WorksheetsPage 1
Total questions: 49
Worksheet time: 25mins
The term "ellipsoid" may be used to describe:
a great circle on the celestial sphere.
the shape of the ecliptic.
the movement of the Earth around the Sun.
the shape of the Earth.
The circumference of the Earth is approximately:
43.200 NM
10.800 NM
21.600 NM
5.400 NM
The Earth is:
a sphere whose centre is equidistant (the same distance) from the Poles and the Equator.
a sphere which has a larger polar circumference than equatorial circumference.
none of the above statements is correct.
considered to be a perfect sphere as far as basic (simple) navigation is concerned.
Seasons are due to the:
Earth's rotation on its polar axis.
Earth's elliptical orbit around the Sun.
variable distance between Earth and Sun.
inclination of the polar axis with the ecliptic plane.
The poles on the surface of the Earth may be defined as:
the points on the surface of the Earth where all meridians intersect at right angles.
the points where the Earth's axis of rotation penetrates the surface of the Earth.
the points at which the vertical lines runs through the centre of the Earth.
the points from where the distance to the equator is equal.
When the Sun's declination is northerly:
midnight Sun may be observed at the south pole.
it is winter on the northern hemisphere.
the sunrise occurs earlier at southern latitudes than the northern latitudes.
the daylight period is shorter in the southern hemisphere than the northern.
Consider the positions (00°N/S, 000°E/W) and (00°N/S, 180°E/W) on the ellipsoid. Which statement about the distances between these positions is correct?
The route via the equator is shorter than the route via the South Pole.
The route via the North Pole is shorter than the route along the equator.
The route via either pole and the route via the equator are of equal length.
The route via the South pole is shorter than the route via the North Pole.
The maximum difference between geocentric and geodetic latitude occurs at about:
60° North and South.
90° North and South.
0° North and South (equator).
45° North and South.
What is difference in latitude from 30°39'S 20°20'E to 45°23'N 40°40'E:
14°44'N
76°2'N
76°2'S
14°44'S
Given: A (56°N, 145°E); B (57°N, 165°W). What is the difference in longitude between A and B?
020°
050°
001°
130°
Which statement about the duration of daylight is true?
Close to the solstices the influence of latitude on the duration of daylight is at its smallest.
In summer the length of the period of daylight decreases with increasing latitude.
Close to the equinoxes the influence of latitude on the duration of daylight is at its smallest.
On September 10th the duration of daylight is longer on the Southern Hemisphere than on the Northern Hemisphere.
The angle between True North and Magnetic North is known as:
deviation.
alignment error.
variation.
dip.
The definition of True North for any observer is:
the direction of the Greenwich meridian to the North Pole.
the direction of the observer's Magnetic North corrected for local variation.
the direction of the observer's meridian to the North Pole.
the reading of the observer's compass corrected for deviation and local variation.
The purpose of establishing a grid is:
to provide a system for directions where a great circle has a constant direction, even if its true direction varies.
minimise the errors introduced when making calculations involving variation.
make a chart covering high latitudes that has the same qualities as the equatorial Mercator chart.
to make the system of latitude and longitude available on a gridded map.
Compass deviation is defined as the angle between:
Magnetic North and compass North.
the horizontal and the total intensity of the Earth's magnetic field.
True North and compass North.
True North and Magnetic North.
The evaluation of your plotting work shows a WCA +3° and a drift 3° left:
your actual position is on the intended track.
the GS was exactly calculated.
the expected W/V and the actual W/V coincide.
the track error is 6°.
You should follow a track due North taking account of a northwesterly wind. You calculated a WCA -8°.
The drift will be 8° left.
A track error of -2° (left) shows a WCA of only -6°.
The drift will be 8° right.
A track error of 2° (right) shows a drift of 10° right.
Given: Course: 040°(T); TAS: 120 kts; Wind speed: 30 kts. Maximum drift angle will be obtained for a wind direction of:
120°
145°
130°
Given: TAS: 132 kts; True HDG: 257°; W/V: 095°(T)/35 kts. Calculate the drift angle and GS.
2°R - 166 kts.
4°R - 165 kts.
3°L - 166 kts.
4°L - 167 kts.
Given: TAS: 472 kts; True HDG: 005°; W/V: 110°(T)/50 kts; Calculate the drift angle and GS.
6°L - 487 kts.
7°R - 487 kts.
7°L - 491 kts.
7°R - 491 kts.
1 nautical mile equals:
5 280 feet.
1 852 metres.
3 081 yards.
0,896 statute mile.
In international aviation the following units shall be used for horizontal distance:
metres, statute miles and nautical miles.
metres, kilometres and nautical miles.
kilometres, statute miles and nautical miles.
kilometres, feet and nautical miles.
What are the initial true course and distance between positions 58°00'N 013°00'W and 66°00'N 002°00'E?
032° - 470 NM.
036° - 638 NM.
042° - 635 NM.
029° - 570 NM.
The distance between positions A and B is 180 NM. An aircraft departs position A and after having traveled 60 NM, its position is pinpointed 4 NM left of the intended track. Assuming no change in wind velocity, what alteration of heading must be made in order to arrive at position B?
8° right
6° right
4° right
2° left
Given: AD = Air distance; GD = Ground distance; TAS = True Airspeed; GS = Groundspeed. Which of the following is the correct formula to calculate ground distance (GD) gone?
GD = (AD - TAS)/TAS
GD = AD x (GS - TAS)/GS
GD = (AD x GS)/TAS
GD = TAS/(GS x AD)
Given: AD: Air distance; GD: Ground distance; TAS: True airspeed; GS: Ground speed. Which of the following is the correct formula to calculate ground distance (GD) gone?
GD = (AD _ GS) ÷ TAS
GD = TAS ÷ (GS x AD)
GD = AD x (GS - TAS) ÷ GS
GD = (AD - TAS) ÷ TAS
An aircraft is flying at FL180 and the outside air temperature is -30°C. If the CAS is 150 kt, what is the TAS?
115 kt
180 kt
195 kt
145 kt
Given: CAS: 230 kts; FL120; OAT: -10°C. What is the TAS?
266 kts
280 kts
273 kts
287 kts
Given: Mach number: 0.8; Flight level: 330; OAT: ISA +15°C; TAS is approximately (compressibility factor 0.94):
420 kts
450 kts
265 kts
480 kts
You are flying at FL80 and air temperature is ISA +15. What CAS is required to make TAS 240 kts?
214 kts
208 kts
226 kts
220 kts
An aircraft is climbing at a constant CAS in ISA conditions. What will be the effect on TAS and Mach number?
TAS increases and Mach No decreases.
Both increase.
TAS decreases and Mach No increases.
Both decrease.
Given: Half way between two reporting points the navigation log gives the following information: TAS 360 kt, W/V 330°/80kt, Compass heading 237°, Deviation on this heading -5°, Variation 19°W. What is the average ground speed for this leg?
354 kt
373 kt
403 kt
360 kt
The Sun moves from East to West at a speed of 15° longitude an hour. What ground speed will give you the opportunity to observe the Sun due South at all times at 60°00'N?
300 kts
450 kts
520 kts
780 kts
You start from P (70°00'N 015°00'E) and fly westward along the parallel of latitude for 2 hours at ground speed 220 kts. What is your position after two hours flight?
006°26'W
021°26'W
007°40'E
006°44'W
Given: Actual HDG: 290°; TAS: 250 kts; Wind: 135/75 kts. What is the ground speed?
300 kts
175 kts
320 kts
An aircraft travels 2,4 statute miles in 47 seconds. What is its ground speed?
183 kts
160 kts
131 kts
209 kts
An aircraft is planned to fly from position A to position B, distance 320 NM, at an average GS of 180 kts. It departs A at 12:00 UTC. After flying 70 NM along track from A, the aircraft is 3 min ahead of planned time. Using the actual GS experienced, what is the revised ETA at B?
1401 UTC
1333 UTC
1340 UTC
1347 UTC
Given: A descending aircraft flies in a straight line to a DME; DME 55,0 NM, altitude 33.000 ft; DME 43,9 NM, altitude 30.500 ft; M = 0,72, GS = 525 kts, OAT = ISA. The descent gradient is:
3,70%
3,90%
3,50%
4,10%
An aircraft is maintaining a 5,2% gradient at 7 NM from the runway, on a flat terrain its height is approximately:
2.210 ft
3.640 ft
1.890 ft
680 ft
You are departing from an airport which has an elevation of 2.000 ft. The QNH is 1013 hPa. 10 NM away there is a waypoint you are required to pass at an altitude of 7.500 ft. Given a groundspeed of 100 kts, what is the minimum rate of climb?
920 ft/min
590 ft/mins
750 ft/min
1.080 ft/min
An aircraft at FL350 is required to commence descent when 85 NM from a VOR and to cross the VOR at FL80. The mean GS for the descent is 340 kts. What is the minimum rate of descent required?
1.900 ft/min.
1.600 ft/min.
1.800 ft/min.
1.700 ft/min.
An aircraft at FL390 is required to descend to cross a DME facility at FL70. Maximum rate of descent is 2.500 ft/min, mean GS during descent is 248 kts. What is the minimum range from the DME at which descent should commence?
63 NM
58 NM
68 NM
53 NM
The outer marker of an ILS with a 3° glide slope is located 4,6 NM from the threshold. Assuming a glide slope height of 50 ft above the threshold, what is the approximate height of an aircraft passing the outer marker (use the 1:60 rule)?
1.400 ft
1.450 ft
1.300 ft
1.350 ft
Construct the triangle of velocities using the following data, and determine the aircraft’s track in this period of time. True Heading: 305°. TAS: 135 kts. W/V: 230°/40. Period of time: from 11:30 to 11:45. What is the track in this period of time?
290°
322°
316°
310°
Given: True course from A to B: 090°; TAS: 460 kts; W/V: 360/100 kts; Average variation: 10°E; Deviation: -2°. Calculate the compass heading and GS.
070° - 450 kts.
102° - 450 kts.
068° - 460 kts.
078° - 450 kts.
Given: TAS = 375 kt, True HDG = 124°, W/V = 130°(T)/55kt. Calculate the true track and GS?
125 - 322 kt
123 - 320 kt
125 - 318 kt
126 - 320 kt
True Heading of an aircraft is 265° and TAS is 290 kt. If W/V is 210°/35kt, what is True Track and GS?
259° and 272kt
271° and 272kt
259° and 305kt
260° and 315kt
Given: TAS: 135 kts; True Heading: 278°; W/V: 140/20 kts. Calculate the True track and GS.
279° - 152 kts.
283° - 150 kts.
275° - 150 kts.
272° - 121 kts.
The DR position represents:
the estimated position taking account of the estimated TAS and wind condition.
the air position corrected by the track error.
the actual position corrected by the track error.
the estimated position in no wind condition.
