Quiz: Performance and Limitations 1

Answer (A) is correct. (FAA-H-8083-25B Chap 11) Under normal conditions, the total landing distance required to clear a 50-ft. obstacle is 1,255 ft. The temperature is standard (32°F), requiring no adjustment. The headwind of 8 kt. reduces the 1,255 by 20% (10% for each 4 knots). Thus, the total distance required will be 1,004 ft. (1,255 × 80%).

Answer (B) is correct. (FAA-H-8083-25B Chap 11) Under standard conditions, the distance to land over a 50-ft. obstacle at 5,000 ft. is 1,195 ft. The temperature is standard, requiring no adjustment. The headwind of 8 kt., however, requires that the distance be decreased by 20% (10% for each 4 kt. headwind). Thus, the landing ground roll will be 956 ft. (80% of 1,195).

Answer (B) is correct. (FAA-H-8083-25B Chap 11) The landing ground roll at a pressure altitude of 1,250 ft. is required. The difference between landing distance at sea level and 2,500 ft. is 25 ft. (470 – 445). One-half of this distance (12) plus the 445 ft. at sea level is 457 ft. The temperature is standard, requiring no adjustment. The headwind of 8 kt. requires the distance to be decreased by 20%. Thus, the distance required will be 366 ft. (457 × 80%).

Answer (B) is correct. (FAA-H-8083-25B Chap 11) The ground roll distance at 5,000 ft. is 495 ft. According to Note 2 in Fig. 38, since the temperature is 60°F above standard, the distance should be increased by 10%. 495 ft. × 110% = 545 ft.

Answer (C) is correct. (FAA-H-8083-25B Chap 11) The total distance to clear a 50-ft. obstacle for a 3,750-ft. pressure altitude is required. Note that this altitude lies halfway between 2,500 ft. and 5,000 ft. Halfway between the total distance at 2,500 ft. of 1,135 ft. and the total distance at 5,000 ft. of 1,195 ft. is 1,165 ft. Since the headwind is 12 kt., the total distance must be reduced by 30% (10% for each 4 kt.). 70% × 1,165 = 816 ft.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) At sea level, the ground roll is 445 feet. The standard temperature needs no adjustment. According to Note 1 in Fig. 38, 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. × 90%).

Answer (B) is correct. (FAA-H-8083-25B Chap 11) 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 pounds. 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 feet.

Answer (C) is correct. (FAA-H-8083-25B Chap 11) 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 pounds. 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 feet.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) To determine the total landing distance, begin at the left side of Fig. 37 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 feet.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) 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 15°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 pounds. 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 1,925 feet.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) If the wind is from the south at 20 knots, runway 14, i.e., 140°, would provide a 40° crosswind component (180° – 140°). Given a 20-knot wind, find the intersection between the 20-knot arc and the angle between wind direction and the flight path of 40°. Dropping straight downward to the horizontal axis gives 13 knots, which is the maximum crosswind component of the example airplane.

Answer (C) is correct. (FAA-H-8083-25B Chap 11) Start on the graph’s horizontal axis at 12 knots and move upward to the 30° angle between wind direction and flight path line. Note that you are almost halfway between the 20 and 30 arc-shaped wind speed lines, which means that the maximum wind velocity for a 30° crosswind is approximately 24 knots if the airplane is limited to a 12-knot crosswind component.

Answer (C) is correct. (FAA-H-8083-25B Chap 11) If the wind is from the north (i.e., either 360° or 0°) at 20 knots, runway 32, i.e., 320°, would provide a 40° crosswind component (360° – 320°). Given a 20-knot wind, find the intersection between the 20-knot arc and the angle between wind direction and the flight path of 40°. Dropping straight downward to the horizontal axis gives 13 knots, which is the maximum crosswind component of the example airplane.

Answer (C) is correct. (FAA-H-8083-25B Chap 11) Start on the bottom of the graph’s horizontal axis at 25 knots and move straight upward to the 45° angle between wind direction and flight path line (halfway between the 40° and 50° lines). Note that you are halfway between the 30 and 40 arc-shaped wind speed lines, which means that the maximum wind velocity for a 45° crosswind is 35 knots if the airplane is limited to a 25-knot crosswind component.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) 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-knot 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 knots.

Answer (A) is correct. (FAA-H-8083-25B Chap 11) The requirement is the crosswind component, which is found on the horizontal axis of the graph. You are given a 30-knot 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-knot wind velocity arc. Then, proceed downward to determine a crosswind component of 19 knots.
Note the crosswind component is on the horizontal axis and the headwind component is on the vertical axis.

Answer (C) is correct. (FAA-H-8083-25B Chap 11) The chart on the right side of Fig. 35 applies to 36°F above standard. At 10,000 ft., TAS is 166 kt. At 12,000 ft., TAS is 163 kt. We can then interpolate these results and assume 11,000 ft. is 164.5 kt. We then interpolate 10,000 (166 kt.) and 11,000 (164.5 kt.) and arrive at the answer of 165.25 kt.

Answer (C) is correct. (FAA-H-8083-25B Chap 11) Refer to Figure 35 and locate the column for –36°F (ISA –20°C). Interpolation will be required to determine the true airspeed (TAS) at 9,500 feet. At 8,000 feet, TAS is 181 MPH, and at 10,000 feet, TAS is 184 MPH; therefore, the difference is 3 MPH.

Answer (C) is correct. (FAA-H-8083-25B Chap 11) 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.

Answer (B) is correct. (FAA-H-8083-25B Chap 11)
1. Refer to the ISA +20°C (+36°F) section of Fig. 35 (because indicated temperature is approximately 20° above ISA).
2. Refer to the 4,000 feet Pressure Altitude row in the ISA +20°C section. IOAT is +29°C, manifold pressure is 21.3" Hg, fuel flow per engine is 11.5 GPH, and TAS is 159.
3. Calculate the time it will take to travel 500 NM at 159 kt.: 500 NM ÷ 159NM/hr = 3.14 hr.
4. Calculate the expected fuel consumption: 3.14 hr. × 11.5 GPH = 36.1 gal.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) 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 knots. Given a calm wind, the 1,000-NM trip will take 6.09 hr. (1,000 NM ÷ 164 knots). 6.09 hr. × 11.5 GPH = 70.1 gal.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) Note that the entire chart applies to 65% maximum continuous power (regardless of the throttle), so use the middle section of the chart, which is labeled a standard day.
The fuel flow at 11,000 feet on a standard day would be 1/2 of the way between the fuel flow at 10,000 feet (11.5 gallons per hour) and the fuel flow at 12,000 feet (10.9 gallons per hour). Thus, the fuel flow at 11,000 feet would be 11.5 – 0.3, or 11.2 gallons per hour.

Answer (A) is correct. (FAA-H-8083-25B Chap 11) Begin with the intersection of the 2,000-ft. pressure altitude curve and 32°C in the left section of Fig. 40. Move horizontally to the right to the first reference line, and then parallel to the closest guideline to 2,500 lb. Then move horizontally to the right to the second reference line, and then parallel to the closest guideline to the right to 20 kt. Then move horizontally to the right, directly to the right margin because there is no obstacle clearance. You will end up at about 650 ft., which is the required ground roll when there is no obstacle to clear.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) The takeoff distance to clear a 50-ft. obstacle is required. Begin on the left side of the graph at standard temperature (as represented by the curved line labeled “ISA”). From the intersection of the standard temperature line and the 4,000-ft. pressure altitude, proceed horizontally to the right to the first reference line, and then move parallel to the closest guideline to 2,800 pounds. From there, proceed horizontally to the right to the third reference line (skip the second reference line because there is no wind), and move upward following equidistantly between the diagonal lines all the way to the far right. You are at 1,750 ft., which is the takeoff distance to clear a 50-ft. obstacle.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) Begin in the left section of Fig. 40 by finding the intersection of the sea level pressure altitude and standard temperature (15°C) and proceed horizontally to the right to the first reference line. Then proceed parallel to the closest guideline, to 2,700 pounds. From there, proceed horizontally to the right to the third reference line. You skip the second reference line because the wind is calm. Then proceed upward parallel to the closest guideline to the far right side. To clear the 50-ft. obstacle, you need a takeoff distance of about 1,400 feet. NOTE: This question was previously released by the FAA and the FAA’s objective is for you to select the “most correct” answer from the choices given. The actual answer is 1,250 feet, but since 1,400 feet is the closest answer, it should be chosen as correct.

Answer (A) is correct. (FAA-H-8083-25B Chap 11) Begin on the left section of Fig. 40 at 38°C (see outside air temperature at the bottom). Move up vertically to the pressure altitude of 2,000 feet. Then proceed horizontally to the first reference line. Since takeoff weight is 2,750, move parallel to the closest guideline to 2,750 pounds. Then proceed horizontally to the second reference line. Since the wind is calm, proceed again horizontally to the right-hand margin of the diagram (ignore the third reference line because there is no obstacle, i.e., ground roll is desired), which will be at 1,150 feet.

Answer (A) is correct. (14 CFR 91.9 and AC 65-32A) The operating limitations for experimental aircraft and aircraft issued a special light-sport airworthiness certificate are a permanent part of the aircraft’s airworthiness certificate and must remain in the aircraft during operation.

Answer (B) is correct. (14 CFR 91.9) An aircraft’s operating limitations may be found in the current, FAA-approved flight manual, approved manual material, markings, and placards, or any combination thereof.