Navigation ATPL Flashcards
Side bands (USB or LSB) are often used with AM radio propagation. An advantage of utilising sideband transmissions is ?
A external noise and interference are reduced
B the carrier does not have to be so powerful
C different information can be transmitted simultaneously on both bands
D all of the above
all of the above
The VOR operates on the principle of ?
frequency differential
amplitude modulation
phase differential
frequency modulation
phase differential
A common error to both the ADF/NDB and VOR navigation systems is ?
vertical polarisation
night effect
coastal refraction
terrain effect
terrain effect
VOR radials are calibrated for the magnetic variation that exists ?
at the station installation
at the aircraft
midway between the station and the aircraft
at the nearer pole
at the station installation
An error or limitation commonly associated with the VOR system is ?
co-channel interference
coastal refraction
night effect
site error
site error
At ranges greater than line of sight, the type of radio wave propagation on which the ADF/NDB
system relies to provide accurate information is ?
the direct wave
the sky wave
the ground wave
a combination of ground and sky wave
the ground wave
Bending and scalloping are terms used when describing errors in ?
VOR
GPS
ADF
DME
VOR
When you select a DME frequency you are ?
selecting a pulsed paired frequency
selecting a VHF phase differential between transmitter and receiver
selecting a VLF wave guide
selecting a VHF primary radar frequency
selecting a pulsed paired frequency
The ADF Sense antenna ?
resolves the 180 degree ambiguity
fine tunes the NDB frequency
identifies the maximum signal position
identifies the null position
resolves the 180 degree ambiguity
An advantage of the secondary radar system over the primary radar system is that the return signal ?
is reflected more strongly by the target
allows contouring of weather returns
is not subject to attenuation
may contain coded information
may contain coded information
You are at FL270 and 30 DME, tracking to a destination along a track which is aligned with an NDB,
LLZ (not approved for extended range fixing) and a VOR radial from the destination. Your aircraft
has both VHF, NAV and ADF equipment. The aid you should use for tracking is ?
the LOC
the VOR
the NDB
the LOC, if you intend to carry out an ILS or LOC approach, otherwise the VOR
The VOR
Refer AIP ENR 1.1 - 27 para 17.4.3 and GEN 1.5 - 5 para 2.2c. The most precise aid is the localiser
but at 30 DME the aircraft is outside the standard 25 nm rated coverage. The AIP ranges beyond 25
nm are for installations that have been nominated for position fixing at ranges beyond 25 nm (para
2.2 GEN 1.5-5)
The theoretical range of a VOR at a receiver height of 12000 ft is ?
134 nm
110 nm
100 nm
150 nm
134 nm
An advantage of the secondary radar system over the primary radar system is that the return signal ?
Is reflected more strongly by the target
allows contouring of weather returns
is not subject to attenuation
eliminates unwanted returns (clutter)
eliminates unwanted returns (clutter)
At ranges up to line of sight, the type of radio wave propagation on which the ADF/NDB system
relies to provide accurate information is ?
the direct wave
the sky wave
the ground wave
a combination of ground and sky wave
The direct wave
The error or limitation commonly associated with the NDB system is ?
vertical polarisation
ground station error
scalloping and bending
terrain effect
Terrain effect
An aircraft is being navigated solely by reference to radio navigation aids. The maximum permissible
time interval between positive radio fixes is ?
one hour
two hours
three hours
a variable time calculated according to expected tracking error
two hours
Refer AIP ENR 1.1 - 25 para 19.1.1 b.
An aircraft is operating under the IFR. It is exempted from the requirement to obtain fixes at the specified intervals when ?
equipped with an approved area navigation system that meets performance
requirements of the intended airspace or route
it carries a three member crew
GPS equipped
using VFR procedures
equipped with an approved area navigation system that meets performance
requirements of the intended airspace or route
Refer AIP ENR 1.1 - 25 para 19.1.1 a and b.
The ADF Loop antenna ?
resolves the 180 degree ambiguity
fine tunes the NDB frequency
identifies the maximum signal position
identifies the null position
identifies the null position
HF radio waves primarily propagate by ?
ground waves
diffracting ground waves
direct waves
single and multi-hop sky waves
single and multi-hop sky waves
Your VOR OBS is set to a required track of 090, and you have a from flag displayed. The CDI has
moved 4 degrees to the right. DME distance from the station is 45 nm. The distance off the
required track you are is ?
2 nm
4 nm
3 nm
6 nm
Using the 1 in 60 rule:
Distance off track = (distance travelled x angle off track)/60
Distance off track = distance travelled x angle off track/ 60 =( 45 nm x 4°)/60 = 3 nm 4° off track at 45 nm is 3 nm off track.
The centre line of the localiser of an ILS is determined by ?
phase comparison of the signal modulation radiated from each side of the centreline
a locus of points of equal signal modulation radiated from each side of the centreline
frequency comparison of signal modulation radiated from each side of the centreline
the doppler shift of the signal modulation radiated from each side of the centreline
a locus of points of equal signal modulation radiated from each side of the centreline
In an ILS display, full deflection of the glide path needle occurs when the aircraft is?
- 5 degrees above or below the glide path
- 7 degrees above or below the glide path
- 0 degrees above or below the glide path
- 4 degrees above or below the glide path
0.7 degrees above or below the glide path
As an aircraft approaches an enroute DME station when overflying, the DME distance readout will ?
reduce at an increasing rate to zero overhead, then increase at an increasing rate
back to the correct value after station passage
reduce at an decreasing rate to read height in nm overhead, then increase at an
increasing rate after station passage
reduce at an increasing rate to read height in nm overhead, then increase at an
decreasing rate after station passage
reduce at an increasing rate to zero overhead, then increase at a decreasing rate
back to the correct value after station passage
reduce at an decreasing rate to read altitude in nm overhead, then increase at an increasing rate after station passage.
Approaching a DME station at altitude, the distance readout reduces at a reducing rate due to slant
error. Past the station, the distance readout increases at an increasing rate as slant error reduces.
The wavelength of a radio signal is defined as ?
the number of wavelengths passing a given point in one second
the length of a radio wave from any point to the same point on the next cycle
the number of wavelengths passing a given point in one minute
amplitude divided by frequency
the length of a radio wave from any point to the same point on the next cycle
The microwave landing system (MLS) has the advantage of ?
curved approaches
multiple runway coverage
greater traffic capacity
GPS coupling
curved approaches
The nearest to the maximum altitude a continuous wave RADALT will usually indicate is ?
20000 ft
10000 ft
5000 ft
2500 ft
2500 ft
Advantages of the gyrostabilised remote indicating compass over the direct reading magnetic
compass are ?
no errors, output can be connected to other flight instruments
low power operation, simple and economical
reliable, reduced errors and not reliant on external power supply
reduced turning and acceleration errors, reduced deviation and slaveable output
reduced turning and acceleration errors, reduced deviation and slaveable output
Advantages of the direct reading magnetic compass over the gyrostabilised remote indicating
compass are ?
minimal errors, effective anywhere on the globe
not reliant on external power, simple construction and economical
output can be slaved to other instruments, stable readout in turbulence
unaffected by latitude, inexpensive and lightweight
not reliant on external power, simple construction and economical
Aircraft weather radar suffers from attenuation. The term ‘attenuation’ means ?
mixing of ground clutter with cloud returns
weakening of the radar transmissions and returns
unwanted returns from side lobes
distortion of the beam width and pulse length
weakening of the radar transmissions and returns
The principal advantage of an IVSI over a conventional VSI is that it ?
is more accurate during a steady climb or descent
does not rely on air pressure from the static source
utilises an electrical servo to assist indication of climb and descent
utilises an inertial system to assist indication of climb and descent
utilises an inertial system to assist indication of climb and descent
One advantage of the gyrostabilised remote indicating compass over the direct reading compass is ?
it is not affected by magnetic variation
turning and acceleration errors are minimised
it is not affected by magnetic influences in the aircraft
the complete system can be mounted behind the instrument panel
turning and acceleration errors are minimised
A Flux Valve for a RMI is usually located ?
in a remote location such as the wing tip or tail section of an aircraft
behind the instrument panel
in the overhead console
in the nose cone
in a remote location such as the wing tip or tail section of an aircraft
Calibrated Airspeed (CAS) is ?
IS corrected for pressure, instrument and compressibility errors
IAS corrected for pressure and instrument errors, for practical purposes IAS
approximates CAS
EAS corrected for density error
TAS corrected for compressibility errors
IAS corrected for pressure and instrument errors, for practical purposes IAS
approximates CAS
With regard to RMI, constant oscillation of the annunciator indicates ?
the gyro cannot be aligned and the system has failed
the gyro is being realigned, so the system can’t be used until realignment is
complete
the flux valve is unable to correctly sense the earth’s magnetic field
the system is operating normally and the gyro is ‘hunting’ about the mean position
the system is operating normally and the gyro is ‘hunting’ about the mean position
A blocked static vent on descent will cause the ASI to ?
overread
underread
read zero
read correctly
overread
PUDSUC.
The worst error in a gyro stabilised magnetic compass is ?
deviation
transport error
turning and acceleration
earth rate wander
Deviation
When using a weather radar, the most turbulent areas are identified by ?
the centre of the avoid area
the closest contours, nearest the avoid zone
the position where the contours are at the greatest spacing
the outer edge of the avoid area
the closest contours, nearest the avoid zone
During a descent the pitot tube of an ASI becomes blocked. As a result the ASI will ?
read zero
underread
continue to read the same airspeed as before the blockage occurred
overread
Under read
The ability of airborne weather radar to distinguish between targets at the same altitude and
bearing but at different ranges (range resolution) is determined by ?
transmitter power antenna size wave length pulse length beam width
Pulse length
You are cruising at A070 directly above a mountain of ELV 6420 ft. The QNH is 1001 HPA with an
OAT of -14 deg C. The Radio/Radar altimeter would indicate
188 ft
972 ft
390 ft
580 ft
188 ft
The altimeter is only accurate in ISA.
If conditions are hotter than ISA the altimeter will underread by approximately 4% of the height
above the QNH reference for each 10 degrees of ISA variation.
If conditions are colder than ISA the altimeter will overread by approximately 4% of the height
above the QNH reference for each 10 degrees of ISA variation.
With -14°C at 7000 ft (7360 ft PA), the variation is ISA-14. This will cause the altimeter to overread
by a little under 6% of the height above the QNH reference. With area QNH, the error will be a little
less than 6% of 7000 ft. Because the temperature is colder than ISA and the altimeter is
overreading, if it indicates 7000 ft, the aircraft would actually be lower than 7000 ft.
6% of 7000 ft is 420 ft but the error will not be quite 420 ft because the ISA variation is less than
15 degrees. The true altitude of the aircraft would be approximately 6580 ft. The mountain
elevation is 6420 ft so a radio altimeter would register the difference between the true altitude of
6580 ft and the mountain elevation of 6420 ft = 160 ft. The closest answer is 188 ft.
To do the calculation on the CR3, set the OAT of -14°C against the altitude of 7000 ft in the true air temperature window (technically the pressure height should be used but the scale is so small that the difference is negligible). Then on the main scale with an indicted height of 7000 ft on the inside, read 6590 ft true altitude on the outside (410 ft less) or 170 ft above the mountain
Two aircraft are operating in the same area. One aircraft is at 7000 ft with 1023 HPA set and the
other is at 8000 ft with 1013 HPA set. The vertical separation of the two aircraft is ?
1000 ft
700 ft
970 ft
1300 ft
1300ft
Separation is worked out based on a common QNH setting. One altimeter needs to be reset to the
subscale setting of the other.
It makes no difference which is adjusted to the other.
Take the 7000 ft reading on a QNH of 1023 down to 1013. If the subscale value decreases, the
indication on the hands of the altimeter will also decrease. In this case the subscale of 1023 could
be wound down to 1013. A reduction of 10 HPA
The accepted relationship between pressure and altitude in the lower levels is and increase of 30
feet of altitude for each Hectopascal reduction in pressure and vice versa. Winding the altimeter
subscale down by 10 HPA will cause the altimeter indication to decrease by 300 ft (10 x 30 ft). The
reading would reduce from 7000 ft to 6700 ft. Compared to 8000 ft indicating in the other aircraft,
this is a difference or separation of 1300 ft.
You are cruising at A070 with the subscale of your aircraft’s altimeter set at 1004 HPA. Another
aircraft is flying in the area with a subscale setting on its altimeter of 1009 HPA. If the other aircraft is flying at A045, your vertical separation if your tracks crossed would be ?
3500 ft
2500 ft
2350 ft
2650 ft
2650 ft
Adjust to a common subscale setting. Wind 7000 ft on a QNH of 1004 up 5 HPA (150 ft) to 7150 ft
on a QNH of 1009 HPA.
Separation= 7150 ft on 1009 HPA-
4500 ft on 1009 HPA
2650 ft
With regard to continuous wave and pulsed wave altimeters ?
Radar Altimeters (pulsed) are used for low level indications, while Radio Altimeters
(continuous wave) are used for high level indications
Radar Altimeters (pulsed) are used for high level indications, while Radio Altimeters
(continuous wave) are used for low level indications
both types have the same useful range of operation
it is the type of terrain (eg. water, sand or snow) that determines the range of the
system, not the method used to transmit the signal
Radar Altimeters (pulsed) are used for high level indications, while Radio Altimeters (continuous wave) are used for low level indications
An aircraft is descending on the glide path of an ILS. The QNH is 1013 HPA and the temp is ISA-15
at all levels.
Assuming that the aircraft is exactly on the glide path, that the equipment is accurate and that the
QNH is correctly set, the altimeter indication in relation to the glide path would be ?
above the glide path
on the glide path
below the glide path
unable to determine without a radar altimeter
Above the glide path
The altimeter is only accurate in ISA. Since it is ISA-15 the altimeter will overread. If the aircraft is
actually on the glidepath then the altimeter would indicate above the path because it overreads.
Aircraft weather radar suffers from attenuation. The term ‘attenuation’ means ?
mixing of ground clutter with cloud returns
weakening of the radar transmissions and returns
unwanted returns from side lobes
distortion of the beam width and pulse length
weakening of the radar transmissions and returns
On the ALICE SPRINGS ILS, the Glide Path intersects the Outer Marker at an altitude of 3120 ft. An
aircraft is making an ILS approach. The local AS QNH of 1028 HPA is set correctly. The surface
temperature is +3°C and the elevation of the airport is 1789 ft.
If this aircraft crosses the Outer Marker with its altimeter indicating the height shown on the
instrument approach, its position in relation to the Glidepath is ?
On the Glide Path
25 ft above the Glide Path
50 ft below the Glide Path
25 ft below the Glide Path
50ft below the Glide path.
The altimeter is only accurate in ISA.
If conditions are hotter than ISA the altimeter will underread by approximately 4% of the height
above the QNH reference for each 10 degrees of ISA variation.
If conditions are colder than ISA the altimeter will overread by approximately 4% of the height
above the QNH reference for each 10 degrees of ISA variation.
With +3°C on the ground at AS (pressure height 1339 ft) , the variation is ISA-9. This will cause the
altimeter to overread by a little under 4% of the height above the QNH reference. With the local AS
QNH, the error will be a little less than 4% of the height of the glidepath above Alice Springs - 1339
ft. Because the temperature is colder than ISA and the altimeter is overreading, if it indicates the
height shown on the approach, the aircraft would actually be below the glidepath.
4% of 1339 ft is 54 ft however the error will not be quite 54 ft because the ISA variation is less
than 10 degrees. The closest answer is 50 ft.
To do the calculation on the CR3, set the OAT of +3°C against the Alice Springs elevation of 1789 ft
in the true air temperature window (technically the Alice Springs pressure height should be used but
the scale is so small that the difference is negligible). Then on the main scale with an indicted
height of 1339 ft AGL on the inside, read 1285 ft true height on the outside (54 feet less)
An aircraft is overflying an aerodrome of elevation 3000 ft. The aircraft altitude is 5000 ft on an
Area QNH of 1013 HPA. OAT at 5000 ft is +25°C.
The true height above the aerodrome is ?
2000 ft
2375 ft
1650 f
5375 ft
2375ft
The altimeter is accurate in ISA. The temperature error is approximately equivalent to 4% of the
height above the QNH reference per 10 degrees of ISA variation.
If conditions are colder than ISA, the altimeter will overread. If conditions are hotter than ISA, the
altimeter will underread.
At 5000 ft, 25°C is ISA+20.
This will cause the altimeter to underread by 8% of 5000 ft (with area QNH the reference is sea
level).
8% of 5000 ft is 400 ft. The altimeter indicates 2000 ft above the aerodrome but it is underreading
by 400 ft so the true height is approximately 2400 ft.
On the CR3, set 25°C at 5000 ft in the true altitude window. On the main scale read the true
altitude of 5375 ft against 5000 ft indicated altitude on the inside. Subtract the elevation of 3000 ft
from 5400 ft true altitude to find the true height of 2375 ft.
On the Alice Springs ILS, the Glide Path intersects the Outer Marker at an altitude of 3120 ft. An
aircraft is making an ILS approach at Alice Springs (ELV of AS is 1789 ft). The local QNH of
1028 HPA is set correctly. The surface temp is +21°C and all equipment is accurate.
If this aircraft crosses the Outer Marker with its altimeter indicating the height shown on the
instrument approach, its position in relation to the GP is ?
on the Glide Path
15 ft below the Glide Path
42 ft below the Glide Path
40 ft above the Glide Path
40ft above the Glide path.
The altimeter is only accurate in ISA.
If conditions are hotter than ISA the altimeter will underread by approximately 4% of the height
above the QNH reference for each 10 degrees of ISA variation.
If conditions are colder than ISA the altimeter will overread by approximately 4% of the height
above the QNH reference for each 10 degrees of ISA variation.
With +21°C on the ground at AS (pressure height 1339 ft), the variation is ISA+9. This will cause
the altimeter to underread by a little under 4% of the height above the QNH reference. With the
local AS QNH, the error will be a little less than 4% of the height of the glidepath above Alice
Springs - 1331 ft.
Because the temperature is hotter than ISA and the altimeter is underreading, if it indicates the
height shown on the approach, the aircraft would actually be above the glidepath.
4% of 1331 ft is 54 feet but the error will not be quite 54 ft because the ISA variation is less than
10 degrees. The closest answer is 40 ft above the glidepath.
To do the calculation on the CR3, set the OAT of +21°C against the Alice Springs elevation of
1789 ft in the true air temperature window (technically the Alice Springs pressure height should be
used but the scale is so small that the difference is negligible). Then on the main scale with an
indicated height of 1331 ft AGL on the inside, read 1372 ft true height on the outside (41 feet
more).
If an aircraft heading South over continental Australia suddenly decelerates, the compass reading
will initially ?
turn East
turn West
will remain steady
oscillate then return to its original heading
will remain steady
If an aircraft heading East over continental Australia suddenly decelerates, the compass reading will
initially ?
turn North
turn South
will remain steady
oscillate then return to its original heading
turn North
In a vertical axis gyroscope at the South pole ?
drift and topple due to Earth rate rotation will be at the maximum values
drift and topple due to Earth rate rotation will be zero
drift will be zero and topple will be maximum
drift will be maximum and topple will be zero
drift and topple due to Earth rate rotation will be zero ?
In a horizontal axis gyroscope at the South Pole ?
drift and topple due to Earth rate rotation will be at the maximum values
drift and topple due to Earth rate rotation will be zero
drift will be zero and topple will be maximum
drift will be maximum and topple will be zero
drift will be maximum and topple will be zero
An aircraft is parked on the ground at a position of 30°S latitude with all electrical and avionics
systems operating. The most correct statement with regards to the error in a directional gyroscope
iS ?
it would have no error
it would have half the error as in flight
it would have the same error as in flight
it would have twice the error as in flight
it would have the same error as in flight
Being airborne or on the ground doesn’t change the apparent precession.
An aircraft with a latitude compensated directional gyro is flown to the equivalent latitude in the
opposite hemisphere. The error experienced at this latitude, compared to the original position will
be ?
the same
doubled
halved
inversely proportional to the square of the cosine of the latitude
Doubled
For accurate altitude measurements, the number of channels a GPS receiver needs is ?
1
2
3
4
4
The minimum number of satellites required to eliminate GPS receiver clock error is ?
4
3
5
6
4
The most significant GPS error is caused by ?
thunderstorms
the ionosphere
solar flares
micro meteors
the ionosphere
The minimum number of satellites needed for an accurate 3D GPS fix is ?
4
Use of the pseudo-random code allows ?
each satellite to transmit on the same frequency
each satellite to transmit on the same code
each satellite to transmit on both AM and FM
each satellite to transmit in turn
each satellite to transmit on the same frequency
In the GPS, the error called ‘geometric dilution of precision’ (GDOP or PDOP) is caused by ?
A wide angle between satellites
a small distance between satellites
ionospheric slowing
receiver aerial at the wrong angle
a small distance between satellites
GPS works on the principle of ?
doppler shift
time signals
distance measuring
time difference
time difference
The GPS pseudo-random code allows for ?
geosynchronous satellite orbits
powerful transmitters
military messages to be transmitted secretly
low power operation
low power operation
The following one which is not a characteristic of the GPS ‘P’ code is ?
it is for use only by US Department of Defence authorised users
it can be encrypted
it is on a higher frequency than the C/A code
it is theoretically more accurate than the C/A code
it is on a higher frequency than the C/A code
GPS receivers can reduce the magnitude of GDOP (PDOP) error ?
by automatic signal reprocessing
by going into memory mode until the error is resolved
by selecting the four best spaced satellites available
they cannot reduce this error
by selecting the four best spaced satellites available.
In relation to background radio noise, the GPS signal is ?
stronger
weaker
the same
on a different frequency
Weaker
Inputs required to be entered into an Inertial Reference System (IRS) by the crew preflight are ?
magnetic variation
initial position
planned groundspeed
TAS
initial position
An Inertial Reference System (IRS) will accept automatic position updates from ?
enroute NDBs at station passage
NBD bearing lines intersecting at not less than 45 degrees within 30 nm
VOR bearing lines intersecting at not less than 45 degrees within rated coverage
2 DMEs within 200 nm
2 DMEs within 200 nm
Flight Crew using Area Navigation Systems for navigation in CTA must advise ATC when times
between updates exceed - (AIP GEN 1.5 8.3.1) ?
2 hours
5 hours for a single unit or 12 hours for multiple units
3 hours for a single unit or 5 hours for multiple units
30 min
3 hours for a single unit or 5 hours for multiple units
The rhumb line track between two points, A (50°25’S 100°E) and B (34°45’S 130°E), is 080T. On a
Lamberts Conformal chart with a convergence factor (n) of 0.8, the straight line track directions at
A outbound and B inbound are ?
A outbound 092T, B inbound 068T
A outbound 068T, B inbound 092T
A outbound 104T, B inbound 056T
A outbound 081T, B inbound 079T
A outbound 092T, B inbound 068T
Calculate the longitude change:
Departure longitude= 100E
Destination longitude= 30E
Longitude change = 30°
Multiply longitude change by (n) to find bearing change Bearing change = longitude change x (n) = 30 × 0.8 =24°
The straight line tracks are the great circle tracks. The rhumb line and the great circle tracks are
equal at the half way mark so the Great Circle track at one end will be more than the rhumb line by
12 degrees (half the total bearing change) and the Great Circle track at the other end will be less
than the rhumb line by 12 degrees (half the total bearing change).
Since this track is Easterly (080T) the destination end bearing will be 12 degrees less than the
rhumb line and the departure end bearing will be 12 degrees more than the rhumb line (east is
least).
Departure end Great Circle track bearing (outbound) = rhumb line track bearing 080 + 12 = 092°
Destination end Great Circle track bearing (inbound) = rhumb line track bearing 080 - 12 = 068 degrees
The great circle track outbound from an airport at 120°E Longitude in the Northern hemisphere is
080T and the inbound great circle track is 115T at your destination airport at 170°E Longitude. The
Longitude your track will pass closest to the pole is ?
145E 1345 127E 120E 170E
134T
Find the convergence factor (n) for the trip. Convergence factor (n) = bearing change/longitude change = 35/50 = 0.7
Determine if the track will pass through a bearing of 090 degrees or due East (270 or due West
if the track was westerly).
In this case the outbound track of 080° (North East) will run through 090° (East) on the way to
the destination end of 115° (South East) so a bearing of 090° is the bearing marking the end of
the North Easterly section of the track and the start of the South Easterly section of the track
and therefore the most Northerly point of the track.
Calculate the bearing change from the departure track (080) to due East (a track of 090°). Due East as a bearing is 10 degrees more than the departure track of 080.
Calculate the longitude change required to produce the bearing change.
10 degrees of bearing change out of the total change of 35 degrees occurs at 10/35ths (28.6%) of the total change. Therefore the meridian which the track crosses at this
time will be the meridian 28.6% of the way along from departure. The total longitude change is
50° (170E - 120E) so the longitude change required to produce a 10° bearing change is
28.6% of 50 (.286 x 50 = 14°).
Another way of determining the longitude change is to divide the bearing change by the (n)
factor. In this case required bearing change to due West is 10° so at a rate of (n) of 0.7
degrees bearing change per meridian of longitude the longitude change will be 10/0.7 = 14°
Apply the longitude change for the desired bearing change to the longitude of the departure
point. For flights with decreasing longitude subtract the change. For flights with increasing
longitude add the change.
In this case the longitude is increasing so the 14° is added to the departure longitude of 120E.
Most Northerly Meridian
= Departure meridian
+ required longitude change
=120E + 14
= 134E
On a Lambert’s conformal conic projection, a straight line is drawn from 35°S 139°E to the destination at 38°S 145°E. The rhumb line track for the same trip would be ?
coincident with the great circle track
left of the straight line track
right of the straight line track
on the straight line track
left of the straight line track
On a Lambert’s conformal conic projection, a straight line is drawn from 35°S 139°E to the destination at 28°S 139°E. The rhumb line track for the same trip would be ?
equal to the great circle track only at half way
left of the straight line track
right of the straight line track
on the straight line track
on the straight line track
On a Lambert’s conformal conic projection, a straight line is drawn from 16°S 128°E to the destination at 27°S 149°E. The outbound straight line track is 116° and the inbound straight line track is 104°. The convergence factor (n) for this trip is ?
.52
.57
.636
.875
.57
The track covers 21 meridians while the bearing changes 12°. The convergence factor (n) is
calculated by dividing the bearing change by the longitude change.
n = B/L
= 12/21
= .57
On a polar stereographic chart, meridians appear as ?
straight lines
curves concave to the pole
curves concave to the equator
curves concave to the point of origin
Straight lines
On a Lambert’s conformal conical projection, a Great Circle track away from the parallel of origin appears ?
concave to the parallel of origin
concave to the nearer pole
straight
concave to the nearer standard parallel
concave to the parallel of origin
On a Mercator’s projection a straight line represents ?
a rhumb line
a great circle
an isogonal
the shortest path between two points
A rhumb line
On a Polar Stereographic projection, rhumb lines are ?
represented by straight lines near the pole, becoming concave to the pole at low latitude curved concave to the equator curved concave to the pole straight
curved concave to the pole
On a Polar Stereographic projection, great circles are ?
represented by straight lines crossing the pole, becoming concave to the pole at low latitude curved concave to the equator curved concave to the pole straight
represented by straight lines at the pole, becoming concave to the pole at low latitude.
In the Northern Hemisphere, the great circle track bearing decreases when flying ?
North
South
East
West
West
On a Lambert’s projection, the distance scale is expanded ?
outside the standard parallels
between the standard parallels
North of the parallel of origin
South of the parallel of origin
outside the standard parallels
On a Lambert’s projection a line is drawn from a point at 26°S 153°E to 26°S 143°E. The standard parallels are 24°S and 34°S. The great circle track for this flight will be ?
North of the straight line track
South of the straight line track
on the straight line track
on the opposite side of the straight line track to the rhumb line
North of the straight line track
On a Lambert’s conformal conic projection, a straight line is drawn from 16°S 128E to the destination at 27°S 149°E. The outbound straight line track is 116° and the inbound straight line track is 104°. What is the rhumb line track for this trip?
139
122
110
116°
110°

The rhumb line is the average of the great circle track at either end, or equal to the great circle
track at half way.
The outbound GC track is 116° and the inbound GC track is 104°. The rhumb line is the average of
these two - 110°
For weather avoidance you divert off track 45° left for 5 minutes and then intercept your original track at 30°. At a groundspeed of 420 kt, the extra distance the diversion covers compared to the original track distance is ?
17 nm
24 nm
31 nm
14 nm
17nm
In a 45 degree triangle, the short sides are 70% of the long side. During the 5 minutes diversion the equivalent distance along track would have taken only 3.5 min (70% of 5 min).
In a 60 degree triangle the shortest side is 50% of the longest side. The shortest side is 3.5 minutes so the longest side will be 7 minutes. The other angle is 30 degrees. With a 30 degree angle, the other side is 86.6% of the long side. The longest side is 7 minutes so the other side will
be 6 minutes long.
The direct track is 3.5 + 6 minutes = 9.5 minutes
The track around the diversion is 5 + 7 minutes
= 12 minutes
The diversion takes 2.5 minutes extra. At 420 kt (7 nm per minute), the extra distance covered is
17.5 nm (2.5 x 7 nm/min).
Your intended track is 045M. Due to low visibility you alter heading 20° to the left for 10 minutes,
then 30° to the right for 30 minutes. Assuming that your groundspeed of 130 kt remains constant, if
you now turn 30° to the left, you will regain track in ?
30 minutes
10 minutes
5 minutes
20 minutes
5 minutes
If the closing angle is half the track error, it will take twice the time (and distance) to regain track.
If the closing angle is twice the track error, it will take half the time (and distance) to regain the
track.
The ETA, for a flight arriving at VANCOUVER (UTC = LST+8) with an ETD from BRISBANE
(UTC = LST-10) of 11052030 LST and an ETI of 13 hours 50 minutes, (answer in LST as an eight
figure group) is ?
11061620
11050020
11051620
11060020
11051620
2030 Departure BN as local time on November 5th
= 2030 local in 24 hour time
= 052030 local as a local date time group
- 1000 hours (local to Z conversion)
= 051030 Z Departure as a Z date time group
+ 13:50 hours flight time
= 060020 Arrival Vancouver as Z date time
- 0800 hours (Z to local conversion)
= 051620 Arrival Vancouver as a local standard date time group
= 11051620
Arrival Vancouver as a local standard month date time group
The local standard date and time of arrival in TOKYO (UTC = LST-9) for a flight of 9 hours duration
which departed ANCHORAGE (UTC = LST+9) at 1000 LST on 14th March is?
03150400
03151300
03141900
03141300
03151300
The estimated flight time from GILLIGAN’S ISLAND (GI) to ADELAIDE (AD) is 7 hours 30 minutes.
NOTAM information indicates that ‘works in progress’ will close ADELAIDE airport to arrivals from
0600 LST on 6th February. You plan to depart GI to arrive AD 1 hour prior to commencement of the
WIP. If GI LST = UTC-10, and AD LST = UTC+9 hours 30 minutes, your GI departure time in LST
as a six figure group is ?
061200 050200 060200 050300 050030
050200
A flight departs MELBOURNE (YMML) at 9:20 pm LST on January 23 and arrives in LOS ANGELES
(KLAX) at 4:00 pm LST on January 23. LST in YMML is UTC+11 and the LST in KLAX is UTC-8.
The flight time for the flight is ?
13 hours 40 minutes
12 hours 40 minutes
12 hours 20 minutes
14 hours 40 minutes
13 hours 40 minutes
A flight departs BRISBANE (YBBN) at 2 am LST on August 31 for TONGA.
The track distance is 1995 nm and the average groundspeed for the flight is 460 kt.
LST YBBN is UTC+10 and LST TONGA is UTC +13.
The ETA for TONGA in LST is ?
300920
310920
300720
310020
310920
The LMT at a position on Longitude 060°W at 0700 UTC is ?
1100 LMT
0300 LMT
2300 LMT
1500 LMT
Calculate time difference from UTC:
Earth rotates at 15 degrees per hour or 4 minutes of time per degree
Time Difference
= Longitude x4 min/degree
= 60 x 4
= 240 minutes
= 4 hours
If longitude is West, UTC is ahead of local.
If longitude is East, UTC is behind local.
At 60°W, local time is 4 hours behind 0700 UTC.
0700UTC- 0400UTC to local conversion at 60° W longitude
= 0300 local
The groundspeed required when flying West at the Equator at Mean Sea Level (MSL) to keep the
sun apparently stationary in the sky is ?
600 kt
750 kt
900 kt
1100 kt
900 kts
Earth rotates at 15 degrees per hour. At the equator one degree of longitude is 60 nm of distance
To keep the sun stationary in the sky, an aircraft would have to fly west at the speed of the Earth’s rotation (15 degrees per hour) which is a speed of 15 degrees per hour x 60 nm = 900 nm/hr or 900 kt.
UTC at a position at Longitude 170°E at 1600 LMT is ?
0440 UTC
0320 UTC
0429 UTC
0340 UTC
Calculate time difference from UTC: Earth rotates at 15 degrees per hour or 4 minutes of time per degree Time Difference=Longitude x4 min/degree = 170 × 4 = 680 minutes = 11.33 hours =11:20
If longitude is West, UTC is ahead of local.
If longitude is East, UTC is behind local.
At 170°E, 1600 local time is 11:20 ahead of UTC.
16:00 local at 170°E longitude
-11:20 local to UTC conversion at 170° longitude
=04:40 UTC
A flight departs HONOLULU (HNL) for SYDNEY (SY) via NADI (NDI).
The flight plan time interval from HNL to NDI is 6 hours 30 minutes and NDI to SY is 3 hours 45
minutes. The flight is planned to remain on the ground at NADI for 30 minutes. LST in HNL is UTC-
10 and LST in SY is UTC+10.
If the flight departs HNL at 8 pm local time on 4 October, the ETA for SY in local time would be ?
1045 on 4 October
0245 on 5 October
1045 on 5 October
0245 on 6 October
0245 on 6 October
8 PM Depart HNL on October 4th
=20:00 HNL local departure in 24 hour time
042000 local as a HNL local date time group
+ 10:00 (local HNL to Z conversion)
050600 Departure HNL as a Z date time group
+10:45 = total flight time + 30 minutes stopover
051645 = Arrival Sydney as a Z date time group
+10:00 (Z to local Sydney conversion)
060245 Arrival Sydney as a local standard date and time.
The time and date is 1225 LMT on 1st of October at a position on 155° 17’ E Longitude. The LMT
and date in Los Angeles (118° 22’ W) is ?
301811 LMT
311811 LMT
311759 LMT
301721 LMT
155° 17’ E = 155.28333 degrees
(17 min of longitude/60 =.28333 degrees Long)
155.28333 degrees/15 = 10.352 hr = 10 hr 21 min (.352 hours/60 = 21 min of time).
118° 22’ W = 118.3666 degrees
(22 min of longitude/60 =.36666 degrees Long)
118.3666 degrees/15 = 7.891 hr = 07 hr 53 min
(.891 hours/60 = 53 min of time).
LMT on 155° 17’ E Longitude
= 01 1225 LMT
- 1021 offset (East longitudes are ahead of UTC)
UTC = 01 0204 UTC convert - 0753 offset (West longitudes are ahead of UTC) LMT on 118° 22' W Long = 30 1811 LMT( 30 days in September)
Refer ERC L2. You are tracking GRIFFITH to MILDURA on W451 with OEI cruise TAS of 170 kt and
a W/V at OEI cruise altitude of 240M/70 kt. Your determination of the one engine inoperative (OEI)
Critical Point for GRIFFITH (GTH) and MILDURA (MIA), as a distance from MIA is ?
139 nm
61 nm
65 nm
142 nm
The CP formula is:
Distance to CP from departure = (Total distance between alternates x Groundspeed home from CP)/
Groundspeed on from CP + Groundspeed home from CP
From ERC L2.
Measure the distance GTH - MIA : 198nm
Measure the track on from CP to MIA: 260M
Measure the track Home from CP to GTH: 080M
On CR3
Calculate groundspeed on with wind: 240M/70kts Groundspeed: ON TAS: 170 Track: 260 W/V: 240/70
Crosswind: 23 kts Drift: 8° Effective TAS: 168kts Headwind: 66kts Groundspeed: 102kts
Calculate groundspeed home with wind Groundspeed: Home TAS:170kts Track: 080 W/V: 240/70 Crosswind:23kts Drift:8 degrees Effective TAS: 168kts Tailwind: 66kts Groundspeed: 234kts Calculate CP with formula: Distance to CP from GTH= (198 nm x 234 kt)/ (102 kt + 234 kt) = 138 nm from GTH Distance to CP from MIA = 198 nm - 138 nm = 60 nm
Refer to ERC L2. Your position is overhead MILDURA VOR (MIA) at 0200 UTC, FL185 enroute to
WHYALLA (WHA) on W448. Another aircraft en-route BROKEN HILL - AD at FL180 on W426, reports
that it is at KONDO (approx 33°S 140°30’E) at 0225 UTC and is estimating VILAD at 0252 UTC. Your
position on the MIA - WHA W448 track at 0225 UTC is DME MIA 70.
Your determination of which aircraft, the Southbound or Westbound, will first cross the intersection
of the MIA - WHA W448 and KONDO - VILAD W426 tracks, and the lateral separation in nm when the intersection is first reached by one of the aircraft is ?
southbound will cross first, lateral separation of 5 nm
westbound will cross first, lateral separation of 3 nm
southbound will cross first, lateral separation of 2 nm
westbound will cross first, lateral separation of 5 nm
southbound will cross first, lateral separation of 5 nm
Calculate your groundspeed:
Over MIA at 0200
70 DME MIA at 0225
70 nm in 25 min = 168 kt
Calculate the Southbound groundspeed:
KONDO at 0225
VILAD at 0252
92 nm in 27 min = 204 kt
Calculate your time to the intersection:
Distance to the intersection from 70 DME MIA fix = 39 nm
Time to the intersection = 39 nm/168 kt x 60 = 13.93 minutes
Calculate Southbound time to intersection:
Distance to the intersection from KONDO = 41 nm
Time to the intersection from KONDO = 41 nm/204 kt x 60 = 12.06 minutes
Southbound arrives first by 1.87 minutes.
Southbound arrives while you are still 1.87 minutes away at 168 kt = 5.2 nm
Refer to ERC L2. Position is overhead GRIFFITH (GTH). TAS 235 kt, safe endurance available
(excluding reserves and alternate fuel) is 160 minutes. If all reserve and alternate fuel must remain
intact, and without making allowance for the descent, your determination of the maximum average
headwind component which could tolerated for the remainder of the flight to WHYALLA (WHA) via MILDURA is ?
64 kt
68 kt
71 kt
81 kt
71 kt
From ERC L2. Distance GTH to WHA via MIA is 436 nm. With 160 minutes safe endurance, the
minimum groundspeed required is 163.5 kt (436 nm/160 min x 60). With a TAS of 235 kt a groundspeed of 163.5 kt would be causes by a headwind of 71.5 kt (235 kt - 163.5 kt = 71.5 kt).
Distance GTH - MIA - WHA = 436nm/
Safe Endurance of 160min = required speed of 2.725 nm/min
Groundspeed = 2.725 x 60= 163.5kts
TAS= 235kts
GS = 163.5kts
Headwind limit = 235-163.5= 71.5kts
Refer ERC L2. Position is tracking W448 direct MILDURA (MIA) - WHYALLA (WHA), at 110 nm GPS
WHA with TAS 240 kt. A fuel feed problem has reduced the safe endurance (excluding reserves) to
45 minutes. Using a W/V of 280M/70 kt, (disregard the descent and assume that clearance for all
airspace is available) your determination of the last point of safe diversion (PSD) to ADELAIDE
(YPAD), as a distance from WHA is ?
32 nm
41 m
62 nm
48 nm
48 nm
Try 48 nm WHA as possible PSD.
Calculate time to PSD from present position and from PSD to ADELAIDE.
Groundspeed to PSD from present position:
Track:279M WV: 280M/70kts TAS: 240kts Crosswind: Nil Drift: Nil Effective TAS:240kts Head/tail wind: -70kts Groundspeed: 170kts Time to PSD : 62nm @ 170kts = 21.88 minutes.
Groundspeed PSD to Adelaide:
Track: 167M W/V: 280M/ 70kts TAS:240kts Crosswind: 64kts Drift: 15° Effective TAS: 232kts Head/tailwind: +25kts Groundspeed: 257kts Time to AD = 98nm @ 257 = 22.88mins
Total = 21.88 min + 22.88 min = 44.76 minutes (safe endurance is 45 min so this distance is the PSD as it is within 1 min of the available Safe Endurance).

You have the following data: Time 0100 UTC - VOR MA 316 (FROM) DME MA 80. Time 0119 UTC - VOR MA 316 (FROM) DME MA 150. HDG of 312M and a TAS of 220 kt have been maintained since 0100 UTC (no charts needed). The actual wind velocity is ?
W/V 220M/15 kt
W/V 045M/15 kt
W/V 245M/18 kt
W/V 190M/22 kt
W/V 220M/15 kt
On CR3:
Track: 316M Heading: 312M TAS: 220kts Groundspeed: 22kts ( 70DME in 19mins) Drift: 4° (316° -312°) Crosswind: 15kts from the left Head/Tail: +1 (Tailwind)
Refer ERC L6. You are planning a flight from ALICE SPRINGS (AS) to DARWIN (DN) via A461. TAS is
220 kt and W/V is 330M/50 kt. The position of the Critical Point for TINDAL (TN) and TENNANT
CREEK (TNK), on the AS-DN track, as a distance from DN is ?
305 nm
285 nm
268 nm
248 nm
268 nm
Draw a line between the two alternates - TINDAL and TENNANT CREEK.
Bisect the line at right angles and extend a bisector to the original A461 track to find the nil wind CP.
Compute the wind effect based on the time to fly from the nil wind CP to either alternate at the TAS.
174 nm CP to alternate at 220 kt = .791 hr
W/V 330/50 for .791 hr = 40 nm wind effect (50 kt x .791 hr).
Plot the wind effect as a 40 nm vector from 330M blowing to the nil wind CP.
From 40 nm along the wind vector extend a line back to the original A461 track parallel to the bisector.
This position on A461 is the CP. 14 nm North of MONIC. 282 - 14 = 268 nm from DN in this example. See the inset in the diagram below.
The time is 0140 UTC. Your position is overhead NGUKURR (waypoint DAMPA) enroute to DARWIN
(DN) on W356. Safe endurance at NGUKURR (excluding reserves) is 89 minutes. TAS is 200 kt. WV
at your cruise level is 300M/15 kt. The latest point of safe diversion (PSD) to TINDAL (TN),
measured as a distance from DN, and your ETA at the PSD are ?
62 nm, ETA 0245
70 nm, ETA 0243
78 nm, ETA 0240
84 nm, ETA 0238
70 nm, ETA 0243
You have 89 minutes of Safe Endurance to fly from NGUKURR to TINDAL via the closest PSD to
Darwin without exceeding 89 minutes total flight time.
Try 70 nm DN as a possible PSD.
Calculate the time to fly from:
NGUKURR to the PSD:
Distance: 192 nm Track: 297M TAS: 200kts W/V: 300M /15kts Groundspeed: 185kts ETI: 192nm @ 185kts = 62.27minutes (ETA 0242)
From PSD to TINDAL
Distance: 91nm Track: 159M TAS: 200kts W/V: 300M/15kts Groundspeed: 211kts ETI: 25.88
Total time NGUKURR to PSD and PSD to TINDAL
= 62.27 min + 25.88 min = 88.15 min (safe endurance 89 min so within 1 min so Ok)
An aircraft is required to track outbound from a radio navigation aid and then return along a
reciprocal track. The rated coverage of the aid is 50 nm. Using a maximum possible tracking error
of 9 degrees either side of track for the entire flight, the distance from the aid (to the nearest 5 nm)
the aircraft be flown, and still be assured of coming within the rated coverage of the aid on the
return flight is ?
330 nm
165 nm
200 nm
75 nm
165nm
Using a tracking error of + 9° either side of track, a track error of 50 nm will occur after 333 nm of
track distance using the 1 in 60 rule.
The 1 in 60 rule can be used to compute the distance off track from the track error angle and vice
versa. At 60 nm of distance travelled, the number of degrees off track is equal to the number of
miles off track. At 60 nm along track, a 9° track error would be 9 nm off track. At 120 nm along
track, 9° off track would be 18 nm off track.
Since 9° goes into 60 at total of 6.66 times, by multiplying the rated coverage of an aid by 6.66, the
along track distance resulting in a track error equal to the rated coverage can be found for a 9°
error.
In this example 60/9 x 50 nm = 333 nm. In this question, the requirement is to arrive back at the
departure aid allowing for a track error of + 9° either side of track both out and on the return, so
the maximum total distance out and back is 333 nm or 166.5 nm each way (333/2).
Refer ERC L7 and L5. Your flight is from ALICE SPRINGS (AS) to MT ISA (MA). TAS is 430 kt and
the W/V is 295M/85 kt. Plot on L7. Your determination of the point on the AS - MA track from
which it will take the same time to either return to AS or to divert to BIRDSVILLE (BDV), measured
as a distance from MA is ?
234 nm
210 nm
180 nm
263 nm
180nm
Bisect the track AS to BV (it does all fit on L7!). Extend the bisector to the original J64 track. This
the nil wind CP. Compute the wind effect:
Wind effect: = ((Nil wind distance from CP to either alternate)/TAS)) x Windspeed
= (272 nm CP to AS/430 kt) x 85
= .632 hr x 85 kt
= 54 nm
From 54 nm on the 295M wind vector, extend a line parallel to the bisector back to the track to find the actual CP. 182 nm from AS or 181 nm from MA.
You are overhead JODEX (285/32 Scone) tracking to QUIRINDI. You are experiencing 8 degrees right drift. The ADF is tuned to Scone NDB. The ADF relative bearing to Scone NDB is ?
073 081 089 106 114
089
The ADF relative bearing is the difference between the aircraft heading (017 with 8 degrees right
drift) and the track to Scone (106). 106 - 17 = 089 relative bearimg
Using a protractor with the aircraft heading as North, the ADF relative bearing to Scone NDB can be
read directly. With 8 degrees right drift the aircraft is being blown to the right with wind from the
left so the heading will be 8 degrees to the left of the 025 track to QUIRINDI.
You are planning to track by day to LEINSTER from KALGOORLIE via LEONORA (S28 52.7 E121 19.2) at A100 ft above overcast cloud. The LEONORA NDB is unserviceable. The
navigation aids which could be planned to be used for radio position fixing overhead LEONORA are ?
Position lines from KALGOORLIE and LEINSTER NDB’s
Position lines from LEINSTER NDB and KALGOORLIE VOR
Position lines from KALGOORLIE DM and LEINSTER NDB
Either A, B or C
none of the above
Refer AIP EN 1.1 para 19.5.1 defines a positive radio fix as: Station passage over a: NDB,VOR, TACAN, marker beacon or a DME site or determined by the intersection of two or more position lines which intersect at angles of not less than 45° and which are obtained from NDB's, VOR's, Localisers, or ME's in any combination. The position lines must be within the rated coverage of the aid with the exception that if a fix is determined entirely by position lines from NDB's, the position lines must be within a range of 30 nm from each of the NDB's. KALGOORLIE and LEINSTER NDB's position lines intersect at 24 degrees and are beyond 30 nm. LEINSTER NDB (range 70 - ERSA) and KALGOORLIE VR (range 120 at A100) are in range but the position lines intersect at 24 degrees (less than the 45 degree minimum required). KALGOORLIE DME is within range 115 nm away. LEINSTER NDB (range 70 - ERSA) is in range At A100 ft the rated coverage is 120 nm for both the VOR and DME (AIP GEN 1.5). So this combination is suitable as the position lines intersect at 66 degrees (331 - 265). (355-90 = 265)
You are tracking 255M from ALPHA to BRAVO. At 0325Z you obtain a fix 90 nm from departure.
Safe Endurance at this point is 118 minutes. Wind has been 220M/45. TAS is 335 kt.
From your 0325Z fix the wind is forecast to be 160M/35.
What is the location of the Point of No Return to ALPHA, as a distance from ALPHA?
327 nm ALPHA 417 nm ALPHA 287 nm ALPHA 377 nm ALPHA 394 nm ALPHA
377 nm ALPHA
First Determine the fuel required to return to ALPHA from the 0325Z fix as this cannot be used in
the PR formula.
The track to ALPHA from the 0325Z position is 075M. (the reciprocal of 255)
Find the ETI for the 90 nm using a groundspeed based on the first wind of 220M/45. From the CR3
the GS for this leg is 371 kt. The ETI is 14.6 min.
Subtract the 14.6 minutes from the safe endurance of 118 min to give 103.4 minutes revised S.E.
Use this revised S.E. to calculate a PNR for your current position using new groundspeeds based on the forecast wind of 160M/35 from 0325Z on.
Groundspeed Out (based on a track of 255M wind of 160M/35 and TAS 335 kt) = 336 kt Groundspeed Home (based on a track of 075M wind of 160M/35 and TAS 335 kt) = 330 kt Time to PNR = 103.4 x 330/ 336 + 330 = 51.23 minutes to the PNR from 0325Z position.
Distance to the PNR = 51.23 minutes x GSO 336 kt / 60 = 287 nm from 0325Z position
Distance to PNR from ALPHA = 287 nm + 90 nm = 377 nm from ALPHA.
Refer to ERC L4
You are overhead INJUNE airstrip tracking 334 degrees magnetic from ROMA on W499. You are experiencing 4 degrees left drift. The ADF is tuned to ROMA NDB.
The ADF relative bearing to ROMA NDB is ?
176 184 154 158 150
176
The heading is 338 degrees due to the 4 degrees of left drift (left drift means wind is from the right
so heading is on the right of the track) 334 + 4 = 338 heading.
The HATS formula shows that Heading + ADF relative bearing = Track to the Station.
It then follows that Relative Bearing
=Track 154 - Heading 338 = -184 (+360)
= 176 Relative Bearing
Remember if the HATS formula gives you a number more than 360 or less than 0, then you need to subtract or add 360 to result in a number between 0 and 360 as a relative bearing.
The error associated with the NDB/ADF is ?
bending and scalloping
vertical polarisation error
site effect error
quadrantal error
Quadrantal error
An NDB station radiates a signal which is ?
in two parts, one constant and the other variable in phase depending on relative
position to the station
omni directional
in the VHF band
calibrated for the magnetic variation at the station
omni directional
An NDB signal is received by an ADF in an inbound aircraft equipped with a fixed card ADF. The navigation information is presented as ?
a magnetic track to the station
a true track to the station
a relative bearing to the station
a radial
a relative bearing to the station
With a fixed card ADF, the needle points to the station relative to the nose of the aircraft. This is referred to as relative bearing.
In a vertical axis gyroscope at the South pole ?
drift and topple due to Earth rate rotation will be at the maximum values
drift and topple due to Earth rate rotation will be zero
drift will be zero and topple will be maximum
drift will be maximum and topple will be zero
drift and topple due to Earth rate rotation will be zero
In a horizontal axis gyroscope at the South Pole ?
drift and topple due to Earth rate rotation will be at the maximum values
drift and topple due to Earth rate rotation will be zero
drift will be zero and topple will be maximum
drift will be maximum and topple will be zero
drift will be maximum and topple will be zero
On a Lambert’s conformal conic projection, a straight line is drawn from 16°S 128°E to the destination at 27°S 149°E. The outbound straight line track is 116° and the inbound straight line track is 104°.
The convergence factor (n) for this trip is ?
.52 .57 .636 .875 
.57
The track covers 21 meridians while the bearing changes 12°. The convergence factor (n) is
calculated by dividing the bearing change by the longitude change.
n = B/L
= 12/21
= .57
An aircraft has flown a track of 332M while holding a heading of 320M with a groundspeed of 445 kt
and a TAS of 420 kt. The actual wind experienced in flight is ?
228M/85 kt 218M/95 kt 265M/90 kt 085M/95 kt 205M/110 kt
218M/95kts
Track: 332M
HDG: 320M Drift = 12° = 87 kt left crosswind
TAS:420kts
ETAS: 411 kt
GS: 445 kt Wind Component = + 34 kt tailwind
When flying at an indicated altitude of 4500 ft on local QNH with a temperature of ISA+10, your
true altitude is approximately ?
4680 ft
4320 ft
4050 ft
4950 ft
4680ft
In an Inertial Navigation System (INS), compensation for Earth rate rotation must be applied ?
according to the aircraft speed over the ground
equally at all points over the Earth’s surface
according to a specific latitude
according to a specific longitude
according to a specific latitude
The required earth rate compensation is a function of latitude since what is being compensated for is the horizontal component of earth rate felt by the gyros, and that varies with latitude. At the equator, this value is 15.04 degrees per hour, and with travel either further north or south, it
reduces in proportion to the Cosine of the Latitude until it becomes zero at the poles.
In an Inertial Navigation System (INS), compensation for transport rate must be applied ?
according to the aircraft speed over the ground
equally at all points over the Earth’s surface
according to a specific latitude
at 15.04°
° per hour
according to the aircraft speed over the ground
Transport rate compensation is developed using the velocity signal. The electronics through which it
is sent contain a term proportional to the earth’s radius. So in reality, the transport rate signal
torquing the gyro is the velocity of the aircraft divided by the earth’s radius.
In an Inertial Navigation System (INS), the platform on which the accelerometers are mounted
must be aligned with ?
magnetic North
true North
the aircraft longitudinal axis
the local parallel of latitude
True North
The INS platform must be aligned with True North so the appropriate compensations can be applied
to the correct axes of the platform and to provide a reference for navigation. Magnetic data is
available by applying the variation for a specific location.
Refer to ERC L3. You are flying from TAMWORTH to OAKEY on W684 and W207 and decide to calculate the position of the PSD for GOONDIWINDI. Your current position is overhead INVERELL (315 151E) with a safe endurance (excluding all reserves) of 50 minutes. TAS is 220 kt and wind from the IRS is 260M/40 kt. The location of the PSD for GOONDIWINDI as a distance from
INVERELL is ?
Potential Answers: 97nm 103nm 112nm 118nm
Correct answer:
103nm
112 nm INVERELL is too far. The ETIs
exceed the safe
endurance by 3.6
minutes.
103 nm INVERELL is correct. The ETIs
add up to within a minute of the safe
endurance of 50 minutes.
Significant diversions for weather avoidance delay your flight by 12 minutes soon after top of climb
on a leg of 480 nm. The wind component on this leg is +50 kt. The delay will move the position of
your previously calculated CP at normal TAS of 480 kt a distance of ?
nil
approximately 10 nm closer to destination
approximately 10 nm further from destination
approximately 5 nm towards the departure point
Nil
The CP position is determined by the groundspeed On and Home from the CP. The time to the CP
(groundspeed out) has no influence on the CP position.
Coastal Refraction is an error which results in ?
VOR signals bending toward the coast when approaching an inland station from
overwater at an angle to the coast
NDB signals bending toward the coast when approaching an inland station from
overwater perpendicular to the coast
NDB signals bending toward the coast when approaching an inland station from
overwater at an acute angle to the coast
VOR signals bending away from the coast when approaching a coastal station from
overwater at an angle to the coast
NDB signals bending toward the coast when approaching an inland station fromoverwater at an acute angle to the coast.
Coastal Refraction is most pronounced when crossing the coast at an angle and when the station is inland from the coast.
The signal is bent towards the coast
As an aircraft approaches an enroute DME station when overflying, the DME ground speed readout
will ?
reduce at an increasing rate to zero overhead, then increase at an increasing rate
back to the correct value after station passage
reduce at an decreasing rate to zero overhead, then increase at an decreasing rate
back to the correct value after station passage
increase at an increasing rate to zero overhead, then decrease at an increasing rate
back to the correct value after station passage
reduce at an increasing rate to zero overhead, then increase at a decreasing rate
back to the correct value after station passage
reduce at an increasing rate to zero overhead, then increase at a decreasing rate back to the correct value after station passage.
If an aircraft pitot tube becomes blocked while in cruise at a constant FL, and a climb is then commenced, the airspeed indicator (ASI) will ?
read correctly
overread
underread
indicate zero airspeed
overread
With a blocked pitot tube, the total pressure at the current flight level is trapped inside the capsule.
If the aircraft then climbs, the dropping static pressure surrounding the capsule will allow the high
total (pitot) pressure to expand the capsule more than normal causing it to overread.
An aircraft climbs at a constant 310 IAS from FL150 to FL250 with an average rate of climb between FL150 and FL250 of 1550 ft/min. The distance covered over the ground during the climb with a wind component of +50 kt and a temperature deviation of ISA-5 is ?
30 mm 35 nm 40 nm 45 nm 50 nm
50 nm
A constant IAS climb results is an increasing TAS and Mach number.
To find the average TAS, calculate the Mach number at 2/3rds of the climb to allow for non linear climb performance.
2/3rds of the way from FL150 to FL250 is FL217.
At a constant IAS of 310 at FL217 the Mach number is 0.695M.
At ISA-5 the temperature at FL217 is -33°C.
This results in a TAS of 419 kt.
With 50 kt tailwind, the ground speed is 469 kt.
10000 ft at 1550 ft per minute takes 6.45 minutes
6.45 minutes at 469 kt = 50 nm
A Doppler VOR installation has the advantage of reducing errors associated with ?
ground station equipment
obstructions in the vicinity of the site
airborne equipment
vertical polarisation
obstructions in the vicinity of the site
A Doppler VOR transmitter almost totally eliminates site effect errors by use of FM instead of AM and the Doppler principle of velocity induced phase shift. No special receiving equipment is needed in the aircraft. Doppler VOR antennae are bigger than conventional ones.
An aircraft climbs at a constant 310 kt IAS until 0.78M is reached and then continues to climb at
0.78M. On an ISA+10 day, the Flight Level at which these two speeds are equal is closest to ?
FL320
FL300
FL280
FL260
FL280
Using the CAS/PH window in the CR3 - position the Mach pointer to 0.78M, then read FL280 against
the CAS of 310.
The purpose of the annunciator in a gyrostabilised remote indicating compass system is to indicate ?
that the magnetic detector unit and gyro are correctly aligned and synchronised
a weakening of the horizontal (H) component of the Earth’s magnetic field
whether or not the electric heading signal is being fed to ancillary equipment
power failure
that the magnetic detector unit and gyro are correctly aligned and synchronised.
The annunciator normally oscillates slightly one side to the other when the flux valve signal and compass card reading are synchronised.
A horizontally polarised radio signal is best received by a ?
vertical antenna
horizontal antenna
circular antenna
linear antenna
horizontal antenna
Refer ERC L5. You fix your position at HUMOK on W540 at 0106 UTC en-route DUBBO to BOURKE
at A080, TAS 180 kt, WV 130M/30 kt and safe endurance (excluding reserves) 40 minutes. The
position of the PSD for BREWARRINA as a distance from BOURKE, and your ETA at this point are ?
16 nm from BOURKE, ETA 0133
95 nm from BOURKE, ETA 0133
16 nm from BOURKE, ETA 0115
25 nm from BOURKE, ETA 0133
16 nm from BOURKE, ETA 0133
A Machmeter is subject to ?
pressure error and density error
instrument error and density error
pressure error and temperature error
pressure error and instrument error
pressure error and instrument error
Mach meters are not affected by compressibility errors since the error depends on speed an and
altitude, both of which are inputs to the instrument and so can be compensated for. All instruments
are affected by instrument error. Pressure error refers to errors in accurately sampling the air
pressures due to placement of the static and pitot sources. All pressure instruments are affected by
this error.
While on a track of 353M you must divert around an unexpected thunderstorm. You alter heading
60° to the left and fly for 7 minutes, before altering heading 120° to the right and flying another 7 minutes. If the wind is 040M/42 kt, your position in relation to the direct track at the end of the diversion is ?
5 miles right of track 7 miles left of track 5 miles left of track 10 miles left of track 7 miles right of track 10 miles right of track
7 miles left of track.
The diversion tracks around two sides of an equilateral triangle. The third side is the original planned track. The average track during the diversion is 353M so the average crosswind during the diversion is the crosswind relative to the average track.
The crosswind on a track of 353M is 30 kt.
During the diversion the aircraft has been exposed to 30 kt crosswind from the right for a total of 14 minutes (7 minutes each side).
30 kt average crosswind for 14 minutes results in
a distance of 7 nm left of track (14/60 x 30 = 7).
Refer ERC L4. You obtain a positive fix overhead BOWLY at 0240 UTC enroute to LONGREACH from
TOWNSVILLE on W660. At 0305 UTC you obtain another fix placing you right of track, over SLASH at F140, TAS 265 kt, forecast WV of 260M/40 kt. You have been maintaining the planned heading and TAS since BOWLY. The actual WV(M) affecting you since BOWLY was ?
090M/22 kt
325M/27 kt
140M/25 kt
270M/22 kt
325M/27kts
Track: 203M Heading: 208M ( Based on original track and forecast wind of 260M/40kts) Drift: 5° Crosswind: 23kts right TAS: 265kts Effective TAS: 264kts Groundspeed: 278kts (116nm in 25 mins) Head/tailwind: +14kts
Refer ERC L2. You obtain a position fix from position lines reference waypoint BORTO on a track of
041M TO BORTO and a track of 117M TO NARACOORTE Airport on the GPS. From this position, the track and distance to waypoint SWELL on W519 is ?
310M/45 nm
320M/45 nm
325M/45 nm
315M/45 nm
310M/45nm
In flight an aircraft has a VOR tuned to a station where the magnetic variation is 10° E. At the aircraft present position the magnetic variation is 8° E. The VOR OBS is selected to 090 and the CDI is centered with the TO flag showing. The MI NBD indication of the track to an NDB at the same location as the VOR station would be ?
090
088
092
270
092
Your VOR OBS is set to a required track of 090, and you have a from flag displayed. The CDI has moved 4 degrees to the right. DME distance from the station is 45 nm. The distance off the required track is ?
2 nm
4 nm
3 nm
6 nm
3nm
Using the 1 in 60 rule.
4° off track at 45 nm is equal to 3 nm. (45/60 x 4° = 3 nm).
The centre line of the localiser of an ILS is determined by ?
phase comparison of the signal modulation radiated from each side of the centreline
a locus of points of equal signal modulation radiated from each side of the centreline
frequency comparison of signal modulation radiated from each side of the centreline
the doppler shift of the signal modulation radiated from each side of the centreline
a locus of points of equal signal modulation radiated from each side of the centreline.
The Localiser transmits a VHF radio signal consisting of two overlapping lobes from an antenna located about 300 mt past the upwind threshold. The runway centreline is distinguished by a locus of equal lobe strength reception of the signal by the cockpit receiver.
One advantage of the gyro stabilised RMI over a conventional direct reading magnetic compass is ?
reduced magnetic variation due to the remote location of the flux valve
stability in turbulence due to the inability of the MI card to rotate in azimuth
elimination of magnetic deviation due to the remote location of the flux valve
reduced turning and acceleration errors due to the low precession rate of the gyro
reduced turning and acceleration errors due to the low precession rate of the gyro.
Since the gyro stabilisation minimises the turning and acceleration errors, the main error remaining is deviation, even though remote locating the flux valve helps reduce this error, it is still present to a small degree.
The UTC time at a position on Longitude 145 E at 0600 local time is ?
2020 UTC
1620 UTC
1700 UTC
2100 UTC
2020 UTC
The earth rotates at a rate of 360° of longitude in approximately 24 hours. Each degree of longitude represents approximately 4 minutes of time. 145 degrees of longitude represents about 145 x 4 minutes = 580 minutes (9 hours 40 minutes). Easterly longitudes are ahead of UTC so at 145°E the
local standard time is about 9 hr 40 minutes ahead of UTC. 0600 local - 9 hr 40 = 2020 UTC.
Refer to the GPWT extract.
On a track of 100T, the pressure level which would result in the highest groundspeed for an aircraft
cruising at a TMN of 0.80M is ?
HPA GPWT
- 2908551
- 2808040
- 2706535
- 2704020
200 HPA
250 HPA
300 HPA
400 HPA
250 HPA
HPA. TAS W/C G/S
- +84. 548
- +80. 555
- +64. 544
- +39. 534
Side bands (USB or LSB) are often used with AM radio propagation. Of the following the one which is not an advantage of utilising sideband transmissions is ?
external noise and interference are reduced
the carrier does not have to be so powerful
different information can be transmitted simultaneously on both bands
the signal does not require fine tuning (clarification)
signal does not require fine tuning (clarification)
The others are all advantages of sideband transmission.
In an ILS display, from full left to full right deflection of the localiser needle indicates a beam width of ?
2.5 degrees
5 degrees
10 degrees
20 degrees
5 degrees
The ILS localiser course deviation is 4 times more sensitive than normal VOR course deviation. The localiser is half a degree per dot compared to 2 degrees per dot for the VOR.
In an ILS display, full deflection of the glide path needle occurs when the aircraft is ?
- 5 degrees above or below the glide path
- 7 degrees above or below the glide path
- 0 degrees above or below the glide path
- 4 degrees above or below the glide path
0.7 degrees above or below the glide path
The glideslope is even more sensitive than the localiser. A total of 1.4 degrees full up to full down or
about 0.7 degrees either side of centre at full scale deflection.
You depart YBAS for YPAD at 0910 UTC, maintaining FL290, TAS 480 kt, WV 250M/60 kt, average track 155M, safe endurance (excluding reserves) is 60 minutes.
The position of the PR as a distance from YBAS is ?
PNR 235 nm from YBAS
PNR 30 nm from YBAS
PNR 225 nm from YBAS
PNR 295 nm from YBAS
PNR 235 nm from YBAS
Time to PNR= Safe Endurance x Groundspeed Home /
Groundspeed out + Groundspeed Home
= 60 minutes x 471 kt /
483 kt + 471 kt
= 29.6 minutes
Distance to PNR = (Time to PNR x Groundspeed Out )/ 60
= 238 nm AS
The ILS Outer Marker signal is ?
1300 Hz, alternate dots and dashes and a flashing amber light
3000 Hz, 6 high pitched dots per second and a flashing white light
400 Hz, 2 low pitched dashes per second and a flashing blue light
600 Hz, 5 high pitched dots per second and a flashing red light
400 Hz, 2 low pitched dashes per second and a flashing blue light
A locator is ?
a 25 Hz signal used to locate the aircraft over the destination aerodrome
the point of commencement of an ILS approach
the left/right guidance beam of an ILS
a low powered NDB used in association with an instrument approach
a low powered NDB used in association with an instrument approach.
The great circle track outbound from an airport at 120°E Longitude in the Northern hemisphere is 080T and the inbound great circle track is 115T at your destination airport at 170°E Longitude.
The Longitude where your track will pass closest to the pole is ?
145 134 127 120 170
134°
Step 1: Find the convergence factor (n) for the trip.
Convergence factor (n)= bearing change/ longitude change
=35/50
= 0.7
Step 2:
Determine if the track will pass through a bearing of 090 degrees or due East (270 or due West
if the track was westerly).
In this case the outbound track of 080° (North East) will run through 090° (East) on the way to
the destination end of 115° (South East) so a bearing of 090° is the bearing marking the end of
the North Easterly section of the track and the start of the South Easterly section of the track
and therefore the most Northerly point of the track.
Step 3: Calculate the bearing change from the departure track (080) to due East (a track of 090°).
Due East as a bearing is 10 degrees more than the departure track of 080.
Step 4: Calculate the longitude change required to produce the bearing change.
10 degrees of bearing change out of the total change of 35 degrees occurs at 10/35ths (28.6%) of the total change. Therefore the meridian which the track crosses at this time will be
the meridian 28.6% of the way along from departure. The total longitude change is 50° (170E - 120E) so the longitude change required to produce a 10° bearing change is 28.6% of 50 (.286 x 50 = 14°)
Another way of determining the longitude change is to divide the bearing change by the (n)
factor. In this case required bearing change to due West is 10° so at a rate of (n) of 0.7
degrees bearing change per meridian of longitude the longitude change will be 10/0.7 = 14°.
Apply the longitude change for the desired bearing change to the longitude of the departure
point. For flights with decreasing longitude subtract the change. For flights with increasing
longitude add the change.
In this case the longitude is increasing so the 14° is added to the departure longitude of 120E.
Most Northerly Meridian=
Departure meridian + required longitude change
=120E + 14
=134E
You are at FL150. The rated coverage of a VOR station located at sea level is ?
90 nm
120 nm
150 nm
180 nm
150 nm
Refer AIP GEN 1.5-5.
The meteorological condition which would give the maximum amount of radar energy reflected from a cloud is ?
mist
freezing fog
dry hail
heavy wet snow
Heavy wet snow
Weather radar energy reflects off liquid water. frozen dry hail or dry snow will not reflect well but wet snow will reflect a strong return.
With regard to airborne weather radar, ‘attenuation’ refers to ?
mixing of ground clutter with cloud returns
weakening of the radar transmissions and returns
unwanted returns from side lobes
beam width and pulse length distortion of returns
weakening of the radar transmissions and returns.
Hooks, fingers, and nodules as may be seen on an airborne weather radar display are primarily associated with ?
wet snow
hail shafts
icing on the radome
supercooled rain
Hail shafts
Strange shapes like hooks, fingers and nodules may indicate the presence of hail. Even though hail does not show up itself, a hook shape with an apparently open area in the middle on the radar screen is not likely to actually be clear but rather full of something which the radar can not see like hail.
You are tracking inbound to YBBN VOR on the 228 radial with a constant heading of 044M and TAS
of 366 kt. You also have the YBBN ME selected and the reading decreases from 77 to 71.2 in 50 seconds.
The actual WV is ?
280M/35 kt
255M/60 kt
250M/100 kt
290M/40 kt
255M/60kts
Track: 048M Heading: 044M Drift: 4° Crosswind: 25kt left TAS: 366kt Effective TAS: 366kt Groundspeed: 418kts ( 5.8nm in 50s) Head/tailwind:+52 kt
You are inbound on runway LLZ heading to your destination that is served by a NDB, VOR and a LLZ. Both the aircraft and yourself are equipped to use all the aids and you are within the rated coverage of all the aids. The aid you would use for tracking purposes is ?
all NDB LLZ VOR LLZ if conducting a LLZ approach otherwise the VOR
LLZ
The LLZ is the most precise tracking aid and should therefore be used if it defines the track - which it does if you are inbound on RWY heading.
With regards to ATC SSR Radar installations and weather, SSR Radar provides ?
thunderstorm avoidance guidance
special WX Radars are needed to give weather related guidance
guidance around thunderstorms if operating in Terminal Area Mode
secondary return of significant weather to allow pilots guidance from ATC
special WX Radars are needed to give weather related guidance.
Pure SSR shows no WXR returns but if a specific weather radar is installed then ATC may have access to the meteorological situation.
On a Lambert’s conformal conic projection, a straight line is drawn from 16°S 128°E to the destination at 27°S 149°E. The outbound straight line track is 116° and the inbound straight line track is 104°. The rhumb line track for this trip is ?
130
122
110
116
110°
The rhumb line is the average of the great circle track at either end, or equal to the great circle
track at half way. The outbound GC track is 116° and the inbound GC track is 104°. The rhumb line is the average of these two - 110°.
The greatest error in an ASI, due to compressibility, is caused by ?
high temperature
high altitude
low temperature
low altitude
High altitude.
As altitude increases, the compressibility error increases so most error will be apparent at high altitude.
When plotting a magnetic radial of a VOR on a chart ?
you should plot the radial direct on the chart using the grid
you must allow for deviation at the station
you must allow for variation at the aircraft’s present position
you add or subtract the variation to give you a true bearing before plotting on the
chart
you add or subtract the variation to give you a true bearing before plotting on the chart.
When plotting a VOR radial using the latitude and longitude, the variation must be taken into account. (If you use the existing magnetic information from existing published radials, then
magnetic radials can be directly drawn).
aircraft is parked on the ground at a position of 30°S latitude with all electrical and avionics systems operating. With regards to the error in a directional gyroscope ?
it would have no error
it would have half the error as in flight
it would have the same error as in flight
it would have twice the error as in flight
It would be the same error as in flight.
Being airborne or on the ground doesn’t change the apparent precession.
An aircraft overflys an aerodrome of 2000 ft elevation. The correct local QNH of 987 HPA is set on the aircraft altimeter. The OAT at the aerodrome is 0°C.
As the aircraft overflys the aerodrome at an indicated altitude of 3500 ft, the approximate aircraft radio altimeter indication will be ?
1500 ft 1560 ft 1440 ft 720 ft 2280 ft
1440ft
At the aerodrome the OAT is O°C. This means that the temperature variation is approximately ISA- 10. This will cause the altimeter to overread by approximately 4% of the height (AGL with local
QNH), in this case 4% of 1500 ft.
On the calculator, 4% of 1500 ft is 60 ft. Since the altimeter is overreading by about 60 ft, the true
altitude (or height) will be 60 ft lower than the indicated altitude (or height).
The radio altimeter indicates true height AGL. With a barometric altimeter indicated altitude of 3500
ft this equates to an indicated height of 1500 ft. The radio altimeter will indicate a true height 60 ft
lower i.e. 1440 ft.
Or
On the whiz wheel in the true altitude window, set 0°C against the aerodrome elevation of 2000 ft
(technically pressure altitude is used - but for practical and exam purposes, altitude will do).
Then on the main inner scale of the whiz wheel find 1500 ft indicated height (use height AGL when
working with local QNH) and read the true height of 1440 ft on the outer main scale.
aircraft is flying on a heading of 194M. The fixed card ADF is showing a relative bearing of 187.
Magnetic Variation is 4W. The true bearing of the aircraft from the NDB is ?
194 021 187 017 197
197
Using the H.A.T.S. formula (heading + ADF Relative bearing = Track to the Station), the track to
the station is 021.
Therefore the aircraft is on the 201 radial (M) or the 201(M) track FROM the station. With 4W variation, the true bearing is 201(M) - 4
= 197(T). (When Variation is WEST,
Magnetic is BEST).
An aircraft transmitting from a vertically orientated antennae would produce ?
a vertically polarised signal in a vertical direction
a horizontally polarised signal in a horizontal direction
a vertically polarised signal in a horizontal direction
a horizontally polarised signal in a vertical direction
a vertically polarised signal in a horizontal direction.
A vertical antenna sends out a signal sideways (horizontally) but with a vertical polarisation.
As the pilot in command you have calculated the position of the Critical Point on a proposed flight. A revised forecast has now become available that shows a strong headwind component in place of the
previously calm forecast winds.
The effect this will have on the time involved from the critical point is ?
the upwind leg will be longer, the downwind leg will be shorter
the upwind leg will be longer, the downwind leg will be longer
nil, it will remain the same
the upwind leg will be shorter, the downwind leg will be shorter
the upwind leg will be shorter, the downwind leg will be longer
Nil, it will remain the same
In CP calculations the effect of wind is to change the POSITION of the equal time point, but the ETI On or Home remains the same (and equal of course).
An advantage of a laser gyro over a mechanical gyro is ?
no error
not affected by perceived precession
higher RPM
no moving parts
No moving parts
An aircraft climbs at a constant 310 IAS from FL150 to FL250 over a distance of 50 nm with a wind
component of -30 kt and a temperature deviation of ISA+10. The average rate of climb between
FL150 and FL250 is ?
900 ft/min 1000 ft/min 1250 ft/min 1350 ft/min 1450 ft/min
1350ft/ min
First convert the 310 IAS into a TAS. The most realistic TAS is one corresponding to 310 IAS at a
level 2/3rds of the way up the climb (21700 ft).
Using 2/3rds of the climb as the level for the average TAS is more representative than the average
of the TAS at 15000 and 25000 ft, or even the TAS at half way since the climb performance of an
aircraft is not linear and reduces with altitude.
The easiest way to convert 310 IAS into a TAS is to first convert it to a Mach number using the CR3.
310 IAS at FL217 is a Mach number of 0.697M then use the temperature to compute the TAS.
ISA +10 at 21700 ft is -18°C (255K)
At this temperature, 0.697M is a TAS of 433 kt (V255K x 39 x.697)
With 30 kt headwind the groundspeed will be 403 kt.
The ETI to cover 50 nm at 403 kt is 7.44 minutes.
7.44 minutes to climb 10000 ft is 1344 ft/min
The number of satellites of a GPS constellation, required in order for a GPS receiver to correct for
receiver clock error is ?
3 4 5 6 2
4
The number of satellites of a GPS constellation, required in order for a GPS receiver to achieve RAIM is ?
3 4 5 6 8
5
Refer ERC L5.
You depart PARKES (33S148E) for OODNADATTA on G222. At 1017 UTC your position is on track
over KADUV with a heading of 280M and a TAS of 350 kt. At 1045 UTC your position is off track over WHITE CLIFFS aerodrome (31S143E). The average wind between the two positions is closest to ?
260M75 kt
340M/90 kt
240M/90 kt
140M/75 kt
240M/90 kts
1017 :KADUV
1045: WHITE CLIFFS
135 nm in 28 minutes = 289 kt Groundspeed
Track 292M HDG: 280M TAS: 350 kt ETAS:343 kt GS: 289 kt Drift 12° = 73 kt crosswind from left Wind comp = -54 kt headwind On CR3 - WV= 240M/90 kt
A flight departs BRISBANE (YBBN) at 2 am LST on August 31 for TONGA. The track distance is 1995 nm and the average groundspeed for the flight is 460 kt. LST YBBN is UTC+10 and LST TONGA is UTC+13.
The ETA for Tonga in LST is ?
300920
310920
300720
310020
310920
Depart BN 310200 LST
-1000 =301600 UTC departure BN
+0420 flight time (1995 nm at 460 kt is a flight time of 4 hours 20 minutes.)
= 302020 UTC arrival
+1300
= 310920 LST arrival Tonga
The LMT at a position on Longitude 060°W at 0700 UTC is ?
1100 LMT
0300 LMT
2300 LMT
1500 LMT
0300 LMT
LMT at 60° West Longitude is 4 hours behind UTC (60/15 = 4)
0700 UTC -0400 =0300 LMT
Refer to the following diagram of a Sine wave.
The letter which represents the dimension of wavelength is ?
B
Refer ERC L8.
You are on a flight en-route between CAIGUNA and NORSEMAN on B469 having obtained a fix over CAIGUNA at 0615 UTC at FL350, 0.82M, with a WV of 220M/90 kt, OAT -41°C. At 0619 UTC you hear another aircraft report SORT and tracking to NORSEMAN then CAIGUNA at FL330, reported
CAS 197 kt. The time you expect to pass the other aircraft is ?
0643 UTC
0640 UTC
0638 UTC
0635 UTC
0635 UTC
Other Aircraft Groundspeed: CAS 197 at FL330 = 0.57 M at -37 OAT (ISA +13) TAS = 341 kt ETAS = 335 kt Wind comp = +60 kt Groundspeed = 395 kt
Your Groundspeed:
0.82M at -41 OAT (ISA+13) TAS = 486 kt ETAS = 481 kt drift = 8° Wind comp = -60 kt Groundspeed = 421 kt
Your position at 0619:
CAIGUNA at 0615
4 minutes later at 421 kt = 28 nm past CAG = 219 nm from SORTU
219 nm between the two aircraft at 0619 will be covered at the combined rate of both groundspeeds.
219 nm/816 kt = .268 hours (16 minutes)
The two aircraft will pass each other 16 minutes after 0619 = 0635.
Refer to ERC L3. You are overhead JODEX (285/32 SCONE) tracking to QUIRINDI. You are experiencing 8 degrees left drift. The ADF is tuned to SCONE NDB. The ADF relative bearing to SCONE NDB is ?
073 081 089 106 114
073
Track to QUIRINDI: 025
Heading at JODEX: 033 (8 degrees to the right of 025 with left drift)
Track to Scone: 106- Heading at JODEX:033 = ADF Relative Bearing: 073
On a Lambert’s projection, straight lines approximate ?
rhumb lines
great circles
parallels of latitude
lines of constant bearing
Great circles
On a Polar Stereographic projection, great circles are approximately straight lines ?
near the equator
near the pole
only on East - West tracks
over the whole chart
Near the pole
On a Polar Stereographic chart, great circles which cross the pole are straight (meridians) but great circles which do not cross the pole are curved. The greater the distance from the pole, the more curved the great circle appears. The equator is a great circle and it appears curved because it does not cross the pole.
In the Northern Hemisphere, the great circle track bearing decreases when flying ?
North
South
East
West
West
On a Lambert’s projection, the distance scale is expanded ?
outside the standard parallels
between the standard parallels
North of the parallel of origin
South of the parallel of origin
outside the standard parallels
On a Lambert’s projection a line is drawn from a point at 26°S 153°E to 26°S 143°E. The standard parallels are 24°S and 34°S. The great circle track for this flight will be ?
North of the straight line track
South of the straight line track
on the straight line track
on the opposite side of the straight line track to the rhumb line
North of the straight line track
While on approach to land on the glideslope of an ILS, the glideslope CDI suddenly fluctuates. The probable cause is ?
unrelated to other aircraft in the vicinity
another aircraft that has just taken off on the same runway
another aircraft landing on an adjacent runway flying across your glidepath.
another aircraft taxiing which inadvertently operates it’s weather radar while pointing
at the GS transmitter
another aircraft landing on an adjacent runway flying across your glidepath.
An aircraft taking off ahead of you on the same runway may cause localiser fluctuations but is unlikely to affect the glideslope.
Weather radar is unlikely to affect the operation of ILS equipment.
Another aircraft on approach to an adjacent runway flying across your glidepath may cause glideslope fluctuations.
You are tracking inbound on the 270 radial, OBS selected to 090 and CDI centred. The RMI is tuned to an NDB co-located with the VOR you are using. The indication on the RMI is a track to the station of 088M. You are experiencing a strong left cross wind. The reason for the discrepancy is ?
drift due to the strong cross wind
precession error in the RMI
the fact that magnetic variation is applied at the station for VOR bearings and at the
aircraft for NDB bearings
scalloping of the NDB signal
the fact that magnetic deviation is applied at the station for VOR bearings and at the
aircraft for NDB bearings
the fact that magnetic variation is applied at the station for VOR bearings and at the aircraft for NDB bearings.
Wind has no influence on the track to a station. One reason a magnetic track can have a different bearing on either end is that the variation may be different at either end of the track. The VOR is
calibrated for the variation which exists at the VOR site, while an MI track to the station is determined according to the variation at the aircraft position.
On a Lambert’s conformal conical projection, a Great Circle track away from the parallel of origin appears ?
concave to the parallel of origin
concave to the nearer pole
straight
concave to the nearer standard parallel
Concave to the parallel of origin.
On a Mercator’s projection a straight line represents ?
a rhumb line
a great circle
an isogonal
the shortest path between two points
A rhumb line
On a Polar Stereographic projection, rhumb lines are ?
represented by straight lines near the pole, becoming concave to the pole at low latitude curved concave to the equator curved concave to the pole straight
curved concave to the pole

On a Polar Stereographic projection, great circles are ?
represented by straight lines at the pole, becoming concave to the pole at low latitude curved concave to the equator curved concave to the pole straight
represented by straight lines at the pole, becoming concave to the pole at low latitude.
An aircraft is maintaining a track of 330M past a VOR located 55 nm to the right of track. The aircraft heading is 340M due to 10° drift.
The OBS selections which would be used with a FROM indication to confirm the aircraft position
when abeam the VOR is ?
240
270
260
230
240
With the receiver located at FL390, the theoretical maximum range of a VOR situated at sea level is ?
240nm
You commence a flight sector with a track of 235M allowing for a forecast W/V of 320M/25 kt. Your
TAS is 180 kt. The actual W/V is 250M/40 kt.
After 25 minutes flight time, the actual position of the aircraft relative to your expected position is ?
Plot both winds on the CR3 with the original track of 235M and 180 TAS at the top. The forecast wind of 320M/25 kt resulted in 3 kt of headwind and 25 kt of crosswind or 8 degrees drift.
The actual wind during the 25 minutes was 250M/40 kt. Relative to the 235M track this wind results
in 38 kt of headwind and 10 kt of crosswind or 3 degrees drift.
In 25 minutes the aircraft has encountered 34 kt more headwind (38 - 4 =34) than expected. 34 kt
more headwind for 25 minutes will position the aircraft 14 nm before the planned position (34 x
25/60 = 14.
In the 25 minutes ETI the aircraft has encountered and 15 kt less crosswind from the right (10
instead of 25) than expected. 15 kt less crosswind for 25 minutes will position the aircraft 6 nm
right of the 235 M track (15 x 25/60 = 6).
A flight leaves BRISBANE (YBBN) for TONGA (NFTF) on 30 September at 0900 EST. LST at NFTF is
UTC+13. LST YBBN is UTC +10. The flight ETI is 4 hours 30 minutes.
The ETA at NFTF in LST as an eight figure Month/Date/time (MMDDhhmm) group is ?
09301630 LST
At a constant FL and Mach Number, if the temperature increases the IAS will ?
Stay the same
The relationship between IAS and Mach number only depends on flight level not temperature. A higher temperature would result in a higher True Air Speed (TAS) for any Mach number but the same IAS if the flight level stays the same. This can be see on the CR3. For example if you set 300 kt IAS (CAS) against FL300 in the CAS/Pressure Height window, the Mach pointer will indicate 0.79M regardless of the OAT.
On a Lambert’s Conformal chart the direct track is drawn between AUCKLAND (37015/17448E) and
HOBART (4251S/14731E). The initial true track from AUCKLAND is 247T and the convergence factor (n) of the chart is 0.515.
The final true track into HOBART is ?
261 T
Refer ERC L5.
You obtain a positive fix overhead LONGREACH en-route to MOUNT ISA on J89 at 0257 UTC.
Cruising level … FL280
TAS…—— 410 kts
WV……. 240M/60 kt
You have a Safe Endurance (excluding reserves) of 90 minutes. Your calculation of the location of
the PSD to TOWNSVILLE, as a distance from MOUNT ISA, is closest to ?
1 PSD 253 nm from MOUNT ISA
2 PSD 60 nm from MOUNT ISA
3 PSD 75 nm from MOUNT ISA
4 PSD 244 nm from MOUNT ISA
PSD 60 nm from MOUNT ISA
Try 60 nm MT ISA as a possible PSD position. Calculate the ETI to fly from LONGREACH to the PSD
then from the PSD to TOWNSVILLE. Add the two ETIs together and compare the total to the 90
minutes safe endurance.
If the total ETI equals 90 minutes (within about one minute) then the PSD is correct.
If the ETI is greater than 90 minutes the PSD is before 60 nm MT ISA.
If the ETI is less than 90 minutes, the PSD is after 60 nm MT ISA.
On a Polar Stereographic projection, meridians appear to be ?
Straight
When descending at a constant indicated airspeed, the true Mach number of an aircraft will ?
Reduce
The relationship between Mach number and IAS only depends on altitude (flight level). If the flight
level decreases, the Mach number for a constant IAS is lower.
Refer ERC L4.
You obtain a positive fix overhead LONGREACH en-route to TAROOM on J89 at 0215 UTC.
Cruise level: FL290 Mach Number ........... 0.82M WV .... 210M/80 kt SAT…..-33 deg C You calculation of the ETA for TAROOM as a 4 figure time group (hhmm)is closest to ?
0255 UTC
You are flying at FL250 with an OAT of -25 deg C and QNH 1013 HPA.
The true altitude of an aircraft is ?
higher than the flight level
When warmer than ISA, altimeter underreads.
Indicating FL250, the true altitude will be 26000 ft.
As an aircraft approaches an enroute ME station before overflying the aid, the DME ground speed and DME distance readout respectively will ?
for groundspeed, reduce at an increasing rate to zero overhead, then increase at a decreasing rate back to the correct value after station passage, and for distance, reduce at an decreasing rate to equal height above the aid in nm overhead, then increase at an increasing rate back to the correct value after station passage.
Refer ERC L5.
You depart PARKES at 0339 UTC for BROKEN HILL cruising via W428.
Cruise level: FL220
TAS: 245 kt
WV: 290M/70 kt
Your calculation of the location of the ETP (Critical Point) for GRIFFITH and BROKEN HILL,
measured as a track distance from BROKEN HILL, and your ETA at that point are closest to ?
1 ETP 100 nm from BHI and ETA 0503 UTC
2 ETP 120 nm from BHI and ETA 0505 UTC
3 ETP 135 nm from BHI and ETA 0515 UTC
4 ETP 142 nm from BHI and ETA 0514 UTC
ETP 100 nm from BHI and ETA 0503 UTC
For an aircraft flying at FL270, the maximum theoretical range of a VOR located at Mean Sea Level is ?
200nm
Refer to the following diagram of a Sine wave. The letter which labels the dimension of wave amplitude is ?
A
Refer to ERC L4.
You are planning a flight from TAROOM to LONGREACH on J89 and have the following inflight data:
Wind at cruise level : 260M/40 kt
TAS: 350 kt
Your calculation of the location along the J89 track of the EP for CHARLEVILLE and LONGREACH,
measured as a distance from LONGREACH, is closest to ?
1 125 nm
2 116 nm
3 107 nm
4 92 nm
116nm
You are to depart an aerodrome in the Southern Hemisphere at longitude 153E and fly to a destination at longitude 133E. You have the following data:
Convergence factor at this latitude: N= 0.75
Great circle track outbound from your departure aerodrome: 225T
The great circle track inbound to your destination is ?
240T
QUESTION ANAV 045
On a Lambert’s projection a line is drawn from a point at 19S/147 to 24S/134E. The outbound straight line track is 245M and the inbound straight line track is 253M.
To three decimal places, the convergence factor (n) for this straight line track is ?
.615
The track crosses 13 meridians. The track changes by 8 degrees (245 to 253) in the same distance.
The convergence factor (n) equals bearing change divided by longitude change = 8/13 = .615
On a Mercator’s projection, straight lines approximate ?
Rhumb Lines
A flight is arriving at BRISBANE (UTC = LST-10) with an ETD from departure point VANCOUVER (UTC = LST+8) of 08312030 LST. The ETI is 14 hours 20 minutes.
As an eight figure group in LST, the ETA for BRISBANE is ?
09020450
You are to depart an aerodrome in the Southern Hemisphere at longitude 173E and fly to a destination at longitude 162W. You have the following data:
Convergence factor at this latitude: n = 0.80
Great circle track outbound from your departure point: 125T
The rhumb line track to your destination is ?
115T
Flying East from 173E to 162W covers 25 meridians (7 up to 180 then another 18 down to 162W).
With a convergence factor (n) of .8, the bearing will change by 25 x .8 = 20 degrees between departure and destination. The departure track is 125°, the destination end track is 105° (East is least in the Southern Hemisphere) and the rhumb line track is the average of the twt = 115°.
Refer ERC L3.
You are on a flight from PORT MACQUARIE to WALGETT via TAMWORTH. You have the following
inflight data overhead TAMWORTH:
Safe endurance (excluding reserves): 40mins
TAS: 290 kts
Wind at cruise level from INS: 280M/40kts
Your calculation of the distance from WALGETT to the PSD for diversion to MORE is closest to ?
1 45 nm
2 35 nm
3 75 nm
4 15 nm
45 nm
Your flight has been forced to divert from the flight planned track of 180M around thunderstorms on
descent into Melbourne. Your present position is on the 015M radial inbound at 40 DME. ATS direct
you to make a 45 degree intercept to resume flight planned track to Melbourne. The additional distance incurred by this track, as compared to the direct track to YMML from your present position (to the nearest nm) is ?
3 nm
Refer ERC L3.
You tracking from SYDNEY to WALGETT on V316 when you obtain a position fix at 0612Z. Position
is on track 100 DME SYDNEY.
At 0628Z you obtain a second fix on the W811 track abeam waypoint ONUMA (approx 031S 148.5E). Your heading since 0612Z has been an average of 321M and TAS has been 450 kt.
The wind velocity affecting you since 0612Z has been ?
080M/57kts
Your inbound flight planned track to an aerodrome is 360M. Due to AT vectoring you deviate from
your flight planned track by 30 degrees left at 60 DME. 4 minutes later fix your position 35 DME
from your destination aerodrome on the 210 radial. At this time ATC dears you to track direct to the
aerodrome from your present position. The extra track miles this diversion incurred, compared to
your original flight plan (to the nearest nm), would be ?
10nm
You calculate the ETP on a track from ALPHA to FOXTROT using the following data:
Average track ……… 157M
Total distance …………… 225 nm
TAS……… 420 kt
Forecast wind …………… 230M/30 kt
If the actual wind is 280M/40 kt, when compared to the original position, the ETP will shift ?
8 nm closer to ALPHA
Refer ERC L6.
You tracking from GROOTE EYLANDT to ALICE SPRINGS. At 0342Z you obtain a positive radio fix from the following position lines:
IBUSU bearing 279M FROM the waypoint
ENRES bearing 346M FROM the waypoint
At 0418Z you fix your position 55 DME TENNANT CREEK on the 078 Radial from the TNK VOR.
Heading has been a constant 180M and TAS 350 kt since 0342Z.
Your calculation of the average wind velocity since 0342Z is ?
Your calculation of the average wind velocity since 0342Z is- 1 085M/55 kt 2 296M/50 kt 3 268M/50 kt 4 108M/50 kt
085M/55
On a Lambert’s conformal conical projection with standard parallels at 27S and 33S, a straight line
is drawn from departure at 25S/153E to destination at 34S/151E.
When crossing latitude 31S, the Great Circle track will be ?
Left of the straight line track.
A great cirde track is virtually a straight line on a Lambert’s chart however technically the true great
dirdle is slightly concave toward the parallel of origin (which is half way between the standard
parallels).
Refer ERC L3.
Your present position is overhead MOREE (S30 00 E149 00) on track TAMWORTH to ROMA via
W298 to ST GEORGE then W499 to ROMA. You have the following inflight data overhead MOREE:
TAS: 290kts
Safe Endurance: 55mins (excluding reserves)
Wind: 200M/50
Your calculation of the position of the PSD to OKEY, measured as a distance from ROMA, is closest to ?
1 50 nm from ROMA
2 20 nm from ROMA
3 110 nm from ROMA
4 80 nm from ROMA
80 nm from ROMA
While planning a flight from VANCOUVER to TOKYO you read a NOTAM imposing a curfew on arrivals at TOKYO restricting landings to between the hours of 6 am and 10 pm
daily TOKYO local time.
Your estimated flight time from VANCOUVER to TOKYO is 9 hours 45 minutes. VANCOUVER time is UTC-8 and TOKYO time is UTC+9.
The earliest and latest LST departure times at which you could depart VANCOUVER to arrive at TOKYO outside the curfew times on 29 August, respectively are ?
1 earliest 280315 and latest 281915
2 earliest 281115 and latest 290315
3 earliest 281815 and latest 291115
4 earliest 290645 and latest 292245
1 earliest 280315 and latest 281915
Refer ERC L2.
At 1832 UTC, a positive fix is obtained flying overhead NATYA (NYA) enroute to ADELAIDE (YPAD) on H247.
You have the following inflight data:
Cruise level: FL140
TAS: 250kts
Wind: 220M/50kts
Your calculation of the ETP for MILDURA and ADELAIDE, expressed as a distance from ADELAIDE, is closest to ?
1 85 nm from YPAD
2 94 nm from YPAD
3 75 nm from YPAD
4 108 nm from YPAD
3 75 nm from YPAD
Determine the nil wind ET by bisecting the track between the alternates (MILDURA and Adelaide)
then extending the bisector to the track you are on. This intersection marks the nil wind ETP location. Plot the wind effect by drawing the wind vector direction toward the nil wind ETP. The length of the vector is determined by the distance from the nil wind ETP divided by the TAS then
multiplied by the wind speed. From the origin of the wind vector, draw a line back to the original track parallel to the bisector - where it intersects the track is the adjusted ETP - 76 nm Adelaide in this case.
You are tracking 030M past a VOR station abeam and to the right of track. The VOR station is located 40 nm from the aircraft. You are holding 10 degrees left drift.
The OBS setting which would result in the CDI centred and a FROM indication when passing abeam
the station is ?
300
When tracking North (000) the abeam position from would be 270. Tracking 030 degrees means the
abeam position is 30 degrees right of 270. 270 + 30 = 300. The aircraft drift is irrelevant for a VOR.
On a Lamberts chart a track is plotted from S38 33 E113 05 to S29 45 E103 30 (a track running north west). The standard parallels for the chart are 20S and 35S.
In relation to the plotted track, the great circle track between these points would be ?
concave to the equator left of track
Refer GPWT extract.
Given:
Track: 090T
Mach number : 0.82M
GPWT Extract HPA 150: 2908556 200: 2808560 250: 2707556 300: 2707047 400: 2706531 500: 2703516
The highest groundspeed would be achieved at pressure level ?
400HPA
For weather avoidance purposes you divert left of track at a 45 degree angle for 5 minutes and then intercept your original track at a 30 degree angle. Groundspeed is a constant 420 kt.
Compared to the original track distance, the extra distance covered by the diversion (to the nearest
nm) is ?
17 nm
In a 45 degree triangle, the short sides are 70% of the long side. During the 5 minutes diversion the equivalent distance along track would have taken only 3.5 min (70% of 5 min).
In a 60 degree triangle the shortest side is 50% of the longest side. The shortest side is 3.5 min so the longest side will be 7 minutes. The other angle is 30 degrees. With a 30 degree angle, the other side is 86.6% of the long side. The longest side is 7 minutes so the other side will be 6 minutes
long.
The direct track is 3.5 + 6 minutes
= 9.5 minutes
The track around the diversion is 5 + 7 minutes
= 12 minutes
The diversion takes 2.5 minutes extra. At 420 kt (7 nm per minute), the extra distance covered is
17.5 nm (2.5 x 7 nm/min).
A Great Circle track is drawn from longitude 120E to 110E in the Southern Hemisphere.
Track inbound to destination: 305 degrees
Chart convergence factor (n): 0.78
The track outbound from departure (to the nearest degree) would be ?
297 degrees
