Inaccurate RYA Teaching : CTS - Quest For a New Method

[Dockhead]

Originally Posted by Seaworthy Lass

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 .

The issue is only that this is not what the RYA say -- you've not understood them.

In the sense that your method DOES work and IS brilliant, then of course it's NOT important how you get it. Even a monkey can use a protractor and doesn't need to understant the mathematical principles being used. But a lot of wrong ideas have been expressed about the nature of the RYA method, which will be an obstacle to the NEXT brilliant idea!

[Paul Elliott]

By the way, remember my comment about apples and oranges? No doubt there's a lot of this going on here.

[Seaworthy Lass]

Folks, could you please concentrate on LJH's post # 326
He applied my method in CPN.

Look at the CPN results:
010 degrees, 12.8 nm and 3 hrs 12 mins

The results using the SWL method were:
009 degrees, 12.8 nm and 3 hrs 12 mins (3.2 hours) SEE POST #229

The RYA method gave 46 degrees when it should have been 9.
This is a whopping 36 degrees off. SEE POST #215

The comment from the senior RYA instructor was:
"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."
The trip is possible and may have been needed to be done in an emergency. Having a correct CTS is absolutely vital if the passage has to be made.

Why continue using the RYA method that can give such wild results when there is a better, not much more onerous method?

My graph from post # 229 is shown again in the next post. It is near identical to the open CPN plot .

Originally Posted by LJH

Good Morning, SWL asked me to show her method used on CPN to find the CTS. (note the A-B course line is not drawn, measured or used)

Steps:
1. Plot A and B (triangles on attachments)
2. Use the measure feature to plot the tide set for each period (magenta on attachments) (A-C1, C1-C2, C2-C3, C3-C4)
3. Use the measure feature from point C4 through B to 16.0 miles and drop a mark there (this is L)
4. Use the measure feature to measure 12 miles from C3 towards B and drop a mark there (this is S)
5. Use the measure feature to measure from L to B to S. This will give you the ration SB/SL = 1.1/6.1)
6. Apply that ratio to the last current to measure .36 miles from point C3 toward C4. Drop a mark there. This is K, the blue diamond on the attachments.
7. Create a route from K to B and view route properties. Enter your boat speed and it will show 010 degrees, 12.8 nm and 3 hrs 12 mins.
8. To draw ground track use the measure function to measure from the tide points parallel to the K-B course (010) for the distance travelled (e.g. From C1 4nm, from C2 8nm, etc). These are the red “X”s.
9. Make a route from A, to B by joining the red Xs in between and you get the ground track (red line).
10. If you wish to you can transfer this route to your chart plotter and monitor you progress hourly relative to the red track, or just monitor you XTE.

[Seaworthy Lass]

Reposting my chart from post # 229

It is essentially identical to the CPN plot

[daddle]

Have been entertained from the sidelines of this thread. Good stuff. Wild turns thru vectors and triangles. Everybody learned something.

Fortunately for sailors this type of navigation is pretty useless. You know what is going to happen halfway to B, right? The wind will shift and you will go off on some bad tack angle and not reach B until sunrise.

[Seaworthy Lass]

Originally Posted by daddle

Have been entertained from the sidelines of this thread. Good stuff. Wild turns thru vectors and triangles. Everybody learned something.

Fortunately for sailors this type of navigation is pretty useless. You know what is going to happen halfway to B, right? The wind will shift and you will go off on some bad tack angle and not reach B until sunrise.

Welcome to the thread Daddle
In this case you are motoring in calm water with at a constant throttle position.
You need to get from A to B and it CAN be done.
My method gives you the answer to within a degree.
The RYA result was 36 degrees off - dangerously off IMHO .

[Paul Elliott]

Originally Posted by Paul Elliott

Sorry, but between two points the proper constant heading will result in the least distance sailed, and thus the fastest passage (ignoring the wind, etc -- we're motoring in these examples). This applies regardless of current or current variations. The vector diagrams make this pretty obvious.

I believe that SWL has correctly claimed that her constant-heading gives the fastest passage.

Sorry for the detour, but this has been bugging me. My answer was correct for the "uniform current field" we have been assuming for the sake of discussion, but for a real-world non-uniform field, with eddies to be used or avoided, etc, then a non-constant heading may very well be faster. Calculating that is close to impossible, and brute-force iterative, monte-carlo, and other methods must be employed. Intuitive local knowledge usually trumps all that anyway, because of the uncertainties in the available predicted data.

[TeddyDiver]

What all of the recommended methods lack, for the route planning purposes, are the considerations of best time to sail and the optimal speed
and in the case of sailboat the best wind direction for the intended trip. Since everybody's throwing in their suggestions here's one more

and sorry about the sloppy drawing, but the principle is to draw the tide speeds on a separate ruler and place it parallel to a line indicating the tide directions (in the case of changing directions just make a broader ruler where the tidal timeline is drawn. Think about throwing a MOB pole in the sea and drawing the drift..)
Slide the tideline so the planned sailing time is at the A then choose the point (the time you plan to arrive to the destination) and draw a line to B. That's your CTS.
Taking into account the wind direction for sailing is pretty forward too as you can see (hopefully) from my drawing.
BR Teddy

[Seaworthy Lass]

Example 4

In the last example I was criticised by the RYA instructor for selecting an unrealistic situation. He said the current was too high (8,6,2 then 2 knots) and this would never be encountered and the only reason the boat could get there was because the tide turned at the end.

So in this example the current is as follows:
Ist hour: 3.5 knots 135 T
2nd hour: 1.5 knots 135 T
3rd hour: 0.5 knots 135 T

So nothing weird about this is there?

B is due east of A, 4.24 nm away
Boat speed = 1.8 knots (see explanation below as to why it is low)

You have a short journey due east of 4.24 nm to get to B, which is a spot you consider safe to be so before you proceed into the harbour.
(I will tell you this distance is actually 4.24 nm so that you can mark it on the chart for the purposes of this calculation, but for the SWL method it is just a spot you want to go to on the chart, the exact distance is irrelevant for my method unless you want to calculate your SMG for some reason).

This is an emergency situation and you must get there ASAP. Conditions are calm, there is no wind and the water is flat.
Unfortunately your prop is damaged and your maximum boat speed is 1.8 knots (your know this from the last trip).

Can you get to your destination to deal with this emergency?
What is the CTS?
What is the time taken?

I will post the RYA results and mine later today, but in the meantime I leave you to think about this and work it out if you can

[Dockhead]

Originally Posted by Paul Elliott

Sorry for the detour, but this has been bugging me. My answer was correct for the "uniform current field" we have been assuming for the sake of discussion, but for a real-world non-uniform field, with eddies to be used or avoided, etc, then a non-constant heading may very well be faster. Calculating that is close to impossible, and brute-force iterative, monte-carlo, and other methods must be employed. Intuitive local knowledge usually trumps all that anyway, because of the uncertainties in the available predicted data.

Yes, this is an important limitation to the entire enterprise here. It's a very good point.

We assume a "uniform current field" -- a very apt phrase for it.

All bets are off if you can take advantage of a favorable eddy or other local effect, or need to avoid an unfavorable local effect.

We are also not considering (Teddy Diver's point in another point) anything concerning wind. We have not been correcting for leeway.

In practical real life, the problem becomes quite complicated when you can't sail on your ideal constant heading, and you have to tack.

English sailors are light years ahead of us on this (since they deal with strong currents every day ); no one there is unaware of the power of constant heading passages. They debate a much more complicated issue -- lee bowing.

The question is whether you get a lift by tacking so that the tide is always on your lee bow.

I confess that I have no idea whether it works or not. There is a ferocious debate which has gone on for decades.

The ground track is dramatically shortened if you use the lee bow technique, which leads to some claims that this gives you a left -- this is obviously bogus to anyone who knows even a little about constant heading steering. But others say that is not the source of the lift.

Another theory is that it increases true wind (the wind will be reduced if you are being swept away from the direction of the wind, by the tide), and you get a lift that way. But we rarely have a shortage of wind in the English Channel; that "lift" would more than likely just mean you have to put a reef in.

So I can't say. It bothers me a lot that I don't know the answer.

[Lodesman]

Originally Posted by Seaworthy Lass

Lodesman, could you please draw a few step by step diagram using the data in example 3, not example 1 to show us how this works? This would be very helpful.

If I get a chance. It may be awhile. It's not an impossible journey, but it is an improbable one. You realize the RYA method assumes you are actually capable of following the course line, and produces an average of CTS to maintain the courseline (yes I realize that you do not actually follow the course line - it's an abstract thought). In this improbable problem, you are incapable of maintaining the course line - the best effort is minimizing the loss and leaving yourself closest to destination when you are able to overcome the set.

And why mark the distance off the rhumb line, why not directly toward B and see how much it overshoots? Then you could try half that amount, then half again etc, until you get close.

Erm, because this is what you do with simple tidal triangles. The more precise method is to determine SMG in order to determine time required to complete the passage - it's basic nav, not rocket-science. Does your system work for a single set solution? I think not. I agree your method works, but it is completely different from the method used for most CTS problems, and is really only useful for a small fraction of the situations any sailor is likely to find himself/herself in. If said sailor does frequent channel crossings, then they might find a use for your method; or they might already be comfortable and happy with the RYA method.

Either way, that would take AGES!!!!!

Not any longer than your method. If time is of essence, then I could have a rough calc and steering it, before you've ratio'ed your SB:LB - I've then got hours of passage left to refine it (though it would take only a minute or two).

How many times do you have to keep stabbing guesses before you arrive at the correct answer of the amount of time you a subjected to the last lot of current?
No need to stab any guesses with my method .

I'm not sure how many times I have to say this - your method seems to require guesswork (eyeballing) - mine uses calculations. In many cases it would be one calculation, but for others it requires refining by recalculating once (maybe twice in really extreme conditions)

I just just do it twice, once for the beginning and once for the end of the last current vector, then I look at the proportion of the undershoot compared to the overshoot. If they are equal for example, then you use half the hour. Simple as that, no trial and error until you can eventually get close as you are suggesting.

I don't use trial and error. I use a rough and possibly a refined calculation, just as you might use in certain astronav problems, or certain tidal calculations. Your method is less precise than mine - I rounded all the figures in my example to whole degrees and cables - inasmuch as that would be all the accuracy that any sailor at a chart-table is likely to be capable of or wanting to achieve.

I haven't seen your answer to what you do when (not 'if') your fix puts you off of your plotted CMG?

[sailr69]

this post is making my head swim. makes me wonder how I got anywhere on the boat.

[Paul Elliott]

For what it's worth, for SWL Example 4 I get a course to steer of 005.46 degrees, and a transit time of 1.9985 hours. The distance traveled (through the water) is 3.5974 miles. This time was close enough to two hours so I didn't bother adjusting for any current over/under.

I did this on my cellphone calculator, not by graphing. Had the numbers come out other than virtually a whole number of hours it would have required interpolation.

Of course I could easily have messed up these calculations. One advantage of a graphical method is that you get an obvious sanity-check.

[Seaworthy Lass]

Originally Posted by Paul Elliott

For what it's worth, for SWL Example 4 I get a course to steer of 005.46 degrees, and a transit time of 1.9985 hours. The distance traveled (through the water) is 3.5974 miles. This time was close enough to two hours so I didn't bother adjusting for any current over/under.

I did this on my cellphone calculator, not by graphing. Had the numbers come out other than virtually a whole number of hours it would have required interpolation.

Of course I could easily have messed up these calculations. One advantage of a graphical method is that you get an obvious sanity-check.

Close

[Dockhead]

Originally Posted by LJH

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.

You are fundamentally not understanding the method or even the reason why Seaworthy is plotting A to K.

The purpose of her exercise is to get an angle -- the course to steer. She's working it with a protractor because it's easier and quicker than math, but the mathematical principle is the same. In order to solve the problem, she is building up a triangle with vectors equal in time.

Strictly speaking, trig needs lines, but it's ok to talk about points if it is easier to understand. If you have three points, you have enough information to do trig and get a CTS. But you need three if you don't know any of the angles; two is not enough.

Two of the three points are A and B. The third point is K. Any two points are a line, in math, which is why it doesn't really matter whether you talk about points or lines.

This is the principle which she, and everyone else, is using to derive that angle which is CTS:

"The law of cosines relates the lengths of the sides of a plane triangle to the cosine of one of its angles. Using notation as in Fig. 1, the law of cosines says
where γ denotes the angle contained between sides of lengths a and b and opposite the side of length c. . . . The law of cosines is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known. "



Law of cosines - Wikipedia, the free encyclopedia

The word "trigonometry" means "triangle-measuring" in Greek, and in fact that is literally what Seaworthy is doing when she builds up her triangle, and then literally measures the angle between two of the sides with her protractor. In any case, whenever you do trig, you are putting a problem into the form of a triangle so you can figure out an angle or the length of a line, or both. There are three lines and three angles in a triangle -- six things, altogether; the Law of Cosines allows you to know any three unknown things if you know the other three things.


When you sum the vectors of the tides, you are finding the position of the end point -- K in this case -- of the last unknown line. The starting point of that line is A, so you must know A to do that one. Once you know the positions of A, B and K then you can get the angle you need by plotting it and measuring the angle with a protractor. It doesn't make any difference whether you think of the points, or of the lines between them; mathematically it's all the same. In any case, you're doing trig -- three angles, and three lines (or if you like, three points), and the relations between them.


We are not getting anywhere near Great Circle paths. No one has mentioned this at all. For the sake of simplicity we are doing the problems in two dimensions (which is plenty close enough at these distances). People are having enough trouble with regular trig; I'm not even going to think about spherical trigonometry -- let's not even go there (it would be useless anyway, since I am clueless about spherical trig ). But joking aside, spherical trig IS used in navigation over large distances.