Applying Human Factors Principles
Additional Questions
Performance
8·1 Which would provide the greatest gain in altitude in the shortest distance during climb after takeoff?
A-Vy.
B-VA·
C-Vx·
8-1. Answer C. GFDPPM 8-16 (AFH) Vx is the best angle of climb. This gives you the greatest gain in altitude for horizontal distance traveled. Answer (A) is wrong because Vy is the best rate of climb. This provides you the greatest gain in altitude over a period of time. Answer (B) is wrong because VA is maneuvering airspeed.
8·2 After takeoff, which airspeed would the pilot use to gain the most altitude in a given period of time?
A-Vy.
B-Vx·
C-VA·
8·2. Answer A. GFDPPM 8-16 (AFH) See explanation for Question 8-1.
8-3 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.
8-3. Answer B. GFDPPM 8-19 (PHB) Because the air is less dense, there is less airflow through the propeller, and the force and efficiency are reduced. Answers (A) and (C) are wrong because the propeller force and efficiency are reduced, not increased.
8·4 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.
8-4. Answer C. GFDPPM 8-8 (PHB) High temperatures increase density altitude with a resulting decrease in aircraft performance. In addition, high humidity reduces engine performance.
8-5 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.
8-5. Answer B. GFDPPM 8-19 (PHB) A high density altitude decreases engine performance with a resulting reduction in climb performance.
8-6 What effect, if any, does high humidity have on aircraft performance?
A - It increases performance.
B - It decreases performance.
C - It has no effect on performance.
8-6. Answer B. GFDPPM 8-8 (PHB) High humidity reduces engine performance by slightly increasing the density altitude of air entering the engine and retarding smooth burning of the fuel.
8·7 (Refer to figure 36.) Approximately what true airspeed should a pilot expect with 65 percent maximum continuous power at 9,500 feet with a temperature of 36°F below standard?
A-178MPH.
B -181 MPH.
C -183 MPH.
8-7. Answer C. GFDPPM 8·22 (PHB) Use the left-hand portion of the table, under ISA -36°F. Interpolate between the TAS values for 8,000 feet (181 MPH) and 10,000 feet (184 MPH). The closest answer is 183 MPH.
8-8 (Refer to figure 36 on page 8-3.) 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 inches Hg
Wind.................................................................... Calm
A - 60.2 gallons.
B - 70.1 gallons.
C - 73.2 gallons.
8-8. Answer B. GFDPPM 8-22 (PHB) The temperature of 22°C is found on the right-hand portion of the table (ISA + 20°C) at 8,000 feet. Read across to find a fuel flow of 11.5 GPH, and TAS of 164 KTS (use knots because the distance is in nautical miles). Now, find the time enroute by dividing 1,000 n.m. by 164 KTS.(Normally you would use groundspeed, but with a calm wind, TAS equals groundspeed.) The time enroute is approximately 6:06 hrs. Then multiply the time by fuel flow. The total fuel consumption is 70.1 gallons.
8-9 (Refer to figure 36 on page 8-3) 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.
8-9. Answer B. GFDPPM 8-22 (PHB) Use the center portion of the table for a standard day. You will need to interpolate to find the fuel flow for 11,000 feet which is halfway between 12,000 and 10,000 feet. The answer is 11.2 (11.5 - 10.9 = .6 + 2 = .3 + 10.9 = 11.2).
8-10 (Refer to figure 36 on page 8-3.) 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°P higher than standard.
A -19.8 inches Hg.
B - 20.8 inches Hg.
C - 21.0 inches Hg.
8-10. Answer C. GFDPPM 8-22 (PHB) The RPM is the same for all altitudes. Therefore, to determine what manifold pressure (MP) is required to achieve 65% maximum continuous power, enter the table under ISA + 36°F. The MP for 6,000 feet is 21.0", and for 8,000 feet it is 20.8". The interpolated MP for 6,500 feet is 20.95". The closest answer is 21.0" Hg.
8-11 (Refer to figure 37 on page 8-5.) 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.
8-11. Answer B. GFDPPM 8-12 (PHB) First, compute the difference between the runway (180°) and the wind (220°). The result is an angle of 40 degrees. Find the intersection of the 40 degree line and the 30 knot wind velocity are, then read across to the left side to find the headwind component of 23 knots.
8-12 (Refer to figure 37 on page 8-5.) 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.
8-12. Answer C. GFDPPM 8-12 (PHB) Start with the crosswind component of 25 knots at the bottom of the chart, and follow the line straight up to where it intersects the 45 degree angle line. This intersection is midway between the 30 and 40 knot wind velocity lines, or 35 knots.
8-13 (Refer to figure 37.) What is the maximum wind velocity for a 30° crosswind if the maximum crosswind component for the airplane is 12 knots?
A -16 knots.
B - 20 knots.
C - 24 knots.
8-13. Answer C. GFDPPM 8-12 (PHS) Start with the crosswind component of 12 knots at the bottom of the chart, and follow the line straight up to where it intersects the 30 degree angle line. This intersection is approximately 24 knots on the wind velocity scale.
8-14 (Refer to figure 37.) With a reported wind of north at 20 knots, which runway (6, 29, or 32) is acceptable for use for an airplane with a 13-knot maximum crosswind component?
A - Runway 6.
B - Runway 29.
C - Runway 32.
8-14. Answer C. GFDPPM 8-12 (PHS) At first glance, Runway 32 is most closely aligned with north (360°). To verify, find the crosswind component for each runway. Runway 32 is 40 degrees from the wind, and since the windspeed is 20 knots, the crosswind component is slightly less than 13 knots, so Runway 32 is acceptable. Runway 6 is 60 degrees from the wind, and the crosswind component is about 17.5 knots. Runway 29 is 70 degrees from the wind, and the crosswind component is about 19 knots. Both Runways 6 and 29 exceed the 13 knot maximum crosswind component.
8-15 (Refer to figure 37 on page 8-5.) With a reported wind of south at 20 knots, which runway (10, 14, or 24) is appropriate for an airplane with a 13-knot maximum crosswind component?
A-Runway 10.
B - Runway 14.
C - Runway 24.
8-15. Answer B. GFDPPM 8-12 (PHS) The same process is used as in Question 8-14. Runway 14 is most closely aligned with the wind and would have the least crosswind. The crosswind angle and component for each runway is: Runway 14, 40 degrees, 12.5 knots; Runway 10, 80 degrees, 19.7 knots; Runway 24, 60 degrees, 17.5 knots. Runway 14 is the only appropriate runway because the crosswind component is less than 13 knots.
8-16 (Refer to figure 37 on page 8-5.) What is the crosswind component for a landing on Runway 18 if the tower reports the wind as 2200 at 30 knots?
A -19 knots.
B - 23 knots.
C - 30 knots.
8-16. Answer A. GFDPPM 8-12 (PHS) The crosswind angle is 40 degrees (220° - 180° = 40°). Find the intersection of 40 degrees and 30 knots. Then read down to find the crosswind component of about 19 knots.
8-17(Refer to figure 38 on page 8-6.) Determine the approximate total distance required to land over a 50foot obstacle.
OAT...................................................................... .90°F
Pressure altitude........... ....... .............. ....... ..... ....4,000 ft
Weight ........ ......... .............. .............. ................ .2,800 lb
Headwind component..........................................l0 kts
A-1,525 feet.
B-1,950 feet.
C - 1,775 feet.
8-17. Answer C. GFDPPM 8-15 (PH B) Start at the lower left at 90°F (32°C), and move up to where it intersects the 4,000-foot pressure altitude line. Go right to the weight reference line and then down and to the right to 2,800 lbs. Go straight right to the wind component reference line, down and to the right to the 10 knot headwind line, and straight right to the obstacle height reference line. Move up and to the right through the obstacle height and read the landing distance over a 50-foot obstacle on the right-hand scale. It is approximately 1,775 feet.
8-18 (Refer to figure 39.) 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.
8-18. Answer B. GFDPPM 8-14, 15 (PH B) Use the table listed under sea level and 59°F, which is the standard temperature. Since you need to find the landing ground roll distance, do not include obstacle clearance. The ground roll is given as 445, but according to Note 1, you need to correct for headwind by decreasing the distance 10% for each 4 knots of headwind. In this case, subtract 10% of 445 (44.5) from 445.
The closest answer is 401 feet.
8-19 (Refer to figure 39.) Determine the total distance " required to land over a 50-foot obstacle.
Pressure altitude ..... ... ........... .... ......................7,500 ft
Headwind............................................................... 8 kts
Temperature...........................................................3 2 of
Runway...... ........... .............. ......................Hard surface
A -1,506 feet.
B - 1,004 feet.
C - 1,205 feet.
8-19. Answer B. GFDPPM 8-14, 15 (PH B) According to the table, the landing distance over a 50foot obstacle is 1,255 feet at 7,500 feet and 32°F. Note 1 says to decrease this distance by 10% for each 4 knots of headwind, so with 8 knots headwind, subtract 20 percent. 1,255 x .80 = 1,004 feet.
8-20 (Refer to figure 39 on page 8-7.) 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.
8-20. Answer B. GFDPPM 8-14, 15 (PHB) Use the table at 5,000 feet and 41°F. The distance to, land over a 50 ft obstacle is 1,195. According to Note 1, decrease the distance by 2.0% (239) for the 8 Knot headwind: (1,195 - 239 = 956 total landing distance).
8-21 (Refer to figure 39 on page 8-7.) Determine the approximate landing ground roll distance.
Pressure altitude............................................5,000 ft
Headwind...........................................................Calm
Temperature... .................................................101ΊF
A - 495 feet.
B - 545 feet.
C - 445 feet.
8-21. Answer B. GFDPPM 8-14, 15 (PHB) At 5,000 feet and 41°F (ISA Standard Temperature), the ground roll distance is 495 feet. According to Note 2, this distance is increased 10% for each 60°F above standard.
8-22 (Refer to figure 39 on page 8-7.) Determine the total distance required to land over a 50-foot obstacle.
Pressure altitude... .... ..... ... ................ .................3,750 ft
Headwind...................... ............ ........ .................. .12 kts
Temperature ............... ............................................. .Std
A - 794 feet.
B - 836 feet.
C - 816 feet.
8-22. Answer C. GFDPPM 8-14, 15 (PHB) 1. At 2,500 feet and Standard ISA temperature, the landing distance over a 50-foot obstacle with zero wind is 1,135 feet. At 5,000 feet this distance is 1 ,195 feet. At 3,750 feet assume the landing distance is half way between 1,135 and 1,195 feet. (1,135 + 1,195) feet + 2 = 1,165 feet. 2. Note 1 says to decrease the distance 10% for each 4 knots of headwind. The headwind is 12 knots. 12 knots x 10% decrease/4 knots = 30% decrease. 1,165 feet x (100% - 30%) = 816 feet.
8-23 (Refer to figure 39 on page 8-7.) 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.
8-23. Answer B. GFDPPM 8-14, 15 (PHB) This problem requires that you interpolate between the ground roll distances at sea level and 2,500 feet PA, Since 1,250 feet is midway between the two values, the ground roll would be 457.5 (470 - 445 = 25 + 2 = 13.5 + 445 = 457.5). To correct for headwind, subtract 20% of the distance (10% for each 4 kts). 20% of 457.5 is 91.5. The landing distance is 457.5 - 91.5 or 366 feet.
8-24 (Refer to figure 41.) 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.
8-24. Answer B. GFDPPM 8-6, 7 (PH B) Since temperature is standard, start at the intersection of the ISA and 4,000 foot pressure altitude line. Move right to the reference line and follow the guide line diagonally downward to the 2,800 pound line. Since winds are calm, move straight across to the obstacle height reference line. Follow the guide line upward to the 50 foot line, which is on the right-hand border. The takeoff distance is approximately 1,700 feet.
8-25 (Refer to figure 41.) Determine the total distance required for takeoff to clear a 50-foot obstacle.
OAT..........................................................................Std
Pressure altitude Sea level
Takeoff weight......... 2,700 1b
Headwind component ........... ..Calm
A -1,000 feet.
B-1,400 feet.
C - 1,700 feet.
8-25. Answer B. GFDPPM 8-6, 7 (PH B) Since temperature is standard, start at the intersection of the ISA line and sea level (S.L.). Move right to the reference line and follow the guide line diagonally downward to the 2,700 pound line. Move straight across to the obstacle height reference line, since winds are calm. Follow the guide line upward to the 50 foot line. The takeoff distance is about 1,400 feet.
8-26 (Refer to figure 41 on page 8-9.) Determine the approximate ground roll distance required for takeoff.
OAT ..100°F
Pressure altitude ..2,000 ft
Takeoff weight ..2,750 1b
Headwind component .Calm
A -1,150 feet.
B - 1 ,300 feet.
C - 1,800 feet.
8-26. Answer A. GFDPPM 8-6, 7 (PHB) Start at 100°F, move up to the 2,000 foot pressure altitude line, then right to the reference line. Follow the guide line down to 2,750 pounds. Since winds are calm, and there is no obstacle, move straight across to the right-hand border. The ground roll is about 1,150 feet.
8-27 (Refer to figure 41 on page 8-9.) Determine the approximate ground roll distance required for takeoff.
OAT . ......90°F
Pressure altitude.... ........ ..... .2,000 ft
Takeoff weight............ .... .................................2,500 lb
Headwind component......................................... .20 kts
A-650 feet
B - 850 feet
C - 1,000 feet
8-27. Answer A. GFDPPM 8-6,7 (PH B) Start at 90°F, move up to the 2,000 foot pressure altitude line, then right to the reference line. Follow the guide line down to 2,500 pounds. Move across to the next reference line, and follow the headwind guide line down to 20 knots. Since there is no obstacle, move straight across to the right-hand border. The ground roll is about 650 feet.
8-28 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.
8-28. Answer A. GFDPPM 8-32 (PHS) The empty weight of an aircraft includes unusable fuel. The term basic empty weight includes full engine oil. On older airplanes, the term licensed empty weight includes only undrainable oil. Therefore, answer (A) is the best choice. Answer (B) is wrong because empty weight also includes full operating fluids and unusable fuel. Answer (C) is wrong because empty weight does not include full fuel tanks.
8-29 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.
8-29. Answer C. GFDPPM 8-33 (PHB) This problem requires converting the weight of fuel to gallons. Since the standard weight of gasoline is e pounds per gallon, divide 110 pounds by 6, to find an answer of 18.33, or 18.4 gallons.
8-30
The CG is located how far aft of datum?
A-CG92.44.
B -CG94.01.
C-CG 119.8.
8-30. Answer B. GFDPPM 8-36 (PH B) First, fill in the table by entering the fuel weight (30 gal x 6 lb/gal = 180 lbs). Then, multiply each weight by the arm to find the moment.
The CG is the total moment divided by the total weight
8-31 (Refer to figures 33 and 34 on pages 8-13 and 8-14.) 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.
8-31. Answer A. GFDPPM 8-39 (PHB) Add up all the weights and you will find the airplane is 45 lbs. underweight. When adding the 45 lbs. of baggage, be sure to verify that the resulting center of gravity (CG) is within limits.
8-32 (Refer to figures 33 and 34 on pages 8-13 and 8-14.) 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 lb
A - 19 pounds overweight, CG within limits.
B - 19 pounds overweight, CG out of limits forward.
C - Weight within limits, CG out of limits.
8-32. Answer C. GFDPPM 8-36, 39 (PHS) First, construct a weight and moment table.
The total weight is at the maximum limit. Divide total moments by total weight to find the CG of 81.0, which is outside the limits.
8-33 (Refer to figure 35 on page 8-16.) What is the maximum amount of baggage that may be loaded aboard the airplane for the CG to remain within the moment envelope?
A - 105 pounds.
B - 110 pounds.
C - 120 pounds.
8-33. Answer A. GFDPPM 8-40 (PHS) Refer to Figure 35 to convert oil and fuel to pounds. Add up the known weights, for a total of 2,195 pounds.
Subtract 2,195 pounds from 2,300 max weight to find the maximum possible baggage weight of 105 pounds. While it appears that choice A is the only correct answer, it is a good idea to check the CG limits. Use the LOADING GRAPH and find the moment for each weight.
Total the moments and locate the maximum weight on the CENTER OF GRAVITY MOMENT ENVELOPE graph. The intersection of the loaded weight and moment is at the upper right-hand corner of the normal category envelope, and is just barely within limits.
8-34 (Refer to figure 35 on page 8-16.) Calculate the moment of the airplane and determine which category is applicable.
A -79.2, utility category.
B - 80.8, utility category.
C - 81.2, normal category
8-34. Answer B. GFDPPM 8-40 (PHB) Complete the table of weights and moments, using the LOADING GRAPH
The total moment is 80.9. Use the CENTER OF GRAVITY MOMENT ENVELOPE graph to find the total weight and total moment. The intersection falls within the upper right-hand corner of the utility category envelope.
8-35 (Refer to figure 35 on page 8-16.) What is the maximum amount of fuel that may be aboard the airplane on takeoff if loaded as follows?
A - 24 gallons.
B - 32 gallons.
C - 40 gallons.
8-35. Answer C. GFDPPM 8-40 (PH B) Complete the table of weights and moments, using the LOADING GRAPH.
The total weight without fuel is 2,060 pounds. This is 240 pounds below the maximum of 2,300 pounds. Dividing by 6 lb/gal, the maximum fuel load is 40 gallons. It is a good idea to check the moments as well. The total moment of 102.4 is within the CG envelope, so 40 gallons is acceptable.
8-36 (Refer to figure 35 on page 8-16.) Determine the moment with the following data:
A - 69.9 pound-inches.
B - 74.9 pound-inches.
C - 77.6 pound-inches.
8-36. Answer B. GFDPPM 8-40 (PHB) Use the LOADING GRAPH to determine the moments. Add these to find the total moment of 74.9.
8-37 (Refer to figure 35 on page 8-16.) Determine the aircraft loaded moment and the aircraft category.
A -78.2, normal category.
B - 79.2, normal category.
C - 80.4, utility category.
8-37. Answer B. GFDPPM 8-40 (PHB) Use the LOADING GRAPH to determine the moments
The total weight is 2,033 pounds, and the total moment is 79.2. Use the CENTER OF GRAVITY MOMENT ENVELOPE graph with the total weight and total moment. The intersection falls within the normal category, and outside the utility category.
8-38 (Refer to figures 33 and 34 on pages 8-13 and 8-14.) 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.
8-38. Answer A. GFDPPM 8-36, 41 (PHS) Use the weight shift formula to determine how far the CG shifts.
The change in CG is approximately 2.93, which is closest to answer (A).
8-39 (Refer to figures 33 and 34 on pages 8-13 and 8-14.) Which action can adjust the airplane's weight to maximum gross weight and the CG within limits for takeoff?
Front seat occupants............................................425lb
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.
8-39. Answer B. GFDPPM 8-39 (PHS) Complete the weight and moment table as shown below.
Using the table in Figure 34, you'll find the total weight and total moment are within the limits.
8-40 (Refer to figures 33 and 34 on pages 8-13 and 8-14.) What effect does a 35-gallon fuel bum (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 CG is unaffected.
C - Weight is reduced to 2,680 pounds and the CG moves forward.
8-40. Answer A. GFDPPM 8-39 (PHS) Use the chart in Figure 33 to find the weight and moment for 35 gallons of fuel (main tanks), and subtract these values from the total weight and moment. The result is the total weight and moment after the fuel burn.
8-41 (Refer to figures 33 and 34 on pages 8-13 and 8-14.) With the airplane loaded as follows, what action can be taken to balance the airplane?
Front seat occupants............................................411lb
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.
8-41. Answer B. GFDPPM 8-36,39,41 (PHB) Construct the table as shown below, find the subtotal, weight and moment, and use the chart in Figure 34. The subtotal moment (2,222.4) at the original weight is less than the minimum (forward) limit.
Since the baggage compartment is in an aft location, adding weight to this part of the airplane will shift the CG aft. Add the baggage weight and moment to the subtotals to find adjusted totals. Check the chart in Figure 34 to ensure that the moment is within limits. Answer (A) is wrong because if the auxiliary wing tanks are filled and the total weight and moment are adjusted, the moment will be less than the minimum. To check answer (C), find the original CG using the subtotals:
Then use the weight shift formula:
The weight of fuel is 10 gal x 6 Lb/gal = 60 Lbs. The distance between arms is 94 - 75 = 19. Since the fuel is transferred from an arm of 75 to an arm of 94, the CG moves aft 0.4 inches.
The new CG is 80.1 (79.7 + 0.4). Then, find the new moment on the chart in Figure 34. The new moment is less than the minimum.
8-41A (Refer to figure 62.) 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.
8-41A. Answer C. GFDPPM 8-35 To solve this problem, you must calculate the moments generated by the weights on each side of the fulcrum, then solve for the unknown weight as shown below.
(50 x 50) + (Y x 25) = (100 x 100)
8-41B · (Refer to figure 61.) 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.
8-41 B. Answer A. GFDPPM 8-41,42 To solve this problem, the moment of the 500 pound weight on the left side of the fulcrum must equal the sum of the moments of the 250 pound weight and the unequal weight of the plank on the right side. (500x) = (250 x 20) + (200 x 15); 500x = 5,000 lb-in + 3,000 lb-in; 500x = 8,000 lb-in; x=16in. Since the 500 pound weight is now sitting at 15 inches from the fulcrum, it must be moved 1 inch to the left.
8-42 (Refer to figure 8 on page 8-23.) What is the effect of a temperature increase from 25 to 500P 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.
8-42. Answer C. GFDPPM 8-9 and 8-56 (PHB) Follow the line above 25°F up to where it intersects 5,000 feet pressure altitude, and read 3,850 feet density altitude on the left scale. Do the same with 50°F, up to 5,000 feet, and then left to read 5,500 feet. The difference is an increase of 1,650 feet.
8-43 (Refer to figure 8 on page 8-23.) 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.
8-43. Answer B. GFDPPM 8-9 and 8-56 (PHB) Using the table on the right side of the chart, interpolate between 28.2 and 28.3 to get a conversion factor of 1,611 (a value 20% of the way between 1,630 and 1,533). Add this to the indicated altitude of 1,380 for a pressure altitude of 2,991 feet. Answer (A) is what you get if you use 28.2 without interpolating. Answer (C) is what you get if you use 28.3 without interpolating.
8-44 (Refer to figure 8 on page 8-23.) Determine the density altitude for these conditions:
Altimeter setting.................................................. .29.25
Runway temperature ...........................................+81ΊF
Airport e1evation......................................5,250 ft MSL
A - 4,600 feet MSL.
B - 5,877 feet MSL.
C - 8,500 feet MSL.
8-44. Answer C. GFDPPM 8-9 and 8-56 (PHB) First find the pressure altitude by using the pressure altitude conversion factor scale and interpolate for 29.25. The conversion factor is 626 (673 - 579 = 94 -;- 2 = 47 + 579 = 626). This is added to 5,250 feet to find a pressure altitude of 5,876 feet. Now, find 81°F on the OAT scale at the bottom of the graph and follow its line vertically to where it intersects with the 5,876-foot pressure altitude line. From this point, follow the horizontal density altitude line to the left scale to find an approximate density altitude of 8,500 feet.
8-45 ( Refer to figure 8.) 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.
8-45. AnswerA. GFCPPM 8-9 and 8-56(PHB) Find the conversion factors for 30.00 and 29.92, and interpolate to find the factor for 29.96 (-73-0=-73χ2 = -36.5). Subtract 36.5 from the elevation of 3,563 feet, to find a pressure altitude of 3,526.5 feet. This is rounded to 3,527 feet
8-46 (Refer to figure 8 on page 8-23.) What is the effect of a temperature increase from 30 to 500P on the density altitude if the pressure altitude remains at 3,000 feet MSL?
A - 900-foot increase.
B - 1,100-foot decrease.
C - 1,300-foot increase.
8-46. Answer C. GFDPPM 8-9 and 8-56 (PH B) First you must find the density altitude (DA) for 30°F. It is 1,600 feet. At 50°F, the DA is 2,900 feet, for an increase of 1 ,300 feet. Keep in mind that an increase in temperature will increase density altitude.
8-47 (Refer to figure 8 on page 8-23.) 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.
8-47. Answer A. GFDPPM 8-9 and 8-56 (PHB) First you must interpolate to find the conversion factor for 29.97 (-73 - 0 = -73 + 8 increments = -9 x 5 increments = -45). Subtract 45 from 1,386 to find the pressure altitude of 1,341 feet.
8·48 (Refer to figure 8 on page 8-23.) What is the effect of a temperature decrease and a pressure altitude increase on the density altitude from 900P and 1,250 feet pressure altitude to 55°P and 1,750 feet pressure altitude?
A -1,700-foot increase.
B - 1,300-foot decrease.
C - 1,700-foot decrease.
8-48. Answer C. GFDPPM 8-9 and 8-56 (PHB)
1. Enter the graph at 90°F on the bottom scale. Draw a line straight up to meet the upsloping 1,250 ft. pressure altitude line (visualize this line or draw it in between the 1,000 and 2,000 ft. lines), then go left to 3,600 feet on the density altitude scale.
2. Repeat for 55°F and 1,750 ft. pressure altitude to get a density altitude of 1,900 feet.
3. The difference is (1,900 - 3,600) feet = -1,700 feet.
8-49 (Refer to figure 21 on page 4-19.) En route to First Plight Airport (area 5), your flight passes over Hampton Roads Airport (area 2) at 1456 and then over Chesapeake Municipal at 1501. At what time should your flight arrive at First Flight?
A-1516.
B -1521.
C-1526.
8-49. Answer C. GFDPPM 8-54 (PHB) This question requires you to calculate groundspeed and then estimated time of arrival.
1. Determine the actual groundspeed (GS) of the aircraft.
a. Measure the distance between Hampton Roads Airport and Chesapeake Municipal (CPK) using your plotter (10 n.m.).
b. Determine the elapsed time (15:01 - 14:56 = 5 min.).
c. Determine the GS (10 n.m. in 5 min. = 2 n.m./min.
X 60 min. = 120 n.m. per hour groundspeed).
2. Determine the estimated time of arrival (ETA) at First Flight Airport.
a. Measure the distance between Chesapeake Municipal and First Flight Airport (50 n.m.).
b. Determine the time enroute between the two points (50 n.m. at 120 knots = approximately 25 minutes).
c. If the aircraft was over Chesapeake Municipal at 15:01, the ETA at First Flight Airport is approx. 15:26. (15:01 + 25 min. = 15:26).
8-50 (Refer to figure 22 on page 4-20.) What is the estimated time en route from Mercer County Regional Airport (area 3) to Minot International (area 1)? The wind is from 330° at 25 knots and the true airspeed is 100 knots. Add 3-1/2 minutes for departure and climbout.
A - 44 minutes.
B - 48 minutes.
C - 52 minutes.
8-50. Answer B. GFDPPM 8-54 (PHS) This question requires you to calculate groundspeed and then complete a time-speed-distance problem.
1. Measure the distance between Mercer County Regional Airport and Minot International (59 n.m.).
2. Determine the True Course (TC) (012°).
3. Determine the groundspeed using your flight computer.
a. Enter the wind direction and speed (330° True at 25 kts.)
b. Enter the TC (012°)
c. Enter the True Air Speed (TAS) (100 kts.)
d. GS = 80 kts.
4. Determine the time enroute using your flight computer (59 n.m. at 80 n.m./hr. = 44 min. 15 sec.)
5. Add departure and climb-out time (3 min. 30 sec. + 44 min. 15 sec. = 47 min. 45 sec.) This is rounded to 48 minutes.
8-51 (Refer to figure 23 on page 4-21.) What is the estimated time en route from Sandpoint Airport (area 1) to St. Maries Airport (area 4)? The wind is from 215° at 25 knots, and the true airspeed is 125 knots.
A - 38 minutes.
B - 34 minutes.
C - 30 minutes.
8-51. Answer B. GFDPPM 8-54 (PHS) This question requires you to calculate groundspeed and then complete a time-speed-distance problem.
1. Measure the distance from Sandpoint Airport to St. Maries Airport (approximately 58 n.m.).
2. Determine the true course (TC = 181°).
3. Determine groundspeed using the flight computer.
a. Enter the wind direction and speed (215° True at 25 kts.).
b. Enter the TC (181°). c. Enter the True Airspeed (125 kts.).
d. GS=103 kts. 4. Determine the time enroute using the flight computer (58 n.m. at 103 n.m./hr. = 34 min.).
8-52 (Refer to figure 23 on page 4-21.) Determine the estimated time en route for a flight from Priest River Airport (area 1) to Shoshone County Airport (area 3).
The wind is from 030 at 12 knots and the true airspeed is 95 knots. Add 2 minutes for climb-out.
A - 27 minutes.
B - 29 minutes.
C - 31 minutes.
8-52. Answer C. GFDPPM 8-54 (PHS) This question requires you to calculate groundspeed and then complete a time-speed-distance problem.
1. Measure the distance from Priest River Airport to Shoshone County Airport (48 n.m.). 2. Determine the true course (TC = 143°).
3. Determine groundspeed using the flight computer.
a. Enter the wind direction and speed (030° True at 12 kts.).
b. Enter the TC (143°).
c. Enter the True Airspeed (95 kts.).
d. GS=99 kts.
4. Determine the time enroute using the flight computer (48 n.m. at 99 n.m./hr. = 29 min.).
5. Add 2 min. for departure and climb-out (2 min. + 29 min. = 31 min.).
8-53 (Refer to figure 23 on page 4-21.) What is the estimated time en route for a flight from St. Maries Airport (area 4) to Priest River airport (area I)? The wind is from 300° at 14 knots and the true airspeed is 90 knots. Add 3 minutes for climb-out.
A - 38 minutes.
B - 43 minutes.
C - 48 minutes.
8-53. Answer B. GFDPPM 8-54 (PH B) This question requires you to calculate groundspeed and then complete a time-speed-distance problem.
1. Measure the distance from St. Maries Airport to Priest River Airport (53 n.m.).
2. Determine the True Course (345°).
3. Determine groundspeed using your flight computer.
a. Enter the wind direction and speed (300° at 14 kts.).
b. Enter the True Course (345°).
c. Enter the TAS (90 kts.).
d. GS = 80 kts.
4. Determine the time enroute using your flight computer (53 n.m. at 80 n.m./hr. = 39 min. 45 sec.).
5. Add 3 min. for departure and climb-out (3 min. + 39 min. 45 sec. = 42 min. 45 sec.). This is rounded to 43 minutes.
8-54 (Refer to figure 24 on page 4-22.) What is the estimated time en route for a flight from Allendale County Airport (area 1) to Claxton-Evans County Airport (area 2)? The wind is from 100° at 18 knots and the true airspeed is 115 knots. Add 2 minutes for climb-out.
A - 33 minutes.
B - 27 minutes.
C - 30 minutes.
8-54. Answer C. GFDPPM 8-54 (PHB) This question requires you to calculate groundspeed and then complete a time-speed-distance problem.
1. Measure the distance from Allendale County to Claxton-Evans County Airport (57 n.m.).
2. Determine the true course (TC = 212°).
3. Determine groundspeed using the flight computer. " a. Enter the wind direction and speed (100° True at .
18 kts.).
b. Enter the TC (212°).
c. Enter the True Airspeed (115 kts.).
d. GS = 121 kts.
4. Determine the time enroute using the flight computer (57 n.m. at 121 n.m./hr. = 28 min. 15 sec.).
5. Add 2 min. for departure and climb-out (2 min + 28 min. 15 sec. = 30 min. 15 sec.). This is rounded to 30 minutes.
8-55 (Refer to figure 24 on page 4-22.) What is the estimated time en route for a flight from C1axtonEvans County Airport (area 2) to Hampton Varnville Airport (area 1)? The wind is from 290° at 18 knots and the true airspeed is 85 knots. Add 2 minutes for climb-out.
A - 35 minutes.
B - 39 minutes.
C - 44 minutes.
8-55. Answer B. GFDPPM 8-54 (PHB) This question requires you to calculate groundspeed and then complete a Time-Speed-Distance problem.
1. Measure the distance from Claxton-Evans County Airport to Hampton Varnville Airport (57 n.m.).
2. Determine the True Course (045°).
3. Determine the groundspeed using your flight computer.
a. Enter the wind direction and speed (290° at 18 kts.).
b. Enter the True Course (045°).
c. Enter the TAS (85 kts.).
d. GS = 91 kts.
4. Determine the time enroute using 'your flight computer (57 n.m. at 91 n.m./hr. = 37 min. 30 sec.).
5. Add 2 minutes for departure and climb-out (2 min. + 37 min. 30 sec. = 39 min. 30 sec.) The closest answer is 39 minutes.
8-56 . (Refer to figure 24 on page 4-22.) While en route on Victor 185, a flight crosses the 248° radial of Allendale VOR at 0953 and then crosses the 216° radial of Allendale VOR at 1000. What is the estimated time of arrival at Savannah VORTAC?
A-1023.
B -1036.
C-1028.
8-56. Answer C. GFDPPM 8-54 (PHS) This question requires you to calculate groundspeed and then estimated time of arrival.
1. Determine actual groundspeed of the aircraft. a. Measure the distance along Victor 185 where it crosses the 2480 radial and the 216° radial of Allendale VOR (10 n.m.).
b. Determine the elapsed time. (10:00 - 09:53 = 7 min.).
c. Determine the groundspeed (10 n.m. in 7 min. = 86 knots).
2. Determine the Estimated Time of Arrival (ETA) over the Savannah VORTAC.
a. Measure the Distance along Victor 185 from the 216° radial of Allendale VORTAC to the Savannah VORTAC (40 n.m.).
b. Determine time enroute (GS is 86 kts.) (40 n.m.
at 86 n.m./hr. = 28 min.) c. Determine estimated time of arrival (10:00 + 28 mins. = 10:28).
8-57 (Refer to figure 26 on page 4-24.) What is the estimated time en route for a flight from Denton Muni (area 1) to Addison (area 2)? The wind is from 2000 at 20 knots, the true airspeed is 110 knots, and the magnetic variation is 7° east.
A - 13 minutes.
B - 16 minutes.
C - 19 minutes.
8-57. Answer A. GFDPPM 8-54 (PHS) This question requires you to calculate groundspeed, then complete a time-speed-distance problem.
1. Measure the distance from Denton Muni to Addison Airport (23 n.m.).
2. Determine True Course (128°).
3. Determine the groundspeed using your flight computer.
a. Enter the wind direction and speed (200° at 20 kts.).
b. Enter the True Course (128°).
c. Enter the TAS (110 kts.).
d. GS = 102 kts.
4. Determine time enroute (23 n.m. at 102 n.m./hr = 13 min. 30 sec.). The closest answer is 13 minutes.
8-58 (Refer to figure 26 on page 4-24.) Estimate the time enroute from Addison (area 2) to Redbird (area 3). The wind is from 3000 at 15 knots, the true airspeed is 120 knots, and the magnetic variation is 7° east.
A - 8 minutes.
B - 11 minutes.
C - 14 minutes.
8-58. Answer A. GFDPPM 8-54 (PHS) This question requires you to calculate groundspeed, then complete a time-speed-distance problem.
1. Measure the distance from Addison Airport to Redbird Airport (17 n.m.).
2. Determine the True Course (186°).
3. Determine the groundspeed using your flight computer.
a. Enter the wind direction and speed (300° at 15 kts.).
b. Enter the TC (186°).
c. Enter the TAS (120 kts.).
d. GS = 125 kts.
4. Determine time enroute (17 n.m. at 125 n.m./hr. = 8 min.8 sec.). The closest answer is 8 minutes.
8-59 If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 knots, the wind would be from
A - 019° and 12 knots.
B - 200° and 13 knots.
C - 246° and 13 knots.
8-59. Answer C. GFDPPM 8-61 .You should use a flight computer to solve for wind direction and velocity. This calculation is essentially the reverse of predicting groundspeed from forecast winds aloft.
8-60 (Refer to figure 63.) In flying the rectangular course, when would the aircraft be turned less than 90°?
A - Comers 1 and 4.
B - Comers 1 and 2.
C - Comers 2 and 4.
8-60. Answer A. GFDPPM 8-58 (AFH) To maintain the rectangular ground track at turn 1, the aircraft is turned less than 90 degrees to establish a right crab into the wind. Approaching turn 4, the aircraft is crabbed left into the wind and is therefore turned less than 90 degrees into the direct headwind.
8-61 (Refer to figure 67.) While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is not completed before crossing the road or reference line. This would most likely occur in turn
A - 1-2-3 because the bank is decreased too rapidly during the latter part of the turn.
B - 4-5-6 because the bank is increased too rapidly during the early part of the turn.
C - 4-5-6 because the bank is increased too slowly during the latter part of the turn.
8-61. Answer B. (AFH) Turn 4-5-6 begins with an upwind leg (Le., traveling into a headwind), resulting in a slower ground speed. The slower ground speed requires the pilot to add bank more slowly than on the downwind leg, so the aircraft has more time to travel equidistant from the "road' as it did on the other side, helping to create a semicircles with the same radii.
8-62 How far will an aircraft travel in 2-1/2 minutes with a groundspeed of 98 knots?
A-2.45 NM.
B-3.35NM.
C-4.08NM.
8-62. Answer C. GFDPPM 8-54 This problem may be solved using a flight computer. The basic formula for time, speed, distance calculations is:
Distance Traveled=Ground Speed x Time Traveled.
Therefore: x = (98 n.m./60 minutes) x 2.5 minutes
8-63 On a cross-country flight, point A is crossed at 1500 hours and the plan is to reach point B at 1530 hours. Use the following information to determine the indicated airspeed required to reach point B on schedule. Distance between A and B 70 NM Forecast wind 310° at 15 kts Pressure altitude 8,000 ft Ambient temperature -10°C True course 270°. The required indicated airspeed would be approximately
A -126 knots.
B -137 knots.
C -152 knots.
8-63. Answer B. GFDPPM 8-56, 8-60 (PHB) This question requires a flight computer to first solve for groundspeed (140 kts.). Second, use each of the indicated airspeeds given in the choices to compute a true airspeed (152 kts.) that will yield a ground speed equal to 140 kts. The answer is 137 kts.