CTS vs Following Rhumb Line - DILUTED THREAD

[Seaworthy Lass]

14 November 2013

Recently the issue of the quickest route between two points with the influence of current cropped up again.

Is it crabbing along the rhumb line (ie a following straight line on the chart plotter) or is it calculating a Course To Steer (CTS) and heading off on the one heading being swept all over the place if the current is variable?

CTS wins hand down (except if the current is constant the whole way, then you are travelling along the rhumb line anyway and the results are the same).

Early in 2013 a couple of threads ran discussing why the CTS was the quickest. Dockhead in particular wrote some excellent posts. The information was well buried in the first thread, so I thought I would copy the most
relevant posts and display them here so anyone wanting info would find it easier to access.

This was the thread the first 50 posts here came from:
http://www.cruisersforum.com/forums/...ame-94879.html

This was an interesting follow up thread:
http://www.cruisersforum.com/forums/...ine-97251.html

If I have missed any posts you think are important, please PM me.

Cheers
SWL

[Dockhead]

......
Although I am a total fan of old-fashioned chart work, hand calculated tidal vectors, etc., etc., etc., I would never in hell "use a current chart to figure out the proper ferrying angle". If the current goes only one way and at one speed during the time of the passage, I would simply adjust the heading until the electronically calculated COG matches bearing to waypoint. I might even put the pilot on "Track" mode (I can hear the shudders of the Luddites). There is no chart-based method which comes close to the accuracy and convenience of these methods, for that situation.

If crossing a tidal body of water with varying speeds and directions of water, you don't calculate a "ferrying angle". You caculate set and drift for each hour of your passage at your estimated passage speed, and add it up to get a CTS -- course to steer -- to your destination (which for me is always one mile uptide of where I actually want to go, to give a margin of error). This is a pretty laborious job -- takes me a good hour to work out a plan for a typical Channel crossing, but I like to do it all by hand without using a calculator (I must sound like I'm contradicting myself, but I prefer to do all the math with a pencil in an analogue manner, because many of the calculations are guesses and approximations and I don't have nearly enough math to do numerical approximations on a computer). I do three complete sets of calculations -- one for 7 knots average, one for 8 knots average, and one for 9 knots average speed.

Then you steer that one course (or much better, put the pilot on that one course in its plain old "heading mode"), and pay no attention as the tide sweeps you back and forth. You check the accuracy of your calculations once every hour against XTE as calculated by the plotter -- if you do your vectors by hand, that's the result you get anyway for every hour's calculation -- cumulative set and drift in relation to the rhumb line should equal XTE as calculated by your plotter. You also must keep an eye on your passage speed -- if you are faster or slower than your assumption, it will throw everything out. I usually re-run the whole thing in the middle of the Channel and calculate a new CTS which, if everything has gone well, will not be more than a degree or two off the original one.

That's the way we do it over here. You use every electronics means at your disposal, but a big part of the job cannot be done by any electronic system whose existence I'm aware of.

[Jammer Six]

When one is making a passage across tidal waters, as in the examples given, does the electronic magic box generate a course that is significantly different from that generated on paper by hand? If so, why?

[Dockhead]

Originally Posted by Jammer Six

When one is making a passage across tidal waters, as in the examples given, does the electronic magic box generate a course that is significantly different from that generated on paper by hand? If so, why?

Absolutely. The electronic box -- at least my ones -- will only tell you what is the rhumb line to your destination. It can only tell you what is the shortest distance to your destination over ground.

If you are crossing tidal waters, you need the shortest distance over water. That is because your boat doesn't sail in the bottom; it sails on water. The shortest distance over water is a constant compass bearing.

[Dockhead]

Originally Posted by goboatingnow

huh, only if the tide is symmetrical.

I for one, only calculate CTS if I have good tidal stream data, in most places you do not. Equally I set a limit on how far my XTE should get as I have found if I get the CTS wrong, I can get out of position, often ending downtide of the destination. Id have to say that most people today are merely using XTE.

in many areas that I am familiar with , I merely run CTS in my head, ie I allow for a degree of tide, over a period, often just compensating for it in my head, then use XTE as a quide to my success. While I have taught the RYA YM night classes to many people, Im not a great fan of the "compute all the vectors " approach. The on water events often throw everything out and force repeated recalculations,

Why would the tide need to be symmetrical? And why only if you have good tidal stream data?

In any case, calculating or even guessing set and drift and adding it up to get CTS is the only way to get across a moving body of water efficiently.

Let me put it another way -- if you were an omniscient navigator, who had tidal stream data accurate to 1 degrees and 0.0001 knots with resolution of one minute, and enough computing power to deal with it all, you would still make your transit by steering a constant compass bearing calculated using just that data. That is the shortest path to your destination. If you have to change course, you are increasing your distance through the water. The ideal path is a straight line through the water. A big S curve over ground, with XTE exceeding 10 miles at times in the case of a typical Channel crossing at springs.

We are not omniscient, and our tidal atlases have resolution of only one hour, and random arrows drawn on the chart, so a vagueish approximation of the streams. But that's good enough -- already gives us a much better CTS than we get from just holding a finger in the air. And we correct it as we go along. What mostly plays hell with these calculations is passage speed. A knot faster than you assumed and you are exposed to the current you thought for that hour for that much less time -- 12% to 15% less depending on that speed.

[Dockhead]

Originally Posted by goboatingnow

Agh I see you were talking about theoretically perfect course to steer calculations. dr cooper I presume.

Dave

Yes.

What I mean is that the shortest way to sail any passage across tidal or otherwise moving water is to point the bow in the same compass direction the entire way, notwithstanding how the water pushes you all over the place with the tide. The problem is knowing what compass direction to point the bow in, more about anon.

Important to note: when I write pushes you all over the place, I meanin relation to the ground. Chart plotters actually make it harder to visualize and understand this, because they show you where you are in relation to the ground, giving us a ground fetish which can be very harmful to good navigation. You are actually sailing a straight line through the water, notwithstanding the crazy "S" curve you see on the plotter.

On such a passage, every change of heading lengthens the distance sailed -- distance through water, which is the only way your boat can sail.

In order to sail the shortest distance, you would have to have an ideal CTS. In order to compute an ideal CTS, you would have to have perfect knowledge about the motion of the water over your course, and of course your own speed.

You do not have this perfect knowledge, so you have to make the best possible approximation.

It is absolutely amazing how well it works considering the low quality of the data. I guess if you do your approximations with feeling, the errors tend to cancel each other out -- I would be interested to know the theory behind it (my brother is a prof of math and physics -- need to ask him). I am rarely more than a mile off at my destination, after a mid-Channel course correction of never more than a degree or two, as long as my estimate of speed is reasonably close, although the XTE gets up to more than 10 miles along the way.

[nigel1]

Originally Posted by Dockhead

You do not have this perfect knowledge, so you have to make the best possible approximation.

It is absolutely amazing how well it works considering the low quality of the data. I guess if you do your approximations with feeling, the errors tend to cancel each other out -- I would be interested to know the theory behind it (my brother is a prof of math and physics -- need to ask him). I am rarely more than a mile off at my destination, after a mid-Channel course correction of never more than a degree or two, as long as my estimate of speed is reasonably close, although the XTE gets up to more than 10 miles along the way.

This is where GPS and chart plotters are so useful. You've done your "proper" nav, and the chart plotter is ideal for monitoring your progress. You'll also get a much better idea of the actual tidal conditions if you have a good heading sensor and accurate log.
Crossing the channel you'll not be wanting to spend too much time below with pencil and rule.

[Seaworthy Lass]

Originally Posted by CaptForce

It's difficult for me to accept that I would stand in such complete opposition to someone that seems to present such intelligent discourse. I sense that we must have some basic agreement, but I can't find it. Basic geography would present the shortest passage between two points crossing over the sea as a straight line and also maintaining a straight line over the ocean floor. It would be the mistake of maintaining a constant compass heading and keeping the bow pointed in the same direction regardless of variations in current set that would result in the inefficient sigmoid curve.

CaptForce, sorry, but Dockhead is correct. You need to look at vector diagrams showing tide, speed over water and speed over ground. The quickest journey is to calculate the correct heading for the passage and stick to this, moving in a straight path relative to the water and a wavy pattern relative to the ground .

[foolishsailor]

Dockhead is correct.

One way to visualize is to think of an ant crossing a treadmill.

The tread mill is one meter wide and 3 meters long. Lets assume a simulation of a symetrical tide for simplicities sake.

The ant starts crossing the treadmill and it starts moving to the left 2 meters. As the ant reaches the mid point of crossing the treadmill moves back to the right 2 meters.

The ant, no matter how far the treadmill moves each way, has only travelled 1 meter across the surface of the treadmill to get to the other side. He is attached to the surface. He has however travelled over ground in a sinewave a substantial added distance.

The movement of a boat in tidal waters is the same, you are attached to the surface. The shortest distance is the distance you travel through the water regardless of how that water shifts you around.

A good way to see this for yourself is to calculate a travel distance, find your cts and then go. Compare your distance travelled as calculated by your knotmeter and your gps. If the tide is symetrical the knotmeter will be very close to the actual distance.

[noelex 77]

My high school maths teacher said first draw a good diagram.

Well i never could draw a good diagram but here is bad vector diagram I drew.

I have assumed the simplistic case where the two currents cancel each other out and the current is at right angles to direction of travel. It's easiest to imagine a constant current that suddenly reverses half way along.

You can see the red line is longer meaning maintaining zero cross track error is the slowest / longest way. The quickest way is to allow the tide to push you to one side then the other.

[Dockhead]

Originally Posted by CaptForce

Ah, now I think I see the problem and why so many seem to stick to an intelligent answer that I can not agree with. It's all because of the "for simplicities sake" and the simple example from the high school math teacher that does not wish to calculate the complexities of the real world. All those that are following the plan of maintaining a constant compass heading while crossing a current may remain satisfied with this procedure "for simplicities sake", but in the real world they will be taking a longer course on the sigmoid curve because the drift, if not the set also, will vary. Let me explain with one of the most common passages taken across a current. If you were to procede on a plan of crossing in a direction of 90 degrees M from Palm Beach, Florida across the Gulfstream to the Bahamas, you might, for simplicities sake determine that you will need to take a constant course of 115 degrees M at five knots, due to the average current set to the north at 2.5 kts, in order to reach your destination in the same manner as many suggest. The method is simple, but during the begining of the trip, when the current is less than the 5 kts at the center of the Gulfstrem, you would be making headway to the south of the rumbline course. Half way across the passage, when you are in the full five kt current, you will meet your rhumbline and then be moved north of your most efficient path. Finally, as you approach your Bahamian landfall and the current has weakened, you will arrive at your destination on the heading of 115M. This method works for "simplicities sake", but if you had elected to remain on the couse determined by your GPS you would arrive sooner and traveled in a straight line as you would constantly be recieving updates to change your required course vector as the strength of the current changed. Kudos to all the math teachers and all those correctly planning to cross an imaginary current that remains at a constant velocity for the passage. There is nothing wrong with taking the longer course if you value the method that gives you a simple answer. By the way, that treadmill analogy is excellent. Now, just imagine that the center of the treadmill is moving faster than the edges!

Nope, sorry, you've got it all wrong, and that such a smart and experienced person like you could fail to grasp it is a good illustration of why this is so hard to teach.

Try to think of it this way -- you sail through water, not over ground. The most efficient course is a direct course through the water -- a straight line. A straight line through water is a constant compass heading. The constant compass heading is emphatically not "for simplicity's sake" -- it is the direct route through water.

Now if the water moves around, that efficient course will describe a rambling path over the bottom. But that does not matter, because your boat is not attached to the bottom -- a mile sailed is a mile sailed through the water.

Let's take a real simple example -- you are sailing Needles to Cherbourg, 60nm, bearing 180. Your average speed is 5 knots, so 12 hours. You set out just as the tide is setting to the West, and it will run West for six hours, then turn East. It runs, let's say, an average of 2 knots in each direction.

Your fastest way to get there is to set your pilot to steer 180 and leave it there for 12 hours whereupon you will arrive in Cherbourg. As the tide runs West, it will sweep you off the rhumb line. 6 hours later, halfway across, you are 12 miles down Channel from the rhumb line. And that is just where you want to be. Because now the tide is turning, and it is sweeping you now directly to Cherbourg. You will see that your COG will now gradually converge with BTW, although your bow is still pointed 180, more or less at the Cap d' Hague. That's the way I will be sailing that passage.

That way one sails 60 miles through the water, which is the shortest possible distance, so 12 hours at 5 knots, and one is in Cherbourg.

Now imagine you tried to do what you just said -- stay on the rhumb line, for the shortest route over ground. You can do this -- just set your pilot on "Track" mode, and your pilot will do it automatically. To counteract 2 knots of current, the pilot will steer you at first about 35 degrees to the East of Cherbourg, a course of about 145, so your bow will be pointed way off to the East, and your SOG towards S will be about 4 knots. You will not get half way across the Channel in 6 hours, because you are crabbing along the rhumb line, stemming the tide. After six hours, you have only made about 24 miles of Southing, because you have wasted 1 mile out of 5 sailed just crabbing towards the rhumb line.

When the tide turns, you are in the wrong place. Because now the tide, which had been sweeping you West, is now trying to take you to the East, but you are directly N of Cherbourg, on the rhumb line. So now you're going to do the opposite of what you did the previous 6 hours -- you will point your bow to the W, about 35 degrees off of S or about 215 when the tide is running 2 knots and your boat speed is 5 knots. You will again crab along the rhumb line making about 4 knots SOG towards Cherbourg. You will not be in Cherbourg after 12 hours, you will have made only about 48 miles made good after 12 hours, with 12 miles still to go. The tide will turn yet again, and again you will point your bow towards the W and again you will stem the tide, crabbing along the rhumb line.

You will have sailed a straight line across the ground, but you will have sailed 15 extra miles through the water, about 75 in all, which will take you 3 extra hours since your boat cannot feel miles over ground.

I will have sailed a huge long distance across the ground, in a long "S" curve, getting up to 12 miles XTE at one point, but I will have sailed exactly 60 miles through the water -- which I say again, is the only way any boat can sail -- boats don't have legs which reach down to the bottom. I will arrive in Cherbourg after 12 hours, 3 hours ahead of you.

Got it now?

In your Gulf Stream example, you will be crabbing back and forth staying on the rhumb line, sailing a longer distance through the water, changing heading all the time as the current gets stronger and weaker, rather than sailing the one true CTS which will let the current take you right to your destination, a shorter distance through the water.


Both of these examples assume perfect knowledge of the currents, and of our average passage speed -- of course we don't have perfect knowledge. Still, we know that sailing along the rhumb line is the wrong way to cross a moving body of water, and becomes a radically wrong way to do it, the more varied the current is. Even approximate knowledge of the motion of the body of water gives us enough to calculate a CTS and sail more efficiently.

I don't think I've been more than 2 miles off in my last 10 Channel crossings, and I'm usually within a mile, after one mid-Channel correction.

[noelex 77]

Let's do some simple calculations for a passage. I am sure doing the calculations will convince the doubters that Dockhead is correct.
We are traveling due north to a destination 60Nm away.
We a motoring at a constant STW of 5 knots.

The current will be as follows
1st hour 1K due west
2nd hour 2k due west
3rd hour 3k due west
4th hour 3k due west
5th hour 2k due west
6th hour 1k due west

7th hour 1k due East
8th hour 2k due East
9th hour 3K due East
10th hour 3K due East
11th hour 2K due East
12th hour 1K due east

After the 12th hour for subsequent hours there is no current.


Calculate the length of time the passage will take for:
A) steering a constant due North compass course ( swinging east and then back on course)
B) steering a constant COG ( with zero cross track error)


OK let's hear those answers.Preferably with a bit of working out to show how you arrived at the number.

[Seaworthy Lass]

Originally Posted by Andrew Troup

......
(I know this doesn't make any difference to the real world answer; it just makes it a bit harder to prove with a simple thought picture)

The easiest thing I can think of is that the shortest distance between two points is in a straight line. When sailing the straight line is relative to the water, not relative to the ground. Therefore for shortest time is taken following one constant heading all the way .

PS I was typing while you posted Dockhead. You explained it well.

[Dockhead]

Originally Posted by Jammer Six

Dockhead, if all this is so, then why did they teach me all about set, drift and laying out a course to compensate for all manner of water movement?

They taught you right - with one set of vectors for set and drift in one direction, that is, with a current with constant speed and direction, then you have one simple calculation to get the right course to steer -- CTS.

It gets interesting when the current varies in speed and/or direction, which is what we're talking about. You still make that same calculation which was taught to you, only you use the SUM of the vectors over the time of your passage, not the single vector of a single current.

People, I didn't make this up myself, as much as I would like to take credit for it , it is the correct and mathematically only answer to the problem.

[noelex 77]

Lets work out the maths hour by hour.
A) steering a constant due North compass course (swinging east and then back on course)
1st hour our boat is 5Nm north and 1Nm east
2nd hour our boat is 10Nm north and 3Nm east
3rd hour our boat is 15Nm north and 6Nm east
4th hour our boat is 20Nm north and 9Nm east
5th hour our boat is 25Nm north and 11Nm east
6th hour our boat is 30Nm north and 12Nm east

Then the current changes

7th hour our boat is 35Nm north and 11Nm east
8th hour our boat is 40 Nm north and 9Nm east
9th hour our boat is 45Nm north and 6Nm east
10th hour our boat is 50Nm north and 3Nm east
11th hour our boat is 55Nm north and 1Nm east
12th hour our boat has arrived at its destination

Journey time 12 hours.

B) steering a constant COG ( with zero cross track error)
1st hour our boat is 4.9Nm on track due north
2nd hour our boat is 9.5Nm on track due north
3rd hour our boat is 13.5Nm on track due north
4th hour our boat is 17.5Nm on track due north
5th hour our boat is 22.1Nm on track due north
6th hour our boat is 27Nm on track due north

Then the current changes

7th hour our boat is 31.9Nm on track due north
8th hour our boat is 36.5Nm on track due north
9th hour our boat is 40.5Nm on track due north
10th hour our boat is 44.5Nm on track due north
11th hour our boat is 49.1Nm on track due north
12 hour our boat is 54 Nm on track due north

There is still 6 Nm to go when the boat that has followed Dockheads plan has arrived and is enjoying a beer
If we assume no current from now on. (any current at right angles would be slower)

13th hour our boat is 59Nm on track due north

Journey time 13hours and 12mins

You have lost 1hour and 12 mins of beer (or even worse French wine) drinking time

( the actual Vmg can be calculated using Pythagoras. Vmg = the square root of 5squared -the current squared.
For example if the current is 2 knots the Vmg will be the square root of 25-4=4.6)
Normally the current would not be exactly at right angles and a simple vector diagram is easier, quicker and accurate enough)

Note the above example is for the simplest case, but same principal applies if the currents do not cancel. Just add add up the currents and make allowance depending on the journey time ( refer to Dockheads post on how to do this)

In practice a lot of other factors may determine the best course, especially when sailing. Factors like expected wind shifts can make a big difference.

A good navigator will put all this information together to determine the best course to steer at any one time, but you must correctly understand the effects of currents and how to optimise you course to take advantage of them, or to minimise there adverse effects.

This illustrates the point that knowing your position is only a small part of navigation, or pilotage if you prefer.