Unit 5: Airport Runway Length

Unit 5: Airport

Runway Length

Syllabus

Airport planning and Airport layout-, Runway orientation, Wind Rose diagram, Basic runway length, corrections for runway length, Airport classification, Geometric design.

Airport capacity, Aircraft parking systems.2018

Airport Classification

ICAO Airport ClassificationAirport

TypeBasic Runway

Length (m)Width of Runway

Pavement (m)

Max Longitudinal

Grade (%)

Max Min

A >2100 2100 45 1.5

B 2099 1500 45 1.5

C 1499 900 30 1.5

D 899 750 22.5 2.0

E 749 600 18 2.0

Code No

Equivalent Single Wheel

Load (kg)

Tire Pressure (kg/cm2)

1 45000 8.5

2 34000 7.0

3 27000 7.0

4 20000 7.0

5 13000 6.0

6 7000 5.0

7 2000 2.5

Example:

An airport B-3 would have basic runway length

ranging between 1500-2099m. Single wheel load

capacity of 27000 with a tire pressure of 7 kg/cm2

Runways

Definition

It is a strip of land used by aircrafts fortake-off and landing operations. It isperhaps the single most important facilityon the airport.

Dr. Mohd Ahmadullah Farooqi

Actual Runway Length =

Basic Runway Length + Corrections

Basic Runway Length (ICAO)

Airport Type

Basic Runway Length (m) Width of Runway

Pavement

Max Longitudinal Grade (%)

Max Min

A >2100 2100 45 1.5

B 2099 1500 45 1.5

C 1499 900 30 1.5

D 899 750 22.5 2.0

E 749 600 18 2.0

Standard Atmospheric Parameters:

• Temperature at MSL = 15C

• Pressure at MSL – 760mm of Hg

• Air Density = 1.225kg/m3

If the standard atmospheric conditions vary dueto any reason - corrections are applied to thebasic runway length to calculate the actualrunway length.


Corrections to basic runway length

There are three main corrections to be appliedto basic runway length to determine the actuallength of runway for an airport. These are:

• Elevation Correction

• Temperature Correction

• Gradient Correction


Elevation Correction

Change in elevation affects air density, atmospheric pressure and temperature. Correction should be applied for change in altitude.

The Elevation Correction is as shown below:

Correction for Altitude: Increase runway length by 7% per 300m altitude above MSL


Temperature Correction

If standard temperature varies, correction to runway length should be applied:


1. Compute Airport Reference Temperature (ART)

2. Compute Standard Temperature at the given Elevation (STE)

3. Compute Increase in ART above STE= ART- STE

4. Apply Correction based on the value obtained in Step-3

Airport Reference Temperature (ART)

ART = 𝑇1 + 1/3(𝑇2 − 𝑇1)

Where,𝑇1 = Monthly mean of average daily temperature for the hottest month of the year (C)

𝑇2 = Monthly mean of maximum daily temperature for the same month (C)


Standard Temperature at Elevation (STE)

STE = Temperature at MSL +/- (rate of change of temperature x elevation)

Rate of change of temperature with height is given as:

-6.5C / km height ---------- Up to 11km height from MSL

Constant at -56.5C --------- 11-20km height (Stratosphere)

+1C / km height ---------- 20-32km height (Troposphere)



Increase basic runway length by 1% for every 1C

rise in Airport Reference Temperature (ART).


Gradient Correction

Longitudinal Gradient:

• If the gradient is steep, it may cause pre-mature lift-off or may cause structural damage

• It will consume more energy and will need longer runway to attain desired ground speed


Effective Longitudinal Gradient: It refers of the average gradient computed by subtracting maximum and minimum elevations along the runway divided by the total length of runway.

Gradient Correction

Runway length is increased at a rate of

20% for every 1% of the effective

gradient

Note:

This correction is applied only if the combined

correction for Elevation and Temperature

remains less than 35%


Summary: Basic Runway Length Corrections

Correction Amount Combined Corrections

1 Elevation Correction 7% per 300m rise above

MSL The combined

correction for Elevation

and Temperature

should NOT exceed

35%

2 Temperature

Correction

1% for every 1C rise in

airport reference

temperature.

3 Gradient Correction 20% for every 1% of the

effective gradient


Numerical Example-1Problem-1:Compute the airport reference temperature if the monthly mean of averagedaily temperature of the hottest month is 27.3C and monthly mean ofmaximum daily temperature for the same month is 43.2C.

Airport Reference Temperature (ART) = 𝑇1 +1

3(𝑇2 − 𝑇1)

Where, 𝑇1= Monthly mean of average daily temperature of the hottest month (C)𝑇2= Monthly mean of maximum daily temperature of the hottest month (C)

𝐴𝑅𝑇 = 27.3 +1

3(43.2 − 27.3)

= 27.3 + 5.3,

𝐴𝑅𝑇 = 32.6°𝐶

Solution


Numerical Example-2

Problem 2:If the airport is located at mean sea level (MSL), and the airport reference temperature is that calculated in Problem 1. Apply temperature correction to the runway length.

Solution

At MSL, the Standard Temperature at Elevation (STE) is given as 15C.

The difference in temperature = (ART-STE) = 32.6-15 = 17.6C

Let the runway length be L meters.

The temperature correction is applied to increase the runway length, L, by 1% for every degree rise in temperature. Hence,

The correction is: L ×1

100× 17.6°𝐶

= 0.176LThe corrected length = L + 0.176L = 1.176L


Given: Airport Reference Temperature : 32.6C.

Numerical Example-3

Problem 3:Compute the corrected length of runway for an airport located at an elevation of170m above MSL if the length of runway under standard conditions is 2100m. Theairport reference temperature is 32.6C and the runway has to be provided with aneffective gradient of 0.23 percent.

Solution:

Elevation Correction:

7

100× 2100 ×

170

300= 83.3m

Corrected length = 2100 + 83.3= 2183.3m


Numerical Example-3

Calculation of standard temperature at elevation of 170m above MSL (STE).= 15° − 0.0065 × 170 = 13.9°𝐶


Rise of temperature above ART = 32.6° − 13.9° = 18.71°C

Correction = 2183 ×1

100× 18.71 = 408.3m

Corrected Length = 2183.3 + 408.3 = 2591.6m

Check for the total correction:2591.6 − 2100

2100× 100 = 23.4% => OK

The combined correction (Elevation + Temperature) should remain below 35%


Numerical Example-3

Gradient Correction:

20

100× 2591.6 × 0.23 = 119m

Corrected Length = 2591.6 + 119 = 2710.81m

Rounding the above value to the nearest 10m, the corrected length of runway is 2720m


End to end Runway Length (m)

Gradient (%)

0 -200 +1

200-600 -1

600-1200 +0.8

1200-1600 +0.2

Numerical Example-4The Longitudinal section of the runway provides the following data

Calculate the Effective Gradient of the runway

Chainage Elevation

0 100

200 100 + 0.01 x 200 = 102.0

600 102.0 - 0.01 x 400 = 98.0

1200 98.0 + 0.008 x 600 = 102.8

1600 102.8 + 0.002 x 400 = 103.6

Max difference in elevation = 103.6 - 98 = 5.6mTotal Runway Length = 1600mEffective Gradient = 5.6/1600 x 100 =0.35

Basic Runway Length

Basic Runway Length

It refers to the length of an airport runway under the following assumptions:Related to runway:

No wind is blowing on runway

Runway is levelled (No effective gradient)

Related to Airport:

Airport is at sea level

The temperature at the airport is 15C (Standard Temperature)

Related to aircraft:

Aircraft is loaded to its capacity

Related to route to destination:

No wind is blowing on the way to destination

Standard temperature prevails along the way


Factors Affecting Basic Runway Length

The following factors affect the calculation of basic runway length:

• Aircraft characteristics

• Airport environmental conditions

• Safety requirements


Aircraft Characteristics

• Power and propulsion system

• Critical aircraft:The aircraft that requires longest runway length for landing and take-off operations. The length of runways for both the operations may be determined from the flight manual of aircraft performance.

• Gross landing and take-off weight of the aircraft

• Aerodynamic and mechanical characteristics


Airport Environment

• Atmosphere

• Temperature

• Surface wind

• Altitude

• Runway Gradient


Safety requirements

• Normal landing case

• Normal take-off case

• Engine Failure Case

Each of the above cases is discussed in the next slides


Normal Landing Case

Normal Landing Case

The aircraft should come to a halt within 60% of the landingdistance. The runway of full strength pavement is provided forthe entire landing distance

StopTouchdown

Point

Runway Threshold

15m

60% of the landing distance

Landing Distance

Full Strength Runway


Normal Landing Calculations

• Field Length (FL) = landing distance (LD)

• LD = Stopping distance (SD) / 0.60 = SD x 1.67

• Length of full strength runway = LD

StopTouchdown

Point

Runway Threshold

15m

60% of the landing distance

Landing Distance



Normal Take-off Case


• The take-off distance (TOD) must be equal to 115% of the actual distance the aircraft uses to reach a height of 10.5m

• TOD should be equal to 115% of the distance to reach a height of 10.5m.

10.5m

Distance to reach a height of 10.5m

Take-off Distance = 115% of the distance to reach 10.5m height


Lift-off distance (LOD)

115% of LOD Clearway > Half this distance

Clearway


Normal Take-off Calculations• Field Length (FL) = Full Strength (FS) runway + Clearway

(CW)• TOD = 1.15 x D10.5m

• CW = 0.5[TOD -1.15(LOD)]• Take-off Run (TOR) = TOD – CW• Length of full strength runway (FS) = TOR


10.5m

TOR=Distance to reach a height of 10.5m

Take-off Distance = 115% of the distance to reach 10.5m height


Lift-off distance (LOD)

115% of LOD Clearway > Half this distance

Clearway


So the runway should look as shown in Figure

Full Strength Runway Clearway

Min 150m

Normal Take-off Runway Composition

• It requires a clearway, as shown in figure below.• The width of clearway should not be less than

150m (500ft) • The clearway ground area should not have any

object protruding a plane inclined upwards at a slope of 1.25% from the end of runway.


Engine Failure Case

Engine Failure Case-Criterion

It is an emergency condition!

This condition applies when the aircraft is speeding up on the runway to take-off and pilots detect some problem in the engine(s):

Two Options exist:

Option 1. To abort the flight (This is permissible only if the speed of aircraft is below the designated speed (engine failure speed), or

Option 2. Proceed with the take-off and turn the aircraft back from the turning zone (This option applies if speed is > engine failure speed).

Option -1 is important from runway length design perspective: The runway should be adequately long to let the plane to de-accelerate and come to a safe halt without running beyond the runway.


…Engine Failure Case

10.5m

Accelerate-Stop Distance

Take-off Distance = Distance to reach 10.5m height


Lift-off distance (LOD) Clearway = Half this distance

Engine Failure

Stopway

Clearway

De-accelerate-Stop Distance


Stopping in Emergency: Calculations-1Engine Failure, take-off proceeded case

Field Length (FL) = FS + CWTOD = D10.5

CW = 0.5[TOD-LOD]TOR = TOD + CWLength of FS runway = TOR


10.5m





Engine Failure

Stopway

Clearway


Stopping in Emergency: Calculations-2

Engine Failure, take-off aborted case

FL = FS + SW

FL = Deaccelerate stop distance (DAS)


10.5m





Engine Failure

Stopway

Clearway


The Required Basic Runway length

Field Distance = max{TOD2, TOD3, DAS, LD}Full Strength Runway (FS) = max{TOR2, TOR3, LD}SW = DAS – max{ TOR2, TOR3, LD}CW = min{(FL - DAS), CL2, CL3}SWmin = 0CWmin = 0CW max = 300m


Clearway

Min 150m

Stopway


Airport Environment

• Atmosphere

• Temperature

• Surface wind

• Altitude

• Runway Gradient


Unit 5: Airport Runway Length

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Transcript of Unit 5: Airport Runway Length