By
Engr. Ghulam Hussain
Lecturer
Civil Engineering Department
The University Of Lahore
 Single
runway
 Parallel runways
 Open-V runways
 Intersecting runways
There are several important aircraft operational
characteristics to know about the aircraft mass.
 Aircraft mass expenditures are significant and thus
need to be accounted for in the air vehicle runway
length analysis.
 OEW = operating empty weight (or mass) is the
weight (or mass) of the aircraft without fuel and
payload (just the pilots and empty seats)
 MTOW = maximum takeoff operating weight (or mass)
- structurally the maximum demonstrated mass at
takeoff for safe flight
(excludes run-up fuel and includes OEW, fuel &
Payload). Longer trips requires more fuel with less
payload

MALW = maximum allowable landing weight (or
mass) is the maximum demonstrated landing
weight (or mass) to keep the landing gear intact
at maximum sink rate (vertical speed)
 MSPW = maximum structural payload weight (or
mass) is the maximum demonstrated payload to
be carried without stressing the aircraft fuselage
 MZFW = maximum zero fuel weight (or mass) is
the sum of the OEW and the MSPW
 MTW = maximum taxi weight (or mass) of the
maximum demonstrated weight (or mass) for
ground maneuvering. Usually slightly more than
MTOW (includes run-up fuel).

 DTW
= desired takeoff weight (or mass) is
the weight of the aircraft considering fuel
(includes reserve), payload and OEW to
complete a given stage length (trip distance)
DTW = PYL + OEW + FW (trip fuel +

reserve)
where:



PYL is the payload carried (passengers and cargo)
OEW is the operating empty weight
FW is the fuel weight to be carried (usually includes
reserve fuel)
 Selecting
the length of runway is perhaps the
important decision which must be made in
the planning of landing area.
 Length of runway mainly depends on





The type of aircraft
Its payload
Trip length (fuel weight)
Altitude and temperature at the airport
Safety regulations
 It
is the length of runway under the following
assumed condition at the airport:






Airport altitude at sea level
Temperature at airport is standard (150 c)
Runway is leveled in the longitudinal direction
No wind is blowing on runway
Runway surface is dry
Aircraft is loaded at its full loading capacity
 Corrections
to basic runway length
Aircraft Type
Takeoff (ft.)
Landing (ft.)
B747-200B
10,500
6,150
DC-10-30
10,490
5,960
Concorde
10,280
8,000
B727-200
10,080
4,800
A300 B4
8,740
5,590
B737-200
6,550
4,290
DC-9-50
7,880
4,680
F28-2000
5,490
3,540
F27-500
5,470
3,290
SD3-30
3,900
3,400
FS
= full strength pavement distance
 CL = clearway distance
 SW = stop way distance
 FL
= field length (FS+SW+CL)
 LOD = lift off distance
 TOR = takeoff run
 TOD
= takeoff distance
 LD
= landing distance
 SD
= stopping distance
 D35 = distance to clear an 11 m (35 ft.)
obstacle
 DAS = distance to accelerate and stop (ASD)

 Runway
in future discussion refers to full
strength pavement (FS) - Support the full
weight of the aircraft
 For turbine aircraft the regulation do not
requires FS for the entire Take-off Distance
(TOD), while for piston aircraft requires FS
for entire TOD
 Runway Field Length (FL) have three basic
components



Full strength pavement (FS)
Clearways (CL)
Stopways (SW)
 Clearway



(CL)
Rectangular area beyond the runway not less
than 500 ft. wide and not longer than 1000 ft.
Extends from end of runway with a slope not
exceeding 1.25 percent above which no object
protrude except for the threshold lights on two
sides of the runway (not higher than 26 inches)
Allowing aircraft to climb safely to clear an
imaginary 11 m (35 ft. obstacle)
 Stopway



(SW)
Area beyond runway, width not less than runway
Paved surface that allows aircraft to stop in
situation of abandoned takeoff (engine failure in
turbine powered is not very common)
Permit use of lesser strength pavement for
turbine powered, while for piston aircraft
requires FS for entire SW
 The
following cases are considered for
determining the basic runway length


Normal takeoff (all engines working fine)
Engine-out takeoff condition
Continued takeoff
 Aborted takeoff


Landing
 The
cases which works out the longest
runway length is finally adopted
 The
normal landing case requires that an air
craft should come to a stop within 60% of the
landing distance. The runway of full strength
pavement is provided for the entire landing
distance.
Stop
Runway
60% of landing distance
Landing distance
 The
landing distance should be 67% longer
than the demonstrated distance to stop an
aircraft
 The
normal take-off case requires a clearway
which is an area beyond the runway and is in
alignment with the centre line of runway.
 The width of clearway is not <150m and is
kept free form obstruction.
 The clearway ground area or any object on it
should not protrude a plane inclined upward
at a slope of 1.25% from the runway.
Normal Takeoff Case
clearway
Clearway ≤ ½ of this distance
10.5m height
Lift-off distance
115% of Lift-off distance
Distance to 10.5m height
115% of distance to 10.5m height ( take-off distance)
Longitudinal section
Plan
Normal Take-off Case
Clearway
Min150m
Runway
Normal Takeoff Case
CL is min. 500 ft. wide with
a grade less than 1.25 deg.
 Dictated

Continued takeoff sub case


by two scenarios:
Actual distance to clear an imaginary 11 m (35 ft)
obstacle D35 (with an engine-out)
Aborted or rejected takeoff sub case

Distance to accelerate and stop (DAS)
Engine Out Analysis
*
* Decision speed is the speed chosen by the aircraft captain in relation to the respective limitations
of the aircraft, the airline operator rules and procedures, runway characteristics and actual
meteorological conditions
Clearway ≤ ½ of this distance
Engine
Failure
Decelerated – stop distance
10.5m
height
Stop way
Lift-off distance
Clear way
Accelerated stop distance
Distance to 10.5m height ( take-of distance )
Longitudinal section
Runway
Plan
Engine Failure Case
Stop
way
Min150m
Clear way

Lengths for the critical aircraft:



Case 1: Normal take-off Case 2: Engine-failure take-off
Case 3: Eng. -failure aborted take-off Case 4: Landing
Final analysis
FL  max(TOD1 , TOD2 , DAS3 , LD4 )
FS  max(TOR1 , TOR2 , LD4 )
SW  DAS  max(TOR1 , TOR2 , LD4 ) SWmin  0
CL  min(FL  DAS4 , CL1,max , CL2,max )
CLmin  0

CLmax  1000 ft.
If both ends of runway are to be used, the field length
components (FS, SW and CL) must exist in each direction.
 Determine
the runway length requirements
for turbine powered aircraft. Following
aircraft performance characteristics are
observed:

Normal take off:



Engine failure:



Lift of distance
= 2460 m
Distance to 11 m height = 2730 m
Engine failure aborted take off:


Lift off distance
= 2100 m
Distance to 11 m height = 2400 m
Accelerate stop distance = 2850 m
Normal landing:

Stop distance
= 1500 m

For normal take off:
TOD1
= 1.15 (D351) = 1.15 X 2400 = 2760 m
 CL1
= 0.5[TOD1-1.15 (LOD1)] = 0.5[2760-1.15 X
2100]=172.5 m
 TOR1
= TOD1 – CL1 = 2760 – 172.5 = 2587.5 m


For engine failure:
TOD2
 CL2
 TOR2


For engine failure aborted take off:


= D35.2 = 2730 m
= 0.5[TOD2 – LOD2] = 0.5[2730 - 2460] = 135 m
= TOD2 – CL2 = 2730 – 135 = 2595 m
DAS
= 2850 m
For normal landing:
LD
 FL4

= 1500/0.667 = 2248 m
= FS4 = LD = 2248 m
 The

FL

FS

SW

CL
actual runway components are:
= max (TOD1,TOD2, DAS, LD)
= max (2760, 2730, 2850, 2500)
= 2850 m
= max (TOR1, TOR2, LD)
= max (2587.5, 2595, 2248)
= 2595 m
= DAS – max (TOR1, TOR2, LD)
= 2850 - max (2587.5, 2595, 2248)
= 2850 – 2595
= 255 m
= min [(FL – DAS), CL1, CL2)
= min [(2850 – 2850), 172.5, 135)
= min (0, 172.5, 135) = 0
 Declared
distances are the distances which
the airport owner declares are available and
suitable for:




Take-off run available (TORA)
Take-off distance available (TODA)
Accelerate stop distance available (ASDA)
Landing distance available (LDA)
 The
basic runway length is for mean sea level
Elevation having standard atmospheric
conditions.
 For any change in elevation, temperature
and gradient for actual site of construction,
necessary corrections are to be applied to
obtain the length of runway.
 The
air density reduces as the elevation increases,
this in turn reduces the lift on the wings of the
aircraft and the aircraft requires greater ground
speed before aircraft becomes airborne. To achieve
greater speed, longer length of runway is required.
 Higher altitude requires longer runway length
 ICAO recommends that basic runway length should
be increased at the rate of 7% per 300m rise in
elevation above MSL.

Increase the required runway length at a rate of
7% for each 300m (1000ft.) airport elevation
above MSL
 Elevation factor, Fe = 0.07*E + 1
 Where, E is airport elevation above MSL in units
of 300m (1000ft.)
 Higher
temperature requires longer runway
 Higher temperatures results in lower air
density, resulting in lower output of thrust
 Increase is not linear with temperature, rate
of increase higher at higher temperatures
 Standard temperature is 59oF (15oC) at sea
level (MSL)
 Increase in length is 0.42% to 0.65% per
degree Fahrenheit

The length corrected for elevation is to be further
increased at a rate of 1% for each degree centigrade by
which the airport Reference temperature exceeds the
Standard temperature at the elevation of airport site.

Reference temperature (T) is defined as:
 T2  T1 
T  T1  

 3 
T1 = Mean of mean Daily Temp. for the hottest month
of the year (Hottest month has the highest mean
daily temperature)
T2 = Mean Maximum Daily Temperature for the same
month


Standard Temperature at the airport site can be
determined by reducing the standard
temperature at MSL (15oC) at a rate of 6.5oC per
1000 m or 1.981oC per 1000 ft. rise in the airport
elevation above MSL.
Temperature correction factor Ft , is computed
through following Equations
Ft  0.01T  15  6.5 * E   1
Ft  0.01T  15  1.981* E   1
(in meters)
(in feet)
Where, E = Airport elevation in 1000 m above
MSL
 Uphill
gradient requires more runway length
than a down-ward gradient
 Increase and decrease in runway length is
linear with change in gradient
 Length increases by 7-10% for each 1 percent
increase in gradient (max gradient is 1.5%)
 Average uniform gradient: straight line
joining the ends of the runway (no point 5 ft.
above the average)
 Effective gradient: difference in elevation
between the highest and lowest points
divided by length of runway

The runway length having been corrected for
elevation & temperature be further increased at
a rate of 10% for each 1% of the runway Effective
Gradient (G). The Effective Gradient is the
maximum elevation differential of runway center
line divided by the total length of runway.
RLmax  RLmin
G
L
Fg  0.1* G  1
Where, RLmax and RLmin are the reduced levels of
highest and lowest points along the runway center
line.
 Greater
the head wind, shorter is the runway
 The direction of the wind also effects the
allowable take-off weight for the airplane
 A 5-kn headwind approximately reduces the
take-off length by 3 percent
 A 5-kn tailwind approximately increases the
take-off length by 7 percent.
 For planning, no wind is considered if light
wind occurs at the airport sight
 Presence
of water or slush (reduce braking
resistance)
 From 0.25 to 0.5 inch, take-off weight must
be substantially reduced to overcome the
retarding force of water and slush
 Velocity at which hydroplaning develops:

For tire pressure range = 120 – 200 psi, Vp ranges
from 110 to 140 mph (usual landing and take-off
speed)
Vp  10* (TirePressure)0.5


Find out the required length for the airport of reference code 4 D,
located at 450 m above mean sea level. The runway effective gradient is
0.5%. The monthly mean maximum and mean daily temperatures of the
hottest month of the year are 27oC and 18oC, respectively.
Solution:
 Field length = 1800 m
 Correction for elevation = 0.07 *450/300 +1 = 1.105
 Corrected runway length = 1800 * 1.105 = 1989 m
 Standard temperature at the airport site = 15- 6.5*450/1000 = 12.08C
 Airport reference temperature = 18 + (27-18)/3 = 21C
 Rise in temperature = 21 – 12.08 = 8.92C
 Correction for temperature = 0.01 * 8.92 + 1 = 1.0892
 Corrected length for temperature = 1989 * 1.0892 = 2166 m
 Correction for gradient = 0.01 * 0.05 + 1 = 1.05
 Final runway length = 2166 * 1.05 = 2275 m
…??