04 Dead Reckoning Navigation Flashcards
ATPL GEN NAV
Given: GS = 236 kt. Distance from A to B = 354 NM. What is the time from A to B?
1 HR 30 MIN
Given: An aircraft is flying at FL100, OAT = ISA - 15ºC. The QNH, given by a meteorological station with an elevation of 100 ft below MSL is 1032 hPa. 1 hPa = 27 ft Calculate the approximate True Altitude of this aircraft.
9900 ft
Given: TAS = 250 kt, HDG (T) = 029°, W/V = 035/45kt. Calculate the drift and GS?
1L - 205 kt
CAS 120 kt, FL 80, OAT +20°C. What is the TAS?
141 kt
An aircraft takes on 320 US gallons of fuel weighing 870 kg.
What is the specific gravity of the fuel?
0.72
Given: TAS = 485 kt, True HDG = 226°, W/V = 110°(T)/95kt. Calculate the drift angle and GS?
9°R - 533 kt
At reference. 1215 UTC LAJES VORTAC (38°46’N 027°05’W) RMI reads 178°, range 135 NM. Calculate the aircraft position at 1215 UTC?
40°55’N 027°55’W
Given:
True Track 095° TAS 160 kts True Heading 087° GS 130 kts Calculate W/V
057°/36 kts
An aircraft takes off from the aerodrome of BRIOUDE (altitude 1 483 FT, QFE = 963 hPa, temperature = 32°C). Five minutes later, passing 5 000 FT on QFE, the second altimeter set on 1 013 hPa will indicate approximately :
6 500 FT
Given:
Groundspeed 240 kts
Distance to go 500 NM
Calculate the time to go:
2 hours 05 min
…………………………………….
To travel 500 NM at a speed of 240 kt will take 2 hr 5 min
Background Information
Simply, divide distance by speed to find time
500 nm ÷ 240 kt = 2 hr 5 min
Given: True HDG = 002°, TAS = 130 kt, Track (T) = 353°, GS = 132 kt. Calculate the W/V?
095/20kt
Given: TAS = 90 kt, HDG (T) = 355°, W/V = 120/20kt. Calculate the Track (°T) and GS?
346 - 102 kt
Given:
True HDG 074° TAS 230 kts Track (T) 066° GS 242 kts Calculate the W/V.
180/35 kts
CAS is: 320 kt Flight level: 330 OAT: ISA +15°C TAS is approximately (compressibility factor 0.939):
530 kt
Given: TAS = 200 kt, Track (T) = 110°, W/V = 015/40kt. Calculate the HDG (°T) and GS?
099 - 199 kt
Given: True course from A to B = 090°, TAS = 460 kt, W/V = 360/100kt, Average variation = 10°E, Deviation = -2°. Calculate the compass heading and GS?
070° - 450 kt
Given: GS = 95 kt. Distance from A to B = 480 NM. What is the time from A to B?
Given: GS = 95 kt. Distance from A to B = 480 NM. What is the time from A to B?
Given: TAS = 375 kt, True HDG = 124°, W/V = 130°(T)/55kt. Calculate the true track and GS?
123 - 320 kt
Maximum allowable crosswind component is 20 kt. Runway 06, RWY QDM 063°(M).
Wind direction 100°(M). Calculate the maximum allowable wind speed?
33 kt
Given: TAS = 190 kt, True HDG = 085°, W/V = 110°(T)/50kt. Calculate the drift angle and GS?
8°L - 146 kt
The following information is displayed on an Inertial Navigation System: GS 520 kt, True HDG 090°, Drift angle 5° right, TAS 480 kt. SAT (static air temperature) -51°C. The W/V being experienced is:
320° / 60 kt
Given: TAS = 370 kt, True HDG = 181°, W/V = 095°(T)/35kt. Calculate the true track and GS?
186 - 370 kt
Given: CAS 120 kt, FL 80, OAT +20°C. What is the TAS?
141 kt
How many NM would an aircraft travel in 1 min 45 secs if the GS is 135 kt?
3.94
Route ‘A’ (44°N 026°E) to ‘B’ (46°N 024°E) forms an angle of 35° with longitude 026°E. Variation at A is 3°E. What is the initial magnetic track from A to B?
322°
…………………………..
A quick diagram indicates that this track from A to B is going north (ish) (from 44ºN to 46ºN) and west (ish) (from 026ºE to 024ºE). This means that the track will be more than 270ºT but less than 360ºT. If the angle between the track and the local meridian is 35º then the track = 360º - 35º = 325ºT variation is 3ºE so the track is 322ºM.
An aircraft is flying at FL180 and the outside air temperature is -30°C. If the CAS is 150 kt, what is the TAS?
195 kt
An aircraft is following a true track of 048° at a constant TAS of 210 kt. The wind velocity is 350° / 30 kt. The GS and drift angle are:
192 kt, 7° right
TAS = 472 kt, True HDG = 005°, W/V = 110°(T)/50kt. Calculate the drift angle and GS?
6°L/490 kt
An aircraft flies at FL 250. OAT = - 45°C. The QNH, given by a meteorological station with an elevation of 2830 ft, is 1033 hPa. Calculate the clearance above a mountain ridge with an elevation of 20410 ft.
4 200 ft
Given: TAS = 140 kt, HDG (T) = 005°, W/V = 265/25kt. Calculate the drift and GS?
10R - 146 kt
Given: GS = 120 kt. Distance from A to B = 84 NM. What is the time from A to B?
00 HR 42 MIN
Given:
TAS: 210 kts
Fuel flow: 42 US Gal/hr
Specific gravity: 0.72
What is the specific fuel consumption?
0.545 kg/NM air distance.
The ICAO definition of ETA is the:
estimated time of arrival at destination
Given: M 0.80, OAT -50°C, FL 330, GS 490 kt, VAR 20°W, Magnetic heading 140°, Drift is 11° Right. Calculate the true W/V?
020°/95 kt
Given: TAS = 485 kt, OAT = ISA +10°C, FL 410. Calculate the Mach Number?
0.825
Given:
FL250, OAT -15 ºC, TAS 250 kt. Calculate the Mach No.?
0.40
How long will it take to travel 284 NM at a speed of 526 km/h?
1 hour ............................. To travel 284 NM at a speed of 526 km/h will take 1 hour. Background Information First convert 284 nm into kilometers 284 nm x 1.852 = 526 km Next, divide distance by speed to find time 526 km ÷ 526 = 1 hr
Given: True HDG 133° TAS 225 kts Track (T) 144° GS 206 kts Calculate the W/V.
075/45 kts
An aircraft departs Waypoint ELSEY (N59° E120°) on a magnetic heading of 005°M and a TAS of 240 kt. W/V = 100°/40. Local magnetic variation = 5°E. What is the approximate position of the aircraft after 15 minutes of flight?
N60° E120°
Given:
Magnetic track = 075°, HDG = 066°(M), VAR = 11°E, TAS = 275 kt. Aircraft flies 48 NM in 10 MIN. Calculate the true W/V °?
340°/45 kt
Given: M 0.80, OAT -50°C, FL 330, GS 490 kt, VAR 20°W, Magnetic heading 140°, Drift is 11° Right. Calculate the true W/V?
020°/95 kt
The reported surface wind from the Control Tower is 240°/35 kt. Runway 30 (300°). What is the cross-wind component?
30 kt
Given: True HDG = 233°, TAS = 480 kt, Track (T) = 240°, GS = 523 kt. Calculate the W/V?
110/75kt
If it takes 132.4 minutes to travel 840 NM, what is the speed in km/hr?
705 ............................... The speed in km/hr is 705 km/hr Background Information First convert 840 nm into kilometers 840 nm x 1.852 = 1,555.68 km Next, convert the time from minutes into hours. 132.4 mins ÷ 60 = 2.21 hr Next, divide distance by time to find speed 1,555.68 km ÷ 2.21 hr = 705 km/hr
Given:
TAS = 470 kt, True HDG = 317°, W/V = 045°(T)/45kt. Calculate the drift angle and GS?
5°L - 470 kt
Given: GS = 345 kt. Distance from A to B = 3560 NM. What is the time from A to B?
10 HR 19 MIN
Consider the following factors that determine the accuracy of a DR position:
- The flight time since the last position update.
- The accuracy of the forecasted wind.
- The accuracy of the TAS.
- The accuracy of the steered heading.
Using the list above which of the following contains the most complete answer?
1, 2, 3 and 4
What is the ISA temperature value at FL 330?
-51°C
…………………….
In an International Standard Atmosphere the temperature is +15ºC at sea level and has a lapse rate of 1.98ºC per 1,000 up to 36,090 feet.
At FL330 (33,000 feet) the temperature is +15º - (33 x 1.98º)
= +15º - 65.34º
= - 50.34ºC
Given: TAS = 170 kt, HDG(T) = 100°, W/V = 350/30kt. Calculate the Track (°T) and GS?
109 - 182 kt
Given: GS = 105 kt. Distance from A to B = 103 NM. What is the time from A to B?
00 HR 59 MIN
Given:
TAS = 290 kt, True HDG = 171°, W/V = 310°(T)/30kt. Calculate the drift angle and GS?
4°L - 314 kt
Given: GS = 122 kt. Distance from A to B = 985 NM. What is the time from A to B?
8 HR 04 MIN
Given:
Maximum allowable tailwind component for landing 10 kt. Planned runway 05 (047° magnetic).
The direction of the surface wind reported by ATIS 210°. Variation is 17°E. Calculate the maximum allowable wind speed that can be accepted without exceeding the tailwind limit?
10 kt
Given: GS = 435 kt. Distance from A to B = 1920 NM. What is the time from A to B?
4 HR 25 MIN
An aircraft is flying at FL100. The OAT = ISA - 15°C. The QNH given by a station at an elevation 3000 ft is 1035hPa. Calculate the approximate True Altitude.
10 200 ft
The accuracy of the, manually calculated, DR-position of an aircraft is, among other things, affected by:
the accuracy of the forecast wind.
Given: TAS = 132 kt, HDG (T) = 053°, W/V = 205/15kt. Calculate the Track (°T) and GS?
050 - 145 kt
Given: Runway direction 083°(M), Surface W/V 035/35kt. Calculate the effective headwind component?
24 kt
Given: TAS = 485 kt, HDG (T) = 168°, W/V = 130/75kt. Calculate the Track (°T) and GS?
174 - 428 kt
Given: True HDG = 054°, TAS = 450 kt, Track (T) = 059°, GS = 416 kt. Calculate the W/V?
010/50kt
Given:
TAS: 154 mph
Fuel flow: 28 Imp Gal/hr
Specific gravity: 0.72
What is the specific range?
1.46 NM air distance/kg
Given: Required course 045°(M); Variation is 15°E; W/V is 190°(T)/30 kt; CAS is 120 kt at FL 55 in standard atmosphere. What are the heading (°M) and GS?
055° and 147 kt
The accuracy of the, manually calculated, DR-position of an aircraft is, among other things, affected by:
the flight time since the last position update.
The QNH, given by a station at 2500 ft, is 980hPa.The elevation of the highest obstacle along a route is 8 000 ft and the OAT = ISA -10°C.
When an aircraft, on route has to descend the minimum indicated altitude (QNH on the subscale of the altimeter) to maintain a clearance of 2000 ft, will be:
10 400 ft