FAA Instrument Rating Airplane (IRA)Navigation SystemsHard
A localizer has a total course width of 5°, and the transmitter antenna is located approximately 5 NM from the outer marker on final approach. What is the approximate total (full-scale) width of the localizer course at the outer marker?
- A5,300 feet
- B2,650 feet
- C1,325 feet
- D875 feet
Show answer & explanationAnswer & explanation
Correct answer: B. 2,650 feet
Convert 5 NM to feet: 5 × 6,076 = 30,380 ft. Half of the 5° course width is 2.5°; half-width = 30,380 × tan(2.5°) ≈ 30,380 × 0.0437 ≈ 1,327 ft. The total (both-sides) course width = 2 × 1,327 ≈ 2,650 feet.
Why the other options are wrong
- A. 5,300 ft overstates the width; it doubles the correct total incorrectly.
- C. 1,325 ft represents only the half-width, not the total course width.
- D. 875 ft is too narrow for the given distance and course angle.
Localizer Course Width
The localizer signal width is calibrated so that full-scale CDI deflection typically represents about 700 feet on each side of centerline at the runway threshold, resulting in an angular course width usually between 3° and 6°.
- Typical full course width: 3°–6° (commonly ~5°)
- Width in feet increases with distance from the antenna
- Formula: half-width (ft) = distance (ft) × tan(half the course angle)
Memory trick: 'Farther out, the beam spreads out.'