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► Find the time to the point of safe return (PSR). Given: Maximum useable fuel 15 000 kg. Minimum reserve fuel 3500 kg. TAS Out 425 kt, head wind component out 30 kt. TAS Return 430 kt, tailwind component return 20 kt. Average fuel flow 2150 kg/h:
Answer: 2 h 51 min
PSR = SE (safe endurance) x GSh / (GSo + GSh)
GS(H) = 450 kt (430 + 20)
GS(O) = 395 kt (425 - 30)
SE = 5.348 ((15000-3500) / 2150)
PSR = 5.348 x 450 / (395 + 450)
PSR = 2h 51 min
(You could also use the GFF formulas to get the answer. Both will give you 2H51)
► An aeroplane is flying at TAS 180 kt on a track of 090°. The W/V is 045°/50kt. How far can the aeroplane fly out from its base and return in one hour?
Answer: 85 NM
SE (safe endurance) = 1 hour
Calculate wind component:
045 - 090 = -45
COS(-45) = 0.7 (Positive value so HWC)
0.7 x 50kt = 35kt HWC
GS out = 145kt
GS home = 215kt
PSR = 1 x 215 / (145 + 215)
PSR = 35 min
D =V x T
D = 145 x 0.59
D = 85 NM
► An aircraft takes off from an airport 2 hours before sunset. The pilot flies a track of 090°(T), W/V 130°/20kt, TAS 100 kt. In order to return to the point of departure before sunset, the furthest distance which may be travelled is:
Answer: 97 NM
We need to be back in 2 hours so that'll be our SE
Find wind component:
130 - 90 = 40
COS(40) = 0.77 (Positive value so HWC)
0.77 x 20kt = 15kt HWC
GS out = 85kt
GS home = 115kt
PSR = 2 x 115 / (85 + 115)
PSR = 1.15 (1h9min)
D = 1.15 x 85kt
D = 97 NM
► A helicopter is on a 150 NM leg to an off-shore oil rig. Its TAS is 130 kts with a 20 kts tailwind, its endurance is 3 hrs 30 min without reserve. Upon reaching destination, it is asked to proceed outbound to locate a ship in distress, on a track which gives a 15 kts tailwind. Maintaining zero reserve on return to the oil rig, the helicopter can fly outbound for a distance of:
Answer: 160,3 NM
Note that the SE is not 03:30 for the formula because we also had to fly to the oil rig first which took (150NM / 150 GS) 1 hour. Safe endurance will therefore be 02:30
GS Out = 130 - 15 = 115kt
GS Home = 130 + 15 = 145kt
PSR = 2.5 x 145 / (115 + 145)
PSR = 1.39
D = T x V
D = 160 NM
► An aircraft is to fly from B to C. The total fuel available is 50.000 kg but the aircraft must land with at least 5.000 kg. From B to C the TAS is 400 kts, wind component 30 kts headwind, fuel flow 7.800 kg/hr. Should the aircraft have to re-turn from the PSR its TAS shall be 380 kts, wind component 30 kts tailwind and fuel flow 7.500 kg/hr. What is the distance and time to the point of safe return (PSR) from B?
Answer: 1.143NM; 185min
Fuel that we can use = 50.000 - 5.000 = 45.000
For this Question we'll be using the Fuel Flow formulas. First calculate the GFF (Gross Fuel Flow) for each leg: GFF = FF / GS on that leg.
B -> C = FF (7.800) / 370 GSo = 21.08GFF
C -> B = FF (7.500) / 410 GSh = 18.29GFF
Next, add both together to find the Total GFF
21.08 + 18.29 = 39.37 Total GFF
To find the distance to the PSR, simply divide the Total fuel available by the Total GFF
45.000 / 39.37 = 1143 NM
1143 NM / 370kt = 3 hours 5 minutes = 185 minutes
► A flight is planned from L to M, distance 850 NM. Using the following data, calculate the time and distance to PSR:
Wind component out: 35 kts (TWC)
TAS: 450 kts
Mean fuel flow out: 2.500 kg/hr
Mean fuel flow inbound: 1.900 kg/hr
Fuel available: 6.000 kg
Answer: 1 hrs 16 minutes; 616 NM
Same as the Question above, first find the GFF for both legs:
L -> M = 2.500 FF / 485 GSo = 5.15 GFF
M -> L = 1.900 FF / 415 GSh = 4.57 GFF
Total GFF = FF1 + FF2
Total GFF = 9.72
Distance to reach PSR = 6.000 / 9.72 = 617NM
T = D (617) / V (485)
T = 1 hour 16 min
Answer: 2 h 51 min
PSR = SE (safe endurance) x GSh / (GSo + GSh)
GS(H) = 450 kt (430 + 20)
GS(O) = 395 kt (425 - 30)
SE = 5.348 ((15000-3500) / 2150)
PSR = 5.348 x 450 / (395 + 450)
PSR = 2h 51 min
(You could also use the GFF formulas to get the answer. Both will give you 2H51)
► An aeroplane is flying at TAS 180 kt on a track of 090°. The W/V is 045°/50kt. How far can the aeroplane fly out from its base and return in one hour?
Answer: 85 NM
SE (safe endurance) = 1 hour
Calculate wind component:
045 - 090 = -45
COS(-45) = 0.7 (Positive value so HWC)
0.7 x 50kt = 35kt HWC
GS out = 145kt
GS home = 215kt
PSR = 1 x 215 / (145 + 215)
PSR = 35 min
D =V x T
D = 145 x 0.59
D = 85 NM
► An aircraft takes off from an airport 2 hours before sunset. The pilot flies a track of 090°(T), W/V 130°/20kt, TAS 100 kt. In order to return to the point of departure before sunset, the furthest distance which may be travelled is:
Answer: 97 NM
We need to be back in 2 hours so that'll be our SE
Find wind component:
130 - 90 = 40
COS(40) = 0.77 (Positive value so HWC)
0.77 x 20kt = 15kt HWC
GS out = 85kt
GS home = 115kt
PSR = 2 x 115 / (85 + 115)
PSR = 1.15 (1h9min)
D = 1.15 x 85kt
D = 97 NM
► A helicopter is on a 150 NM leg to an off-shore oil rig. Its TAS is 130 kts with a 20 kts tailwind, its endurance is 3 hrs 30 min without reserve. Upon reaching destination, it is asked to proceed outbound to locate a ship in distress, on a track which gives a 15 kts tailwind. Maintaining zero reserve on return to the oil rig, the helicopter can fly outbound for a distance of:
Answer: 160,3 NM
Note that the SE is not 03:30 for the formula because we also had to fly to the oil rig first which took (150NM / 150 GS) 1 hour. Safe endurance will therefore be 02:30
GS Out = 130 - 15 = 115kt
GS Home = 130 + 15 = 145kt
PSR = 2.5 x 145 / (115 + 145)
PSR = 1.39
D = T x V
D = 160 NM
► An aircraft is to fly from B to C. The total fuel available is 50.000 kg but the aircraft must land with at least 5.000 kg. From B to C the TAS is 400 kts, wind component 30 kts headwind, fuel flow 7.800 kg/hr. Should the aircraft have to re-turn from the PSR its TAS shall be 380 kts, wind component 30 kts tailwind and fuel flow 7.500 kg/hr. What is the distance and time to the point of safe return (PSR) from B?
Answer: 1.143NM; 185min
Fuel that we can use = 50.000 - 5.000 = 45.000
For this Question we'll be using the Fuel Flow formulas. First calculate the GFF (Gross Fuel Flow) for each leg: GFF = FF / GS on that leg.
B -> C = FF (7.800) / 370 GSo = 21.08GFF
C -> B = FF (7.500) / 410 GSh = 18.29GFF
Next, add both together to find the Total GFF
21.08 + 18.29 = 39.37 Total GFF
To find the distance to the PSR, simply divide the Total fuel available by the Total GFF
45.000 / 39.37 = 1143 NM
1143 NM / 370kt = 3 hours 5 minutes = 185 minutes
► A flight is planned from L to M, distance 850 NM. Using the following data, calculate the time and distance to PSR:
Wind component out: 35 kts (TWC)
TAS: 450 kts
Mean fuel flow out: 2.500 kg/hr
Mean fuel flow inbound: 1.900 kg/hr
Fuel available: 6.000 kg
Answer: 1 hrs 16 minutes; 616 NM
Same as the Question above, first find the GFF for both legs:
L -> M = 2.500 FF / 485 GSo = 5.15 GFF
M -> L = 1.900 FF / 415 GSh = 4.57 GFF
Total GFF = FF1 + FF2
Total GFF = 9.72
Distance to reach PSR = 6.000 / 9.72 = 617NM
T = D (617) / V (485)
T = 1 hour 16 min
