5 Airplane Performance and Weight and Balance

62. (Refer to Figure 32 on page 231 and Figure 33 on page 231.) Determine if the airplane weight and balance is within limits. Front seat occupants = 340 lb Rear seat occupants %3D = 295 lb Fuel (main wing tanks) = 44 gal Baggage = 56 īb A. 20 pounds overweight, CG aft of aft limits. B. 20 pounds overweight, CG within limits. C. 20 pounds overweight, CG forward of forward limits.

Answer B is correct. DISCUSSION: see pae 230

12. (Refer to Figure 8 on page 203.) What is the effect of a temperature increase from 30 to 50 °F on the density altitude if the pressure altitude remains at 3,000 feet MSL? A. 1,000-foot increase. B. 1,100-foot decrease. C. 1,300-foot increase.

Answer (C) is correct. DISCÚSSION: Increasing the temperature from 30°F to 50°F, given a constant pressure altitude of 3,000 ft., requires you to find the 3,000-ft. line on the density altitude chart at the 30°F level. At this point, the density altitude is approximately 1,650 feet. Then move up the 3,000-ft. line to 50°F, where the density altitude is approximately 2,950 feet. There is an approximate 1,300-ft. increase (2,950 - 1,650 feet). Note that 50°F is just about standard and pressure altitude is very close to density altitude. Answer (A) is incorrect. A 1,000-ft. increase would be caused by a temperature increase to 45°F. Answer (B) is incorrect. A decrease in density altitude would be caused by a decrease, not an increase, in temperature.

1. What are the standard temperature and pressure values for sea level? A. 15°C and 29.92" Hg. B. 59°C and 1013.2 millibars. C. 59°F and 29.92 millibars.

Answer (A) is correct. DISCÚSSION: The standard temperature and pressure values for sea level are 15°C and 29.92" Hg. This is equivalent to 59°F and 1013.2 millibars of mercury. Answer (B) is incorrect. Standard temperature is 59°F (not 59°C). Answer (C) is incorrect. Standard pressure is 29.92" Hg (not 29.92 millibars).

7. As air temperature increases, density altitude will A. decrease. B. increase. C. remain the same.

Answer (B) is correct. DISCUSSION: Increasing the temperature of a substance decreases its density, and a decrease in air density means a higher density altitude. Therefore, with an increase in temperature the air density decreases, providing a higher density altitude. Answer (A) is incorrect. As temperature increases the density altitude will increase, not decrease. Answer (C) is incorrect. Density varies inversely with temperature. Increasing the temperature of a substance decreases its density, and a decrease in air density means a higher density altitude.

31. (Refer to Figure 36 on page 210.) What is the headwind component for a landing on Runway 18 if the tower reports the wind as 220° at 30 knots? A. 19 knots. B. 23 knots. C. 26 knots.

Answer (B) is correct. DISCUSSION: 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 incorrect. This is the crosswind (not headwind) component. Answer (C) is incorrect. This would be the headwind component if the wind were 30° (not 40°) off the runway.

28. (Refer to Figure 35 on page 208.) 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? A. 178 MPH. B. 181 MPH. C. 183 MPH.

Answer (C) is correct. DISCUSSION: 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. 10,000 feet - 8,000 feet = 2,000 feet 9,500 feet- 8,000 feet = 1,500 feet 1,500 feet + 2,000 feet = .75 .75x3 MPH = 2.25 MPH 181 MPH + 2.25 MPH = 183.25 MPH Answer (A) is incorrect. The expected TAS at 6,000 feet is 178 MPH. Answer (B) is incorrect. The expected TAS at 8,000 feet is 181 MPH.

51. See page 216

Answer (C) is correct. DISCUSSION: See page 216

49. If an aircraft is loaded 90 pounds over maximum certificated gross weight and fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained? A. 10 gallons. B. 12 gallons. C. 15 gallons.

Answer (C) is correct. DISCUSSION: Since fuel weighs 6 lb./gal., draining 15 gal. (90 lb. + 6) will reduce the weight of an airplane that is 90 lb. over maximum gross weight to the acceptable amount. Answer (A) is incorrect. Fuel weighs 6 (not 9) lb./gallon. Answer (B) is incorrect. Fuel weighs 6 (not 7.5) lb./gallon.

6. Which combination of atmospheric conditions will reduce aircraft takeoff and climb performance? A. Low temperature, low relative humidity, and low density altitude. B. High temperature, low relative humidity, and low density altitude. C. High temperature, high relative humidity, and high density altitude.

Answer (C) is correct. DISCUSSION: Takeoff and climb performance are reduced by high density altitude. High density altitude is a result of high temperatures and high relative humidity. Answer (A) is incorrect. Low temperature, low relative humidity, and low density altitude all improve airplane performance. Answer (B) is incorrect. Low relative humidity and low density altitude both improve airplane performance.

32. (Refer to Figure 36 on page 210.) Determine the maximum wind velocity for a 45° crosswind if the maximum crosswind component for the airplane is 25 knots. A. 25 knots. B. 29 knots. C. 35 knots.

Answer (C) is correct. DISCUSSION: 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 (A) is incorrect. This would be the maximum wind velocity for a 90° (not 45°) crosswind. Answer (B) is incorrect. This would be the maximum wind velocity for a 60° (not 45°) crosswind.

23. (Refer to Figure 40 page 206.) Determine the approximate ground roll distance required for takeoff. OAT = 32°C Pressure altitude = 2,000 ft Takeoff weight = 2,500 lb Headwind component = 20 kts A. 650 feet. B. 850 feet. C. 1,000 feet.

Answer (A) is correct. DISCÚSSION: 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 incorrect. The ground roll distance required if the wind were calm is 850 ft. Answer (C) is incorrect. The ground roll distance required at maximum takeoff weight is 1,000 ft.

43. (Refer to Figure 38 on page 215.) Determine the approximate landing ground roll distance. Pressure altitude = 5,000 ft Headwind = Calm Temperature = 101°F A. 495 feet. B. 545 feet. C. 445 feet.

Answer (B) is correct. DISCUSSION: 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. x 110% =D 545 ft. Answer (A) is incorrect. This would be ground roll if the temperature were 41°F, not 101°F. Answer (C) is incorrect. This is obtained by decreasing, not increasing, the distance for a temperature 60°F above standard.

20. (Refer to Figure 40 on page 206.) Determine the approximate ground roll distance required for takeoff. OAT = 38°C Pressure altitude = 2,000 ft Takeoff weight = 2,750 lb Headwind component = Calm A. 1,150 feet. B. 1,300 feet. C. 1,800 feet.

Answer (A) is correct. DISCUSSION: 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 (B) is incorrect. This would be the ground roll distance required at maximum takeoff weight. Answer (C) is incorrect. This would be the total distance required to clear a 50-ft. obstacle.

66. (Refer to Figure 32 on page 233 and Figure 33 page 233.) What effect does a 35-gallon fuel burn (main tanks) have on the weight and balance if the airplane weighed 2,890 pounds and the MOM/100 was 2,452 at takeoff? A. Weight is reduced by 210 pounds and the CG is aft of limits. B. Weight is reduced by 210 pounds and the OCG is unaffected. C. Weight is reduced to 2,680 pounds and the CG moves forward.

Answer (A) is correct. DISCUSSION: The effect of a 35-gal. fuel burn on weight balance is required. Burning 35 gal. of fuel will reduce weight by 210 lb. and moment by 158. At 2,680 lb. (2,890 - 210), the 2,294 MOM/100 (2,452 - 158) is above the maximum moment of 2,287; i.e., CG is aft of limits. This is why weight and balance should always be computed for the beginning and end of each flight. Answer (B) is incorrect. Intuitively, one can see that the CG would be affected. Answer (C) is incorrect. Although the moment has decreased, the CG (moment divided by weight) has moved aft.

30. (Refer to Figure 36 on page 210.) What is the crosswind component for a landing on Runway 18 if the tower reports the wind as 220° at 30 knots? A. 19 knots. B. 23 knots. C. 30 knots.

Answer (A) is correct. DISCUSSION: 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 (B) is incorrect. Twenty-three knots is the headwind (not crosswind) component. Answer (C) is incorrect. Thirty knots is the total wind (not crosswind component).

53. (Refer to Figure 60 on page 219.) How should the 500-pound weight be shifted to balance the plank on the fulcrum? A. 1 inch to the left. B. 1 inch to the right. C. 4.5 inches to the right.

Answer (A) is correct. DISCUSSION: To find the desired location of the 500-Ilb. weight, compute and sum the moments left and right of the fulcrum. Set them equal to one another and solve for the desired variable: left = right 500 lb.(X) 250 lb.(20 in.) + 200 lb.(15 in.) 500X = 8,000 X = 16 in. The 500-lb. weight must be 16 in. from fulcrum to balance the plank. The weight should be shifted 1 in. to the left. Answer (B) is incorrect. The 500-lb. weight should be 16 in. from the fulcrum; thus, it must be moved 1 in. to the left, not right, to balance the plank. Answer (C) is incorrect. Shifting the 500-lb. weight 4.5 in. to the right would cause the plank to be heavier on the right side.

46. (Refer to Figure 38 on page 215.) Determine the total distance required to land over a 50-foot obstacle. Pressure altitude = 7,500 ft Headwind = 8 kts Temperature = 32°F Runway = Hard surface A. 1,004 feet. B. 1,205 feet. C. 1,506 feet.

Answer (A) is correct. DISCUSSION: 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 willi be 1,004 ft. (1,255 x 80%). Answer (B) is incorrect. The figure of 1,205 ft. results from incorrectly assuming that an adjustment for a dry grass runway is necessary and then applying that adjustment (an increase of 20%) to 1,004 ft. than to the total landing distance required to clear a 50-ft. obstacle as stated in Note 3, which is 1,255 ft. Answer (C) is incorrect. The figure of 1,506 ft. is obtained by increasing, not decreasing, the distance for the headwind.

13. (Refer to Figure 8 on page 203.) What is the effect of a temperature increase from 35 to 50°F on the density altitude if the pressure altitude remains at 3,000 feet MSL? A. 1,000-foot increase. B. 1,100-foot decrease. C. 1,300-foot increase.

Answer (A) is correct. DISCÚSSION: Increasing the temperature from 35°F to 50°F, given a constant pressure altitude of 3,000 ft., requires you to find the 3,000-ft. line on the density altitude chart at 1,950 feet. Then 35°F level. At this point, the density altitude is approximately move up the 3,000-ft. line to 50°F, where the density altitude is approximately 2,950 feet. There is an approximate 1,000-ft. increase (2,950 - 1,950 feet). Note that 50°F is just about standard, and pressure altitude is very close to density altitude. Answer (B) is incorrect. A 1,100-ft. decrease would require a temperature decrease to 54°F. Answer (C) is incorrect. A 1,300-ft. increase would be caused by a temperature increase to 58°F.

15. (Refer to Figure 8 on page 205.) Determine the pressure altitude at an airport that is 3,563 feet MSL with an altimeter setting of 29.96. A. 3,527 feet MSL. B. 3,556 feet MSL. C. 3,639 feet MSL.

Answer (A) is correct. DISCÚSSION: Note that the question asks only for pressure altitude, not density altitude. Pressure altitude is determined by adjusting the altimeter setting to 29.92" Hg, i.e., adjusting for nonstandard pressure. This is the true altitude plus or minus the pressure altitude conversion factor (based on current altimeter setting). On the chart, an altimeter setting of 30.0 requires you to Asubtract 73 ft. to determine pressure altitude (note that at 29.92, nothing is subtracted because that is pressure altitude). Since 29.96 is halfway between 29.92 and 30.0, you need only subtract 36 (-73/2) from 3,563 ft. to obtain a pressure altitude of 3,527 ft. (3,563 - 36). Note that a higher-than-standard barometric pressure means pressure altitude is lower than true altitude. Answer (B) is incorrect. You must subtract 36 (not 7) from 3,563 ft. to obtain the correct pressure altitude. Answer (C) is incorrect. You must subtract 36 (not add 76) from 3,563 ft. to obtain the correct pressure altitude.

16. (Refer to Figure 8 on page 205.) Determine the pressure altitude at an airport that is 1,386 feet MSL with an altimeter setting of 29.97. A. 1,341 feet MSL. B. 1,451 feet MSL. C. 1,562 feet MSL.

Answer (A) is correct. DISCÚSSION: Pressure altitude is determined by adjusting the altimeter setting to 29.92" Hg. This is the true altitude plus or minus the pressure altitude conversion factor (based on current altimeter setting). Since 29.97 is not a number given on the conversion chart, you must interpolate. Compute 5/8 of -73 (since 29.97 is 5/8 of the way between 29.92 and 30.0), which is 45. Subtract 45 ft. from 1,386 ft. to obtain a pressure altitude of 1,341 feet. Note that if the altimeter setting is greater than standard (e.g., 29.97), the pressure altitude (i.e., altimeter set to 29.92) will be less than true altitude. Answer (B) is incorrect. You must subtract 45 ft. (not add 65) from 1,386 ft. to obtain the correct pressure altitude. Answer (C) is incorrect. You must subtract 45 ft. (not add 176) from 1,386 ft. to obtain the correct pressure altitude.

11. (Refer to Figure 8 on page 203.) Determine the density altitude for these conditions: Altimeter setting = 30.35 Runway temperature = +25°F Airport elevation = 3,894 ft. MSL A. 2,000 feet MSL. B. 2,900 feet MSL. C. 3,500 feet MSL.

Answer (A) is correct. DISCÚSSION: With an altimeter setting of 30.35" Hg, 394 ft. must be subtracted from a field elevation of 3,894 to obtain a pressure altitude of 3,500 feet. Note that the higher-than-normal pressure of 30.35 means the pressure altitude will be less than true altitude. The 394 ft. was found by interpolation: 30.3 on the graph is -348, and 30.4 was -440 feet. Adding one-half the -92 ft. difference (-46 ft.) to -348 ft. results in -394 feet. you Once have found the pressure altitude, use the chart to plot 3,500 ft. pressure altitude at 25°F, to reach 2,000 ft. density altitude. Note that since the temperature is lower than standard, the density altitude is lower than the pressure altitude. Answer (B) is incorrect. This would be the density altitude if you added (not subtracted) 394 ft. to 3,894 feet. Answer (C) is incorrect. This is pressure (not density) altitude.

8. You have planned a cross-country flight on a warm spring morning. Your course includes a mountain pass, which is at 11,500 feet MSL. The service ceiling of your airplane is 14,000 feet MSL. After checking the local weather report, you are able to calculate the density altitude of the mountain pass as 14,800 feet MSL. Which of the following is the correct action to take? A. Replan your journey to avoid the mountain pass. B. Continue as planned since density altitude is only a factor for takeoff. C. Continue as planned because mountain thermals will assist your climb.

Answer (A) is correct. DISČÚSSION: Because the density altitude through the mountain pass is higher than the service ceiling of the aircraft, it will be impossible to fly through the pass given the current conditions. You must replan your journey to avoid the mountain pass. Answer (B) is incorrect. Density altitude affects all aspects of aircraft performance, not just takeoff performance. Answer (C) is incorrect. Mountain thermals cannot be relied upon to safely carry you through the mountain pass.

47. Which items are included in the empty weight of an aircraft? A. Unusable fuel and undrainable oil. B. Only the airframe, powerplant, and optional equipment. C. Full fuel tanks and engine oil to capacity.

Answer (A) is correct. DISCUSSION: The empty weight of an airplane includes airframe, engines, and all items of operating equipment that have fixed locations and are permanently installed. It includes optional and special equipment, fixed ballast, hydraulic fluid, unusable fuel, and undrainable oil. Answer (B) is incorrect. Unusable and undrainable fuel and oil and permanently installed optional equipment are also included in empty weight. Answer (C) is incorrect. Usable fuel (included in fuli fuel) and full engine oil are not components of basic empty weight.

26. (Refer to Figure 35 on page 208.) 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 A. 31.4 gallons. B. 36.1 gallons. C. 40.1 gallons.

Answer (B) is correct. DISCUSSION: 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 + 159 NM = 3.14 hr. hr. 4. Calculate the expected fuel consumption: 3.14 hr. x 11.5 GPH = 36.1 gal. Answer (A) is incorrect. The amount of 31.4 gallons is the expected fuel consumption for a 500-NM flight with a true airspeed of 183 (not 159) kt. Answer (C) is incorrect. The amou of 40.1 gallons is the expected fuel consumption for a 500-NM flight with a true airspeed of 143 (not 159) kt. 10

40. If an emergency situation requires a downwind landing, pilots should expect a faster A. airspeed at touchdown, a longer ground roll, and better control throughout the landing roll. B. groundspeed at touchdown, a longer ground roll, and the likelihood of overshooting the desired touchdown point. C. groundspeed at touchdown, a shorter ground roll, and the likelihood of undershooting the desired touchdown point.

Answer (B) is correct. DISCUSSION: A downwind landing, in an emergency or other situation, will result in a faster groundspeed at touchdown, which means a longer ground roll, which in turn increases the likelihood of overshooting the desired touchdown point. Answer (A) is incorrect. The airspeed will probably be the same even though the groundspeed is greater, and the control during the landing roll will be less due to the high groundspeed. Answer (C) is incorrect. The ground roll is longer, not shorter, and there is a greater likelihood of overshooting, not undershooting, the touchdown point due to the faster groundspeed.

41. (Refer to Figure 38 on page 215.) Determine the approximate landing ground roll distance. Pressure altitude = Sea level Headwind = 4 kts Temperature = Std A. 356 feet. B. 401 feet. C. 490 feet.

Answer (B) is correct. DISCUSSION: 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. x 90%). Answer (A) is incorrect. This would be the ground roll with an 8-kt., not a 4-kt., headwind. Answer (C) is incorrect. Ground roll is reduced, not increased, to account for headwind.

35. (Refer to Figure 36 on page 210.) With a reported wind of south at 20 knots, which runway is appropriate for an airplane with a 13-knot maximum crosswind component? A. Runway 10. B. Runway 14. C. Runway 24.

Answer (B) is correct. DISCUSSION: 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 (A) is incorrect. Runway 10 would have a crosswind component of 20 knots. Answer (C) is incorrect. Runway 24 would have a crosswind component of approximately 17 knots.

24. (Refer to Figure 35 on page 208.) What fuel flow should a pilot expect at 11,000 feet on a standard day with 65 percent maximum continuous power? A. 10.6 gallons per hour. B. 11.2 gallons per hour. C. 11.8 gallons per hour.

Answer (B) is correct. DISCUSSION: Note that the entire chart applies to 65% maximum continuous power (regardless of the throttle), so use the middle section of the chat, 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 incorrect. You must add (not subtract) 0.3 to 10.9 to obtain the correct fuel flow. Answer (C) is incorrect. You must subtract (not add) 0.3 from 11.5 to obtain the correct fuel flow.

36. (Refer to Figure 37 on page 212.) Determine the total distance required to land. OAT = Std Pressure altitude = 10,000 ft Weight = 2,400 lb Wind component = Calm Obstacle = 50 ft A. 750 feet. B. 1,925 feet. C. 1,450 feet.

Answer (B) is correct. DISCUSSION: 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 (A) is incorrect. This is the total distance required to land with a 30-kt. headwind, not a calm wind, and without an obstacle, not with a 50-ft. obstacle. Answer (C) is incorrect. This is the approximate total distance required to land at a pressure altitude of 2,000 ft., not 10,000 ft., and a weight of 2,300 lb., not 2,400 pounds.

44. (Refer to Figure 38 on page 215.) Determine the approximate landing ground roll distance. Pressure altitude = 1,250 ft Headwind = 8 kts Temperature = Std A. 275 feet. B. 366 feet. C. 470 feet.

Answer (B) is correct. DISCUSSION: 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 x 80%). Answer (A) is incorrect. The distance should be decreased by 20% (not 40%). Answer (C) is incorrect. This is the distance required at 2,500 ft. in a calm wind.

34. (Refer to Figure 36 on page 210.) What is the maximum wind velocity for a 30° crosswind if the maximum crosswind component for the airplane is 12 knots? A. 16 knots. B. 20 knots. C. 24 knots.

Answer (C) is correct. DISCUSSION: 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 (A) is incorrect. Sixteen knots would be the maximum wind velocity for a 50° (not 30°) crosswind. Answer (B) is incorrect. Twenty knots would be the maximum wind velocity for a 40° (not 30°) crosswind.

14. (Refer to Figure 8 on page 203.) What is the effect of a temperature decrease and a pressure altitude increase on the density altitude from 90°F and 1,250 feet pressure altitude to 55°F and 1,750 feet pressure altitude? A. 1,700-foot increase. B. 1,300-foot decrease. C. 1,700-foot decrease.

Answer (C) is correct. DISCÚSSION: The requirement is the effect of temperature decrease and a pressure altitude increase on density altitude. First, find the density altitude at 90°F and 1,250 ft. (approximately 3,600 feet). Then find the density altitude at 55°F and 1,750 ft. pressure altitude (approximately 1,900 feet). Next, subtract the two numbers. Subtracting 1,900 ft. from 3,600 ft. equals a 1,700-ft. decrease in density altitude. Answer (A) is incorrect. Such a large decrease in temperature would decrease, not increase, density altitude. Answer (B) is incorrect. Density altitude would decrease 1,300 ft. if the temperature decreased to 60°F, not 55°F.

17. (Refer to Figure 8 on page 205.) What is the effect of a temperature increase from 25 to 50° F on the density altitude if the pressure altitude remains at 5,000 feet? A. 1,200-foot increase. B. 1,400-foot increase. C. 1,650-foot increase.

Answer (C) is the best answer. DISCUSSION: Increasing the temperature from 25°F to 50°F, given a pressure altitude of 5,000 ft., requires you to find the 5,000-ft. line on the density altitude chart at the 25°F level. At this point, the density altitude is approximately 3,800 ft. increase (5,400 Then move up the 5,000-ft. line to 50°F, where the density altitude is approximately 5,400 ft. There is about a 1,600-ft. ft. - 3,800 ft.). As temperature increases, so does density altitude; i.e., the atmosphere becomes thinner (less dense). Because a 1,600-foot increase is not an answer choice, 1,650-foot increase would be the best answer. Answer (A) is incorrect. A 1,200-ft. increase would result increase of 18°F (not 25°F). Answer (B) is increase would result from a temperature from a temperature incorrect. A 1,400-ft. increase of 20°F (not 25°F).

60. (Refer to Figure 67 on page 227.) Determine the condition of the airplane: Pilot and copilot = 316 lb Passengers Fwd position = 130 lb Aft position %3D = 147 lb Baggage = 50 lb Fuel = 75 gal A. 197 pounds under allowable gross weight; CG 83.6 inches aft of datum. B. 163 pounds under allowable gross weight; CG 82 inches aft of datum. C. 197 pounds under allowable gross weight; CG 84.6 inches aft of datum.

Answer A is correct DISCUSSION: seepage 226

64. (Refer to Figure 32 on page 233 and Figure 33 on page 233.) What is the maximum amount of baggage that can be carried when the airplane is loaded as follows? Front seat occupants = 387 lb Rear seat occupants = 293 lb Fuel = 35 gal A. 45 pounds. B. 63 pounds. C. 220 pounds.

Answer A is correct. DISCUSSION: see pae 232

65. (Refer to Figure 32 on page 233 and Figure 33 on page 233.) Upon landing, the front passenger (180 pounds) departs the airplane. A rear passenger (204 pounds) moves to the front passenger position. What effect does this have on the CG if the airplane weighed 2,690 pounds and the MOM/100 was 2,260 just prior to the passenger transfer? A. The CG moves forward approximately 3 inches. B. The weight changes, but the CG is not affected. C. The CG moves forward approximately 0.1 inch.

Answer A is correct. DISCUSSION: see pae 232

58. (Refer to Figure 67 on page 225.) Determine the condition of the airplane: Pilot and copilot = 375 lb Passengers aft position = 245 lb Baggage = 65 lb Fuel = 70 gal A. 185 pounds under allowable gross weight; CG is located within limits. B. 162 pounds under allowable gross weight; CG is located within limits. C. 162 pounds under allowable gross weight; CG is located aft of the aft limit.

Answer A is correct. DISCUSSION: see page 224

63. (Refer to Figure 32 on page 231 and Figure 33 on page 231.) Calculate the weight and balance and determine if the CG and the weight of the airplane are within limits. Front seat occupants = 350 lb Rear seat occupants = 325 lb Baggage = 27 lb Fuel = 35 gal A. CG 81.7, out of limits forward. B. CG 83.4, within limits. C. CG 84.1, within limits.

Answer B is correct. DISCUSSION: see pae 230

59. (Refer to Figure 67 on page 225.) Determine the condition of the airplane: Pilot and copilot = 400 lb Passengers -- aft position = 240 lb Baggage = 20 lb Fuel = 75 gal A. 157 pounds under allowable gross weight; CG is located within limits. B. 180 pounds under allowable gross weight; CG is located within limits. C. 180 pounds under allowable gross weight, but CG is located aft of the aft limit.

Answer B is correct. DISCUSSION: see page 224

61. (Refer to Figure 32 on page 229 and Figure 33 on page 229.) With the airplane loaded as follows, what action can be taken to balance the airplane? Front seat occupants = 411 lb Rear seat occupants = 100 lb Main wing tanks = 44 gal A. Fill the auxiliary wing tanks. B. Add a 100-pound weight to the baggage compartment. C. Transfer 10 gallons of fuel from the main tanks to the auxiliary tanks.

Answer B is correct. DISCUSSION: see page 228

68. (Refer to Figure 32 on page 235 and Figure 33 on page 235.) Which action can adjust the airplane's weight to maximum gross weight and the CG within limits for takeoff? Front seat occupants = 425 lb Rear seat occupants = 300 lb Fuel, main tanks = 44 gal A. Drain 12 gallons of fuel. B. Drain 9 gallons of fuel. C. Transfer 12 gallons of fuel from the main tanks to the auxiliary tanks.

Answer B is correct. DISCUSSION: see page 234

67. (Refer to Figure 32 on page 235 and Figure 33 on page 235.) Determine if the airplane weight and balance is within limits. Front seat occupants = 415 lb Rear seat occupants = 110 lb Fuel, main tanks = 44 gal Fuel, aux. tanks = 19 gal Baggage = 32 Ilb A. 19 pounds overweight, CG within limits. B. 19 pounds overweight, CG out of limits forward. C. Weight within limits, CG out of limits.

Answer C is correct. DISCUSSION: see page 234

22. (Refer to Figure 40 on page 206.) Determine the total distance required for takeoff to clear a 50-foot obstacle. OAT = Std Pressure altitude = 4,000 ft Takeoff weight = 2,800 lb Headwind component = Calm A. 1,500 feet. B. 1,750 feet. C. 2,000 feet.

Answer (B) is correct. DISCUSSION: 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 (A) is incorrect. This would be the total distance MW-required with a 10-kt. headwind. Answer (C) is incorrect. This would be the total distance required at maximum takeoff weight.

37. (Refer to Figure 37 on page 212.) 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 A. 1,525 feet. B. 1,775 feet. C. 1,950 feet.

Answer (B) is correct. DISCUSSION: To determine 37 on the 4.000-ft. pressure altitude line at the total landing distance, begin at the left side of Fig, the intersection of 90°F. Proceed horizontally to the right to the Tirst reference line. Proceed 2,800 lb., and then straight across to the second reference line. parallel to the closest guideline to parallel to the proceed parallel to the closest guideline for obstacles to find the to the right, to the third reference line. Given a 50-ft. obstacle, closest headwind quideline to the 10-kt. Since the headwind component is 10 kt., proceed line. Then move directly total distance of approximately 1,775 feet. the total distance required with an 18-kt. headwind, not a Answer (A) is incorrect. A distance of 1,525 ft. would be 10-kt. headwind. Answer (C) is incorrect. A distance of 1,950 ft. would be the total distance required with calm wind conditions, not with a 10-kt. headwind.

39. (Refer to Figure 37 on page 212.) Determine the total distance required to land. OAT = 32°F Pressure altitude = 8,000 ft Weight = 2,600 Ib Headwind component = 20 kts Obstacle = 50 ft A. 850 feet. B. 1,400 feet. C. 1,750 feet.

Answer (B) is correct. DISCUSSION: 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 (A) is incorrect. This would be the ground roll with no obstacle. Answer (C) is incorrect. This would be the total distance required at maximum landing weight.

45. (Refer to Figure 38 on page 215.) Determine the total distance required to land over a 50-foot obstacle. Pressure altitude = 5,000 ft Headwind = 8 kts Temperature = 41°F Runway = Hard surface A. 837 feet. B. 956 feet. C. 1,076 feet.

Answer (B) is correct. DISCUSSION: 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 (A) is incorrect. The distance should be decreased by 20% (not 30%). Answer (C) is incorrect. The distance should be decreased by 20% (not 10%).

52. (Refer to Figure 61 innoage 218.) If 50 pounds of weight is located at point X and 100 pounds at point Z, how much weight must be located at point Y to balance the plank? A. 30 pounds. B. 50 pounds. C. 300 pounds.

Answer (C) is correct. DISCUSSION: Compute and sum the moments left and right of the fulcrum. Set them equal to one another and solve for the desired variable: left = right 50 lb.(50 in.) + Y(25 in.) = 100 lb.(100 in.) 2,500 + 25Y = 10,000 25Y = 7,500 Y = 300 lb. Answer (A) is incorrect. The weight of 30 pounds in the place of Y would cause the plank to be heavier on the right side. Answer (B) is incorrect. The weight of 50 pounds in the place of Y would cause the plank to be heavier on the right side.

48. An aircraft is loaded 110 pounds over maximum certificated gross weight. If fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained? A. 15.7 gallons. B. 16.2 gallons. C. 18.4 gallons.

Answer (C) is correct. DISCUSSION: Fuel weighs 6 lb./gallon. If an airplane is 110 lb. over maximum gross weight, 18.4 gal. (110 lb. + 6) must be drained to bring the airplane weight within limits. Answer (A) is incorrect. Fuel weighs 6 (not 7) Ib./gallon. Answer (B) is incorrect. Fuel weighs 6 (not 6.8) lb./gallon.

33. (Refer to Figure 36 on page 210.) With a reported wind of north at 20 knots, which runway is acceptable for use for an airplane with a 13-knot maximum crosswind component? A. Runway 6. B. Runway 29. C. Runway 32.

Answer (C) is correct. DISCUSSION: 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 (A) is incorrect. Runway 6 would have a crosswind component of approximately 17 knots. Answer (B) is incorrect. Runway 29 would have a crosswind component of 19 knots.

50. See page 216

Answer (B) is correct. DISCUSSIONS: See page 216

18. (Refer to Figure 8 on page 205.) Determine the pressure altitude with an indicated altitude of 1,380 feet MSL with an altimeter setting of 28.22 at standard temperature. A. 3,010 feet MSL. B. 2,991 feet MSL. C. 2,913 feet MSL.

Answer (B) is correct. DISCÚSSION: Pressure altitude is determined by adjusting the altimeter setting to 29.92" Hg, i.e., pressure. This is the indicated altitude of 1,380 ft. plus adjusting for nonstandard the pressure altitude conversion factor (based on the current altimeter setting). factor schedule. Add 1,533 On the right side of Fig. 8 is a pressure altitude conversion ft. for an altimeter setting of 28.30 and 1,630 ft. for an altimeter you must subtract 20% of the difference from 1,630 ft. (1,630 - 1,533 = 97 × setting of 28.20. Using interpolation, between 28.3 and 28.2 .2 = 19). Since 1,630 - 19 = 1,611, add 1,611 ft. to 1,380 ft. to get the pressure altitude of 2,991 feet. by adding the conversion Answer (A) is incorrect. The figure of 3,010 feet is obtained factor for an altimeter setting of 28.20, not an altimeter setting of 28.22, to the indicated altitude Answer (C) is incorrect. The figure of2,913 feet is obtained by adding the conversion factor for an alumeter setting of 28.30. not an altimeter setting of 28.22, to the indicated altitude

4. What effect does high density altitude, as compared to low density altitude, have on propeller efficiency and why? A. Efficiency is increased due to less friction on the propeller blades. B. Efficiency is reduced because the propeller exerts less force at high density altitudes than at low density altitudes. C. Efficiency is reduced due to the increased force of the propeller in the thinner air.

Answer (B) is correct. DISCÚSSION: The propeller produces thrust in proportion to the mass of air being accelerated through the rotating propeller. If the air is less dense, the propeller efficiency is decreased. Remember, higher density altitude refers to less dense air. Answer (A) is incorrect. There is decreased, not increased, efficiency. Answer (C) is incorrect. The propeller exerts less (not more) force on the air when the air is thinner, i.e.., at higher density altitudes.

25. (Refer to Figure 35 on page 208.) 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 A. 60.2 gallons. B. 70.1 gallons. C. 73.2 gallons.

Answer (B) is correct. DISCÚSSION: 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. x 11.5 GPH = 70.1 gal. Answer (A) is incorrect. The figure of 60.2 gallons is the expected fuel consumption for a 1,000-NM flight with a true airspeed of 189 (not 164) knots. Answer (C) is incorrect. The figure of 73.2 gallons is the expected fuel consumption for a 1,000-NM flight with a true airspeed of 157 (not 164) knots.

3. Which factor would tend to increase the density altitude at a given airport? A. An increase in barometric pressure. B. An increase in ambient temperature. C. A decrease in relative humidity.

Answer (B) is correct. DISCÚSSION: When air temperature increases, density altitude increases because, at a higher temperature, the air is less dense. Answer (A) is incorrect. Density altitude decreases as barometric pressure increases. Answer (C) is incorrect. Density altitude decreases as relative humidity decreases.

2. What effect, if any, does high humidity have aircraft performance? A. It increases performance. B. It decreases performance. C. It has no effect on performance.

Answer (B) is correct. DISČÚSSION: As the air becomes more humid, it becomes less dense. This is because a given volume of moist air weighs less than the same volume of dry air. Less dense air reduces aircraft performance. Answer (A) is incorrect. High humidity reduces (not increases) performance. Answer (C) is incorrect. The three factors that affect aircraft performance are pressure, temperature, and humidity.

5. What effect does high density altitude have on aircraft performance? A. It increases engine performance. B. It reduces climb performance. C. It increases takeoff performance.

Answer (B) is correct. DISČÚSSION: High density altitude reduces all aspects of an airplane's performance, including takeoff and climb performance. Answer (A) is incorrect. Engine performance is decreased (not increased). Answer (C) is incorrect. Takeoff runway length is increased, i.e., reduces takeoff performance.

21. (Refer to Figure 40 on page 206.) Determine the total distance required for takeoff to clear a 50-foot obstacle. OAT = Std Pressure altitude = Sea level Takeoff weight 2,700 lb Headwind component = Calm A. 1,000 feet. B. 1,400 feet. C. 1,700 feet.

Answer (B) is the best answer. DISCUSSION: 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 incorrect. This would be the total distance required at 2,200 lb. takeoff weight. Answer (C) is incorrect. This would be the total distance required at maximum takeoff weight.

29. (Refer to Figure 35 on page 208.) Approximately what true airspeed should a pilot expect with full throttle at 10,500 feet with a temperature of 36°F above standard? A. 190 KTS. B. 159 KTS. C. 165 KTS.

Answer (C) is correct. DISCUSSION: 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 (A) is incorrect. A TAS of 190 kt. cannot be found on this chart. The maximum speed for this aircraft is 166 kt., found at 10,000 ft. pressure altitude. Logically, a TAS of 190 kt. is not between 163 kt. and 166 kt., where our TAS at 10,500 ft. will be found. Answer (B) is incorrect. The expected TAS of 159 kt. would be appropriate around 14,000 ft., not at 10,500 ft. with full throttle.

9. A pilot and two passengers landed on a 2,100 foot east-west gravel strip with an elevation of 1,800 feet. The temperature is warmer than expected and after computing the density altitude it is determined the takeoff distance over a 50 foot obstacle is 1,980 feet. The airplane is 75 pounds under gross weight. What would be the best choice? A. Taking off into the headwind will give the extra climb-out time needed. B. Try a takeoff without the passengers to make sure the climb is adequate. C. Wait until the temperature decreases, and recalculate the takeoff performance.

Answer (C) is correct. DISCUSSION: The majority of pilot-induced accidents occur during the takeoff and landing phases of flight. In this instance, the pilot in command this aircraft has an important decision to make. The takeoff distance over a 50-foot obstacle appears on initial inspection to be possible (1,980 feet on a 2,100-foot runway). It is important to remember, however, the performance charts are based on ideal conditions and created by testing brand new aircraft with optimal performance and highly experienced test pilots at the controls. It would be ill-advised for this pilot to attempt to take off. The pilot should wait for the temperature to decrease and recalculate the takeoff performance. Answer (A) is incorrect. There are no winds provided in this question and no guarantee the takeoff performance into a headwind would be improved in any way. Answer (B) is incorrect. The decision to attempt a takeoff without the passengers and ensure climb performance is flawed in a few ways. Charts provided in the aircraft information manual should be used to determine climb performance prior to a flight. Coming to the realization the climb performance is not sufficient to clear terrain features and obstacles once airborne is a position no pilot wants to find him/herself in. If the pilot did attempt to take off without and the climb performance was adequate, there the passengers is absolutely no reason to believe the performance would be sufficient when the passengers are added and the weight of the aircraft is increased.

27. (Refer to Figure 35 on page 208.) 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. A. 19.8" Hg. B. 20.8" Hg. C. 21.0" Hg.

Answer (C) is correct. DISCUSSION: 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 (A) is incorrect. The setting of 19.8" Hg. would achieve 65% power at 36°F below (not above) standard temperature. Answer (B) is incorrect. The manifold pressure at 6,500 ft. is closer to 21.0 than 20.8.

42. (Refer to Figure 38 on page 215.) Determine the total distance required to land over a 50-ft. obstacle. Pressure altitude = 3,750 ft Headwind = 12 kts Temperature = Std A. 794 feet. B. 836 feet. C. 816 feet.

Answer (C) is correct. DISCUSSION: 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% x 1,165 = 816 ft. Answer (A) is incorrect. This would be the total distance to land at a pressure altitude of 2,500 ft., not 3,750 ft., with a 12-kt. headwind and standard temperature. Answer (B) is incorrect. This would be the total distance to land at a pressure altitude of 5,000 ft., not 3,750 ft., with a 12-kt. headwind and standard temperature.

38. (Refer to Figure 37 on page 212.) 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 A. 1,450 feet. B. 1,550 feet. C. 1,725 feet.

Answer (C) is correct. DISCUSSION: 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 (A) is incorrect. This would be the total distance required with a 20-kt., not 10-kt., headwind. Answer (B) is incorrect. This would be the total distance required at a pressure altitude of 2,000 ft., not 3,000 feet.

10. If the outside air temperature (OAT) at a given altitude is warmer than standard, the density altitude is A. equal to pressure altitude. B. lower than pressure altitude. C. higher than pressure altitude.

Answer (C) is correct. DISCUSSION: When temperature increases, the air expands and therefore becomes less dense. This decrease in density means a higher density altitude. Pressure altitude is based on standard temperature. Thus, density altitude exceeds pressure altitude when the temperature is warmer than standard. Answer (A) is incorrect. Density altitude equals pressure altitude only when temperature is standard. Answer (B) is altitude when incorrect. Density altitude is lower than pressure the temperature is below standard.

19. (Refer to Figure 8 on page 205.) Determine the density altitude for these conditions: Altimeter setting = 29.25 Runway temperature = +81°F Airport elevation = 5,250 ft MSL A. 4,600 feet MSL. B. 5,877 feet MSL. C. 8,500 feet MSL.

Answer (C) is correct. DISCUSSION: With an altimeter setting of 29.25" Hg, about 626 ft. (579 plus 1/2 the 94-ft. pressure altitude conversion factor difference between 29.2 and 29.3) must be added to the field elevation of 5,250 ft. to obtain the pressure altitude, or 5,876 feet. Note that barometric pressure is less than standard and pressure altitude is greater than true altitude. Next, convert pressure altitude to density altitude. On the chart, find the point at which the pressure altitude line for 5,876 ft. crosses the 81°F line. The density altitude at that spot shows somewhere in the mid-8,000s of feet. The closest answer choice is 8,500 feet. Note that, when temperature is higher than standard, density altitude exceeds pressure altitude. Answer (A) is incorrect. This would be pressure altitude if 650 ft. were subtracted from, not added to, 5,250 ft. MSL. Answer (B) is incorrect. This is pressure altitude, not density altitude.

See pages 220-223 to view question 54-57.

See page 220-223 for answers and figures.