Flight Planning Flashcards
Ram Air Temperature (RAT) is defined as ?
defined as the sum of the OAT (ambient Outside Air Temperature),
plus some of the ram temperature rise.
Whats a Temperature probe sensor recovery factor ?
Represents the amount of ram temperature rise indicated by the probe.
OAT temp = -50c sensory recovery factor of 0.8, Ram temp rise at a given Mach number is 30°C calculate RAT ?
OAT of -50°C + 0.8 (80%) of 30°C (24°C) = -26°C RAT
Total Air Temperature (TAT) ?
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.
Calculate TAT if OAT is -50c and ram temp rise is 30c ?
OAT of -50°C + the full 30°C temperature rise = -20°C TAT
Standard temperature at FL110 (11000 ft) is ?
15°C at MSL - [11 x1.98°C] = -6.78°C)
the pressure rise also
causes the airspeed indicator (ASI) to ?
Over read at high speeds
EAS is simply ?
IAS corrected for compressibility error
In the B727 the corrections for instrument and position error are
less than 1 kt so for practical purposes,
in this aircraft IAS = ?
CAS
find EAS for an aircraft flying at 300 KCAS at FL300 ?
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.
For practical purposes, the Indicated Mach Number is the ?
True Mach Number TMN
The Formula used to compute the speed of sound in knots is ?
Mach 1.0 (in kt) = Square root of OAT in degrees K x 39 -1
To then compute the TAS for any Mach Number other than 1?
TAS = Square root of OAT in degrees K x 39 x M -1 (where M = Mach Number)
To convert the Celsius temperature to Kelvin, simply add ?
simply add 273 to the Celsius
temperature.
find the speed of sound at MSL in ISA: (15°C) ?
M1.0 = (15C + 273 K) = V288 K x 39 -1 = 661 kt
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 = ?
V288 K x 39 x .8 -1 = 528 kt
Find the speed of sound at the tropopause in ISA: (-56°C) ?
M1.0 = (-56C + 273 K) = V217 K x 39 -1 = 574 kt
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 = ?
V217 K x 39 x .8 -1 = 459 kt
find the TAS at 0.80M at FL310 at a temperature of ISA+10 ?
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
find the Mach Number at a TAS of 455 kt with an OAT of -50°C ?
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).
find the OAT at a TAS of 420 kt and a Mach Number of 0.70M ?
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)
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) ?
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
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 ?
300s i.e FL300-FL350
Will weather, winds or trip length effect what cruising level you pick in the exam ?
No
How does thrust limits effect cruising levels ?
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.
What is optimal flight level ?
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.
How does buffet limits effect flight level ?
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.
To ensure that the aircraft does not exceed its buffet
boundaries, draw ?
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.
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?
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.
The optimum FL is not necessarily the same as the maximum FL why ?
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.
What is the EMZW ?
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).
The optimum level for a flight at a particular Mach Number is ?
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).
EMZW = ?
EMZW = BRW- [(distance x 10 kg/nm)/2 + 1600 kg].
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 ?
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.
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
O may rise by up to 30°C or more above OAT due to compression
350 CAS at 20,000 ft PA. equals what EAS (CAS)?
338 TAS
340 TAS
343 TAS
347 TAS
340 TAS refer to table (F correction Factors for TAS ) page 1-6 in the 727 Handbook
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
Set altimeter to 1013.2 hPa
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
OAT of -40°C + (0.8 x 30°C) = -40°C + 24°C = -16°C RAT
At CAS 160KT and P.A. 35,000 what is the Mach Number?
0.49
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
a speed ratio of how fast the aircraft is going compared to the speed of sound in the current ambient conditions
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
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 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
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.
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
the pressure rise results in an over-read error in AS from the total pressure rise
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
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
CAS
Using your maps only. Calculate the direct distance Darwin NT (YPDN) to Cairns QLD (YBCS)?
875 NM
906 NM
990 NM
830 NM
906 NM
What is the standard temp at FL450?
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)
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
0.82
the cockpit TAT indicates -12°C at FL330 at 0.82M; The OAT is ?
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
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
the current aircraft CAS and Flight Level / Pressure Altitude
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
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.
Using your maps only. Calculate the direct distance Melbourne VIC (YMML) to RAF Curtin WA (YCIN)?
1648 NM
1520 NM
1725 NM
1864 NM
1648 NM
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
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.
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
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.
What is TAS at M0.78 at 32,000 feet in knots?
© 455 kts
© 464 kts
© 471 kts
© 450 kts
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
If only considering fuel economy, the most efficient cruise method is ?
LRC
cruise climb
step climb cruise
constant mach cruise
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.
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?
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
What is TAS at M0.8 at 25,000 feet in knots?
0 472 kts
O 480 kts
O 468 kts
0 464 kts
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
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?
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
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
-46°C
For climb calculation winds from RSWT will most commonly be used from ?
O FL240
© FL340
O FL180
O FL300
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.
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
500 hPa, 400 hPa, 300 hPa, 250 hPa, 200 hPa, 150 hPa
What is the wind direction from YPAD to YPOD at FL340?
310°
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
480kts
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
-350
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
YMML/YMIDG FL340
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
- 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.
- 65,500 kg round up to 70,000
- Enter correct descent profile table on page 4 - 3 (standard is .80M/280/250 KIAS) and go down to FL for TOD
- Not advised otherwise, so use .80M/280/250 KIAS
- 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).
- Use FL180 ISA ddssstt 22 037 13, ISA-22, OAT -13, ISA+9, Wind 220/37KT
- Refer to descent profile for distance and time, so you can calculate TAS
- FL350 distance = 119 nam
- Time = 23 minutes
- TAS = 119 / 23 x 60 = 310.43 kt =310 kt
- Use formula GNM = GS x time / 60
- GS = 310 + -37kt HWC = 273kt
- GNM = 273 x 23 / 60 = 104.65 GNM = 105 GNM
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
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.
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
FL330
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
74,300 kg TOCGW
- BRW to nearest 2000 kg = 78,000 (as per note 11 on page 2-2 B727 handbook)
- 2/3rds of climb height is 31,000ftx 0.666666667 = 20,666 ft
- So FL180 is the most similar level
- 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
- Interpolate between 76,000 kg and 80,000 kg to get 78,000 kg data
- Subtract climb fuel to get TOCGW = 77,400 kg BRW - 3100 kg = 74,300 kg
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
74,000 kg TOCGW and 175 nm
- BRW to nearest 2000 kg = 78,000 (as per note 11 on page 2-2 B727 handbook)
- 2/3rds of climb height is 33,000 ft x 0.66666667 = 21,999 ft
- So FL240 is the most similar level
- 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
- Interpolate between 76,000 kg and 80,000 kg to get 78,000 kg data
27.5/3400/182 - 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
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
74,425 kg TOCGW
- BRW to nearest 2000 kg = 78,000 (as per note 11 on page 2-2 B727 handbook)
- 2/3rds of climb height is 31,000 x .666666667 = 20,666 ft
- So FL180 is the most similar level
- 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
- 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
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
97 GNM
- 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. - 65,000 kg round up to 70,000
- Enter correct tiescent profile table on page 4 - 3 (standard is.80M/280/250 KIAS) and go down to FL for TOD
- Not advised otherwise, so use .80M/280/250 KIAS
- 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).
- Use FL180 ISA ddssstt 22 037 13, ISA-22, OAT -13, ISA+9, Wind 220/37KT
- Refer to descent profile for distance and time, so you can calculate TAS (interpolate if necessary)
- FL320 distance = interpolate (114 nam - 107 nam) / 2 + 107 nam = 110.5 nam
- Time (interpolate) = 21.5 minutes
- TAS = 110.5 / 21.5 x 60 = 308.37 kt =308 kt
- Use formula GNM = GS x time / 60
- GS = 308 + -37kt HWC = 271 kt
- GNM = 271 x 21.5 / 60 = 97.108 GNM = 97 GNM
The upper level winds are ?
northerly
southerlv
easterly
westerly
example: 2309064, where 23 = 230°, 090 = 90 knots, 64 = -64°C
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
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
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
9.5
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
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
For a TAS of 420 KT find ETAS with 20° drift?
394-396
For a TAS of 380 KT find ETAS with 13° drift?
369-371
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
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
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
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
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
82,900 kg
Climb estimate for 150 nm is BRW 86,000 kg - 3100 kg = 82,900 kg
Correct Answer: 82900
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
O FL330, TODGW 75,966 kg, TOD 114 nm from YMHB
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
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
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°
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
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°
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°€
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°
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
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
75,023 kg
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
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
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°
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
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°
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
