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The total correction for elevation temperature and gradient for a runway, should NOT be more than:1. 35%2. 25%3. 15%4. 10% |
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Answer» Correct Answer - Option 1 : 35% Concept: Correction for basic Runway length (l) i) Correction for elevation ICAO recommends that basic runway length should be increased at the rate of 7% per 300 m rise in elevation above mean sea level. ∴ Correction for elevation \( = \frac{7}{{100}} \times runway\;length \times \frac{{Airport\;elevation}}{{300}}\) ii) Temperature correction Airport reference Temperature (ART) \({\rm{T}} = {{\rm{T}}_{\rm{a}}} + \frac{{{{\rm{T}}_{\rm{m}}} - {{\rm{T}}_{\rm{a}}}}}{3}\) Tm → monthly mean of maximum daily temp of hottest month Ta → monthly mean of the average daily temp of the hottest month Rise in temperature = ART - SAT Where, ART is Atmospheric reference temperature SAT Standard atmospheric Temperature As per ICAO, Basic runway length after correction for elevation should be further increased at the rate of 1% for every 1° C rise of the airport reference temperature correction for temperature \(= Correctedlength \times \frac{1}{{100}} \times Rise\;in\;temperature\) Note: As per ICAO, the cumulative connection for elevation and temperature together should not be ≯ 35%. Cumulative (%) correction \(= \frac{{{{\rm{l}}_2} - {{\rm{l}}_1}}}{{\rm{l}}} \times 100 \not > 35\% \) If exceeds modify, l2 = 1.35 × l iii) Gradient correction For every 1% effective gradient, runway length will be increased by 20% \({\rm{Final\;length\;}}\left( {{{\rm{l}}_3}} \right) = {{\rm{l}}_2} + \left( {20 \times \frac{{{{\rm{G}}_{{\rm{effective}}}}}}{{100}}{\rm{\% }}} \right){{\rm{l}}_2}\) \({{\rm{G}}_{{\rm{effective}}}}\left( {\rm{\% }} \right) = \frac{{{\rm{R}}{{\rm{L}}_{{\rm{highest\;point}}}} - {\rm{R}}{{\rm{L}}_{{\rm{lower\;point}}}}}}{{{\rm{Runway\;unit\;}}\left( {{{\rm{l}}_2}} \right)}}\). |
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