Why an out-and-back never breaks even
Set a 7 knot boat against a 2 knot current, out 9 miles and back 9 miles. Out, the boat makes good 5 knots and takes 1 hour 48. Back, it makes good 9 knots and takes 1 hour. Total 2 hours 48. In still water the same 18 miles takes 2 hours 34. The current cost fourteen minutes despite giving back every knot it took.
The reason is that time is distance over speed, and averaging speeds is not the same as averaging times. You are stuck in the slow half for longer than you enjoy the fast half, so the slow half has more weight in the total. The relationship is clean enough to memorise:
Round-trip time = still-water time ÷ (1 − (current ÷ boat speed)²)
| Current | As a fraction of a 7 kt boat | Out 9 nm | Back 9 nm | Total | Penalty |
|---|---|---|---|---|---|
| 0 kt | 0% | 1 h 17 | 1 h 17 | 2 h 34 | — |
| 1 kt | 14% | 1 h 30 | 1 h 08 | 2 h 38 | +2.1% |
| 2 kt | 29% | 1 h 48 | 1 h 00 | 2 h 48 | +8.9% |
| 3 kt | 43% | 2 h 15 | 0 h 54 | 3 h 09 | +22.5% |
| 4 kt | 57% | 3 h 00 | 0 h 49 | 3 h 49 | +48.5% |
| 5 kt | 71% | 4 h 30 | 0 h 45 | 5 h 15 | +104% |
| 6 kt | 86% | 9 h 00 | 0 h 42 | 9 h 42 | +277% |
The shape of that column is the useful part. Up to about a third of boat speed the penalty is small enough to ignore in planning. Past half, it stops being a nuisance and starts being the thing that decides whether the trip happens. At 6 knots against a 7 knot boat, nearly the entire day is the outbound leg.
What counts as the current for a leg
This page wants the along-track component: the part of the current parallel to the leg, positive with you and negative against. If the current runs at an angle to the leg, that component is drift × cos(angle between the set and the track), and it is smaller than the drift itself. A 2 knot current 60 degrees off your track only contributes 1 knot along it. The remaining sideways part does not slow you down; it makes you steer, which is a separate calculation.
The temptation is to enter the full drift and be done. On a leg that runs straight up or down the stream that is correct. On anything else it overstates the effect, sometimes badly — at 75 degrees off, only a quarter of the current is doing anything to your speed.
Legs are for tide changes, not just for turns
Most people break a passage into legs at waypoints, which is natural because that is where the course changes. The more useful reason to add a leg is that the tide changed. A six-hour passage through a tidal stream spends part of it with the stream and part against, and modelling it as one average is exactly the mistake the penalty table above warns about — averaging currents understates the time.
A practical division is one leg per hour of tidal stream where the stream matters, taking the rate for that hour from the atlas. Four legs is enough for a short day. Beyond that the honest answer is that the forecast granularity has run out before the arithmetic has.
Stops are the part people forget
The running time is arithmetic and the elapsed time is a plan. Between them sit the things that are not motion: waiting for a bridge, taking on fuel, an hour at anchor for lunch, the twenty minutes it takes to get the anchor up and stowed. Adding them at the end of each leg is crude but it is closer than leaving them out, and it moves the arrival time by more than most people expect over a four-leg day.
What it does not model is a wait that depends on the clock rather than the plan — a bridge that opens on the half hour, a lock that runs to a schedule, a bar you can only cross near high water. Those pull the whole plan around them rather than adding to it, and they are usually the reason a departure time gets chosen in the first place.
Questions people ask
Why does a round trip through a current take longer when I get the current back on the way home?
Because you spend more hours in the slow half than in the fast half, so the slow half counts for more in the total. The clean statement is that the round-trip time is the still-water time divided by one minus the square of the current as a fraction of boat speed. For a 7 knot boat in a 2 knot stream that factor is 1.089, so the trip takes about nine percent longer. The squaring is why the effect is negligible for a fast boat and dominant for a slow one: the same 2 knots against a 25 knot boat costs less than one percent, and against a 4 knot boat it costs a third of the day.
Should I enter the full current or only part of it?
Only the part that runs along the leg. That is the drift multiplied by the cosine of the angle between the current set and your track, so a 2 knot current 60 degrees off the leg contributes 1 knot and a 2 knot current 30 degrees off contributes 1.73. The sideways remainder does not change your speed along the track in this model; it changes the heading you have to steer, which is a separate piece of trigonometry and lives on the course to steer page. Entering the full drift on a leg that crosses the stream will overstate its effect, sometimes by a factor of three or four.
What does the calculator do when the current is stronger than the boat?
It reports that the leg does not complete rather than printing a huge or negative time. A leg with a foul current at or above your speed through the water has a speed over ground of zero or less, and the arithmetic for time gives infinity or a negative number, neither of which means anything. The honest output is that the plan as entered has no arrival time. In practice this is what happens in a narrow inlet at peak flow with a small engine, and the answer is to go at a different state of tide rather than to look for a better number.
Is the average speed here the same as the average of my leg speeds?
No, and this is the same trap as the round trip. Average speed made good is total distance divided by total running time, which weights each leg by the time it took, not by its length or by how many legs there are. A plan with a 9 mile leg at 5 knots and a 9 mile leg at 9 knots has a simple speed average of 7 knots and a real average made good of 6.43. The page reports the second one, because that is the one that predicts the arrival time. The simple average predicts nothing.
Does this include fuel reserve?
No. If a burn rate is entered it multiplies by the running time to give fuel for the plan as written, with nothing held back and no allowance for the passage taking longer than it says. What you carry, how much of the tank is actually usable, and what fraction you keep untouched are separate decisions, and there is no reserve rule stated anywhere on this site as a fact, because reserve is a judgement about the water you are in and the boat you are in. The boat fuel range calculator takes the reserve percentage as your input and works range around it.