PLOTTING
QUESTIONS
1.
Overhead
Cookhouse HDG 030º (C), Compass DEV +2º East, Drift 10º Right, VAR. 22º West. Track Made Good is:
a.
000º T
b.
016º T
c.
020º T.
* c * l * a * 1 * Track Made Good * Use CDMVT
for this one, and you get a True Hdg of 010°. If you have corrected for 10° Right drift, this would have been 10° to the left. Drift is always from Hdg to Track so 010
Hdg = 10° R Dr = 020° Track. Answer C. *
2.
Airplane overhead
HARARE at FL070 and climbing to FL180 at a mean rate of climb of 600 ft/min.
RAS 160 kts, Temp DEV = ISA +10ºC. The
mean TAS on the climb is:
a.
193 kts
b.
197 kts
c.
202 kts.
* c * l * c * 2 * Mean TAS on climb * Climbing
at constant RAS needs 2/3 of the height change. Height change is 11000, so 2/3
is 7333 ft plus the 7000 at the start. The height at which TAS must be
calculated is therefore 14333 ft. The temp at that height in ISA should be
-13,4°C, so with a deviation of
+10 the temp to use is -3,4°C. Enter the info into the Pathfinder and you get a TAS of 202,3 kts.
Answer C. *
3.
An aircraft flies
from A to B in 2 hrs 20 min. Dist 588 nm, TAS 288 kts. In order to fly from B
to A in 2 hrs 27 min, the TAS for the return flight should be:
a.
240 kts
b.
252 kts
c.
204 kts.
* c * l * c * 2 * Required TAS * The flight
time of 2 h 20 m over a distance of 588 nm gives a GS of 252 kts. If TAS is 288
kts, this means that there is a headwind of 36 kts. On the return this will be
a tailwind. If the return journey is to be accomplished in 2 h 27 m the GS must
be 240 kts. If you have a TWC of 36 kts, the required TAS is 204 kts. Answer C.
*
4.
Airplanes A and B
are both flying on the same track and are initially 72 nm apart. GS of A is 288
kts. Airplane A passes B after 1 hr 20 min. The GS of B is:
a.
248,7 mph
b.
433,6 kph
c.
216,0 kts.
* c * m * b * 2 * Relative Velocity GS * This
is a manipulation of the closing speed formula, which is closing
distance/closing speed = time to close. Therefore 72/x = 1 h 20 m. 1:20x = 72,
so x = 72/1:20 = 54 nm. This means that B is slower by 54 kts. 288 - 54 = 234
kts. This makes C wrong. Converted to mph would give you 269,3, making A wrong
as well. Converting to kph gives you 433,4. Answer B. *
5.
An airplane
leaves A {S12:02 W157:04) on July 31st at 09:00 LMT on a flight to B
(S41:20 E174:48) at an average GS of 312 kts, and arrives at B at 11:19 LMT on
August 1st, having flown a distance of:
a.
1308 nm
b.
1414 nm
c.
725 nm.
* c * m * a * 2 * Distance flown *
Depart A: 09:00 LMT 31 Jul
Arc W157:04 10:28 West, so UT Best (add)
19:28 UT
Arrive B: 11:19 LMT 1 Aug
Arc E174:48 11:39 East, so UT least (subtract)
23:40 UT 31 Jul
The
difference between UT Depart and UT Arrive will be flight time, which is 23:40
- 19:28 = 4h 12m. At a GS of 312 kts this gives a distance flown of 4.2 x 312 =
1310,4 nm. Answer A. *
6.
On a Mercator chart
the scale is 1:2 000 000 at N35:00. The scale is equal to 1:1 000 000 at
Latitude:
a.
N65:49
b.
N40:35
c.
N33:33
* c * l * a * 1 * Scale on Mercator * Use
ABBA. Scale denominator at A x Cos Lat B - Scale at B x Cos A.. In this case 2
000 000 x X = 1 000 000 x Cos 35, or X = 1 000 000 x Cos 35/2 000 000, or
0,409576. Inverse Cos = 65:49:18. Answer A. *
7.
Aircraft A, GS
425 kts, passes over NL at 13:45Z. Aircraft B, at FL310, OAT ‑40º C, passes
over NL at 13:57Z, overtaking A at 15:39Z. Aircraft B relative closing speed
was:
a.
50 kts
b.
75 kts
c.
95 kts.
* c * l * a * 1 * Closing speed calculation *
At 1357 aircraft A would have flown a distance of 85 nm since 1345 (12 min at
425 kts). This is the required closing distance. The time to overtake is 1h 42m
(1.7h). Substitute these into the formula: Closing distance/closing speed =
time, so 85/X = 1.7, or 1.7 x X = 85, so X = 85/1.7 = 50. Answer A. *
8.
A to B, Distance
159 nm, FL130, OAT‑10º C. Tailwind 25 kt. The required RAS to fly from A to B
in 46 min is:
a.
150 kts
b.
170 kts
c.
190 kts.
* c * l * a * 1 * Required RAS calculation *
To fly 159 nm in 46 min will require a GS of 207,4 kts. With a tailwind of 25
kts, the TAS will be 182,4 kts. Use the REQ CAS page on the Pathfinder and
enter the relevant information at the prompts. You will get a Req CAS of 149,8
kts. Answer A. *
9.
Forecast wind
velocities for a climb to cruise altitude are: W/V 270/20; W/V 315/30; W/V
360/35. The mean climb W/V is:
a.
324/24
b.
315/28
c.
330/30
* c * l * a * 1 * Mean wind calculation *
Either plot the 3 vectors on a sheet of paper, or use the Hdg/TAS page on the
Pathfinder. Enter the first 2 winds, and you get a THdg of 297,2 and a TAS of
46,4. Jot these down, and then enter them again in the top 2 spaces. Then enter
the third wind and the result is 323,7/69.7. remember that there are 3 winds,
so divide 69,7 to get the wind speed (23,23) and the direction will be 323,7.
Answer A. *
10.
Standard time of
sunrise at PARIS, FRANCE (N48:44 E002:22) on 17 February is:
a.
07:56
b.
07:05
c.
06:56
11.
An airplane at DR
position (S33:55 E028:15) Var 22°, obtains a RMI QDM of 310° from an NDB (S29:30
E020:30) Var 19°W. The bearing to plot from the NDB on a Lambert’s Chart (CCF 0,5) is:
a.
108°T
b.
112°T
c.
115°T
* c * l * b * 1 * Bearing to plot * The QDM of
310 becomes a QUJ of 288. The reciprocal of 288 is 108, but chart convergency
has to be added before it can be plotted from the station. CC = Ch Long x CCF,
or 7,75 x 0,5 = 3,875, or 4°. 108 + 4 = 112 QTE. Answer B. *
12.
Position of A is
S31:35 E029:46. Position B is on the same parallel of latitude. The great
circle bearing of B from A is 262°. The longitude of B is:
a.
E000:47
b.
E000:24
c.
W000:47
* c * l * c * 1 * Longitude calculation * If
the two places are on the same longitude, the rhumb line track will be 270°. Ca is therefore 8°, so convergency is 16°. Use the formula: C = Ch Long x Sin Lat, or Ch Long =
C/Sin Lat. So X = 16/Sin 31:35 = 30° 32' 58" (30,5496). This is to the west of
position A so E029:46 - 30:33 = - 0:47, which puts position B west of the
Greenwich meridian, or W000:47. Answer C. *
13.
A route facility chart has a scale of 1:2 600
000. One cm on the chart is equal to:
a.
12 nm
b.
14 nm
c.
16 nm.
* c * l * b * 1 * Earth distance * A chart
length of 1 cm on a chart with a scale of 1 : 2 600 000 will give an earth distance
of 2 600 00 cm. Convert this to nm, using 2,54 cm to the inch, 12 in to the
foot, and 6080 ft to the nm, or use a figure of 185300 cm to the nm. Either way
you get just over 14 nm. Answer B. *
14.
13:00 Overhead SISHEN NDB SS (S27:42 E023:00), TAS 230 kts.
14:00 Overhead VICTORIA WEST VOR VWV (S31:24 E023:09).
At 14:00 the aircraft returns to SISHEN. The ETA at SS
is:
a.
14:49
b.
14:55
c.
15:00
* c * l * b * 1 * ETA calculation VWY - SS *
The distance on the chart between the two places is about 219 nm. This flown in
one hour will also be the GS. If TAS was 230, then there was a HWC of 11 kts.
This will be a TWC on the return, so the GS return will be 230 + 11 = 242.
219/241 = 54m 31s, or 55 min. ETA will be 1400 + 55 = 1455. Answer B. *
15.
09:43Z VICTORIA WEST (S31:24 E023:08) set heading 089º(T) for NEWCASTLE
(S27:45 E030:00), FL110. TAS 160 kts.
11:19Z BLOEMFONTEIN VOR BLV (S29:06 E026:19) Radial 110, DME range 17.
ETA for NEW CASTLE is:
a.
12:42
b.
13:04
c.
12:56
16.
On track between
ALEXANDER BAY (28:34 E016:31) and UPINGTON (S28:24 E021:31). ALEXANDER BAY
radio requires report abeam KEETMANSHOOP VOR KTV (S26:33 E018:00). Variation at
KTV is 17,5°W. When abeam KTV the
radial is:
a.
196°
b.
178°
c.
220°
17.
At 10;48z
aircraft is heading 169°(M) inbound to EAST LONDON ELV (S33:02 E027:53). The ELV VOR/DME indicates radial 349° DME 12 nm. At this point a diversion to PORT
ELIZABETH PEV (S33:58 E025:36) is
planned. Assume no wind and variation 23°W. The distance and initial heading is:
a.
130 nm, 240°M
b.
126 nm, 262°M
c.
130 nm, 251°M
18.
While en route
from GEORGE VOR GGV (S34:00 E022:22) to SIR LOWRY’S PASS NDB SP (S34:09 E018:58), GS 120 kts, a relative bearing of
083° is obtained from
SUTHERLAND NDB SL (S32:24 E020:40). From this position the flight time to SP NDB is:
a.
51 min
b.
39 min
c.
43 min.
19.
Aircraft overhead
EAST LONDON at 09:30Z en route to DURBAN, mean track 041°T, distance 246 nm, mean W/V 270/45, FL110, OAT -7°C. The RAS that has to be maintained in order to
arrive overhead DURBAN at 10:30Z is:
a.
186 kts
b.
195 kts
c.
204 kts.
* c * m * a * 2 * Required RAS EL - DN * The
distance between EL and DN is about 245 nm, and since the flight time is one
hour, this is the required GS. To find RAS we need TAS which we don’t have, so
use the answer options as the come. A RAS of 186 kts, with OAT and P Alt will
give a TAS of 218,7 kts. With W/V and Track this will give a GS of 245,6 kts.
Options B and C will give a much higher TAS, and therefore higher GS, so they
are wrong. Answer A. *
20.
15:15 Overhead SELEBI-PIKWE SP (S22:03 E027:48) TAS 240 kts, heading 280°M.
16:30 Overhead GHANZI GZ (S21:41 E021:40). The
mean W/V:
a.
125/44
b.
058/46
c.
236/44
* c * l * a * 2 * Mean Wind SP - GZ * The
distance between the two places is 340 nm and the measured required track is
273°T. GS will be 340.1:15 =
272. True heading will be 280°M - 12° W Var = 268°T. Enter the four values into the Unknown Wind page. Answer A is
closest. *
21.
11:21 Overhead GHANZI GZ (S21:41 E021:40), FL180,
TAS 170 kts, W/V 250/55 on direct track to NIEUWOUDTVILLE NVV (S31:21 E019:03).
ETA at CAPE TOWN FIR (S27:30) is:
a.
16:36Z
b.
13:50Z
c.
14:02Z
22.
13:42Z Overhead COOKHOUSE (S32:46 E025:46) on a
direct track to LICHTENBURG LG (S26:10 E026:10).
14:49Z VOR KMV, radial 106°, DME 35 nm.
At 14:49Z the alteration of heading required to arrive
overhead LICHTENBURG is:
a.
10° right
b.
14° right
c.
19° right
23.
08:15 Overhead WALVIS BAY NDB RK (S23:00 E014:37), TAS 240 kts, heading 017°(T).
08:49 Alter heading 053°(T).
09:29 Overhead GROOTFONTEIN NDB (S19:37 E018:06).
The mean W/V is:
a.
276/36
b.
309/31
c.
354/37
24.
Two airplanes
depart at 13:00Z from aerodromes 400 nm apart. Airplane A, Track 063° (T), GS 188 kts. Airplane B, Track 243°(T), GS 155 kts. Time at which they cross will be:
a.
14:10Z
b.
14:05Z
c.
14:00Z
* c * l * a * 1 * Time to cross * Closing time
= distance/closing speed, or X = 400/(188 + 155) = 1,166h, or 1h 10m. 13:00 +
1:10 = 1410. Answer A. *
25.
An aircraft
departs from N60:00 E005:00 on a rhumbline track 270°T. After flying 265 nm it arrives at:
a.
E013:50
b.
E003:50
c.
W003:50
* c * l * c * 1 * New longitude * Departure
formula: Dep (in nm) = Ch Long (in minutes) x Cos Lat, or Ch Long = Dep/Cos
Lat, or 265/Cos 60 = 530 minutes/60 = 8° 50'. On track of 270° this will take you 3° 50 past the Greenwich meridian, or W000:50. Answer C.
*
26.
From EUROPE ISLAND
(S22:20 E040:21) to CAPE TOWN (S33:58 E018:36), departure time 01:30Z 2nd
January. ETA CAPE TOWN 17:30 SAST 2nd January. The total distance is
4850 statute miles. The average groundspeed is:
a.
301 kts
b.
346 kph
c.
346 kts.
* c * l * a * 2 * Average groundspeed *
Departure is at 0130 UT. Arrival is at 17:30 SAST. Convert this to UT by
subtracting the STF (East so UT least). SA STF = 2:00, so ETA is 15:30 UT.
Flight time is ETA - ATD = 14 hours. The distance flown is 4850 sm, so convert
to nm = 4214,5 nm. 4215/14 = 301 kts. Answer A. *
27.
From CLAN
WILLIAM NDB CW to KIMBERLEY VOR KMV,
distance 364 nm. At 116 nm from CW the airplane is 11 nm right of planned
track. To fly direct to KMV from this position, heading must be changed:
a.
6° left
b.
8° left
c.
12° left.
28.
On a MERCATOR
chart:
a.
Distance is
measured along a straight line because that is the shortest distance between
two points
b.
The shortest
earth distance between two points can be represented by a curved line
c.
Parallels are
concave to the equator.
* t * l * b * 1 * Chart characteristics * A is
wrong as a straight line on a Mercator is a rhumb line and therefore not the
shortest distance. C is wrong as parallels are parallel to the equator. A
curved line is a great circle and is the shortest distance between two points.
Answer B. *
29.
The great circle
distance between position A (N68:00
E118:32) and position B (N68:40
W061:28) is:
a.
2840 nm
b.
2640 nm
c.
2440 nm
* c * l * b * 1 * Great circle distance * B is
on the anti-meridian of A, as the two longitudes add up to 180. The shortest
distance between them will be the great circle route via the pole. 22° up from N68:00 to the NP, then 22° down to N68:00 gives a total of 44°. At 60 nm per degree this is 2640 nm. Answer B. *
30.
An airplane,
groundspeed 472 kts, is approaching checkpoint CHARLIE, ETA 2117. ATC request
the airplane to cross CHARLIE at 2120. To accomplish this, the airplane will
reduce its goundspeed to 415 kts at:
a.
20:52Z
b.
20:55Z
c.
20:58Z
* c * l * b * 1 * Reducing groundspeed * The
formula is Delay (minutes) x New GS x Old GS/Difference in GS x 60, or (3 x 472
x 415)/(57 x 60) = 171,8 nm. Divide this by Old GS and you get 22 minutes.
21:17 - 00:52 = 20:55. Answer B.
31.
Sector Track Distance TAS Drift Time
A to B 209°T 107
nm 230 kts 7° left 29
min
B to C 267°T 96
nm 230 kts
Assuming the W/V remains constant, the groundspeed
from B to C is:
a.
190 kts
b.
202 kts
c.
214 kts.
* c * m * b * 2 * Groundspeed calculation *
Find the GS for A to B, 107 nm/29 min = 221,4 kts. The True heading A to B is
209° T Hdg + 7° left drift = 216°T. Enter the information on the Unknown Wind page. W/V
= 285,2/28,9. On Hdg/GS enter to figures for B to C. GS = 202,4 kts. Answer B.
*
32.
An aircraft
passes overhead NDB A at 10:00Z, TAS 180 kts, heading 157°M. NDB A indicates a relative bearing of 186° when aircraft passes over NDB B, 49 nm from A at
10:20Z. Variation through out is 8° West. The wind direction and speed is:
a.
229/16
b.
110/24
c.
125/37
* c * m * c * 2 * Calculation of W/V * Convert
heading 157°M to true 157 - 8 = 149°T. A relative bearing of 186 at B indicates that there
is 6° of drift to the right,
hence the correction to left. This makes the track 149 + 6 = 155°T. 49 nm in 20 minutes gives a GS of 147 kts. Enter
the details into the Unknown Wind page and you get a W/V of124,6/37,1. Answer
C. *
33.
A LAMBERT’S chart
has the parallel of origin at S27:00. The difference between earth convergence
and chart convergence at S34:00 is:
a.
0,56 degrees
b.
0,45 degrees
c.
0,11 degrees
* c * l * c * 1 * Earth convergency vs CC *
The chart convergency is Sin 27, or 0,4539905. Earth convergency at S34 is Sin
34, or 0,559929. The difference between the two is 0,105, or 0,11 degrees.
Answer C. *
34.
An airplane is
maintaining a track of 127°(T), heading 151°(M), Var 16° W. The heading to fly the reciprocal track is:
a.
315°M
b.
323°M
c.
331°M
* c * l * a * 1 * Reciprocal heading to steer
* Convert heading 151°M to true, 151 - 16° W Var = 135°T. this gives a drift of 8° Left. On the return the drift will be 8° right, and the required track the reciprocal of 127,
namely 307°T. add Var 16 W and Mag
Track is 323°M. Apply 8° right drift and required heading is 315°M. Answer C. *
35.
An airplane at FL 330 is request to commence
descent at 60 nm from a VOR/DME and to cross the VOR at FL 070. The mean
groundspeed during descent is 277 kts. The minimum rate of descent required is:
a.
1800 ft/min
b.
1900 ft/min
c.
2000 ft/min
* c * l * c * 1 * Rate of Descent required *
60 nm covered at a GS of 277 kts will take 13 min. during that time 26000 ft
must be lost, so 26000/13 = 2000 ft/min. Answer C. *
36.
The length of the parallel of latitude on a chart
between (N49:14 E175:53) and B (N49:14 W168:42) is 24,87 cm. The scale of the
chart expressed as a representative fraction is:
a.
1 : 4 500 000
b.
1 : 6 900 000
c.
1 : 5 700 000
* c * m * a * 2 * Scale calculation * The
Change Longitude between the two places is 15°25'. But this is at N49:14, so you need the departure
formula to calculate the earth distance. (15°25' x 60) x Cos 49:14 = 604 nm. Chart length is 24,87
cm. Convert ED to cm and divide by 24,87. Answer A is closest. Answer A. *
37.
The standard time
of sunrise at OSLO, NORWAY (N59:45
E010:37) on 22nd February is:
a.
06:37
b.
07:20
c.
07:37
38.
11:47Z Overhead COOKHOUSE. (S32:46 E025:45), TAS 220
kts, Hdg 005° (T).
12:10Z Observed drift 14° right.
12:21Z Crossing BURGERSD()RP VOR BDV (S30:59
E026:17) radial 125.
Mean W/V is:
a.
305/55
b.
315/65
c.
295/55
39.
09:20Z Overhead SISHEN (S27:42 E023:00) set heading 094°M, FL 110, OAT 0°C, RAS 108 kts.
10:37Z Overhead WELKOM (S28:00 E026:42).
10:43Z Using W/V 330/55 heading for KROONSTAD
(S27:40 E027:I6) is:
a.
353°M
b.
011°M
c.
011°T
40.
An airplane
overhead BLOEMFONTEIN (S27:05 E026:18)
at 0630Z on a direct track to GEORGE (S34:00 E022:22), heading 247°M. At 0750Z VICTORIA WEST NDB bears 090° relative by ADF. The ETA overhead GEORGE is:
a.
0842Z
b.
0849Z
c.
0855Z
41.
1510Z Overhead EAST LONDON ELV (S33:02 E027:53), heading 010°T, TAS 180 kts.
155,5Z Alter heading onto 045°T, TAS 200 kts.
1634Z Overhead PIETERMARITZBURG NDB PM
(S29:40 E030:24).
The mean W/V that has affected the airplane since
1510Z is:
a.
257/35
b.
282/29
c.
328/21
42.
0630Z Overhead HARTEBEESPOORTDAM HBV (S25:42
E027:53), TAS 150 kts, W/V 240/35. On a direct track to BULAWAYO VBU (S20:05)
E028:42). The ETA at the HARARE FIR boundary (S21:38 E028:29) is:
a.
0757
b.
0751
c.
0745
43.
0945Z Overhead NIEUWOUDTVILLE (S31:21 E019:03), FL270, TAS 340 kts, on a direct
track to GHANZI (S21:41 E021:39).
1018Z UPINGTON NDB UP (S28:23 E021:16) QDM is 124
1032Z UPINGTON VOR UPV radial 358
The ETA GHANZI is:
a.
1122Z
b.
1128Z
c.
1134Z
44.
0900Z Overhead CAPE TOWN (S33:58 E018:35) on a direct track to WINDHOEK WH
(S22:29 E017:27), heading 349°T, TAS 460 kts, GS 420 kts.
0933Z ALEXANDER BAY ABV (S28:34 E016:31) DME range 116 nm.
0950Z ALEXANDER BAY ABV DME range 116 nm.
Heading to steer at 0950Z for WINDHOEK is:
a.
330°T
b.
335°T
c.
341°T
45.
1500 Overhead SUTHERLAND VOR SLV (S32:23 E020:40), TAS 270 kts.
1600 Overhead
ULCO VOR UCV (S28:23 E024:20).
At 1600 airplane returns to SUTHERLAND. ETA SLV is:
a.
1700
b.
1710
c.
1719
46.
1000 Overhead BEIRA (S19:46 E034:55) on a flight to HARARE (S15:55 E031:05).
1010 Overhead MUTARE (S19:00 E032:26).
At 1010 the alteration of heading required to arrive
overhead HARARE is:
a.
9° right
b.
13° right
c.
22° right
47.
The great circle
distance from A (N78:13 E156:27) to B (N69:27 W023:33) is:
a.
1827 nm
b.
1940 nm
c.
2175 nm
* c * l * b * 1 * Great circle distance * B is
on the anti-meridian of A, as the two longitudes add up to 180. The shortest
distance between them will be the great circle route via the pole. 11° 47' up from N78:13 to the NP, then 20° 33' down to
N69:27 gives a total of 32°20'. At 60 nm per degree this is 1940 nm. Answer B. *
48.
X to Y distance
107 nm, FL120, OAT +5°C, headwind 15 kts. The required RAS to fly from X to Y in
37 minutes is:
a.
129 kts
b.
140 kts
c.
154 kts
* c * l * c * 1 * Required RAS * 107 nm in 37
minutes = 173,5 kts GS. With a HWC of 15 kts this would require a TAS of 188,5
kts. Use the Req CAS page, and you get an answer of 153,6 kys. Answer C. *
49.
Standard time of
sunset at LISBON, PORTUGAL (N38:45 W009:07) on 10 March is:
a.
1738
b.
1838
c.
1938
50.
An airplane
maintaining a track of 141°M at a groundspeed of 180 kts
obtains the following bearings from UPINGTON VOR UPV (S28:24 E021:16).
0917Z UPINGTON VOR radial 277
0934Z UPINGTON VOR radial 266
0949Z UPINGTON VOR radial 185
a.
S29:29 E021:33
b.
S28:55 E020:24
c.
S29:48 E021:40
51.
An airplane is
overhead BLOEMFONTEIN (S29:05 E026:18)
at 0630Z on a direct track to GEORGE airport (S34:00 E022:22), heading 247°M. At 0750Z VICTORIA WEST NDB VW bears 090° relative by ADF. The ETA overhead GEORGE is:
a.
0842Z
b.
0849Z
c.
0855Z
52.
An airplane at
FL330, groundspeed 480 kts is 376 nm from it's destination at 1735Z. Mean rate
of descent is 2000 feet/min, mean descent groundspeed 285 kts. The latest time
to commence the descend to arrive overhead the destination at 6000 feet is:
a.
1814Z
b.
1822Z
c.
1827Z
* c * m * a * 2 * Latest time to descend * The
height to be lost in the descent is 33000 - 6000 = 27000. At 2000 fpm this will
require 13,5 minutes. At a GS of 285 kts this will cover a distance of 64,1 nm.
The remaining cruise distance is 376 - 64,1 = 311,9 nm. At a GS of 480 kts this
will require 39 minutes. 17:35 + 00:39 = 1814Z. Answer A. *
53.
Aircraft flies
the direct track from VICTORIA WEST VWV (S31:24
E023:08) to PORT ELIZABETH PEV (S33:58
E025:36), track 163°M, heading 170°M. When abeam the COOKHOUSE NDB CH (S32:44 E025:45) the ADF will
indicate:
a.
263 relative
b.
270 relative
c.
277 relative
54.
0630 Overhead BULAWAYO VBL (S20:05 E028:42),
TAS 150 kts, W/V 240/35, on a direct track‑to HARTEBEESPOORTDAM HBV
(S25:42 E027:53) the ETA at FAJS FIR boundary (S27:38 E028:29) is:
a.
0701
b.
0707
c.
0713
55.
0645 Overhead GHANZI GZ (S21:41 E021:39) FL280, TAS 340 kts, on a direct
track to NIEUWOUDTVILLE NVV (S31:21 E019:03).
0748 UPINGTON VOR UPV (S28:23 E021:16) radial 358.
0803 UPINGTON NDB UP QDMI 124
The ETA NVV is:
a.
0836Z
b.
0843Z
c.
0850Z
56.
1037Z Overhead DURBAN VOR DNV (S29:56 E031:00) on a direct flight to PHALABORWA
(S23:56 E031:07).
1049Z MAPUTO VOR VMA (S25:52 E032:34) radial 281, DME 65 nm.
At 1049Z the alteration of heading required to arrive
overhead PHALABORWA is:
a.
5° left
b.
11° left
c.
17° left
57.
0932 Overhead SUTHERLAND VOR SLV (S32:23
E020:40) heading 059°T, TAS 2'60 kts.
1000 Overhead VICTORIA WEST VOR VWV
(S31:24 E023:08), The mean W/V is:
a.
278/50
b.
206/55
c.
305/45
58.
0730Z Overhead GROOTFONTEIN NDB GF (S19:36 E018:06), TAS 210 kts, heading 158°T.
0805Z Alter heading 107°T.
0845Z Overhead GHANZI NDB GZ (S21:41 021:40). The
mean W/V since 0730Z is:
a.
173/34
b.
215/42
c.
247/42
59.
1927 Overhead DIAZ POINT (S26:40 E014:37) heading
037°T, TAS 252 kts, W/V
195/60.
2009 Alter heading for WALVIS BAY (S23:00
E014:37)
The heading to steer for WALVIS BAY at 2009 is:
a.
286°M
b.
293°M
c.
300°M
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