3.2 Multi-Hour Courses to Steer & Changing Tidal Streams

3.2 Multi-Hour Courses to Steer & Changing Tidal Streams

When planning a passage across open sounds, bights, or channels, the skipper must balance steering efficiency against spatial safety. Tidal streams cannot simply be averaged by taking an arithmetic mean of direction and speed. They must be resolved as vectors, either hour-by-hour or as a cumulative vector sum.

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Method 1: Stepped Hourly Courses to Steer

In the stepped method, the navigator divides the voyage into 1-hour segments and calculates a fresh Course to Steer for each consecutive hour:

1. Hour 1: From departure Waypoint A, lay off Hour 1 tide. Swing 1 hour STW to intersect the ground track at Ground Waypoint 1 (GW1). Derive CTS for Hour 1.
2. Hour 2: From GW1, lay off Hour 2 tide. Swing 1 hour STW to intersect the ground track at Ground Waypoint 2 (GW2). Derive CTS for Hour 2.
3. Consecutive Hours: Repeat for each subsequent hour until reaching destination B.

Practical Advantages and Limitations

• Track-Keeping: With the assumed constant vectors, each step follows the planned line. In practice, changing tide, leeway and steering errors require position checks and adjustments between calculated hourly positions.
• Planned Clearances: Check each leg against hazards and position uncertainty; the plotted line alone does not ensure the actual yacht maintains its clearance.
• Operational Cost: Requires course alterations every hour, resetting compass headings, and re-trimming sails or re-balancing helm on sailing vessels.

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Method 2: Single Composite Vector Triangle

When navigating across wide, deep waters clear of hazards, computing a single composite course avoids hourly alterations. The composite method sums consecutive tidal vectors into one resultant vector before swinging the total distance run through water.

Construction Steps (3-Hour Passage)

1. Draw Total Ground Track: Connect departure Point A to destination Point B with a line marked with two arrowheads (>>).
2. Sum Tidal Vectors End-to-Tail:
   • From departure Point A, lay off tidal vector for Hour 1 to point $T_1$.
   • From point $T_1$, lay off tidal vector for Hour 2 to point $T_2$.
   • From point $T_2$, lay off tidal vector for Hour 3 to Point C.
   • Line AC represents the Net Cumulative Tidal Vector. Mark line AC with three arrowheads (>>>).
3. Calculate Total Distance Through Water: $\text{Total Water Run} = \text{STW (knots)} \times \text{Total Duration (hours)}$ For example, 5.0 knots × 3 hours = 15.0 NM.
4. Swing Total Distance Arc: Place dividers set to 15.0 NM at Point C (cumulative tide tip), and swing an arc to intersect the Ground Track line at Point D.
5. Derive the Single Water Track: Draw line CD, marked with one arrowhead (>). Direction from C to D is the single True Water Track to steer for the entire 3 hours.
6. Apply Leeway, Variation, and Deviation: Correct this True course through CADET to establish the compass course.

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The Mid-Passage Cross-Track Hazard (The "Banana Curve")

The fundamental danger of composite CTS is that even when the duration has been solved to reach the destination, the yacht need not follow the straight ground track during intermediate hours. Instead, the yacht traces a curved trajectory over the seabed (the "banana curve") as varying hourly tides sweep it laterally.

If the tidal stream sets strongly east during the first two hours, the yacht will be swept several nautical miles east of the track before a reversing tide in the third hour carries it back west. If an offshore shoal, wind farm, or shipping lane lies 1 to 2 miles east of the drawn track line, a yacht on a composite CTS will wander directly into danger.

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Worked Numerical Comparison: Stepped vs. Composite CTS

Passage Parameters

• Track: 15.0 NM due South (Ground Track = 180° True).
• Yacht STW: 5.0 knots (Nominal 3-hour water run = 15.0 NM).
• Hourly Tidal Streams:
  • Hour 1: Set = 090°(T), Drift = 2.5 knots (2.5 NM East)
  • Hour 2: Set = 090°(T), Drift = 1.5 knots (1.5 NM East)
  • Hour 3: Set = 270°(T), Drift = 1.0 knot (1.0 NM West)

Composite CTS Resolution

• Net Cumulative Tidal Vector (AC): $\text{Net Drift} = +2.5\text{ NM} + 1.5\text{ NM} - 1.0\text{ NM} = +3.0\text{ NM East (090°T)}$
• Vector Triangle Geometry: From Point C (3.0 NM East), swing a 15.0 NM arc across the 180°(T) ground track: $\sin(\theta) = \frac{3.0\text{ NM}}{15.0\text{ NM}} = 0.2000 \implies \theta = 11.54^\circ \approx 11.5^\circ$ Because net tide sets East, steer 11.5° West of South into the stream: $\mathbf{\text{Composite True CTS} = 191.5^\circ\text{(T)} \approx 192^\circ\text{(T)}}$
• Southward Ground Progress in 3 Hours: Because the ground track (180°T) and tidal stream (090°T) are perpendicular, the achieved ground progress is the adjacent side of the right-angled vector triangle: $\text{Ground Progress}_{\text{composite}} = \sqrt{15.0^2 - 3.0^2} = \sqrt{225 - 9} = \sqrt{216} \approx \mathbf{14.70\text{ NM South (14.697 NM)}}$ Average Speed Over Ground: $\text{SOG}_{\text{avg}} = \frac{14.70\text{ NM}}{3\text{ hr}} = 4.90\text{ knots}$.
• Cross-Track Position at Hour 2:
  • Steaming 2 hours on 191.5°(T) at 5.0 kn moves the boat $10.0 \times \sin(11.54^\circ) = 2.0\text{ NM West}$ through water.
  • Cumulative tidal drift during Hours 1 and 2 is $2.5 + 1.5 = 4.0\text{ NM East}$.
  • Net lateral displacement over ground: $+4.0\text{ NM} - 2.0\text{ NM} = \mathbf{+2.0\text{ NM East of track line!}}$
  • If an offshore sandbank lies 1.5 NM East of the planned line, the composite course causes grounding before the tide turns.

Stepped CTS Resolution

• Hour 1 (Tide 2.5 kn East): $\sin(\theta_1) = \frac{2.5}{5.0} = 0.5000 \implies \theta_1 = 30.0^\circ \implies \mathbf{\text{CTS}_1 = 210^\circ\text{(T)}}$ $\text{Ground Progress}_1 = \sqrt{5.0^2 - 2.5^2} = \sqrt{18.75} \approx \mathbf{4.33\text{ NM South (4.3301 NM)}} \quad (\text{SOG}_1 = 4.33\text{ kn})$ Cross-track offset = 0.0 NM (pinned to 180°T line at GW1).
• Hour 2 (Tide 1.5 kn East): $\sin(\theta_2) = \frac{1.5}{5.0} = 0.3000 \implies \theta_2 = 17.5^\circ \implies \mathbf{\text{CTS}_2 = 197.5^\circ\text{(T)} \approx 198^\circ\text{(T)}}$ $\text{Ground Progress}_2 = \sqrt{5.0^2 - 1.5^2} = \sqrt{22.75} \approx \mathbf{4.77\text{ NM South (4.7697 NM)}} \quad (\text{SOG}_2 = 4.77\text{ kn})$ Cumulative southward progress after 2 hours: $4.3301 + 4.7697 = 9.10\text{ NM South}$. Cross-track offset = 0.0 NM.
• Hour 3 (Tide 1.0 kn West): $\sin(\theta_3) = \frac{1.0}{5.0} = 0.2000 \implies \theta_3 = 11.5^\circ \implies \mathbf{\text{CTS}_3 = 168.5^\circ\text{(T)} \approx 168^\circ\text{(T)}}$ $\text{Ground Progress}_3 = \sqrt{5.0^2 - 1.0^2} = \sqrt{24} \approx \mathbf{4.90\text{ NM South (4.8990 NM)}} \quad (\text{SOG}_3 = 4.90\text{ kn})$ Cross-track offset = 0.0 NM.
• Total Stepped Progress in 3 Hours: $\text{Total Ground Run}_{\text{stepped}} = 4.3301 + 4.7697 + 4.8990 = \mathbf{14.00\text{ NM South (13.999 NM)}}$

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The Essential Reality: Duration Trial vs. Ground Progress

Why Ground Progress Falls Short

For the purely east-west tides in this example, crabbing reduces the southward component of the water velocity:

• In a purely beam tide, holding the ground-track direction makes SOG less than STW. A tide with a favourable along-track component can instead make SOG exceed STW; use the complete vector sum.
• In Stepped CTS, the heavy 2.5-knot beam tide in Hour 1 forces a 30° crab angle, slashing SOG to 4.33 knots. Over 3 hours, cumulative ground progress is only 14.00 NM, leaving the yacht 1.00 NM short of Waypoint B.
• In Composite CTS, a constant 11.5° crab angle produces an average SOG of 4.90 knots. Cumulative ground progress is 14.70 NM, leaving the yacht 0.30 NM short of Waypoint B.

How Navigators Solve the Remaining Distance

The three-hour data do not specify the tide after Hour 3, so they do not determine an arrival time. Obtain that tide before extending either plot.

For a completed exercise extension, assume the tide remains 270°T at 1.0 knot throughout Hour 4, with STW still 5.0 knots and no leeway. Both boats are on the planned line after three hours. A fresh triangle gives a water track of 168.5°T and southward SOG of 4.899 knots:

1. Stepped arrival: Remaining distance is $15.0-13.9988=1.0012$ NM. Time is $1.0012/4.899 \times 60=12.3$ minutes, giving a total of about 3 hours 12 minutes.
2. Composite arrival: Remaining distance is $15.0-14.6969=0.3031$ NM. Alter from the composite 191.5°T water track to 168.5°T. Time is $0.3031/4.899 \times 60=3.7$ minutes, giving a total of about 3 hours 4 minutes.
3. Planning before departure: Iterate the trial duration using the tidal displacement over that entire interval. Check the intermediate positions as well as the final endpoint. If the fourth-hour tide differs from this exercise assumption, recalculate both the final course and ETA.

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Comparison Table & Skippering Decision Rules