Design of Airports MCQ Quiz - Objective Question with Answer for Design of Airports - Download Free PDF

Explanation:

Maximum Longitudinal and Effective Gradients for 'A' Type Airport Runways

The International Civil Aviation Organization (ICAO) has set guidelines for the design and construction of airport runways. One of the important aspects to consider in runway design is the gradient, which affects the performance and safety of aircraft during landing and takeoff. The maximum allowable gradients help ensure that the aircraft can safely maneuver on the runway without excessive effort or risk.

The ICAO recommends that the maximum longitudinal gradient for 'A' type airport runways should be 1.50%. This ensures that the runway slope is not too steep for safe aircraft operations.

The maximum effective gradient, which considers the overall slope of the runway, should be 1.00%. This helps in maintaining a consistent and manageable slope for aircraft movement

Concept:

According to Horonjeff, the radius of a taxiway for a supersonic transport aircraft is given by:

\( R = \frac{0.388 \cdot W^2}{\left( \frac{T}{2} - S \right)} \)

Where:

W = Wheel base of aircraft = 30 m

T = Width of taxiway pavement = 22.5 m

S = Distance from midpoint of gear to pavement edge = \( 6 + \frac{\text{wheel tread}}{2} = 6 + \frac{6}{2} = 9 \, \text{m} \)

Step-by-step Calculation:

\( \frac{T}{2} = \frac{22.5}{2} = 11.25 \, \text{m} \)

\( \left( \frac{T}{2} - S \right) = 11.25 - 9 = 2.25 \, \text{m} \)

Now, \( R = \frac{0.388 \cdot 30^2}{2.25} = \frac{0.388 \cdot 900}{2.25} = \frac{349.2}{2.25} = 155.2 \, \text{m} \)

Concept:

According to ICAO, the basic runway length should be increased by 7% for every 300 m rise in elevation above mean sea level.

\(Correction = \frac{7}{100} \times \text{Runway Length} \times \frac{\text{Airport Elevation}}{300}\)

Given:

Runway length = 1840 m

Airport elevation = 250 m

Calculation:

\(Correction = \frac{7}{100} \times 1840 \times \frac{250}{300}\)

\(= 0.07 \times 1840 \times 0.8333 = 107.52 \, \text{m}\)

The turning radius is influenced by the speed of the aircraft (V) and the friction factor (f) between the aircraft tires and the taxiway surface. The formula provided (V²/125f) suggests a relationship where the turning radius increases with the square of the aircraft's velocity and decreases with higher friction. Understanding this relationship helps in designing taxiways that accommodate various aircraft speeds while ensuring safety during turns.

Explanation:

Basic Runway length:

It is the length of runway under the following assumed conditions at the aircraft

(i) Airport altitude at sea level.

(ii) Temperature at the airport is standard (15° C)

(iii) Runway is leveled in the longitudinal direction.

(iv) No wind is blowing on runway.

(v) Aircraft is loaded to its full loading capacity.

(vi) There is no wind blowing enroute to the destination.

(vii) Enroute temperature is standard.

Correction for Elevation, Temperature and Gradient:

(i) Correction for Elevation: Basic runway length is increased at the rate of 7% per 300 m rise in elevation above the mean sea level.

(ii) Correction for Temperature: 

Airport reference temperature = \({T_a} + \frac{{{T_m} - {T_a}}}{3}\)

Where, Ta = Monthly mean of average daily temperature

Tm = Monthly mean of the maximum daily temmperature for the same month of the year

Total correction for elevation plus temperature ⇒ 35% of basic runway length

(c) Correction for Gradient:

(i) Steeper gradient results in greater consumption of energy and as such longer length of runway is required to attain the desired ground speed.

(ii) After having been corrected for elevation and temperature should be further increased at the rate of 20% for every 1% of effective gradient.

Concept:

Airport reference temp. (T_r) \(= {T_a} + \frac{1}{3}\left( {{T_m} - {T_a}} \right)\)

T_a = Avg. temp of the hottest month

T_m = Monthly mean of maximum daily temp. of same month.

Calculation:

T_A = 25°C

T_m = 40°C

\({T_\pi } = {T_a} + \frac{1}{3}\left( {{T_m} - {T_a}} \right)\)

\(= 25 + \frac{1}{3}\left( {40 - 25} \right)\)

= 25 + 5 = 30°C

Explanation:

Effective length of runway:  

For takeoff, effective length of the runway means the distance from the end of the runway at which the takeoff is started to a point at which the obstruction clearance plane associated with the other end of the runway intersects the runway centerline. 

Additional Information

Basic Runway Length:

• It refers to the length of an airport runway under the following assumptions:​
  • ​No wind is blowing on runway;
  • Runway is levelled (No effective gradient);
  • Airport is at sea level and Standard Temperature at the airport is 15°C;
  • Aircraft is loaded to its capacity;
  • No wind is blowing on the way to destination Standard temperature prevails along the way.

Factors affecting basic runway length
Aircraft characteristics	Power and propulsion system Type of an aircraft
Safety requirements	Normal landing Normal take-off case Engine failure case while take-off
Airport environment	Atmosphere Temperature wind Location and condition of the runway Altitude Runway gradient

Important point:

Basic runway length is calculated with the following assumptions:

i) It is calculated at Mean sea level (MSL)

ii) It is calculated for standard temp 15°C at MSL.

iii) The gradient is assumed to be zero.

Due to the variance of these assumptions, some corrections are applied.

i) Elevation Correction: 7% increase in basic runway length for every 300 m rise above MSL

ii) Temperature correction: 1% increase for every 1°C rise in airport reference temperature with respect to standard temperature at elevation.

Where,

Standard temperature at elevation = Temperature at MSL – 0.0065 × Elevation

This increase is made on the already corrected runway length for elevation.

iii) Gradient correction: 20% increase for 1% of effective gradient.

Explanation:

Taxiway:

• Taxiways as defined by ICAO are a defined path on a land aerodrome established for the taxiing of aircraft and intended to provide a link between one part of the aerodrome and another in an expeditious manner.

Design criteria for taxiway:

1) Longitudinal gradient as per ICAO

Type or Class of airport	Permissible gradient
A/B	1.5 %
C/D/E	3 %

2) Transverse gradient as per ICAO

Type/Class of airport	Permissible gradient
A/B/C	1.5 %
D/E	2 %

3) Rate of change of longitudinal gradient as per ICAO

Rate of change	Class of airport
1% / 30 m of vertical curve	A/B/C
1.2% / 30 m of vertical curve	D/E

Concept:

Airport reference temperature (ART) \(= {T_a} + \frac{{{T_m} - {T_a}}}{3}\)

Where,

Ta = monthly mean of the average daily temperature for the hottest month of the year

Tm = monthly mean of the maximum daily temperature for the same month

Calculation:

Ta =  30°C

Tm = 45°C

\(ART = 30 + \frac{{45 - 30}}{3}\)

∴ ART = 35 ° C

Explanation:

Correction for basic Runway length (I)

1) 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.

2) Temperature correction

Airport reference Temperature (ART)

\(T_a + {1\over3}(T_m-T_a)\)

Tm = Monthly mean of the maximum daily temperature of the hottest month.

Ta = Monthly mean of the average daily temperature of the hottest month

Rise in temperature = ART - SAT

Where,

SAT is Standard Atmospheric Temperature

NOTE: Due to the elevation, there is a decrease in standard temperature at the rate of -6.50 c per 1000 m rise.

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

Note: As per ICAO, the cumulative connection for elevation and temperature Cumulative (%) correction together should not be ≯35%

If exceeds modify, I2 = 1.35 x I

3) Gradient correction

For every 1% effective gradient, runway length will be increased by 20%

\(Gradient = {Elevation \space difference\over Original \space length \space of \space runway}\)

Concept:

Correction for basic Runway length (l)

i) Elevation correction

For every 300-meter rise above MSL, the length will be increased by 7%

Corrected length \(\Rightarrow {{\rm{l}}_1} = {\rm{l}} + \left( {\frac{{\rm{x}}}{{300}} \times \frac{7}{{100}}} \right) \times {\rm{l}}\)

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 average daily temp of hottest month

For energy 1°C rise of ART above (SATx i.e. Standard Airport Temperature) length will be increased by 1%

Corrected length \( \Rightarrow {{\rm{l}}_2} = {{\rm{l}}_1} + \left( {\frac{{{\rm{\Delta T}}}}{{100}}} \right){{\rm{l}}_1}\)

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%

\({\rm{Final\;length\;}}\left( {{{\rm{l}}_3}} \right) = {{\rm{l}}_2} + \left( {20 \times \frac{{{{\rm{G}}_{{\rm{effective}}}}}}{{100}}{\rm{\% }}} \right){{\rm{l}}_2}\)

\({{\rm{G}}_{{\rm{effective}}}}\left( {\rm{\% }} \right) = \frac{{{\rm{R}}{{\rm{L}}_{{\rm{highest\;point}}}} - {\rm{R}}{{\rm{L}}_{{\rm{lower\;point}}}}}}{{{\rm{Runway\;unit\;}}\left( {{{\rm{l}}_2}} \right)}}\).

Explanation

As the airport is at MSL and the airport temperature is the same, the elevation and temperature correction is zero.

Along the runway the gradient is different, Determining the entire gradient correction 

Gradient correction:

Effective rise or fall

\( = \frac{1.2}{{100}} \times 300 - \frac{{0.7}}{{100}} \times \left( {600 - 300} \right) + \frac{{0.6}}{{100}} \times \left( {1100 - 600} \right) - \frac{0.8}{{100}} \times \left( {1400 - 1100} \right) - \frac{{1.0}}{{100}} \times \left( {1700 - 1400} \right)\)

 = 3.6 - 2.1 + 3 - 2.4 - 3

= - 0.9

 ∴ Cumulative rise/fall is - 0.9 m 

End – to – end runway (m)	Gradient (%)	Rise/Fall	Cumulative
0 to 300	+ 1.2	+ 3.6	+ 3.6
300 to 600	- 0.7	- 2.1	+ 1.5
600 to 1100	+ 0.6	3	+ 4.5
1100 to 1400	- 0.8	- 2.4	+ 2.1
1400 to 1700	- 1.0	-3	- 0.9

 

\(Effective\;gradient = \frac{{Maximum\;Difference\;in\;level}}{{Total\;Length}}\)

\( = \frac{{4.5 - \left( { - 0.9} \right)}}{{1700}} \times 100 = 0.3176\;\% \)

∴   The effective gradient is 0.3176 %

Explanation:

Basic runway length is based on some assumptions:-

a) Derived at Mean Sea level

b) Derived for a standard temperature of 15°C

c) The gradient must be zero.

The actual runway need not follow all these guidelines and hence we must apply some corrections.

The sequential correction is applied in the basic runway length.

The first correction is altitude correction, the second correction is temperature correction and the last correction is gradient correction.

Altitude Correction:

Basic runway length is increased by 7% for every 300 m rise above mean sea level.

This is done due to the decrease in drag force on the aircraft owing to the decrease in density of air as the altitude increases.

Given:

Height above MSL = 300 m, and Basic Runway length = 1500 m

After applying elevation correction,

\({\rm{Correction}} = 1500 \times \frac{7}{{100}} \times \frac{{300}}{{300}} = 105{\rm{\;m}}\)

Corrected Runway Length = 1500 + 105 = 1605 m

Important Points

Temperature correction and gradient corrections are applied on the subsequently corrected runway length. 

Explanation:

Correction for elevation: 7% increase per 300 m

So, correction \(= \frac{7}{{100}} \times \frac{{535}}{{300}} \times 2000\)

= 249.66 m

Corrected length = 2000 + 249.66 = 2249.66 m

Correction for temperature:

Standard atmospheric temperature = 15 – 0.0065 × 535 = 11.5225°C

Rise of temp = 22.65°C – 11.523°C = 11.127°C

\(Correction = \frac{{2249.66}}{{100}} \times 11.127 = 250.320\;m\)

Correct length = 2249.66 + 250.320 = 2499.98 m

Check for total correction for elevation plus temperature

\(Total\;correction\;\% = \frac{{2499.98 - 2000}}{{2000}} \times 100 = 24.99\%\)

According to ICAO, this should not exceed by 35%.