For a VFR routing along a mountainous region with the information given, determine the lowest usable flight level for this leg:
Elevation of nearby airport: 3400 ft
Grid-MORA: 5400 ft
Obstacle Clearance required: 2000 ft
QNH 983 hPa
Temperature ISA-12
Magnetic Course: 220°(M)
For a VFR routing along a mountainous region with the information given, determine obstacle clearance in the situation:
Elevation of nearby airport: 8800 ft
Cruising Level: FL 120
Minimum Grid-Area-Altitude: 10500 ft
QNH 983 hPa
Temperature: ISA-12
Minimum Safe Altitude (MSA) 8500 ft
OAT -20°C at FL 85
QNH 1003 hPa
Elevation 4000 ft
The minimum safe Pressure Altitude, rounded to 10 ft, is:
For a VFR routing along a mountainous region with the information given, determine the lowest usable flight level for this leg:
Elevation of nearby airport: 1300 ft
Grid-MORA: 6500 ft
Obstacle Clearance required: 2000 ft
QNH 993 hPa
Temperature ISA-18
Magnetic Course: 120°(M)
Departure elevation 1500 ft
Cruise Level FL075
QNH 1023 hPa
OAT ISA
1 hPa 30 ft
What vertical distance do you have to climb to Cruise Level?
Descent from 7500 ft AMSL to 1000 ft AMSL, 6 NM from a VORTAC, GS 156 kt, Rate of Descent 800 ft/min.
The distance between the VORTAC and Top of Descent (TOD) is:
Cruise Level FL95
GS 155 kt
TAS Constant
Wind component Negligible
OAR ISA
QNH 1030 hPa
Rate of Descent 500 ft/min
To arrive at 2000 ft overhead the airfield, descent must commence at a distance of:
With the following information given, calculate the TAS:
Cruising Level: FL 370
Mach number: M 0.80
Temperature: ISA-10
Wind component: -65 kt
Compass Heading 252°
Variation 22°E
Deviation 3°W
Drift Angle 9° right
The True Track is:
True course (TC) 017°
W/V 340°/30 kt
True air speed (TAS) 420 kt
Calculate the wind correction angle (WCA) and ground speed (GS):
+2°, 406 kt
-2°, 436 kt
+2°, 396 kt
-2°, 396 kt
-2°, 396 kt
Given a heading of 220 at a TAS of 130 kt. and wind/velocity 280/30, calculate (1) drift angle and (2) ground speed:
(1) 13°R, (2) 142 kt.
(1) 13°R, (2) 118 kt.
(1) 13°L, (2) 142 kt.
(1) 13°L, (2) 118 kt.
(1) 13°L, (2) 118 kt.
Given a heading of 060 at a TAS of 140 kt. and wind/velocity 170/20, calculate (1) drift angle and (2) ground speed:
(1) 7°L, (2) 132 kt.
(1) 7°R, (2) 148 kt.
(1) 7°L, (2) 148 kt.
(1) 7°R, (2) 132 kt.
(1) 7°L, (2) 148 kt.
Time to cruising level 36 min.
Still Air Distance 157 NAM
With an average 60 kt tailwind the ground distance is:
193 NM
216 NM
155 NM
128 NM
193 NM
Still air climb distance 189 Nautical Air Miles (NAM), time to climb 30 min.
Calculate the ground distance with a 30 kt head wind:
184 NM
174 NM
196 NM
205 NM
174 NM
FL 370: M 0.80, SAT -60° C, Headwind 15 kt
FL 330: M 0.78, SAT -60° C, Headwind 5 kt
FL 290: M 0.80, SAT -55° C, Headwind 15 kt
FL 270: M 0.76, SAT -43° C, Wind Zero
The Flight Level which will allow for the maximum ground speed is:
FL 270
FL 370
FL 330
FL 290
FL 270
QNH 1025 HPa
ISA +10 deg. C
IFR flight
135° magnetic course airway
MSA 7800 ft
The minimum flight level is:
FL90
Given the following information, calculate the minimum flight level to be flown under IFR:
MOCA: 4400 ft
Magnetic course: 233(M)
Elevation of nearby airport: 1500 ft
Magnetic variation: 12°W
Temperature: ISA-18
QNH reported at airport: 993 hPa
Transition Altitude: 5000 feet
Cruise Level FL370
Cruise Regime M 0.74
OAT -47° C
The TAS in kt is:
Distance 300 NM
Wind (true) 340/40
Magnetic course 300°
Variation 20°E
TAS 440 kt
Calculate the flight time:
160 Imp.GAL
Fuel density: 0.81 kg/L
The fuel mass is:
Fuel mass 180 kg
Fuel density: 0.80 kg/L
The fuel volume is:
Fuel mass 240 kg
Fuel density: 0.78 kg/L
The fuel volume is:
In case a crew plans to pre-position fuel at the next airport to re-fill a quantity of 1200 lbs, how many of 40-Imperial gallon fuel drums have to be ordered, assuming a specific gravity of 0.78 kg/L?
At 0.8 relative fuel density the fuel flow of a turbine engine is 200 kg/h.
With all other in-flight conditions constant, at 0.75 density the fuel flow will be: