Freq:   ADF – 190 TO 1750* KHz   NDB – 190-450 KHz (* To tune AM Radio Freq)

Band:   High LF to Low MF

 Bearing:  Three types (a) True  (b)  Magnetic  (c) Relative (wrt ac fore and aft axis)


 Principle.   Bearing by Loop Theory.

 Working.  (Basic)

1. In the aircraft two antennas are fitted i.e. loop and sense antenna.

2. Loop antenna has got directional properties and the polar diagram of its transmission is a figure of  eight.

3.  A signal from ground Tx is received in different phases at the vertical axis of the loop antenna.

4. When loop antenna is aligned with the ground Tx induced emf is maximum and when it is at 90° emf is zero.  Therefore values of induced emf vary as Cosine of the antenna angle.

5.  A simple loop antenna has two null points and two maxima, therefore it suffers from 180 deg ambiguity.

6.  This 180° ambiguity is resolved by a sense aerial whose polar diagram is a circle.

7.  Combined polar diagram of both aerials is a cardioid which has only one null position.

8.  This null point of cardioid is used to rotate the pointer of the RBI.

Simplified Principles of Airborne Direction Finding

The original method of airborne D/F involved a ‘loop’ aerial to receive the signals. The shape of the ‘loop’ could be anyone of several, including rectangular similar to Fig a.

In Fig ‘a’ below the loop aerial consists of two vertical members, A and B, connected in the form of a loop by horizontal members. If a vertically polarized radio wave is incident upon the loop, it will induce voltages in the vertical members of the loop of value Va and Vb.

Consider a wave incident at an angle θ to the plane of the loop (Fig ‘b’ below). Distance AB is insignificant compared with the range from the transmitter to the loop, so both A and B receive the same signal strength. However, as the signal travels a different distance to each, there is a phase difference at A and B given by AB cos θ.

Since AB is constant, the value of the resultant voltage in the loop is proportional to θ, giving the ‘figure of eight’ polar diagram for a loop aerial shown in Fig ‘c’. The plus and minus signs show the sign of cos θ, and hence the resultant voltage, Vr, in both lobes of the loop. The horizontal polar diagram has two sharply defined minima at θ = 90° and θ = 270°, and two poorly defined maxima at θ = 0° and θ = 180°.

If a loop aerial which is receiving a wave from a transmitter is rotated, the resultant voltage in the loop will vary as θ varies. When θ = 90° or 270° the resultant voltage is zero. When θ = 0° or 180° the resultant voltage is a maximum. As the minima are more sharply defined, these are used for direction finding. To take a manual loop bearing, the loop is rotated until a minimum signal, or null, is found, when the transmitter must, subject to certain errors mentioned later, be on the line normal to the plane of the loop. However, there is no indication on which side of the loop the transmitter is sited. The process of resolving this ambiguity is known as sensing.
























Sensing. If a vertical omni-directional aerial is placed midway between the two vertical members of the loop, the voltage induced in it by an incident wave will be midway in phase between the voltages induced in the vertical members of the loop (Fig ‘a’). It can be shown that the phase of the voltage in the sense aerial, Vs, is always 90° removed from the phase of the resultant voltage, Vr, in the loop.

If the incident wave comes from the left of the normal to the plane of the loop (direction X in Fig ‘b’), Vs leads Vr by 90°, while if it comes from the right (direction Y), Vr leads Vs by 90°. By permanently incorporating suitable components in the sense aerial circuit, the phase of Vs can be retarded by 90°. If this is done Vs will be in phase with Vr if the wave comes from the left of the normal and in anti phase if the wave comes from the right. The aerials are designed so that the value of Vs is equal to the maximum value of Vr.



















If Vs and Vr are combined, the polar diagram shown in Fig below, will result. The figure of 8 is the polar diagram for the loop alone, the circular polar diagram is for the sense aerial alone, and the heart-shaped or cardioid polar diagram is for the loop and sense aerial combined (to the right of the normal to the loop (CD), Vs and Vr are in antiphase, and cancel each other while to the left they are additive). In many installations the sense aerial is located away from the mid-point of the loop, but the principle is unchanged.

Generation of Cardioid

















Factors Affecting Accuracy of ADF/Range/Errors of ADF.

  1. Night Effect. During night the electron density reduces and the radio waves are refracted even in the MF band the sky wave from ionosphere interferes with the horizontal arm of the loop aerial, because of this sky wave, ADF pointer hunts or oscillates, thereby affecting accuracy.

                                           Sky Wave Interference during Night

 Methods to Reduce Night Effect

 Use a beacon which 70 nm from the aircraft position since first sky wave return is received at a skip distance of 70 nm or more.

  1. Use a beacon which is operating at a lower frequency because atmospheric attenuation will be high.
  2. Use a beacon which is transmitting at a higher power because ground wave will be stronger

Coastal Error.  Density of air over land is higher than that over the sea, so a wave leaving from land towards the sea bends towards the coast or away from normal.  This error is called coastal error.  Coastal error causes aircraft position shown closer to coast.


 Methods to Reduce Coastal Error

 Use a beacon whose signal is leaving coast line at 90° because bending will be the least.

  1. Which of the beacons will cause max coastal error for signal leaving at (a) 30° (b) 60° (c) 90°    A. (a). Acuter the angle larger the coastal error.
  1. Fly as high as possible since at higher altitudes the density of air over land is almost the same as that above sea.
  1. Use a beacon closer to the coast. This will ensure signal is stronger and medium change is taking place for a lesser distance so bending will be less.

4..  Use a beacon which is transmitting at high power since weaker signals bend more..


Quadrantal Error.  Due to the electromagnetic field of the aircraft any signal reaching at loop antenna undergoes direction change, the error caused due to this is termed as Quadrental Error.  The error is maximum when the signal is approaching from quadrental headings with respect to aircraft fore and aft axis and is minimum along the cardinal headings wrt to aircraft heading.  Errors can be calculated and calibrated and a correction card is provided in the aircraft for reference.

Q) An aircraft is flying on Hdg 045 which of the following signals reaching loop antenna will give max quadrantal error

(a) 225° (b)315°  (c) 180°  (d) 135°       A. (c)

Interference Error.  When an aircraft is operating in an area having more than one ground station, interference may cause ADF to give erroneous reading, this is called interference error.  To prevent Interference error a Designated Operational Coverage (DOC) is given in AIP which denotes the distance till which it is guaranteed that there will be no interference from any nearby operating NDB. DOC is also referred to as Protected Range.  DOC of NDB is valid during  day time only.

 Loop Misalignment Error.  This is due to misalignment of the loop antenna during installation on the aircraft.

Static Interference Error.   This is due to weather phenomenon like thundery activity, lightning etc.

  1. Which of the following errors are most predominant in ADF operation. (a) Static Interference  (b) Night Effect  (c)  Quadrantal Error  (d)  Coastal Effect  A.  (c)

Factors Affecting Range of NDB

  1. Transmission Power Range ⇒ √ Tx Power         (Two double increase 4 times)
  1. Frequency – lower the frequency (less attenuation hence higher range)
  1. Night Effect – causes NDB range to reduce to 70 nm (max).
  1. Terrain – Range over sea will be more than land due to lower attenuation.

–  Range over plains will be more than hilly terrain.

Types of NDB.    (a) Locator (10-25 Nm)  (b) Homing/Holding (50 Nm)  (c) Long Range

  1. Locator NDB. Low powered NDB normally co-located with ILS markers (IM, MM and OM) Range is 10 -15 Nm.
  2. Homing/Holding NDB. It is used for homing or holding over the station.  Range is approx 50 Nm.
  3. Long Range/Enroute. These are high powered NDB and range depends on the requirement.

Some Interesting Facts about ADF
If there are two NDBs, one on the coast and the other fairly inland, and if the coastal refraction for both propagations is the same when the bearings are taken, then when you plot the position lines, you will find that the NDB which is further inland gives greater error.

When flying over water if you take two / three bearings to make a fix, and if errors due to coastal refraction are present, then the fix you make will put you coastward from your true position. Do a little plotting exercise and check it. First plot a fix from three position lines unaffected by coastal refraction and then plot another fix from the lines with error.

In Fig below an aircraft is flying a track of 060° (T) in no wind conditions. When in position A it obtains a relative bearing of 050° R from NDB. Now if it waits until the bearing has changed to twice the original value, i.e. 100°R, its position at that time is at B. In this situation, we have a triangle ABC in which side AB equals side BC. This enables the pilot to calculate his distance from NDB, when at B. This distance is equal to distance AB which he can calculate from knowledge of his ground speed. When there is wind, allowance must be made for the drift applied in the ADF reading.















Problems on RBI

 An ac is homing on to a station on Hdg 090°, what will be the RBI indication?

  1. 000°
  1. An ac is homing on to a station on Hdg 090° with 10° S drift what will be the RBI indication? A. 010°


  1. An ac is homing on to a station on Trk 090° with 15° P drift what will be the RBI indication? A. 345° Track

Track 090° RBI 15° to Left (345°)


  1. An ac is tracking out from a station on Hdg 200° with 10° S drift what will be the RBI indication? A. 190° 190

Track 200° RBI 10° to Rt (190°)

  1. An ac is flying constant Hdg with 10° S drift and is making good a track 5 Nm to the right of the centre line of the airway but parallel to it. Find ADF reading from an NDB situated on the centre of the airway 30 Nm ahead? A. 340 (Closing angle to Stn =(5 x 60)/30 = 10°)

                                                                                     Track 090° RBI 20° to Right (020°) to Stn

  1. An ac undergoes a wing tip bearing change of 10° in 6 min while crossing abeam a ground stn. Find (a) GS of ac if distance of beacon is 60 nm. Fuel consumed to reach stn if fuel consumption is 100 gals/hr?  A 108 nm, 60 gals

 GS =180 K Hence 3nm/min in 6 min 18 nm.  10=18 x 60/D 0r D= 18 x 6 =108 NM

Time to 108 nm @180 K =36 mins, Hence fuel consumed =36 x 60/180 = 60 gals

q)  An ac undergoes a wing tip bearing change of 45° in 18 min while crossing abeam a ground stn. Find (a) Distance from Gd stn when GS is 240 K (b) Fuel consumed to reach stn if fuel consumption is 300 kgs/hr? A. 72 Nm, 900 Kgs (Dist to Stn = 240x (18/60) = 72, FC =300×18/60)

Q)  At 0900h an ac measures RB of 025, at 0910 RB changes to 050. Find (a) Distance from Gd stn when GS is 200 K. A. 33.3 Nm ( Time to Stn = 10 min; Dist = 200×10/60 = 33.3 Nm)

 Q) At 1000h aircraft measures RB of 045. At 1006 RB measured is 090°.  If GS is 180 Kts, find the distance of aircraft from the ground station. A. 18 Nm  (Time to Stn =06m; D= 180×6/60=18)

 Q) At 0800h aircraft measures RB of 030°. At 0806 RB measured is 060°.  If GS is 180 Kts, find the distance of aircraft from the ground station. If fuel consumption is 90 gals/hr, find fuel required A. 18 Nm, 9 gals. (Time to Stn =6m; D= 180×6/60= 18Nm, F Used = 90x 6/60 = 9 gals)

 Q) At 0900h aircraft measures RB of 025°. At 0910 RB measured is 050°. Find (a) the distance of aircraft from the ground station if GS is 200 Kts (b) If distance is 48 nm find GS. A. 33 Nm, 288 K

(D = 200×10/60=33.3Nm, GS = 48 x 60/10 = 288 K)


Sample Questions


  1. Which of the following are all factors affecting the accuracy of ADF indications?


(a)        Night effect, tropospheric ducting, coastal refraction, hill effect.

(b)       Coastal refraction, static interference, loop alignment, quadrantal error.

(c)        Quadrantal effect, site error, night effect, station interference.

(d)       Thunderstorm interference, loop alignment, duct propagation, hill effect.


  1. Which of the following are all factors affecting the range of an ADF?


  • Night effect, loop alignment, static interference
  • Transmission power, static interference, type of emission
  • Frequency, transmission power, night effect
  • Terrain attenuation, frequency, loop alignment


  1. Given all other factors equal, which of the following- signals and emission types will give the greatest range from an NDB?
  • A vertically polarised NON AlA signal
  • A horizontally polarised NON A2A signal
  • A vertically polarised A2A signal
  • A vertically polarised NON A2A signal


  1. If an NDB emits a NON A2A signal, what should the pilot do with the BFO switch?
  • BFO on all the time
  • BFO on during tuning and identification
  • BFO on during tuning only
  • BFO on during identification only


  1. Which of the following frequencies might be used by an NDB?


  • 40 kHz
  • 400kHz
  • 40 MHz
  • 400 MHz



  1. An aircraft is heading 240(M). The RBI indicates 030°. What is the magnetic bearing of the aircraft from the NDB?


  • 030°
  • 090° 270             30°          090
  • 210° Brg of Ac from NDB    (Ac Hdg 240)
  • 270°


  1. A coastline runs North-South. At which aircraft position will coastal refraction most affect its ADF?


  • 000° from the beacon
  • 030° from the beacon
  • 060° from the beacon
  • 090° from the beacon


  1. How can a pilot reduce the effect of coastal refraction when flying over the sea?


  • By flying at high altitude and using NDB well inland
  • By flying at low altitude and using NDB well inland
  • By flying at high altitude and using NDB close to the coast
  • By flying at low altitude and using NDB close to the coast


  1. An aircraft is heading a constant 045°(M). On which magnetic bearing from the aircraft will the ADF be most subject to quadrantal error?


  • 045°
  • 075°
  • 090°
  • 135°


  1. The protected range for an NDB is valid:
    • at all times?
    • only at night?
  • only during the day?
  • only over the sea?


  1. Given all other factors equal, which of the following will be received at the greatest range during the day?
  • A signal of frequency 250 kHz over the sea
  • A signal of frequency 450 kHz over the sea
  • A signal of frequency 250 kHz over the desert
  • A signal of frequency 450 kHz over the desert


  1. What is the maximum range at which ADF can be safely used at night?
    • The published protected range
    • Half the published protected range
    • 70 nm
    • 200 nm

Leave a comment

Your email address will not be published. Required fields are marked *

5 thoughts on “NDB/ ADF

  • Arunaksha Nandy Post author

    Guys, if you find anything useful in the site kindly leave a comment. In case there is any suggestion to improve then those are also welcome. Thank you

    • Arunaksha Nandy Post author

      Thank you Yash for bringing it to my notice. I made an error while typing it. Now I have explained the full thing above, you can have a look and see if it satisfies your doubt.

  • YAsh

    Can you explain the logic used behind questions on Wing Tip Brg ? Are those questions complete ? I guess there is insufficient information provided

    • Arunaksha Nandy Post author

      Now you can have a look at the explanation of the wing tip bearing with diagram. Hope that satisfies your query If anymore explanation needed do let me know. Thank you Yash.