FLIGHT PLANNING EXAM 4 WORKING

Copyright Avfacts 1999. Page 1. FLIGHT PLANNING EXAM 4 WORKING Q 1. This is a STEP CLIMB question. Refer to B727 manual 3 engine Altitude Capability section on page 2-14. ISA Step 1. Find maximum arrival weight at FL370. Step 2. Add 50 kg/1, 000 ft climbed to the 69, 900 kg found in step 1 to find the start of climb GW. In this case 200 kg + 69, 900 kg = 70, 100 kg start of climb GW. Step 3. Find required to lower aircraft weight to start of climb GW, then find (1/2 way between current and start of climb GW). In this case 71, 000 kg (nearest). FL370 ISA FL370 Mach 0.79 M 0.79 69, 900 kg 200 Start climb GW 70, 100 kg 53 kt/isa Mach 0.82 71T DUMAV 477 kt 530 kt FF 4, 355 SGR 8.216 Current GW 71, 300 kg FL330 Step 4. Now fill in cruise data. Refer flight profile. Then divide required by the SGR to get distance to the acceleration point. In this case 146 nm. Answer a best. Req d 146 nm 1, 200 kg

Copyright Avfacts 1999. FP exam 4 working/page 2. Q 2. Refer B727 manual page 2-14 Altitude Capability. Step 1. Find Max GW that the aircraft can accelerate directly to cruise at Mach 0.84 in ISA +10 conditions. FL Mach No. ISA +10 FL310 0.84 74, 600 kg 76T TUNGO 0810 UTC 75 kt/isa +10 M 0.84 Earliest acceleration point 74, 600 kg Mach 0.82 491 kt 416 kt From Hobart FL310 Step 2. Find, which is half way between the current GW at MIDOG, and the acceleration point GW. In this case it is 75, 950 kg (say 76T). FF 4, 683 SGR 11.692 required 2, 700 kg Step 3. Fill the in profile up to SGR, then divide the fuel required to be burnt by the SGR. The result is the distance to the acceleration point. In this case 231 nm. Answer c is closest! 231 nm Q 3. Refer B727 manual page 4-4 (3 Engine holding data), and page 5-4, para 4 (Gear down policy). 65, 000 62, 500 60, 000 FL150 1, 210 kg/hr 1, 170 kg/hr 1, 130 kg/hr So 3 engine gear up ISA FF is 1, 170 x 3 = 3, 510 kg/hr. ISA - 5 FF = 3, 510 x 0.99 = 3, 475 kg/.hr. Correct for gear down by add 50%. 3, 475 kg x 1.5 = 5, 212 kg/hr gear down. Answer a best!

Copyright Avfacts 1999. FP exam 4 working/page 3. Q 4. Fuel Summaries Normal Ops Depress Ops 1 Eng Inop Ops FF 12, 000 VR 1, 200 FR 3, 300 INTER B 2, 000 Traffic Hold 1, 333 Taxi IN 100 Taxi OUT 150 Min FOB 20, 083 kg FF 13, 000 VR Nil FR 2, 250 Wx Hold Not Required Traffic Hold Not required Taxi IN 100 Taxi OUT 150 Min FOB 15, 500 kg FF 12, 500 VR 1, 250 FR 1, 500 TEMPO A 4, 000 Traffic Hold Nil Taxi IN 100 Taxi OUT 150 Min FOB 19, 500 kg Note: 1. Normal ops assumes you get to your destination without problems, so carry whatever holding fuel applies to your destination only. 2. If you lose an engine enroute, you can either return to departure airport or continue ON to destination. Carry whichever airport has the greatest weather holding to preserve that option. In this case the TEMPO at A. Do NOT carry any traffic holding when considering 1 Inop as you will have priority ATC clearance. 3. No weather holding fuel for depressurised provided the airports are above landing minima. Do not carry any traffic holding - you will have priority ATC clearance. In this case the 1 Inop situation requires the most fuel due to the 4, 000 kg TEMPO requirement on A. Answer a best!

FP exam 4 working/page 4. Q 5. The question asks for the minimum FOB at the Darwin ramp for NORMAL OPERATIONS. You do NOT have to assess the minimum fuel required for depressurised or 1 engine inop operations! FL330 Ave 98kt/ISA+4 72T FL220 18 kt/+10 LRC Mach 0.792 464 kt FL185 43 kt BRW 78, 000 kg 28 min 3, 182 anm 562 kt FF 4, 189 SGR 7.454 22 min 670 kg 114 anm Darwin 4, 338 kg Cairns 190 gnm 582 gnm 130 gnm Ave Track 101 M/Total Distance 902 gnm Normal ops fuel summary FF 8, 808 VR 881 FR 3, 300 INTER B 2, 000 Traffic Hold 1, 000 Taxi IN 100 Taxi OUT 150 Min FOB 16, 239 kg Answer d best!

FP exam 4 working/page 5. Q 6. The question asks for the minimum FOB at the Darwin ramp for Yaw Damper Inop operations. You do NOT have to assess the minimum fuel required for depressurised or 1 engine inop operations! FL290 Ave 34kt/ISA+14 72T TopC 74, 642 kg FL200 11 kt/+16 CAS 280 kt Mach 0.730 FL185 9 kt 450 kt BRW 78, 000 kg 27.5 min 3, 358 kg 173 anm 484 kt FF 4, 182 21 min 640 kg 101 anm SGR 8.641 Darwin 6, 023 kg Moresby 178 gnm 697 gnm 104 gnm Ave track 075 M/Total Distance 979 gnm Normal operations fuel summary FF 10, 421 VR 1, 042 FR 3, 300 INTER B 2, 000 Traffic Hold Not required Note: As yaw damper is classed as an abnormal operation, we do NOT need to carry traffic holding fuel (refer manual page 1-21). Taxi IN 100 Taxi OUT 150 Min FOB 17, 013 kg Answer c best!

FP exam 4 working/page 6. Q 7. The question asks for the minimum (ie: flight fuel) from overhead Darwin for 1 Engine Inop operations to the alternate of Tindal. Make sure you apply the 1 Eng Inop reserves correctly. Darwin 67, 000 kg FL185 7 kt/+10 FL130 7 kt/isa +10 TopC 65, 463 kg 1, 500 ft 66, 600 kg 10.5 min 1, 137 kg 51 air nm LRC 1 Inop Mach 0.560 65T 360 kt 353 kt FF 4, 251 SGR 12.042 13 min 520 kg 50 anm FL185 7 kt 638 kg Tindal 50 gnm 53 gnm 49 gnm Distance 152 gnm Climb data ISA+10, pretend BRW 67, 000 kg. 68, 000 kg 67, 000 kg 64, 000 kg SL to FL130 N/ops 6.5 min/1, 050 kg/26 anm 6.5 min/1, 025 kg/26 anm 6 min/950 kg/23 anm Less SL to 1, 500 kg 2 min/388 kg/0 anm N/ops 1, 500 ft to FL130 4.5 min/637 kg/26 anm +2 Eng penalty page 2-2a 6 min/500/25 anm 1 Eng Inop net climb data 10.5 min/1, 137 kg/51 anm Total flight fuel is 2, 695 kg. Answer c best!

FP exam 4 working/page 7. Q 8. The question asks for the minimum (ie: flight fuel) from the 1200 UTC fix to landing at Perth normal operations. You are in-flight and so fixed reserve is 2, 250 kg. You must calculate hold fuel accurately here. FL240 40 kt/isa FL185 Nil wind 1, 758 kg Mach 0.80 68T FL275 50 kt/isa 483 kt 2.5 min 45 kg FL310 1200 UTC 69, 000 kg 443 kt 18.5 min 615 kg 84 air nm FF 5, 091 SGR 11.492 1, 712 kg Perth LW 64, 470 kg 84 gnm 201M/149 gnm 20 gnm 253 gnm Step 1. Step Descent to FL240 data using GW 70, 000 kg. Min air nm FL310 to SL 21 660 107 - FL240 to SL 18.5 615 85 FL310 to FL240 2.5 min 45 kg 22 Step 2. Correct step descent air nm to get 20 gnm using wind at FL275 (ie: half way down). Step 3. Find for FL240 cruise sector to overhead Ballidu VOR, then data (1, 712 kg). Step 4. Find GW on arrival overhead Ballidu VOR, then using an approx holding rate of 4, 000 kg/hr, calculate the GW of the aircraft when it is half way through it s 30 minute holding period (ie: 1, 000 kg less than the arrival GW at Ballidu of 67, 243 kg, which produces a GW of approx 66T. Use this weight to get holding FF, and from that the in holding pattern for 30 minutes. In this case 1, 758 kg). Step 5. Find descent from FL240 to SL, plus the circuit fuel, as a straight in approach was NOT quoted. Step 6. Add up the s, and you will get 4, 530 kg. Answer d best!

FP exam 4 working/page 8. Q 9. The question asks for the minimum (ie: flight fuel) from overhead Darwin for normal operations to the alternate of Melbourne. Make sure you apply the normal ops reserves correctly (eg: the in-flight FR 2, 250). FL330 37 kt/isa +3 TopC 65, 650 kg 65T FL235 22 kt/isa LRC Mach 0.780 457 kt FL185 Nil wind 494 kt 1, 500 ft 67, 600 kg 15 min 1, 950 kg 104 anm FF 3, 781 SGR 7.654 1018 kg 22 min 670 kg 109 air nm Adelaide 68, 000 kg Melbourne 110 gnm 133 gnm 109 gnm Ave Track 106M/Total Distance 352 gnm Step 1. Climb data ISA, pretend BRW 68, 000 kg. 68, 000 kg Normal ops fuel summary SL to FL330 N/ops 17 min/2, 300 kg/104 anm Less SL to 1, 500 kg N/ops 1, 500 ft to FL330 2 min/350 kg/0 anm 15 min/1, 950 kg/104 anm Step 2. Find an approx LW at Melbourne, then base your descent data on it. Step 3. Find an cruise, then get LRC Mach No/FF from in-flight tables. Step 4. Add up flight fuel (4, 038 kg) then apply in-flight reserves. Remember, No final taxi as you are in-flight. FF 4, 038 VR 404 FR 2, 250 INTER YMML 2, 000 Taxi IN Nil Taxi OUT Nil Min FOB 8, 692 kg Answer b best! d: fpwork4 Plane Logic

FP exam 4 working/page 9. Q 10. VILOL 70, 000 kg NONET 67, 785 FL310 60 kt/isa-3 67T FL185 50 kt TopD 61, 803 kg FL270 54 kt/isa +5 FL270 68 kt/isa +5 FL270 68 kt/isa +5 62T 64T 66T Mach.680 Mach.687 Mach.693 Fuel in Box 1, 431 kg 67T 20 min 630 kg 91 anm 410 kt 464 kt FF 3, 819 SGR 8.231 414 kt 482 kt FF 3, 949 SGR 8.192 418 kt 486 kt FF 4, 079 SGR 8.393 66, 345 kg FL270 52 kt/isa-4 Brisbane LW 60, 773 1, 078 kg 1, 794 kg 1, 679 kg 108 gnm 131 gnm YROM VILOL 77 gnm 219 gnm 200 gnm Step 1. Find FOB at BR (ie: 21, 600 - taxi out 150 = 21, 450 kg). Step 2. Subtract reserves to find flight fuel. FOB at BR 21, 450 FR 1, 500 INTER YBBN 2, 000 Traffic Nil Taxi IN 100 110 % 17, 850 100% 16, 227 kg Step 3. Find LW (ie: BR of 77, 000 kg - flight fuel 16, 227 kg = 60, 773 kg). Continued on next page...

FP exam 4 working/page 10. Step 4. Find 1 Engine Inop altitude capability at the PNR. The GW at the PNR can be estimated as being half-way between the BRW of 77, 000 kg, and the LW back at Brisbane of 60, 773 kg. In this case it is 68, 887 kg (say 69 Tonne). Now enter 1 Engine Inop altitude capability tables on manual page 5-6. Typically the Inop cruise levels for the B727 are between FL230 and FL280. Using the FL235 forecast temperatures usually works out best. In this case FL270 is highest IFR hemispherical level HOME with 1 Engine Inop. Step 5. Find descent data into Brisbane from FL270 using LW of 70T. Refer flight profile. Step 6. Find 1 Inop data for segment YROM to TopD, and from it a GW at YROM on way HOME. Step 7. Find 1 Inop data for segment VILOL to YROM, and from it the GW at VILOL on way HOME. Step 8. Find 1 Inop data for segment NONET to VILOL, and from it a GW at NONET on way HOME. Note: The box was defined as starting/ending at NONET. This is because the PNR 1 Inop must be at least at, or beyond the ETP 1 Inop, which I calculated as being 611 nm from Brisbane. VILOL was NOT picked to start the box as it s position is before the ETP 1 Engine Inop. You do NOT want to find yourself in the position of having a box which straddles two met block regions, as it is impossible to average the winds OUT and HOME accurately. Step 9. Define the amount of fuel in the box, by taking the aircraft GW at NONET (HOME) from the GW at NONET (OUT). In this case the 67, 785 kg (OUT) given in the question, less the found GW (HOME) of 66, 354 kg. The fuel in the Box is therefore 1, 431 kg. Step 10. Find the SGR OUT and SGR HOME, and add these together. Step 11. Divide the Fuel in the box (1, 431 kg by the sum of the SGR OUT + SGR HOME. This defines the length of the box (ie: the distance to the PNR from NONET). Remember, the question asked for the position of the PNR 1 Inop as a distance from Alice Springs. In this case 331 nm. Answer a is best!

Copyright Avfacts 1998. FP exam 4 working/page 11. Q 11. FL235 45 kt/+5c YBTL 75, 000 kg 142 gnm TopC 71, 981 kg 25 min 3, 019 kg 161 anm Mach 0.79 462 kt 390 kt FF 4, 068 SGR 10.430 70T 4, 276 kg FL350 72 kt/+7c ETP 67, 705 kg IAS 310 kt Mach = 0.59 370 kt 337 kt FF 4, 543 SGR 13.481 65T 5, 554 kg FL130 HWC ON 33 kt/-3c 13 min 520 kg FL185 33 kt YPDN 410 gnm 412 gnm 43 gnm 552 gnm 455 gnm Fuel Summary Flight fuel 13, 769 Fixed reserve 2, 250 Final taxi 100 Initial taxi 150 Min FOB at ramp 16, 269 kg

FL330 12 kt/isa -5 FP exam 4 working/page 12. Q 12. From Darwin 1440 UTC 72, 500 kg 72T ETP Dep 71, 050 kg Mach 0.82 472 kt 460 kt FF 4, 338 IAS 69T 310 kt Mach 0.59 FL130 5 kt/isa -5 SGR 9.430 1, 462 kg 369 kt 364 kt FF 4, 606 13 min 520 kg 50 anm SGR 12.654 200 gnm 155 gnm 3, 746 kg Alice Springs Darwin 296 gnm 49 gnm This question is giving you a normal operations position fix enroute, and asking you for the minimum amount of usable fuel which must be on board at the 1440 UTC fix for Depressurised operations. You should plan the flight normal operations OUT to the ETP, then assume depressurised operations either ON to Alice Springs, or HOME to Darwin. The flight fuel ON or HOME will be the same, as the ETP is also the equi-fuel point. For diagram, and thinking simplicity the data ON from the ETP has been calculated and shown. Step 1. Calculate the NORMAL OPS fuel burn to the given ETP position. Step 2. Find descent data FL130 into Alice Springs. It is always 13 min/520 kg/50 anm irrespective of weights. Step 3. Find to TopD YPAS. 310 KIAS speed schedule at FL130 equates to a Mach No. of 0.59. Use this Mach No. to find. Step 4. Add up the normal ops and Dep ops fuel burns, and add the 2, 250 kg FR to get an answer of 8, 378 kg. Answer c is best! Remember: 1. NO final taxi fuel is required in this case as we were in-flight when we did the calculation. 2. No holding fuel as long as airport (s) are above landing minima.

FP exam 4 working/page 13. Q 13. Refer to B727 manual page 6-3 Driftdown 2 Eng Inop Step 1. Step 2. 75, 000 kg 6, 000 kg GW at Fix 75, 000 kg - to overhead 6. 000 kg - Circuit (TAF only) Wind 50 kt TAIL Nil Wind LW at Destination Answer a best! 68, 600 kg 400 nm Q 0921 UTC 70, 250 kg 69T APOMA 67, 956 kg FL330 53 ktisa-5 LRC Mach 0.787 453 kt To Sydney 506 kt FF 3, 895 Answer d is best! Alice Springs 299 gnm SGR 7.698 2, 294 kg Track 121M/298 gnm

FP exam 4 working/page 14. Q 15. Step 1. Find minimum required. Refer Nav computer set-up. 95 minutes 666 nm Opposite hour marker read of 420 kt. Step 2. Find head wind component (HWC). In this case 61 kt. Step 3. Minimum is the minimum + the HWC. In this case 481 kt. Step 4. Use Nav computer set-up below to get minimum Mach No. Set Temp (-53) against Mach index arrows. Opposite the minimum of 481 kt on the outside scale, read the minimum Mach No. of M 0.833 on the inside scale. Answer b best! Q 16. This question, is designed to see if you can use the buffet boundary charts, and from that find a, groundspeed and ETA at a given waypoint. Step 1. Find minimum steady flight speed from the NIL turbulence buffet boundary charts on manual page 2-11. Minimum IAS is 195 kt. Step 2. Convert this IAS to a Mach No. at FL330. Refer set-up below Set 195 kt IAS opposite FL330. In Mach window read M 0.562 Step 3. Find using OAT given (ISA-5 at FL330). Set OAT of -56C opposite Mach index arrows, and read of 322 kt opposite M 0.562 Step 4 & 5. Find groundspeed (398 kt), then divide distance of 295 nm by the to get minutes to TopD (ie: 44.5 ). Add this time to the fix time of 1015 UTC to get ETA at TopD of 1059.5 UTC. Answer c is best!

FP exam 4 working/page 15. Q 17. 1442 UTC Fix 74, 500 kg YROM 69, 580 kg Answer b best! 72T IAS 242 kt Mach.70 397 kt 451 kt FF 3, 620 SGR 8.027 Alice Springs 214 nm 4, 920 kg Track 094M Dist 613 nm Refer to manual page 4-4 3 Eng Holding. FL 75, 000 kg 72, 000 kg 70, 000 kg FL350 248 kt/1410 kg/hr 239 kt/1260 kg/hr FL330 247 kt/1374 kg/hr 242/1296 kg/hr 238 kt/1244 kg/hr FL300 245 kt/1320 kg/hr 236 kt/1220 kg/hr Correct FF for non ISA day (1296 x 3 Eng) x 0.98 = 3, 810 kg/hr Correct FF for straight line holding 3, 810 x 0.95 = 3, 620 kg/hr. Find SGR and to YROM... refer flight profile above. END OF WORKING FILE 4.