CTS vs Following Rhumb Line - DILUTED THREAD

[Dockhead]

Originally Posted by goboatingnow

I think everyone excepts in a basically symmetrical tide, that calculating overall CTS is the best method. I can't see from a purely speed perspective why anyone would argue otherwise.

However on a unidirectional tide is there any difference. Once an autopilot essentially sails a series of miniature course triangles , its something I've never done the maths on.


Dave

Constant compass heading = straight water track. So that's always the shortest and fastest way to get anywhere, assuming of course you know the correct heading. That's in the RYA training materials, even.

If the tide is unidirectional with a constant speed, your CTS = COG to your waypoint -- the two approaches converge. So your ideal constant heading happens to take you over the rhumb line.

If the tide is unidirectional with varying speeds, like the Gulf Stream for example, then you will go wrong with steering to COG -- that is, staying on the rhumb line. Here you get back to the classical CTS calculation to get your ideal way across, which is a straight line through the water, which is a constant compass heading.

And the approach is exactly the same -- you just add up the set and drift for every hour and calculate a CTS to get you to the right spot at the edge of the treadmill, to use the ant analogy, to get to the right spot by the shortest possible path.

If you change headings every hour, then you are sailing an erratic course through the water and losing miles. However straight the line may appear over ground. The key thing is really to just forget about the ground, which is not relevant when you are crossing a moving body of water, even if it is moving in only one direction.

[Dockhead]

Originally Posted by noelex 77

The simple rule is that is always quicker, all other things being equal, to calculate the average current for the voyage and then maintain that constant heading as the current changes.
Relying on the chart plotter to maintain zero cross track error is always less efficient.

The only exceptions occur when the current is the same thought the voyage. (Then the average current is identical to instantaneous current and the constant heading will maintain zero cross track error, so the results are identical).The second exception is when there is no sideways component to the current and once again a constant heading will maintain zero cross track error so the results are identical.

Indeed, but those are not exceptions. Those are simply two cases where the efficient track through the water coincides with a straight track over ground. There is no exception to the rule that the straight track through the water will be the most efficient way there -- it's an axiom -- depending only on the assumption that you are dealing with a body of water moving all together.

A case not like that -- where all bets are off -- is if for example you are navigating a river with a strong adverse current in the middle and less current along the sides. You need to bug out of the channel and skirt the banks. But that's already a completely different problem.

[Dockhead]

Originally Posted by Andrew Troup

I agree with this constant compass heading solution for all the examples given so far in this thread*, where the current is flowing one way, then the other (at 180deg to the original direction) but not necessarily at the same speed, or for the same time.

I also agree when it is flowing in the same direction at different rates, as you just alluded to.

*Except in the real world situation corresponding to the revised model I introduced in my second-to-last post, where the conveyor belt could swing to intermediate angles, say 70 deg :

I think this scenario invalidates the constant compass heading tactic.
It is no longer consistent with the ant following a straight line on the belt.

That straight line, viewed from above, is now at a new angle with respect to North, whereas simply speeding up, slowing down, or reversing the belt does not affect that angle.

-------

From your other post, you make the point "But not necesssarily perpendicularly across." ... I was never under that illusion - I understood that the perpendicular solution only works when the current sets cancel out over the duration of the passage.

What I'm saying is that if the belt slews through an intermediate angle for part of the crossing (lets say it slews 30 deg CW), in order to follow a straight line drawn on the belt, an ant who is unfortunate enough to be relying on a compass will have to correct his heading by adding 30 degrees for the duration of the slew.

No, not indeed.

In any case, however the belt slews around, the ant needs to walk a straight path across the belt. Any wandering around in relation to the belt loses him miles (inches).

His navigational task, which is exactly the same as ours, is to calculate the sum of all the vectors, however wild they are, which apply during his odyssey. He will get a point on the edge of the belt which will be in front of his destination upon arrival. He needs to steer directly for that point and not deviate from a straight line however the belt slews around.

That's really exactly what we do when we calculate our CTS. Our CTS is the bearing to that bit of water which will be in front of our destination when we get there. That is the bit of water we need to get to. It matters not whether the water moves towards us, away from us, left, right, up, or down, or any combination of any of those. The object of calculating CTS is really just finding the spot in the water which will be in front of our destination upon arrival, so that we can steer towards it on a constant heading.

To give another example, proving that this method is not limited to a case with tides running both ways:

You have a high speed mobo which can make Cherbourg from Needles in one tide, in six hours.

You depart Needles at slack tide.

The first hour you will be set W for 0.25 miles
The second hour you will be set W for 1 miles
The third hour you will be set W for 1.5 miles
The fourth hour you will be set W for 2.5 miles
The fifth hour you will be st W for 1 miles
The sixth hour there is no set.

To navigate correctly you add up all the motion -- 6.25 miles.

That means if you steer towards a point which is presently 6.25 miles E of Cherbourg, the sum of all the streams will set you W cumulately 6.25 miles, and if you motor a constant bearing towards where that point was at the beginning of your passage, you will arrive right in front of Cherbourg 6 hours later. So you calculate a course to steer + 6.25 degrees from your bearing to Cherbourg or 186.25 (in reality, you will of course aim to be a mile or so uptide of Cherbourg to give yourself a margin for error). You steer that 186.25 or probably rather 188 the whole way.

If however you change course every hour, you will sail a bunch of wasted miles. Especially during the fourth hour, you will be uselessly crabbing hard against the tide in order to get to a spot at the end of the hour you don't want to be. Your VMG towards Cherbourg will fall, and for nothing.

So you see -- still exactly the same as the first example. It's still true, as it always is, that the shortest distance to your destination is a straight line through the water, that is, a constant heading.

If the stream is not perpendicular to your course, it's still the same, but I bet everyone has got it by now, surely!!

[Dockhead]

Originally Posted by goboatingnow

I don't beleive anyone argues the theory Dockhead. I do think that the practice ( vector summing) is rarely used in any sort of complicated tidal situation. I've certainly not seen it routinely done.

I think crossing the Channel or crossing the Gulf Stream, this is the only reasonable way to do it. The consequence of getting it wrong is ending up downtide of your destination with a hard, hard slog to finally get there. So it behooves every navigator to know how to calculate a single CTS for such a passage.

The maximum XTE on such a passage is really nothing other than the distance uptide you need to be when the tide changes in order to stay on your efficient course. For a margin of error, you actually need a bit more, not less, XTE at the maximum, to ensure that you really will be uptide, and not downtide, when you get there allowing for some variations in speed. I don't know why you are nervous about big XTE's -- there is one right XTE, which if you fail to achieve, will mean a very unpleasant end to your passage. I can tell you from the experience of many, many Channel crossings, that it is one thing to glide into Cherbourg or Roscoff with the tide bringing you home, and entirely another to have to claw your way uptide at the end of an 8 hour or 10 hour sail, inevitably bringing with it also having to harden up on the wind -- ugh. That's what you get by not being far enough uptide when the tide changes -- that is, not having enough XTE in the middle of the passage.

If you're tacking, then of course it becomes much more complicated, and you also have to account for leeway.

What you might have in mind as a "complicated tidal situation" is something like the passage from St. Malo to St. Heliers in the Channel Islands. Whew, that is complicated. The tides don't just sweep back and forth, they sweep all around in all different directions in a rotary manner. Picking the right time to depart, much less finding the right CTS, and with all kinds of obstacles like the Minkies, makes my brain overheat.

As you suggest, I do this entirely by feel, by hook, by crook, which is really not the right way to do it, but I don't have any other tools. I do it in a way which I instinctively know puts me uptide of the destination as I approach the end of the passage. What is really needed for this is a good computer program with tidal streams in it which can run a large number of scenarios to find the best departure time and the best route. Here a computer is really needed because you really need to crunch a ton of numbers in order to have a clue about the right way to sail, much more than you could feasibly do by hand.


I am actually right now on the Red Funnel ferry to the IOW with an entire new B&G navigation system in my suitcases, on my way to my boat to start installing it. One thing I am really looking forward to playing with is the capability of this system to calculate and display set & drift in real time based on a real time comparison between heading and water speed with COG & SOG. For this to be reasonably accurate, you need good heading data (which you need for so many other things as well) and you need very accurate STW data. I have bought the Airmar 2183 three-axis gyro stablized heading sensor (after toying with the idea of buying the Furuno GPS heading sensor, rejecting the idea finally because it is too physically huge to mount anywhere on my boat), and I am hoping that the new Airmar ultrasonic N2K speed, depth, temp sensor will be shipping, prior to my splashing the boat. It should be interesting.

[Stu Jackson]

This has been fascinating. I have been sailing with a friend for many years. He's an engineer, too. i explained "current sailing" to him, and he first disagreed, thinking that using the GPS heading would get us there quicker.

So we did an example: heading east 3 nm across a current that was running north.

In all of the examples that have been discussed here, some "nitpicking" seems to have creeped in, and Dockhead had repeatedly and correctly come back and given explanations of how varying currents and directions do NOT change the basic facts, that these are simply VARIABLES in the basic calculations of current sailing.

Back to the example. Regardless of whether the current changes during the voyage, for THIS example, simply assume it is CONSTANT. If we followed the GPS heading - and we did - we ended up sailing a longer course since it "sagged" - think about it. If we "compensated" for the current, we got there quicker.

Try re-reading Dockhead's great explanations again, and do some "homework" on "current sailing."

Great subject, thanks. Remember, all the "buts..." are simply changes in the variables, which need to be part of the basic current vector/calculations.

[Andrew Troup]

The problem with applying a conveyor belt representation to multi-directional scenarios is that it needs to be slewed to represent current movement in a new direction.

This slewing makes it look as though the constant heading does not produce a straight track through the water, but I've persuaded myself the problem is entirely with the belt representation. In order to travel in a new direction, bodies of water do not slew, like some sort of mobile whirlpool.

Here's how I convinced myself that the constant heading solution DOES work even for currents travelling at multiple angles. I hope it might work for someone else.

The notional substitute for a belt was inspired by Seaworthy Lass' movable plane. (Thanks Lass, for the idea, and for your strenuous and thought-provoking brainstorming of alternative analogies, above and beyond the calls of duty and slumber)

Imagine a cheap plastic, semi-see-through outdoor picnic tablecloth from the fifties, with checkerboard squares (our one was red and white; other colours are available)

By hanging suitable weights on one of more edges of the tablecloth (or some other more realistic means), we can get it to move steadily in any direction we want. This is done in such a way that the lines of the pattern remain parallel with the edges of the table, meaning there is no slew.

We have trained an ant, and equipped it with crampons. These leave a trail of punctures in the tablecloth and dents in the table.

We pick from one hat a departure and destination point, expressed as coordinates on the table, the unit of measure being the same as one square on the cloth.

From another hat, we pick a series of tablecloth displacements, expressed as squares per minute.
A sample might read: (move cloth) 2 squares to the right for first minute, then 2 squares up for the next minute.

Say the departure we are given happens to be directly above the destination, at a distance of 5 squares.

To simplify things a bit, we'll assume the ant adjusts its speed to arrive over the destination after the predetermined period of time. The speed is assumed constant.

Here's how it plays out:

We prepare the tablecloth for the trial run as follows: Mark the position of the departure point. Move the cloth 2 squares right and 2 up. Mark the position of the arrival point (on the tablecloth, in both cases).
By simple subtraction and addition, we can mentally work out that the destination point on the tablecloth will be 3 squares up and 2 left of the departure point.

Draw a straight line on the cloth between these points.
Have the ant walk the line. The length will be the square root of: (3 squared) plus (2 squared), IOW sqrt 13, or 3.6 squares, and the ant has two minutes to do this, so it must walk at a rate of 1.8 squares per minute.

Now we can run the simulation.
The ant sets off along the line we drew, and we move the cloth according to the instructions. After two minutes, the ant is over the destination, having walked the shortest distance possible across the cloth.

At the end of the first minute, the ant will be half way along the line, 1.5 squares up and 1 square left on the tablecloth
The cloth will have moved 2 squares right.

We can either add these together to find out whereabouts on the table the ant has got to, or simply look at the crampon marks in the table.
Either method will show that during that first minute the ant travelled 1.5 square up and one square right relative to the table.

During the second minute, the ant moves a further 1.5 squares up and 1 square left on the tablecloth.
However the tablecloth is now moving upwards 2 squares, so the ant travels 3.5 squares up and 1 square left relative to the table.

By adding the two results, we see that the ant travels up five squares (1.5 + 3.5) and sideways no squares (1-1) as required.

The job of the ant is to get between two points on the tablecloth in a given time, at the lowest possible speed. The ant doesn't need to know anything about how the tablecloth is moving over the table, so it's a very simple abstraction from the point of view of the ant, and even an ant would probably recognise that a straight line is the best solution to their simplified problem.

Who am I to argue with an ant?

So this is one way to demonstrate not only that is a straight line through the water the optimum solution (which most people can be persuaded of), but that a constant compass heading achieves this result. Even when the current is from various random directions.

(Note that this current progression is conceptually identical to the one I raised for a boat in post #266, which lead to me expressing a bit of doubt in #269 about the constant-heading solution in such a case. I have now erased those doubts.

The "optimal solution" I proposed in that case, although I didn't realise it, was a straight rhumb line up the table. If the ant was required to do this, it would have had to walk two squares left during the first minute, and as many squares up as it could get.
During the remaining minute the ant would walk up the remaining squares.

Solving this requires striking two arcs of equal radius from departure and arrival points on cloth such that they intersect on the vertical line 2 squares left of the departure. (Because the ant travels the same distance across the cloth during the first minute as the second). The radius of those arcs is about 2.15 squares

This requires the ant to crawl at 2.15 squares per minute and is so far above the optimal 1.8 figure as above (it's about 20% more distance) that I'm embarrassed to have suggested it.

And I apologise to anyone who was persuaded by my previous line of reasoning, including one who posted to that effect based on my initial raising of some doubt in post #257.)

[Seaworthy Lass]

Originally Posted by Andrew Troup

............
Imagine a cheap plastic, semi-see-through outdoor picnic tablecloth from the fifties, with checkerboard squares (our one was red and white; other colours are available)
[/INDENT]By hanging suitable weights on one of more edges of the tablecloth (or some other more realistic means), we can get it to move steadily in any direction we want. This is done in such a way that the lines of the pattern remain parallel with the edges of the table, meaning there is no slew.

We have trained an ant, and equipped it with crampons. These leave a trail of punctures in the tablecloth and dents in the table.

We pick from one hat a departure and destination point, expressed as coordinates on the table, the unit of measure being the same as one square on the cloth.

From another hat, we pick a series of tablecloth displacements, expressed as squares per minute.
A sample might read: (move cloth) 2 squares to the right for first minute, then 2 squares up for the next minute.

Say the departure we are given happens to be directly above the destination, at a distance of 5 squares.
..........
So this is one way to demonstrate not only that is a straight line through the water the optimum solution (which most people can be persuaded of), but that a constant compass heading achieves this result. Even when the current is from various random directions.

Perfect model!
The constant heading regardless of any incremental changes in direction of current was not in doubt, but it is satisfying to have a model to fit it.

I was hung up on how to fix the departure and arrival points on the plane (equivalent to the edges of the treadmill), but of course they are not on the plane at all but on the "table" .

I would have the ant leave droppings that instantaneously seep through to the table though, rather than wearing crampons (unnecessary drag ). Maybe think of an ant who has had too much Mexican food the night before departing on his journey?

You are the ant's pants!

[Dockhead]

Originally Posted by CaptForce

I agree that the dabate is not a question of position fixes or the use of navigation tools or of ant behavior. The debate is about calculating elapsed time. We can not dispute that the variables of speed and distance are direcly proportional to time. If speed and distance are increased or decreased proportionally, then time remains the same.

The debate is wether the choice of the constant heading or the shortest distance course GPS track results in a disproportional change in speed an distance.

I remain with the opinion that the time elapsed would be the same; however, as with any valid endeavor, I would be eager to hear convincing opposing opinions supported by good data.

All right, let's try again.

1. Distance sailed through the water does not necessarily equal distance over the ground. If the water is moving, it will ALWAYS be different.

2. The shortest path between any two points is a straight line.

3. Since we sail in water, we care about distance through the water, not over ground. Therefore, the shortest distance across moving water for a boat is a straight line through the water.

4. A straight line through water is a constant heading. NOT a constant COG, which will vary according to the motion of the water.

Therefore, the shortest distance across moving water is a constant heading calculated to put you in that bit of water which will be in front of your destination when you get there.

A corollary to this -- a straight line across the ground, following the GPS approach, will ALWAYS be the wrong way across, unless either (a) the water is not moving; or (b) the water is moving at a constant speed and direction for the entire passage (in both cases, constant heading and straight track across ground coincide).


OK, is that more clear? It boils down to this -- you have a choice to go straight with regard to the ground, or straight with regard to the water. You need to choose the latter if you want to sail less distance and get there sooner. Sailing a straight line over ground means sailing a crooked path through water, which will add miles to your passage. Except only exceptions (a) and (b) above, where the shortest paths through water and over ground are in the same place.


There are specific examples in Post 253 by Noelex, with calculations showing the difference in time and distances:

http://www.cruisersforum.com/forums/...ml#post1126538

And in Post 237: http://www.cruisersforum.com/forums/...ml#post1126273


Let me give another example showing how different water and ground distances can be:

You are crossing a river one mile wide with a 6 knot current which runs due North and South. You are on the W bank of the river in a boat capable of making 6 knots. You want to sail to a dock on the E bank of the river which is one mile downstream from your point of departure.

You point your bow directly across the stream, and in 10 minutes you are there. You sailed 1.4 miles over ground but only 1 mile through the water, which is why you got there in 10 minutes although you were only going 6 knots.

That is to illustrate the difference in water distance and ground distance. In that example, you never feel the 0.4 miles extra over ground -- you don't sail in the bottom.

Now another example -- exaggerating the effect to underline -- let's suppose you're crossing a tidal body of water, 60 miles across, on Mars, say, with the tide exactly perpendicular to your course. On Mars the tides run only for two hours at a time, then an hour pause, then two hours in the other direction, then 10 hours of slack water. They run at 10 knots. You are sailing W to E and the tide runs N and S.

What's the fastest way to get there? Go full speed ahead against the tide, when it's running, in order to stay closer to the rhumb line and minimize distance over ground? Or forget about the tides since they cancel each other out? If you hold a constant heading of 90, you will sail 60 miles through water and get there in 10 hours at 6 knots, although you will have sailed 100 miles over ground as the tide swept you back and forth. Sailing to your GPS and staying on the rhumb line you will sail a shorter distance over ground, but longer through the water. If you can make 10 knots at full revs and stem the tide when its running, then you will have sailed 60 miles over the ground, but you will have added 40 miles of useless sailing through the water, at redline to boot, and you will get there 4 hours later, by following the GPS approach.


In Post 237, you have a choice between sailing 60 miles through the water on a constant heading (although this is a huge "S" curve over ground, taking you 12 miles off the rhumb line at one point), or 75 miles through water on the rhumb line and 60 miles over ground. Since your boat can only sail in water, the constant heading saves you 15 miles and three hours of sailing, compared to staying on the rhumb line over the shortest distance over ground following the GPS approach.


Again, the shortest distance over ground will NEVER be the shortest distance through the water except for the two cases above. You never feel ground distance; you only feel distance through the water since that's what your boat sails in. The shortest distance across a moving body of water is ALWAYS a constant heading.

[Dockhead]

Originally Posted by CaptForce

How can I possibly be motivated to read the rest of this post when you begin by changing the definitons of distance and speed? A boat does not transit a greater distance because it's speed is reduced while traveling against a current. Without an agreed operational definiton of speed and distance that conforms with the international norms of physics we have no means to communicate.

I'll add again that, even if you were to be correct, there's no convincing if we are speaking a different language.

A boat's speed through the water is not reduced while travelling against a current. Its speed over ground is.

Enlightenment in this question lies in understanding that our boats move through water, not over ground. It's absolutely essential to understand the concept of distance through water versus distance over ground. I thought there were some examples which really well illustrated this difference.

So definitions of speed and distance:

SOG, COG, and distance over ground in miles -- all ground-referenced

STW, heading, and distance through water in miles -- all water-referenced.

You cannot navigate efficiently without the water-referenced concepts. Because -- once again -- that's all our boats know how to do.

Like I said in the last post -- distance through the water and distance over ground are always different if the water is moving. You can navigate in order to minimize the distance over ground -- that would be putting your pilot on "track mode" and letting it keep you on the rhumb line, sailing in a straight line from "A" to "B" referenced to the ground. Result -- minimum distance over ground. But you will sail further than necessary through the water.

OR, you can choose to minimize the distance through water -- that would be calculating CTS and sailing a constant heading. You will sail the shortest possible distance through water, but the distance over ground will be increased. Since your boat sails through water, you will get there faster by choosing to sail the shortest distance through water. You will sail further over ground, but you don't care in the least. The ground distance is irrelevant if your boat doesn't have legs which reach down to the bottom.

[Andrew Troup]

I frankly fail to see how anyone could remain unconvinced by Dockhead's great explanations, but everyone's brain is wired slightly differently, and often it's a matter of finding the exact key to unlock a hidden assumption which is blocking things, so here's another angle which just occurred to me:

0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0 – 0

What's the quickest way to swim across a river: is it to stay on the shortest line between where you start and the opposite side? This would mean staying on a perpendicular, which is not the same as swimming at a perpendicular.
Consider a river which is flowing almost as fast as you can swim. If you are trying to stay on a perpendicular, you will have to swim almost due upstream. You will edge out into the river, staying over a notional perpendicular line drawn on the bottom.

If you get tired or the current is stronger in the middle, you will have to angle directly upstream to try and stay on the line, until the point is reached where it's all you can do to stay over the line, and you are no longer crossing the river at all.

Whereas if you simply swim perpendicular to the current, it will take the same time to swim across the river as it would if it stopped flowing.

This is true even if the river flows much faster than you can swim, whereas by adopting the other strategy, you cannot make any headway at all across a river which merely matches your speed.

On the perpendicular heading (rather than perpendicular 'Course over Ground') you will cover a much greater distance across the ground than the width of the river, but that increased distance is entirely due to the river's flow: you have not had to contribute anything to covering the increased distance. I

f, on reaching the other side, you encounter another river flowing at the same speed in the opposite direction, you should adopt the same strategy.

In other words, swim at the same heading which worked last time: perpendicular to the current. You will be swept downstream by the new current, which incidentally will spit you out on the bank directly opposite the point at which you entered the first river. But on the map it will look as though you covered a great deal of unnecessary distance in travelling across the equivalent of a tidal current which reversed when you were half way across.

[Seaworthy Lass]

CaptForce, given the strength of your convictions, you are probably thoroughly sick of reading all this by now, but I will have a go at illustrating an example, as I think it is a very important concept.

The details:
- You and Dockhead have a channel to cross that is 20nm wide going directly south to north (magnetic) from A to B.

- The current is flowing exactly east-west (magnetic) at 2 knots for the first two hours of the journey, then it instantly flips and is going 2 knots the other way for several hours.

- You and Dockhead are both motoring at 2400 revs and in still water this gives each of you a speed of 5 knots (I know that's slow, but lets make the numbers simple). Neither of you changes revs for the entire journey.

- You are sticking to the straight line between A and B thinking this is the quickest route. You will actually find you need to initially motor at a compass heading of 24 degrees to be able to plot a straight course on the chartplotter.

Dockhead had worked out that a compass heading of 360 (0) degrees will get him from A to B the quickest.

I have drawn your tracks as shown by the chartplotter and marked your positions at the 1,2, 3 and 4 hour marks.

At the one hour mark the current would have swept Dockhead off to the west, and increased his speed over ground as he is not fighting against it. He has made progress of 5 nm towards the other shore, but not directly towards B.
You have had to point into the current and to maintain a straight line towards B on the chartplotter and this has slowed you down to 4.58 knots as the VMG on the chartplotter.

At the two hour mark Dockhead is even further from the line between A and B, but he is actually closer to the other shore than you are. Suddenly the current flips and is going 2 knots the other way.
Dockhead placidly does nothing, and keeps his compass heading 360.
You frantically need to alter your compass heading from 24 degrees to 336 degrees to maintain that straight line on the chart plotter.

At the three hour mark Dockhead is being swept back towards your line. His speed over ground is still better than 5 knots as the current is pushing him back to the rhumb line. You are 13.75 nm from your departure point at this stage and your VMG has been a steady 4.58 knots the entire way.

At the four hour mark Dockhead has just arrived. You still have 1.68 nm to go. You arrive about 22 minutes after Dockhead.

This is the charplotter track for each of you:

[Seaworthy Lass]

Originally Posted by goboatingnow

Seaworthy. I understand DH vectors. How did you compute CF progress.
Dave

CF's speed relative to the water is 5 knots (fixed throughout the journey as he is keeping constant revs). The current is pushing the boat at 2 knots perpendicular to the line between A and B. So we can use Pythagoras' triangle to work out CF's VMG (see diagram below).

The VMG squared plus two squared = five squared.
So VMG squared = 25 - 4 = 21
So VMG = 4.58

The formula for the angle of deviation (and therefore the compass heading in this case as the destination point is magnetically north of the departure point) is as follows:
Sinθ = 2/5
where θ is the angle
This works out to be 24 degrees.

[Andrew Troup]

Originally Posted by Seaworthy Lass

...I will have a go at illustrating an example, as I think it is a very important concept.....

This is the charplotter track for each of you:

Bravo, Lass

I'm not sure if I can spell irrefutable, but I know it when I see it !

The little pictures showing CF's boat pointing elsewhere than the destination for the entire duration of the trip might seem a small touch, but I'm hoping it will be the telling detail.

People get used to doing this in the situation where the current is always from one direction, but your depiction makes it clear why this is not a winner for reversing current.

I personally think there are now so many different, coherent and persuasive explanations that it would make more sense for anyone who is still not persuaded to mull over what's already on record, than for us to continue generating new ones.

I think this is true even for the more tricky situation, where rather than simply reversing, the current flows from several non-orthogonal directions.

The point to take from all this is:
This particular scenario is not able to be optimised using just a GPS (or any other position fixing method)
You need to know, in advance, what the current rates and directions will be, in order to be able to work out what heading to put the boat on. You also need to know how long it will take to get there, so the problem means going round the mulberry bush several times

I guess a mathematician or software designer would call it a recursive problem. Until you've worked out what course to steer, you don't know when you'll get there, but unless you know when you'll get there, you don't know what course to steer.

If I found myself either needing to (unlikely) or curious to (eminently possible) optimise such a trip, I'd set up a spreadsheet, and use the 'ant on a checkered tablecloth' concept, substituing "Northing" and "Easting" in nautical miles instead of checkers up and checker right, to come up with a computed heading based on a guess at arrival time. The computed heading would derive simply from the ratio of Northing to Easting needed through the water.

The spreadsheet would also compute the (constant) speed needed to cover that distance through the water in the time available.

I would then adjust the arrival time, based on how fast I expected to be able to go in relation to the figure computed, and re-run the simulation.

If conditions changed early in the transit, I'd swap to one of the other (discarded) plans; if they changed later, I'd run a new simulation.

Given I hate motoring, I'm not likely to profit much from such a procedure. Different headings would probably imply different speeds through the water, and apparent windspeed would be another consideration. If part of the plan involves tacking, it could get very complicated indeed.

However, thinking through how I'd do it helps cement the learnings from discussing this with so many deeply knowledgeably and wonderfully communicative people. Thanks to all.

I've done similar spreadsheets in the past, but only to find the optimum time to arrive at the start point for a tricky passage, say round North Cape, through Cook or Foveaux Strait or Current Basin, all in NZ in order not just to trim as much as twelve hours from a passage, but more importantly to avoid wind against tide in places where, to quote the NZ Pilot, "Prayer may be of assistance" when both are strong.

I've never used a spreadsheet to calculate a heading to steer, but I think it would be much preferable to doing it on paper, given a complex sequence of strong current vectors in relation to available boatspeed, and the recursive (if that's the word) nature of the calculation.

[Seaworthy Lass]

Originally Posted by Andrew Troup

......The little pictures showing CF's boat pointing elsewhere than the destination for the entire duration of the trip might seem a small touch, but I'm hoping it will be the telling detail.

People get used to doing this in the situation where the current is always from one direction, but your depiction makes it clear why this is not a winner for reversing current.

I struggled to find an explanation that might get through to the disbelievers. So many posts had already given great examples but not got the message through.

I thought a diagram of what the charplotter would show for the track and positions of the boats (and the angle of the boats as well) may make it easier to view and understand .

Originally Posted by Andrew Troup

I personally think there are now so many different, coherent and persuasive explanations that it would make more sense for anyone who is still not persuaded to mull over what's already on record, than for us to continue generating new ones.

I'm done. It was a final attempt

Originally Posted by Andrew Troup

The point to take from all this is:
This particular scenario is not able to be optimised using just a GPS (or any other position fixing method)
You need to know, in advance, what the current rates and directions will be, in order to be able to work out what heading to put the boat on. You also need to know how long it will take to get there, so the problem means going round the mulberry bush several times
.......

Yes, neither a GPS or chartplotter will give you the compass heading you need to follow to make your journey in the shortest possible time. Accurate calculations are hard as the exact current is usually unknown and the boat speed is only a guess as well, but at least it can be factored in.

Been an interesting discussion. Thanks all.

[Dockhead]

Originally Posted by goboatingnow

Andrew , it can only be done two ways. Either you have advance tidal information , hence you just draw the vectors on your chart and add them up and derive a single course wide CTS. Alternatively you can run hourly or multi hourly CTS based on assumptions of what you see the boat being subjected too.

Often one has some information but not enough conclusive information. Equally,in long journeys it is useful to break them up into several legs and re compute CTS

I don't think anyone disputes the symmetrical tidal model and CTS. What would be interesting is to compute an error graph for various non symmetrical but partly reversing currents or currents that don't flow perpendicular to the direction of travel. Hence it would allow one to see how much or how little advantage the single CTS method gives.

Dave

It's all the same whether the tide is symmetrical or not. Seaworth Lass' excellent drawing shows what happens when you sail the hypotenuse of the triangle -- it's the longer leg. Anytime the water is moving, you have a choice of sailing a straight line through the water, or not. If you deviate from the straight line through the water -- the constant heading -- whether through ignorance or through not having enough data to compute a good CTS, then you are adding miles. Only if the water is moving at exactly the same speed and direction over the whole passage, will the straight line through the water coincide with a straight line over ground.

Andrew Troup had a subtle insight about the recursiveness of the calculation of CTS. It is very true, and it means that even with perfect information about tidal streams, the calculation can be very complex and indeed impossible to do by hand. So far we have assumed that we could know our arrival time with reasonable certainty. Of course we don't know very well -- in a sailboat, your speed is always changing.

But more importantly -- and this is really important -- we don't know the distance. If we are sailing to Cherbourg from Needles with a symmetrical tide, so that CTS approximately equals bearing to waypoint, then we know it's close to 60 miles. But if there is a big difference between CTS and BTW then your passage through water will be more than 60 miles. More or less time in passage means more or less time exposed to different currents, so will change the correct CTS. So yes -- it's recursive -- you have to guess, then run the numbers, then refine the number (time on passage, once you understand the distance better), then run the numbers again.

With very complicated rotary currents like around the Channel Islands, it becomes too complicated for my feeble brain. Need a good computer program to do it. For crossing the Gulf Stream or crossing the English Channel, it is surprising what amazingly good results you get from rather approximate information and fairly vague guesses. I am rarely more than a mile off. There are some complicated mathematical principles at work, I suspect, involving perhaps fuzzy numbers and approximations (otherwise known as "magic" to those of us who are uninitiated). I'll try to remember to ask my brother about it (a math & physics prof).

To step three steps back -- the purpose of this part of the discipline of navigation is, to put it very crudely but perhaps, aptly, is to avoid arriving downtide of your destination. That is really the main point, isn't it? To work with the currents instead of fighting them, to steer in a way that lets them sweep you to your destination, instead of crabbing along the rhumb line, always sailing the hypotenuse of Seaworthy Lass' excellent triangle. Arriving uptide even by several miles is not a big problem (so you want to err on that side when you are making your guesses and approximations). Even if you only realize it an hour out from your destination, you just bear off a little and let the tide bring you in. Even a single mile downtide, on the other hand, can be a total b*tch, especially off the coast of Normandy where the currents are faster than on the English coast. They really rip when you get close to C-bourg, and in a small boat in light winds and at springs you might not even be able to get there at all, if an hour out you are even a little too far downtide (that is, if you are not far enough uptide of the rhumb line -- if you don't have enough XTE). In which case you divert to Alderny

By the way, that's another good demonstration of CTS versus GPS track sailing across currents. If the current is faster than your boat speed, there comes a point where you will never even arrive by GPS track navigation, when you can make it comfortably by steering constant CTS. The difference between the approaches is exaggerated, the higher the current speed is compared to your boat speed. We get 6 knots at springs in much of the English Channel; up to 12 around Alderny, so this is not at all a hypothetical question at all, if we are talking about a small boat making 5 knots.