Celestial Navigation · Amplitudes — Formula
Using sin(amplitude) = sin(dec) ÷ cos(lat), for declination N18° at latitude 45°N the amplitude is about:
- AE18°N
- BE26°N✓ Correct
- CE45°N
- DE13°N
Explanation
sin A = sin 18° ÷ cos 45° = 0.3090 ÷ 0.7071 = 0.4370, so A = 25.9° ≈ 26°. Named from east toward the declination's pole: E26°N (rising bearing ≈ 064°T).
Authority: Bowditch (Pub. No. 9), Amplitudes
Practice the full Celestial Navigation bank
Free spaced-repetition quizzing across 2550 USCG exam questions — it schedules your reviews so the ones you miss come back until they stick.
Related Celestial Navigation questions
- Amplitudes — Formula
The amplitude of a celestial body is found from sin(amplitude) = sin(declination) ÷ cos(latitude). For dec N20° at latitude 40°N, the amplitude is approximately:
- Time
The Nautical Almanac tabulates the positions of celestial bodies against which time standard?
- Time
A vessel in west longitude has a zone description of +5. To convert zone time to UT you must:
- Coordinates
How is the GHA of a star obtained from the Nautical Almanac?
- Coordinates
Sidereal Hour Angle (SHA) is best defined as:
- Coordinates
Declination of a celestial body is the celestial equivalent of: