TM 1-1510-224-10 1. The    obstacle    horizontal    distance    from reference zero equals the obstacle distance    from    brake    release    less    the accelerate-go   distance   to   50   feet   AGL (16,294  ft  5389  ft)  =  10,905  feet  =  1.79 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 5389 ft = 102.4 feet = 102 feet The total height required to clear the obstacle is:  88 ft + 102 = 190 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: 1.29%. 4. Read  the  scheduled  net  gradient  of  climb from  the  Net  Takeoff  Flight  Path  Second Segment  Flaps  Approach  graph  (fig.    7A- 30) 2.32%. 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 600 feet. Obstacle Distance from Brake Release ..............10.71 nm 1. Obtain   the   accelerate-go   distance   to   50 feet AGL..........................................................................  5389  feet  (0.89 nm). 2. Read   the   scheduled   distance   from   the Horizontal  Distance  From  Reference  Zero To  Third  Segment  Climb  Flaps  Approach graph (fig.  7A-31)   5.65 nm. 3. Add  the  results  of  steps  I  and  2  to  obtain total   distance   to   start   of   third   segment climb, (0.89 nm + 5.65 nm) = 6.54 nm. 4. Distance   to   obstacle   from   start   of   third segment  climb  is  obtained  by  subtracting results  of  step  3  from  10.71  nm.    (10.71 6.54) = 4.17 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 5389 ft = 102.4 feet = 102 feet The total height required to clear the obstacle is:  600 ft + 102 ft = 702 feet. 6. Required  climb  gradient  to  clear  obstacle is obtained using the following formula: % Gradient = (RH) x (F) / (D) Where: RH = Required Height (in feet) above 500 feet F = A units conversion factor of 0.0165 D = Distance (in nautical miles) to obstacle from start of third segment Therefore: % Gradient = (702 500) (0.0165) + 4.17 = 0.80% 7. Obtain  (from  the  Net  Take-Off  Flight  Path Third    Segment    One-Engine    Inoperative graph,   fig.      7A-34)   the   scheduled   third segment  net  gradient  of  climb  of  2.07%. Since  this  gradient  exceeds  the  required gradient  of  0.80,  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  -4C, 9000  feet  pressure  altitude,  15,500  pounds,  and  obtain the following results: Climb Two Engines (flaps up) .............. 2237 ft/min Climb Gradient........................................... 11.3% Climb Two Engines (flaps approach) .... 2053 ft/min Climb Gradient........................................... 10.7% I. Climb One Engine Inoperative.    Enter  the  graph at -4C, 9000 feet pressure altitude, 15,500 pounds, and obtain the following results: Climb One Engine Inoperative ............... 358 ft/min Climb Gradient............................................. 2.0% 7A-13