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► An aircraft is flying at MACH 0.84 at FL 330. The static air temperature is -48°C and the headwind component 52 kt. At 1338 UTC the controller requests the pilot to cross the meridian of 030W at 1500 UTC. Given the distance to go is 570 NM, the reduced MACH No. should be:
Answer: 0.80
We need to cover 570NM in 1 hour 22 minutes. 570/1h22min = 417 GS
417 GS + 52 WC = 469 TAS
On your navigation computer align the mach number index with the temperature (-48). On the outer scale find 469 TAS and on the inner scale read MN 0.80.
► How many feet do you have to climb to reach FL 75. Given: FL 75, departure aerodrome elevation 1500ft. QNH = 1023 hPa, temperature = ISA; 1 hPa = 30 ft.
Answer: 6300 ft
1) Go from pressure altitude to indicated altitude:
(Difference between QNH and 1013) x 30 = 300ft
7500 + 300 (1023 level lies lower than 1013) = 7800ft
2) Aircraft does not start at the QNH level but at aerodrome (1500 above that level)
7800ft - 1500ft = 6300ft
► An aeroplane flies at an airspeed of 380kt. It flies from A to B and back to A. Distance AB : 480NM. When going from A to B it experiences a headwind component = 60kt. The wind remains constant. The duration of the flight will be:
Answer: 2h 35min
GS Out = 320kts
GS Home = 440kts
Leg A - B = 480NM / 320kt = 1:30
Leg B - A = 480NM / 440kt = 1:05
Total = 2:35
► Flight planning chart for an aeroplane states that the time to reach the cruising level at a given gross mass is 36 minutes and the distance travelled is 157 NM (zero-wind). What will be the distance travelled with an average tailwind component of 60kts?
Answer: 193NM
157NM / 36min = 261 GS (= also TAS because of the "zero-wind")
Now consider 60kts tailwind
261 TAS + 60kt = 321kt GS
36min x 321kt = 193NM
► You are flying a constant compass heading of 252°. Variation is 22°E, deviation is 3°W and your INS is showing a drift of 9° right. True track is?
Answer: 280°
252 CH - 3W = 249 MH + 22E = 271 TH + 9Right Drift = 280°
(Drift right means that the track is right of the heading)
► In the cruise at FL 155 at 260 kt TAS, the pilot plans for a 500 feet/min descent in order to fly overhead MAN VOR at 2.000 feet (QNH 1030). TAS will remain constant during descent, wind is negligible , temperature is standard. The pilot must start the descent at a distance from MAN of:
Answer: 120NM
! Don't forget to apply the difference in Hpa (1013 and 1030), it makes it 1490ft and not 2000ft
Difference in altitude (14010) / (500ft/min) = 28 minutes
28min x 260 TAS = 121 NM
► A multi engine piston aeroplane is on an IFR flight. The fuel plan gives a trip fuel of 65 US Gallons. The alternate fuel, final reserve included, is 17 US Gallons. Contingency fuel is 5% of the trip fuel. The usable fuel at departure is 93 US Gallons. At a certain moment the fuel consumed according to the fuel gauges is 40 US Gallons and the distance flown is half of the total distance.
Answer: The remaining fuel is not sufficient to reach the destination with reserves intact
40USG used and we are halfway, we need another 40USG for the second part.
How much fuel do we have available (excluding the reserves?)
93 USG - 40 USG (used for halfway) = 53 USG Left
53 USG - 17 USG (reserves) = 36 USG (not enough for the second half)
Other way to look at it: We have used 40 USG so are left with (93-40) 53 USG. For the second half we need another 40 USG so we would be left with (53-40) 13 USG which means we would be using our reserves (17 USG).
► After flying for 16 min at 100kt TAS with a 20 kt tail wind component, you have to return to the airfield of departure. You will arrive after:
Answer: 24 min
16min x 120 GS = 32NM distance
32NM / 80 GS (tailwind changed into a headwind) = 24 minutes
► A sector distance is 450 NM long. The TAS is 460 kt. The wind component is 50 kt tailwind. What is the still air distance?
Answer: 406 nautical air miles (NAM)
NAM = (NGM / GS) x TAS
NAM = (450 / 510) x 460
NAM = 406
► At 1000 UTC Shanwick Oceanic clears you to enter the Oceanic control area at 47°00minN 008°W, 220 track miles from your current position, at time 1033 UTC. At FL200, ISA conditions and with TAS 430kts, headwind 30kts, FMS being unavailable, what is the required mach number to comply with this instruction?
Answer: M 0.70
We need to cover 220NM in 33 minutes
220NM / 33min = 400 GS
400 GS + 30kt HWC = 430 TAS
On your navigation computer align the mach number index with the OAT (standard at 20.000ft so use "15°C - (2x20) = -25°C"). On the outer scale find TAS 430 and read the MN on the inner scale = 0.70
► A descent is planned from FL340 so as to arrive at FL100 at a distance 6 NM from a VORTAC. With a GS of 280 kts and a rate of descent of 1.200 ft/min. The distance from the VORTAC when descent is started is:
Answer: 99 NM
Altitude difference (24.000) / 1.200 ft/min = 20 minutes
20min x 280kts = 93NM
BUT we need to be at that altitude 6NM BEFORE the VORTAC so 93NM + 6NM = 99NM
► During an IFR flight in a Beech Bonanza the fuel indicators show that the remaining amount of fuel is 100 lbs after 38 minutes. The total takeoff fuel at departure was 160 lbs. For the alternate fuel, 30 lbs is necessary. The planned fuel for taxi was 13 lbs. Final reserve fuel is estimated at 50 lbs. If the fuel flow remains the same, how many minutes can be flown to the destination with the remaining fuel?
Answer: 12 minutes
Determine the fuel flow:
We used 60 lbs in 38 minutes = 95 lbs per hour
We have 100 lbs left in the tanks. Subtract from that the reserve fuels (30 and 50) = 20 lbs that can be used
20lbs / (95lbs/h) = 12 minutes
► During a VFR flight at a navigational checkpoint the remaining usable fuel in the tanks is 60 US Gallons. The reserve fuel is 12 US Gallons. According to the flight plan the remaining flight time is 1h35min. Calculate the highest acceptable rate of consumption:
Answer: 30.3 USG/hr
60 – 12 = 48 USG
48 / 1h35 = 30.3
► Given: CAS/RAS of 130 kts, OAT 0°C at 10.000ft , trip distance of 240 NGM, track 275° (T) and W/V 030/30kts . What is your true heading and time enroute?
Answer: 285° and 88 minutes
1) Find the TAS via the navigation computer by aligning 0°C and 10.000ft in the airspeed window. Find 130 CAS on inner scale and read 152 TAS on the outer scale.
TAS = 152kts
2) Find wind component:
030° - 275° = -245
COS(-245) = -0.422 (negative number so TAILWIND)
0.422 x windspeed (30) = 13kts tailwind component
GS = 165kts
240 NGM / 165 GS = 88 minutes
(Only 1 answer has 88 minutes so we can already select the correct one and save time on the exam)
Answer: 0.80
We need to cover 570NM in 1 hour 22 minutes. 570/1h22min = 417 GS
417 GS + 52 WC = 469 TAS
On your navigation computer align the mach number index with the temperature (-48). On the outer scale find 469 TAS and on the inner scale read MN 0.80.
► How many feet do you have to climb to reach FL 75. Given: FL 75, departure aerodrome elevation 1500ft. QNH = 1023 hPa, temperature = ISA; 1 hPa = 30 ft.
Answer: 6300 ft
1) Go from pressure altitude to indicated altitude:
(Difference between QNH and 1013) x 30 = 300ft
7500 + 300 (1023 level lies lower than 1013) = 7800ft
2) Aircraft does not start at the QNH level but at aerodrome (1500 above that level)
7800ft - 1500ft = 6300ft
► An aeroplane flies at an airspeed of 380kt. It flies from A to B and back to A. Distance AB : 480NM. When going from A to B it experiences a headwind component = 60kt. The wind remains constant. The duration of the flight will be:
Answer: 2h 35min
GS Out = 320kts
GS Home = 440kts
Leg A - B = 480NM / 320kt = 1:30
Leg B - A = 480NM / 440kt = 1:05
Total = 2:35
► Flight planning chart for an aeroplane states that the time to reach the cruising level at a given gross mass is 36 minutes and the distance travelled is 157 NM (zero-wind). What will be the distance travelled with an average tailwind component of 60kts?
Answer: 193NM
157NM / 36min = 261 GS (= also TAS because of the "zero-wind")
Now consider 60kts tailwind
261 TAS + 60kt = 321kt GS
36min x 321kt = 193NM
► You are flying a constant compass heading of 252°. Variation is 22°E, deviation is 3°W and your INS is showing a drift of 9° right. True track is?
Answer: 280°
252 CH - 3W = 249 MH + 22E = 271 TH + 9Right Drift = 280°
(Drift right means that the track is right of the heading)
► In the cruise at FL 155 at 260 kt TAS, the pilot plans for a 500 feet/min descent in order to fly overhead MAN VOR at 2.000 feet (QNH 1030). TAS will remain constant during descent, wind is negligible , temperature is standard. The pilot must start the descent at a distance from MAN of:
Answer: 120NM
! Don't forget to apply the difference in Hpa (1013 and 1030), it makes it 1490ft and not 2000ft
Difference in altitude (14010) / (500ft/min) = 28 minutes
28min x 260 TAS = 121 NM
► A multi engine piston aeroplane is on an IFR flight. The fuel plan gives a trip fuel of 65 US Gallons. The alternate fuel, final reserve included, is 17 US Gallons. Contingency fuel is 5% of the trip fuel. The usable fuel at departure is 93 US Gallons. At a certain moment the fuel consumed according to the fuel gauges is 40 US Gallons and the distance flown is half of the total distance.
Answer: The remaining fuel is not sufficient to reach the destination with reserves intact
40USG used and we are halfway, we need another 40USG for the second part.
How much fuel do we have available (excluding the reserves?)
93 USG - 40 USG (used for halfway) = 53 USG Left
53 USG - 17 USG (reserves) = 36 USG (not enough for the second half)
Other way to look at it: We have used 40 USG so are left with (93-40) 53 USG. For the second half we need another 40 USG so we would be left with (53-40) 13 USG which means we would be using our reserves (17 USG).
► After flying for 16 min at 100kt TAS with a 20 kt tail wind component, you have to return to the airfield of departure. You will arrive after:
Answer: 24 min
16min x 120 GS = 32NM distance
32NM / 80 GS (tailwind changed into a headwind) = 24 minutes
► A sector distance is 450 NM long. The TAS is 460 kt. The wind component is 50 kt tailwind. What is the still air distance?
Answer: 406 nautical air miles (NAM)
NAM = (NGM / GS) x TAS
NAM = (450 / 510) x 460
NAM = 406
► At 1000 UTC Shanwick Oceanic clears you to enter the Oceanic control area at 47°00minN 008°W, 220 track miles from your current position, at time 1033 UTC. At FL200, ISA conditions and with TAS 430kts, headwind 30kts, FMS being unavailable, what is the required mach number to comply with this instruction?
Answer: M 0.70
We need to cover 220NM in 33 minutes
220NM / 33min = 400 GS
400 GS + 30kt HWC = 430 TAS
On your navigation computer align the mach number index with the OAT (standard at 20.000ft so use "15°C - (2x20) = -25°C"). On the outer scale find TAS 430 and read the MN on the inner scale = 0.70
► A descent is planned from FL340 so as to arrive at FL100 at a distance 6 NM from a VORTAC. With a GS of 280 kts and a rate of descent of 1.200 ft/min. The distance from the VORTAC when descent is started is:
Answer: 99 NM
Altitude difference (24.000) / 1.200 ft/min = 20 minutes
20min x 280kts = 93NM
BUT we need to be at that altitude 6NM BEFORE the VORTAC so 93NM + 6NM = 99NM
► During an IFR flight in a Beech Bonanza the fuel indicators show that the remaining amount of fuel is 100 lbs after 38 minutes. The total takeoff fuel at departure was 160 lbs. For the alternate fuel, 30 lbs is necessary. The planned fuel for taxi was 13 lbs. Final reserve fuel is estimated at 50 lbs. If the fuel flow remains the same, how many minutes can be flown to the destination with the remaining fuel?
Answer: 12 minutes
Determine the fuel flow:
We used 60 lbs in 38 minutes = 95 lbs per hour
We have 100 lbs left in the tanks. Subtract from that the reserve fuels (30 and 50) = 20 lbs that can be used
20lbs / (95lbs/h) = 12 minutes
► During a VFR flight at a navigational checkpoint the remaining usable fuel in the tanks is 60 US Gallons. The reserve fuel is 12 US Gallons. According to the flight plan the remaining flight time is 1h35min. Calculate the highest acceptable rate of consumption:
Answer: 30.3 USG/hr
60 – 12 = 48 USG
48 / 1h35 = 30.3
► Given: CAS/RAS of 130 kts, OAT 0°C at 10.000ft , trip distance of 240 NGM, track 275° (T) and W/V 030/30kts . What is your true heading and time enroute?
Answer: 285° and 88 minutes
1) Find the TAS via the navigation computer by aligning 0°C and 10.000ft in the airspeed window. Find 130 CAS on inner scale and read 152 TAS on the outer scale.
TAS = 152kts
2) Find wind component:
030° - 275° = -245
COS(-245) = -0.422 (negative number so TAILWIND)
0.422 x windspeed (30) = 13kts tailwind component
GS = 165kts
240 NGM / 165 GS = 88 minutes
(Only 1 answer has 88 minutes so we can already select the correct one and save time on the exam)
