Chapter 5

Réussis tes devoirs et examens dès maintenant avec Quizwiz!

What is the maximum amount of baggage that may be loaded aboard the airplane for the CG to remain within the moment envelope?WEIGHT (LB)MOM/1000Empty weight1,35051.5Pilot and front passenger 250---Rear passengers 400--- Baggage --- --- Fuel, 30 gal.--- --- Oil, 8 qt. ---- 0.2 A 105 pounds. B 120 pounds. C 110 pounds.

A 105 pounds. Your answer is CORRECT. (FAA-H-8083-25B Chap 10)For the amount of weight left for baggage, compute each individual moment by using the loading graph and add them up. First, compute the moment for the pilot and front seat passenger with a weight of 250 lb. Refer to the loading graph and the vertical scale at the left side and find the value of 250. From this position, move to the right horizontally across the graph until you intersect the diagonal line that represents pilot and front passenger. From this point, move vertically down to the bottom scale, which indicates a moment of about 9.2. To compute rear passenger moment, measure up the vertical scale of the loading graph to a value of 400, horizontally across to intersect the rear passenger diagonal line, and down vertically to the moment scale, which indicates approximately 29.0. To compute the moment of the fuel, recall that fuel weighs 6 lb. per gal. The question gives 30 gal., for a total fuel weight of 180 lb. Now move up the weight scale on the loading graph to 180, then horizontally across to intersect the diagonal line that represents fuel, then vertically down to the moment scale, which indicates approximately 8.7. To get the weight of the oil, see Note 2 at the bottom of the loading graph section of Fig. 34. It gives 15 lb. as the weight with a moment of -0.2. Now total the weights (2,195 lb. including 15 lb. of engine oil). Also total the moments (98.2 including engine oil with a negative 0.2 moment). With this information, refer to the center of gravity moment envelope chart. Note that the maximum weight in the envelope is 2,300 lb. The amount of 2,300 lb. - 2,195 lb. already totaled leaves a maximum possible 105 lb. for baggage. However, you must be sure 105 lb. of baggage does not exceed the 109 moments allowed at the top of the envelo

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 183 MPH. B 181 MPH. C 178 MPH.

A 183 MPH. Your answer 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.10,000 feet - 8,000 feet=2,000 feet9,500 feet - 8,000 feet=1,500 feet1,500 feet ÷ 2,000 feet=.75.75 × 3 MPH=2.25 MPH181 MPH + 2.25 MPH=183.25 MPH

Determine the density altitude for these conditions:Altimeter setting = 30.35Runway temperature = +25°FAirport elevation = 3,894 ft. MSL A 2,000 feet MSL. B 3,500 feet MSL. C 2,900 feet MSL.

A 2,000 feet MSL. Your answer is CORRECT. (FAA-H-8083-25B Chap 11)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. Once you 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.

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 3,639 feet MSL. C 3,556 feet MSL.

A 3,527 feet MSL. Your answer is CORRECT. (FAA-H-8083-25B Chap 11)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 subtract 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.

What is the maximum amount of fuel that may be aboard the airplane on takeoff if loaded as follows?WEIGHT (LB)MOM/1000Empty weight1,35051.5Pilot and front passenger 340 --- Rear passengers 310 ---Baggage 45 --- Oil, 8 qt.--- --- A 40 gallons. B 32 gallons. C 24 gallons.

A 40 gallons. Your answer is CORRECT. (FAA-H-8083-25B Chap 10)To find the maximum amount of fuel this airplane can carry, add the empty weight (1,350), pilot and front passenger weight (340), rear passengers (310), baggage (45), and oil (15), for a total of 2,060 pounds. (Find the oil weight and moment by consulting Note 2 on Fig. 34. It is 15 lb. and -0.2 moments.) Gross weight maximum on the center of gravity moment envelope chart is 2,300. Thus, 240 lb. of weight (2,300 - 2,060) is available for fuel. Since each gallon of fuel weighs 6 lb., this airplane can carry 40 gallons of fuel (240 ÷ 6 lb. per gallon) if its center of gravity moments do not exceed the limit. Note that long-range tanks were not mentioned; assume they exist. Compute the moments for each item. The empty weight moment is given as 51.5. Calculate the moment for the pilot and front passenger as 12.8, the rear passengers as 22.5, the fuel as 11.5, the baggage as 4.0, and the oil as -0.2. These total to 102.1, which is within the envelope, so 40 gallons of fuel may be carried.

Determine the approximate landing ground roll distance.Pressure altitude = 5,000 ftHeadwind = CalmTemperature = 101°F A 545 feet. B 495 feet. C 445 feet.

A 545 feet. Your answer 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.

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 12 gallons. B 15 gallons. C 10 gallons.

B 15 gallons.Your answer is CORRECT. (FAA-H-8083-25B Chap 10)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. C 10 gallons. BACK NEXT All Questions Progress: 20/20 MARK Contact Us Copyright ©1995-2023 Gleim Publications, Inc. and/or Gleim Internet, Inc.

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 165 KTS. C 159 KTS.

B 165 KTS. Your answer 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.

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 40.1 gallons. B 36.1 gallons. C 31.4 gallons.

B 36.1 gallons. Your answer 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= 3.14 hr.hr.4.Calculate the expected fuel consumption: 3.14 hr. × 11.5 GPH = 36.1 gal.

Determine the approximate landing ground roll distance.Pressure altitude = 1,250 ftHeadwind = 8 ktsTemperature = Std A 275 feet. B 366 feet. C 470 feet.

B 366 feet. Your answer 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%).

Determine the approximate landing ground roll distance.Pressure altitude = Sea levelHeadwind = 4 ktsTemperature = Std A 356 feet. B 401 feet. C 490 feet.

B 401 feet. Your answer 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%).

Determine the approximate ground roll distance required for takeoff.OAT = 32°CPressure altitude = 2,000 ftTakeoff weight = 2,500 lbHeadwind component = 20 kts A 1,000 feet. B 650 feet. C 850 feet.

B 650 feet. Your answer 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.

Calculate the weight and balance and determine if the CG and the weight of the airplane are within limits.Front seat occupants = 350 lbRear seat occupants = 325 lbBaggage = 27 lbFuel = 35 gal A CG 81.7, out of limits forward. B CG 83.4, within limits. C CG 84.1, within limits.

B CG 83.4, within limits. Your answer is CORRECT. (FAA-H-8083-25B Chap 10)Total weight, total moment, and CG must all be calculated. As in most weight and balance problems, you should begin by setting up the schedule as shown below. Next, go to the "Moment limits vs. weight" chart (Fig. 33), and note that the maximum weight allowed is 2,950, which means that this airplane is 23 lb. under maximum weight. At a total moment of 2,441, it is also within the CG limits (2,399 to 2,483) at that weight. Finally, compute the CG. Recall that Fig. 32 gives moment per 100 inches. The total moment is therefore 244,100 (2,441 × 100). The CG is 83.4 (244,100 ÷ 2,927).

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 Transfer 12 gallons of fuel from the main tanks to the auxiliary tanks. B Drain 9 gallons of fuel. C Drain 12 gallons of fuel.

B Drain 9 gallons of fuel. Your answer is CORRECT. (FAA-H-8083-25B Chap 10)First, determine the total weight to see how much must be reduced. As shown below, this original weight is 3,004 pounds. Fig. 33 shows the maximum weight as 2,950 pounds. Thus, you must adjust the total weight by removing 54 lb. (3,004 - 2,950). Since fuel weighs 6 lb./gal., you must drain at least 9 gallons. To check for CG, recompute the total moment using a new fuel moment of 158 (from the chart) for 210 pounds. The plane now weighs 2,950 lb. with a total moment of 2,437, which falls within the moment limits on Fig. 33.

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

B High temperature, high relative humidity, and high density altitude. Your answer is CORRECT. (FAA-H-8083-25B Chap 11)Takeoff and climb performance are reduced by high density altitude. High density altitude is a result of high temperatures and high relative humidity.

Determine if the airplane weight and balance is within limits.Front seat occupants = 415 lbRear seat occupants = 110 lbFuel, main tanks = 44 galFuel, aux. tanks = 19 galBaggage = 32 lb A 19 pounds overweight, CG out of limits forward. B Weight within limits, CG out of limits. C 19 pounds overweight, CG within limits.

B Weight within limits, CG out of limits. Your answer is CORRECT. (FAA-H-8083-25B Chap 10)Both the weight and the total moment must be calculated. Begin by setting up the schedule shown below. The fuel must be separated into main and auxiliary tanks, but weights and moments for both tanks are provided in Fig. 32. Since 415 lb. is not shown on the front seat table, simply multiply the weight by the arm shown at the top of the table (415 lb. × 85 in. = 35,275 lb.-in.) and divide by 100 for moment/100 of 353 (35,275 ÷ 100 = 352.75). The rear seat moment must also be multiplied (110 lb. × 121 in. = 13,310 pound-inches). Divide by 100 to get 133.1, or 133 lb.-in. ÷ 100. The last step is to go to the "Moment limits vs. weight" chart (Fig. 33). The maximum weight allowed is 2,950, which means that the airplane weight is within the limits. However, the CG is out of limits because the minimum moment/100 for a weight of 2,950 lb. is 2,422.

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

B increase. Your answer is CORRECT. (FAA-H-8083-25B Chap 4)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.

Determine the pressure altitude at an airport that is 1,386 feet MSL with an altimeter setting of 29.97. A 1,451 feet MSL. B 1,562 feet MSL. C 1,341 feet MSL.

C 1,341 feet MSL. Your answer is CORRECT. (FAA-H-8083-25B Chap 11)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.

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

C It reduces climb performance. Your answer is CORRECT. (FAA-H-8083-25B Chap 11)High density altitude reduces all aspects of an airplane's performance, including takeoff and climb performance.

Determine the total distance required for takeoff to clear a 50-foot obstacle.OAT = StdPressure altitude = Sea levelTakeoff weight = 2,700 lbHeadwind component = Calm A 1,000 feet. B 1,400 feet. C 1,700 feet.

Determine the total distance required for takeoff to clear a 50-foot obstacle.OAT = StdPressure altitude = Sea levelTakeoff weight = 2,700 lbHeadwind component = Calm A 1,000 feet. B 1,400 feet. C 1,700 feet.


Ensembles d'études connexes

Early U.S. History Chapter 5 Study Guide

View Set

Teaspoons, Tablespoons, cups CUT IT IN HALF

View Set

General Insurance Practice Questions

View Set

Writing a Narrative Application Essay 70%

View Set

Supply Chain Management Exam 3(9-12, 14,15)

View Set

Anti-Discrimination Law: Title VII of the Civil Rights Act of 1964

View Set

International Business Chapter 9

View Set