Maritime Questions › Tides Celestial

You have taken a morning sun sight and obtained a corrected true altitude of 42°18'N. From your DR position you have calculated the computed altitude (Hc) as 42°32'N and the azimuth (Zn) as 096°T. Describe how you plot the position line.

A. Intercept (Marcq St. Hilaire) method plotting: (1) INTERCEPT DISTANCE: Intercept = Hc − Ho (computed altitude − observed altitude). Hc = 42°32' and Ho = 42°18'. Intercept = 42°32' − 42°18' = +14' of arc. Since Hc > Ho: the body is FURTHER from the observer than the DR position. The vessel's actual position line is AWAY from the body by 14 nautical miles (1' of arc = 1 nautical mile on the Earth's surface); (2) AZIMUTH: Zn = 096°T — the Sun bears approximately East from the DR position; (3) PLOTTING: Plot the DR position on the chart. From DR, draw a line in the direction of Zn (096°T). Measure 14nm AWAY from the Sun (i.e., in the direction 096° + 180° = 276°T, because Hc > Ho = position is away from the body). Mark the intercept point 14nm from DR in the 276° direction; (4) POSITION LINE: Through the intercept point, draw a line PERPENDICULAR to the azimuth line. This perpendicular line is the Sun Line (position line) — the vessel lies somewhere along this line.
B. Plot the intercept distance toward the Sun if Ho > Hc; away from the Sun if Hc > Ho. In this case Hc > Ho so plot 14nm toward the Sun (096°) from DR.
C. The intercept = Ho − Hc = 42°18' − 42°32' = −14'. Plot 14nm toward the Sun from DR. The position line runs through the DR position, not 14nm from it.
D. The azimuth line IS the position line. Draw a line from the DR position in the direction 096°T — the vessel lies on this line at the point where the intercept distance (14nm) is measured.
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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?
A. 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.
B. The charted clearance of 8.2m is at MHWS — since the tide at 1030 is not at MHWS, the clearance will always be greater than 8.2m. The vessel with 7.4m air draught is always safe.
C. Apply the seasonal correction directly to the air draught: 7.4 + 0.1 = 7.5m. Since 7.5m < 8.2m clearance, the passage is safe at any state of tide.
D. Without access to the full Admiralty Tide Tables the calculation cannot be completed. Use the bridge clearance at chart datum (which is always the stated value) as the worst-case figure.
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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?
A. The charted depth of 8.5m is greater than the vessel's 7.8m draught — the vessel always has positive UKC in this channel regardless of tide.
B. 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.
C. The tide is falling — entering on a falling tide is never permitted. The master must wait for the next high water.
D. At 0800 the tidal height is 4.2m (same as HW at 0600) because it takes at least 3 hours for the tide to begin falling.
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