OK, so here’s the deal.

Capt Force’s scenario is:

1.

* Current stronger in the middle, slack at the beginning and end.

* Average force 2 knots

3. * 4 hours passage.

4. * Speed through the water of constant 5 knots

That’s not actually enough information to calculate any passage times. We need to know the distance, and we need to know the streams hour by hour. Calculations made without that data are meaningless -- in fact, the missing data points to corruption in the logic -- conclusions are being presumed ("the vectors

*must *add up, because that's the only way the two approaches can give the same result").

I’m going to make up the essential missing data. I will calculate every half hour, and I’m going to use the following streams:

0 hour - 0:30 0 knots

0:30 – 1:00 2 knots

1:00 – 1:30 2 knots

1:30 – 2:00 4 knots

2:00 – 2:30 4 knots

2:30 - 3:00 2 knots

3:00 – 3:30 2 knots

3:30 – 4:00 0 knots

OK, average 2 knots, as per the assumptions. The current is perpendicular to the rhumb line. Let’s say BTW is 180 and current runs E to W.

First, let’s calculate a valid CTS. Without doing this correctly the whole exercise is useless. In this case, it’s a simple calculation – we simply add up the set of the current, which amounts to 8 miles to the W. So we need a CTS which will put us 8 miles further E.

Next, we will need to know the distance, which Capt Force did not give us. The distance over ground is not 20 miles (sorry Andrew), because we cannot make 5 knots made good towards the waypoint with a unidirectional current across our path. Average VMG towards the waypoint must be less than boat speed. I can calculate how far I can get in 4 hours by solving a right triangle with hypotenuse of 20 miles, and C side of 8 miles – simples. So the rhumb line has a length of 18.33 miles, and we have a distance to waypoint which can be sailed in the presumed 4 hours at the presumed 5 knots, provided we sail the correct path.

We also have my course correction 23.58 degrees, and angle of the apex of the triangle. Subtract that number from 180, the bearing to waypoint, and we have our CTS.

So now we have a complete picture of the constant heading passage:

* CTS constant heading of 156.42

* Distance made good towards waypoint 18.33 miles

* Distance covered through water 20 miles

* Time on passage 4 hours.

Now let’s calculate the

GPS track passage. Since this vessel is moving in a straight line over the ground, this calculation is much easier if we do it in a ground-referenced manner.

0 hour – 0:30 HDG 180 SOG 5.0 Distance covered over ground 2.5

0:30 – 1:00 HDG 156.42 SOG 4.58 Distance covered over ground 2.29

1:00 - 1:30 HDG 156.42 SOG 4.58 Distance covered over ground 2.29

1:30 – 2:00 HDG 126.87 SOG 3.0 Distance covered over ground 1.5

2:00 – 2:30 HDG 126.87 SOG 3.0 Distance covered over ground 1.5

0:30 – 1:00 HDG 156.42 SOG 4.58 Distance covered over ground 2.29

1:00 - 1:30 HDG 156.42 SOG 4.58 Distance covered over ground 2.29

0 hour – 0:30 HDG 180 SOG 5.0 Distance covered over ground 2.5

So in four hours the

GPS track boat has sailed the same 20 miles through the water as the constant heading boat (obviously -- 4 hours * 5 knots), but has only covered 17.16 miles over ground along the rhumb line, which is its ground track. So after four hours, it has not arrived yet – it has another 1.17 miles to go. Because of the longer passage time, it will be exposed to that much more set of the current, so we can approximate with some accuracy that it will make 4.58 knots SOG along the rhumb line during this extra time (that is your SOG staying on the rhumb line with a perpendicular current of 2 knots, which is the average). So the GPS track boat will need 0.26 hours more to get there. So now we have a complete picture of the GPS track boat's passage.

Results:

* Constant heading boat gets there in 4 hours

* GPS track boat gets there in 4.26 hours

* Both boats cover the same distance made good – 18.33 miles

* The constant heading boat

sails 20 miles through water

* The GPS track boat

sails 21.3 miles through water.

* The constant heading boat sails 20 miles through water.

* The constant heading boat has an advantage of 6.5%

QED.

The difference reflects the crooked path through the water produced by the GPS track approach. It matters not whether the changing heading is a smooth progression or a series of instant changes. The varying headings themselves represent an indirect path through water. I know Capt Force doesn’t like the idea of distance through water, but not liking it doesn’t mean that it doesn't exist, just like repeating over and over again that the world is flat doesn’t make that so. We sail in water – our speed is speed through water, the constant in all of our equations, so the fastest way to get anywhere is that way which puts the smallest number of miles of water between you and your destination, which optimizes your distance through water. Exactly water – your ground path, as Dave keeps reminding us, is totally irrelevant. You cannot get away from that fact. A straight path is shorter than a crooked one, a curved one, or whatever shape, any shape at all, which varies from a straight line. Any change in heading, any single isolated change, any set of changes of any number, any constantly changing heading, whether smoothly or in jumps, any change whatsoever, means that you are deviating from the most efficient passage, that is, the shortest and fastest way to get where you are going.

The GPS and modern chart plotter are wonderful devices, but they simply do not calculate the most efficient path over a body of moving water. The navigator has to do that (the original theme of this thred). Electronic devices cannot make such calculations without current data over the whole passage.

The advantage of the constant heading approach increases with increasing current differential, and especially with reversing currents. On an

English Channel crossing, the difference is huge, truly dramatic, which is why these principles are second nature to us

English Channel sailors

As Dave said.

For anyone who STILL doesn't get it, then just picture this:

The same scenario, the same 18.33 miles, the same 4 hours and 5 knots, BUT -- the current is slack the whole way except for the last hour, when it runs at 8 knots (not absurd -- we have such currents in the Channel Islands).

So the constant bearing boat will sail exactly the same CTS and will arrive in exactly the same about of time -- 4 hours.

How about the GPS track boat?

Well, it will NEVER arrive. Never! Because at the beginning of the last hour, the GPS track boat will be on the rhumb line 5 miles away from the destination, which bears 180. But now the GPS track boat is making 5 knots through the water in an 8 knots current. Its maximum VMG towards the waypoint will be on course 090, stemming the current, but its VMG towards the waypoint will be -- negative.

NOW can anyone still hang on to the belief that the two approaches bring the same results? And there is no, zero, difference between this scenario and Capt Force's scenario, in terms of the total vectors, as witnessed by the identical strategy of the constant heading boat, bringing identical results.

Let's say we make it a little milder -- two hours of slack water followed by two hours of four knots.

The constant heading boat will still use the same CTS and will still arrive in four hours.

The GPS track boat will make 5 knots SOG for the first two hours, but then will only make 3 knots SOG at a heading of 132.8. With 9.17 miles to go, he's got more than 3 hours to go before arrival. His total journey time will be more than five hours, and he will arrive more than an hour later than the constant heading boat.