Quiz: Performance and Limitations 2

Answer (B) 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.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) The standard empty weight of an airplane includes airframe, engines, and all items of operating equipment that have fixed locations and are permanently installed. It includes unusable fuel, full operating fluids, and full engine oil.

Answer (B) is correct. (FAA-H-8083-25B Chap 10) Both the total weight and the total moment must be calculated. As in most weight and balance problems, you should begin by setting up a schedule as below. Note that the empty weight in Fig. 67 is given as 2,110 with a moment/100 in. of 1,652 (note the use of moment/100 on this chart), and that empty weight includes the oil. Be aware that some table values do not result in an accurate mathematical answer. You should use the table as a guide, but do not neglect to check your math.
The next step is to compute the moment/100 for each item. The pilot and copilot together weigh 316 lb., and their moment/100 is 268.6 lb.-in. (316 lb. × 85 in. ÷ 100). The passengers (forward position) moment/100 is 144.3 lb.-in. (130 lb. × 111 in. ÷ 100). The passengers (aft position) moment/100 is 199.92 lb.-in. (147 lb. × 136 in. ÷ 100). The baggage moment/100 is 75 lb.-in. (read directly from the table). The 75-gal. fuel weight is 450 lb., and the moment/100 is 338 lb.-in. (read directly from the table). Note that the gross weight of 3,203 is within the 3,400-lb. maximum allowable by 197 lb., which is within the CG envelope.

Answer (B) is correct. (FAA-H-8083-25B Chap 10) Both the total weight and the total moment must be calculated. As in most weight and balance problems, you should begin by setting up a schedule as below. Note that the empty weight in Fig. 67 is given as 2,110 with a moment/100 in. of 1,652 (note the use of moment/100 on this chart), and that empty weight includes the oil.
The next step is to compute the moment/100 for each item. The pilot and copilot moment/100 is 340 lb.-in. (400 lb. × 85 in. ÷ 100). The passengers (aft position) moment/100 is 326.4 lb.-in. (240 lb. × 136 in. ÷ 100). The baggage moment/100 is 30 lb.-in. (read directly from the table). The 75-gal. fuel weight is 450 lb., and the moment/100 is 338 lb.-in. (read directly from the table). Note that the gross weight of 3,220 lb. is within the 3,400 lb. maximum allowable by 180 lb., and that this moment/100 of 2,686.4 lb.-in. is within the moment envelope at the intersection with 3,220 lb.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) Both the total weight and the total moment must be calculated. As in most weight and balance problems, you should begin by setting up a schedule as below. Note that the empty weight in Fig. 67 is given as 2,110 with a moment/100 in. of 1,652 (note the use of moment/100 on this chart), and that empty weight includes the oil.
The next step is to compute the moment/100 for each item. The pilot and copilot moment/100 is 318.75 lb.-in. (375 lb. × 85 in. ÷ 100). The passengers (aft position) moment/100 is 333.2 lb.-in. (245 lb. × 136 in. ÷ 100). The baggage moment/100 is 97.5 lb.-in.(65 lb. × 150 in. ÷ 100). The 70-gal. fuel weight is 420 lb., and the moment/100 is 315 lb.-in. (read directly from the table). Note that the gross weight of 3,215 lb. is within the 3,400 lb. maximum allowable by 185 lb., and that the moment/100 of 2,716.45 is within the moment envelope at the intersection with 3,215 lb.

Answer (B) 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.

Answer (B) is correct. (FAA-H-8083-25B Chap 10) 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 correct. (FAA-H-8083-25B Chap 10) The requirement is the effect of a change in loading. Look at Fig. 32 for occupants. Losing the 180-lb. passenger from the front seat reduces the MOM/100 by 153. Moving the 204-lb. passenger from the rear seat to the front reduces the MOM/100 by about 74 (247 – 173). The total moment reduction is thus about 227 (153 + 74). As calculated below, the CG moves forward from 84.01 to 81.00 inches.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) The maximum allowable weight on the “Moment limits vs. weight” chart (Fig. 33) is 2,950 pounds. The total of the given weights is 2,905 lb. (including the empty weight of the airplane at 2,015 lb. and the fuel at 6 lb./gal.), so baggage cannot weigh more than 45 pounds.
It is still necessary to compute total moments to verify that the position of these weights does not move the CG out of CG limits. The total moment of 2,459 lies safely between the moment limits of 2,422 and 2,499 on Fig. 33, at the maximum weight, so this airplane can carry as much as 45 lb. of baggage when loaded in this manner.

Answer (A) 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).

Answer (A) is correct. (FAA-H-8083-25B Chap 10) Both the total weight and the total moment must be calculated. As in most weight and balance problems, you should begin by setting up a schedule as below. Note that the empty weight in Fig. 32 is given as 2,015 with a moment/100 in. of 1,554 (note the change to moment/100 on this chart) and that empty weight includes the oil.
The next step is to compute the moment/100 for each item. The front seat occupants’ moment/100 is 289 (340 × 85 ÷ 100). The rear seat occupants’ moment/100 is 357 (295 × 121 ÷ 100). The fuel (main tanks) weight of 264 lb. and moment/100 of 198 is read directly from the table. The baggage moment/100 is 78 (56 × 140 ÷ 100). The last step is to go to the “Moment limits vs. weight” chart (Fig. 33) and note that the maximum weight allowed is 2,950, which means that the plane is 20 lb. over. At a moment/100 of 2,476, the plane is within the CG limits because the moments/100 may be from 2,422 to 2,499 at 2,950 pounds.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) You need to calculate the weight and moment of the loaded airplane. The weight of the empty plane, including oil, is 2,015 lb., and it has a moment of 1,554. The 411 lb. in the front seats has a total moment of 349.35 [411 × 85 (ARM) = 34,935 ÷ 100 = 349.35]. The rear seat occupants have a weight of 100 lb. and a moment of 121.0 [100 × 121 (ARM) = 12,100 ÷ 100 = 121.0]. The fuel weight is given on the chart as 264 lb. with a moment of 198. On the Fig. 33 chart, the acceptable moment/100 range for 2,790 lb. is 2,243 to 2,374. Thus, the CG of 2,222.35 is forward of the acceptable moment/100 range. At 2,890 lb. (2,790 + 100) and a moment/100 of 2,362.35 (2,222.35 + 140), the new loaded airplane is within the acceptable moment/100 range of 2,354 to 2,452.

Answer (C) is correct. (FAA-H-8083-25B Chap 10) The moments for the pilot, front passenger, fuel, and oil are on the loading graph in Fig. 34. Total all the moments and the weight as shown in the schedule below.
Now refer to the center of gravity moment envelope graph. Find the gross weight of 2,033 on the vertical scale, and move horizontally across the graph until intersecting the vertical line that represents the 79.2 moment. A moment of 79.2 lb.-in. falls into the normal category envelope.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) To find the CG moment/1000, find the moments for each item and total the moments as shown in the schedule below. For the fuel, the loading graph shows the maximum as 38 gal. for standard tanks. Note that the reference point for 38 gal. of fuel is not depicted correctly in the FAA figure. Use the fuel weight of 228 lb. for the calculation (38 gal. × 6 lb. = 228 pounds). Find the oil weight and moment by consulting Note 2 on Fig. 34; it is 15 lb. and –0.2 moments. These total 75.1, so this answer is correct.

Answer (C) is correct. (FAA-H-8083-25B Chap 10) First, total the weight to get 1,999 lb. Note that the 38 gal. of fuel weighs 228 lb. (38 gal. × 6 lb./gallon).
Find the moments for the pilot and front seat passenger, rear passengers, and fuel by using the loading graph in Fig. 34. Find the oil weight and moment (15 lb. and –0.2 moments) by consulting Note 2. Note the reference point for 38 gal. of fuel is not depicted correctly. Use the fuel weight of 228 lb. for the calculation. Total the moments as shown in the schedule below. Now refer to the center of gravity moment envelope. Find the gross weight of 1,999 lb. on the vertical scale, and move horizontally across the chart until intersecting the vertical line that represents the 80.8 moment. Note that a moment of 80.8 lb.-in. falls into the utility category envelope.

Answer (C) 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 envelope. On the loading graph, 105 lb. of baggage indicates approximately 10 moments. Thus, a total of 108.2 moments (98.2 + 10) is within the 109 moments allowed on the envelope for 2,300 lb. of weight. Therefore, baggage of 105 lb. can be loaded.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) To find the desired location of the 500-lb. weight, compute and sum the moments left and right of the fulcrum. Set them equal to one another and solve for the desired variable. The 500-lb. weight must be 16 in. from the fulcrum to balance the plank. The weight should be shifted 1 in. to the left.

Answer (C) is correct. (FAA-H-8083-25B Chap 10) Compute and sum the moments left and right of the fulcrum. Set them equal to one another and solve for the desired variable

Answer (C) 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.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) ( ? )
To compute the CG, you must first multiply each weight by the arm to get the moment. Note that the fuel is given as 30 gallons. To get the weight, multiply the 30 by 6 lb. per gal. (30 × 6) = 180 pounds. Now add the weights and moments. To get CG, you divide total moment by total weight (193,193 ÷ 2,055.0) = a CG of 94.01 inches.

Answer (B) 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.

Answer (C) is correct. (FAA-H-8083-25B Chap 10) 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 (B) is correct. (FAA-H-8083-25B Chap 10) Tail-heavy loading produces very light control forces, which makes it easy for the pilot to inadvertently overstress the aircraft.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) Airplanes become less stable at all speeds as the center of gravity is moved backward. The rearward center of gravity limit is determined largely by considerations of stability.

Answer (A) is correct. (FAA-H-8083-25B Chap 10) The recovery from a stall in any airplane becomes progressively more difficult as its center of gravity moves backward. Generally, airplanes become less controllable, especially at slow flight speeds, as the center of gravity is moved backward.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) Pressure altitude is determined by adjusting the altimeter setting to 29.92" Hg, i.e., adjusting for nonstandard pressure. This is the indicated altitude of 1,380 ft. plus or minus the pressure altitude conversion factor (based on the current altimeter setting).
On the right side of Fig. 8 is a pressure altitude conversion factor schedule. Add 1,533 ft. for an altimeter setting of 28.30 and 1,630 ft. for an altimeter setting of 28.20. Using interpolation, you must subtract 20% of the difference between 28.3 and 28.2 from 1,630 ft. (1,630 – 1,533 = 97 × .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.

Answer (A) 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.

Answer (A) 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.

Answer (A) is correct. (FAA-H-8083-25B Chap 11) 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 the 35°F level. At this point, the density altitude is approximately 1,950 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 1,000-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 (C) is correct. (FAA-H-8083-25B Chap 11) 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 correct. (FAA-H-8083-25B Chap 11) 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 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.

Answer (C) 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.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) 2High density altitude reduces all aspects of an airplane’s performance, including takeoff and climb performance.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) 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 (B) is correct. (FAA-H-8083-25B Chap 11) When air temperature increases, density altitude increases because, at a higher temperature, the air is less dense.

Answer (B) is correct. (FAA-H-8083-25B Chap 11) 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 correct. (FAA-H-8083-25B Chap 12) 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 (A) is correct. (FAA-H-8083-25B Chap 8) On warm days, the atmospheric pressure levels are higher than on cold days. Your altimeter will indicate a lower than true altitude. Remember, “low to high, clear the sky.”

Answer (A) is correct. (FAA-H-8083-25B Chap 8) The airplane will be lower than the altimeter indicates when flying in air that is colder than standard temperature. Remember that altimeter readings are adjusted for changes in barometric pressure but not for changes in temperature. When one flies from warmer to cold air and keeps a constant indicated altitude at a constant altimeter setting, the plane has actually descended.

Answer (C) is correct. (FAA-H-8083-25B Chap 8) In air that is warmer than standard temperature, the airplane will be higher than the altimeter indicates. Said another way, the altimeter will indicate a lower altitude than actually flown.

Answer (B) is correct. (FAA-H-8083-25B Chap 8) When flying from higher pressure to lower pressure without adjusting your altimeter, the altimeter will indicate a higher than actual altitude. As you adjust an altimeter barometric setting lower, the altimeter indicates lower.

Answer (C) is correct. (FAA-H-8083-25B Chap 8) When an altimeter setting is at a lower value than the correct setting, the altimeter is indicating less than it should and thus would be showing lower than the actual altitude above sea level.

Answer (B) is correct. (FAA-H-8083-25B Chap 8) Pressure altitude equals true altitude when standard atmospheric conditions (29.92" Hg and 15°C at sea level) exist.

Answer (C) is correct. (FAA-H-8083-25B Chap 8) Pressure altitude and density altitude are the same when temperature is standard.

Answer (B) is correct. (FAA-H-8083-25B Chap 8) Indicated altitude (what you read on your altimeter) approximates the true altitude (distance above mean sea level) when standard conditions exist and your altimeter is properly calibrated.

Answer (C) is correct. (FAA-H-8083-25B Chap 8) Altimeter setting is the value to which the scale of the pressure altimeter is set so that the altimeter indicates true altitude at field elevation.

Answer (B) is correct. (FAA-H-8083-25B Chap 8) Pressure altitude is the airplane’s height above the standard datum plane of 29.92" Hg. If the altimeter is set to 29.92" Hg, the indicated altitude is the pressure altitude.

Answer (B) is correct. (FAA-H-8083-25B Chap 8) Density altitude is the pressure altitude corrected for nonstandard temperature.

Answer (A) is correct. (FAA-H-8083-25B Chap 8) True altitude is the actual altitude above mean sea level, i.e., MSL.

Answer (B) is correct. (FAA-H-8083-25B Chap 8) Absolute altitude is altitude above the surface, i.e., AGL.