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Question 1 of 33
1. Question
The take off climb speed of an aircraft is required to be able to climb at a gradient of at least ____% at take off power.
Correct
20.7.4 of the Civil Aviation Orders
7 TAKE-OFF CLIMB PERFORMANCE
7.1 In the take-off configuration with landing gear extended, an aeroplane must have the ability to achieve a climb gradient of 6% at take-off safety speed, without ground effect, and with all engines operating at take-off power.
Incorrect
20.7.4 of the Civil Aviation Orders
7 TAKE-OFF CLIMB PERFORMANCE
7.1 In the take-off configuration with landing gear extended, an aeroplane must have the ability to achieve a climb gradient of 6% at take-off safety speed, without ground effect, and with all engines operating at take-off power.
Question 2 of 33
2. Question
Approach speed must not be below ____ times stall speed.
Correct
Refer CAO 20.7.4 Para 10
Incorrect
Refer CAO 20.7.4 Para 10
Question 3 of 33
3. Question
In full landing configuration, an aircraft should be able to climb at 3.2% gradient in take off power, at a speed below _____ times stall speed.
Correct
20.7.4 of the Civil Aviation Orders
9 Landing climb performance
9.1 In the landing configuration with all engines operating at take-off power an aeroplane must have the ability to climb at a gradient of 3.2% in standard atmospheric conditions at a speed not exceeding 1.3VS.
Incorrect
20.7.4 of the Civil Aviation Orders
9 Landing climb performance
9.1 In the landing configuration with all engines operating at take-off power an aeroplane must have the ability to climb at a gradient of 3.2% in standard atmospheric conditions at a speed not exceeding 1.3VS.
Question 4 of 33
4. Question
If using a take off weight chart, which of the following is likely to be most true?
Correct
Incorrect
Question 5 of 33
5. Question
When using a landing performance chart, what can generally be said in regards to aircraft weight?
Correct
Incorrect
Question 6 of 33
6. Question
CAR 92 allows aircraft to land in many different locations that are not necessarily an airfield. These may include an appropriate paddock or other location such as private land.
Correct
20.7.4 of the Civil Aviation Orders
aircraft landing areameansan area of land suitable for use for the landing or take-off of aircraft.
CAR 92 Use of aerodromes
(1) A person must not land an aircraft on, or engage in conduct that
causes an aircraft to take off from, a place that does not satisfy
one or more of the following requirements:
(a) the place is an aerodrome established under the Air
Navigation Regulations;
(b) the use of the place as an aerodrome is authorised by a
certificate granted, or registration, under Part 139 of
CASR;
(c) the place is an aerodrome for which an arrangement under
section 20 of the Act is in force and the use of the
aerodrome by aircraft engaged in civil air navigation is
authorised by CASA under that section;
(d) the place (not being a place referred to in paragraph (a),
(b) or (c)) is suitable for use as an aerodrome for the
purposes of the landing and taking-off of aircraft;
and, having regard to all the circumstances of the proposed
landing or take-off (including the prevailing weather
conditions), the aircraft can land at, or take-off from, the place
in safety.
Incorrect
20.7.4 of the Civil Aviation Orders
aircraft landing areameansan area of land suitable for use for the landing or take-off of aircraft.
CAR 92 Use of aerodromes
(1) A person must not land an aircraft on, or engage in conduct that
causes an aircraft to take off from, a place that does not satisfy
one or more of the following requirements:
(a) the place is an aerodrome established under the Air
Navigation Regulations;
(b) the use of the place as an aerodrome is authorised by a
certificate granted, or registration, under Part 139 of
CASR;
(c) the place is an aerodrome for which an arrangement under
section 20 of the Act is in force and the use of the
aerodrome by aircraft engaged in civil air navigation is
authorised by CASA under that section;
(d) the place (not being a place referred to in paragraph (a),
(b) or (c)) is suitable for use as an aerodrome for the
purposes of the landing and taking-off of aircraft;
and, having regard to all the circumstances of the proposed
landing or take-off (including the prevailing weather
conditions), the aircraft can land at, or take-off from, the place
in safety.
Question 7 of 33
7. Question
The _____ is the length of runway declared available and suitable for the ground run of a landing aircraft.
Correct
20.7.4 of the Civil Aviation Orders
landing distance available means the distance specified by CASA as being the effective operational length available for use by aircraft for landing at certified or registered aerodromes or the distance available for landing on an aircraft landing area.
Incorrect
20.7.4 of the Civil Aviation Orders
landing distance available means the distance specified by CASA as being the effective operational length available for use by aircraft for landing at certified or registered aerodromes or the distance available for landing on an aircraft landing area.
Question 8 of 33
8. Question
The length of runway declared and available for the ground run of an aircraft is known as:
Correct
Refer CAO 20.7.1B – Definitions
Incorrect
Refer CAO 20.7.1B – Definitions
Question 9 of 33
9. Question
The length of runway declared and available for the ground run of an aircraft plus any clearway is known as:
Correct
take-off distance available means the distance specified by CASA as being the effective operational length available for use by aircraft for take‑off at certified or registered aerodromes or the distance available for take-off on an aircraft landing area. This includes the clearway.
Incorrect
take-off distance available means the distance specified by CASA as being the effective operational length available for use by aircraft for take‑off at certified or registered aerodromes or the distance available for take-off on an aircraft landing area. This includes the clearway.
Question 10 of 33
10. Question
The length of runway declared and available for the ground run of an aircraft, plus any clearway plus ______ ft is known as TO*A.
Note * is used so that it does not answer another question. The important aspect here is that you know how many feet are included.
Correct
Refer CAO 20.7.4 Para 6
Incorrect
Refer CAO 20.7.4 Para 6
Question 11 of 33
11. Question
Take-off safety speed shall not be less than ____ times the stall speed.
Correct
take-off safety speed means the speed specified on the aeroplane take-off chart being the minimum speed to which an aeroplane must be accelerated in establishing the take-off distance required. This is 1.2 for most GA aircraft. Check the flight manual.
Incorrect
take-off safety speed means the speed specified on the aeroplane take-off chart being the minimum speed to which an aeroplane must be accelerated in establishing the take-off distance required. This is 1.2 for most GA aircraft. Check the flight manual.
Question 12 of 33
12. Question
Find the pressure height for QNH 1003 hPa, elevation 4000 ft.
Correct
PH = elev + (1013 – QNH) x 30 = 4000 + (1013 – 1003) x 30 = 4000 + 300 = 4300 ft
Incorrect
PH = elev + (1013 – QNH) x 30 = 4000 + (1013 – 1003) x 30 = 4000 + 300 = 4300 ft
Question 13 of 33
13. Question
Find the pressure height for QNH 1003 hPa, elevation 3000 ft.
Correct
PH = elev + (1013 – QNH) x 30 = 3000 + (1013 – 1003) x 30 = 3000 + 300 = 3300 ft
Incorrect
PH = elev + (1013 – QNH) x 30 = 3000 + (1013 – 1003) x 30 = 3000 + 300 = 3300 ft
Question 14 of 33
14. Question
If the pressure height for QNH 1003 hPa, elevation 3000 ft = 3300 ft,
and the pressure height for QNH 1003 hPa, elevation 4000 ft = 4300 ft.
Then, the pressure height for QNH 1003 hPa, elevation 2000 ft must be 2300 feet!
So what then is the simple rule of thumb formula for pressure height?
Correct
Incorrect
Question 15 of 33
15. Question
If the pressure height for QNH 1003 hPa, elevation 3000 ft = 3300 ft,
Without using a calculator find pressure height for QNH 993 hPa, elevation 4000 ft.
Correct
If the PH is changed by 300 ft for 10 hpa, then 20 hpa will simply double that change. PH = 4000 + (1013 – 993) x 30 = 4600 ft
Incorrect
If the PH is changed by 300 ft for 10 hpa, then 20 hpa will simply double that change. PH = 4000 + (1013 – 993) x 30 = 4600 ft
Question 16 of 33
16. Question
Which of the following is most true?
Correct
Density height adds the variation due to pressure (pressure height) and the variations due to temperature.
Incorrect
Density height adds the variation due to pressure (pressure height) and the variations due to temperature.
Question 17 of 33
17. Question
Within normal variations, does temperature or atmospheric pressure have more influence on density altitude?
Correct
If we assume that the temp varies by about ± 10 °C and the QNH varies ± 10 hPa
A QNH variation of 10 hPa makes a 300 ft variation to pressure height
A temp variation of 10° C from ISA temp = 1200 feet variation in density height
Incorrect
If we assume that the temp varies by about ± 10 °C and the QNH varies ± 10 hPa
A QNH variation of 10 hPa makes a 300 ft variation to pressure height
A temp variation of 10° C from ISA temp = 1200 feet variation in density height
Question 18 of 33
18. Question
This is a bizarre question, but it will make you learn something if you have not already.
We don’t know the QNH, but we know the ISA deviation is 10° C warmer than standard at our takeoff aerodrome.
The variation in density altitude due to ISA deviation will be equal to:
Correct
Each 1 degree of variation from ISA affects density altitude by 120 ft. If the temperature is 10 degrees warmer than standard, then the density altitude will be (+10 x 120 = 1200 ft) higher.
Incorrect
Each 1 degree of variation from ISA affects density altitude by 120 ft. If the temperature is 10 degrees warmer than standard, then the density altitude will be (+10 x 120 = 1200 ft) higher.
Question 19 of 33
19. Question
The takeoff is completed when:
Correct
Incorrect
Question 20 of 33
20. Question
VTOSS is published in the aircraft’s take-off chart and is usually not less than:
Correct
Incorrect
Question 21 of 33
21. Question
When an aircraft finishes its take-off run, the next step is to:
Correct
The primary objective as soon as the take-off run finishes is to accelerate and attain safe airspeed and attain VTOSS by 50 feet. This is called the acceleration segment.
Incorrect
The primary objective as soon as the take-off run finishes is to accelerate and attain safe airspeed and attain VTOSS by 50 feet. This is called the acceleration segment.
Question 22 of 33
22. Question
The length of runway declared suitable for the ground run part of a take-off is the:
Correct
ERSA INTRO Approx Pg 21
Incorrect
ERSA INTRO Approx Pg 21
Question 23 of 33
23. Question
The length of surface suitable for wheels to run on during the ground roll is:
Correct
Refer CAO 20.7.1B Definitions
Incorrect
Refer CAO 20.7.1B Definitions
Question 24 of 33
24. Question
The length of runway available and any clearway available is known as:
Correct
Incorrect
Question 25 of 33
25. Question
TODA allows for:
Correct
CAO 20.7.4.6 Take off distance required is the distance to accelerate from a standing start with all engines operating and to achieve take off safety speed at a height of 50 ft above take-off surface.
Incorrect
CAO 20.7.4.6 Take off distance required is the distance to accelerate from a standing start with all engines operating and to achieve take off safety speed at a height of 50 ft above take-off surface.
Question 26 of 33
26. Question
What is a declared clearway?
Correct
Refer CAO 20.7.1B Definitions
Incorrect
Refer CAO 20.7.1B Definitions
Question 27 of 33
27. Question
The clearway:
Correct
Incorrect
Question 28 of 33
28. Question
The above extract is from ERSA RDS. What is the declared clearway length for YSCB?
Correct
Clearway = TODA – TORA
= 1739 – 1679
= 60 m
Incorrect
Clearway = TODA – TORA
= 1739 – 1679
= 60 m
Question 29 of 33
29. Question
The ERSA RDS declares and LDA of 1679 m for YSCB. This distance:
Correct
Incorrect
Question 30 of 33
30. Question
The ASDA includes the length for takeoff run:
Correct
Incorrect
Question 31 of 33
31. Question
You are calculating the TODR for an aircraft weighing 1900 kg being used for Aerial work. You calculate the distance to accelerate from a standing start with all engines operating and to achieve take-off safety speed at a height of 50 feet above the take-off surface to be 1000 m. What is the TODR you must select at a minimum to use?
Correct
20.7.4 of the Civil Aviation Orders
6 Take-off distance required
6.1 Subject to paragraph 6.3, the take-off distance required is the distance to accelerate from a standing start with all engines operating and to achieve take-off safety speed at a height of 50 feet above the take-off surface, multiplied by the following factors:
(a) 1.15 for aeroplanes with maximum take-off weights of 2 000 kg or less;
(b) 1.25 for aeroplanes with maximum take-off weights of 3 500 kg or greater; or
(c) for aeroplanes with maximum take-off weights between 2 000 kg and 3 500 kg, a factor derived by linear interpolation between 1.15 and 1.25 according to the maximum take-off weight of the aeroplane.
Incorrect
20.7.4 of the Civil Aviation Orders
6 Take-off distance required
6.1 Subject to paragraph 6.3, the take-off distance required is the distance to accelerate from a standing start with all engines operating and to achieve take-off safety speed at a height of 50 feet above the take-off surface, multiplied by the following factors:
(a) 1.15 for aeroplanes with maximum take-off weights of 2 000 kg or less;
(b) 1.25 for aeroplanes with maximum take-off weights of 3 500 kg or greater; or
(c) for aeroplanes with maximum take-off weights between 2 000 kg and 3 500 kg, a factor derived by linear interpolation between 1.15 and 1.25 according to the maximum take-off weight of the aeroplane.
Question 32 of 33
32. Question
Given a TAS of 150 KT and a cross wind component of 35 knots, find effective TAS (ETAS).
Correct
CASA could give you this type of question to ensure you know high drift angles effect you ETAS.
You would only need to calculate ETAS for strong crosswinds components relative to TAS, ie. high drift angles about 10 degrees and higher.
Eg
Given a PNR with a 150 kt TAS and 35 kt crosswind component find pnr etc.
You can use your E6B or CR to get drift
But let’s approximate with
CWC x 60 / TAS = WCA
35 x 60 / 150 =14° drift
ETAS = 150 x cos (14°) = 145.5 kt
A 4.5 kt or 3% variation between TAS and ETAS
To highlight the importance in really strong cross winds imagine this:
Given a PNR with a 150 kt TAS and 100 kt cross wind component find pnr etc.
CWC x 60 / TAS = WCA
100 x 60 / 150 =40° drift
ETAS = 150 x cos (40°) = 100 kt
And in slight cross winds
Given a PNR with a 150 kt TAS and 3 kt cross wind component find pnr etc.
CWC x 60 / TAS = WCA
3 x 60 / 150 = 1° drift
ETAS = 150 x cos (1°) = 149.97 kt
The magic number seems to be anything above about a 10 degree drift, which is anything with a cross wind above about 16% of TAS
Given a PNR with a 150 kt TAS and 16% cross wind component or 25 kt crosswind component find pnr etc.
CWC x 60 / TAS = WCA
25 x 60 / 150 =10° drift
ETAS = 150 x cos (10°) = 147.7 kt
Notice how at 10% , now we are starting to effect TAS to the point where our calcs could be out.
Where as with a 15 kt cross wind which is 10% of TAS 150
It doesn’t seem viable to need to calculate ETAS.
CWC x 60 / TAS = WCA
15 x 60 / 150 =6° drift
ETAS = 150 x cos (6°) = 149.1 kt
For those using CR model flight computers , this is the importance of the little black section to the left of the triangle on the wind side.
On the CR models, we use this whenever drift is above 10°
Incorrect
CASA could give you this type of question to ensure you know high drift angles effect you ETAS.
You would only need to calculate ETAS for strong crosswinds components relative to TAS, ie. high drift angles about 10 degrees and higher.
Eg
Given a PNR with a 150 kt TAS and 35 kt crosswind component find pnr etc.
You can use your E6B or CR to get drift
But let’s approximate with
CWC x 60 / TAS = WCA
35 x 60 / 150 =14° drift
ETAS = 150 x cos (14°) = 145.5 kt
A 4.5 kt or 3% variation between TAS and ETAS
To highlight the importance in really strong cross winds imagine this:
Given a PNR with a 150 kt TAS and 100 kt cross wind component find pnr etc.
CWC x 60 / TAS = WCA
100 x 60 / 150 =40° drift
ETAS = 150 x cos (40°) = 100 kt
And in slight cross winds
Given a PNR with a 150 kt TAS and 3 kt cross wind component find pnr etc.
CWC x 60 / TAS = WCA
3 x 60 / 150 = 1° drift
ETAS = 150 x cos (1°) = 149.97 kt
The magic number seems to be anything above about a 10 degree drift, which is anything with a cross wind above about 16% of TAS
Given a PNR with a 150 kt TAS and 16% cross wind component or 25 kt crosswind component find pnr etc.
CWC x 60 / TAS = WCA
25 x 60 / 150 =10° drift
ETAS = 150 x cos (10°) = 147.7 kt
Notice how at 10% , now we are starting to effect TAS to the point where our calcs could be out.
Where as with a 15 kt cross wind which is 10% of TAS 150
It doesn’t seem viable to need to calculate ETAS.
CWC x 60 / TAS = WCA
15 x 60 / 150 =6° drift
ETAS = 150 x cos (6°) = 149.1 kt
For those using CR model flight computers , this is the importance of the little black section to the left of the triangle on the wind side.
On the CR models, we use this whenever drift is above 10°
Question 33 of 33
33. Question
Refer to ERSA for answer.
Wind angle to runway is 30° at 25 KT. What is the cross wind component?
Type answer as a 2 digit number only.