Gleim 1.X.XX Flashcards
To hold an airplane in level flight at airspeeds from very slow to very fast, a pilot must coordinate thrust and…
Angle of attack
(Refer to Figure 20)
Which instrument shows the aircraft in a position where a straight course after a 90 degree left turn would result in intercepting the 180 radial?
3
RMI 3 shows the airplane on R-135 of the VOR (southeast on a heading of 300 degrees). A 90 degree turn to 210 would cause the airplane to fly southwest and thus intercept the R-180.
(Refer to Figure 20)
Which instrument shows the aircraft in a position where a 180 degree turn would result in the aircraft intercepting the 150 radial at a 30 degree angle?
6
RMI 4 is on R-105 of the VOR (north-northeast on a heading of 360). A 180 degree turn would cause the airplane to intercept R-150 at a 30 degree angle (180-150=30)
(Refer to Figure 17)
Which illustration indicates that the airplane will intercept the 060 radial at a 75 degree angle outbound, if the present heading is maintained?
5
The present magnetic heading of the airplane in illustration 5 is 345 degrees, so you will cross R-060 at a 75 degree angle. The TO indication indicates you are east of the 330-150 degree radials. The right deflection on the 240 degree OBS selection means you are south of the 060 radial.
(Refer to Figure 17)
Which illustration indicates that the airplane will intercept the 60 degree angle inbound, if the present heading is maintained?
6
Illustration 6 indicates that the airplane is east of the station on a heading of 300 degrees. The course selector is set to 240 and a TO indication. On a heading of 300 degrees, the airplane will intercept the 060 radial inbound, which is 240 degrees TO the station, at a 60 degree angle (300-240=60).
(Refer to Figure 17)
Which illustration indicates that the airplane should be turned 150 degrees left to intercept the 360 degree radial at a 60 degree angle inbound?
1
By turning the airplane as indicated in illustration 1, 150 degree left you would be heading 240 degrees. This would be a 60 degree interception to the 360 radial.
(Refer to Figure 2)
Select the correct statement regarding stall speeds.
Power-on stalls occur at lower airspeed in shallower banks.
(Refer to Figure 2)
Select the correct statement regarding stall speeds. The airplane will stall…
10 knots higher in a power-on, 60 degree bank, with gear and flaps up, than with gear and flaps down.
(Refer to Figure 4)
What is the stall speed of an airplane under a load factor of 2.5 G’s if the unaccelerated stall speed is 60 knots?
96 knots.
First, find 2.5 G’s on the far left vertical scale and move horz. to the right to the load factor curve, which intersects at about a 68 degree bank. Then move vertically up from that point to the intersection of the stall speed increase curve. Next, move left horz. to the vertical axis to determine a 60% increase in stall speed. (60kts x .60 = 36, 36+60=96kts)
(Refer to Figure 15)
Airport pressure altitude 2,000 ft
Airport temperature 20C
Cruise pressure altitude 10,000 ft
Cruise temperature 0C
What will be the fuel, time and distance required to climb to cruise altitude under the given conditions?
5 gallons, 9 minutes, 13 NM
Go up from 0C cruise temp to the 10,000 ft cruise pressure altitude. From there, proceed horz. to the right to the intersection of each of the curves (fuel, time, distance). From there, proceed down to 6 gal, 11 min, 16NM.
Since the airport pressure alt. is 2,000 ft, go back to the 20C and up to the 2,000 ft pressure alt line. Then horz. to each of the curves and move down to determine 1 gal, 2 min, 3NM.
6 - 1 = 5 gal, 11 - 2 = 9 min, 16 - 3 = 13NM
(Refer to Figure 9)
Using a normal climb, how much fuel would be used from engine start to 12,000 ft pressure altitude?
Aircraft weight 3,800 lbs
Airport pressure alt 4,000 ft
Temperature 26C
58 pounds
At 3,800 lbs, 51 lbs of fuel is required to climb from sea level to 12,000 ft (according to Fig. 9). From sea level to 4,000 ft, only 12 lbs is required. The net difference is 39 lbs.
The air temp of 26C is 19C over standard temp at 4,000 ft ( [4x2=8 standard lapse rate] -15C = 7C). There is a 1% increase for each 1 degree above standard (19C = 1.19). 36 lbs x 1.19 = 46.41 lbs of fuel. Finally, add 12 lbs to cover taxi, takeoff, etc.
46 + 12 = 58 lbs of fuel.
(Refer to Figure 9)
Using a normal climb, how much fuel would be used from engine start to 10,000 ft pressure altitude?
Aircraft weight 3,500 lbs
Airport pressure alt 4,000 ft
Temperature 21C
35 pounds
At 3,500 lbs, 31 lbs of fuel is required to climb from sea level to 10,000 ft (according to Fig. 9). From sea level to 4,000 ft only 11 lbs is required. The net difference is 20 lbs to climb from 4,000 ft pressure alt to 10,000 ft pressure alt.
The air temp of 21C is 14C over standard temp.( [4x2=8C standard lapse rate] - 15C = 7C). There is a 1% increase for each 1 degree above standard (14C = 1.14). 20 lbs x 1.14 = 22.8 lbs of fuel. Finally, add 12 lbs to cover taxi, takeoff, etc.
23 + 12 = 35 lbs of fuel.
(Refer to Figure 10)
Using a maximum rate of climb, how much fuel would be used from engine start to 6,000 ft pressure altitude?
Aircraft weight 3,200 lbs
Airport pressure alt 2,000 ft
Temperature 27C
24 pounds
At 3,200 lbs, 14 lbs of fuel is required to climb from sea level to 6,000 ft (according to Fig. 10). From sea level to 2,000 ft only 4 lbs is required. The net difference is 10 lbs to climb from 2,000 ft pressure alt to 6,000 ft pressure alt.
The air temp of 27C is 16C over standard temp.( [2x2=4C standard lapse rate] - 15C = 11C). There is a 1% increase for each 1 degree above standard (16C = 1.16). 10 lbs x 1.16 = 11.6 lbs of fuel. Finally, add 12 lbs to cover taxi, takeoff, etc.
12 + 12 = 24 lbs of fuel.
