TM 55-1510-222-10
15°C FAT, 3499 feet pressure altitude, 16,000 pounds
takeoff weight, 1.9% downhill runway gradient, and a 10-
knot headwind component.
(1) Example 1 - Close-In Obstacle Clearance:
Given:
Obstacle Height Above Aircraft
at Brake Release .........................................184 feet
Obstacle Distance from Brake Release......... 15,437 feet
1.
The
obstacle
horizontal
distance
from
Reference
Zero
equals
the
obstacle
distance from brake release less the
accelerate-go distance to 50 feet AGL
(15,437 ft - 4500 ft) = 10,937 feet = 1.8
nautical miles.
2.
Determine the total height required to clear
the obstacle by adding to the obstacle
height the decrease in aircraft altitude
during the takeoff procedure due to a
downhill runway gradient.
1.9% gradient x 4500 ft = 85.5 ft = 86 feet
The total height required to clear the obstacle is: 184 ft
+ 86 ft = 270 feet
3.
Obtain the required gradient to clear the
obstacle from the DISTANT TAKEOFF
FLIGHT PATH graph using the obstacle
distance from Reference Zero found in step
1, and the total height determined in step 2:
2.0%.
4.
Read the scheduled net gradient of climb
from the NET TAKEOFF FLIGHT PATH -
SECOND SEGMENT - FLAPS APPROACH
graph: 2.4%.
Thus, the calculations indicate that a takeoff weight of
16,000 pounds will result in a net climb gradient greater
than that required to clear the obstacle, even if an
engine should fail at the most critical takeoff point.
(2) Example 2 - Obstacle Clearance above 500
Feet: Given:
Obstacle Height Above Aircraft
at Brake Release .........................................860 feet
Obstacle Distance from Brake Release................6.4 nm
1.
Obtain the takeoff distance to 50 feet
AGL............................. 4500 feet (0.74 nm)
2.
Read the scheduled distance from the
HORIZONTAL
DISTANCE
FROM
REFERENCE ZERO TO THIRD SEGMENT
CLIMB graph .....................................2.8 nm
3.
Add the results of steps 1 and 2 to obtain
total distance to start of third segment climb.
(0-74 nm + 2.8 nm) = 3.54 nm
4.
Distance to obstacle from start of third
segment climb is obtained by subtracting
results of step 3 from 6.4 nm. (6.4 - 3.54) =
2.86 nm
5.
Add to the obstacle height above the aircraft
at brake release any decrease in aircraft
altitude during the takeoff resulting from a
downhill runway gradient.
The sum is the total height required to clear
the obstacle:
(1.9% gradient + 100) x 4500 ft + 85.5 ft +
86 feet
The total height required to clear the
obstacle is: 860 ft + 86 ft = 946 feet
6.
Required climb gradient to clear obstacle is
obtained using the following formula:
% Gradient = (Required Height Above 500
feet x 0.0165) + (Distance to Obstacle from
Start of Third Segment in NM)
% Gradient = ((946 ft - 500 ft) x 0.0165) +
2.86 NM = 2.58 %
7.
Obtain the scheduled third segment net
gradient of climb of 3.0%. Since this
gradient exceeds the required gradient of
2.58%, the calculations indicate that the
obstacle will be cleared at a takeoff weight
of 16,000 pounds even if an engine should
fail at the most critical takeoff point.
k.
Climb - Two Engines. Enter the graphs at 15°C,
3499 feet pressure altitude, 16,000 pounds, and obtain
the following results:
Climb - Two Engines (FLAPS UP)................. 2310 ft/min
Climb - Two Engines (FLAPS APPROACH)..2240 ft/min
I.
Climb - One Engine Inoperative. Enter the
graph at 15°C, 3499 feet pressure altitude, 16,000
pounds, and obtain the following results:
Climb - One Engine Inoperative ...................... 520 ft/min
7-6
