CTS vs Following Rhumb Line - DILUTED THREAD

[Seaworthy Lass]

Originally Posted by Dockhead

........
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.

Some excellent points have been raised in this post underlining the importance of making at least rough calculations about compass heading required for a leg of a passage.

For the example I gave in post # 340, many people would shrug their shoulders and say "what does it matter if I arrive 22 minutes after Dockhead, I am cruising, not racing".

But if the current had changed direction a bit earlier and started increasing significantly, CaptForce may not have reached his destination for hours after Dockhead had arrived.

Making these calculations at least roughly is not difficult if the current is approximately perpendicular to the line between departure and arrival points - I just assume constant speed and look at the expected current at each hour and then cancel out opposing current and see what I am left with for how long. It's then just one lot of figures I need to work on.

As you said, importantly this also allows you to plan the passage to avoid adverse current (or take advantage of it when it is favourable) .

[Dockhead]

Originally Posted by Seaworthy Lass

For the example I gave in post # 340, many people would shrug their shoulders and say "what does it matter if I arrive 22 minutes after Dockhead, I am cruising, not racing".

A valid point, and no doubt many people will just say -- "Screw it, let the plotter calculate my route, and let the pilot steer it, while I lounge in the cockpit."

We do it for fun, so if that makes you happier, then go for it.

But for many other people, getting sail trim just right, for example, is not primarily in order to arrive 22 minutes earlier, but for the satisfaction of doing it well.

Same thing applies to navigation. "Screw it, I'll just crab across along the rhumb line, it's only an hour extra" is one way to get there, and maybe it doesn't matter to some people. But if you sail for fun, including the fun of learning complex disciplines and doing it well, surely there is no satisfaction in that approach. Surely it's a lot more fun to do it right and get there with the help of the current, instead of fighting it the whole way.


And of course some times even cruisers really do need to get somewhere sooner rather than later. That's certainly often the case when I'm cruising, especially with my father on board. We don't drink underway, but 17:00 cocktail hour is a sacred event. So my father gets really cross if the anchor is not down by 16:45. He doesn't care about much else about my passage plans, but he stringently checks the ETA, to make sure there is no conflict with cocktail hour. So it behooves me not to screw up the navigation!

[Dockhead]

Originally Posted by goboatingnow

It's important to remember here that we are comparing vectors. Neither the GPS method or the CTS have anything to go with the ground. Because in effect all the GPS does is sail small CtS triangles.

So there's no advantage in attempting to compare actual ground tracks.

The fact is that the sum of the hypotenuse of a number of smaller vector triangles is always equal of greater them the hypotenuse of the larger combined vector triangle its has nothing to do with symmetrical tides ( that's merely the poster child )

It doesn't matter whether you use time or distance or speed since they are linear functions of each other.

The actual ground track in each method is utterly irrelevant. This is a mistake many people make when drawing vectors on a physical chart. The actual ground track especially with vectors at non right angles can be quite difficult to work put. The maximum xte are not at least

Yes! The ground tacks are irrelevant indeed, and thinking about with a fixation on the ground will lead to errors.

The constant in the problem is speed through the water. Ergo, that approach which produces the shortest distance through the water will get you there faster. The shortest distance between two points is a straight line. A straight line through the water can only be sailed on a constant heading. Ergo, the fastest way between any two points across moving water will always be on a constant heading. All of that is irrefutable.

A GPS track approach will require frequent changes of heading. Thus you are not sailing a straight line through the water and thus you will always be sailing a greater distance through the water. At a constant speed through water you will thus always get there later, except the two cases where the two approaches coincide (slack water, and constant speed and direction of current). In these two cases, the GPS track approach will not require any changes of heading; thus you are sailing a straight line through the water exactly as if you had figured it out using a constant heading approach.


What is really essential is an understanding of the concept of distance through the water. Anyone hung up on distance over ground and ground tracks will find it very hard to visualize the problem.


In practice, of course, as you say, the difference will be small if the difference in current is small, so in many cases not worth the hassle of calculating. The other problem is information -- the less information you have about the currents and about your own speed on passage, the less accurate your CTS calculation will be and the less valuable it will be.

[Seaworthy Lass]

Originally Posted by CaptForce

......
I never heard the idiom before, but I like it. "The penny dropped for me" when I accepted that the GPS track IS a summation of vectors.

OH, the penny has just dropped for me CaptForce! I know now why you are going wrong.
Vectors DON'T sum up as you think they do.

Take a very simple example:
- Track is north magnetic
- Boat speed in still water is 5 knots for both your boat and Dockhead's
- Perpendicular current from the east, 2 knots first hour all the hour, then 4 knots from the same direction for the entire next hour.
- You follows a straight line on the chartplotter
- Dockhead works out the angle needed for a total of 6 knots of current over two hours and follows that as his compass heading.

You allows for 2 knots of current in the first hour and travel 4.58 nm along the straight chartplotter line (north)
In the second hour with 4 knots of current you travel 3 nm.
Total progress = 7.58 nm

Dockhead travels allows for 3 knots of current in the first hour and his shift to the north is 4 nm
This is repeated in the second hour.
Total progress north = 8 nm and he end up on your line at this point.

I repeat again, vectors don't sum up as you have assumed! This is where you are going wrong.

This is the diagram for the speed vectors for each of you (the chartplotter diagrams are shown in the next post, I can't get them to attach together:

PS it is a woman's prerogative to pick up the towel again at any time . I will give a big cheer at this end when you realise why constant heading is always the quickest way of travelling between two points.

[Seaworthy Lass]

The chartplotter diagrams for the two boats over two hours:

[Seaworthy Lass]

Originally Posted by Andrew Troup

I've done a quick and dirty mathematical analysis of CaptForce's own scenario, assuming a distance across of twenty nautical miles.

I broke the crossing into multiple strips and applied simple trig. There was no preconception built into the analysis as to what the track across the ground would be for the Dockhead constant heading transit, but it turned out to be exactly as CaptForce has predicted.

I was not surprised because I had already convinced myself he was right on this point.

I thought of several different ways of explaining it to myself, because it was a counterintuitive proposition at first.

One way of working out what heading to adopt, for a Constant Heading transit, is to steam at the correct angle to exactly cancel the average tidal current, in other words, to make our track across the ground exactly at right angles to the average current, when we encounter it.

In this case, that's 2 knots, and we'll encounter that much current one quarter of the way across.

Until then, we'll arc upstream. But at the quarter transit mark, we'll be making way parallel to the rhumbline. It cannot be otherwise, because we've calculated our heading to make that exact thing come true.

For the next quarter of the crossing, the tide increases, and just as an additional two knots of tide was enough to bring our detour upstream to an end in the first sector, a further two knots is just enough to cancel the excursion and bring us back to the rhumb line. Because the current is a maximum at this point, when we cross the line we are now heading more steeply downstream than at any other time before or after.

(The track across the ground for the remainder of the passage is a mirror image, with an instant double flip about E-W and then N-S.)

Initially we are losing badly and get swept downstream, but we pluckily maintain the same heading as before, and sure enough by 3/4 of the way across, the crabbing due to our angled heading is once again sufficient to cancel the 2 knot current at this location and we're back parallel to the rhumb line, as far downstream as we were upstream at the maximum extent.

From now on, we're winning against the decreasing current. 7/8 across, we're parallel to our track 1/8 across, angling steeply up, almost making good our heading.
By the time we touch the far side, our track has angled up even further, exactly on our heading.

When you think about it carefully, it does make perfect sense.

The question of which tactic wins the race is a different question, but I would prefer to leave that until there's substantial agreement on the question of the track (whether CF and I are right or wrong) - because if I'm wrong, the rest of my analysis must also be wrong.

The analysis is not a trivial issue, if you want to be able to plot the ground track accurately, because the current in this scenario is increasing or decreasing every inch of the way.

My calculus was never great and it's virtually non existent now, so I resorted to brute force and used an iterative method with a spreadsheet. It was not difficult in a quiet office on land, but I wouldn't be quite so keen on tackling it when the 'office' was leaning over and leaping off waves.

OH, you are right about the track being sygmoidal for current coming constantly from one direction but varying in strength.

It is soooooooooooo very counterintuitive.
Well done for working this out.

I have realised my speed vectors are displayed a bit wrongly so will post a new diagram in a sec (end result is still the same though ).

[Seaworthy Lass]

Sorry folks, my last speed diagram was a little incorrect. Net result is the same, I have just shown the vectors correctly here:

[Seaworthy Lass]

Originally Posted by goboatingnow

to my understanding that is not what happens in Dockheads case, your diagram suggest a heading change. Dockhead does not change heading.

The speed vector diagram below shows absolutely no change in Dockhead's heading (the thick red lines 5 long are parallel). Only the ground track direction changes after an hour.

Originally Posted by goboatingnow

All these attempts to resolve the matter to a ground track completely miss the point, its irrelevant.
dave

It is critical to be able to calculate and plot your ground track - how else do you work out if you are going to hit any obstacles along the way

[Seaworthy Lass]

Originally Posted by Dockhead

Peace, y'all!

Dave just means irrelevant for purposes of calculating an efficient passage. Naturally, we also need to have an idea of where we are going in relation to obstacles, in order to avoid them. No one would disagree with that. You two are not disagreeing with each other.

I must confess to having had a brain f*art about the shape of the ground path in a unidirectional current -- YES, it is also sinusoidal, and yes, you end up downcurrent from the rhumb line for a little bit at the end of your passage. Sorry, my mistake -- I take back my statement. A momentary lapse in visualization.

I'm getting on the plane to Helsinki and will have some numbers to post when I land!

Yes, I had a brain freeze too LOL, apologies to CaptForce for my error. But the shape of the curve does not alter the fact that following the constant heading is the quicker route!

Look forward to some figures on time to get across the channel. Yours on a straight heading are easy (once we know the width of the channel, we already know average current and you log speed), but I find CaptForce's speed is impossible to calculate without knowing exactly how the current varies across the channel he describes.

Have a good trip. Make sure the pilot examines his likely crosswind and calculates the constant heading needed for the quickest flight LOL .

PS You don't end up downstream for a little bit of the journey, but for half if the current is symmetrical

[Dockhead]

OK, here are the first numbers. I analyzed Capt Force's scenario based on these parameters:

5 knots constant STW

2 knots average current, perpendicular to BTW, varying linearly from 0 to 5 knots to 0

18.33 miles over ground to waypoint (longest possible distance which can be reached given the previous parameters).

Here is what the GPS track boat would do if we use granularity one hour -- that is, we analyze one hour at a time based on the average current and average heading during that hour:

Constant speed through water:5,00Distance over ground to waypoint:18.33Average current:2,00period:av current during that periodSOG (VMG to waypoint)distance made goodhourly11,004,904,9023,004,004,0033,004,004,0041,004,904,9017,80

OK, so the GPS track boat will travel 17.8 miles during four hours. I have not yet solved the problem of current varying by distance rather than by time. Varying by distance I will have to stretch out the array of current speeds (which will further slow down the GPS track boat). Right now we are seeing the scenario as if the current varies by time regardless of how far the boat has gotten.

Remember that the constant heading boat will travel 18.33 miles made good towards the waypoint, and will arrive, in any combination whatsoever of currents which add up to an average of 2 knots.


Now here is what the GPS track boat looks like if we analyze every 5 minutes. That gives us 48 periods of 5 minutes each. I have analyzed the SOG and therefore distance made good towards the waypoint for each period, using a spreadsheet:

period:av current during that periodSOG (VMG to waypoint)distance made good10,005,000,4220,175,000,4230,354,990,4240,524,970,4150,704,950,4160,874,920,4171,044,890,4181,224,850,4091,394,800,40101,574,750,40111,744,690,39121,914,620,38132,094,540,38142,264,460,37152,434,370,36162,614,270,36172,784,150,35182,964,030,34193,133,900,32203,303,750,31213,483,590,30223,653,410,28233,833,220,27244,003,000,25254,003,000,25263,833,220,27273,653,410,28283,483,590,30293,303,750,31303,133,900,32312,964,030,34322,784,150,35332,614,270,36342,434,370,36352,264,460,37362,094,540,38371,914,620,38381,744,690,39391,574,750,40401,394,800,40411,224,850,40421,044,890,41430,874,920,41440,704,950,41450,524,970,41460,354,990,42470,175,000,42480,005,000,4217,52Checksum average current:2,00

Et voila!! Eureka! Now the GPS track boat travels only 17.52 miles towards the waypoint, compared to 18.33 for the constant heading boat.

This is an additional proof of the constant heading principle -- the smaller the triangles and more frequent the changes of heading, the slower the GPS track boat will go over ground. This trend will continue until the heading changes are continuous, at which point there will be maximum disadvantage of the GPS track compared to constant heading.

The reason is that every "grain" of our analysis, in this case one hour, and then five minutes, is a micro passage which is sailed on an ideal constant heading course. The larger these grains are, the more the GPS track boat will deviate from the rhumb line, and the faster the GPS track boat goes over ground.

I have to run off to a meeting now, but I will do another set of numbers analyzing every minute (240 periods). We will see that the distance achieved by the GPS track boat will fall further.

Next task will be to correct the current-over-time verus current-over-distance problem.

If anyone wants to check my formulae, I can email the spreadsheet.

[Seaworthy Lass]

Originally Posted by Dockhead

I made the same mistake, but realized (with help from some on here) that indeed you will always, in any case, be downtide for at least a short bit at the end of any CTS passage. It's still the most efficient way.

Think about an 8 hour passage with 6 knots of tide for just the first hour, and slack tide for the rest. On the ideal passage, which is as always a constant heading, you will only make up 1/8 of the 6 miles you are knocked off during that first hour when the current flows. The other 7 hours you will stay on the same heading and make up the rest of your XTE as you go, and you will approach from "downtide", based on the sum of tides you were overcoming.


This thread has been enormously valuable because no matter how well you think you know this stuff, more and more complex implications and nuances which are often counterintuitive keep coming out of the woodwork. I have learned a tremendous lot; hope others have too.

Enormously valuable for me too! I aways knew one compass heading was most efficient , but I never thought too much about the ground track. It is obviously really important to do so as there may be obstructions in the way.

[Dockhead]

So here is my workup of Capt Force's scenario using constantly changing current which starts at 0 increasing to 4 then decreasing back to 0, with average of 2.

The current increases and decreases in a linear manner and is at 0 for both first periods and at 4 for both middle periods. This is rather artificial but I think it works ok (if anyone objects, please let me know and suggest a better approach.

Attachment 53003

The main formula is for SOG for a given period:

=SQRT(SUM(POWER5,2)-(POWER"X",2) where "X" is the set of the current for the given period.

I have run checks on average current and linearity of current change, which are shown in the spreadsheet.

I ran three scenarios for the GPS track boat:

1. hourly
2. five minutes (48 periods analyzed)
3. one minute (240 periods analyzed)

SOG was calculated for each period from which I derived distance made good towards the waypoint.

The results of course confirm that sailing a GPS track, however you analyze it, is considerably slower than sailing the constant heading, something I think everyone here except Capt Force accepts.

Unfortunately, the results did not correspond to my hypothesis that the smaller the analysis period (the smaller the vector triangles we calculate), the slower the GPS track boat will go. I am perplexed by this and will welcome any insight into why. The only thing I can imagine is that my current scenario, while perfectly linear at the given level of granularity, is not really -- because the beginning, ending, and two middle periods are all at the maximum and minimum values, not at the average value which would be achieved is the "line" went all the way to instantaneous ends and middle of the graph. So perhaps the slightly faster passage with one-minute analysis reflects the shorter period of time at maximum current -- two minutes instead of 10 minutes -- as between the one-minute and five-minute analyses. I don't know if this is enough to explain that; insights welcome.

The results are as follows:

Constant heading 18.33 miles made good in four hours
GPS track, hourly analysis, 17.7980 miles made good in four hours
GPS track, five minutes analysis, 17.5217 miles made good in four hours
GPS track, one minute analysis, 17.5778 miles made good in four hours

The most serious failing of my model is that the currents to which the GPS track boat is exposed to under various scenarios is the same as the currents the CTS boat is exposed to. And so I could only run 4 hours of the GPS track boat's passage, which does not get the boat into port.

This is flattering to the GPS track boat, as in reality, going more slowly (more slow VMG towards the waypoint) it will be explosed to more current, slowing it down further. So my analysis exaggerates the speed of the GPS track boat by that whatever that bit.

If anyone can suggest how I can overcome that -- "stretch" the current graph to match the pace of the boat in each scenario, I will be grateful. Anyone can download and hack on the model, just please rename it every time so that we don't get versions mixed up.

[Seaworthy Lass]

In response to Dockhead Andrew posted this:

Originally Posted by Andrew Troup

......Have you already forgotten that we all made prize dicks of ourselves through having not the faintest idea what the ground track looked like for the constant heading transit?

You are being very diplomatic here .

It was only me and Dockhead and GoBoatingNow who made prize dicks of ourselves LOL. You were the only one who realised CaptForce was correct with his symmetrical sinusoidal shape for the chartplotter track .

[Dockhead]

Originally Posted by Seaworthy Lass

In response to Dockhead Andrew posted this:


You are being very diplomatic here .

It was only me and Dockhead and GoBoatingNow who made prize dicks of ourselves LOL. You were the only one who realised CaptForce was correct with his symmetrical sinusoidal shape for the chartplotter track .

Ha, ha. There were plenty of brain farts all around on here. I think I probably get the prize for the most!

But my darling -- Capt Force was not correct about the symmetrical sinusoid shape for the chart plotter track for a constant heading boat in a unidirectional current. It's symmetrical only in a symmetrical changing current. For a unidirectional current, the vast majority of the passage is spent on the upstream side of the rhumb line. You fall off downstream of the rhumb only just at the end -- if the current is unidirectional like in Capt. Force's scenario.

I think that was one of the least egregious of the brain farts on here, certainly I have much worse ones When you imagine a constant heading passage, you just don't need to think about the ground track -- it's irrelevant to the mathematical problem (NOT of course to avoiding obstacles, as Dave pointed out ).

[Seaworthy Lass]

Originally Posted by Dockhead

Ha, ha. There were plenty of brain farts all around on here. I think I probably get the prize for the most!

But my darling -- Capt Force was not correct about the symmetrical sinusoid shape for the chart plotter track for a constant heading boat in a unidirectional current. It's symmetrical only in a symmetrical changing current. For a unidirectional current, the vast majority of the passage is spent on the upstream side of the rhumb line. You fall off downstream of the rhumb only just at the end -- if the current is unidirectional like in Capt. Force's scenario.

I think that was one of the least egregious of the brain farts on here, certainly I have much worse ones When you imagine a constant heading passage, you just don't need to think about the ground track -- it's irrelevant to the mathematical problem (NOT of course to avoiding obstacles, as Dave pointed out ).

LOL Dockhead. Dont make the same mistake I did. Plot it out!
If the current is unidirectional but a mirror image in strength, your chartplotter track WILL actually be a mirror image track (crossing the rhumb line at the halfway point - cross my heart LOL)

Check it out. I have learned a lot from this thread on what your ground track will be like following a constant heading to compensate for variable current .