Inaccurate RYA Teaching : CTS - Quest For a New Method

[Lodesman]

Originally Posted by Seaworthy Lass

Easy as that! No extrapolations (that make invalid assumptions and can gives results that have significant errors), and I think the method is based on solid principles, unlike any method utilising the rhumb line.

The method I proposed uses what you call the rhumb line, doesn't require interpolation/extrapolation, and easily portions the final vector without guesstimation - yet you haven't commented on it

[LJH]

Originally Posted by Seaworthy Lass

Yes, it does seem to preached that it is a good approximation (or by some instructors even that it is a precise way of getting you to B). It can have significant errors. I will make my next example one were it is about 30 degrees out .

Will have to draw step by step diagrams at some stage to try and make my method more comprehensible.

If anyone is capable of doing the RYA method, they would be capable of doing mine I think. I probably just haven't explained it well .

The RYA method will get you nicely to D, which may be a significant distance from B. Your method would probably be easier to illustrate if you end with a significant crossing current so that the vector through S is well separated from the vector through L.

[LJH]

Originally Posted by Wotname

OK. don't want to derail SWL thread so will keep this brief and then get back to regular programming

Originally Posted by Dockhead
I bet it's because unlike tidal currents, winds are not very predictable. So probably pilots do a single vector with solutions for COG and SOG and work out from that whether or not they will arrive or crash, out of fuel

Forecast winds should be good but aren't always. They predicted hour to hour at various levels; you use the most recent forecast and update as you go.

If they are on a multi-hour flight and know more or less that they will have an hour of these winds, then an hour of those winds, it might still be very profitable for them to calculate a single CTS. I would be surprised if this art does not exist in the body of air navigation disciplines.

Yes, almost always you calculate a single CTS. Not up to date but I guess it is now done with small hand held programmed computer / calculator. I learnt on and used the E-B6 whizz wheel, no doubt it would handle the the problem under discussion. . If I was keen (and I'm not), I would break it out, relearn it and adapt it to SWL examples. Here is a good description of the device E6B - Wikipedia, the free encyclopedia

Yes I accept the importance in your backyard and agree fully. I also accept that is a mark of any good seaman to know this stuff even if they don't use it; however I expect most recreational sailors don't care unless they need to use it in their backyard.

Full marks to you and the other posters for strongly debating the issue.

I just happen to have my old E6B in my nav table. I might just be keen enough to blow the dust off it. I think most EFIS systems will do this, but Air Traffic Separation Schemes and traffic density will cut down on the ability to actually fly the optimum CTS.

[Dockhead]

Originally Posted by Andrew Troup

Dockhead:

My comments are threaded through your recent post to me:

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

Thank you, and ! Whew ! , in that order.


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.

I do plan to address that, but I wanted to make sure we were on the same page on the how and why of getting to that point.

It will work perfectly if the average tide during the uncalculated partial hour equals average tide of the calculated part of the passage.

That's what I acknowledged in the post you replied to: under that singular circumstance, I can well imagine it is correct.

I think you sell 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.

See above

So you can describe it either as a shifted destination, or a fudge, with equal accuracy, I guess.


I agree: to me, conceptually, it's a shifted destination; functionally, it also works out to be a useful fudge, in the typical case.

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.

Huh? and which Andrew moight that be, to be sure, so it is? <wink>
And when you've cleared that up, which mistake?

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.

<grin> how much do ya wanna bet?

If you're not going to finish the calculations, as the RYA would have us do,

And I've underlined your word 'not' - I'm hoping you'll tell me it was a frayed not. I mean, a superfluous not. I mean, not a not.

... you have to have the course line to calculate some vector triangle.

OK, here you need to throw me a rope. "Course line" ? Please clarify what this means, and why I have to have it.

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.

Extending your answer to my previous question should clear this up too (pretty please?)

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.

I'll leave Dave to you guys, he's above my pay scale.

I think you have brilliantly disproven this showing a nearly 10% error of the RYA method in a real case.

Are you getting me mixed up again, this time with the Worthy SeaLass?

- - - - -

Now I need to remind you that, once you've answered my questions and we've tidied up the matters arising, I will still want you to address my points a) and b) from an earlier post, insofar as they still seem relevant.

OK?

Sorry, I was unwittingly aggregating answers to you and answers to a SWL post -- a bit of a mess. I edited my previous post a bit.

So your points a) and b), if I can correctly identify them:

I guess that would be these?

"Please address these objections: if you understand them, please refute them; if not, please seek clarification from me.

" a) .... the RYA method ... use(s) the rhumb line as a geometric construction aid.

This isn't a big problem in the usual case, but in unusual cases, their construction technique, to interpolate* the final hour of tide, is simply not possible, using the rhumb line. In such cases, it needs to be swung through 90 degrees.

b) They .... shift ... the destination to some arbitrary place on the rhumb line.

....This ... is not a big problem in the usual case, but like most dumbed-down rules, it misleads the user by diverting them away from an understanding based on first principles. "


OK, here's my comment -- I think you don't quite get the method (which as I said I am sympathetic to -- I also couldn't get it for a d*mned long time of breaking my head over it).

The "rhumb line" (let's better call it the "course line", as the RYA do, because it has a different function) cannot be "swung", or you lose the vector triangle. It is an essential element of the calculation, even in SWL's case (she just doesn't need to draw it).

The course line is one leg of your vector triangle -- don't forget. With a better method like SWL's, you don't much use it for anything, and don't need to draw it on the chart, because the lines converge on the destination. But the RYA needs this line, and in fact, you are instructed to extend it out beyond the destination, even onto land. That is so the vector triangle can be formed out of whole hour calculations where the water track line will not intersect with the destination.

So the destination is not "shifted to some arbitrary place on the rhumb [course] line". You know the length of your water track line -- it's a function of time as a whole number of hours and your assumed average boat speed. You swing THAT until the end of it touches the course line. That spot of intersection is not arbitrary -- that's your whole hours vector triangle. Got it now? The essential output of the RYA method is the angle of the water track line after you've swung it from the end of all your tidal vectors to the point where it touches the course line. That's your CTS (subject to further adjustment for leeway and variation, of course).


So with the RYA method you end up with a perfectly (mathematically) constructed vector triangle, only it is not for your exact passage, since it has only counted whole hours (and it might count through the whole hour BEYOND your destination, if that is closer). It does not interpolate the uncalculated last partial hour, it merely "inflates" or "deflates" the whole hour vector triangle until it corresponds to your destination. So the uncalculated last partial hour is basically whitewashed over -- as if it corresponds to the average of the calculated parts of the passage. Since you will round up or down to the nearest whole hour, the maximum size of the uncalculated period is 30 minutes. If the uncalculated period is only 15 minutes or less, then the result will be pretty good in any case.

Do you get it now? It's actually very clever. It is designed to be "close enough for government work"; however Seaworthy has forcefully proven that it is not, actually, close enough.


Perhaps another way of looking at it is this: Maybe users of the RYA method, instead of "inflating" the vector triangle, if it is smaller than the real one, just sail to that point on the rhumb line and then "track mode" steer from there. It will be less than an hour of sailing and with less than an hour to go, no one is steering the constant heading anyway -- you're already homing in on your destination by other methods. If it is used in real practice like this, then I guess it's somewhat usable.

However, I still don't like to be at the mercy of random outcome of the uncalculated partial hour, the size of which can be anything from 1 minute to 30, and variation from the average tide can be anything. I prefer to have an exact CTS which I calculate not to the destination, but a mile or so uptide, to center the possible outcomes there with a mile or so on either side as perfectly acceptable outcomes, rather than right at the destination where a few cables downtide is already a problem, even while a couple of miles uptide is fine.

The destination is the edge of the circle of acceptable outcomes, so it seems to me to be a fundamentally wrong aiming point. You need to rather be aiming at the center of that circle of acceptable outcomes, and that will necessarily be someplace uptide of the destination. So looking at it again from the point of view of practical, usable, practice, I am still convinced that an exact CTS calculation is very useful, and I don't like at all the wide range of random outcomes from the RYA method.

[LJH]

Originally Posted by Lodesman

The method I proposed uses what you call the rhumb line, doesn't require interpolation/extrapolation, and easily portions the final vector without guesstimation - yet you haven't commented on it

As you said in your post (#710) of Re: Distinct Activities: Shackled by a Common Name? your method is optimum from a big ship perspective. If I am trying to track a rumbline in a situation where the tidal flow and leeway are not in my favour then my VMG along the rumb line almost comes to a stand still. If I use the CTS method then I will be up current and will maintain a significant VMG. Much more efficient overall. If I am a 5-6 knot speed range then I want to have as much as possible working in my favour.

[jackdale]

SWL - the current vectoring is used to solve for both CTS and Speed Made Good. How are you solving SMG?

I have been going through your solutions and the RYA solutions, as well as a solution based on the mean current.

[LJH]

Originally Posted by jackdale

SWL - the current vectoring is used to solve for both CTS and Speed Made Good. How are you solving SMG?

I have been going through your solutions and the RYA solutions, as well as a solution based on the mean current.

The VMG along the rhumb line will be he cosine of the angle between the rhumb line and the CTS times your boat speed.

[jackdale]

Originally Posted by LJH

The VMG along the rhumb line will be he cosine of the angle between the rhumb line and the CTS times your boat speed.

The standard vectoring method solves for SMG without using a calculator or log tables.

[Seaworthy Lass]

Originally Posted by jackdale

The SWL method is faulty.

The rhumb line is crucial.

Example

Travelling from Cautionary buoy to Entrance Island
Current 000 T
Current 1st hour 2 knots
Current 2nd hour 1 knot
Current 3rd hour 0 knot

Givens
Distance = 14.69 miles
Boat speed = 5 knots
CMG = 239 T

To calculate
SMG = 4.4 knots
CTS = 229 T

Chart work

Hi Jackdale
I have finally had a chance to compute the example you set using my method.

Results using my method:
CTS = 230 degrees
Time taken = length KB divided by 5 = 16.4 / 5 = 3.3 hours
SMG = course length divided by time taken = 14.69 / 3.3 = 4.45 knots (rounded off to 4.5 knots)

I have marked the ground track as a dotted red line.
I have heavily dotted the points to make them visible in the photograph, but all work was done with a super sharp pencil.

Examples with zero current for the last hour of travel are truly easy with my method. You still need to mark the distance vectors from the beginning and end of what you think is the last hour of travel to check they fall either side of B so you know you have the correct number of current segments marked.
There is, however, no need to work out what proportion to use, as the last current vector is zero.
So you very simply mark K at the end of the current displacement vectors and join that to B.

The angle of KB is the CTS = 230 degrees
KB = distance travelled in water = 16.4 nm
Time taken = 16.4 / 3 = 3.3 hours
SMG = length course divided by time taken = 4.45 knots (4.5 rounded off )

My results are almost the same to yours. That is because:
1. D falls close to B. If D coincides with B the RYA method has you heading to where you want to go, so there is less scope for error.
2. We are only dealing with a small amount of current at right angles to the course line (the actual course is 239 so if there was absolutely no current the CTS would only be 9 degrees different, again not much scope for error)

For examples like this the RYA method will give close to the correct result.
The RYA method was only one degree off in this case.
230 degrees is the correct course to steer

[Seaworthy Lass]

Originally Posted by jackdale

SWL - the current vectoring is used to solve for both CTS and Speed Made Good. How are you solving SMG?

I have been going through your solutions and the RYA solutions, as well as a solution based on the mean current.

SMG = course length divided by time taken
Time taken = length KB divided by boat speed

It is easier to compute than with the RYA method as there are no extrapolations to plot on the chart as there is with the RYA method.

Glad my method is under more serious consideration now.

[Seaworthy Lass]

Originally Posted by jackdale

The standard vectoring method solves for SMG without using a calculator or log tables.

Abacus used?

[jackdale]

In these examples I have used your data from post 34

Boat speed is constant at 4 knots throughout the journey.
Destination is 8.5 nm due east.
Current is always from the north:
1st hour 3 knots
2nd hour: 2 knots
3rd hour: 0.5 knots

In the first one I used the mean current 1.8 knots (5.5 /3)

And got



The distance measured from A to D is 3.5 miles which equates to a SMG of 3.5 knots.

In the second I used the cumulative current 5.5 knots



The distance measure from A to D is 10.75 miles which equates to a speed of 10.75/3=3.5 knots

In both cases the CTS is 62T

I did not use a calculator to measure do the math. I can still do long division. I also did not need a cosine.

The key to the RYA method is that it provides an SMG is just as important as a CTS.

In addition if you need to dead reckon you can do so be drawing a DR line of 62 T from your point of departure and use the current information available to establish an estimated position.

I am busy tonight. But tomorrow I am going to re-do my first example with the RYA method but with a different current direction for the second hour.

[Andrew Troup]

Dockhead

Thanks for your thoughtful and detailed post
I'll change to "Course line" in discussing the RYA method, if you need me to.

I think I'll probably stick to rhumb line for all other methods-- unless someone can give me a good reason to use a different term.
Rhumb line is an exactly correct description of a straight line on the chart between two points.

Because the ocean is not flat, it is not actually the shortest distance across the water, but this only matters, for short passages, if you're shooting for a Nobel prize as a pedant.

(Forgive the obligatory ant tie-in)

Your statement:

"The .... "course line" .... cannot be "swung", or you lose the vector triangle. It is an essential element of the calculation, even in SWL's case (she just doesn't need to draw it)."

leaves me gasping, and scratching my head.

Vector solution, particularly addition and subtraction, can be done in many ways.

Triangles are a tool, an optional tool, of vector maths.

Summation of (orthogonal) coordinates is another.
(eg 'Northing' & 'Easting')


Vector maths, in turn, is an optional tool for vector solution, particularly addition and subtraction.

Graphical solution is another. No triangles are involved.


A triangle is just one way of representing a vector, much as a painting, a sketch, and a verbal description are ways of representing a lamp-post. They are not a lamp-post. The lamp-post does not rely on them.

In saying that a graphical solution collapses if an admittedly unused leg of an unused triangle is unavailable, you're promoting a low-level, optional tool to the status of high-level imperative.

Aren't you?

- - - -

If you consider that Seaworthy's method relies on the rhumb line to deliver a CTS and distance through water to the true destination, you need to remember that this is harder to prove than disprove. (A bit like the proposition that 'all rabbits are brown').

The best you can do by way of proof, it seems to me, is to try to disprove it, and discover that you cannot.

I, on the other hand, think I can. Fairly simply.

[Wotname]

Originally Posted by LJH

I just happen to have my old E6B in my nav table. I might just be keen enough to blow the dust off it. I think most EFIS systems will do this, but Air Traffic Separation Schemes and traffic density will cut down on the ability to actually fly the optimum CTS.

Here is a link to act as primer in case (like me) you might need it .
http://www.stefanv.com/aviation/flight_computers.html

SWL, you might like to have a quick look at this also as in my mind, the wind triangle is essentially the same as a current triangle.

[DeepFrz]

Yes, I agree that the wind triangle is essentially the same as a current triangle.

I am working on a real world example of why the rhumb line is important.

As a pilot I was taught to create a wind/course triangle and CTS. Then when in the air to use features along the rhumb line to correct the CTS to bring me back to the rhumb line. It was always stressed that it was very important to stay on/close to the rhumb line as this was the track that had been studied for hazards. Of course we checked for hazards that were within a certain distance of the rhumb line, I believe it was on an angle from A of 15 deg. on both sides of the rhumb line.

Staying on the rhumb line was also stressed in my boating navigation courses and the importance of always knowing where your were, as near as could be calculated by using different methods, was always stressed. If a sailor gets used to drifting off the rhumb line on a constant CTS, unless he plots his track points (as has SWL) he will not know how far off a hazard he may be. She will also not be sure of her position should she have to start a DR plot in the case of sudden fog. I think it is very important, if not for speed reasons, but for safety reasons to stay as close as possible to the rhumb line while sailing in any unfamiliar water. Maybe this sense of importance comes from learning to sail in Jacks waters, which can be extremely hazardous, as can so many other areas of the world.

As well as learning to sail on the Wet Coast of Canada I used to do a lot of backwoods Canoeing. Knowing where you were at all times was extremely important in the wilderness.