Page 118 - Planning and Design of Airports

Air craft Characteristics Related to Airport Design 87

where SW is zero.
min
CL = min [(FL − DAS), (CL ), (CL )] (2-14)
1max 2max
where CL is zero and CL is 1000 ft.
min max
If operations are to take place on the runway in both directions, as
is the usual case, the field length components must exist in each direc-
tion. Example Problem 2-1 illustrates the application of these require-
ments for a hypothetical aircraft.
Example Problem 2-1 Determine the runway length requirements according to
the specifications of FAR 25 and FAR 121 for a turbine-powered aircraft with the
following performance characteristics:
Normal takeoff:
Liftoff distance = 7000 ft
Distance to height of 35 ft = 8000 ft
Engine failure:
Liftoff distance = 8200 ft
Distance to height of 35 ft = 9100 ft
Engine-failure aborted takeoff:
Accelerate-stop distance = 9500 ft
Normal landing:
Stop distance = 5000 ft
From Eq. (2-4) for a normal takeoff
TOD = 1.15 D35 = (1.15)(8000) = 9200 ft
1 1
CL = 0.50[TOD − 1.15(LOD )] = (0.50)[9200 − 1.15(7000)] = 575 ft
1max 1 1
TOR = TOD − CL = 9200 − 575 = 8625 ft
1 1 1max
From Eq. (2-5) for an engine-failure takeoff
TOD = D35 = 9100 ft
2 2
CL = 0.50(TOD − LOD ) = 0.50(9100 − 8200) = 450 ft
2max 2 2
TOR = TOD − CL = 9100 − 450 = 8650 ft
2 2 2max
From Eq. (2-6) for an engine-failure aborted takeoff DAS = 9500 ft
From Eq. (2-7) for a normal landing

LD = SD = 5000 = 8333 ft
.
.
060 060
Using the above quantities in Eqs. (2-8) through (2-11), the actual runway com-
ponent requirements become
FL = max [(TOD ), (TOD ), (DAS), (LD)]
1 2
= max [(9200), (9100), (9500), (8333)] = 9500 ft
FS = max [(TOR ), (TOR ), (LD)]
1 2
= max [(8625), (8650), (8333)] = 8650 ft