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► If CAS is 190 kts. Altitude 9000 ft. Temp ISA - 10°C. True course (TC) 350°. W/V 320/40. Distance from departure to destination is 350 NM. Endurance 3 hours, and actual time of departure is 1105 UTC. The distance to the Point of Equal Time (PET) is:
Answer: 203 NM
PET = Distance x GS(H) / (GS(O) + GS(H))
Distance = 350 NM
GS(H) = ?
GS(O) = ?
Find the TAS with the navigation computer (align OAT with 9000ft and read TAS on the outer scale against CAS 190kts on the inner scale)
TAS = 214kts
Find the wind component by the formula;
Wind direction (320) - Course (350) = -30
COS(-30) = 0.866 (positive number so HEADWIND)
0.866 x windspeed (40) = 35kts HWC
GS(O) = 179kts
GS(H) = 249kts
PET = 350NM x 249kts / (179kts + 249kts)
PET reached after = 203NM
► If CAS is 190 kts. Altitude 9000 ft. Temp ISA - 10°C. True course (TC) 350°. W/V 320/40. Distance from departure to destination is 350 NM. Endurance 3 hours, and actual time of departure is 1105 UTC. The Point of Equal Time (PET) is reached at:
Answer: 1213 UTC
Same Question as above. The answer you'll get from the PET formula is a distance (NM). If you were asked for a time, use the T = D / V where V(speed) is the ground speed out.
T = 203 NM / 179 kts
T = 01:08
Start was at 11:05
11:05 + 01:08 = 12:13
► Find the distance from waypoint 3 (WP 3) to the critical point. Given: distance from WP 3 to WP 4 = 750 NM, TAS out 430 kt, TAS return 425 kt, tailwind component out 30 kt, head wind component return 40 kt:
Answer: 342 NM
PET = Distance x GS(H) / (GS(O) + GS(H))
Distance = 750 NM
GS(H) = (425 - 40) = 385 kt
GS(O) = (430 + 30) = 460 kt
PET = 750 x 385 / (460 + 385)
PET = 342 NM
► Distance from A to B is 1000 NM, all engine TAS is 400kts , engine failure TAS is 350kts. Wind component between A and PET is 50kts tailwind and wind component between PET to B is 30kts tailwind. What is the distance and time to the engine failure PET from A?
Answer: 441 NM, 59 min
Important for engine failure Questions on PET:
- For the general formula, BOTH ground speed out and home are the engine failure GS
- For the final "T = D/V" formula we need to use the all engine GS
GSo (1 engine) = (350 + 30) = 380 kts
GSh (1 engine) = (350 - 50) = 300 kts
1000 x 300 / (300 +380) = 441 NM
D (441) / V (all engine GSo) (400 + 50) = 59 minutes
With these Questions you have to picture yourself AT the PET with a failed engine and wondering if it will be quicker to turn around and fly back or continue to the destination. The Groundspeed (1engine) OUT (continue) is therefore the +30kts from the second part and not the +50kts from the first part.
► You fly from C to D, a distance of 450 NM. The WC C - D is +30, and the WC D - C is -40. TAS is 160 kt and reduced TAS is 130 kt. The fuel flow is 165 kg/hr, and the safe endurance when overhead C is 4 hours. Calculate PET between C and D, based on reduced TAS for the flight from PET to C/D. What is the flying time from C to PET?
Answer: 0:51
Same explanation as the Question above. The only difference being that here they refer to a 'reduced TAS' (= TAS for an engine failure).
GS OUT (continue with the engine failure from PET to D) = (130 + 30) = 160kt
GS HOME (turn around at PET and fly back to C) = (130 - 40) = 90kt
GS OUT ALL ENGINES = (160 + 30) = 190kt
PET = 450 x 90 / (160 + 90)
PET = 162 NM
T = D (162) / V (190)
T = 51 min
Answer: 203 NM
PET = Distance x GS(H) / (GS(O) + GS(H))
Distance = 350 NM
GS(H) = ?
GS(O) = ?
Find the TAS with the navigation computer (align OAT with 9000ft and read TAS on the outer scale against CAS 190kts on the inner scale)
TAS = 214kts
Find the wind component by the formula;
Wind direction (320) - Course (350) = -30
COS(-30) = 0.866 (positive number so HEADWIND)
0.866 x windspeed (40) = 35kts HWC
GS(O) = 179kts
GS(H) = 249kts
PET = 350NM x 249kts / (179kts + 249kts)
PET reached after = 203NM
► If CAS is 190 kts. Altitude 9000 ft. Temp ISA - 10°C. True course (TC) 350°. W/V 320/40. Distance from departure to destination is 350 NM. Endurance 3 hours, and actual time of departure is 1105 UTC. The Point of Equal Time (PET) is reached at:
Answer: 1213 UTC
Same Question as above. The answer you'll get from the PET formula is a distance (NM). If you were asked for a time, use the T = D / V where V(speed) is the ground speed out.
T = 203 NM / 179 kts
T = 01:08
Start was at 11:05
11:05 + 01:08 = 12:13
► Find the distance from waypoint 3 (WP 3) to the critical point. Given: distance from WP 3 to WP 4 = 750 NM, TAS out 430 kt, TAS return 425 kt, tailwind component out 30 kt, head wind component return 40 kt:
Answer: 342 NM
PET = Distance x GS(H) / (GS(O) + GS(H))
Distance = 750 NM
GS(H) = (425 - 40) = 385 kt
GS(O) = (430 + 30) = 460 kt
PET = 750 x 385 / (460 + 385)
PET = 342 NM
► Distance from A to B is 1000 NM, all engine TAS is 400kts , engine failure TAS is 350kts. Wind component between A and PET is 50kts tailwind and wind component between PET to B is 30kts tailwind. What is the distance and time to the engine failure PET from A?
Answer: 441 NM, 59 min
Important for engine failure Questions on PET:
- For the general formula, BOTH ground speed out and home are the engine failure GS
- For the final "T = D/V" formula we need to use the all engine GS
GSo (1 engine) = (350 + 30) = 380 kts
GSh (1 engine) = (350 - 50) = 300 kts
1000 x 300 / (300 +380) = 441 NM
D (441) / V (all engine GSo) (400 + 50) = 59 minutes
With these Questions you have to picture yourself AT the PET with a failed engine and wondering if it will be quicker to turn around and fly back or continue to the destination. The Groundspeed (1engine) OUT (continue) is therefore the +30kts from the second part and not the +50kts from the first part.
► You fly from C to D, a distance of 450 NM. The WC C - D is +30, and the WC D - C is -40. TAS is 160 kt and reduced TAS is 130 kt. The fuel flow is 165 kg/hr, and the safe endurance when overhead C is 4 hours. Calculate PET between C and D, based on reduced TAS for the flight from PET to C/D. What is the flying time from C to PET?
Answer: 0:51
Same explanation as the Question above. The only difference being that here they refer to a 'reduced TAS' (= TAS for an engine failure).
GS OUT (continue with the engine failure from PET to D) = (130 + 30) = 160kt
GS HOME (turn around at PET and fly back to C) = (130 - 40) = 90kt
GS OUT ALL ENGINES = (160 + 30) = 190kt
PET = 450 x 90 / (160 + 90)
PET = 162 NM
T = D (162) / V (190)
T = 51 min
