Course to Steer to Offset a Known Current

What the rule requires

The current triangle is stated in one line: course steered plus the drift vector equals course made good, and the same triangle inverted gives the course steered required to make a given track over the ground Bowditch Ch. 24 §2402. Every course-to-steer problem is that inversion. Nothing new is added; what changes is which leg you know and which one you are solving for.

Three legs, and each one is a direction paired with a speed:

• Course steered and speed through the water. What you pass to the helm. This is the leg the dead reckoning plot is built from, since a DR position advances the vessel on course steered and distance run through the water alone Bowditch Ch. 8 §801.
• Set and drift. Set is the direction toward which the current flows; drift is its speed Bowditch Ch. 8 §803.
• Course and speed made good. The resultant. In this problem its direction is given to you as the intended track, and its magnitude — the speed of advance along that track — comes out of the construction Bowditch Ch. 23 §2302.

In the forward problem you know the first two legs and want the third. In the course-to-steer problem you know the direction of the third leg and the whole of the current leg, and you want the direction of the first. That is why the solution needs an arc rather than a straight vector sum: the steered leg's length is fixed by your speed through the water, but its direction is the unknown, so you swing that length until it lands on the track.

All three legs must be laid off over the same time interval. The current leg is laid off for the elapsed time at the drift , so if you use one hour of current you must use one hour of the vessel's run. One hour is the working standard because the vector lengths in nautical miles then equal the speeds in knots and nothing has to be scaled twice.

Directions are true. The plotted course line carries the true course above it as C 090 and the speed below as S 8, with each position labelled by 24-hour time Bowditch Ch. 8 §802. A set taken as magnetic from one source and a track scaled as true from the compass rose will not close correctly.

The steered course lies upcurrent of the intended track. The current leg displaces the head of the triangle toward the set, so the closing leg must lean the other way to bring the resultant back onto the track. Say it as a direction check before you plot: if the current sets across your track to starboard, the course to steer is to port of the track.

Your intended track is 000°T. The current sets 090°T at 2.0 knots. Before plotting anything, is the course to steer greater or less than 000°?

Less than 000° — that is, in the 340s or 350s. The current sets across the track toward the east (starboard side), so the course to steer must fall to the west of the track, upcurrent. Any answer option to the east of 000° is wrong before you compute a single degree, which on a multiple-choice question usually deletes two of the four.

Telling it apart: which leg is the unknown

The two current problems use the same triangle and differ in one thing only — the leg you are solving for. That determines where the current vector is laid off, and getting that backwards is the single most expensive error on this topic.

• Course to steer (this lesson). The unknown is the direction of the steered leg. The track is fixed, so the current vector is laid off from the starting fix and the steered leg closes the triangle onto the track line by arc Bowditch Ch. 24 §2402.
• Estimated position. The unknown is where you will be. The course steered is fixed, so the current is laid off from the DR position for the elapsed time, and the end of that vector is the EP, plotted as a dot inside a square Bowditch Ch. 8 §803.
• Set and drift as a result. Neither is given; you derive them by comparing a DR position with a fix for the same time, the DR having ignored current and leeway from the outset Bowditch Ch. 8 §801.

The vocabulary trips people as reliably as the geometry. Two checks are in general circulation and both hold against the definitions: set is the direction the current is set toward, and drift is the only one of the pair measured in knots. Current is also given by the direction it flows toward, which is the opposite convention from wind.

Working a question

You are steaming out of an approach channel. Your 0900 fix is on the chart, the next waypoint bears 340°T and lies 12.0 NM off, and you intend to make good the straight track between the two. Speed through the water is 10.0 knots. The current sets 040°T at 3.0 knots. Find the course to steer, the speed of advance, and the ETA.

1. Draw the intended track from the 0900 fix, 340°T, and extend it past the waypoint. This line is what you must make good; it is not yet a course line and carries no C and S label, because it is not what you will steer.
2. From the same fix, lay off the current: direction 040°T, length 3.0 NM. Direction 040 because set is the direction toward which the current flows Bowditch Ch. 8 §803, and 3.0 NM because that is one hour of drift. Note where the current is relative to the track: 040 is 60° clockwise of 340, so the current has a component across the track to starboard, and the course to steer will be to port of 340.
3. Set the dividers to 10.0 NM — one hour of speed through the water — and from the head of the current vector swing an arc to cut the track line. Speed through the water, not speed of advance, because the steered leg represents motion through the water Bowditch Ch. 8 §801. If the arc will not reach the track line at all, the across-track component of the current exceeds your speed through the water and the track cannot be made good at that speed.
4. Read the direction from the head of the current vector to the arc intersection: 325°T. That is the course to steer. It is 15° to port of the track, on the upcurrent side, which is the check you made at step 2.
5. Measure the fix to the intersection along the track: 11.2 NM. Laid off over one hour, that is a speed of advance of 11.2 knots. It exceeds 10 knots because the current has a component along the track as well as across it; had the current set 040 against a track of 220, the same 3.0 knots would have cost you speed instead.
6. Time to run 12.0 NM at 11.2 knots is 1 h 05 m, so the ETA is 1005. ETA comes off speed of advance. Using 10 knots here would put the ETA at 1112 and lose the point.
7. Label the course line you will actually run C 325 above and S 10 below, and label positions by 24-hour time Bowditch Ch. 8 §802. Positions advanced along 325 at 10 knots are DR positions and will plot to port of the intended track, because DR ignores the current entirely.

Same problem, but the drift is 6.0 knots instead of 3.0. Roughly what happens to the course to steer and the speed of advance?

The offset roughly doubles and the speed of advance rises: course to steer approximately 309°T, speed of advance about 11.6 knots. The current vector doubles in length, so the arc of 10.0 NM has twice as far to lean back to reach the track, and the along-track component of the current doubles as well. Both legs of the answer move, which is why a question that changes only the drift can still change the correct course and the ETA.

Where candidates lose the point

Laying the current off in the reciprocal. The candidate treats set the way wind is reported and draws the vector from 040 toward 220. The triangle then closes on the wrong side and the answer offered as 355°T looks correct. Set is the direction toward which the current flows Bowditch Ch. 8 §803, so a set of 040 is drawn toward 040.

Steering downcurrent. With the current on the starboard bow, the answer 355°T is offered alongside 325°T and attracts anyone who applied the offset arithmetically without looking at the plot. The resultant of a course steered to starboard of the track and a current setting to starboard of it cannot lie on the track Bowditch Ch. 24 §2402.

Swinging the arc with the wrong speed. Using the required speed of advance, or the intended speed over the ground, in place of speed through the water. The steered leg is through-water motion, which is the only motion the DR plot recognises Bowditch Ch. 8 §801. A question that gives both a speed through the water and a speed of advance is testing exactly this.

Mismatched time bases. One hour of drift against half an hour of run, or a current tabulated over some other interval used at face value. The current leg is laid off for the elapsed time ; if the legs do not share an interval the triangle is meaningless even when the plotting is neat.

Computing the ETA on speed through the water. Distance along the track is made good at the speed of the resultant, which is the vector sum of the vessel's motion and the current Bowditch Ch. 23 §2302. In the worked example that is 11.2 knots, not 10.

Using a set from the wrong hour. Tidal currents reverse direction with the tide Bowditch Ch. 24 §2401, so a set lifted from a different stage of the tide than the run will put the offset on the wrong side of the track and produce an error of twice the drift angle.

Plotting DR positions along the intended track. Having found the course to steer, the candidate then advances positions up the track line and labels them as DR. A DR position is generated from course steered and distance run through the water, ignoring current ; the position you expect to occupy on the track is the DR corrected for set and drift, which is the EP, a dot inside a square .

Check yourself

Your intended track is 090°T at 12 knots through the water. The current sets 180°T at 3 knots. Is the course to steer north or south of 090°, and why?

North of 090 — in the 070s. The set of 180 is 90° clockwise of the track, so the current carries the vessel to starboard of it, and the steered course must lie upcurrent, to port of the track. The offset is set by the across-track component of the current against the speed through the water, here 3 against 12, which comes out at about 14°: course to steer roughly 076°T, speed of advance about 11.6 knots.

A question gives you the course steered, the speed, the set and the drift, and asks for the vessel's position in one hour. Which construction applies?

The forward triangle, not the course-to-steer construction. Advance the DR from the fix on the course steered at the speed through the water, then lay the set and drift off from that DR position for the elapsed time; the end of the current vector is the estimated position, plotted as a dot inside a square. The intended track plays no part, and no arc is swung.

Track 200°T, speed through the water 8 knots, set 200°T at 2 knots. What is the course to steer and the speed of advance?

Course to steer 200°T and speed of advance 10 knots. The current is dead astern with no across-track component, so the arc lands on the track line without any offset and the triangle collapses into a straight line. The distractors here are the ones that offset the course anyway; a fair current changes the ETA, not the heading.

Set 315°T at 4.0 knots, intended track 045°T, speed through the water 3.5 knots. What do you report?

That the track cannot be made good at that speed. The current sets 90° across the track, so the whole 4.0 knots must be offset by the across-track component of a 3.5-knot leg, which is impossible: the arc of 3.5 NM never reaches the track line. Increasing speed through the water above 4.0 knots is what makes the triangle close.

You are told to make good 010°T, the current sets 010°T at 2 knots, and you must cover 15 NM in one hour. Your speed through the water is 12 knots. Does the passage work?

Yes, with about a knot to spare. The current is fair and along the track, so course to steer is 010°T and the speed of advance is 12 plus 2, or 14 knots — short of 15. Speed through the water of 13 knots would be required. The point being tested is that the resultant's magnitude, not the vessel's speed through the water, is what covers ground.

Your course line is labelled `C 325` and `S 10`, and the intended track is 340°T. At 1100 you take a fix that plots on the track. What does the fix tell you about the current used in the solution?

That the set and drift you allowed for were close to correct. The DR position for 1100, advanced on 325 at 10 knots, lies to port of the track; the fix lying on the track means the current moved the vessel from that DR back onto it as predicted. Had the fix plotted off the track, the difference between the DR position and the fix for the same time gives the actual set and drift, which is then used for the next leg.