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Question 188-1 : 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 in ? [ Question security ]
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 188-2 : 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%
Admin .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 188-3 : 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
Admin .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 188-4 : 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 188-5 : The correct formula for climb/descent gradient in % is .gradient in % = ?
Vertical distance x 100 / ground distance
Admin .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 188-6 : The correct formula for climb/descent gradient in ° is .gradient in ° = ?
Vertical distance ft / 100 / ground distance nm
Admin .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 188-7 : 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 ?
The wind at fl120
Question 188-8 : 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°
Admin .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 188-9 : 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 188-10 : 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 188-11 : 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 188-12 : 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 188-13 : 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 188-14 : 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 188-15 : 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 188-16 : 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 188-17 : 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 188-18 : 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
Question 188-19 : The advantage of the use of slotted antennas in modern radar technology is to ?
Virtually eliminate lateral lobes and as a consequence concentrate more energy in the main beam
.the main advantage of a slotted antenna is that it produces a much narrower beam and eliminates most of the lateral lobes therefore more energy can be concentrated in the main beam or you can use less power to transmit
Question 188-20 : The frequency which corresponds to a wavelength of 12 cm is ?
2500 mhz
.frequency mhz = 300 / wavelength m .frequency mhz = 300 / 0 12 m = 2500 mhz
Question 188-21 : A cumulonimbus cloud in the vicinity of an aeroplane can cause certain navigation systems to give false indications this is particularly true of the ?
Adf
Question 188-22 : A radio altimeter employing a continuous wave signal would have ?
A directional aerial for transmission and another one for reception
Admin .a radio altimeter is a self contained on board aid which indicates the true height or the lowest wheels with regard to the ground .a continuous wave fm radio beam in the shf band 4200 4400 mhz is directed towards the ground in a 30° cone fore and aft and 30° athwartships across the ship from side to side the signal is reflected back to the aircraft .continuous wave as opposed to pulse radar eliminates minimum target reception range since the time delay for a pulsed signal would be too small to measure properly as well the antennae cannot switch between transmit and receive that quickly as the potential time interval is very small you need separate transmitter and receiver aerial
Question 188-23 : Which statement relating to the stabilization of airborne weather radar antennae is true ?
They are stabilized with respect to the pitch and rollaxis but not with respect to the yaw axis
Question 188-24 : The antennae of modern airborne weather radars are stabilized by means of ?
Inputs from the aircraft's attitude system
.automatic antenna stabilization as employed in today's weather avoidance radar systems consists of an electro mechanical means of maintaining a selected beam scan relative to the earth's horizon during moderate aircraft maneuvers to accomplish this a reference is established by the aircraft's vertical gyro usually a component of the autopilot or integrated flight control system
Question 188-25 : The type of modulation used for the ils frequency carrier is ?
Amplitude modulation
.amplitude modulation the information that is impressed onto the carrier wave by altering the amplitude of the carrier
Question 188-26 : The quadrantal error of an adf ?
Is caused by the refraction from the aircraft's fuselage and is compensated for
.the quadrantal error is a distortion of the incoming signal from the ndb station by re radiation from the airframe this is corrected for during installation of the antenna
Question 188-27 : The skip distance of hf transmission will increase with ?
Higher frequency and higher position of the reflecting ionospheric layer
.skip distance the distance between the transmitter and the point on the surface of the earth where the first sky return arrives .it will increase for a higher frequency .it will increase for a higher position of the reflecting ionospheric layer
Question 188-28 : Which statement about the errors and effects on ndb radio signals is correct ?
The mountain effect is caused by reflections onto steep slopes of mountainous terrain which may cause big errors in the bearing
.mountain effect mountain areas can cause reflections and to a lesser extent diffractions and lead to the error direction reading by adf .this effect similar to night effect is obtained in mountainous areas where the energy received from an ndb consists of two or more waves one of them direct and others by reflection from the mountains bearing indications are found to change rapidly until the affected area is passed .for information night effect radio waves take two paths to the radio compass receiver the first and normal path is along the earth's surface if only these waves were received the compass would point directly to the ndb the second path is via one or more wave refracting layers above the earth the ionosphere returning to earth to mix with direct waves complete changes in the nature of the waves take place on this path and produce errors in direction
Question 188-29 : What causes the so called night effect ?
A change in the direction of the plane of polarisation due to reflection in the ionosphere
Admin .'night effect' is the influence of sky waves and ground waves arriving at the adf receiver lead at a difference of phase and polarisation which introduce bearing errors at night .night effect predominates around dusk and dawn
Question 188-30 : An amplitude modulation is shown in the figure . 2069 ?
A
Question 188-31 : In his basic type a dipole antenna adapted for a frequency of 110 mhz will have a wire of length of ?
136 cm
.wavelength in meter = 300 / frequency in mhz .wavelength in meter = 300 / 110 = 2 72 m 272 cm .a dipole antenna is a wire of length equal to one half of the wavelength .272 cm / 2 = 136 cm
Question 188-32 : How is the unit hertz hz defined ?
The number of electromagnetic oscillations per second
.frequency is the number of cycles occurring in one second in a radio wave expressed in hertz hz frequency is normally given in hertz hz where 1 hz = 1 cycle/second .a cycle is defined as a complete series of values of a periodical process . 2549
Question 188-33 : An ndb transmits on 427 khz the corresponding wavelength is ?
702 5 m
.frequency khz = speed of light / wavelength m .427 khz = 300000 / wavelength m .wavelength m = 300000 / 427 = 702 57 m
Question 188-34 : An electromagnetic wave consists of an oscillating electric field e and an oscillating magnetic field h their propagation speed is ?
The speed of light
.propagation speed is the speed at which radio wave radiates through space .depending upon the electrical nature of the medium through which it travels the speed varies slightly we can assume a constant speed of light value of 300 000 000 meters 162 000 nm per second
Question 188-35 : The electromagnetic waves refracted from the e and f layers of the ionosphere are called ?
Sky waves
Question 188-36 : The simplest type of antenna construction is a ?
Dipole antenna which is a wire of length equal to one half of the wavelength
. 2550. dipole antenna
Question 188-37 : An electromagnetic wave consists of an oscillating electric field e and an oscillating magnetic field h which statement is correct ?
The e and h fields are perpendicular to each other
Question 188-38 : The frequency which corresponds to a wavelength of 8 25 m is ?
