Inaccurate RYA Teaching : CTS - Quest For a New Method

[bewitched]

Originally Posted by Seaworthy Lass

ROFL, why else do you want to sail a single CTS?
Because you want to make the FASTEST passage LOL. A couple of decades racing/cruising in Australia's tidal waters has taught me nothing if not that!

I do a lot of racing too and at quite a high level too. If it was simply a case of calculating a single course to steer to get the fastest passage, racing would be really boring.

Consider the following: A to B is 90 deg and 5nm.

What is the fastest course to steer in each of these scenarios? (ignoring all variables other than tide)

1. Zero tide
2. A constant north south tide at 1 kt
3. A constant north south tide at 2 kts

The answer to 1. is obviously a CTS of 90 deg
The answer to 2. will be somewhere less than 90deg
And the answer to 3. will be somewhere less than 90deg and also less than answer 2.

So what does this very simple exercise tell us? Well it tells me that if the tidal flow rate varies, the CTS will have to change to remain optimal. Maintaining a single CTS, derived from an average tide state over the course, really is not the way to get there quickly.

[Seaworthy Lass]

Originally Posted by bewitched

I do a lot of racing too and at quite a high level too. If it was simply a case of calculating a single course to steer to get the fastest passage, racing would be really boring.

Consider the following: A to B is 90 deg and 5nm.

What is the fastest course to steer in each of these scenarios? (ignoring all variables other than tide)

1. Zero tide
2. A constant north south tide at 1 kt
3. A constant north south tide at 2 kts

The answer to 1. is obviously a CTS of 90 deg
The answer to 2. will be somewhere less than 90deg
And the answer to 3. will be somewhere less than 90deg and also less than answer 2.

So what does this very simple exercise tell us? Well it tells me that if the tidal flow rate varies, the CTS will have to change to remain optimal. Maintaining a single CTS, derived from an average tide state over the course, really is not the way to get there quickly.

Winning many races?

Suggest you review your strategy .

[Seaworthy Lass]

Originally Posted by jackdale

This trip is impossible. There is absolutely no way that it can be done. These currents are way too strong. If you method shows that the route is feasible, it is fatally flawed.

Sigh ......

You announce the route is infeasible and that my method is fatally flawed if it gives me a CTS to get there (at a constant speed through water of four knots) but you are unwilling to bet a bottle of Scotch on it?

Well, no matter. Perhaps the RYA will send me a case of Scotch for providing them with a vastly superior technique for calculating the CTS, as it allows computation for ALL situations, regardless of your current or boat speed, as long as the AVERAGE adverse current is not greater than your boat speed.

Even my method will not help you then LOL!

[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)

[cal40john]

Originally Posted by jackdale

I am re-doing the numbers after a re-think. But there is a serious problem,

An important question.

How are you determining speed. Knotmeter, GPS, ?

I'd spend 2 hours relaxing on the boat at the marina then motor at approx. 90 degrees heading for under 3 hours. If I plotted the ground track correctly for the example as stated, the RYA method leaves you almost 4 miles short of your intended destination, so if no current the last hour you have to motor another hour. We both get there at the same time, but I spent 2 hours of my time reading, at the bar, etc.

Kirk's Kobayashi maneuver.


Yes the example is contrived and I wouldn't go out, but I can't see that the method was applied incorrectly. You don't go backwards, but you do a lot of sideways.

John

[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 cal40john

I'd spend 2 hours relaxing on the boat at the marina then motor at approx. 90 degrees heading for under 3 hours. If I plotted the ground track correctly for the example as stated, the RYA method leaves you almost 4 miles short of your intended destination, so if no current the last hour you have to motor another hour. We both get there at the same time, but I spent 2 hours of my time reading, at the bar, etc.

Kirk's Kobayashi maneuver.

Yes the example is contrived and I wouldn't go out, but I can't see that the method was applied incorrectly. You don't go backwards, but you do a lot of sideways.

John

The trip is not unfeasible by any means.
The speed made good is 3.4 knots (11 nm / 3.2 hours).

Using the RYA method would have shown the trip was not possible.
This is CERTAINLY not the case, although it seems I am not going to be enjoying any single malt anytime soon. Pity as it is rare as hen's teeth on the Greek islands .

[jackdale]

Originally Posted by cal40john

Yes the example is contrived and I wouldn't go out, but I can't see that the method was applied incorrectly. You don't go backwards, but you do a lot of sideways.

John

John I had a brain f*rt and did a rethink. You are correct. That sideways motion is so severe that it will not let make the rhumb line (CMG, whatever), until very late in the voyage. I will be posting my solution in a couple of minutes.

[goboatingnow]

Jeepers I leave you all alone for few days >.....


Im glad to see another RYA instructor here as well.

I think whats getting mixed up are the comparisons between the models and the reality , and then comparing the models


First Seaworthy as youve thrown a fair bit of mud at me.

The RYA method is correct mathematially , thats does not mean it doesnt accurately reflect reality or not. For time based vector addition , given the known vectors proportionately inflating each vector is correct. Thats simple time based vector addition.

Its means the CTS read off is correct mathematically , its doesnt mean its correct in real life .

The RYA method inflates or deflates each tidal vector by the same extension or reduction in time.

Yes there are inaccuracies when this method is compared to real life , I have always said that and I laid out some of those inaccuracies.

If you want better accuracies, you can plot 30, 15, or 10 minute tides if you have them The accuracy of the RYA method compared to real life improves.

I repeated that The RYA properly taught always examines where D falls, If D B is greater then 30 minutes that there are inaccuracies in inflating the triangle over more then 30 minutes. If that is the case plot a further hour of tide and "deflate" the tides. A good instructor points out these issues in a classroom

It is ridiculous to suggest your method is more accurate, whether the RYA it is or isnt, its within the accuracy of the underlying data, well within it in fact. taking extreme examples really proves nothing. The RYA method is simple, does not require calculations and most people want a vague CTS. ( SInce they cant hold anything like it in practice).

Your whole argument rests on the treatment of how the remaining time is treated, whether it is propotional by time over all the tides being plotted ( including the last one) or whether you proportion the last one ( if I understand what u do). Yet neither method takes into account what real tides do and the real issues encoutered on passages. Both provide data that has significant errors when compared to real life

The conclusion , Either models are flawed. But at least the RYA one is easy (easier).


I cant understand why you critise the training web page showing a reversing sinusodial tide as incorrect. The ground track is correct, for a non reversing as we pointed out there is a rhumb line cross over. You howled your disapproval but never actually laid out what that is.

PS I measure my example I proved with a protractor, I got the result I did, I shall recheck it tomorrow, but my results where closer then you computed.

As I said the RYA method remains in general within the bounds of accuracy is less complex that SWLs method. Yes of course there are errors in compared to real life , but so what , there are huge amounts of other errors which dawrf that calculation anyway.

In practice you will have to repeatedly re compute CTS, in effect you end up with a one hour method anyway.

Dave

[nigel1]

SWL, in your last example showing the RYA (traditional) method, you are using 4 hours worth of tidal vectors, and 4 hours worth of boat speed, which puts point D well beyond B. Anyone trying to work out the CTS would then re work the vectors, but use only the first 3 hours of tidal vectors, which by eyeball would provide a CTS with much more north in it. If that comes well short of B, then re work using a small portion of the final hours tide.
You cannot really compare your method to the RYA method, when you use 3.2 hours of vectors for one, and 4 for the other.

[Seaworthy Lass]

Originally Posted by jackdale

John I had a brain f*rt and did a rethink. You are correct. That sideways motion is so severe that it will not let make the rhumb line (CMG, whatever), until very late in the voyage. I will be posting my solution in a couple of minutes.

Jackdale, why would you want to be on the rhumb line before you reach B?

This is truly a huge flaw with the RYA method. All calculations should be made to get to B, not some other arbitrary point.

A method is not very good unless it can deal with ANY data thrown at it.

Do you teach students that the method is severely flawed if you cannot get D close to B in your computations? When you evaluate other instructors do you check they stress this to their students?

The instructor in the previous thread even kept insisting for ages that the RYA method took you directly to B and that D was an artificial point you would never cross!
He persisted in thread after thread telling me how wrong and confused I was when I suggested the RYA method was just an approximation, insisting it was mathematically perfect, it was only limited by the data imput. And since he thought he was going to B not D, that means mathematically perfect to get to B .

In the first example I gave in this thread where the RYA method was out by five degrees you may have said was "good enough" if you were seeking just an approximate result.

The RYA result in the second example was 14 degrees out - not an error I would find acceptable when computing a CTS.

The RYA result in the third example was 37 degrees out !!!!
In my opinion that is just absolutely useless.

Yes, the method may cover lots of situations, but if it also fails in lots of situations, yet it's validly seems to have never been questioned.

My technique is not much harder than the RYA method and it gets you there each and every time regardless of what data you throw at it .

[jackdale]

Originally Posted by Seaworthy Lass

Jackdale, how about a wager of a bottle of fine single malt?
I am game if you are.

And by the way, do you agree that that a CTS of 46 degrees is the one you get using the RYA method?

I agree 046 T is the CTS.



As I mentioned your scenario is completely unrealistic; the only that saved it was the last change in current, which itself is not realistic as there would have been a turn resulting in 0 current. Had that not occurred the route is impossible.

CTS = 046 T
SMG = 21/4 = 5.25 knot (please show me how you calculated your SMG.

After one hour the closest you would be to the AB line is C1

After 2 hours the closest you would be is D2

After three hours the closest you would be is E2

In reality you would be farther away from the CMG line.

John is correct. You are getting strong boost from the current.

Your scenario that has no counterpart in reality. A 4 knot shift in current over one hour will only occur in a very narrow channel.

Would it be possible for some one with access to currents for the English Channel to provide some real numbers for current velocity and direction.

[Seaworthy Lass]

Originally Posted by nigel1

SWL, in your last example showing the RYA (traditional) method, you are using 4 hours worth of tidal vectors, and 4 hours worth of boat speed, which puts point D well beyond B. Anyone trying to work out the CTS would then re work the vectors, but use only the first 3 hours of tidal vectors, which by eyeball would provide a CTS with much more north in it. If that comes well short of B, then re work using a small portion of the final hours tide.

Ah, yes, seems logical doesn't it!

That is NOT what the RYA method does though. It plots the WHOLE number of hours to get to what they call the "course line" implying that is where you want to be.

Originally Posted by nigel1

You cannot really compare your method to the RYA method, when you use 3.2 hours of vectors for one, and 4 for the other.

Precisely

My method is not like the RYA method at all, it is far superior as it DOES allow me to work with fractions of the final hour of current and it does aim to get me to B, not some point on the line through A and B (either side of it in fact) that can only be reached in a WHOLE number of hours.

[nigel1]

Originally Posted by Seaworthy Lass

My method is not like the RYA method at all, it is far superior as it DOES allow me to work with fractions of the final hour of current and it does aim to get me to B, not some point on the line through A and B (either side of it in fact) that can only be reached in a WHOLE number of hours.

When I was studying coastal nav. we were taught that if your point D was a significant distance from your destination, and if tidal variations were significant, then to re work the problem using additional vectors to get closer to your destination.
I dont really know what the RYA teach as I have never done any of there courses, but the solution to working out CTS is based on the same methods used by the RYA.
Navigation is not just about knowing calculations etc, its also about using a bit of common sense

[Seaworthy Lass]

Originally Posted by jackdale

I agree 046 T is the CTS.

Thank you for confirming I had computed the CTS correctly for the RYA method.

Originally Posted by jackdale

As I mentioned your scenario is completely unrealistic; the only that saved it was the last change in current, which itself is not realistic as there would have been a turn resulting in 0 current. Had that not occurred the route is impossible.

And why can't current reverse at the end of a journey .
And remember you are shifting through 11 nm, so the current may not necessarily be zero anywhere along your path .

I can in fact show it reversing for longer and produce a similar result. As a matter of interest I will draw this up during the next few days .

Originally Posted by jackdale

CTS = 046 T
SMG = 21/4 = 5.25 knot (please show me how you calculated your SMG.

The SMG using my method is the distance between A and B divided by the total time taken to get from A to B.
SMG = 11 / 3.2 = 3.4 knots, not an unfeasible journey at all for a boat only doing 4 knots.

Originally Posted by jackdale

After one hour the closest you would be to the AB line is C1
After 2 hours the closest you would be is D2
After three hours the closest you would be is E2
In reality you would be farther away from the CMG line.

Sadly if you followed the RYA CTS of 46 degrees, yes this is where you would be.

If you followed the correct CTS of 9 degrees (which my method does give) you would always be less than 3 nm of the rhumb line. I will post my course over ground shortly.

This is not some unfeasible journey!!

Originally Posted by jackdale

Your scenario that has no counterpart in reality. A 4 knot shift in current over one hour will only occur in a very narrow channel.

Would it be possible for some one with access to currents for the English Channel to provide some real numbers for current velocity and direction.

So the RYA method is only relevant to currents in the English channel?
Do you teach students that?

I can produce thousands of examples that show the RYA method arriving at an incorrect CTS. I could keep going endlessly. In fact if I don't have to write explanations I am now quite quick at it.