The Turn and Slip IndicatorNavigation — Instrumentation — DGCA CPL practice questions

All 6 questions — The Turn and Slip Indicator

Navigation — Instrumentation · DGCA CPL. The correct option is marked on each.

1. Q1. The rate of turn indicator uses (i) ............... which spins (ii)...................

   • A.(i) a space gyroscope — (ii) up and away from the pilot
   • B.(i) a tied gyro — (ii) anticlockwise when viewed from above
   • C.(i) a rate gyro — (ii) up and away from the pilot✓
   • D.(i) an earth gyro — (ii) clockwise

   Why: The turn indicator uses a rate gyro (one gimbal, spring restrained, measuring precession caused by yaw rate). The rotor spins "up and away from the pilot" in British instruments (anticlockwise when viewed from the front — but the question refers to the direction of spin visible from the pilot's view, which is up and away). See Section 2 . — Rate gyro is the key gyro type for the TBI. It has only one gimbal (unlike two in the DGI and AH), and uses precession (not rigidity) to measure rate of turn.

2. Q2. The gyro in a rate of turn indicator has (i) ....................... operating speed than the gyros used in other instruments because (ii)……………........

   • A.(i) a lower — (ii) a higher rigidity is not required✓
   • B.(i) the same — (ii) it uses the property of rigidity
   • C.(i) a higher — (ii) a low precession rate gives a greater operating range
   • D.(i) variable — (ii) more than one rate of turn is desired

   Why: The TBI rate gyro uses the property of precession (not rigidity) to measure rate of turn. High rigidity would be undesirable because a stiffer gyro requires a larger torque to precess, making the instrument less sensitive to small rates of turn. Low rotor speed = lower rigidity = easier to precess = more sensitive rate measurement. See Section 2 and Section 4 . — This is a classic exam question. Contrast with DGI and AH which require high rigidity. The TBI deliberately uses low rigidity for precession-based rate sensing.

3. Q3. The TBI shown alongside indicates: [image shows needle left of centre at Rate 1, ball left of centre]

   • A.a rate of turn to the left, slipping in✓
   • B.an aircraft taxiing and turning starboard
   • C.that the aircraft will complete a turn in one minute
   • D.the aircraft is yawing to the right

   Why: Needle left of centre = turning left. Ball left of centre (inside the turn) = slipping in (too much bank for the TAS and rate of turn). See Section 8 . — Rate 1 = 2 minutes for 360°. Rate 2 = 1 minute for 360°. "Slipping in" = ball inside the turn.

4. Q4. When the pointer of a rate of turn indicator shows a steady rate of turn:

   • A.the calibrated spring is exerting a force about the lateral axis equal to the rate of turn
   • B.the force produced by the spring is producing a precession equal to but opposite to the rate of turn is correctly banked
   • C.the spring is providing a force which produces a precession equal to the rate of turn (in the opposite direction)
   • D.the spring is providing a force which produces a precession equal to the rate of turn (in the correct direction)✓

   Why: At equilibrium, the spring tension provides a secondary torque that generates a secondary precession equal in rate to the aircraft's rate of turn. This precession is in the direction that maintains the equilibrium (it is in the same direction as the turning tendency that produced the primary precession). See Section 3 . — The equilibrium concept: spring tension ↔ rate of turn. Greater turn rate = larger spring stretch = larger secondary torque = larger secondary precession = larger gimbal tilt = larger needle deflection.

5. Q5. If the filter of the air driven rate of turn indicator becomes partially blocked:

   • A.the aircraft will turn faster than indicated✓
   • B.the instrument will over-read
   • C.the rate of turn indicated will be unaffected
   • D.the radius of the turn will decrease

   Why: A partially blocked filter reduces suction → rotor underspeeds → reduced gyroscopic rigidity → the same rate of turn requires less spring tension → smaller gimbal tilt → instrument under-reads. The actual rate of turn is higher than the indicated rate — so the aircraft is turning faster than the indicator shows. See Section 5 . — "Blocked filter → underreads → aircraft turns FASTER than shown." This is a safety-critical point. The pilot may think they are at Rate 1 but are actually turning faster.

6. Q6. The radius of a turn at rate 1, and TAS 360 kt is:

   • A.10 NM
   • B.5 NM
   • C.7.5 NM
   • D.2 NM✓

   Why: Rate 1 = 3°/sec = 360° in 2 minutes. At 360 kt, distance in 2 min = 360 × (2/60) = 12 NM = circumference. Diameter = 12 ÷ π = 12 × 7/22 ≈ 3.8 NM. Radius = 3.8 ÷ 2 ≈ 1.9 NM ≈ 2 NM . See Section 10 . — Radius ≈ TAS(kt) ÷ 190 (in NM). So 360 ÷ 190 ≈ 1.9 NM ≈ 2 NM. Diameter ≈ TAS ÷ 95.