Inaccurate RYA Teaching : CTS - Quest For a New Method

[Seaworthy Lass]

Originally Posted by DeepFrz

I get it now. I also think that you are basically using the same method of triangulation and that the instructor was saying something I don't get. Maybe I should go back to the other thread, as I put it on ignore...

I am definitely not using the same method of triangulation.

Look at the ground tracks on the two diagrams I posted.
I get to B.
The RYA method does not get to B until you at some point realise this and change heading.
You then haven't had a constant CTS.

And I am not talking about arriving early and then reassessing. In the above example the RYA method has you approaching B more than a nm to the south of it.
How is that efficient or safe?

[DeepFrz]

Proof does not come from convincing someone of something, proof is in the proving that your hypothesis holds up to scrutiny. How can it be that by steering a course that puts you on a direct, straight ground track from point A to point B is less efficient than steering a course that takes you out of your way in a curved line. The curved line adds time and distance to your journey. Time and distance means fuel wasted if motoring. But perhaps you can convince me.

[Seaworthy Lass]

COURSE OVER GROUND TRACK
= red dotted line

MY METHOD
Example 2
Note that I arrive at the destination on one single heading at a constant travelling at a speed of 3 knots.
Because I am plotting this on a chart normally, I can easily see if there are any dangers along the way .

[Seaworthy Lass]

COURSE OVER GROUND TRACK
= red dotted line

RYA METHOD
Example 2
Note that unless you realise the problem and change course at some stage, you would never arrive at B.

When you realise your error (and I am talking about the error in the method, not in the data or the calculations) hopefully the current is not sufficiently adverse that you are unable to make it!!!!!

This is an important point and as I have allowed for any current at the end of the journey I would not have any problems at all arriving at B and doing so on one heading

[Dockhead]

Originally Posted by Andrew Troup

Dockhead

Once we've come to a common understanding on this point, I hope you will address any residual validity in my points a) and b)

OK?

I think your explanation and terminology seeks to explain the behaviour of the RYA method, whereas mine seeks to understand the underlying concepts.

(I don't think our explanations are at odds, FWIW)

It's as if you're describing what happens to the revs when I move the 'throttle' lever on a diesel engine.

whereas I'm saying "This is why it happens".

I think the latter description is more useful when the former description does not happen as expected.

- - - - -

Here's my attempt to explain the underlying process in conceptual terms, leading up to the phase you are describing:

(for simplicity, I'm assuming no errors in data or execution, a vessel motoring, and current data expressed as mean values for each hour):

The vectors for x whole hours of current are added, from the start point.

When enough vectors have been stacked that the last vector's endpoint (lets call it "C after x hrs", or "C...") is close enough that we can get within an hour of the true destination in x hours of steaming, a position "D" is marked on the rhumb line (a line passing through both the departure and the true destination.)

This mark is made at a distance from "C..." which corresponds to x hours of steaming.

(Sometimes this will not be possible, for geometric reasons, but that's for another discussion)

A line is now drawn from C... through D, and this is taken as the Course To Steer.

The length of "C...D" corresponds to the distance through the water from the departure point to D, after a period of x hours

My explanation pauses at this point, because this is the end of the conceptually simplest phase of the process, and the information we've now got is potentially useful. (eg "That's near enough; we can eyeball it from here")

A shortened version of what we've done so far, sounds like this to me:

"Work out the point D on the rhumb line which is nearest to the true destination, reachable after x whole hours. Work out the CTS for that destination"

(optional: Work out the distance through the water to that adjusted destination)

The next step is to get us to the true destination.

The distance through the water to the true destination is arrived at by what you call 'inflation' of the optional distance mentioned above. I don't think I need to go into that in any detail, because there is only one thing about the remainder of the process which seems relevant to my point, and that is this:

If we follow the same CTS we worked out for the adjusted destination, we will only pass through the true destination under the special circumstances you describe.

Under those circumstances, both the course to steer and the distance through the water will be correct.

Under any other circumstances, neither will be correct

But ...... IN ALL INSTANCES we will first pass through the 'adjusted' destination, at the time (and distance) we expect to.

So I think it's helpful to treat it as an interim destination. Sometimes it will be useful to use it as such, but conceptually I think it is always valid to think of it as such.

If we can't remember the fiddly bit at the end, we've still got something solid and comprehensible (and correct), and it's back on the rhumb line.

Which (as long as we KNOW that's what to expect) is generally going to be a good thing.



Please tell me which parts of this make sense to you, and which do not.

Very well explained -- yes, I get it all and agree with it.

The only thing you left out is the "inflation" (or "deflation") of the vector triangle which gets you conceptually to your destination according to the RYA method. It will work perfectly if the average tide during the uncalculated partial hour equals average tide of the calculated part of the passage.

I think SWL sells the RYA method just a teeny bit short by leaving this out, although it's basically a rhetorical point -- you do describe the practical effects correctly, I think.

So you can describe it either as a shifted destination, or a fudge, with equal accuracy, I guess. I think it's probably more correct to think of it as a fudge, but probably there is no real difference in the consequences of this frame of reference. I think maybe if you think of it as a fudge, you can more easily avoid mistakes like Andrew's.

One other small point: I really don't think that it's any "obsession with the rhumb line" which causes the flaws in the RYA method. I bet you'll agree with me if you think about it. If you're not going to finish the calculations, as the RYA would have us do, you have to have the course line to calculate some vector triangle.

Using the course line to calculate some vector triangle is the best way to do it if we consider it impractical or useless to calculate the last partial hour.


And that's the next battle we will have to fight -- it's Dave's defense (now, I think) that it's useless to calculate the last partial hour because we can't get close enough anyway.

SWL has brilliantly disproven this showing a nearly 10% error of the RYA method in a real case.

Nevertheless we have brigades insisting that a useful CTS is unknowable given the kind of data we have about tides and currents. Nobody knows the relationship between precision and accuracy of tidal data we are using and precision and accuracy of calculated CTS, but we can find it out mathematically, plus we can do a real comparison with a precise CTS passage model done with the high resolution Admiralty data. We are all guessing at this point, but I think it is pretty reasonable to assert that a 10% error in CTS is going to be noticeably wrong even with very poor tide data. I would be really surprised if we get an error of anywhere near 10% with typical hourly tide data; I'm guessing a couple of degrees will be typical. We shall see.


Now I have another completely new and different theory about all of this.

I think that we don't even need hourly (much less 5-minute-ly) tide data to do a usefully accurate CTS calculation.

I think we only need the quanta and timing of the limits of the currents. I think that since tides vary according to sine waves and that we will know to a high degree of accuracy what the currents will be if we only know the limits. There are some anomalies as we know -- double high water (Solent), high water stands, indeterminate low water time (Poole) -- but these are the result of reflections in constricted waters and will not be typical of the behavior of currents in open water.

I think if we have the right formula, we can calculate very precise tidal vectors without any more data than the quanta and limits of the tides.

This is beyond my meagre math skills so I'm going to ask my brother (nuclear physicist and former math prof) to help. I hope he has time.

[Seaworthy Lass]

Originally Posted by DeepFrz

Proof does not come from convincing someone of something, proof is in the proving that your hypothesis holds up to scrutiny. How can it be that by steering a course that puts you on a direct, straight ground track from point A to point B is less efficient than steering a course that takes you out of your way in a curved line. The curved line adds time and distance to your journey. Time and distance means fuel wasted if motoring. But perhaps you can convince me.

DeepFrz, please go back to the other navigation thread and see for yourself. Or perhaps Dockhead and AndrewTroup can step in at this point and point you to the relevant posts in that thread.

This has already been proven in the other thread.

You will save fuel following a single CTS NOT following a straight ground track.

[DeepFrz]

Maybe I should have added, in a classroom situation. Often, as in the Vendee Globe, the participants have to because of wind, waves or some other situation deviate far from the Great Circle route to find a more efficient means of getting to a specific point. However if circumstances allowed the great circle route would have been much more efficient. In a shorter track like we are talking about a straight line (rhumb line) will be the most efficient way to get from point A to point B.

Anyway, it is 02:00 here and I have to get up early. Ciao...

[Seaworthy Lass]

Originally Posted by Dockhead

The only thing you left out is the "inflation" (or "deflation") of the vector triangle which gets you conceptually to your destination according to the RYA method. It will work perfectly if the average tide during the uncalculated partial hour equals average tide of the calculated part of the passage.
.....

This inflation or deflation can lead to serious error.
The RYA method for determining CTS was 9% out in my first example and 14% out in the second.

I do not think these are not acceptable errors. Do you?

[Seaworthy Lass]

Has anyone checked out the ground tracks I posted details of for the two examples using the two methods to calculate the CTS?

The RYA method does not get you to B on one heading.
You could be in trouble once you realise if the current is too adverse at that stage to get you to B.

My method takes me to B.

Why pick a method inherently flawed if there are other options?

[cal40john]

Originally Posted by DeepFrz

Maybe I should have added, in a classroom situation. Often, as in the Vendee Globe, the participants have to because of wind, waves or some other situation deviate far from the Great Circle route to find a more efficient means of getting to a specific point. However if circumstances allowed the great circle route would have been much more efficient. In a shorter track like we are talking about a straight line (rhumb line) will be the most efficient way to get from point A to point B.

Anyway, it is 02:00 here and I have to get up early. Ciao...

One of the unrealistic examples from the other thread to prove the point was you have a 4 knot current perpendicular to your rhumb line for 2 hours, then it becomes 3 knots for an hour, then 1 for an hour. You can do 4 knots boatspeed. You spend 2 hours motoring straight into the current to maintain yourself on the rhumbline making zero progress towards your destination. The other boat has done a multi vector CTS calculation to steer the shortest distance through the WATER, not the shortest over the bottom, he is steering at some angle but he is making progress to the destination.

The English channel examples were good also. Say it takes 12 hours to cross the channel, 6 hours of tide nearly perpendicular one way, then 6 hours the other. If you crab into the current you travel a greater distance through the water to stay on the rhumb line and make less forward progress. If you chose the original compass course and held it the tide would carry you off rhumb one side then the other, but you wind up where you wanted to go while traveling at best speed.

Hope that was easier than finding 2 posts out of 800.

John

[Dockhead]

Originally Posted by DeepFrz

Proof does not come from convincing someone of something, proof is in the proving that your hypothesis holds up to scrutiny. How can it be that by steering a course that puts you on a direct, straight ground track from point A to point B is less efficient than steering a course that takes you out of your way in a curved line. The curved line adds time and distance to your journey. Time and distance means fuel wasted if motoring. But perhaps you can convince me.

This is Capt Force all over again.

You don't sail over ground, so your track and distance over ground it not relevant.

What is relevant is the distance you sail through water. That is evident from this proposition: time of passage is a function of speed through water, which is constant, not speed over ground. To solve time of passage you need distance through water. Since the shortest path between two points is a straight line, it means the quickest passage through moving water is a constant heading -- that's going straight through the water.

That is the whole purpose of CTS calculations.

Another formulation of the situation:

If the water is moving, your ground track and water track will be different. If the water is moving at other than a constant rate and direction, either your ground track, or your water track will be crooked. You can sail a straight ground track by putting your pilot on "track" mode and constantly correcting your heading to keep you crabbing along the rhumb line. In this case, however, your heading will change constantly and you will be sailing a crooked path through water.

OR, you can sail a straight path through water on a constant heading. In this case, your ground track will be crooked.

But since your speed is speed through water, the fastest way to get there is with the straight water track and crooked ground track.

Does that make sense?

We had a very long, vigorous argument with Capt Force about this (and we are all greatly indebted to him for forcing us to think through all of the concepts). He posed a problem -- a boat on a four hour passage over water, motoring at a constant 5 knots, subject to a one-sided current varying continuously from 0 to 4 knots and back to 0. What's the fastest way to get there -- "track mode" over the rhumb line, or on a constant heading? We proved to his satisfaction that the fastest way is on a constant heading. Here is the calculation:

captforcescenario_distancesliced1.xls

You will see that not only does the boat sailing a constant heading get there faster, but when you calculate the GPS track boat's passage -- the smaller the vector triangles, the slower it goes. This is the result of the fact that if you roughly follow the rhumb line but change course less often, you can sail something close to a small, ideal CTS passage between your course corrections -- little sinusoidal ground tracks which are, however, straight water tracks, which speed you up comparing to sticking closer to the straight ground track. It's another proof of the constant heading principle.

[Dockhead]

Originally Posted by Seaworthy Lass

This inflation or deflation can lead to serious error.
The RYA method for determining CTS was 9% out in my first example and 14% out in the second.

I do not think these are not acceptable errors. Do you?

Completely agree. I think you have conclusively proven that the RYA method is actually f*cked, and my hat has been off to you since day before yesterday, and remains off .

But either cleverly or fortuitously, you have chosen extreme examples. Try to run a scenario where the last partial hour runs at the average of the analyzed full hours. Your boat will arrive at the destination on the same constant heading. The idea is that it's "close enough", and in some cases it will be. But you have convinced me that it is disastrously wrong in many cases. Although we have not yet done our work on the precision and accuracy translation, I know for d*mned sure that I am never off by as much as 10% in ANY of my CTS calculations. I have enough data to know that my method is much better than that.

[Seaworthy Lass]

Originally Posted by Dockhead

Completely agree. I think you have conclusively proven that the RYA method is actually f*cked, and my hat has been off to you since day before yesterday, and remains off .

But either cleverly or fortuitously, you have chosen extreme examples. Try to run a scenario where the last partial hour runs at the average of the analyzed full hours. Your boat will arrive at the destination on the same constant heading. The idea is that it's "close enough", and in some cases it will be. But you have convinced me that it is disastrously wrong in many cases. Although we have not yet done our work on the precision and accuracy translation, I know for d*mned sure that I am never off by as much as 10% in ANY of my CTS calculations. I have enough data to know that my method is much better than that.

Thanks Dockhead. Your support is really, really appreciated xxxxxxx.

Yes, of course, I have deliberately chosen examples of where the RYA method fails. In lots of situations it is a good approximation, but the method does fail and it fails badly sometimes as I will show either tonight or tomorrow.

It is actually is not only if the average of the current vectors for X number of hours is significantly different to the average for X+1 (where you arrive in the interal between X and X+1), but also if the last current vector points exactly at the destination. That is where the RYA method truly falls to bits.

Yes, the RYA method gives a reasonable approximation in some circumstances, but why choose a method that will only work with certain data.
And to top it off, this limitation is not recognised by the organisation promoting it.
Some instructors are promoting it as being "mathematically precise".

We now have an experienced instructor with the following qualifications:
ISPA Yachtmaster Offshore Instructor Evaluator
CYA Advanced Cruising Instructor
IYT Yachtmaster Coastal Instructor

telling me my method is faulty and that there is nothing wrong with the RYA method (in other words it always gives a good approximation of the CTS).

My next example will hammer home my point even more strongly.

Why use a method that can fail if another method is available?

The only reason I can think of is "Because that's the way it has always been done around here".

[Seaworthy Lass]

Originally Posted by Dockhead

Completely agree. I think you have conclusively proven that the RYA method is actually f*cked, and my hat has been off to you since day before yesterday, and remains off .

But either cleverly or fortuitously, you have chosen extreme examples. Try to run a scenario where the last partial hour runs at the average of the analyzed full hours. Your boat will arrive at the destination on the same constant heading. The idea is that it's "close enough", and in some cases it will be. But you have convinced me that it is disastrously wrong in many cases. Although we have not yet done our work on the precision and accuracy translation, I know for d*mned sure that I am never off by as much as 10% in ANY of my CTS calculations. I have enough data to know that my method is much better than that.

Thanks Dockhead. Your support is really, really appreciated xxxxxxx.

Yes, of course, I have deliberately chosen examples of where the RYA method fails. In lots of situations it is a good approximation, but the method does fail and it fails badly sometimes as I will show either tonight or tomorrow.

It is actually is not only if the average of the current vectors for X number of hours is significantly different to the average for X+1 (where you arrive in the interal between X and X+1), but also if the last current vector points exactly at the destination. That is where the RYA method truly falls to bits.

Yes, the RYA method gives a reasonable approximation in some circumstances, but why choose a method that will only work with certain data.
And to tip it off, this limitation is not recognised by the organisation promoting it.

We now have an experienced instructor with the following qualifications:
ISPA Yachtmaster Offshore Instructor Evaluator
CYA Advanced Cruising Instructor
IYT Yachtmaster Coastal Instructor

telling me my method is faulty and that there is nothing wrong with the RYA method (in other words it always gives a good approximation of the CTS).

My next example will hammer home my point even more strongly.

Why use a method that can fail if another method is available?

The only reason I can think of is
"Because that's the way it's always been done around here".

[Wotname]

Over 130 posts in total and more than 100 posts in the past 24 hours.

I can't keep up and work at the same time and so on.

Quick scratch pad analysis:

"RYA method" very imprecise but good enough to teach to new sailing students; it will get them close but they will never use it anyway, just turn on the chartplotter and follow the rhumb line. They will not understand the details of why that is not always the best way.

SWL method, much better, more accurate especially in the classroom and interesting for the handful who want to have a better understanding of the principles behind the method

Some posters (myself included) can't be a*rsed to work out all the examples - but I salute SWL and Dockhead among others for doing the hard work.

Other posters don't (yet) understand why ground track is unimportant in these examples, track through water is everything - apart from knowing that bottom dangers might lie in the ground track.

Interestingly (to me at least) is that the problems of the water moving while trying to get from point A to point B is almost (if not identical to) the same problems that aircraft have when the air moves about while they fly (i.e. wind). Basic pilot navigational training covers these problems in a far less complicated way. I suspect it is not so important for the average sailor to really know this stuff as mostly it doesn't really cause a problem whereas it is a serious problem for the pilot if he/she doesn't arrive at a suitable air field while there is still enough fuel.

Fight the good fight Seawortly