Email
info@vayudootaviation.com
Phone No
+91 7276526726
become a pilot
Home
About us
courses
CPL
ATPL
RTR(A)
Airline Cadet Program
IGRUA Entrance Prep
B.Sc. Aviation
Pilot Training
INDIA
USA
CANADA
SOUTH AFRICA
NEW ZEALAND
EUROPE
Aircraft Type-Rating
Airbus A320
Boeing 737
Blogs
contact
CPL Test Series
Question Bank
Questions for MEASUREMENT OF ELEMENTS
Answer the following questions
Test Mode
Training Mode
1. From "A" to "B" distance 2,000 nm. Track 080°, TAS 180K WIND 290/20 for first 1000 nm & 320/20 for rest of flight, endurance 15 Hrs. find a) Time of flight. b) Distance to PNR c) ETA to PNR (ETD 1000 IST)
- [A) 10 HRS 21 MIN B) 1336 NM C) 16 HRS 50 MIN]
- [A) 10 HRS 21 MIN B) 1336 NM C) 16 HRS 40 MIN]
2. Find G/S and Drift. Trk - 090°, TAS(kts) - 160, W/V - 330/18, G/S - ________ , HDG - 0.84°, DRIFT - ________ .
- [G/S - 168, DRIFT - 06°]
- [G/S - 168, DRIFT - 09°]
3. On a flight from A to B: Tr 317 WIND 205/35 TAS 185 K DPNR 732 nm Fligh Fuel 1200 lbs find: a) Fuel consumption b) SAR c) If CP is reached 45 min before PNR, find excess fuel carried on board.
- [A) 149 LBS/HR B) 1490 NM C) 233 LBS]
- [A) 149 LBS/HR B) 1490 NM C) 236 LBS]
4.
Dist = 1,250 nm. TRK = 030° FOB = 1,400 kg, Winds = 360/30 kts, Reserve = 150 kg TAS = (N) = 298 kts, F/C (N) = 200 kg/hr. (R) = 250 kts F/C (R) = 150 kg/hr. Find DCP and TCp ; DPNR and TPNR (ii) SAR ; Fuel required to go to destination.
- [SAR = 1,400/200 × 298 = 2086 NM FLIGHT TIME = 1.250/272 = 4 HRS 35 MIN. FUEL TO DESTINATION = 4.59 × 200 = 915 KG.]
- [SAR = 1,400/200 × 298 = 2086 NM FLIGHT TIME = 1.250/272 = 4 HRS 35 MIN. FUEL TO DESTINATION = 4.59 × 200 = 919 KG.]
5.
Dist. = 800 nm TRK 330° W/V = 270°/30 kts TAS (N) = 240 kts TAS (R) = 180 kts DCP and Tcp = ?
- [DCP = DH/O+H = 800 × 193/193 + 163 = 1.54,400 / 356 = 434 NM, TCP = 434 / 224 = 1 HRS 56 MIN]
- [DCP = DH/O+H = 800 × 193/193 + 163 = 1.54,400 / 356 = 434 NM, TCP = 434 / 224 = 1 HRS 36 MIN]
6.
Dist = 2000 nm Track = 090° W/V = 090°/30 kts for first 1,000 nm for remaining . = 330°/18 kts TAS (4 engine) = 380 TAS (3 engine) = 260 kts FOB = 2,500 kg Reserve = 350 kg FC (4 engine) = 360 kg/hr FC (3 engine) = 280 kg/hr.
- [I) DCP = DH/O+H= 1,000×16/16+446 = 34.6 NM, DCP = 1,035 NM. (II) TIME = 1,000/350 + 35/389=2.85 + 0.089 = 2.93 = 2 HR 56 MINS. (III) DPNR --- FROM TO GO A-X - TRK - 090°, TAS - 380, W/V - 090/30, G/S - 350, DIST - 1000, FT - 2.85, F/C - 360, FF - 1026. FROM TO GOHX-B - TRK - 270°, TAS - 260, W/V - 090/30, G/S - 290, DIST - 1000, FT - 3.44, F/C - 280, FF - 963. FROM TO GOX-B - TRK - 090°, TAS - 380, W/V - 330/18, G/S - 389, DIST - 1000, FT - 2.57, F/C - 360, FF - 925. FROM TO GHB-X - TRK - 270°, TAS - 260, W/V - 330/18, G/S - 251, DIST - 1000, FT - 3.98, F/C - 280, FF - 1115 = TOTAL = 4029.IF FF = 4029 DPNR = 2000, FF = 2500 - 350 = 2150KG DPNR = (2000×2150) ÷ 4029, = 1,067 NM. TPNR = 1,000/250 + 67/389 = 2.85 + 0.172 = 3.02 = 3 HR 01 MIN.]
- [I) DCP = DH/O+H= 1,000×16/16+446 = 34.6 NM, DCP = 1,035 NM. (II) TIME = 1,000/350 + 35/389=2.85 + 0.089 = 2.93 = 2 HR 56 MINS. (III) DPNR --- FROM TO GO A-X - TRK - 090°, TAS - 380, W/V - 090/30, G/S - 350, DIST - 1000, FT - 2.85, F/C - 360, FF - 1026. FROM TO GOHX-B - TRK - 270°, TAS - 260, W/V - 090/30, G/S - 290, DIST - 1000, FT - 3.44, F/C - 280, FF - 963. FROM TO GOX-B - TRK - 090°, TAS - 380, W/V - 330/18, G/S - 389, DIST - 1000, FT - 2.57, F/C - 360, FF - 925. FROM TO GHB-X - TRK - 270°, TAS - 260, W/V - 330/18, G/S - 251, DIST - 1000, FT - 3.98, F/C - 280, FF - 1115 = TOTAL = 4029.IF FF = 4029 DPNR = 2000, FF = 2500 - 350 = 2150KG DPNR = (2000×2150) ÷ 4029, = 1,067 NM. TPNR = 1,000/250 + 67/389 = 2.85 + 0.172 = 3.02 = 3 HR 10 MIN.]
7. The effect of an increasing strength of wind at 90° to track on the time and distance to the PET will be
Distance - none, time - increase
Distance - decrease, time - decrease
Distance - none, time - none
8. From P to Q distance 2000 nm. Track 090, TAS 4 Eng. 180 K TAS 3 Eng. 150 K, WIND 090/20 for first 1000 nm. & 150/20 for rest of the flight. FOB 1600 gal, reserve 200 gal. fuel consumption 4 engine 100 GPH, 3 Eng 80 GPH. Find: a) Time & distance to CP b) Time & distance to PNR, return on 3 Eng. a) Time & distance to CP b) Time & distance to PNR, return on 3 Eng.
- [A) 1097 NM, 6 HRS 49 MIN B) 1278 NM, 7 HRS 53 MIN]
- [A) 1097 NM, 6 HRS 49 MIN B) 1278 NM, 7 HRS 54 MIN]
9. Distance 1015 nm. TAS 180 K, head wind 10 K, fuel cons 1200 lbs/hr Max AUW 90,000 lbs operational mass 50,000 lbs reserve fuel 3,000 lbs (to be added for PNR) find a) Time & distance to CP b) Time & distance to PNR
- [A) 536 NM, 3 HRS 9 MIN B) 760 NM, 4 HRS 28 MIN.]
- [A) 536 NM, 3 HRS 9 MIN B) 760 NM, 3 HRS 28 MIN.]
10. Find G/S and Drift. Trk - 220°, TAS(kts) - 310, W/V - 330/42, G/S - ________ , HDG - 227°, DRIFT - ________ .
- [G/S - 322, DRIFT - 07° .]
- [G/S - 322, DRIFT - 05° .]
11.
Dist = 1,500 nm TAS (4 engines) = 360 kts TAS (3 engines) = 300 kts TRK = 090° W/V = 33°/30 kts FOB = 4,500 kg Reserve = 500 kg F/C (4 engines) = 800 kg/ hr F/C (3 engines) = 600 kg / hr Find
- [FROM TO - GOA-B, TRK - 090°, TAS - 360, W/V - 330/30, G/S - 3744, DIST - 1500, FT - 04:00, FC - 800, FUEL REQ - 3208. FROM TO - GHB-A ,TRK - 270°, TAS - 300, W/V - 330/30, G/S - 284, DIST - 1500, FT - 05:16, FC - 600, FUEL REQ - 3160. = 6368. IF FF IS 6,368 KGS DPNR = 1,500 NM. IF FF IS 4,000 KGS DPNR = 1,500/6,368 × 4,000 = 942 NM, TPNR = 942/374 = 2 HRS 21 MINS.]
- [FROM TO - GOA-B, TRK - 090°, TAS - 360, W/V - 330/30, G/S - 3744, DIST - 1500, FT - 04:00, FC - 800, FUEL REQ - 3208. FROM TO - GHB-A ,TRK - 270°, TAS - 300, W/V - 330/30, G/S - 284, DIST - 1500, FT - 05:16, FC - 600, FUEL REQ - 3160. = 6368. IF FF IS 6,368 KGS DPNR = 1,500 NM. IF FF IS 4,000 KGS DPNR = 1,500/6,368 × 4,000 = 942 NM, TPNR = 942/374 = 2 HRS 31 MINS.]
12. From "A" to "B" distance 1500 nm. Track 347, TAS 4 Eng - 175 K TAS 3 Eng. 152K WIND 270/20 for first 1000 nm. & 245/20 for rest of the flight. FOB 1200 G. Reserve 100G, fuel cons 4 Eng. 120 GPH, 3 Eng 100 GPH. Find a) Time & distance to CP b) Time & distance to PNR return on 3 Engines
- [A) 756M 4228 B) 811, 3:48]
- [A) 756M 4228 B) 811, 4:48]
13. On a flight from A to B dist. 1100 nm. Track020° WIND 220÷°×/30. F/C 1200 lbs/hr FOB 10000 lbs reserve 1500 lbs, TAS 210 K (4-eng) TAS 185 K (3 eng) find: a) Time & distance to CP b) Time distance to PNR c) SAR
- [A) 465 NM 1HR 55 MIN B) 730 NM 3HR 4MINS C)1750]
- [A) 465 NM 1HR 55 MIN B) 730 NM 3HR 4MINS C)1780]
14.
Trk = 270°, TAS = (N) = 240 kts (R) = 150 kts , W/V = 090°/20 up to CP = 330×°/38 after CP , Dist = 1,100 nm FOB = 1,800 kg, Res = 200 kg FC (N) = 450 kg/hr, FC(R) = 300 kg/ hr.
- [I) DCP = DH/O+H = 1,100×130 / 130+127 = 556.4 NM. TCP = DCP/G = 556.4/260 = 2 HRS 08 MIN. FROM TO GOA-CP - TRK - 270°, TAS - 240, W/V - 090/20, G/S - 260, DIST - 556, FT - 2:08, F/C - 450, FF - 960. FROM TO GHCP-A - TRK - 090°, TAS - 150, W/V - 090/20, G/S - 130, DIST - 556, FT - 4:15, F/C - 300, FF - 1280. = 2240. NOTE : IN THIS CASE FUEL REQUIRED FOR PNR TO BE AT CP IS MORE THAN THE FUEL ON BOARD. THEREFORE PNR LIES BEFORE CP. PRESUME DPNR TO BE AT CP: WHEN FF IS 2240 DPNR = 556 NM. WHEN FF IS 1 DPNR = 556/2240 NM. WHEN FF IS 1,600 DPNR = (556×1,600) ÷ 2,240 = 394 NM. ALTERNATELY USING GROSS FUEL CONSUMPTION (GFC) FORMULA: GFC (UNITS/GNM) = FUEL CONSUMPTION (UNITS/HR) / GROUND SPEED (KNTS) = FC/GS GFC = 45 ÷ 260 + 300 ÷130 = 1.73 + 2.31 = 4.04 KG/GNM. DPNR = FF ÷ CFC = 1600 ÷ 4.03 = 396 NM. TPNR = 396 ÷ 260 = 1 H 31 MIN.]
- [I) DCP = DH/O+H = 1,100×130 / 130+127 = 556.4 NM. TCP = DCP/G = 556.4/260 = 2 HRS 08 MIN. FROM TO GOA-CP - TRK - 270°, TAS - 240, W/V - 090/20, G/S - 260, DIST - 556, FT - 2:08, F/C - 450, FF - 960. FROM TO GHCP-A - TRK - 090°, TAS - 150, W/V - 090/20, G/S - 130, DIST - 556, FT - 4:15, F/C - 300, FF - 1280. = 2240. NOTE : IN THIS CASE FUEL REQUIRED FOR PNR TO BE AT CP IS MORE THAN THE FUEL ON BOARD. THEREFORE PNR LIES BEFORE CP. PRESUME DPNR TO BE AT CP: WHEN FF IS 2240 DPNR = 556 NM. WHEN FF IS 1 DPNR = 556/2240 NM. WHEN FF IS 1,600 DPNR = (556×1,600) ÷ 2,240 = 394 NM. ALTERNATELY USING GROSS FUEL CONSUMPTION (GFC) FORMULA: GFC (UNITS/GNM) = FUEL CONSUMPTION (UNITS/HR) / GROUND SPEED (KNTS) = FC/GS GFC = 45 ÷ 260 + 300 ÷130 = 1.73 + 2.31 = 4.04 KG/GNM. DPNR = FF ÷ CFC = 1600 ÷ 4.03 = 396 NM. TPNR = 396 ÷ 260 = 1 H 21 MIN.]
15. Flight distance 1500nm, wind component out +40kt, back - 50kt, TAS 320kt, endurance - flight time +15%. The distance to the PNR will be
738nm
750nm
786nm
16.
In question 12, FF = 3,000 kg, Calculate Drives and True.
- [TOTAL TIME (TPNR) = 556/260 + 196 / 219 = 2.14 + 0.90 = 3.04 = 3 HR 02 MIN]
- [TOTAL TIME (TPNR) = 556/260 + 196 / 219 = 2.14 + 0.90 = 3.04 = 3 HR 09 MIN]
17.
Trk = 330° W/V = 290°/25 kts up to GP, 330°/15 kts after CP TAS (N) = 220 kts TAS(R) = 160 kts, Dist = 750 nm, POB = 1,500 kg Res = 100 kg, FC (N) = 180 kg/hr. FC (R) = 120 kg/hr. (i) DCP and TCP (ii) DPR and TPNR (assume engine failure at PNR)
- [GFC UP TO CP = 180/200 + 120/178 = 0.9 +6.74 = 1.57, FUEL REQUIRED UP TO CP = 1.58 × 413 = 648.4 LBS, FUEL AVAILABLE AFTER CP = 1400 - 648.4 = 751.6 LBS, GFC AFTER CP = 180/205 + 120 / 175 = 1.56, DPNR FROM CP = 751.6/1.56 = 481.4, DPNR FROM A = 413 + 482 = 895 NMS, TPNR = TA-CP+TCP-PNR = 413 + 200 + 483 + 205 = 2 HR 04 MIN +2 HR 21 MIN = 4 HR 55 MIN.]
- [GFC UP TO CP = 180/200 + 120/178 = 0.9 +6.74 = 1.57, FUEL REQUIRED UP TO CP = 1.58 × 413 = 648.4 LBS, FUEL AVAILABLE AFTER CP = 1400 - 648.4 = 751.6 LBS, GFC AFTER CP = 180/205 + 120 / 175 = 1.56, DPNR FROM CP = 751.6/1.56 = 481.4, DPNR FROM A = 413 + 482 = 895 NMS, TPNR = TA-CP+TCP-PNR = 413 + 200 + 483 + 205 = 2 HR 04 MIN +2 HR 21 MIN = 4 HR 25 MIN.]
18. Full TAS 500 kt. Engine failure TAS 400 kt. Total distance 1000 nm. Along track wind component 40 kt. The distance to the engine failure CP will be
450nm
550nm
450 or 550nm depending on the wind direction
19. On a flight from Y to X : Track 210, WIND 270/35. TAS 200 kts FF 1100 lbs DPNR 785 nm. Find a) fuel consumption b)SAR c) If CP is reached 45 min before PNR, find excess fuel carried on board
- [A) 137 LBS/HR. B) 1600 NMS C) 194 LBS]
- [A) 137 LBS/HR. B) 1600 NMS C) 193 LBS]
20. On a flight from A to B: Track 155, TAS 200 K WIND 260/35 DPNR 1050 nm, F/C 141 GPH find: a) fuel on board b) SAR c) If CP lies one hour 15 min before PNR, find find excess fuel carried on board
- [A) 1506 GAL B) 2136 GAL C) 370 LBS ]
- [A) 1506 GAL B) 2136 GAL C) 367 LBS ]
21. In flight it is discovered that the winds used have been reversal, the effect will be
Give the distance of the PET from the base ahead instead of from the base behind.
No effect on the distance to the PET but the ETA at it will be affected.
Change the distance tot he PNR
22. On a flight from S to T dist. 1250 nm. Track 135 TAS (4 eng) 250 k TAS (3 eng) 215 K WIND 090/30 FOB 12000 lbs reserve 2000 lbs F/C (4 eng) 1200 lbs/hr F/C (3 eng) 1000 lbs/hr find: a) Time& distance to CP b) Time & distance to PNR c) Time & distance to PNR (return on 3 engine) d) SAR
- [A) DCP - 686 TCP - 3 HRS. B) DPNR 1028NM 4432' C) DPNR - 1048 TPNR 4437' D) 2500NMS.]
- [A) DCP - 686 TCP - 3 HRS. B) DPNR 1028NM 4432' C) DPNR - 1049 TPNR 4437' D) 2500NMS.]
23.
Dist = 2,400 nm. Trk = 080° W/V = 030°/25 kts TAS (N) = 440 kts TAS (R) = 360 kts FOB = 10,000 kg, Reserve = 1,000 kg, FC (N) = 1,400 kg/hr, FC(R) = 1,000 / kg / hr.
- [FROM TO GOA-B - TRK - 080°, TAS - 440, W/V - 030/25, G/S - 424, FT - 5.66, FC - 1400, FF - 7925. FROM TO GH-PNR-A - TRK - 260°, TAS - 360, W/V - 030/25, G/S - 376, FT - 6.38, FC - 1000, FF - 6383. = 14308. FF = 14,308 FOR DPNR = 2,400. FF = 9,000 FOR DPNR = 2,400/14,308 × 9,000 = 1,509.6 NM TPNR = 1,510 / 434 = 3 HRS 23 MIN.]
- [FROM TO GOA-B - TRK - 080°, TAS - 440, W/V - 030/25, G/S - 424, FT - 5.66, FC - 1400, FF - 7925. FROM TO GH-PNR-A - TRK - 260°, TAS - 360, W/V - 030/25, G/S - 376, FT - 6.38, FC - 1000, FF - 6383. = 14308. FF = 14,308 FOR DPNR = 2,400. FF = 9,000 FOR DPNR = 2,400/14,308 × 9,000 = 1,509.6 NM TPNR = 1,510 / 434 = 3 HRS 33 MIN.]
24.
Distance = 1,200 nm TRK = 090° TAS (Normal) 200 kts TAS (Reduced) = 150 kts Winds = 090°/20 kts. Find Dcp Tcp
- [DCP = DH/O+H = (1200×170) / (130+170) = 680 NMS, OR BY UNITARY METHOD DCP = (1200×554) (554+424) = 680 NMS, TCP = 680 / 180 = 3HRS 55 MINS.]
- [DCP = DH/O+H = (1200×170) / (130+170) = 680 NMS, OR BY UNITARY METHOD DCP = (1200×554) (554+424) = 680 NMS, TCP = 680 / 180 = 3HRS 46 MINS.]
25. On a flight from K to L dist 1500 nm. TAS 310 K track 350 WIND 240/30 FOB 12000 lbs. reserve 2000 lbs F/C 1800 lbs/hr find: a) Time & distance to PNR b) Time & distance to CP c) SAR
- [A) 856 NM 2HR 40 MIN B) 724 NM, 2HR 16 MIN C) 2066 NM]
- [A) 856 NM 2HR 40 MIN B) 724 NM, 2HR 16 MIN C) 2069 NM]
26. On a flight distance 1073 nm. TAS 170 kts, head wind 20 Kts. a) Find time & distance to CP. b) FOB is just sufficient for flight find the time & distance to PNR.
- [A) 600 NM 4 HRS B) 600 NM 4 HRS ]
- [A) 600 NM 4 HRS B) 600 NM 3 HRS ]
27. From 'X' to "Y" distance 1250 nm. track 120, TAS 4 Eng. 180 K TAS 3 Eng. 150 K, WIND 090/30 up to CP & 060/30 after CP. FOB 1500 G fuel consumption 4 Eng 100 GPH, 3 Eng 80 GPH. Find a) Time & distance to CP in case of engine failure. b) Time & distance to PNR with 200 G of reserve. Return on 3 engnles
- [A) 710 NM, 4 HR 38 MIN B) 1174 NM 7 HRS 28 MIN]
- [A) 710 NM, 4 HR 38 MIN B) 1174 NM 7 HRS 24 MIN]
28.
TEK = 0500 W/V 330°/18 kts DPNR = 800 Nm FOB = 1,540 kg Reserve Reqd. = 150 kg TAS = 200 kts. Find (i) Fuel consumption ; SAR (ii) If CP is reached 30 min before PNR, Find excess fuel carried on board on reaching destination.
- [702 = D×202 / 202+196 , D = 1,383 NM, FLIGHT TIME TO DESTINATION = 1,383/196 = 7:03, TOTAL FUEL CONSUMED = 7:03×175 = 1,220 KG, FUEL REQUIRED = 1,220+150=1,370 FOB = 1,540 EXTRA FUEL = 1540 - 1370 = 270 KG.]
- [702 = D×202 / 202+196 , D = 1,383 NM, FLIGHT TIME TO DESTINATION = 1,383/196 = 7:03, TOTAL FUEL CONSUMED = 7:03×175 = 1,220 KG, FUEL REQUIRED = 1,220+150=1,370 FOB = 1,540 EXTRA FUEL = 1540 - 1370 = 170 KG.]
29.
For flight fuel of 2000 Kg calculated Dpnr = 500 nm. If Flight Fuel is increased by 200 Kg, Dpnr will be: (a) 520 nm (b) 550 nm (c) 450 nm Methods to Solve Questions on CP and PNR: Dcp (Distance to CP) DH / O+H nm Dpnr = E ×GO× GH / Go × GH = NM
- [(I) DCP ONE LEG (RETURN ON 3 ENGINE) DH/O+H (CP IS ALWAYS CALCULATED ON 3 ENGINES OR REDUCED TAS) (II) DCP : WINDS CHANGING AT CP: (IV) PNR : IF FF OF FC CHANGES BY X%, DPNR WILL CHANGE BY SAME PERCENTAGE. IF FF INCREASES, DIST. TO PNR INCREASES, IF RESERVE FUEL INCREASES DPNR DECREASES. IF FUEL CONSUMPTION INCREASES, DPNR DECREASES. (V) PNR: (NO DISTANCE GIVEN) ASSUME ANY DISTANCE AND CALCULATE PNR.]
- [(I) DCP ONE LEG (RETURN ON 3 ENGINE) DH/O+H (CP IS ALWAYS CALCULATED ON 3 ENGINES OR REDUCED TAS) (II) DCP : WINDS CHANGING AT CP: (IV) PNR : IF FF OF FC CHANGES BY X%, DPNR WILL CHANGE BY SAME PERCENTAGE. IF FF INCREASES, DIST. TO PNR INCREASES, IF RESERVE FUEL INCREASES DPNR DECREASES. IF FUEL CONSUMPTION INCREASES, DPNR DECREASES. (V) PNR: (NO DISTANCE GIVEN) ASSUME ANY DISTANCE AND CALCULATE SNR.]
30. On a flight from X to Y : track 250, TAS 195, dist 975 nm WIND 330/35, FOB 900 gal reserve 100 gal F/C 150 GPH find: a) Time & distance to PNR b) Time & distance to CP c) SAR
- [A) 512 NM 2 HR 45 MIN B) 503 NM 2 HRS 42 MIN C) 1170NM]
- [A) 512 NM 2 HR 45 MIN B) 503 NM 2 HRS 42 MIN C) 1130NM]
31. From A to B distance 1450 nm. track 132, TAS 4 Eng 190 K, TAS 3 Eng 160 K, WIND 260/40 up to CP & 350/60 after CP, FOB 18000 lbs, reserve 1000 lbs fuel consumption 4 Eng 1850 lbs/hr 3 Eng. 1450 lbs/hr. find a) Time & distance to CP b) Time & distance to PNR return on 3 Eng
- [A) 571 NM, 2 HRS 41 MIN B) 840, 3 HRS 49 MIN]
- [A) 571 NM, 2 HRS 41 MIN B) 840, 3 HRS 44 MIN]
32. On a simple flight from A to B, the distance to the PET is 1200nm. The PNR endurance is equal to the flight time +10%. The distance to the PNR will be
1320nm provided the same winds apply to both PET and PNR calculations
1320nm provided the wind is not at 90° the track
1260nm provided the same winds apply to both PET and PNR calculations.
33.
In question 5, fuel on board 1,000 kg, F/C = 1550 kg/ hr, Reserve = 100kg. Find Dpnr and Tpnr. WV = 270/30 A ↔ B 800 nm PNR
- [FOB = 1,000 KG RESERVE = 100KG FF = 900 KG F/C (N) = 150 KG/ HR DPNR (DISTANCE TO PNR) = E×GO×GH/GO+GH = NMS E = 900/150 = 6HR DPNR = (6×224×254) / 224+ 254) = 714 NM. TPNR = 714 / 224 = 3 HR 11 MIN. SAR (STILL AIR RANGE): IT IS THE MAXIMUM DISTANCE UP TO WHICH AN A/C CAN FLY IN NIL WIND WHILE CONSUMING ALL FUEL ON BOARD. TAS = G/S (IN WINDS BOTH ARE SAME) SAR = FOB /FC × TAS ]
- [FOB = 1,000 KG RESERVE = 100KG FF = 900 KG F/C (N) = 150 KG/ HR DPNR (DISTANCE TO PNR) = E×GO×GH/GO+GH = NMS E = 900/150 = 6HR DPNR = (6×224×254) / 224+ 254) = 714 NM. TPNR = 714 / 224 = 3 HR 11 MIN. SAR (STILL AIR RANGE): IT IS THE MAXIMUM DISTANCE UP TO WHICH AN A/C CAN FLY IN NIL WIND WHILE CONSUMING ALL FUEL ON BOARD. TAS = G/S (IN WINDS BOTH ARE SAME) SAR = FOB /FC × FAS ]
34. On a flight out to the PET, the GS are found to be less then expected. The PET will be moved along track
Into wind
In to wind provided the wind change continues throughout the flight.
Downwind provided the wind changes continues throughout the flight.
35. The distance to the PNR will be greatest
In still air conditions
If there is a tailwind component going out
If there is a headwind component going out
36. For head wind component of 50 Kts calculated DCP is 750 nm for the total route dist of 1200 nm. In actual flight wind component is found to be 50 Kts of tail wind, new position of DCP will be
450 nm
525 nm
600 nm
37. On a flight from A to B dist. 1015 nm. track 083, TAS 190 K on 4 Engines, 160 K on 3 Engines, WIND 190/45 up to CP& 110/40 after CP. FOB 15000 lbs, reserve 2000 lbs, F/C 1800 lbs/hr on 4 engines & 1400 lbs/hr on 3 engines. Find a) Time& distance to CP b) Time & distance to PNR (return on Engines)
- [A) 544NM 2:46 B) 683NM, 2:39]
- [A) 544NM 2:46 B) 683NM, 3:39]
38. From X to Y distance 1200 nm TAS 160 K, Tail wind of 10 K for first 600 nm. & head wind of 10 K for rest of flight find time & distance to CP
- [600NM, 3:32]
- [600NM, 2:32]
39. In above question find SAR.
- [SAR = 1,000/150 × 240 = 2,600 NM]
- [SAR = 1,000/150 × 240 = 1,600 NM]
40.
TRK = 350° W/V = 120°/28, TAS (4 engine) = 450 kts (3 engine) = 300 kts, Dist. = 1,500 nm. Find Dcp and Tcp
- [DCP = DH/O+H = 1500×281/317+281 = 705 NM, TCP = 705 / 467 = 2 HRS 30 MIN.]
- [DCP = DH/O+H = 1500×281/317+281 = 705 NM, TCP = 705 / 467 = 1 HRS 30 MIN.]
41. An engine failure PNR is calculated to enable the pilot to make the correct decision in the event of
An engine failing after reaching the mi-point
the destination and its alternates closing any time up to the PNR
An engine failing before reaching the mid-point
42. On a flight from A to B : Track 153, TAS 210 K Dist. 925 nm WIND 200/30, FOB 1000 lbs, reserve 200 lbs. F/C 180 lbs/hr. find: a) Time & distance to PNR b) Time & distance to CP c) SAR.
- [A) 508 NM, 2 HRS 42 MIN B) 457 NM 2 HRS 26 MIN C) 1167.0]
- [A) 508 NM, 2 HRS 42 MIN B) 457 NM 2 HRS 26 MIN C) 1169.0]
43. Find G/S and Drift. Trk - 120°, TAS(kts) - 180, W/V - 180/15, G/S - ________ , HDG - 124°, DRIFT - ________ .
- [G/S - 172, DRIFT - 04°.]
- [G/S - 172, DRIFT - 05°.]
44.
In question 12, is fuel sufficient to go to B?
- [NOTE: IF PNR IS BEFORE CP, FUEL CARRIED IS INSUFFICIENT TO GO TO HIGH. ]
- [NOTE: IF PNR IS BEFORE CP, FUEL CARRIED IS INSUFFICIENT TO GO TO DESTINATION. ]
45. Find G/S and Drift. Trk - 315°, TAS(kts) - 220, W/V - 090/22, G/S - ________ , HDG - 319°, DRIFT - ________ .
- [G/S - 235, DRIFT - 03°.]
- [G/S - 235, DRIFT - 04°.]
Submit
Make New Set