1.

The total correction for elevation temperature and gradient for a runway, should NOT be more than:1. 35%2. 25%3. 15%4. 10%

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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