Figure 36 
(Refer to figure 36.) What is the expected fuel consumption for a 
1,000-nautical mile flight under the following conditions?  

Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . 8,000 ft
Temperature 
. . . . . . . . . . . . . . . . . . . . . . . . . 22°C
Manifold pressure 
. . . . . . . . . . . . . . . . . . . . . . . . . 20.8" Hg
Wind 
. . . . . . . . . . . . . . . . . . . . . . . . . Calm





   ANSWER: 70.1 gallons. 

   To determine the fuel consumption, 
   you need to know the number of hours the flight will last 
   and the gallons per hour the airplane will use. The chart is 
   divided into three sections. They differ based on air 
   temperature. Use the right section of the chart as the 
   temperature at 8,000 ft. is 22°C.

   At a pressure altitude of 8,000 ft., 20.8" Hg manifold 
   pressure, and 22°C, the fuel flow is 11.5 GPH and the true 
   airspeed is 164 kt. Given a calm wind, the 1,000-NM trip will 
   take 6.09 hr. (1,000 NM ÷ 164 kt).

         6.09 hr. x 11.5 GPH = 70.1 gal. 

Figure 36 
(Refer to figure 36.) What is the expected fuel consumption for a 
500-nautical mile flight under the following conditions?  

Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . 4,000 ft
Temperature 
. . . . . . . . . . . . . . . . . . . . . . . . . +29°C
Manifold pressure 
. . . . . . . . . . . . . . . . . . . . . . . . . 21.3" Hg
Wind 
. . . . . . . . . . . . . . . . . . . . . . . . . Calm





   ANSWER: 36.1 gallons. 

   At 4,000 ft., 21.3" Hg manifold 
   pressure, and 29°C (use the section on the right), the fuel 
   flow will be 11.5 GPH, and the true airspeed will be 159 kt. 
   The 500-NM trip will take 3.14 hr. (500 NM ÷ 159 kt).

         3.14 hr. x 11.5 GPH = 36.1 gal. 

Figure 36 
(Refer to figure 36.) Determine the approximate manifold pressure 
setting with 2,450 RPM to achieve 65 percent maximum continuous 
power at 6,500 feet with a temperature of 36°F higher than standard. 





   ANSWER: 21.0" Hg. 

   The part of the chart on the right is for 
   temperatures 36°F greater than standard. At 6,500 ft. with a 
   temperature of 36°F higher than standard, the required 
   manifold pressure change is 1/4 of the difference between 
   the 21.0" Hg at 6,000 ft. and the 20.8" Hg at 8,000 ft., or 
   slightly less than 21.0. Thus, 21.0 is the best answer given. 
   The manifold pressure is closer to 21.0 than 20.8. 

Figure 36 
(Refer to figure 36.) Approximately what true airspeed should a 
pilot expect with 65 percent maximum continuous power at 9,500 
feet with a temperature of 36°F below standard? 





   ANSWER: 183 MPH. 

   The left part of the chart applies to 
   36°F below standard. At 8,000 ft., TAS is 181 MPH. At 
   10,000 ft., TAS is 184 MPH. At 9,500 ft., with a temperature 
   36°F below standard, the expected true airspeed is 75% 
   above the 181 MPH at 8,000 ft. toward the 184 MPH at 10,000 
   ft., i.e., approximately 183 MPH. 

Figure 37 
(Refer to figure 37.) What is the crosswind component for a landing 
on Runway 18 if the tower reports the wind as 220° at 30 knots? 





   ANSWER: 19 knots. 

   The requirement is the crosswind 
   component, which is found on the horizontal axis of the 
   graph. You are given a 30-kt. wind speed (the wind speed is 
   shown on the circular lines or arcs). First, calculate the angle 
   between the wind and the runway (220° - 180° = 40°). Next, 
   find the intersection of the 40° line and the 30-kt. headwind 
   arc. Then, proceed downward to determine a crosswind 
   component of 19 kt.

   Note the crosswind component is on the horizontal axis and 
   the headwind component is on the vertical axis. 

Figure 37 
(Refer to figure 37.) What is the headwind component for a landing 
on Runway 18 if the tower reports the wind as 220° at 30 knots? 





   ANSWER: 23 knots. 

   The headwind component is on the 
   vertical axis (left-hand side of the graph). Find the same 
   intersection as in the preceding question, i.e., the 30-kt. wind 
   speed arc, and the 40° angle between wind direction and 
   flight path (220° - 180°). Then move horizontally to the left 
   and read approximately 23 kt. 

Figure 37 
(Refer to figure 37.) Determine the maximum wind velocity for a 45° 
crosswind if the maximum crosswind component for the airplane is 
25 knots. 





   ANSWER: 35 knots. 

   Start on the bottom of the graph's 
   horizontal axis at 25 kt. and move straight upward to the 45° 
   angle between wind direction and flight path line (half-way 
   between the 40° and 50° lines). Note that you are half-way 
   between the 30 and 40 arc-shaped wind speed lines, which 
   means that the maximum wind velocity for a 45° crosswind is 
   35 kt. if the airplane is limited to a 25-kt. crosswind 
   component. 

Figure 37 
(Refer to figure 37.) With a reported wind of north at 20 knots, 
which runway (6, 29, or 32) is acceptable for use for an airplane 
with a 13-knot maximum crosswind component? 





   ANSWER: Runway 32. 

   If the wind is from the north (i.e., 
   either 360° or 0°) at 20 kt., runway 32, i.e., 320°, would 
   provide a 40° crosswind component (360° - 320°). Given a 
   20-kt. wind, find the intersection between the 20-kt. arc and 
   the angle between wind direction and the flight path of 40°. 
   Dropping straight downward to the horizontal axis gives 13 
   kt., which is the maximum crosswind component of the 
   example airplane. 

Figure 37 
(Refer to figure 37.) What is the maximum wind velocity for a 30° 
crosswind if the maximum crosswind component for the airplane is 
12 knots? 





   ANSWER: 24 knots. 

   Start on the graph's horizontal axis at 
   12 kt. and move upward to the 30° angle between wind 
   direction and flight path line. Note that you are almost 
   half-way between the 20 and 30 arc-shaped wind speed 
   lines, which means that the maximum wind velocity for a 30° 
   crosswind is approximately 24 kt. if the airplane is limited to a 
   12-kt. crosswind component. 

Figure 37 
(Refer to figure 37.) With a reported wind of south at 20 knots, 
which runway (10, 14, or 24) is appropriate for an airplane with a 
13-knot maximum crosswind component? 





   ANSWER: Runway 14. 

   If the wind is from the south at 20 kt., 
   runway 14, i.e., 140°, would provide a 40° crosswind 
   component (180° - 140°). Given a 20-kt. wind, find the 
   intersection between the 20-kt. arc and the angle between 
   wind direction and the flight path of 40°. Dropping straight 
   downward to the horizontal axis gives 13 kt., which is the 
   maximum crosswind component of the example airplane. 

Figure 38 
(Refer to figure 38.) Determine the total distance required to land.  

OAT 
. . . . . . . . . . . . . . . . . . . . . . . . . . Std
Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . 10,000 ft
Weight 
. . . . . . . . . . . . . . . . . . . . . . . . . . 2,400 lb
Wind component 
. . . . . . . . . . . . . . . . . . . . . . . . . . Calm
Obstacle 
. . . . . . . . . . . . . . . . . . . . . . . . . . 50 ft.





   ANSWER: 1,925 feet. 

   The landing distance graphs are very 
   similar to the takeoff distance graphs. Begin with the 
   pressure altitude line of 10,000 ft. and the intersection with 
   the standard temperature line which begins at 20°C and 
   slopes up and to the left; i.e., standard temperature 
   decreases as pressure altitude increases. Then move 
   horizontally to the right to the first reference line. Proceed 
   parallel to the closest guideline to 2,400 lb. Proceed 
   horizontally to the right to the second reference line. Since 
   the wind is calm, proceed horizontally to the third reference 
   line. Given a 50-ft. obstacle, proceed parallel to the closest 
   guideline to the right margin to determine a distance of 
   approximately 1,900 ft. 

Figure 38 
(Refer to figure 38.) Determine the approximate total distance 
required to land over a 50-ft. obstacle.  

OAT 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90°F
Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4,000 ft
Weight 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2,800 lb
Headwind component 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 kts





   ANSWER: 1,775 feet. 

   To determine the total landing 
   distance, begin at the left side of Fig. 38 on the 4,000-ft. 
   pressure altitude line at the intersection of 90°F. Proceed 
   horizontally to the right to the first reference line. Proceed 
   parallel to the closest guideline to 2,800 lb., and then straight 
   across to the second reference line. Since the headwind 
   component is 10 kt., proceed parallel to the closest headwind 
   guideline to the 10-kt. line. Then move directly to the right, 
   to the third reference line. Given a 50-ft. obstacle, proceed 
   parallel to the closest guideline for obstacles to find the total 
   distance of approximately 1,775 ft. 

Figure 38 
(Refer to figure 38.) Determine the total distance required to land.  

OAT 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90°F
Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3,000 ft
Weight 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2,900 lb
Headwind component 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 kts
Obstacle 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 ft





   ANSWER: 1,725 feet. 

   To determine the total landing 
   distance, begin with pressure altitude of 3,000 ft. (between 
   the 2,000- and 4,000-ft. lines) at its intersection with 90°F. 
   Proceed horizontally to the right to the first reference line, 
   and then parallel to the closest guideline to 2,900 lb. From 
   that point, proceed horizontally to the second reference line. 
   Since there is a headwind component of 10 kt., proceed 
   parallel to the closest headwind guideline down to 10 kt. and 
   then horizontally to the right to the third reference line. 
   Given a 50-ft. obstacle, proceed parallel to the closest 
   guideline for obstacles to find the landing distance of 
   approximately 1,725 ft. 

Figure 38 
(Refer to figure 38.) Determine the total distance required to land.  

OAT 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32°F
Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8,000 ft
Weight 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2,600 lb
Headwind component 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 kts
Obstacle 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 ft





   ANSWER: 1,400 feet. 

   To determine the total landing 
   distance, begin with the pressure altitude of 8,000 ft. at its 
   intersection with 32°F (0°C). Proceed horizontally to the first 
   reference line, and then parallel to the closest guideline to 
   2,600 lb. From that point, proceed horizontally to the second 
   reference line. Since there is a headwind component of 20 kt., 
   follow parallel to the closest headwind guideline down to 20 
   kt., and then horizontally to the right to the third reference 
   line. Given a 50-ft. obstacle, proceed parallel to the closest 
   guideline for obstacles to find the landing distance of 
   approximately 1,400 ft. 

Figure 39 
(Refer to figure 39.) Determine the approximate landing ground roll 
distance.  

Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sea level
Headwind 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 kts
Temperature 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Std





   ANSWER: 401 feet. 

   At sea level, the ground roll is 445 ft. 
   The standard temperature needs no adjustment. According 
   to Note 1 in Fig. 39, the distance should be decreased 10% 
   for each 4 kt. of headwind, so the headwind of 4 kt. means 
   that the landing distance is reduced by 10%. The result is 
   401 ft. (445 ft. x 90%). 

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