Question 1

Your GPS and radar have failed mid-ocean. You have a sextant, chronometer, and Nautical Almanac. Describe how you would take a meridian passage sun sight to obtain your latitude.

Accepted Answer

Meridian passage (noon sight) procedure: (1) PREPARATION: Calculate from DR position the approximate LMT (Local Mean Time) of meridian passage (local apparent noon, LAN), when the Sun's GHA equals your DR longitude. From the Almanac, extract the Sun's GHA at 1200 GMT, then calculate LAN = 1200 − (GMT−LMT correction for DR longitude); (2) OBSERVATION: From approximately 15 minutes before calculated LAN, observe the Sun through the sextant — track the Sun as it rises toward its maximum altitude, "chasing" it by continually adjusting the sextant drum to keep the limb on the horizon; (3) MERIDIAN PASSAGE: The Sun's altitude stops rising and begins to fall — note the maximum altitude reading from the sextant (this is the altitude at culmination); (4) CORRECTION: Apply index error, dip (height of eye), semi-diameter (lower limb sight), and refraction corrections to get true altitude; (5) ZENITH DISTANCE: ZD = 90° − true altitude; (6) DECLINATION: Extract Sun's declination at LAN from Almanac; (7) LATITUDE: Lat = ZD ± Declination (same hemisphere: add; different hemisphere: subtract). Result = latitude at the moment of meridian passage. Celestial navigation remains in the MCA OOW syllabus (MIN 653 S3) because it is the ultimate backup navigation method — GPS denial, electronic failure, or Arctic passage where GPS accuracy degrades. The meridian passage sight is the most fundamental celestial technique because it gives latitude directly without requiring accurate time or position. At culmination (meridian passage), the body is due S (or N) of the observer and the relationship simplifies to: Latitude = Zenith Distance ± Declination. The "running fix" using two sun sights transferred along the vessel's course track is also examined — the AM sight (giving a morning LOP) is advanced along course and speed to the noon sight and crossed to give an estimated position. Corrections to the sextant altitude: (a) Index error — the residual error of the sextant instrument (may be on or off the arc, typically ±1-5'); (b) Dip — depression of the sea horizon below the horizontal due to height of eye (√height of eye in metres × 1.76 ≈ dip in minutes); (c) Main corrections — refraction (the bending of light in the atmosphere, always negative) and semi-diameter (half the angular diameter of the Sun, +ve for LL, −ve for UL). All corrections are tabulated in NP314 and the Almanac.

Question 2

The charted clearance under a bridge at MHWS is 8.2m. Your vessel has an air draught of 7.4m. HW at the standard port today is 4.8m at 1400. LW before it was 0.6m. The bridge is at a secondary port; seasonal correction +0.1m. You need to pass at 1030. Is it safe?

Accepted Answer

Step-by-step tidal calculation using the Rule of Twelfths (or use Admiralty Tide Tables co-tidal and range factor): (1) Range at standard port = HW 4.8m − LW 0.6m = 4.2m; (2) Duration of rise: LW to HW — need LW time. Assume LW was 0730 (HW 1400 = roughly 6h 30m duration — typical semi-diurnal). Duration ≈ 6h 30m; (3) Time from LW to 1030 = 3 hours; (4) Using Rule of Twelfths — first hour rises 1/12, second 2/12, third 3/12: in 3h the tide has risen (1+2+3)/12 × 4.2m = 6/12 × 4.2m = 2.1m; (5) Height at 1030 = LW + rise = 0.6 + 2.1 = 2.7m; (6) Charted MHWS = 8.2m = height of bridge above Chart Datum (CD). Bridge clearance at 1030 = 8.2 − 2.7 = 5.5m (height of bridge above 1030 water level); (7) Apply seasonal correction: +0.1m raises water level slightly: clearance = 5.5 − 0.1 = 5.4m; (8) Air draught 7.4m vs clearance 5.4m — NOT SAFE by 2.0m. The vessel cannot safely pass at 1030. Tidal clearance calculations appear in virtually every OOW oral. The examiner specifically tests whether the candidate understands that MHWS is the MINIMUM clearance marked on the chart (because that is when the water level is highest, reducing the gap to the bridge). At non-MHWS times, the water is lower than MHWS and the clearance is GREATER. However, this question reverses the trap — the water at 1030 is NOT at MHWS but still high enough to reduce clearance below the air draught. Rule of Twelfths: the tidal rise/fall in a semi-diurnal tide follows the pattern 1:2:3:3:2:1 twelfths per hour. This is an approximation valid for most semi-diurnal ports. For critical calculations (air draught, UKC), always use the Admiralty Tide Tables (NP201 series) and the tidal prediction curves for the standard port, applying secondary port differences (time and height corrections). Seasonal corrections: applied to the tabulated heights to account for seasonal variation in mean sea level (meteorological effects, barometric pressure). In UK waters the seasonal correction to MHWS is typically ±0.1-0.2m and must not be ignored for critical clearances.

Question 3

Your vessel has a maximum draught of 7.8m. The charted depth of the entrance channel is 8.5m below Chart Datum. HW was at 0600 with a height of 4.2m. You enter at 0800. The port requires 0.6m minimum UKC. Is there sufficient water?

Accepted Answer

Depth of water calculation: (1) Time from HW to 0800 = 2 hours — the tide is FALLING; (2) Range: HW 4.2m, assume LW (next) approximately 0.8m (typical UK 3.4m range). Falling tide — using Rule of Twelfths from HW: first hour falls 1/12, second hour falls 2/12; in 2 hours tide has fallen (1+2)/12 × 3.4m = 3/12 × 3.4 = 0.85m; (3) Height at 0800 = HW − fall = 4.2 − 0.85 = 3.35m above CD; (4) Actual depth of water = charted depth + tidal height = 8.5 + 3.35 = 11.85m; (5) UKC = depth of water − draught = 11.85 − 7.8 = 4.05m; (6) Required UKC = 0.6m; available UKC = 4.05m — SAFE with considerable margin. Note: in a real tidal approach, the pilot would calculate on the minimum tidal window and dynamic UKC accounting for squat at approach speed. UKC calculation integrating tidal heights and charted depths is a core OOW competency. The fundamental formula: Actual depth = Charted depth (below CD) + Tidal height above CD. UKC = Actual depth − Vessel draught. The confusion in option B is a classic trap — the charted depth alone does not define whether the vessel can navigate safely; the charted depth at chart datum combined with the tidal height at the time gives the actual available water depth. Entering on a falling tide: option C is incorrect — entering on a falling tide is standard practice; the critical factor is whether there is sufficient UKC throughout the entry. What IS prohibited is entering at a state of tide where the minimum UKC cannot be maintained throughout the transit. Squat: in real port approaches, squat (the dynamic sinkage of a vessel underway, typically 0.1-0.4m depending on speed and canal effect) must be subtracted from the available UKC. A 15-knot vessel in a shallow channel may squat 0.3-0.5m — this must be accounted for in the tidal calculation. The minimum UKC (0.6m in this question) is a port-specific requirement often specified in port authority sailing directions or the pilot card.