 9613 standard. The key equations involve determining distances dss and dsr based on the height hs and angles θ and ∅. The user attempts to solve the equations dss = hs/(sin[θ]) and dsr = hs/(sin[∅]), alongside the path difference Z = 7.81 m and total distance d = 50.25 m. The problem culminates in a quadratic equation derived from the relationship between the distances and heights.

PREREQUISITES
  • Understanding of trigonometric functions and their applications in geometry
  • Familiarity with the ISO 9613 standard for sound attenuation calculations
  • Ability to manipulate quadratic equations and solve for unknown variables
  • Basic knowledge of distance measurement in acoustics
NEXT STEPS
  • Study the derivation of the ISO 9613 equations for sound propagation
  • Learn how to apply trigonometric identities in real-world scenarios
  • Explore methods for solving quadratic equations effectively
  • Investigate the impact of barrier height on sound attenuation in various environments
USEFUL FOR

Acoustic engineers, students studying environmental science, and professionals involved in noise control and sound attenuation design will benefit from this discussion.

jhmz
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Homework Statement



http://img543.imageshack.us/img543/4598/photosfr.jpg

Find dss, dsr and thus hs.

Homework Equations



trig

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)
 
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http://img543.imageshack.us/img543/4598/photosfr.jpg

OP's image.
 
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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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