TM 1-1510-224-10 reference zero equals the obstacle distance    from    brake    release    less    the accelerate-go   distance   to   50   feet   AGL (16,294  ft  5069  ft)  =  11,225  feet  =  1.85 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 5069 ft = 96.3 feet = 96 feet The total height required to clear the obstacle is: 88 ft + 96 = 184 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: 1.19%. 4. Read  the  scheduled  net  gradient  of  climb from  the  Net  Takeoff  Flight  Path  Second Segment  Flaps  Approach  graph  (fig.    7- 30):  2.53%. 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.................... 5069 feet (0.83 nm). 2. Read   the   scheduled   distance   from   the Horizontal  Distance  From  Reference  Zero To Third Segment Climb - Flaps Approach graph (fig.  7-31) .......... 5.07 nm. 3. Add  the  results  of  steps  1  and  2  to  obtain total   distance   to   start   of   third   segment climb, (0.83 nm + 5.07 nm) = 5.9 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 5.9) = 4.81 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  5069  ft  =  96.3  feet  =  96 feet The total height required to clear the obstacle is: 600 ft + 96 ft = 696 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 = (696 - 500) (0.0165) + 4.81 = 0.67% 7. Obtain  (from  the  Net  Take-Off  Flight  Path Third    Segment    One-Engine    Inoperative graph,    fig.        7-34)    the    scheduled    third segment  net  gradient  of  climb  of  2.33%. Since  this  gradient  exceeds  the  required gradient of 0.67%, 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) ............... 2250 ft/min Climb Gradient.............................................. 11.3% Climb - Two Engines (flaps approach) ..... 2068 ft/min Climb Gradient................................................... 3% 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                     386 ft/min Climb Gradient                           2.1%. 7-13