Inaccurate RYA Teaching : CTS - Quest For a New Method

[Seaworthy Lass]

Originally Posted by Wotname

There is that dry SOH again

Since bottles will have to be sent to a few CF members for lending their support in the face of great adversity (this really is a case of David vs Goliath), I am really now hoping it will be a couple of cases. I am forever the optimist .

PS All jokes aside, they actually COULDN'T afford me. It would actually need a few hundred thousand a year to get me out of retirement and even then I would think twice

[Seaworthy Lass]

Start with a cage containing five monkeys.

In the cage, hang a banana on a string and put stairs under it. Before long, a monkey will go to the stairs and start to climb towards the banana.

As soon as he touches the stairs, spray all of the monkeys with cold water.

After a while, another monkey will make an attempt with the same response - all of the monkeys are sprayed with cold water. Keep this up for several days.

Turn off the cold water.

If, later, another monkey tries to climb the stairs, the other monkeys will try to prevent it even though no water sprays them.

Now, remove one monkey from the cage and replace it with a new one.

The new monkey sees the banana and wants to climb the stairs. To his horror, all of the other monkeys attack him. After another attempt and attack, he knows that if he tries to climb the stairs, he will be assaulted.

Next, remove another of the original five monkeys and replace it with a new one. The newcomer goes to the stairs and is attacked. The previous newcomer takes part in the punishment with enthusiasm.

Replace the third original monkey with a new one. The new one makes it to the stairs and is attacked as well. Two of the four monkeys that beat him have no idea why they were not permitted to climb the stairs, or why they are participating in the beating of the newest monkey.

After replacing the fourth and fifth original monkeys, all the monkeys which have been sprayed with cold water have been replaced. Nevertheless, no monkey ever again approaches the stairs.

Why not?

"Because that's the way it's always been done around here".

There are lots of bananas for us to share (and more importantly for future generations to share) if you allow this little monkey to climb up the stairs

[Dockhead]

Originally Posted by Seaworthy Lass

NOOOOOOOOOOOO. I do not use the same vector triangles as the RYA.
And I admitted that you could think of vectors as triangles for the purposes of mathematical calculations. But you could also think of them as rectangles or rhomboids or whatever you fancied (not that this would help with mathematically solving anything LOL).

The only time I work with triangles is right at the end after I have computed the CTS and I want to plot my ground track .



The rhumb line has EVERYTHING to do with the shortcomings of the RYA method. Try working through example 3 using the rhumb line as a reference - you are stuffed if you do.

My lines from the current are directed at B, the destination. There is no D in my method at all. I ignore the rhumb line totally.

I think what you are doing when you use my method, is the procedure I followed in the first thread where I was looking at the proportion of the under or overshoot when the distance vector was arced off at the rhumb line. This method will fail. You need to join the current vector to B and then arc off the distance travelled and then compare the under and overshoot distances.

THERE IS NO RHUMB LINE IN MY METHOD!

Oh, dear, we will still need to work with you on this. I had thought you had gotten this already, but I see not.

You are really in error here, and it's not actually a trivial error.

Your whole method is a vector triangle calculation, just like Paul Elliot said. It's precisely the same method as RYA uses. What you are doing is analogue trigonometry -- solving triangles with your plotter -- and the course line is fundamental, even if you don't draw it or see it. Your plot is nothing more and nothing less than a vector triangle, and the Triangle Law of Vector Addition is the mathematical principle you are using to get your CTS. There are three points on your drawing and three lines between them, whether or not you draw them. Without the length and angle of the course line, you can't calculate anything. This is the bottom of your vector triangle, which is plainly seen in your own drawing.

You cannot do it with a trapezoid or a rhomboid or a circle or whatever. You need trigonometry, which is solving triangles.

Your vector triangle calculation is exactly the same method as RYA -- there's no magic. You've just fixed the partial hour, substituting a much better fudge (average of last hour) for the RYA fudge (average of the whole passage).

[Andrew Troup]

Originally Posted by Paul Elliott

The positions of all three points (A, B, and the end of the combined current vector) are determined by distance and angle.

The position of Point A could be Lat/Long from the GPS

The position of B could be Lat/Long from a list of waypoints, or a point picked out on a chart and described by Lat/Long

There is no need to compute the distance between them.

I think people are getting hung up on the distance being knowable.
That doesn't mean the same thing as being known.

Originally Posted by Paul Elliott

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? ,

Your second sentence contradicts the first.
The first is valid, at the beginning of the procedure;
the second l cannot make sense of.

At no time in the SwL procedure is it necessary to know the positions of A and B simultaneously, even in Lat/Long format.

If this is true, then it is axiomatic that the line between them cannot be essential to the method, because you cannot draw a line between two points when you only know one of them at any given time.

If you can demonstrate how SwL's procedure requires both positions to be known at the same time, then I will gladly concede the point.

[Seaworthy Lass]

Originally Posted by Dockhead

Your vector triangle calculation is exactly the same method as RYA -- there's no magic. You've just fixed the partial hour, substituting a much better fudge (average of last hour) for the RYA fudge (average of the whole passage).

Dockhead, please forget about the triangles, they are irrelevant. Arguing about this is clouding the issue. I submit to you about triangles if you will just focus on the bigger picture.

The RYA method has given the wrong result by 5 the 14 then a whopping 37 degrees in the 3 examples I gave. They are not isolated cases, I can churn out thousands of examples like this.

Results using the SWL method were correct within a degree.

Concentrate on ditching the rhumb line and eating some bananas .

[Dockhead]

Originally Posted by Andrew Troup

Dockhead said, some time ago, that her method could not possibly relinquish the RYA reliance on the rhumb line, because it forms one leg of a triangle which is indispensable to a solution.

My contention is that the triangle he refers to, in fact ANY triangle, is only indispensable, in the SwL method, to those for whom triangles are inseperably bound up with vectors. (To be completely clear: I am NOT such a person)

And no one has yet shown any flaw in my whimsical proof of that, involving a small dinghy puttering across the chart equipped only with a tape measure, a watch, a protractor and their eyes, exactly duplicating the indispensable elements of Seaworthy's method, with zero recourse to triangles. Or the rhumb line.

The participants do not at any stage KNOW where the rhumb line is, or how long it is, and yet they solve the CTS, the elapsed time for the passage, and the distance through the water.

The flaw in your whimsical proof is that you're not doing navigation at all. Once you've gotten to "B", point "A" disappears. You don't need navigation at all if you have a magic tape measure.


If you need a Course to Steer, then you need to get an angle which will take you to your destination despite the movement hither and thither of the water.

That angle is derived from two things:

1. The sum of the water vectors

measured against

2. The two points "A" and "B" in relation to each other (which is nothing other than the course line, and strictly IS the course line mathematically speaking -- any two points in relation to each other).


with your protractor. That is called trigonometry -- triangle-solving. You can do it in an analogue way with a protractor. Or you can crunch the numbers, much easier now than when vector triangle calculations with protractors were invented, using tools like this: Triangle Calculator -- EndMemo

Pay attention to this, and it is very enlightening to this discussion. Here is the digital way to do this:



You can calculate all 3 angles and all 3 sides IF AND ONLY IF you have any 3 items. That's a mathematical law and the foundation of trigonometry.


What Seaworthy is doing is building a triangle so that she has the length of three sides. The water track line is determined by boat speed and time (TIME is constant here -- very important). The sum of vectors line is determined by plotting (can also be calculated digitally). Then you MUST have the course line to complete the calculation. THEN the mathematical laws spit out the angle which is your CTS. You can do it with a protractor, or you can do it with numbers, but it's exactly the same calculation. It's a real classic, and it's not rocket science -- it's called a vector triangle calculation.


Now in this long and fascinating exploration of this issue we have every one of us put our foot in it from time to time and said something stupid. We learn by others calling us on it -- if two heads are better than one, then 10 heads are 10 times better! Those of us who are actually learning are taking two steps forward (two brilliant ideas) and one step back (one stupid one!). The sooner you recognize when you've made a step back, the more valuable the whole process will be to you! Seaworthy has proven herself to be by far the most brilliant of us!! But the idea that she is doing what she is doing without trigonometry is -- well -- what can I say -- really stupid!!

[LJH]

Originally Posted by Dockhead

Oh, dear, we will still need to work with you on this. I had thought you had gotten this already, but I see not.

You are really in error here, and it's not actually a trivial error.

Your whole method is a vector triangle calculation, just like Paul Elliot said. It's precisely the same method as RYA uses. What you are doing is analogue trigonometry -- solving triangles with your plotter -- and the course line is fundamental, even if you don't draw it or see it. Your plot is nothing more and nothing less than a vector triangle, and the Triangle Law of Vector Addition is the mathematical principle you are using to get your CTS. There are three points on your drawing and three lines between them, whether or not you draw them. Without the length and angle of the course line, you can't calculate anything. This is the bottom of your vector triangle, which is plainly seen in your own drawing.

You cannot do it with a trapezoid or a rhomboid or a circle or whatever. You need trigonometry, which is solving triangles.

Your vector triangle calculation is exactly the same method as RYA -- there's no magic. You've just fixed the partial hour, substituting a much better fudge (average of last hour) for the RYA fudge (average of the whole passage).

Dockhead,

Post 326 hows the SWL method being used in OpenCPN. If you review the steps you will see that at no point did I need to use the AB line or distance.

It is a done with a series of vectors from A for the tide set. and then vectors towards B for the CTS. The calculation between the points K to B does not require the line AB or the distance AB. However, if you wish to monitor your progress from A-B using XTE from the rhumb line you can use the AB vector for your Vmg. If you want to monitor along the ground track you do not need it.

[Andrew Troup]

Dockhead: thanks for fronting up and rescinding the "vectors are triangles" assertion; I wondered what you were on, for a moment there. It sounded expensive

I don't recall your proving to Seaworthy that she relied on triangles, I got the impression you repeated your confident assertions
(including the authoritative sounding, since-rescinded assertion above)
so convincingly and so frequently that she eventually threw up a white rag
(makes a change from dropping her towel!).

ON EDIT Since I wrote this, she posted as above, confirming both what I wrote above and the following sentence:

I don't think the triangle thing is an issue for her, or indeed for anyone else apart from you and me, so it was not a major concession.

It still puzzles me that you were so adamant that she must be (unknowingly) using triangles, at a time when, judging from subsequent posts, you were still largely in the dark as to what her method entailed.

I wonder if that reflects a belief in the "Doctrine of the Imperative Triangle", which exists outside of the rules of evidence?

Let me put on the record the simplest statement I can make:

THE ADDITION OF A SUCCESSION OF VECTORS
GRAPHICALLY
DOES NOT REQUIRE OR INVOLVE TRIANGLES

In the attached diagram, the five black vectors are added to produce the red vector. What you see is the entire story. The addition process is completely depicted. All the vectors can be described without recourse to triangles.

If you can see any triangles in this diagram, or imagine some indispensable service they provide, that can only be the product of some belief on your part.

But I hereby abandon any further attempt to persuade you on this point. Seaworthy deserves better than this endless and fruitless side-show.

I've tried everything I can think of, and while I believe the implications of shaking off "triangulism" are not trivial for anyone who aspires to innovate navigational methods, the specific instance is extremely trivial.

[Dockhead]

Originally Posted by LJH

Dockhead,

Post 326 hows the SWL method being used in OpenCPN. If you review the steps you will see that at no point did I need to use the AB line or distance.

It is a done with a series of vectors from A for the tide set. and then vectors towards B for the CTS. The calculation between the points K to B does not require the line AB or the distance AB. However, if you wish to monitor your progress from A-B using XTE from the rhumb line you can use the AB vector for your Vmg. If you want to monitor along the ground track you do not need it.

That method is called a vector triangle calculation. It uses trigonometry in order to work. Don't be deceived by the fact that you don't explicitly calculate the length -- when you do it by plotting this is inherent in the position of B. That doesn't change the math.

[Andrew Troup]

Dockhead posted while I was typing, as follows:

If you need a Course to Steer, then you need to get an angle which will take you to your destination despite the movement hither and thither of the water.

That angle is derived from two things:

1. The sum of the water vectors

measured against

2. The two points "A" and "B" in relation to each other (which is nothing other than the course line, and strictly IS the course line mathematically speaking -- any two points in relation to each other).

- - - - - -

Not true.

That angle is derived from three things:

1) The endpoint of the 'water vectors'

2) The true destination, B

3) The direction of North

I'm done on this point too, in the context of hijacking SwLs thread, but would be happy to continue the discussion off line via PM

Thanks for hanging in there.

[Seaworthy Lass]

Originally Posted by Andrew Troup

Dockhead:
I don't think the triangle thing is an issue for her, or indeed for anyone else apart from you and me, so it was not a major concession.

It is not an issue at all LOL.

Originally Posted by Andrew Troup

But I hereby abandon any further attempt to persuade you on this point. Seaworthy deserves better than this endless and fruitless side-show.

Many thanks, this argument about triangles is totally clouding the issue.

Whether I use triangles or not is truly irrelevant to the use of my method. The fact that I have ditched all reference to the rhumb line and have the distance vector heading to B (which the RYA say is a cardinal sin and NEVER NEVER NEVER should be done unless D exactly coincides with B) is the critical bit .

[LJH]

Originally Posted by Dockhead

What Seaworthy is doing is building a triangle so that she has the length of three sides. The water track line is determined by boat speed and time (TIME is constant here -- very important). The sum of vectors line is determined by plotting (can also be calculated digitally). Then you MUST have the course line to complete the calculation. THEN the mathematical laws spit out the angle which is your CTS. You can do it with a protractor, or you can do it with numbers, but it's exactly the same calculation. It's a real classic, and it's not rocket science -- it's called a vector triangle calculation.

Seaworthy's method offsets position A to position K using the tide set. To calculate the distance and bearing from K to B you do not need the vector from A to B. This is true if you calculate as either a great circle route or a rhumb line. You have X, Y and Z for both points (Lat, Long and Earth's radius) and trig does the rest. Or the protractor and dividers on a chart.

For the actual navigation you may want to plot the rhumb line from A to B, or you may want to plot a series of rumb lines that reflect your planned track.

[Dockhead]

Originally Posted by Andrew Troup

Let me put on the record the simplest statement I can make:

THE ADDITION OF A SUCCESSION OF VECTORS
GRAPHICALLY
DOES NOT REQUIRE OR INVOLVE TRIANGLES


In the attached diagram, the five black vectors are added to produce the red vector.

CORRECT.

Originally Posted by Andrew Troup

What you see is the entire story. The addition process is completely depicted. All the vectors can be described without recourse to triangles

FALSE.

You can get the end point of the vector sum, but you cannot get the angle -- the CTS -- without solving the triangle formed from three points (if you prefer them to lines) -- 1. Origin; 2. Destination; 3. End of the vector addition lines.


There's a real easy way to settle this --

Since you don't see the trignometry in the process of plotting (it's analogue trigonometry, but you don't understand that), let's go digital.

Solve any CTS problem using any method with math, without trigonometry.

You can't. And that's the answer to the question. If you can, I will eat my hat (and send you a 25-year bottle of single malt). But you can't. Because it is a trig problem, nothing more or less.


It's not trivial, because it has lead to serious errors in understanding the differences between RYA method and SWL method.

[Dockhead]

Originally Posted by Andrew Troup

Dockhead posted while I was typing, as follows:

If you need a Course to Steer, then you need to get an angle which will take you to your destination despite the movement hither and thither of the water.

That angle is derived from two things:

1. The sum of the water vectors

measured against

2. The two points "A" and "B" in relation to each other (which is nothing other than the course line, and strictly IS the course line mathematically speaking -- any two points in relation to each other).

- - - - - -

Not true.

That angle is derived from three things:

1) The endpoint of the 'water vectors'

2) The true destination, B

3) The direction of North

Oh, brother! You forgot the Origin!!

[Paul Elliott]

Originally Posted by Andrew Troup

The position of Point A could be Lat/Long from the GPS

The position of B could be Lat/Long from a list of waypoints, or a point picked out on a chart and described by Lat/Long

There is no need to compute the distance between them.

OK, you don't compute this distance. You draw it, or it's been drawn for you.

I think people are getting hung up on the distance being knowable.
That doesn't mean the same thing as being known.

Your second sentence contradicts the first.
The first is valid, at the beginning of the procedure;
the second l cannot make sense of.

I apologize for the confusion. In some earlier post mention was made that the distance to "A" wasn't necessary, given that we had "B". My point is that we *need* to know the locations of A B, and (whatever the current vector endpoint is called, I will call it "C" here).

At no time in the SwL procedure is it necessary to know the positions of A and B simultaneously, even in Lat/Long format.

If this is true, then it is axiomatic that the line between them cannot be essential to the method, because you cannot draw a line between two points when you only know one of them at any given time.

A and B are known. If you take them from the chart, then the distance/location/bearing is provided by that. No, you don't need to personally measure them, since that has already been done for you by the chartmakers. Use lat/lon? Same thing. If you're going to construct the third side of the triangle (heading and speed) then you must know the (relative) positions of A, B, and C. How you know them is irrelevant, but know them you must.

You may not think of it as a triangle, you may not use the trigonometric functions, but there are three points here, so as far as I'm concerned that's a triangle. Assuming they don't all fall on the same line.

If you can demonstrate how SwL's procedure requires both positions to be known at the same time, then I will gladly concede the point.

"A" is the point of departure, "B" is the destination, correct? Are you really asking me to explain why we need to know their positions?

You will note that I haven't mentioned the mechanical construction techniques, nor the details of the final interpolation. Until we agree on *what* it is that we are doing, I'm not too interested in digging into *how* we do it.