Inaccurate RYA Teaching : CTS - Quest For a New Method

[Dockhead]

Originally Posted by Seaworthy Lass

As I don't use the distance between A and B in any of my calculations (only for gathering tide data) and since the number of degrees I need to steer relative to my rhumb line does not depend on the angle of the rhumb line, I am not sure I will wave the white flag yet LOL.

Sure, I would have to draw triangles to compute this mathematically, but I am not computing the angle mathematically at all. I am simply charting it with, unlike the RYA method, not a triangle in sight.

The destination can be anywhere as long at the current is acting on it in the same relative way .

Although that is just an argument about nomenclature LOL.
The big thing is that I do not need to draw a line between A and B for my method at all and I don't.
In the RYA method it is essential.

But you do use the distance, because that is what determines the position of B which allows you to measure an angle to it with your protractor.

You are exactly solving a triangle; you are just using an analogue method instead of crunching sines and square roots. I am sorry to be a pedant about it, but a lot of bombastic but erroneous things were said which really need to be cleared up.

This:



Is called a vector triangle. This is the "triangle in sight" in your method. It is nothing more and nothing less than that. It is a mathematical construct which is an analogue method of solving trigonometric (i.e. triangular) functions invented in the days when there were no digital means to easily crunch the numbers.

Your vector triangle consists of three lines, none of which bears any direct relationship to your passage. These three lines meet at three points. The lines are abstractions, every one. The course line defines the range and bearing between origin and destination. The tide vector line defines the sum of the individual tide vectors (themselves each a triangular function) -- the line defines the sum of all your work with the individual currents, and again it consists of range and bearing -- range is the net displacement, and bearing is the net direction. The water track line defines the distance you sail in the time frame of the triangle -- every leg corresponds to exactly the same time frame, which is THE essential relationship between them.

So when you read off the angle of the water line, you are receiving the output from these triangular functions and you get the pot at the end of that rainbow -- your CTS.

The principle you have used to draw all of this is called the "Triangle Law of Vector Addition" -- see Vector Addition

[Paul Elliott]

Originally Posted by Dockhead

You are exactly solving a triangle
[... great explanation, see above ...]
The principle you have used to draw all of this is called the "Triangle Law of Vector Addition" -- see Vector Addition

Dockhead, that was well and correctly stated.

[Seaworthy Lass]

Originally Posted by Dockhead

But you do use the distance, because that is what determines the position of B which allows you to measure an angle to it with your protractor.

You are exactly solving a triangle; you are just using an analogue method instead of crunching sines and square roots. I am sorry to be a pedant about it, but a lot of bombastic but erroneous things were said which really need to be cleared up.

This:

Attachment 53407

Is called a vector triangle. This is the "triangle in sight" in your method. It is nothing more and nothing less than that. It is a mathematical construct which is an analogue method of solving trigonometric (i.e. triangular) functions invented in the days when there were no digital means to easily crunch the numbers.

Your vector triangle consists of three lines, none of which bears any direct relationship to your passage. These three lines meet at three points. The lines are abstractions, every one. The course line defines the range and bearing between origin and destination. The tide vector line defines the sum of the individual tide vectors (themselves each a triangular function) -- the line defines the sum of all your work with the individual currents, and again it consists of range and bearing -- range is the net displacement, and bearing is the net direction. The water track line defines the distance you sail in the time frame of the triangle -- every leg corresponds to exactly the same time frame, which is THE essential relationship between them.

So when you read off the angle of the water line, you are receiving the output from these triangular functions and you get the pot at the end of that rainbow -- your CTS.

The principle you have used to draw all of this is called the "Triangle Law of Vector Addition" -- see Vector Addition

OK, OK I surrender LOL. You win .

But you don't need to draw them for this computation and you certainly don't need to draw a line between A and B.
Surely there can be no disagreement about that as I have done all my computations for all the examples without any line drawn between A and B?

I suggest one is not drawn as it just distracts people from thinking about where their ground track is (their intended course over ground in other words). It encourages them to plot the track, which I feel is essential, rather than assume their track over ground lies somewhere close to the line between A and B.

[Seaworthy Lass]

Example 3
I have computed the CST for the example above using the SWL method.

Here is the data again.
Boat speed is constant at 4 knots throughout the journey.
You are motoring in flat water.
Destination is 11 nm due east.
Current:
1st hour: 8 knots 135 T
2nd hour: 6 knots 150 T
3rd hour: 2 knots 170 T
4th hour: 2 knots 10 T

What is the CTS and time taken?

Using the SWL method
Computations are plotted in the diagram below (note, these are usually made on a chart).
This was done according to the instructions in my second post.

The position of K on the last vector is determined by the proportion of SB relative to SB + BL.
Measuring this is comes to 1.1 / 1.1 + 4.9 = 0.2
As the last current vector displacement is 2 nm, it means you are subjected to it for 2 x 0.2 nm = 0.4 nm. Mark K at that spot on the vector.

Join K to B.

Measure the angle of the line KB. This is 9 degrees
CTS = 9 degrees true

Time taken is simply 3 hours plus 0.2 hours = 3.2 hours
(No need to even measure KB and divide by the boat speed as I said before to determine that, but if you want to double check by doing it:
Time taken = KB / boat speed = 12.8 / 4 = 3.2 hours)

[Lodesman]

Originally Posted by Seaworthy Lass

Example 3
I have computed the CST for the example above using the SWL method.

Here is the data again.
Boat speed is constant at 4 knots throughout the journey.
You are motoring in flat water.
Destination is 11 nm due east.
Current:
1st hour: 8 knots 135 T
2nd hour: 6 knots 150 T
3rd hour: 2 knots 170 T
4th hour: 2 knots 10 T

What is the CTS and time taken?

Using the SWL method
Computations are plotted in the diagram below (note, these are usually made on a chart).
This was done according to the instructions in my second post.

The position of K on the last vector is determined by the proportion of SB relative to SB + BL.
Measuring this is comes to 1.1 / 1.1 + 4.9 = 0.2
As the last current vector displacement is 2 nm, it means you are subjected to it for 2 x 0.2 nm = 0.4 nm. Mark K at that spot on the vector.

Join K to B.

Measure the angle of the line KB. This is 9 degrees
CTS = 9 degrees true

Time taken is simply 3 hours plus 0.2 hours = 3.2 hours
(No need to even measure KB and divide by the boat speed as I said before to determine that, but if you want to double check by doing it:
Time taken = KB / boat speed = 12.8 / 4 = 3.2 hours)

Not to burst your bubble, but plotting the ground track shows your course of 009º overshooting your destination by 2 cables; and you actually pass it at 3h8m, vice 3h12.

Calculating using my method required replotting the CMG, but I had my course of 008º (and some fractions, I rounded down) with just one refinement. The rough first course was 010º.

It was a rather tortured example to showcase your method - a prudent sailor probably would have waited 2 hours then drove straight East for 2h33m to arrive (you didn't give the 5th hour's tidal set - I would like to see what sort of set you would envision that could not be mitigated in the first two hours where you can actually go faster than the stream?).

[Seaworthy Lass]

Originally Posted by Lodesman

Not to burst your bubble, but plotting the ground track shows your course of 009º overshooting your destination by 2 cables; and you actually pass it at 3h8m, vice 3h12.

Well I would not have overshot it LOL, I would have stopped when I got there. If your calculations are correct, being 4 minutes out with my time estimate is insignificant. I have rounded figures off to the closest 0.1 nm as this was a limitation of my measurements, so that is how the small discrepancy occured.

Had my boat not been violently knocked about by 30+ knot gusts (which it still is) maybe I would have had a better plotting accuracy.

Originally Posted by Lodesman

Calculating using my method required replotting the CMG, but I had my course of 008º (and some fractions, I rounded down) with just one refinement. The rough first course was 010º.

Sounds like it has produced a good result in this case. Would you like to write up a step by step description with diagrams so that we can follow what you do?

All contributions are welcome .
I am on a quest to find a better method than the RYA's and to have it considered, not to promote my method.

Originally Posted by Lodesman

It was a rather tortured example to showcase your method - a prudent sailor probably would have waited 2 hours then drove straight East for 2h33m to arrive (you didn't give the 5th hour's tidal set - I would like to see what sort of set you would envision that could not be mitigated in the first two hours where you can actually go faster than the stream?).

Had the 'prudent' sailor waited two hours I think he would not have made any progress against the current (in fact he would have gone backwards).

[Lodesman]

Originally Posted by Seaworthy Lass

Well I would not have overshot it LOL, I would have stopped when I got there. If your calculations are correct, being 4 minutes out with my time estimate is insignificant. I have rounded figures off to the closest 0.1 nm as this was a limitation of my measurements, so that is how the small discrepancy occured.

I thought the whole point of your method over RYA's was that you wouldn't need to adjust your CTS

Stopping short wouldn't help as you're approaching from the South, you pass 2C East of the destination.

Originally Posted by Seaworthy Lass

Had the 'prudent' sailor waited two hours I think he would not have made any progress against the current (in fact he would have gone backwards).

He would have made great progress during hours 3 and 4. You never stated the current in the 5th hour.

[Dockhead]

Originally Posted by Seaworthy Lass

OK, OK I surrender LOL. You win .

But you don't need to draw them for this computation and you certainly don't need to draw a line between A and B.
Surely there can be no disagreement about that as I have done all my computations for all the examples without any line drawn between A and B?

I suggest one is not drawn as it just distracts people from thinking about where their ground track is (their intended course over ground in other words). It encourages them to plot the track, which I feel is essential, rather than assume their track over ground lies somewhere close to the line between A and B.

Sure -- no need to draw it if you're going to work out the last partial hour and lay the destination. And for damn sure we've seen a lot of people confused by that line -- thinking it has something to do with our passage I fully agree with you there.

But at the same time it's important not to be confused about the ways in which the RYA method is the same as yours. It is fundamentally the same vector triangle calculation; just theirs is incomplete and yours is finished.

I think missing this point has led to another error -- which is to think that this abstract line is what has led the RYA astray -- it really is not that, and it's not the fact that their water track line fall on the course line that causes the problem. It is the size of the triangle and the lack of some essential data input into it which screws up the RYA method. The line is nothing special; it's right there in your method too, mathematically, as we finally agree .


I think it's important to remember that in most cases -- not enough cases, but most -- the RYA method will give a good result. The example of a really wacky ground track which has you approaching your destination from an angle wildly different than the original bearing from your origin does not show an error other than the basically wrong CTS from the lack of essential data about the partial hour. No part of the error comes from your aspect to the destination (this was the "swinging the course line" conversation). And in reality it is nearly impossible to get such an extreme aspect. In nearly all real cases, the RYA method will get us into the range where we anyway go over to "track mode" steering, and in those cases it has really done its job. So I think the theoretical drawbacks of the method are more dramatic than the practical ones. STILL, this is great work, and useful work. I am all in favor of making an exact CTS calculation so that I can properly design my uptide fudge. Your observation about aiming at the right place is right on. I want to aim at a point far enough uptide to exclude anyplace downtide from the likely range of outcomes of my passage. The RYA method is not quite the right tool for that job.

I still don't understand your method but now have time to study it -- did you write a concise but comprehensive instruction? Can you point me to it? I'm lost in the hundreds of posts. I don't have access to a protractor until next week but will try to understand it abstractly.

I would do the partial hour recursively. An accurate calculation would be tedious but I think for me a decent approximation would be pretty easy (and I really think that's what the RYA should be teaching). But I am prepared to adopt your method, if it's easier -- looking forward to studying it.


To give perspective to this -- we English Channel sailors don't use any of these methods. We have more or less perpendicular tides, so we have a much easier and very reliable method which works like this. It's laid out in all of the English Channel pilot books. We estimate passage time, then mark in pencil dots on the tidal atlas representing where we expect to be at every hour of the passage. We look at the tidal diamonds nearest to each of these dots and work out the rate (depending on where we are in the neaps-springs cycle) applicable at that moment in that place. We add up the set East or West of each hour. We definitely calculate partial hours and add the set of that partial hours to the whole pile. Thus we get a net set East or West over the whole passage (I actually run three sets of these numbers for 7, 8, and 9 knots average speed -- it's REAL handy to have them at hand for any mid-Channel correction due to deviation in passage speed).

THEN, we take that total set and correct with the bone-simple method of one degree of course correction for every mile of set at 60 miles. The aiming point is always a mile or two uptide, not the destination itself.

It works BRILLIANTLY well, which is why all English Channel sailors do it this way. I consistently get within a mile or so, and never correct course mid-Channel unless there has been a big deviation from planned passage speed (which can result from the wind, as you know). It's much better than the RYA method as it has no built-in errors, and especially not any built-in random errors, which is the bane of the RYA method.

It's harder for me to apply because I have a fast boat and always do it in less than two tides, so I always have a fairly big steering correction. A small boat making 5 knots will do Cherbourg in exactly two tides, and can usually just aim the bow at Cherbourg and drink tea while the tide does its business. My passage, on the other hand, are 90% between 8 and 9 knots . One was over 10 (and all my calculations were off! ).

That's why I've never done it with a protractor and had forgotten what I learned at uni 30 years ago. This thread has been SUPERBLY educational and has really got me back "im Schwung".

[Seaworthy Lass]

Originally Posted by Dockhead

I still don't understand your method but now have time to study it -- did you write a concise but comprehensive instruction? Can you point me to it? I'm lost in the hundreds of posts. I don't have access to a protractor until next week but will try to understand it abstractly.

It was explained in post # 2

[Dockhead]

Originally Posted by Seaworthy Lass

It was explained in post # 2

I'm still not entirely sure I understood it - still working on it. But if it consists of what I think it does, then it is truly brilliant. And if MISUNDERSTOOd it, then I have the makings of a truly brilliant system of my own .

It exactly uses the analogue method which has existed for hundreds of years to overcome the lack of number crunching power in the pre-digital age . Brilliant simplicity. Fantastic; truly inspired. Unbelievable, really. I'm in awe (pending final confirmation of my understanding of it ).

What it does is simply finds the proportionate average set and drift of the full hour of which the final partial hour is a part, and uses that as the final jink of the vector sum leg.

It is - again, if I understood it correctly - mathematically perfect IF we assume that any part of a whole hour is average. This is definitely not true in reality, but it's as good as we can get without assuming something about the shape of the current development through time.

In any case, it is simple to apply, and the built-in error is approximately an order of magnitude smaller than what is produced by the RYA method. This is really good stuff

I tank we might do even better with a sine curve model of tidal currents, without any hourly analysis at all. But that still has to be developed. . .

Bravo, Seaworthy

[Seaworthy Lass]

Originally Posted by Dockhead

I'm still not entirely sure I understood it - still working on it. But if it consists of what I think it does, then it is truly brilliant. And if MISUNDERSTOOd it, then I have the makings of a truly brilliant system of my own .

It exactly uses the analogue method which has existed for hundreds of years to overcome the lack of number crunching power in the pre-digital age . Brilliant simplicity. Fantastic; truly inspired. Unbelievable, really. I'm in awe (pending final confirmation of my understanding of it ).

THANKS Dockhead
It sounds like you have understood it!

[Seaworthy Lass]

Originally Posted by Dockhead

It exactly uses the analogue method which has existed for hundreds of years to overcome the lack of number crunching power in the pre-digital age . Brilliant simplicity. Fantastic; truly inspired. Unbelievable, really. I'm in awe (pending final confirmation of my understanding of it ).

What it does is simply finds the proportionate average set and drift of the full hour of which the final partial hour is a part, and uses that as the final jink of the vector sum leg.

It is - again, if I understood it correctly - mathematically perfect IF we assume that any part of a whole hour is average. This is definitely not true in reality, but it's as good as we can get without assuming something about the shape of the current development through time.

In any case, it is simple to apply, and the built-in error is approximately an order of magnitude smaller than what is produced by the RYA method. This is really good stuff

Bravo, Seaworthy

Yes Dockhead it is as easy as that. It is nothing really complicated.

And yes, it does assume the current is reasonably constant during that hour, but once you get good with the method you could tweek that proportion yourself according to whether the current was increasing or decreasing.

If you didn't want to measure the proportion, then just eyeball it. I have tried that for a few examples and I am never more than a degree off in the final result, compared to measuring the overshoot and undershoot with a ruler and calculating the proportion.

And the accuracy is always good (ignoring the limitations of the data) if you use a very sharp pencil and are very careful with the vector placement. With a bit of practice one to two degree accuracy could be expected.

And this works for ANY data you throw at it, not just limited conditions of low current coming from certain directions and not reversing at the end of the journey and ..........

Unlike the RYA method you will ALWAYS be within a couple of degrees and there will never be any figures that are wildly out .

[goboatingnow]

Originally Posted by Seaworthy Lass

Yes Dockhead it is as easy as that. It is nothing really complicated.

And yes, it does assume the current is reasonably constant during that hour, but once you get good with the method you could tweek that proportion yourself according to whether the current was increasing or decreasing.

If you didn't want to measure the proportion, then just eyeball it. I have tried that for a few examples and I am never more than a degree off in the final result, compared to measuring the overshoot and undershoot with a ruler and calculating the proportion.

And the accuracy is always good (ignoring the limitations of the data) if you use a very sharp pencil and are very careful with the vector placement. With a bit of practice one to two degree accuracy could be expected.

And this works for ANY data you throw at it, not just limited conditions of low current coming from certain directions and not reversing at the end of the journey and ..........

Unlike the RYA method you will ALWAYS be within a couple of degrees and there will never be any figures that are wildly out .

IU dont mind you arguing for your own method, I really dislike your characterisation of the RYA one. YOUR method is not real world precise, no more then any CTS is real world precise.

Ive demonstrated repeatedly how the RYA method is applied and in that situation as I said above, it could be simply visually interpolated, ( I got 58,5 degrees CTS as a result)

IN the real world , the vector addition results in far more obvious closer situations, in which case the RYA provides the same answer

The RYA method has stood the test of real navigators in the real world, within its limitations , like all limitations it is as accurate as any, apply with intelligence and as real world sailors it will be fine

[Andrew Troup]

Originally Posted by Dockhead

But you do use the distance, because that is what determines the position of B which allows you to measure an angle to it with your protractor.

Dockhead

The position of B is not determined by distance.

It's determined by being a point we are trying to reach.

Once again, you're substituting one representation ('this is how far it is from us') for the thing itself.

B can be specified in numerous ways which make NO reference to where we are when we decide to go to B.

Lat/Long is one.

This fixation with where we have departed from is presumably what Seaworthy battled with, and dropping that fixation was (she tells us) crucial to developing the method you have hailed as brilliant.

I wish my whimsical parable had worked for you, because it spelled out the chartwork of the Seaworthy method in a way I hoped would help others make the same mental leap she has evidently made.

Once the first hour's tidal vector is put on the chart, the point of departure is of NO FURTHER INTEREST in the solution.

(We're NOT solving for ground track, which DOES rely on the point of departure; that's a separate procedure her method does not address)

The remainder of the procedure is carried out with no reference to it, and if that part of the chart got masticated by the ship's baby, it wouldn't matter a damn.

Originally Posted by Dockhead

You are exactly solving a triangle; you are just using an analogue method instead of crunching sines and square roots. I am sorry to be a pedant about it, but a lot of bombastic but erroneous things were said which really need to be cleared up.

This:

Attachment 53407

Is called a vector triangle.
......

The principle you have used to draw all of this is called the "Triangle Law of Vector Addition" -- see Vector Addition

Dockhead

I'm sorry that you wore Seaworthy down on this point, because it seems to me she was correct.

The link you posted says this:



Common methods adding coplanar vectors (vectors acting in the same plane) are

• the parallelogram law
• the triangle rule
• trigonometric calculation

------
They don't say "only methods", they say "common methods"

Another common method, available only to hominids with opposable thumbs and pencils, is a graphical method based on the purer concept of a vector spelled out in my last post.

The triangles you see in the graphical method are imported to it by your mindset.

In your terms, the triangles your mind's eye conjures are a construct.

They are sufficient to solve vectors, but they are NOT necessary.


You are welcome to carry on thinking in triangles; but for those with opposable thumbs there is the simpler option of thinking in straight lines.

[Paul Elliott]

Originally Posted by Andrew Troup

Dockhead

The position of B is not determined by distance.

It's determined by being a point we are trying to reach.

The positions of all three points (A, B, and the end of the combined current vector) are determined by distance and angle. You can't have one of these without all the others. The current vectors start at "A", so you have to know where "A" is. You have to know the distance between "A" and "B", otherwise how do you know where to start the current vector? Claiming anything else is pretty silly. I don't care if you draw the line AB or not, you still need to know the distance.

All this should be quite obvious, and no doubt to many of us it is, but we have some people arguing about the underlying mathematic/geometric principles, and others arguing about the implementation, or construction. They are arguing apples and oranges, without knowing it.

The discussion of "inflation" of the entire passage vs interpolation or extrapolation of the last hour seems to be the root issue here (as well as the effects of the underlying accuracy of the available data). This can be considered without getting into the mechanical details of the solution. The actual pencil and paper technique for arriving at the solution, and appropriate simplifications, approximations, and shortcuts, should be isolated from the "higher level" discussion.

Just a suggestion,