The only part of the current that costs you a heading
Split the current into two pieces relative to the line you want to travel. One piece runs along that line, forwards or backwards. The other runs across it. The along-track piece changes your speed and nothing else — there is no steering answer to it, you simply arrive earlier or later. The across-track piece is the whole problem, and cancelling it is the whole calculation.
The split is a cosine and a sine of the angle between the current set and your track. At twenty degrees off the bow, a current is 94 percent along and 34 percent across. At ninety degrees it is all across and none along. This is why crossing a tidal stream at right angles is the expensive case and running up or down one is free.
Cancelling the cross-track part is then a right-angle problem: the boat has to point far enough upstream that its own sideways component equals the current's, and nothing more. The angle it takes is arcsin(cross-track drift ÷ boat speed through the water). That is the crab angle. Notice what is not in it: the along-track current, the distance, and the time. None of them change the heading.
Six knots of boat against two knots of current
The table is one boat, one current, and the angle between the current and the track varied through a half circle. Track east, boat doing 6 knots through the water, current 2 knots.
| Current relative to track | Across the track | Crab angle | Speed over ground |
|---|---|---|---|
| 0° — dead astern, pushing | 0.00 kt | 0.0° | 8.00 kt |
| 30° | 1.00 kt | 9.6° | 7.65 kt |
| 45° | 1.41 kt | 13.6° | 7.25 kt |
| 60° | 1.73 kt | 16.8° | 6.74 kt |
| 90° — on the beam | 2.00 kt | 19.5° | 5.66 kt |
| 120° | 1.73 kt | 16.8° | 4.74 kt |
| 180° — dead ahead, foul | 0.00 kt | 0.0° | 4.00 kt |
Two things fall out of it. The crab angle is symmetric about the beam — the same 16.8 degrees at 60 and at 120 — while the speed over ground is not remotely symmetric, 6.74 one way and 4.74 the other. And the beam case, which needs the most steering, is not the slowest case; the foul case, which needs none at all, is.
Set is stated backwards from wind
A northerly wind blows from the north. A current setting north flows toward the north. There is no good reason for the two conventions to disagree and they have disagreed for centuries, and it produces a specific and expensive mistake: entering the current 180 degrees out, steering the wrong way, and doubling the offset instead of cancelling it. If the tidal atlas arrow points at the top of the page, the set is 000, not 180.
The check that catches it costs nothing. Look at whether the answer steers you into the current or away from it. The heading should always be upstream of the track — into the set, never with it. If the calculator moves you downstream of the track, the set is entered backwards.
When the current is stronger than the boat
Below a certain boat speed the sine you need is greater than one and there is no heading that holds the track. This is not a failure of the arithmetic; it is a real and quite common situation for a dinghy, a kayak or a sailing boat in light air crossing a strong stream.
What exists instead is a wedge. Every ground track you can achieve lies within arcsin(boat speed ÷ drift) either side of the direction the water is going. A 2 knot boat in a 4 knot stream can only make good a course within 30 degrees of the set — a 60 degree wedge, and everything outside it is unreachable no matter how you steer. The page reports the two edges of that wedge and the heading that reaches the edge nearest the course you wanted.
Speed over ground at the edge is drift × cos(half angle), which for the 2-in-4 case is 3.46 knots — you go somewhere quickly, just not where you asked.
What this does not account for
Leeway is not in it. A boat with topsides and a rig gets pushed sideways through the water by wind as well as carried sideways by the current, and the two are separate problems: leeway is an angle between where the boat points and where it moves through the water, while set is the water itself moving. Adding a leeway allowance means adding degrees to windward on top of the answer here, and how many degrees is a property of your boat that only sailing it tells you.
Compass deviation is not in it either. The magnetic conversion offered here applies variation from the chart and stops there. Deviation is specific to your compass in your boat on that heading, and it comes from a deviation card, not a formula.
Nor is the current constant. One vector for a whole leg is a working simplification. On a leg of any length the tide turns underneath you, and the standard answer to that is to break the passage into shorter legs and solve each one with the vector that applies to it.
Questions people ask
Do I steer into the current or away from it?
Into it, always, and by exactly enough to cancel the sideways component and no more. If the water is setting you to starboard of your track, the heading goes to port of the track. The amount is arcsin of the cross-track drift divided by your speed through the water, which for a 2 knot beam current on a 6 knot boat is about 19.5 degrees. A useful sanity check on any answer, including this one, is whether the heading came out upstream of the track. If it did not, the set has almost certainly been entered as the direction the current comes from rather than the direction it goes, which is the single most common error with these numbers because it is the opposite of the wind convention.
Why does the calculator ask for speed through the water rather than my GPS speed?
Because the GPS figure already contains the current, and feeding it back in would count the current twice. Speed over ground is the answer this page produces, not an input to it. What goes in is what the boat does in still water at the throttle setting or sail trim you plan to hold, which is what a paddlewheel or ultrasonic log reads, or what you know from experience of the boat on a slack tide. If you only have a GPS, the usual workaround is to note the speed over ground on two reciprocal courses through the same water and average them, which cancels the along-track current, though it does nothing about a cross-track one.
How is speed over ground higher than my boat speed on some headings and lower on others?
Because the along-track component of the current adds to the boat vector or subtracts from it, and that component is the drift times the cosine of the angle between the set and your track. Astern of you it adds the full drift; on the beam it adds nothing; ahead of you it takes the full drift off. There is also a smaller effect that surprises people: crabbing into the current costs you speed even when the current is exactly on the beam, because the boat is no longer pointing along the track. A 6 knot boat crabbing 19.5 degrees only contributes 5.66 knots along the track, which is where the missing third of a knot goes.
What happens if the current is faster than the boat?
The track becomes unreachable and the page says so rather than returning a nonsense heading. What remains is a wedge of achievable ground tracks centred on the direction the water is going, half a wedge wide of arcsin(boat speed divided by drift) on each side. A 2 knot boat in a 4 knot stream can make good anything within 30 degrees of the set and nothing outside it. The page reports the wedge, the edge nearest the course you wanted, the heading that reaches it and the speed over ground there. What to do about it is a judgement about waiting for the tide to turn, taking a longer route through slacker water or not going, and none of that is arithmetic.
Does the crossing time change the heading?
No, and that is worth understanding because the other common way of solving this problem makes it look as though it does. The vector triangle drawn on a chart usually uses one hour of current against one hour of boat speed, which produces a triangle whose sides are distances, and people reasonably conclude that the time matters. It does not. Scale both vectors by any duration you like and the angles are unchanged, because both sides scale together. Time only enters the answer at the last step, when the speed over ground is divided into the distance. Change the distance and the heading stays exactly where it was.