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Question 201-1 : You are departing from an airport which has an elevation of 1500 ft. the qnh is 1003 hpa..15 nm away there is a waypoint you are required to pass at an altitude of 7500 ft..given a groundspeed of 120 kt, what is the minimum rate of climb ? [ Exam pilot ]
800 ft/min.
.7500 1500 = 6000 ft..6000 / 7.5 = 800 ft/min.
Question 201-2 : An aircraft is flying at fl200..the qnh, given by a meteorological station at an elevation of 1300ft is 998.2 hpa..oat = 40°c..the elevation of the highest obstacle along the route is 8 000 ft...calculate the aircraft's approximate clearance above the highest obstacle on this route ?
10 500 ft.
.find the qnh altitude 1013 998.2 = 14.8 x 27 = 400 ft.altitude is 19600ft qnh...with aviat 617 computer.against altitude pressure = 20, put °c oat = 40..then read in the inner circle the altitude 19600, the.on the outer circle 18400 true altitude...18400 8000ft = 10400ft = approximate clearance over the obstacle...for information 061 general navigation learning objectives states for questions involving height calculation 30 ft/hpa is to be used unless another figure is specified in the question.
Question 201-3 : 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.
.we need to be at 10000 ft to avoid the obstacle by 2000 ft...temperature correction formula 4° x 10 x 10° = 400 ft..the altimeter over reading in cold air, and if we flew exactly at 10000 ft indicated our true altitude would be 9600 ft...we need to cruise at 10000 + 400 = 10400 ft indicated in order to maintain a clearance of 2000 ft.
Question 201-4 : An aircraft is departing from an airport which has an elevation of 2000 ft..outside temperature is 0°c, qnh = 1013 hpa..it is planned to climb to fl 320 where outside temperature is 60°c..the average cas during climb will be 200 kt, compressibility is negligeable..mean tas during climb is ?
276 kt.
. by convention, at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude...temperature is 0°c at 2000 ft...approximately 4° at sea level we 'gain' 2°c per 1000 ft while descending...21000 ft x 2°/1000ft = 42°c.4°+ 42° = 38°c.temperature is around 38°c at fl210...on computer, in airspeed window, set press alt '21' in front of coat °c ' 38°c', on the outer scale, in front of cas 200 kt, you can read tas 272 kt.
Question 201-5 : During visual navigation in freezing conditions, after heavy snowfall, which of the following landmark will give the best reference for a visual checkpoint ?
A large river.
.after heavy snowfall, roads will not have been cleared by snow ploughs, neither country road or railway. a large river may freeze, but you will always be able to distinguish its path.
Question 201-6 : During a climb at a constant cas below the tropopause in standard conditions ?
Both tas and mach number will increase.
.for those questions, use the very simple 'ertm' diagram.. 1037.the cas line is vertical because the question states climb at a constant calibrated airspeed cas... ertm for e as/ r as rectified air speed or cas / t as/ m ach.
Question 201-7 : An aircraft is descending down a 12% slope whilst maintaining a gs of 540 kt. the rate of descent of the aircraft is approximately ?
Question 201-8 : The departure airfield is at 2000 ft elevation. temperature at the field is +20°c, qnh 1013 hpa. the plan is to climb to fl 290, where outside air temperature is 40°c..the average tas in the climb should be calculated using what fl and temperature ?
Fl 200 with temperature 20°c.
.by convention, at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude...29000 2000 = 27000 ft..2/3 de 27000 = 18000 ft...18000 + 2000 = 20000 ft fl200...température à 20000 ft = 20°c + 20 x 2°c = 20°c.
Question 201-9 : The departure is from an airfield at 2000 ft elevation. temperature at the field is +20°c, qnh 1013 hpa. the plan is to climb to fl 290, where outside air temperature is 40°c. the cas in the climb is 180 kt, compressibility negligible..the average tas in the climb is ?
249 kt.
.1. 29000 2000= 27000ft..2. 27000* 2/3 = 18000..3. 18000+2000 = 20000 your average alt, in climb patern..4. if oat at 2000 alt is +20 so at 20000 will be +20 reference figure 2*18 18 height , explanation 2*18 because temp decrease 2 deg per 1000 ft..5. align temp 16 with 22000 ft in th air speed window cr 3 or iwa 11092 and read oposit 180 kt your 249 tas at outer scale..this is only the way to solve tasks like this.
Question 201-10 : Given.w/v at arrival aerodrome at 1000 ft amsl is 230°/15kt, w/v at tod at fl 130 is 280°/45kt. average track after tod is 220°. isa conditions. descent speed ias = 170 kt..find the gs during the descent ?
163 kt.
.at fl130, isa condition.ias=170kt => tas=206kt..we have w/v 280°/45kt => drift 12°l so mh=232° to get an average track of 220°..so gs= 178kt with computer..at 1000ft amsl, isa condition..ias=170kt => tas=172kt..we have w/v 230°/15kt => drift 1°l so mh=221° to get an average track of 220°..so gs= 157kt with computer..as a result, the average gs for the descent is 157+178 /2 = 167,5kt.
Question 201-11 : Given.w/v at arrival aerodrome at msl is 200°/20kt, w/v at tod at fl 100 is 260°/50kt. average track after tod is 190°..isa conditions. descent speed ias = 150 kt..find the gs during the descent ?
135 kt.
Question 201-12 : An aircraft is cruising in fl180 and thereafter descends to ground level. the following wind information is given.ground level 260°/25 kt.fl030 270°/30 kt.fl060 270°/35 kt.fl090 270°/40 kt.fl120 280°/50 kt.fl150 285°/55 kt.fl180 290°/55 kt.the wind to be used for descent calculation is ?
270°/40 kt.
.by convention, average wind velocity used for climb problems is wind velocity at the altitude 2/3 of the cruising altitude..average wind velocity used for descent problems is wind velocity at the altitude 1/2 of the descent altitude.
Question 201-13 : The distance between two waypoints is 150 nm, to calculate compass heading, the pilot used 2°e magnetic variation instead of 2°w..assuming that the forecast w/v applied, what will the off track distance be at the second waypoint ?
10 nm.
.for each degree of error that you have, at every 60 nm of travel, you will be 1 nm off track..you have 150/60 2.5 nm off track for each degree of error..total error is 4° from 2°e to 2°w.4° x 2.5 nm = 10 nm...using goniometric functions.tan4° = / 150. = tan4° x 150. = 10 nm.
Question 201-14 : True track 085°.groundspeed 180 kt.wind 290°/30kt.variation 4°e..the aircraft is 1.5 nm left of track after 12 minutes...what is the track angle error tke ?
2.5° l.
.tke = distance off track x 60 / distance along track.tke = 1.5 nm x 60 / 36 nm = 2.5°..the aircraft has drifted to the left, therefore tke is 2.5° left.
Question 201-15 : With only a visual straight line as visual cue a canal for example , this line of position must be selected ?
More or less perpendicular to our track.
Question 201-16 : Given.a descending aircraft flies in a straight line to a dme.dme 55 nm, altitude 33000 ft.dme 43.9 nm, altitude 30500 ft.m = 0.72, gs = 525 kt, oat = isa..the descent gradient is ?
Question 201-17 : The descent gradient of an aircraft with the following data is. 60 nm norths of vor xyz, fl350. 10 nm south of vor xyz, fl120 ?
5.4%
.total ground distance is 60+10 = 70 nm..altitude difference is 35000 12000 = 23000 ft...gradient in % = altitude difference in ft x 100 / ground distance in ft...ground distance in feet 70 nm x 6080 ft = 425600 ft..gradient in % = 23000 x 100 / 425600 = 5,4%.
Question 201-18 : The average tas climbing from 1500 ft to fl180, with a given temperature of isa +15°c, a cas of 230 kt and qnh 1032 hpa, is ?
283 kt.
. by convention, at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude...1032 1013 = 19 hpa.19 hpa x 30 ft = 570 ft...18000 + 570 = 18570 ft.2/3 of 18570 = 12380 ft...to convert cas to tas 1% for each 600 ft and 0.2% for each degree of isa deviation..tas = cas x 12380/600 + 0.2% x 15°c = 230 x 20% + 3% = 282.9 kt.
Question 201-19 : An aircraft is turning on a final approach to intercept a 3° glide slope, which is located at an altitude of 700 ft amsl. assuming the turn is made at 4 nm from the threshold, what is a suitable altitude to intercept the glide slope ?
1916 ft.
.1 in 60 rule is a rule of thumb. 3° x 4 nm /60 = 0.2 nm..0.2 nm x 6080 ft = 1216 ft...add 700 ft since we are looking for an altitude = 1216 + 700 = 1916 ft.
Question 201-20 : Given.tas 220 kt.cruising level fl180.track during climb 080°.wind at msl 260°/25kt.wind at fl180 290°/55 kt.from msl to cruising level, find the gs during climb ?
262 kt.
.we have to use the wind at the altitude 2/3 of the cruising altitude.fl180 x 2/3 = fl120..wind changes by 30° from ground to fl180, an speed increases from 30 kt...mean wind at fl120 is 260°+20° and 25kt+20kt = 280°/45kt...with your nav computer you will find 262 kt.
Question 201-21 : An aircraft climbs from ground level to fl180. the following wind information is given.ground level 260°/25 kt.fl030 270°/30 kt.fl060 270°/35 kt.fl090 270°/40 kt.fl120 280°/50 kt.fl150 285°/55 kt.fl180 290°/55 kt.the wind to be used to solve climb problems. e.g. the calculation of the gs from tas ?
280°/50 kt.
.by convention, average wind velocity used for climb problems is wind velocity at the altitude 2/3 of the cruising altitude..average wind velocity used for descent problems is wind velocity at the altitude 1/2 of the descent altitude.
Question 201-22 : An aircraft descends from fl240 to fl040 for the final approach..cas = 220 kt.oat = isa +10°c.the average tas in the descent is ?
273 kt.
.at the exam, average tas used for descent problems is calculated at the altitude 1/2 of the descent altitude..at fl120, isa temperature = 15°c 2°c x 12 = 9°c...oat is isa +10°c, thus oat is +1°c at fl120...on the computer, in airspeed window put +1ºc next to fl120, go to cas 220 kt on inner scale and read tas on outer scale 273 kt.
Question 201-23 : When flying a visual navigation exercise in controlled airspace, it is confirmed the aircraft is exactly on track. atc instructions are to turn left 30 degrees to avoid conflicting traffic. after two minutes they advise, 'you are now two miles left of your original track, turn right to regain track ?
034°.
Question issue de ae this question asks about the on in sixty rule. this rule states that after 60 nm a drift of 1° corresponds to an off track distance of 1 nm 2° = 2 nm, 3° = 3 nm, etc.... the question mentions an off track distance of 2 nm and if you cut the flight distance into half 30 nm instead of 60 nm , angle and off track distance must double 30 nm => 4° => 4 nm. the question states that the controller wants us to regain our original track after 30 nm so the correction angle should be 4° + 30° = 34°.
Question 201-24 : An aircraft in cruise at fl120 is cleared to descend to 3000 ft..the distance to go is 25 nm..calculate the descent gradient. ?
6%.
.we have to descend 9000 ft.25 nm in ft is 25 nm x 6000 ft/nm = 151900 ft... 9000 / 150000 x 100 = 6%.
Question 201-25 : Given.descent from 15000 ft to 3000 ft msl.glide path angle during descent 3°.ground speed 180 kt in 15000 ft.ground speed 150 kt in 3000 ft.calculate the rate of descent ?
It decreases from 900 ft/min to 750 ft/min.
.rate of descent 3° = ground speed kt x 10/2..at 15000 ft, with a ground speed of 180 kt, rate of descent = 180 x 10/2 = 900 ft/min..approaching 3000 ft, with a decelerating speed to reach 150 kt, rate of descent = 150 x 10/2 = 750 ft/min.
Question 201-26 : Which formula can be used to calculate the rate of climb/descent.rate of climb/descent ft/min = ?
Groundspeed kt x gradient ft/nm / 60
.we must know the groundspeed to calculate a climb/descent gradient...calculate rate of descent rod on a given glide path angle or gradient using the following rule of thumb formulae.rod ft/min = gp degrees x gs nm/min x 100.or.rod ft/min = gp per cent x gs kt...calculate climb/descent gradient ft/nm, per cent and degrees , gs or vertical speed according to the following formula.vertical speed ft/min = gs kt x gradient ft/nm / 60.
Question 201-27 : The correct formula for climb/descent gradient in % is.gradient in % = ?
Vertical distance x 100 / ground distance.
.estimate average climb/descent gradient per cent or glide path degrees according to the following rule of thumb..gradient in % = vertical distance ft / 60 / ground distance nm.or.gradient in % = vertical distance x 100 / ground distance...gradient in degrees = arctan altitude difference ft / ground distance ft..or.gradient in degrees = vertical distance ft / 100 / ground distance nm..n.b. these rules of thumb approximate 1 nm to 6 000 ft and are based on the 1 60 rule.
Question 201-28 : The correct formula for climb/descent gradient in ° is.gradient in ° = ?
Vertical distance ft / 100 / ground distance nm.
.estimate average climb/descent gradient per cent or glide path degrees according to the following rule of thumb..gradient in % = vertical distance ft / 60 / ground distance nm.or.gradient in % = vertical distance x 100 / ground distance...gradient in degrees = arctan altitude difference ft / ground distance ft..or.gradient in degrees = vertical distance ft / 100 / ground distance nm..n.b. these rules of thumb approximate 1 nm to 6 000 ft and are based on the 1 60 rule.
Question 201-29 : An aircraft climbs from ground level to fl180. the following wind information is given.ground level 260°/25 kt.fl030 270°/30 kt.fl060 270°/35 kt.fl090 270°/40 kt.fl120 280°/50 kt.fl150 285°/55 kt.fl180 290°/55 kt.the wind to be used to solve climb problems. e.g. the calculation of the gs from tas ?
The wind at fl120.
Question 201-30 : An aircraft is flying according the flight log at the annex. after 15 minutes of flying with the planned tas and true heading, the aircraft is 3 nm south of the intended track and 2.5 nm ahead of the dead reckoning position..to reach destination b from this position, the true heading should be. 2504 ?
078°.
.15 min / 60 = 0,25h..0,25 x 130 kt gs = 32,5 nm..32,5 nm + 2,5nm ahead of dr = 35 nm..50 nm dist. 35 nm = 15 nm..tke = 3 x 60/35 and 3 x 60/15..tke = 5° and 12° ca = 5° + 12° = 17°..as we are south of 095° to get back we need to fly more to the north, therefore 095 17° = 078°.
Question 201-31 : The night effect which causes loss of signal and fading, resulting in bearing errors from ndb transmissions, is due to ?
Skywave distortion of the null position and is maximum at dawn and dusk.
.navigation using an adf to track ndbs is subject to several common effects. for night effect , radio waves reflected back by the ionosphere can cause signal strength fluctuations 30 to 60 nautical miles 54 to 108 km from the transmitter, especially just before sunrise and just after sunset more common on frequencies above 350 khz.
Question 201-32 : Quadrantal errors associated with aircraft automatic direction finding adf equipment are caused by ?
Signal bending by the aircraft metallic surfaces.
. quadrantal error.ndb signals may reach the receiver aerial directly and also after being reflected by the aircraft body. due to electrical circuits and current flowing through them there is an electromagnetic field surrounding the aircraft, in general alignment with its body..this causes the incident radio waves to deflect near the adf receiver aerial. the mixed signal affects the null position and the bearing indicated may be with large error...the maximum effect is at quadrantal relative bearings.045°, 135°, 225° and 315° relative to heading...modern installations are compensated for this error.
Question 201-33 : Errors caused by the effect of coastal refraction on bearings at lower altitudes are maximum when the ndb is ?
Inland and the bearing crosses the coast at an acute angle.
. coastal refraction or shoreline effect.low frequency radio waves will refract or bend near a shoreline, especially if they are close to parallel to it..least when bearings normal to coastline...radio waves passing the coastline at small angles suffer refraction due to different conducting and reflecting properties over land and sea. a false bearing indication is obtained at aircraft flying over sea and taking bearings from ndb located over land. the effect is less for an ndb on coast than one inland and on a bearing 90° to coastline then at an oblique angle. hence, given the choice use beacon at coast and rely on bearings perpendicular to the coastline.
Question 201-34 : Transmissions from vor facilities may be adversely affected by ?
Uneven propagation over irregular ground surfaces.
.due to reflections from terrain, radials can be bent and lead to wrong or fluctuating indications which is called scalloping.
Question 201-35 : If vor bearing information is used beyond the published protection range, errors could be caused by ?
Interference from other transmitters.
.maximum range and altitude published for a vor guaranteed the reception free from harmful interference from other vors when you are within this airspace.
Question 201-36 : What is the wavelength of an ndb transmitting on 375 khz ?
800 m.
.wavelength m = 300 / f mhz..wavelength m = 300000 / f khz..wavelength = 300000 / 375 khz = 800 m.
Question 201-37 : Phase modulation is ?
A modulation form used in gps where the phase of the carrier wave is reversed.
.for the ranging codes and navigation messages to travel from the satellite to the receiver, they must be modulated onto a carrier frequency, by varying the phase of the signal phase modulation. as soon as a data signal will be modulated onto the carrier wave, the phase of the carrier wave is reversed by 180°. the receiver will detect this reversal and is able to reconstruct the data.
Question 201-38 : Factors liable to affect most ndb/adf system performance and reliability include ?
Static interference night effect absence of failure warning system.
.navigation using an adf to track ndbs is subject to several common effects.. night effect radio waves reflected back by the ionosphere can cause signal strength fluctuations 30 to 60 nautical miles 54 to 108 km from the transmitter, especially just before sunrise and just after sunset more common on frequencies above 350 khz... terrain effect high terrain like mountains and cliffs can reflect radio waves, giving erroneous readings magnetic deposits can also cause erroneous readings... electrical effect or static interference electrical storms, and sometimes also electrical interference from a ground based source or from a source within the aircraft can cause the adf needle to deflect towards the electrical source... shoreline effect or coastal refraction low frequency radio waves will refract or bend near a shoreline, especially if they are close to parallel to it... bank effect when the aircraft is banked, the needle reading will be offset...it doesn't provide information of failure or malfunction.
Question 201-39 : Due to doppler effect an apparent decrease in the transmitted frequency, which is proportional to the transmitter's velocity, will occur when ?
The transmitter moves away from the receiver.
.because the radio signals travel at a constant speed assuming they are not refracted by the atmosphere , the receiver can calculate exactly how far away it is from the transmitter..speed is most often calculated by the receiver using the doppler effect, which is the process by which the frequency of a signal changes due to the relative motion of the transmitter. frequency decreases when the transmitter moves away from the receiver.
Question 201-40 : Which one of the following disturbances is most likely to cause the greatest inaccuracy in adf bearings ?
Local thunderstorm activity.
.static emission energy from a cumulonimbus cloud may interfere with the radio wave and influence the adf bearing indication..thunderstorm cause the greatest inaccuracy...coastal effect causes relatively small inaccuracy..the quadrantal error is caused by the refraction from the aircraft's fuselage and is compensated for...precipitation interference is irrelevant.
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