Worksheetsgnav 101-150
Total questions: 50
Worksheet time: 25mins
Which of the aeronautical chart symbols indicates a VOR/DME?
2
6
1
7
What is the meaning of aeronautical chart symbol No. 15?
Aeronautical ground light
Lighthouse
Hazard to aerial navigation
Visual reference point
What is the meaning of aeronautical chart symbol No. 16?
Shipwreck showing above the surface at low tide
Off-shore helicopter landing platform
Off-shore lighthouse
Lightship
A pilot receives the following signals from a VOR DME station: Radial: 180° ± 1°; Distance: 200 NM. What is the approximate error?
± 1 NM
± 3.5 NM
± 7 NM
± 2 NM
At 10:00 hours an aircraft is on the 310° radial from a VOR/DME, at 10 nautical miles range. At 10:10 the radial and range are 040°/10 NM. What is the aircraft's track and ground speed?
080° / 85 knots
085° / 85 knots
085° / 90 knots
080° / 80 knots
An aircraft is over position HO (55°30'N 060°15'W), where YYR VOR (53°30'N 060°15'W) can be received. Magnetic variation is 31°W at HO and 28°W at YYR. What is the radial from YYR?
028°
332°
031°
208°
At 0020 UTC an aircraft is crossing the 310° radial at 40 NM of a VOR/DME station. At 0035 UTC the radial is 040° and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speed are:
090° - 232 kt
085° - 226 kt
088° - 232 kt
080° - 226 kt
What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1) to position N5430 W00900?
214° - 26 NM
358° - 36 NM
169° - 35 NM
049° - 45 NM
What is the radial and DME distance from BEL VOR/DME (N5439.7 W00613.8) to position N5410 W00710?
223° - 36 NM
320° - 44 NM
236° - 44 NM
333° - 36 NM
Which statement is correct about the apparent solar day?
The duration of the apparent solar day is constant throughout a year due to the constant velocity of the earth in its orbit around the sun.
The duration of the apparent solar day is constant throughout a year due to the constant rotational speed of the earth around its axis.
The apparent solar day is the period between two successive transits of the mean sun through the same meridian.
The apparent solar day is the period between two successive transits of the true sun through the same meridian.
For 1st February the Air Almanac lists the following data: Latitude: 66°00'N; Morning civil twilight: 07:56; Sunrise: 09:00; Sunset: 15:28; Evening civil twilight: 16:32. The duration of morning twilight at 66°00'N is:
7 hours 56 minutes and starts at 09:00 UTC.
1 hour 4 minutes and starts at 09:00 UTC.
1 hour 4 minutes and starts at 07:56 LMT.
8 hours 36 minutes and starts at 07:56 UTC.
Position Elephant Point is situated at (58°00'N, 135°30'W). Standard time for this location is listed in the Air Almanac as UTC −8. If sunset occurs at 00:57 UTC on 21st January, what is the time of sunset in LMT?
15:55 on January 20th.
16:57 on January 20th.
09:59 on January 21st.
08:57 on January 21st.
On 27 Feb, at 52°10'S 040°00'E, the sunrise is at 02:30 UTC. On the same day, at 52°10'S 035°00'W, the sunrise is at:
05:10 UTC
02:30 UTC
21:30 UTC
07:30 UTC
Daylight Saving Time:
is used to extend the sunlight period in the evening.
is used in some countries.
is introduced by setting the standard time forward by one hour.
all answers are correct.
Position Elephant Point is situated at (58°00'N, 135°30'W). Standard time for this location is listed in the Air Almanac as UTC −8. If sunset occurs at 00:57 UTC on 21st January, what is the time of sunset in LMT?
15:55 on January 20th.
09:59 on January 21st.
16:57 on January 20th.
08:57 on January 21st.
The countries having a standard time slow on UTC:
will generally be located at westerly longitudes.
will often experience sunrise earlier than the sunrise occurs at the Greenwich meridian.
will generally be located at easterly longitudes.
will often have an earlier standard date than the UTC date.
Refer to almanac: The UTC of sunrise at (66°48'N, 095°26'W) on 27th of January is:
1541 UTC
0927 UTC
1549 UTC
0814 UTC
On 4th February the Air Almanac lists 19:41 as the time of sunset at 50°00'S. An observer registers sunset at 21:13 UTC this day. What is the observer's position?
50°00'S 022°00'E
50°00'S 010°35'W
50°00'S 010°35'E
50°00'S 023°00'W
You have calculated point of no return (PNR) on a flight, having all negative WCs in the flight plan. During the flight you experience that the W/V is stronger but coming from the same direction as in the flight plan. Consider the following statements.
the PNR will not change because neither TAS nor fuel flow has changed.
a recalculated PNR will move toward the place of departure.
the PNR will, if recalculated, move toward the no-wind PNR.
you will arrive at the PNR at a later time than flight planned.
The distance from A to B is 2.368 NM. If outbound ground speed is 365 kts and homebound ground speed is 480 kts and safe endurance is 8 hrs 30 min, what is the time to the PNR?
290 minutes.
190 minutes.
219 minutes.
209 minutes.
Given: Distance A to B: 1.973 NM; Ground speed OUT: 430 kts; Ground speed BACK: 385 kts. The time from A to the point of equal time (PET) between A and B is:
130 min.
162 min.
181 min.
145 min.
Given: Distance A to B: 2.484 NM; Mean ground speed OUT: 420 kts; Mean ground speed BACK: 500 kts; Safe endurance 08 hrs 30 min. The distance from A to the point of safe return (PSR) A is:
1.940 NM
1.736 NM
1.630 NM
1.908 NM
An aircraft was over 'A' at 1435 hours flying direct to 'B'. Given: Distance 'A' to 'B' 2900 NM; True airspeed 470 kt; Mean wind component 'out' +55 kt; Mean wind component 'back' -75 kt; Safe endurance 9 HR 30 MIN. The distance from 'A' to the Point of Safe Return (PSR) 'A' is:
1759 NM
1611 NM
2844 NM
2141 NM
An aircraft is planned to fly from position 'A' to position 'B', distance 480 NM at an average GS of 240 kt. It departs 'A' at 1000 UTC. After flying 150 NM along track from 'A', the aircraft is 2 MIN behind planned time. Using the actual GS experienced, what is the revised ETA at 'B'?
1153
1203
1157
1206
Complete line 1 of the 'FLIGHT NAVIGATION LOG'; positions 'A' to 'B'. What is the HDG°(M) and ETA?
282° - 1128 UTC
268° - 1114 UTC
268° - 1128 UTC
282° - 1114 UTC
Complete line 2 of the 'FLIGHT NAVIGATION LOG', positions 'C' to 'D'. What is the HDG°(M) and ETA?
HDG 193° - ETA 1249 UTC
HDG 188° - ETA 1229 UTC
HDG 183° - ETA 1159 UTC
HDG 193° - ETA 1239 UTC
Complete line 3 of the 'FLIGHT NAVIGATION LOG', positions 'E' to 'F'. What is the HDG°(M) and ETA?
HDG 095° - ETA 1155 UTC
HDG 105° - ETA 1205 UTC
HDG 115° - ETA 1145 UTC
HDG 106° - ETA 1215 UTC
Complete line 4 of the 'FLIGHT NAVIGATION LOG', positions 'G' to 'H'. What is the HDG°(M) and ETA?
HDG 354° - ETA 1326 UTC
HDG 344° - ETA 1336 UTC
HDG 344° - ETA 1303 UTC
HDG 034° - ETA 1336 UTC
Complete line 5 of the 'FLIGHT NAVIGATION LOG', positions 'J' to 'K'. What is the HDG°(M) and ETA?
HDG 320° - ETA 1412 UTC
HDG 337° - ETA 1322 UTC
HDG 337° - ETA 1422 UTC
HDG 320° - ETA 1432 UTC
Complete line 6 of the 'FLIGHT NAVIGATION LOG', positions 'L' to 'M'. What is the HDG°(M) and ETA?
HDG 075° - ETA 1452 UTC
HDG 064° - ETA 1449 UTC
HDG 070° - ETA 1459 UTC
HDG 075° - ETA 1502 UTC
Given: TAS = 472 kt; True HDG = 005°; W/V = 110°(T)/50k. Calculate the drift angle and GS.
6°R/490 kt
6°L/490 kt
6°R/462 kt
6°L/402 kt
Given: True Track 245°; Drift 5° right; Variation 3° E; Compass Hdg 242°; Calculate the deviation.
1° E
5° E
5° W
11° E
Given: TAS: 150 kts; Actual HDG: 270°; Wind: 245/12 kts. What is the wind correction angle?
2° to the left.
2° to the right.
4° to the left.
4° to the right.
Given: TAS: 470 kts; True HDG: 317°; W/V: 045°(T)/45 kts. Calculate the drift angle and GS.
3°R - 470 kts.
5°L - 470 kts.
5°R - 475 kts.
5°L - 475 kts.
Given: TAS: 270 kts; True HDG: 270°; Actual wind: 205° (T)/30 kts. Calculate the drift angle and GS.
6°L - 256 kts.
6°R - 259 kts.
8°R - 259 kts.
6°R - 251 kts.
An aircraft equipped with an Inertial Navigation System (INS) flies with INS 1 coupled with autopilot 1. Both inertial navigation systems are navigating from way-point A to B. The inertial systems' Central Display Units (CDU) shows: - XTK on INS 1 = 0; - XTK on INS 2 = 8L (XTK = cross track). From this information it can be deduced that:
only inertial navigation system No. 2 is drifting
only inertial navigation system No. 1 is drifting
at least one of the inertial navigation systems is drifting
the autopilot is unserviceable in NAV mode
An aircraft is flying with the aid of an inertial navigation system (INS) connected to the autopilot. The following two points have been entered in the INS computer: WPT 1: 60°N 030°W; WPT 2: 60°N 020°W. When 025°W is passed the latitude shown on the display unit of the inertial navigation system will be:
60°05.7'N
60°11.0'N
59°49.0'N
60°00.0'N
As the INS position of the departure aerodrome, coordinates 35°32.7'N; 139°46.3'W are input instead of 35°32.7'N 139°46.3'E. When the aircraft subsequently passes point 52°N 180°W, the longitude value shown on the INS will be:
099° 32.6'W
080° 27.4'W
080° 27.4'E
099° 32.6'E
The following points are entered into an inertial navigation system (INS). WPT 1: 60°N 30°W; WPT 2: 60°N 20°W; WPT 3: 60°N 10°W. The inertial navigation system is connected to the automatic pilot on route (1-2-3). The track change when passing WPT 2 will be approximately:
zero
a 9° increase
a 9° decrease
a 4° decrease
An aircraft has a TAS of 300 kts and a safe endurance of 10 hrs. If the wind component on the outbound leg is 50 kts head, what is the distance to the point of safe endurance?
1.458 NM
1.544 NM
1.622 NM
1.500 NM
An aircraft at position 27°00'N 170°00'W travels 3000 km on a track of 180° (T), then 3000 km on a track of 090° (T), then 3000 km on a track of 000° (T), then 3000 km on a track of 270° (T). What is its final position?
27°00'N 170°00'W
27°00'N 173°18'W
00°00'N/S 170°00'W
27°00'N 143°00'W
Given: Runway direction: 083° (M); Surface W/V: 035/35 kts. Calculate the effective headwind component.
31 kts
27 kts
34 kts
24 kts
If the headwind component is 50 kts, the FL is 330, temperature ISA -7 °C and the ground speed is 495 kts, what is the Mach number?
0,78
0,98
0,75
0,95
Given: Pressure Altitude: 27.000 feet; OAT: -35 °C; Mach number: 0,45; W/V: 270°/85; Track: 200° (T). What is drift and ground speed?
15R / 235 knots
18L / 285 knots
17R / 287 knots
17L / 228 knots
An aircraft at FL370, M 0,86, OAT -44 °C, headwind component 110 kts, is required to reduce speed in order to cross a reporting point 5 min later than planned. If the speed reduction were to be made 420 NM from the reporting point, what Mach number is required?
M 0,81
M 0,75
M 0,73
M 0,79
By what amount must you change your rate of descent given a 10 knot increase in headwind on a 3° glideslope:
50 feet per minute increase.
30 feet per minute increase.
50 feet per minute decrease.
30 feet per minute decrease.
An island is observed to be 15° to the left. The aircraft heading is 120° (M), variation 17° W. The bearing from the aircraft to the island is:
268° (T)
122° (T)
302° (T)
088° (T)
Given: TAS: 197 kts; True course: 240°; W/V: 180/30 kts; Descent is initiated at FL220 and completed at FL40. Distance to be covered during descent is 39 NM. What is the approximate rate of descent?
800 ft/min
1.400 ft/min
1.500 ft/min
950 ft/min
Given: Distance A to B: 475 NM; Planned GS: 315 kt; ATD: 1000 UTC; 1040 UTC - fix obtained 190 NM along track. What GS must be maintained from the fix in order to achieve planned ETA at B?
300 kt
360 kt
340 kt
320 kt
What is the effect on the Mach number and TAS in an aircraft that is climbing with constant CAS?
Mach number decreases; TAS decreases.
Mach number increases; TAS increases.
Mach number increases; TAS remains constant.
Mach number remains constant; TAS increases.
