Flight Planning

Question 1

Ram Air Temperature (RAT) is defined as ?

Answer 1

defined as the sum of the OAT (ambient Outside Air Temperature),
plus some of the ram temperature rise.

Question 2

Whats a Temperature probe sensor recovery factor ?

Answer 2

Represents the amount of ram temperature rise indicated by the probe.

Question 3

OAT temp = -50c sensory recovery factor of 0.8, Ram temp rise at a given Mach number is 30°C calculate RAT ?

Answer 3

OAT of -50°C + 0.8 (80%) of 30°C (24°C) = -26°C RAT

Question 4

Total Air Temperature (TAT) ?

Answer 4

is the sum of the OAT (which is known sometimes as the True Outside Air
Temperature [TOT]) and the full theoretical ram temperature rise due to compressibility effects.

Question 5

Calculate TAT if OAT is -50c and ram temp rise is 30c ?

Answer 5

OAT of -50°C + the full 30°C temperature rise = -20°C TAT

Question 6

Standard temperature at FL110 (11000 ft) is ?

Answer 6

15°C at MSL - [11 x1.98°C] = -6.78°C)

Question 7

the pressure rise also
causes the airspeed indicator (ASI) to ?

Answer 7

Over read at high speeds

Question 8

EAS is simply ?

Answer 8

IAS corrected for compressibility error

Question 9

In the B727 the corrections for instrument and position error are
less than 1 kt so for practical purposes,
in this aircraft IAS = ?

Answer 9

CAS

Question 10

find EAS for an aircraft flying at 300 KCAS at FL300 ?

Answer 10

Enter the F factor table at 300 kt and FL300 and obtain the F Factor of 0.95.
Step 2
Multiply the IAS by the F factor to obtain EAS: 300 x .95 = 285 EAS.

Question 11

For practical purposes, the Indicated Mach Number is the ?

Answer 11

True Mach Number TMN

Question 12

The Formula used to compute the speed of sound in knots is ?

Answer 12

Mach 1.0 (in kt) = Square root of OAT in degrees K x 39 -1

Question 13

To then compute the TAS for any Mach Number other than 1?

Answer 13

TAS = Square root of OAT in degrees K x 39 x M -1 (where M = Mach Number)

Question 14

To convert the Celsius temperature to Kelvin, simply add ?

Answer 14

simply add 273 to the Celsius
temperature.

Question 15

find the speed of sound at MSL in ISA: (15°C) ?

Answer 15

M1.0 = (15C + 273 K) = V288 K x 39 -1 = 661 kt

Question 16

example, find the speed of sound at MSL in ISA: (15°C)
M1.0 = (15C + 273 K) = V288 K x 39 -1 = 661 kt
Similarly, Mach 0.80M at the same temperature = ?

Answer 16

V288 K x 39 x .8 -1 = 528 kt

Question 17

Find the speed of sound at the tropopause in ISA: (-56°C) ?

Answer 17

M1.0 = (-56C + 273 K) = V217 K x 39 -1 = 574 kt

Question 18

Find the speed of sound at the tropopause in ISA: (-56°C)
M1.0 = (-56C + 273 K) = V217 K x 39 -1
= 574 kt
Similarly, 0.80M at the same temperature = ?

Answer 18

V217 K x 39 x .8 -1 = 459 kt

Question 19

find the TAS at 0.80M at FL310 at a temperature of ISA+10 ?

Answer 19

The OAT must be computed by applying the ISA variation to the ISA OAT at FL310 from page
3-106 of the B727 POH.
ISA OAT at FL310 is -46°C.
ISA+10 therefore will be a temperature of -36°C. convert the OAT to Kelvin. In this example -36°C = 237K (-36°C + 273) M0.8 = V237 K x 39 x .8 -1 = 479 kt

Question 20

find the Mach Number at a TAS of 455 kt with an OAT of -50°C ?

Answer 20

Calculate the TAS for 1.00M at -50°C ({V223K [-50 + 273]} x 39 -1 = 581 kt).

2
Divide the TAS given of 455 by the TAS of 1.00M of 581 (455/581 = 0.783 M).

Question 21

find the OAT at a TAS of 420 kt and a Mach Number of 0.70M ?

Answer 21

Calculate the TAS of 1.00M for 420 kt at 0.70M (420+1/0.7 = 600 kt). The 1kt accounts for the
overread error of the formula.
Divide the TAS of 600 kt by 39, then square the answer and subtract 273k to find the OAT
of -36°C (420+1/0.7 = [600 kt/39) squared - 273 = -35°C)

Question 22

For example, refer to the B727 POH page 2-8. A B727 has a BRW of 72000 kg and a climb is to be made to
FL350. From the table it can be seen that with a BRW of 72000 kg the climb to FL350 will take 22 minutes
and cover 136 nm in nil wind (ANM) while consuming 2750 kg of fuel at a published TAS of 418 kt. The forecast wind 2/3 of the way up results in a headwind component for the climb of -30 kt.
Follow these steps to determine how many miles will be covered over the ground (GNM) ?

Answer 22

Divide the 136 ANM by the time interval (ETI) of 22 minutes and multiply by 60 to find climb
TAS. 136/22 x 60 = 371 kt (this is far less than the published figure which should be ignored)
Step 2
Add the wind component to the TAS to find the climb groundspeed. 371 + -30 = 341 GS
Step 3
Multiply the Groundspeed by the ETI and divide by 60 to find the GNM.
341 x 22/60 = 125 GNM

Question 23

we need to look at how to choose a particular cruise level for a flight. The
B727 is designed to cruise at FL’s in the mid to low ?

Answer 23

300s i.e FL300-FL350

Question 24

Will weather, winds or trip length effect what cruising level you pick in the exam ?

Answer 24

No

Question 25

How does thrust limits effect cruising levels ?

Answer 25

This refers to the ability of an aircraft to reach a particular level for a given gross weight and speed. It is based on the available engine thrust, from the forecast ISA deviation at the proposed level. For exam purposes the forecast level from the RSWT nearest to the proposed level is used to determine the ISA deviation.

Question 26

What is optimal flight level ?

Answer 26

This refers to the Flight Level, which will be the most economical level in terms of fuel efficiency for a particular gross weight and speed. This will not necessarily be the maximum level, which the aircraft can reach. Page 2-14 also lists optimum levels.

Question 27

How does buffet limits effect flight level ?

Answer 27

This refers to the maximum level achievable aerodynamically, considering stall margins and high-speed Mach buffet at various weights and turbulence intensity. These margins are graphically presented on pages 2-11, 2-12 and 2-13.

Question 28

To ensure that the aircraft does not exceed its buffet boundaries, draw ?

Answer 28

draw a line across the graph at the Flight Level under consideration and ensure that the SZW will reach or exceed that FL. If it does, the gap between the high and low speed sides is the available speed range.

Question 29

For example, with a top of climb weight of 82000 kg (82t) what is the maximum FL achievable for an Easterly flight at 0.82M with temperatures of ISA+10 at top of climb with moderate turbulence forecast?

Answer 29

Note the possible IFR hemispherical levels for the flight (East or West) and note your Mach Number schedule and the temperature deviation from ISA at the various levels from the RSWT. In this example, the flight is to the EAST so the possible options are FL290, FL330 or FL370 (FL410 is very unlikely in this aircraft). Choose an IFR level and check that your Start Zone Weight or Top of Climb weight does not exceed the maximum permitted thrust limited weight for the ISA deviation at top of climb and speed it can be seen that at FL290 our SZW of 82t is less than the new maximum weight of 83.2t at ISA+10 so this level is the highest FL available. The final step is to ensure that at our chosen FL we are within the aerodynamic envelope of the buffet boundaries.

Question 30

The optimum FL is not necessarily the same as the maximum FL why ?

Answer 30

On a hot day an aircraft may not be capable of achieving the optimum FL, while on a cold day the maximum FL may be well above the optimum. This is because the maximum (thrust limited) FL depends on temperature, weight and speed, however the optimum FL depends only on weight and speed. Optimum weights for each FL are shown on page 2-14.

Question 31

What is the EMZW ?

Answer 31

Since the weight of an aircraft is not constant, but reduces as fuel burns during a flight, the choice of optimum level is not made on the basis of the Brakes Release Weight (BRW) or Landing Weight (LW), but rather at the average weight for the flight (half way along) which we call the Mid-Zone Weight (MZW). Usually the MZW is not known before a flight plan and so the MZW must be estimated. In this case it is known as the Estimated Mid-Zone Weight (EMZW).

Question 32

The optimum level for a flight at a particular Mach Number is ?

Answer 32

the level, which provides the closest match between the optimum weight and the EMZW (refer to the ATPL(A) Examination Information Booklet page 29 para 2.3).

Question 33

EMZW = ?

Answer 33

EMZW = BRW- [(distance x 10 kg/nm)/2 + 1600 kg].

Question 34

Why do we need to add an extra 1600 kg of fuel to the first half of the flight to calculate a more accurate EMZW ?

Answer 34

A typical climb consumes around 3100 kg and covers around 150 nm. 10kg/nm over 150 nm is 1500 kg. This is 1600 kg less than the total climb fuel of about 3100 kg.

Question 35

In high-speed and high altitude flight, the air temperature around the leading edge of airframe surfaces over or around 250 kts ? O may rise by up to 30°C or more above OAT due to compression O may rise by up to 10°C or more above OAT due to decompression O may rise by up to 15°C or more above OAT due to compression O may rise by up to 20°C or more above OAT due to decompression

Answer 35

O may rise by up to 30°C or more above OAT due to compression

Question 36

350 CAS at 20,000 ft PA. equals what EAS (CAS)? 338 TAS 340 TAS 343 TAS 347 TAS

Answer 36

340 TAS refer to table (F correction Factors for TAS ) page 1-6 in the 727 Handbook

Question 37

In the absence of an accurate forecast or METAR, how would you determine the pressure height of the aerodrome? None of these Set altimeter to density height Set altimeter to ELEV Set altimeter to 1013.2 hPa

Answer 37

Set altimeter to 1013.2 hPa

Question 38

If the OAT is-40°C and the temperature probe has a recover factor of 0.8, with a RAM rise of 30°C; what RAT will the probe read? -10°C RAT -30°C RAT -28°C RAT -20°C RAT -16°C RAT

Answer 38

OAT of -40°C + (0.8 x 30°C) = -40°C + 24°C = -16°C RAT

Question 39

At CAS 160KT and P.A. 35,000 what is the Mach Number?

Answer 39

0.49

Question 40

The Mach Number is ? a speed in knots converted to a percentage of the speed of sound at ISA a speed ratio of how fast the aircraft is going compared to the speed of sound but has no relevance to the current ambient conditions a direct speed number relative to the speed of sound and does not change with the current ambient conditions a speed ratio of how fast the aircraft is going compared to the speed of sound in the current ambient conditions

Answer 40

a speed ratio of how fast the aircraft is going compared to the speed of sound in the current ambient conditions

Question 41

Given Mach 0.82 FL310, ACTUAL TAT = -14, ISA TAT = -16: Find the ISA variation and OAT ? ISA Difference +2 and OAT 44°C ISA Difference +4 and OAT -42°C ISA Difference +5 and OAT -41°C ISA Difference +6 and OAT -38°C

Answer 41

Given ACTUAL TAT = -14 and ISA TAT = -16: ISA Difference is + 2 ISA OAT from page 3 - 106 = -46 OAT = ISA OAT + ISA Difference = -46 + 2 = -44°C Correct Answer: ISA Difference +2 and OAT -44°C

Question 42

Question 10 of 20 What is the standard temp at A090? Type your answer correct to 2 decimal places with no symbols or spaces. e.g. for 11.2°C, just type: 11.20

Answer 42

Standard temperature at A090 (9000 ft): 15°C at MSL - (9x 1.98°C) = 15 - 17.82° = -2.82°C Correct Answer: -2.82 I think it's important you can calculate these, but they are listed on page 3 - 106 of the B727 handbook in the "O" column.

Question 43

In high speed flight (above 250 kts) ? the pressure fall results in an under-read error in AS from the total pressure fall the pressure rise results in an under-read error in AS from the total pressure fall the pressure fall results in an over-read error in IAS from the total pressure rise the pressure rise results in an over-read error in AS from the total pressure rise

Answer 43

the pressure rise results in an over-read error in AS from the total pressure rise

Question 44

In regards to the difference between TAS and CAS which is truest? CAS and TAS are closest at lower altitudes CAS equals TAS CAS and TAS have a constant difference irrespective of altitude CAS and TAS are closest at higher altitudes

Answer 44

CAS and TAS are closest at lower altitudes Refer to table (F Correction Factors For TAS) page 1 - 6 in the B727 handbook. You can see at any given CAS the correction factor decreases with altitude. At 200KT CAS, at 10,000 feet; TAS equals CAS Yet at 200 KT at 40,000 feet; TAS equals 0.96 x CAS

Question 45

CAS

Question 46

Using your maps only. Calculate the direct distance Darwin NT (YPDN) to Cairns QLD (YBCS)? 875 NM 906 NM 990 NM 830 NM

Answer 46

906 NM

Question 47

What is the standard temp at FL450?

Answer 47

Standard temperature at FL450 (45,000 ft): 15°C at MSL - (45 x 1.98°C) = 15 - 89.1° = -74.10°C However above 36,090 AMSL feet (the Tropopause), the temperature remains at -56°C up to 66,000 AMSL I think it's important you can calculate these, but they are listed on page 3 - 106 of the B727 handbook in the "O" column. So in a current 2019 version jet airliner you will never have OAT below -56°C. Correct Answer: -56.00 (we did accept 56.00 as well, but CASA won't)

Question 48

At CAS 280KT and P.A. 35,000 what is the Mach Number? Please type your answer to two decimal places with no symbols or spaces. e.g. for Mach Number 0.755 just type 0.76

Answer 48

0.82

Question 49

the cockpit TAT indicates -12°C at FL330 at 0.82M; The OAT is ?

Answer 49

From page 3 - 106 of the B727 handbook FL330 at Mach 0.82 should have an ISA TAT of -20 Bu -12°C is + 8°C warmer than the ISA TAT of -20°C The OAT from column "O" on page 3-106 -5 -50°C + 8°C warmer than ISA OAT of -50°C = -42°C Correct Answer : -42

Question 50

Mach meter uses_________to determine the Mach Number ? the current aircraft TAS and OAT the current aircraft CAS and Density Height the current aircraft TAS and Flight Level the current aircraft CAS and Flight Level / Pressure Altitude

Answer 50

the current aircraft CAS and Flight Level / Pressure Altitude

Question 51

Pilots need to be cautious of aircraft Mach numbers ? particularly at high altitude where high TAS result in low Mach numbers particularly at low altitude where high TAS result in high Mach numbers Particularly at high altitude where high TAS result in high Mach numbers particularly at low altitude where high TAS result in low Mach numbers

Answer 51

Particularly at high altitude where high TAS result in high Mach numbers Flying too close to Mach 1.0 can be extremely dangerous without the correct design features to reduce buffeting and improve control- ability.

Question 52

Using your maps only. Calculate the direct distance Melbourne VIC (YMML) to RAF Curtin WA (YCIN)? 1648 NM 1520 NM 1725 NM 1864 NM

Answer 52

1648 NM

Question 53

What wind is to be used to determine accurate climb figures in your CASA ATPL exams? Use the met data, wind and temperature deviation, closest to 2/3rds of the height of the initial cruise level Use the met data, wind and temperature deviation, closest to one half of the height of the initial cruise level Interpolate the met data, wind and temperature deviation, closest to 1/4 and 2/4 and 3/4 of the height of the initial cruise level Interpolate the met data, wind and temperature deviation, closest to 1/3 and 2/3 of the height of the initial cruise level

Answer 53

Use the met data, wind and temperature deviation, closest to 2/3rds of the height of the initial cruise level Climb. Candidates should use the met data, wind and temperature deviation, closest to 2/3rds of the height of the initial cruise level; also refer to the Handbook, page 2-2, paragraph 9 et seq. For a climb from one level to another, e.g. FL220 to FL310, then the winds and temperatures should be used from the height 2/3rds between the two levels. In his case FL.280.

Question 54

In a constant Mach cruise ? engine RPM will need to be reduced during cruise engine RPM will constantly change during cruise engine RPM will need to be increased during cruise engine RPM will need to remain constant during cruise

Answer 54

engine RPM will need to be reduced during cruise As fuel is burned off, less lift is required, so the AofA reduces. This increases cruise speed and requires a reduction in RPM to keep a constant Mach cruise.

Question 55

What is TAS at M0.78 at 32,000 feet in knots? © 455 kts © 464 kts © 471 kts © 450 kts

Answer 55

32,000 feet OAT = 15 -32 x 1.98 = 15 - 63 = -48°C TAS = VOAT in degrees Kelvin x (39 x Mach Number) -1 M 1.0 =V (-48°C + 273 ) x (39 x 0.78) - 1= V225 x (39 x 0.78) -1 = 455 kts

Question 56

If only considering fuel economy, the most efficient cruise method is ? LRC cruise climb step climb cruise constant mach cruise

Answer 56

cruise climb The cruise climb maintains the most efficient engine RPM and the most efficient AofA. To do this the aircraft climbs slowly during the cruise. In reality though this method is rarely used due to conflicting traffic and the clearances that would be required for constant level changes. A step climb is usually used as a more practical alternative.

Question 57

How many ground nautical miles (GNM) will it take to climb in the B727 to FL330 at BRW of 68,000 kg with a headwind of 40 knots at ISA +10?

Answer 57

Use Table 2.3 page 2-9 Look up table 21 minutes and convert to hours 21/60 = 0.35 hours Multiply time by wind speed per hour = 0.35 x -40 kt wind = -14 nm Look up table 131 nm and add -14 nm = 131 + -14 = 117 GNM Correct Answer: 117.00

Question 58

What is TAS at M0.8 at 25,000 feet in knots? 0 472 kts O 480 kts O 468 kts 0 464 kts

Answer 58

25,000 feet OAT = 15 - 25 x 1.98 = 15 - 49.5 = -34.5 = -35°C TAS = VOAT in degrees Kelvin x (39 x Mach Number) -1 M 1.0 =V (-35°C + 273 ) x (39 x 0.8) - 1= V238 x (39 x 0.8) -1 = 480 kt

Question 59

How many ground nautical miles (GNM) will it take to climb in the B727 to FL310 at BRW of 76,000 kg with a headwind of 20 knots at ISA +10?

Answer 59

Use Table 2.3 page 2-9 Look up table 23 minutes and convert to hours 23/60 = 0.383333333 hours Multiply time by wind speed per hour = 0.38333333 x-20 kt wind = -7.666666666 nm Look up table 146 nm and add -7.666666 nm = 146+ -7.6666666 = 138.33 GNM Correct Answer: 138.33

Question 60

Given: CAS 280 kts, Pres Alt 38,000 ft, Ind. Air Temp -10°C; Find True Air Temperature ? Use your CR computer. -46°C -40°C -42°C -52°C -38°C

Answer 60

-46°C

Question 61

For climb calculation winds from RSWT will most commonly be used from ? O FL240 © FL340 O FL180 O FL300

Answer 61

FL240 Winds for climb are to be used at 2/3rds of the climb height. The most common cruise levels will be above FL300 FL180 + 2/3 = 27000 feet = FL270 FL240 + 2/3 = 36000 feet = FL360 Halfway between these two is FL210 which has no RSWT 21000 + 2/3 = 31500 feet On the balance of probability at ISA or close to ISA the aircraft will be climbing above 31,500 feet. Therefore FL240 is the most common cruise level. You will not be required to calculate this. We have included calculations to reduce arguments about this answer, since most textbooks are wrong and say FL235, which does not have weather forecasts.

Question 62

According to your textbooks and the calculations, the old RSWT charts were prepared for flight levels FL185, FL235, FL300, FL340, FL385, FL445, which correspond with pressures of ? 600 hPa, 500 hPa, 400 hPa, 300 hPa, 200 hPa, 100 hPa 350 hPa, 300 hPa, 250 hPa, 200 hPa, 150 hPa, 100 hPa 450 hPa, 400 hPa, 350 hPa, 250 hPa, 200 hPa, 150 hPa 500 hPa, 400 hPa, 300 hPa, 250 hPa, 200 hPa, 150 hPa

Answer 62

500 hPa, 400 hPa, 300 hPa, 250 hPa, 200 hPa, 150 hPa

Question 63

In a Long Range Cruise (LRC) the ? AofA decreases and the Mach Number increases during the cruise AofA decreases and the Mach Number decreases during the cruise AofA remains constant and the Mach Number increases during the cruise AofA decreases and the Mach Number increases during the cruise AofA remains constant and the Mach Number decreases during the cruise

Answer 63

AofA remains constant and the Mach Number decreases during the cruise The angle of attack for best range is maintained, however as fuel is burned off less airspeed is required to maintain that AofA.

Question 64

For operations at FL 165, use winds ? at FL160 to the nearest 10 degrees and 5 knots at FL150 to the nearest 10 degrees and 5 knots at FL210 to the nearest 10 degrees and 5 knots at FL185 to the nearest 10 degrees and 5 knots

Answer 64

at FL185 to the nearest 10 degrees and 5 knots Operations Below FL 185. Use data at FL 185 [that is, assume the FL 185 wind applies at all levels below FL 185, and the temperature deviation from ISA (not the temperature itself) is constant below FL 185]. Extract wind to the nearest 10 degrees and 5 knots.

Question 65

The forecast for take-off and climb to FL330 is a headwind of 30 knots. This will ? decrease the time between takeoff and FL330 decrease the GNM between takeoff and FL330 increase the time between takeoff and FL330 increase the GNM between takeoff and FL330

Answer 65

decrease the GNM between takeoff and FL330 A headwind will not alter the time taken to climb to FL330, however it will reduce the ground nautical miles covered on climb to FL330.

Question 66

What is the wind direction from YPAD to YPOD at FL340?

Answer 66

310°

Question 67

What is TAS at M0.9 at 31,000 feet in knots? © 527 kts © 534 kts © 531 kts © 515 kts

Answer 67

31,000 feet OAT = 15 -31 x 1.98 = 15 - 61 = -46°C TAS = VOAT in degrees Kelvin x (39 x Mach Number) -1 M 1.0 =V (-46°C + 273 )x (39 x 0.9) - 1= V227 x (39 x 0.9) -1 = 527 kts

Question 68

The time and distance in the climb for the B727 Handbook should be ? used with rounding to 2 decimal places used without any rounding used with standard rounding used with rounding to 1 decimal place used with rounding up to whole numbers

Answer 68

used without any rounding 3.2.1 Climb. The time and distance in the climb should be used as it is in the B727 Handbook without any rounding. If interpolation is required, and as a result a time interval or a distance value is determined that is not a whole minute or whole anm, then the fraction(s) of minutes and/or anm, should be used to determine the climb performance. The TAS for the climb should be determined using a combination of the exact time in the climb with the exact am in the climb. For climbs, the candidate should use the met data closest to the 2/3rd height for the climb, as per paragraph 3.1.1.

Question 69

there is no RSWT forecast for your cruising level ? use the average of the level above and below interpolate from the level above and below use the next closest level use the mean of the level above and below

Answer 69

use the next closest level

Question 70

Given: Mach Number .80, True Air Temp -35°C; Find True Air Speed ? Use your CR computer. 0 498 kts 0 480 kts O 520 kts O 512 kts

Answer 70

480kts

Question 71

When making in-flight temperature adjustment to fuel-burn, for ISA +10 assume ? ISA+8 ISA+10 ISA+9 ISA+12

Answer 71

ISA+9 When making temperature adjustment to fuel-burn, round out the deviation from ISA to the nearest (multiples of) 3 degrees, that is, for ISA+10 assume ISA+9, and for ISA+14 assume ISA+15.

Question 72

The en-route climb fuel adjustment for a take-off from an aerodrome ELEV 6000 ft for ISA+10 is ? -350 -400 -250 -200 -300

Answer 72

-350

Question 73

When determining optimum cruise levels for most efficient RPM, a determinant is the OAT. If we segregated conditions into hotter than usual, colder than usual and ISA; we would say when it's hotter fly _______ when it’s colder fly __________ ? when hotter fly lower and when colder fly higher when hotter fly higher and when colder fly lower when either hotter or colder than ISA fly higher when either hotter or colder than ISA fly lower

Answer 73

when hotter fly lower and when colder fly higher

Question 74

The optimum flight level for a constant Mach Number cruise at a constant flight level is ? select speed for EMZW and cruise level for temp and GW at the point of engine failure select cruise level for EMZW and speed for temp and GW 5% below point of engine failure select cruise level for EMZW and speed for temp and GW at the point of engine failure select speed for EMZW and cruise level for temp and GW 5% below point of engine failure

Answer 74

select speed for EMZW and cruise level for temp and GW at the point of engine failure From the ATPL Exam information booklet, an astonishing statement to select cruise level for gross weight and temp at the point of engine failure. Select speed at the estimated mid zone weight. 3.4.2 Selection of Cruisk Level and Speed. Cruise level should be selected at the gross weight and temperature at the point of engine failure. Speed should be selected at the estimated mid-zone weight of a cruise sector.

Question 75

At what location and level does the inversion occur? O YPAD/YPOD/YMHB FL390 O YSSY/YMCO/YMHB FL290 O YMML/YSSY FL340 © YMML/YBBN FL340 © YMML/YMIDG FL340

Answer 75

YMML/YMIDG FL340

Question 76

Using the RSWT and descent profile provided, TRK 220°T calculate descent GNM, for landing weight (LW) of 65,500 kg from FL 350 ? 107 GNM 103 GNM 109 GNM 105 GNM

Answer 76

1. Determine LW or you will be given TODGW (do not interpolate - round to nearest 10,000 kg see ATPL exam info booklet para 3.2.9) e.g. for 64,999 kg use 60,000 kg, and for 65,000 kg use 70,000 kg. 1. 65,500 kg round up to 70,000 2. Enter correct descent profile table on page 4 - 3 (standard is .80M/280/250 KIAS) and go down to FL for TOD 1. Not advised otherwise, so use .80M/280/250 KIAS 3. Refer to RSWT 1/2 way between FL for TOD and ELEV of aerodrome (in every "normal" question for the B727 this will be FL180). 1. Use FL180 ISA ddssstt 22 037 13, ISA-22, OAT -13, ISA+9, Wind 220/37KT 4. Refer to descent profile for distance and time, so you can calculate TAS 1. FL350 distance = 119 nam 2. Time = 23 minutes 3. TAS = 119 / 23 x 60 = 310.43 kt =310 kt 5. Use formula GNM = GS x time / 60 1. GS = 310 + -37kt HWC = 273kt 2. GNM = 273 x 23 / 60 = 104.65 GNM = 105 GNM

Question 77

Determine initial maximum .82M cruise, FL for Boeing 727 BQ 77, TOCGW 76,000 kg, Mod Turbulence, Melbourne (37.8136° S, E) to Sydney (33.8688° S, 151.2093° E) FL340 FL290 FL360 FL359 FL330

Answer 77

FL330 TOCGW 76,000 kg at ISA from Table 2.5 page 2-14 the thrust limit for .82M is FL350, meaning FL330 or FL290 will be within the .82 thrust limit. Now let's check the buffet boundaries on page 2- 12. The blue line represents .82 Mach and the red curve represents 76,000kg You are within the limits of the high and low speed at both FL330 and FL290. Therefore FL330 is the maximum, even though FL290 has a lot more room to the edges of its safe envelope.

Question 78

When cruising IF tracking 050° in the B727 the most fuel efficient flight levels will most often be ? FL310, FL330, FL370 FL300, FL320, FL340 FL310, FL320, FL340 FL290, FL330

Answer 78

FL290, FL330 This question assumes you will use the cruising levels as per the table in the exam booklet. The most fuel efficient flight levels are FL300 -FL350 most often for the B727. For east the most fuel efficient levels for the B727 will include FL310, FL330, FL350 as in most cases you will not be able to get to FL370 or higher unless you are getting down below 61,000 kg Note all questions relate to IFR in your exams unless otherwise stated. Be carefull Read these instructions from the ATPL exams booklet and read the question carefully. If they do not mention the AIP then choose from the table in the workbook. 2 SELECTION OF FLIGHT LEVELS 2.1 IFR Levels. All questions relate to operations under the IFR. Unless otherwise stated in a question, all operations, with the exception of depressurised cruise, will cruise at altitudes* in accordance with the AIP ENR 1.7 (6) Table of Cruising Levels ENR 1.7 (6) South of 80° S for IF flights, as per the table of Cruise Levels on page 1 of this book. Candidates are expected to recall and apply the appropriate levels for different sectors. In this case you only have FL290, FL330 and FL370 to choose from. Looking at page 1-1 limitations Max BRW is 89,700 kg Looking at page 2-8 climb fuel at ISA to FL290 heavier weights ranges from 2400 kg to 3150 kg Looking at page 2-8 climb fuel at ISA to FL330 heavier weights ranges from 2800 kg to 4050 kg If we take off at Max BRW we will reach FL290 at approx 87200 kg max and FL330 at approx. 86,300 kg max Meaning any time we take off near max weight FL290 will be most suitable based on table 2.5 page 2 - 14 We can only get to FL370 at lighter weights Looking at the table 2.5 on page 2 - 14 FL 330 sits right in the middle of the available weight FL for optimal weights This narrows down our choices to FL290 or FL330

Question 79

If the SZW is 80,000 kg, what would be the highest altitude (rounding to the nearest 1000 feet) within the buffet limit? Forget about directions and levels for this answer. (Refer to 727 Manual, moderate turbulence) FL370 FL330 FL340 FL350 FL310

Answer 79

FL330

Question 80

The B727 will most often achieve best cruise fuel consumption in the range of ? © FL300 - FL400 O FL350 - FL400 O FL250 - FL350 © FL300 - FL350

Answer 80

FL300 - FL350

Question 81

In the B727 with a BRW of 77,400 kg determine climb fuel to find TOCGW to FL310 for TRK 300°T Given RSWT Chart of ? O 74,300 kg TOCGW © 76,000 kg TOCGW O 73,850 kg TOCGW © 74,000 kg TOCGW O 75,150 kg TOCGW

Answer 81

74,300 kg TOCGW 1. BRW to nearest 2000 kg = 78,000 (as per note 11 on page 2-2 B727 handbook) 2. 2/3rds of climb height is 31,000ftx 0.666666667 = 20,666 ft 3. So FL180 is the most similar level 4. ISA temp variation to nearest 5 degrees at FL180 (2/3rds height of climb) = ISA + 8, so use ISA + 10 table. Table 2.3 page 2-9 5. Interpolate between 76,000 kg and 80,000 kg to get 78,000 kg data 6. Subtract climb fuel to get TOCGW = 77,400 kg BRW - 3100 kg = 74,300 kg

Question 82

The standard descent profile (of .80 M / 280 / 250 KIAS unless otherwise advised) mens ? descend at .80M until FL where 0.8M = 280 IAS, then 280 IAS until 10,000 ft, then 250 KT below 10,000 ft descend at .80M until FL where 0.8M = 280 IAS, then 280 IAS until the circuit, then 250 KT in the circuit descend at .80M until 10,000 ft, then 280 IAS until the circuit, then 250 KT in the circuit descend at .80M until FL280, then 250 KT below FL280

Answer 82

descend at .80M until FL where 0.8M = 280 IAS, then 280 IAS until 10,000 ft, then 250 KT below 10,000 ft Refer to page 4-1 B727 handbook.

Question 83

In the B727 with a BRW of 77,400 kg determine climb fuel to find TOCGW to FL330 for TRK 300ºT and also find GNM for climb. Given RSWT Chart of ? 74,450 kg TOCGW and 175 nm 74,350 kg TOCGW and 145 nm 74,250 kg TOCGW and 149 nm 74,000 kg TOCGW and 150 nm 74,000 kg TOCGW and 175 nm

Answer 83

74,000 kg TOCGW and 175 nm 1. BRW to nearest 2000 kg = 78,000 (as per note 11 on page 2-2 B727 handbook) 2. 2/3rds of climb height is 33,000 ft x 0.66666667 = 21,999 ft 3. So FL240 is the most similar level 4. ISA temp variation to nearest 5 degrees at FL240 (2/3rds height of climb) = ISA +8, so use ISA + 10 table. Table 2.3 page 2-9 5. Interpolate between 76,000 kg and 80,000 kg to get 78,000 kg data 27.5/3400/182 6. Using whiz wheel, HWC of 16 kts 16/60 = 0.26666 X27.5mins = 7 nm Subtract 7 nm from 182 nm = 175 nm Subtract 3400 kg from 77,400 to get TOCGW = 74,000 kg

Question 84

In the B727 with a BRW of 77,400 kg determine climb fuel to find TOCGW to FL300 for TRK 240°T Given RSWT Chart of ? 74,850 kg TOCGW 74,690 kg TOCGW 74,425 kg TOCGW 74,250 kg TOCGW

Answer 84

74,425 kg TOCGW 1. BRW to nearest 2000 kg = 78,000 (as per note 11 on page 2-2 B727 handbook) 2. 2/3rds of climb height is 31,000 x .666666667 = 20,666 ft 3. So FL180 is the most similar level 4. ISA temp variation to nearest 5 degrees at FL240 (2/3rds height of climb) = ISA + 8, so use ISA + 10 table. Table 2.3 page 2-9 5. Interpolate between 76,000 kg and 80,000 kg and FL290, FL310 to get 78,000 kg, FL300 data 23.25/2975/145.5 Subtract climb fuel to get TOCGW 77.400 - 2975 = 74,425 kg

Question 85

Using the RSWT and descent profile provided, TRK 220°T calculate descent GNM, for landing weight (LW) of 65,000 kg from FL 320. 101 GNM 100 GNM 99 GNM 97 GNM

Answer 85

97 GNM 1. Determine LW or you will be given TODGW (do not interpolate - round to nearest 10,000 kg see ATPL exam info booklet para 3.2.9) e.g. for 64,999 kg use 60,000 kg, and for 65,000 kg use 70,000 kg. 1. 65,000 kg round up to 70,000 2. Enter correct tiescent profile table on page 4 - 3 (standard is.80M/280/250 KIAS) and go down to FL for TOD 1. Not advised otherwise, so use .80M/280/250 KIAS 3. Refer to RSWT 1/2 way between FL for TOD and ELEV of aerodrome (in every "normal" question for the B727 this will be FL180). 1. Use FL180 ISA ddssstt 22 037 13, ISA-22, OAT -13, ISA+9, Wind 220/37KT 4. Refer to descent profile for distance and time, so you can calculate TAS (interpolate if necessary) 1. FL320 distance = interpolate (114 nam - 107 nam) / 2 + 107 nam = 110.5 nam 2. Time (interpolate) = 21.5 minutes 3. TAS = 110.5 / 21.5 x 60 = 308.37 kt =308 kt 5. Use formula GNM = GS x time / 60 1. GS = 308 + -37kt HWC = 271 kt 2. GNM = 271 x 21.5 / 60 = 97.108 GNM = 97 GNM

Question 86

The upper level winds are ? northerly southerlv easterly westerly

Answer 86

example: 2309064, where 23 = 230°, 090 = 90 knots, 64 = -64°C

Question 87

no turbulence ? buffet boundaries will most likely be far less limiting than thrust limits none of these buffet boundaries will most likely be more limiting than thrust limits buffet boundaries will not affect limits

Answer 87

buffet boundaries will most likely be far less limiting than thrust limits

Question 88

For an enroute climb from FL310 to FL350 in ISA conditions with nil wind, with a BRW 76,000 kg; What will be the time, fuel burnt and distance travelled? 6 minutes, 500 kg, 44 nm 8 minutes, 600 kg, 33 nm 7 minutes, 500 kg, 43 nm 9 minutes, 400 kg, 43 nm

Answer 88

6 minutes, 500 kg, 44 nm Enter table 2.2 at FL310 76,000 kg and read: 19/2550/113 Enter table 2.2 at FL350 76,000 kg and read: 25/3050/157 Subtract the differences which gives for minutes/fuel/dist: 6/500/44

Question 89

In the B727 with a BRW of 76,200 kg determine climb fuel to find TOCGW to FL330 for TRK 300°T and find GNM. Given RSWT Chart of ? 73,000 kg TOCGW and 161 GNM 73,950 kg TOCGW 135.6 GNM 73,400 kg TOCGW and 162 GNM 73,950 kg TOCGW 133.2 GNM

Answer 89

73,000 kg TOCGW and 161 GNM BRW = 76,200 kg, round to nearest even = 76,000 kg 2/3rds of climb height = FL220. Use FL240 RSWT data ISA + 8 Round to ISA + 10. 25 min/3200 kg/168 ANM Fuel burn to FL330 ISA + 10 = 3200 kg TOCGW = 76,200 - 3200 = 73,000 kg Wind component: 240/32 kts TRK 300 degrees 16 kt HW divide by 60 and times by 25 mins in climb = 7 nm Reduce climb distance by 7 nm. = 168 - 7 = 161 gnm

Question 90

A heavy B727 from FL300, in nil wind, will glide approximately ? 100 nm 60 nm 120 nm 80 nm

Answer 90

100 NM 1 nm = 6076 feet 20:1 approximate glide ratio 30,000 feet / 6076 = 4.937 nm 20 x4.937 = 98.7 nm

Question 91

Thrust limits will be more limiting than buffet boundaries ? in no turbulence in moderate turbulence in heavy turbulence none of these

Answer 91

in no turbulence

Question 92

In the B727 at ISA using the LRC schedule for FL330 with a drift angle of 1°, the difference between TAS and ETAS will be ? 2 kt 1 kt 4kt 3 kt 0 kt

Answer 92

0kts At Cruise Speeds 5° x7 / 20 = 1.75 kt = 2 kt 4° x 7 /20 = 1.4 kt = 1 kt 3° x7 / 20 = 1.05 kt = 1 kt 2° x 7 / 20 = 0.7 kt = 1 kt 1° x 7 / 20 = 0.35 kt = 0 kt At Descent Speeds 5° x 6 / 20 = 1.5 kt = 2 kt 4° x 6 /20 = 1.2 kt = 1 kt 3° x 6 / 20 = 0.9 kt = 1 kt 2° x 6 / 20 = 0.6 kt = 1 kt 1° x 6 / 20 = 0.3 kt = 0 kt

Question 93

For a TAS of 400 KT find ETAS with 15° drift? Type your answer as a whole number only with no spaces, decimals or symbols.

Answer 93

386- 388

Question 94

The B727 cruise fuel flow is changed by which of the following? Select all correct responses. OOAT O GW O Mach Number O Altitude O Take-off ELEV O Anti-ice O Landing ELEV O Bleed decisions

Answer 94

OAT GW Mach Number Altitude Anti- Ice Bleed decisions

Question 95

What is the Specific Ground Range of an aircraft weighing 75T in an ISA environment with no wind while cruising at FL350 at M0.79? 9.5 9 10. 5 10

Answer 95

9.5

Question 96

Using only the information in this question and commonly known "estimated" fuel flows for the B727; With a BRW of 85,400 kg, estimate GW at position indicated by Delta. Type your response as numbers only, with no spaces, decimals or symbols. e.g. for 95,200.6 kg just type: 95201

Answer 96

Climb estimate for 150 nm is 3100 kg Zone 1 estimate = 300 nm x 10 kg = 3000 kg Half of Zone 2 estimate = 150 nm x 10 kg = 1500 kg 85,400 kg - 3100 - 3000 - 1500 = 77,800 kg Correct Answer: 77800

Question 97

At a an average sector weight between 70,000 kg and 80,000 kg, with no wind, the B727 have fuel flow closest to ? 9 kg / nm 10 kg / nm 12 kg / nm 11 kg / nm 8 kg / nm

Answer 97

10 kg / nm

Question 98

At an average sector weight above 80,000 kg, with no wind, the B727 have fuel flow closest to ? 11 kg / nm 9 kg / nm 12 kg / nm 10 kg / nm 8 kg / nm

Answer 98

11 kg / nm

Question 99

For a TAS of 420 KT find ETAS with 20° drift?

Answer 99

394-396

Question 100

At an average sector weight between 60,000 kg and 70,000 kg, with no wind, the B727 have fuel flow closest to ? 8 kg / nm 9 kg / nm 12 kg / nm 10 kg / nm 11 kg / nm

Answer 100

9 kg / nm 50 kt HWC leads to SGR increase of 1 kg / nm

Question 101

On the CR model flight computers, the difference between ETAS and TAS is ? All of these interpolated between 5° and 10° drift virtually the same below 5° drift very different above 10° drift

Answer 101

All of these

Question 102

For a TAS of 380 KT find ETAS with 13° drift?

Answer 102

369-371

Question 103

At an average sector weight between 83,000 kg, with a 50 kt headwind, the B727 have fuel flow closest to ? 9 kg / nm 10 kg / nm 12 Kg/ nm 11 kg / nm 8 kg / nm

Answer 103

12 Kg/ nm Add 1 kg for the headwind and another 1 kg for the weight above 80,000 kg

Question 104

At a an average sector weight between 60,000 kg and 70,000 kg, with no wind, at FL 380 the B727 have fuel flow closest to ? 9 kg / nm 12 kg / nm 8 kg / nm 10 kg / nm 11 kg / nm

Answer 104

9 kg / nm At this weight and at FL380 with no wind, the SGR will be about 9 kg/nm

Question 105

The LRC cruise reduces range by approximately how much compared to speed for max range? 0.8% 1.6% 1.9% 1.2% 1%

Answer 105

1%

Question 106

At an average sector weight between 60,000 kg and 70,000 kg, with a 50 kt head wind, at Mach 0.82, the B727 have fuel flow closest to ? 8 kg / nm 11 kg / nm 9 kg / nm 12 kg / nm 10 kg / nm

Answer 106

11kg / nm 50 kt HWC leads to SGR increase of 1 kg / nm Rough SGRs (adjusted by 0.02 x HW/TW component): LRC / Straight line holding - 9.2 Mach 0.79 / 0.8 -9.5 Mach 0.82 - 10.0 Mach 0.84 - 10.2 1 engine inop - 10.7 Depressurised-12.5 Gear down - 20.0

Question 107

Using only the information in this question and commonly known "estimated" fuel flows for the B727; With a BRW of 85,400 kg, estimate GW at position indicated by Foxtrot ? Type your response as numbers only, with no spaces, decimals or symbols. e.g. for 95,200.6 kg just type: 95201

Answer 107

Climb estimate for 150 nm is 3100 kg Zone 1 estimate = 300 nm x 10 kg = 3000 kg Zone 2 estimate = 300 nm x 10 kg = 3000 kg Half of Zone 3 estimate = 150 nm x 10 kg = 1500 kg 85,400 kg - 3100 - 3000 - 3000 - 1500 = 74,800 kg Correct Answer: 74800

Question 108

The SZW at Alpha is 83,200 kg, estimate EMZW at position indicated by Bravo at FL330, ISA, wind -40, .80M. (3 Mark Question) 80,919 kg 82,319 kg 81,879 kg 81,490 kg

Answer 108

81,490 kg The SW at Alpha is 83,200 kg, Half of Zone 1 estimate = 150 nm x 10 kg = 1500 kg 83,200 kg - 1500 = 81,700 kg (we will use this number to ensure our calculation is accurate) Refer to page 3 - 92 Enter table at column 82 and row 330 - Read the bottom number in cell which is fuel flow for one engine = 1615 kg / hr Multiply 1615 x 3 to get fuel flow per hour for the 3 engine B727 = 4845 kg / hr OAT for ISA at FL330 = -50°C Using CR computer, convert 0.80M at -50°C to TAS = 465 kts Alternatively, use SoS equation. TAS 465 kts - 40 kt HW = 425 kts Distance/G5x fuel per hour = Zone Fuel 300/425 x 4845 = 3420 kg of fuel for zone EMZW = TOCGW - (3420/2) = 83,200 - 1710 = 81,490 kg

Question 109

Using only the information in this question and commonly known "estimated" fuel flows for the B727; With a BRW of 86,000 kg, estimate GW at position indicated by Alpha ? 82,900 80,000 81,000 84,500

Answer 109

82,900 kg Climb estimate for 150 nm is BRW 86,000 kg - 3100 kg = 82,900 kg Correct Answer: 82900

Question 110

ETAS is ? can be faster, slower or the same as TAS always faster than TAS always slower than TAS none of these

Answer 110

always slower than TAS

Question 111

A headwind or tailwind outbound will ? reduce the distance to the PNR from the departure aerodrome increase the distance to the PNR from the departure aerodrome none of these may reduce or increase the distance to the PNR from the departure aerodrome, depending on the wind direction

Answer 111

reduce the distance to the PNR from the departure aerodrome

Question 112

In the last 100 nm inbound to Alpha 70% of the track is 240° at 280 KIAS, while 30% of the track is 200° 250 KIAS. RSWT 2404026. Find the average track ?

Answer 112

RSWT: 240° 040 KT-26°C Average Track: (70 x 240 + 30 x 200) / 100 = (16,800 + 6000) /100 = 228° Correct Answer: 228

Question 113

The magnetic variation for 80 nm of the zone 1 is 010° and the remaining 220 nm is 011°. The variation to be used in calculations is ? 10.73° 10.5° 10.64° 11° 10°

Answer 113

Average variation is: (80 x 10 + 220 x 11) /300 = (800 + 2420 ) /300 = 10.73 ° = 11° For zones with only two variations, this can actually be done without any maths. If you fly through two areas of variation, the area what is more than half of the zone is the variation used for that zone.

Question 114

The magnetic variation for 80 nm of the zone 1 is 010°, 70 nm is 011° and the remaining 120 nm is 012°. The variation to be used in calculations is ? 12.02° 11° 10° 12° 11.41°

Answer 114

Average variation is: (80 x 10 + 70 x 11 + 120 x 12) / 270 = (800 + 770 + 1440 ) /270 = 11.14 ° = 11°

Question 115

The approach manoeuvre fuel allowance in the B727 is how many kg?

Answer 115

400kg

Question 116

The taxi shut down fuel flow allowance in the B727 is how many kg?

Answer 116

100 kg

Question 117

Refer to RSWT and ERC-H Plan a flight from Adelaide(YPAD) to Hobart( YMBH) Use takeoff and distance YPAD to DRINA 45 nm and track 121° Assume takeoff, then straight line on track from YPAD to DRINA Straight in approach to YMHB with standard manoeuvre allowance The BRW is limited to BRW 80,500 kg for this flight. Adjusted BW is 47,210 kg Cruise at maximum height and maximum constant Mach cruise and use ERC-H route. No abnormal operational considerations or alternates necessary for this question. Use 2 zones for cruise at .80 Mach, estimated climb fuel, ATPL exam info cruise levels. Expected holding up to 15 min Weather is CAVOK Find: Flight Level. TOD Estimated GW. TOD location. © FL290, TODGW 75,966 kg, TOD 101 nm from YMHB O FL330, TODGW 75,966 kg, TOD 114 nm from YMHB © FL350, TODGW 75,966 kg, TOD 119 nm from YMHB © FL350, TODGW 75,726 kg, TOD 119 nm from YMHB © FL330, TODGW 74,601 kg, TOD 136 nm from YMHB © FL290, TODGW 75.726 kg, TOD 101 nm from YMHB

Answer 117

O FL330, TODGW 75,966 kg, TOD 114 nm from YMHB

Question 118

Given: Wind 090°M/55, Magnetic variation 0°, TAS 450 kt, Track 180° Find heading, ground speed and drift angle ? Heading 187°, GS 445 kts, Drift Angle 7° Heading 175°, GS 447 kts, Drift Angle -5° Heading 185°, GS 445 kts, Drift Angle 5° Heading 173°, GS 447 kts, Drift Angle -7

Answer 118

Heading 173°, GS 447 kts, Drift Angle -7 55 kts of direct crosswind from the left = 7 degrees of drift 7 degrees of drift = ETAS of 447 kts No headwind or tailwind component, GS = 447 kts Take away 7 degrees of drift: HDG = 173 degrees

Question 119

For à 1 engine inoperative or depressurised PNR, the calculation is made using ? fly normal ops out to the PNR, then from PNR back home with abnormality; arriving back with the minimum abnormal operations reserve fuel fly normal ops out to the PNR, then from PNR back home with abnormality; arriving back with the minimum reserve fuel for normal ops fly normal ops 75% out to the PNR, then from the 75% to PNR plus back home with abnormality, arriving back with the minimum abnormal operations reserve fuel fly normal ops 75% out to the PNR, then from the 75% to PNR plus back home with abnormality; arriving back with the minimum reserve fuel for normal ops

Answer 119

fly normal ops out to the PNR, then from PNR back home with abnormality; arriving back with the minimum abnormal operations reserve fuel Refer Pages 1-16, 1-16A of B727 Handbook

Question 120

Given: Track 100, TAS 450, R$WT 21055045; Find heading, ground speed and drift angle ? Magnetic Variation: Nil Heading 107°, GS 466 kt Drift 7° Heading 093°, GS 458 kt, Drift -7° Heading 107°, GS 458 kt, Drift 7° Heading 093°, GS 466 kt, Drift -7°

Answer 120

Heading 107°, GS 466 kt Drift 7° 52 kts of crosswind from the right = 7 degrees of drift Degrees of drift means ETAS = 447 19 kt Tailwind, GS = 447 + 19 = 466 kts Add on 7 degrees of drift: HDG = 107 degrees example: 2309064, where 23 = 230°, 090 = 90 knots, 64 = -64°C RSWT: 210°, 055 knots, -45°C

Question 121

Given: Track 295, TAS 420, RSWT 0506556; Find heading, ground speed and drift angle. Magnetic Variation: Nil Heading 303°, GS 443 kt, Drift 8° Heading 303°, GS 430 kt, Drift -8° Heading 278°, GS 447 kt, Drift 8° Heading 278°, GS 430 kt, Drift -8°

Answer 121

Heading 303°, GS 443 kt, Drift 8° 59 kts of crosswind from the right = 8 degrees of drift 8 degrees of drift mean ETAS = 415 28 kt Tailwind, GS = 447 + 19 = 443 kts Add on 8 degrees of drift: HDG = 303 degrees example: 2309064, where 23 = 230°, 090 = 90 knots, 64 = -64°C RSWT: 050°. 065 knots, -56°€

Question 122

The depressurised fuel flow rate in the B727 is how many kg per nm?

Answer 122

13

Question 123

Given: Wind 030°M/35, Magnetic variation 0°, TAS 450 kt, Track 180°, Find heading, ground speed and drift angle ? Heading 178°, GS 445 kts, Drift Angle 2° Heading 182°, GS 476 kts, Drift Angle 2° Heading 178°, GS 480 kts, Drift Angle -2° Heading 179°, GS 480 kts, Drift Angle 1°

Answer 123

Heading 178°, GS 480 kts, Drift Angle -2° 17 kts of crosswind from the left = 2 degrees of drift 2 degrees of drift means ETAS is not a factor 30 kt tailwind, GS = 450 + 30 = 480 kts Take away 2 degrees of drift: HDG = 178 degrees

Question 124

Given: LW = 58,500 kg, climb dist 150 nm, zone 1 dist 500 nm, zone 2 dist 500 nm, zone 3 dist 280 nm, descent 115 nm, Track 200° WY 020/72. Av Variation 10°E, FL 350 Find: Estimated Brake Release Weight. 75,130 kg 74,380 kg 73,230 kg 79,560 kg

Answer 124

75,830 EBRW = LW or EZW + [(total dist. × fuel flow kg/nm) + 1600] EBRW = 58,500 + (1430 x 11 kg) + 1600 EBRW = 58,500 + 17,330 EBRW = 75,830 kg

Question 125

Given: LW = 63,480 kg, climb dist 150 nm, Distance covered in cruise is 750 nm, Track 200°, WV 195/72. true in the cruise, Av Variation 10°E FL 350 M0.80 and ISA conditions exist during climb and cruise. Find: Estimated Brake Release Weight ? 73,010 kg 75,023 kg 77,160 kg 76,215 kg 75,490 kg

Answer 125

75,023 kg

Question 126

If there is a significant direction change enroute, to attain the most accurate figures, your planning should ? divide either side of the waypoint into separate zones average the conditions either side of the waypoint and apply equal weighting to both for the zone average none of these average the conditions either side of the waypoint on a pro-rata amount for the zone

Answer 126

divide either side of the waypoint into separate zones Using averages is allowable, but for the most accurate answers splitting the route into zones at any significant change in direction or conditions will give more accurate results. In your exam, CASA will generally give you some strong hint or instruction on how they want it done. The golden rule is: more segments = increased accuracy.

Question 127

Refer to RSWT and ERC-H2 Plan a flight from Brisbane (YBBN) to Alice Springs( YBAS) Use takeoff and distance YBBN to ROMA as 250 nm and track 271° Assume straight line departure from YBBN to ROMA (YROM - ROM) Straight in approach to YBAS. Due to undisclosed limitations the BRW is limited to BRW 81,000 kg for this flight. Adjusted BW is 47,210 kg Cruise at maximum height and maximum constant Mach cruise and use ERC-H route. No abnormal operational considerations or alternates necessary for this question. Use 3 zones for cruise, estimated climb fuel, ATPL exam info cruise levels. Find: Find Mach Number cruise speed (in your exam this will normally be given). Estimated TOC weight. Estimated TOD weight. ET for entire flight. 0.82M, 77,900 kg. 70,362 kg, 149 min 0.82M, 77,700 kg, 70,862 kg. 168 min 0.79M, 77,900 kg. 70,362 kg. 149 min 0.80M, 77,900 kg. 70,362 kg, 149 min 0.80M, 77,700 kg, 70,862 kg, 168 min 0.79M, 77,700 kg, 70,862 kg. 168 min

Answer 127

The best RSWT is in the WEST. Make sure you have downloaded and learned all the RSWT locations. It's in our Facebook Group under files or here: RSWT Maps (opens in new tab) Step 1 Identify route waypoints from ERC-H: YBBN to YROM 250 nm 271°, YROM to VILOL 219 nm 269°, VILOL to NONET 200 nm 272°, NONET to PULOL 245 nm 274°, PULOL to YBAS 163 nm 278°. Step 2 Identify the RSWT to use and where the weather changes on the RSWT. Weather changes at YROM and NONET Step 3 Identify and divide your plan into 3 Zones based on the above. YBBN to TOC 271°, Zone 1 TOC to YROM 271°, Zone 2 YROM to NONET 270°, Zone 3 NONET to TOD 276°, TOD to YBAS 278° Step 4 RSWT for 2/3 of TOC is FL240 = 220° 025 KT -21°C, ISA for 240 is 15 - 24 x 1.98 = -33, which is ISA +12 Step 5 From page 2 - 14 Altitude Capability for ISA + 10 Assuming 3100 kg for climb (81,000 - 3100 = 77,900) thrust limit 77.9 in a WEST direction FL310 with any cruise speed except .84M, so we go with 0.82M Step 6 Extract RSWT winds and temps for FL310 for each zone - use FL 300 ISA for FL300 is 15 - 30 x 1.98 = -44°C YBBN to YROM 230° 041 KT -37°C . is ISA +7 YROM to NONET 230° 057 KT -32°C. is ISA +12 NONET to YBAS 260° 039 KT -30°C. is ISA +14 Step 7 Use RSWT winds and CR computer and convert .82 Mach Number to TAS, heading and drift. YBBN to YROM Track 271° Av. Var = 10°E, RSWT 230° 041 KT -37°C, is ISA +7, Magnetic Wind 220° = 465, GS = 433 Heading = 268 Drift =-3° YROM to NONET Track 270° Av. Var = 9°E, RSWT 230° 057 KT-32°C, is ISA +12, Magnetic Wind 221° = 469, GS = 430 Heading = 265 Drift =-5° NONET to YBAS Track 276° Av. Var = 7°E, RSWT 260° 039 KT-30°C, is ISA +14, Magnetic Wind 253° = 472, GS = 436 Heading = 274 Drift =-2° Step 8 Calculate climb Figures from table 2.3 page 2-9 Enter column 80,000 and column 31,000; Time 26 Fuel 3250 Distance 165 TAS 420 = Groundspeed 393, Heading 267, Drift -4° Step 9 Estimate TODGW to get descent distance from page 4 - 3 TOCGW = 77,900, Zone 1 = 85 nm = 850 kg, EZ1W = 77,900 - 850 = 77,050 kg Zone 2= 419 nm = 4190 kg EZ2W = 77,050 - 4190 = 72,860 kg End of Zone 3 weight will be closest to 70,000 kg, so distance for descent is 107 nm, 660 kg fuel, 21 min Zone 3 = 194 + 163 - 107 = 250 nm, EZ3W = 72,860 - 2500 = 70,362 kg Estimated TOD weight = 70,362 Don't forget to add or subtract 1 kg for winds above 50 kt or gross weights above 80T and below 70T or FL above 370 Step 10 Calculate time and distance each zone Estimate EMZW for 3 zones: Time = distance / speed Climb 26 min 165 nm from above Zone 1 TOC to YROM , 250 nm - 165 nm = 85 nm, Time = 85 / 433 x 60 = 11.77 min Zone 2 YROM to NONET, 419 nm, Time = 419 / 430 x 60 = 58.46 min Zone 3 NONET to TOD = 194 + 163 - 107 = 250 nm, Time = 250 / 472 x 60 = 31.78 min Descent 21 min 107 nm from above ETI for flight = 26 + 11.77 + 58.46 + 31.78 + 21 = 149 minutes

Question 128

Given: Track 260, TAS 450, RSWT 0603256; Find heading, ground speed and drift angle. Magnetic Variation: Nil Heading 259°, GS 468 kt, Drift 1° Heading 259°, GS 478 kt, Drift -1° Heading 261°, GS 480 kc, Drift 1° Heading 2612°, GS 488 kt, Drift 2°

Answer 128

Heading 261°, GS 480 kc, Drift 1° 10 kts of crosswind from the right = 1 degree of drift 1 degrees of drift means ETAS is no factor 30 kt Tailwind, GS = 450 + 30 = 480 kts Add on 1 degrees of drift: HDG = 261 degrees

Question 129

Given: Wind 220°M/75, Magnetic variation 0°, TAS 450 kt, Track 180°, Find heading, ground speed and drift angle ? Heading 186°, GS 390 kts, Drift Angle 6° Heading 174°, GS 392 kts, Drift Angle- 6° Heading 186°, GS 388 kts, Drift Angle -6° Heading 171°, GS 394 kts, Drift Angle 9°

Answer 129

Heading 186°, GS 390 kts, Drift Angle 6° 49 kts of crosswind from the right = 6 degrees of drift 6 degrees of drift means ETAS = 449 kts 59 kt Headwind, GS = 449 - 39 = 390 kts Add on 6 degrees of drift: HDG = 186 degrees

Question 130

How to calculate SZW ?

Answer 130

BRW- climb fuel

Question 131

Do buffet boundaries need to be considered for both optimum and maximum flight levels ?

Answer 131

Yes

Question 132

A B727 is on a westerly track, cruising at 0.80M under ISA+10 conditions at top of climb. Start Zone Weight (SW) for the sector is 82000 kg. Moderate turbulence is forecast. If the aircraft Estimated Mid Zone Weight (EMZW) is 79800 kg, the optimum Flight Level that complies with the IFR rules is ? FL350 FL370 FL310 FL280

Answer 132

FL310

Question 133

If turbulence is moderate, the optimum westerly Flight Level that complies with the IF rules for a B727. SZW 80000 kg and an EMZW of 75500 kg, cruising at 0.80M under ISA conditions at top of climb is ? FL310 F1370 FL280 FL350

Answer 133

FL310

Question 134

If turbulence is forecast as moderate for a B727 maintaining FL310 at 0.79M under ISA-5 conditions at top A of climb, the maximum GW at which the aircraft could maintain the next highest IFR cruise level is ?

Answer 134

74000 kg

Question 135

If heavy turbulence is forecast and the Mach Number is to be calculated by referring to the turbulence penetration speed, the highest GW at which the aircraft could maintain FL350 is ? 60000 kg 64400 kg 58100 kg 67600 kg

Answer 135

60,000 kg

Question 136

A B727 is flying an easterly track. SW is 85000 kg and EMZW is 80600 kg. Cruising at 0.79M under ISA+5 conditions at top of climb, with no turbulence, the optimum level that complies with the IFR rules is ? FI330 FL370 FL290 FL310

Answer 136

FL290

Question 137

With a current GW of 64500 kg and cruising at 0.80M, a B727 captain is searching for the optimum easterly Flight Level. The SW was 67000 kg. If conditions at top of climb are ISA+5 and there is moderate turbulence forecast, the Flight Level at which he/she should fly is ? FL370 FL290 FL330 FL350

Answer 137

FL370

Question 138

Q7 There is no turbulence forecast for a flight from PERTH to BRISBANE with a Brakes Release Weight (BRW) of 73000 kg. The cruise Mach Number will be 0.79M and the temperature variation is ISA+5°C at all levels. The maximum initial Flight Level is ?

Answer 138

FL330

Question 139

There is no turbulence forecast for a flight from BRISBANE to ADELAIDE with a Brakes Release Weight (BRW) of 73000 kg. The cruise Mach Number will be 0.82M and the temperature variation is ISA+10°C at all levels. The maximum initial Flight Level is ? FL370 FL350 FL330 FL310

Answer 139

FL350

Question 140

There is no turbulence forecast for a flight from DARWIN to SYDNEY with a Brakes Release Weight (BRW) of 86000 kg. The level is FL 290 the temperature variation is ISA+15°C at all levels. The maximum initial Mach number is ? 0.80 0.82 0.84 0.79

Answer 140

0.79

Question 141

The highest Outside Air temperature (OAT) which would allow a B727 to cruise at FL350 with a gross Weight (GW) of 76000 kg at 0.80M without exceeding cruise thrust limits is ? -26 -54 0 -49

Answer 141

-54

Question 142

With an estimated mid zone weight (EMZW) for a flight of 73500 kg and a cruise Mach Number of 0.82, the optimum Flight Level for a flight from ALICE SPRINGS to CAIRNS is ? FL330 FL310 FL290 FL270

Answer 142

FL330

Question 143

Q12 At a gross weight of 80000 kg at 0.80M at FL330 flying in moderate turbulence, the speed margins in knots above the low speed buffet and below the high-speed buffet are ? 7 kt below the high-speed buffet and 22 kt above the low speed buffet 22 kt below the high-speed buffet and 7 kt above the low speed buffet 10 kt below the high-speed buffet and 10 kt above the low speed buffet 260 kt and 290 kt

Answer 143

7 kt below the high-speed buffet and 22 kt above the low speed buffet

Question 144

When the B727 POH says to 'use average data' it means you should ?

Answer 144

do the working on a single sector basis (find a single EMZW from a positive fix to TOD).

Question 145

How does gross weight effect SGR ?

Answer 145

at average weights from 70 - 80t the B727 will burn close to 10 kg/nm at heavier average weights (above 80t) the rate is closer to 11 kg/nm at highter average weights from 60-70t the rateis closer to 9 kg/nm Less than 60T = 8 kg/ nm

Question 146

How does wind effect SGR ?

Answer 146

with 50 kt of head wind the SGR will increase by an extra 1 kg/nm with 50 kt of tail wind the SGR will decrease by 1 kg/nm

Question 147

How does altitude effect SGR ?

Answer 147

If the FL is below FL280, add an extra 1 kg/nm If the FL is above FL370, subtract 1 kg/nm

Question 148

Are the SGR adjustments cumulative?

Answer 148

Yes

Question 149

Finding EMZW from a BRW requires a climb allowance of 1600 kg to be added to the fuel estimate. What’s the equation?

Answer 149

EMZW = BRW - [(total distance x 10 kg/nm)/2 + 1600 kg]

Question 150

If the SW weight is known then no climb allowance is needed when estimating Mid Zone Weight. What’s the equation required ?

Answer 150

EMZW = SZW - (zone distance x ?? kg/nm)/2

Question 151

Calculate the zone fuel. Equation ?

Answer 151

Zone fuel = Distance/GS x fuel flow

Question 152

Calculate the Estimated Time Interval in minutes (ETI). Equation?

Answer 152

(ETI = Distance/GS x 60)

Question 153

Calculate the End Zone Weight (EZW). Equation? .

Answer 153

EZW = SZW - Zone Fuel EZW = SW 72400 kg - 4927 kg zone fuel = 67473 kg EZW. Enter 67473 into the table

Question 154

How to calculate the actual MZW ?

Answer 154

MZW = (SW + EZW)/2

Question 155

How can we confirm that the calculated MZW is accurate with the EMZW ?

Answer 155

Confirm that the EMZW was within 1000 kg of the flight planned mid zone weight. If the EMZW is the same as the actual mid zone weight (to the nearest 1000 kg) then the estimate was accurate and the fuel flow based on that estimate is also accurate sq no further adjustments are needed.