Inaccurate RYA Teaching : CTS - Quest For a New Method

[jackdale]

Originally Posted by jackdale

Happy

here is why mine is 5.

http://www3.telus.net/jackdale/navle...S/detercts.pps

For those with no PowerPoint

http://www3.telus.net/jackdale/navle...g%20Course.pdf

The sixth step is to establish a DR which is used for calculating EPs.

The current calculation is essentially the same for a single or multiple current, it is the expected current over the voyage.

The key in vectoring is that the length of the line from the end of the current line is the boat speed (either over an hour or the length of the voyage).

[Lodesman]

Originally Posted by Seaworthy Lass

My method involving determining the proportion of the last bit of current applicable by looking at SB/SB+BL is more accurate than determining D1B/D1B+BD2 as the latter will vary depending in the angle of the last bit of current relative to the angle of the rhumb line. My proportion will not change as it does not refer to the rhumb line at all.

I started to respond to this, then realized a picture would be worth a thousand words - hence the delay in responding. To just compare the two methods I used the following scenario - destination (B) is 7nm due north; boat speed is 5kts; tidal set is 2kts to the east for the first hour and 3kts to the north for the second hour. On the left I show the 1-hour triangle (AC¹D¹) - and I broke down the second hour's tidal vector (C¹C²) into five segments of 12 minutes each. From these marks I arced the respective boat speed vectors (5nm at 1 hour, 6nm at 1h12m, 7nm at 1h24m, and so forth). You can see the respective arcs along the "rhumbline" evenly marking out the line from D¹ to D² - this portion of the line is very clearly seen to be one whole hour long, and it is very clear to see that B lies at 1h18m or very close to this; and the resultant line from the 1h18m mark of the tidal vector gives the answer. This is established vector mathematics.

To the right, in your method, we have the first hour boat speed vector marked at S on the line C¹B - if you complete the vector triangle (AC¹S; not drawn) you see the Made Good course takes you to S not the destination. You also see the section of line SB is 2.3nm long, representing at boat speed 5kts, another 28 minutes. The second mark L established by marking a 10nm boatsp vector from C² through B, establishes that the segment BL is 5.5nm long, representing 66 minutes at boat speed. The vector triangle (AC²L) has a Made Good vector that also doesn't go to destination. SB + BL does not equate to an hour, but if you compare the lengths of SB (2.3nm) to SB+BL (7.8nm) and multiply by 60 you do get the right answer of roughly 18 minutes.
It works, but I have no idea why. Perhaps one of the resident mathematicians could explain it. If it can't be proven mathematically, then how do you know it always works and is not just a coincidence? And how do you know how accurate it is? I noted that the two methods did not give exactly the same answer - they were some seconds apart - so one appears less accurate than the other, but I don't know which one.

Originally Posted by Seaworthy Lass

Yes, the only purpose of the technique is for calculating a CTS with variable cross current. It gives a dependable method for doing this.

For the average YM who might go for years without using a multi-tide CTS, trying to pull a one-trick-pony solution from the dark recesses of their memories is bound to go all wrong. While the RYA solution may not be as accurate, it is based on the standard tidal triangle solution that competent YMs would use frequently - the only thing they need to remember from their training is that you can add multiple tidal vectors together, then solve as for the standard solution.

[Seaworthy Lass]

Originally Posted by jackdale

Given a 5 step model to use and teach versus an 11 step model, I will go with the 5 step model. That is a significant advantage.

Jackdale, the number of steps you give does not include checking to see that the number of hours of current you have selected is the correct number to get you closest to B. You need to check that arcing off the next lot of current takes you past B. (By the way, I would strongly discourage picking D if it lies past B though, as the ground track may skirt B an unacceptable amount in some cases).

If you do the above last essential step (and I think we both agree this is essential), then the SWL method has only one step more than the RYA method (it has one step less at the beginning as there is no need to draw a line between A and B if you have variable cross current, and two steps more at the end). But, I do agree it is more onerous.

There is no doubt the SWL technique is more accurate (or at least equally accurate), but I do agree that there is a lot to be said for simplicity. There are several modifications that could to be made to the "rough and ready" RYA method though to improve it .

[Seaworthy Lass]

Originally Posted by Lodesman

I started to respond to this, then realized a picture would be worth a thousand words - hence the delay in responding. To just compare the two methods I used the following scenario - destination (B) is 7nm due north; boat speed is 5kts; tidal set is 2kts to the east for the first hour and 3kts to the north for the second hour. On the left I show the 1-hour triangle (AC¹D¹) - and I broke down the second hour's tidal vector (C¹C²) into five segments of 12 minutes each. From these marks I arced the respective boat speed vectors (5nm at 1 hour, 6nm at 1h12m, 7nm at 1h24m, and so forth). You can see the respective arcs along the "rhumbline" evenly marking out the line from D¹ to D² - this portion of the line is very clearly seen to be one whole hour long, and it is very clear to see that B lies at 1h18m or very close to this; and the resultant line from the 1h18m mark of the tidal vector gives the answer. This is established vector mathematics.

This method works only if the final bit of current is parallel to the rhumb line (as you have drawn). Try to repeat this for various other angles of current and you will find the proportion will vary with the angle! A proportion that varies like this is inherently inaccurate.

In the SWL method, the proportion does not depend on the angle of the final lot of current.

PS Edited to add:
Sorry, not a good explanation. Trying to think how to word it better.

[Seaworthy Lass]

Originally Posted by Lodesman

For the average YM who might go for years without using a multi-tide CTS, trying to pull a one-trick-pony solution from the dark recesses of their memories is bound to go all wrong. While the RYA solution may not be as accurate, it is based on the standard tidal triangle solution that competent YMs would use frequently - the only thing they need to remember from their training is that you can add multiple tidal vectors together, then solve as for the standard solution.

The aim of the standard tidal triangle solution is to keep you on the rhumb line. This is NOT appropriate for a situation with variable cross current so solving "as for the standard solution" is unfortunately inappropriate as well.

[Lodesman]

Originally Posted by Seaworthy Lass

This method works only if the final bit of current is parallel to the rhumb line (as you have drawn). Try to repeat this for various other angles of current and you will find the proportion will vary with the angle! A proportion that varies like this is inherently inaccurate.

In the SWL method, the proportion does not depend on the angle of the final lot of current.

OK, done. With most of the situations it makes no difference. I admit there's a small amount of distortion at the extreme example (on the far right), but it's so small as to be unimportant. No doubt you could find a ridiculously absurd imagination of extreme tidal change that would make this amount measurable, and I invite you to show me such a case.

Do you have mathematical proof of your last statement?

[Lodesman]

Originally Posted by Seaworthy Lass

The aim of the standard tidal triangle solution is to keep you on the rhumb line. This is NOT appropriate for a situation with variable cross current so solving "as for the standard solution" is unfortunately inappropriate as well.

This is a nonsense statement. The tidal triangle solves for the resultant - a composite tidal track begets a composite track made good.

[Seaworthy Lass]

Originally Posted by Lodesman

OK, done. With most of the situations it makes no difference. I admit there's a small amount of distortion at the extreme example (on the far right), but it's so small as to be unimportant. No doubt you could find a ridiculously absurd imagination of extreme tidal change that would make this amount measurable, and I invite you to show me such a case.

Do you have mathematical proof of your last statement?

Hi Lodesman
I fell asleep playing with figures so a delayed response here too .
Swivelling the last bit of current 90 degrees in your example (so it is due east like the first hour of current) produces 0.59 of an hour using the proportion of D1 vs D2, and 0.62 for the proportion of SB vs BS. Not much difference in CTS is produced, but the proportion of 0.62 gets you closer to B.

I had to laugh at your comment about me finding a "ridiculously absurd imagination of extreme tidal change". Do you consider your example of 2 knots of current due east followed by 3 knots due north perfectly normal LOL? Am I the only one not allowed to play with various data?

Anyway, the method you have now taking a liking to is simply the one I proposed in mid January in post #564 on:
http://www.cruisersforum.com/forums/...-94879-15.html
(I first used the term D1 and D2 in post# 623 of that thread).

Yes, I produces a reasonable result in most cases, but it does not work well at the extremes (and it is not possible to work out a result at all in the example in post #772 of this thread). That is why I came up with the latest method.

The amount of work is exactly the same for the two methods I proposed (in drawing the lines and working out proportions), so why choose the one that does not produce as good a result .

PS Edited to add: No mathematical proof. The correct proportion needs to be worked out with a differential equation. My final method is just an approximation as well, but a better approximation

[Seaworthy Lass]

Originally Posted by Lodesman

This is a nonsense statement. The tidal triangle solves for the resultant - a composite tidal track begets a composite track made good.

Which bit is nonsense? If you are following a CTS with variable cross current why should you be aiming for any point on the rhumb line? You are not following the rhumb line, you are just getting to B the quickest possible way.

You do not have to be on the rhumb line to solve a tidal triangle!

[Lodesman]

Originally Posted by Seaworthy Lass

Hi Lodesman
I fell asleep playing with figures so a delayed response here too .
Swivelling the last bit of current 90 degrees in your example (so it is due east like the first hour of current) produces 0.59 of an hour using the proportion of D1 vs D2, and 0.62 for the proportion of SB vs BS. Not much difference in CTS is produced, but the proportion of 0.62 gets you closer to B.

OK, B is at 1h36m along the "rhumbline"; taking 1h36m boatsp vector (8nm) from the end of 1h36m tidal vector (2nm + 1.8nm) closes the triangle at B - can't get any closer than bang on.

I had to laugh at your comment about me finding a "ridiculously absurd imagination of extreme tidal change". Do you consider your example of 2 knots of current due east followed by 3 knots due north perfectly normal LOL? Am I the only one not allowed to play with a various data?

No I don't think it's normal - I just pulled two fairly round figures out of the air that could be used to show what I was saying. I didn't claim it was realistic or not.

Anyway, the method you have now taking a liking to is simply the one I proposed in mid January in post #564 on:
http://www.cruisersforum.com/forums/...-94879-15.html
(I first used the term D1 and D2 in post# 623 of that thread).

I haven't taken a liking to it - I would tend to use my original method if I found it necessary, for the same reasons I mentioned at post #699 of the mentioned thread. I guess I didn't recognize the "original SWL method" as I came into that thread at the same time you had binned it, in favour of your present method.

PS Edited to add: No mathematical proof. The correct proportion needs to be worked out with a differential equation. My final method is just an approximation as well, but a better approximation

So you have no way of proving that it is the better approximation? How can you keep saying that it is?

[HappySeagull]

You do not have to be on the rhumb line to solve true, But it WILL be a side when you do solve, even if you don't draw the line.Points in a line are infinitely small but a seagull's eyes see them quite well.
As to drift vectors, however they dodge about hour to hour- they are a single side too, invisible, but still present, and resolve to a single line in the solution. (That might inspire another way to do this...)

so,...in effect, finding the ratio of the last drift required,with whatever method, and even where angles vary hour to hour,is about finding that invisible single drift vector in length and angle. The correct length is when ...

Time will be equal to the other two sides with Destination B at one corner.

because, Time is equal on all sides. That is the whole point of the effort.That is another way to look at the solution. Lodesman-RYA and SWL both use a more-less ranging mechanism to discover the very same triangle. by applying different time triangles....

but lodesman-RYA method chooses to use the same angle as the RhumbA->B has on one side of the triangle while ranging Destination mark B that happens to lie on it.This has a certain limitation.
SWL "ranges" by targeting the same point Destination B but isn't constrained by the Rhumb's angle in the ranging shots.

But Rhumb angle and length StartA->DestinationB will be one side of the(solved) triangle. It is the time that is equal on all sides which is being discovered.

Simple.

[Seaworthy Lass]

Originally Posted by Lodesman

So you have no way of proving that it is the better approximation? How can you keep saying that it is?

Simply because the expected ground track when following the CTS using my method gets you consistently very close to the destination, regardless of the data you choose to throw at it

There are always limitations with the data, but you could select the tidal stream data for the last partial hour much more accurately if you know roughly what proportion of an hour was needed (that is a bit more laborious though). if the data is correct, my method will give very close to the correct result very consistently, unlike other methods - I won't name them LOL, as I have been chastised for doing that (by the way who makes up the rules here? )

An easy way to check for flaws in the method is to throw some extreme data at it. I cannot find any situation that my method will not work for, but would love some feedback if you have found anything to the contrary. This would send me back to the drawing board, but that is a help, not a hindrance .

[Lodesman]

Originally Posted by Seaworthy Lass

Which bit is nonsense? If you are following a CTS with variable cross current why should you be aiming for any point on the rhumb line? You are not following the rhumb line, you are just getting to B the quickest possible way.

You do not have to be on the rhumb line to solve a tidal triangle!

You're not aiming at "any point on the rhumb line" - as a planning process it has no more expectation that the boat will stay on the "rhumbline" than your method - that is, you have a start point and an end point, and the vector that joins those two points is the resultant desired by you, whether the boat tracks along it or not. The vector mathematics for the single CTS is solved in the same manner as the standard triangle.

Regardless of the method used, I trust that you will plot your EP with the actual set, rather than the average set for the whole trip - in this case you will expect to track along the track made good (which is distinct from the planned track (rhumbline)).

[Lodesman]

Originally Posted by Seaworthy Lass

Simply because the expected ground track when following the CTS using my method gets you consistently very close to the destination, regardless of the data you choose to throw at it
.

I noted at post #249 in this thread that you were farther out at the end of plot, than I was with my original method. It's an approximation, as are all the methods ultimately - I don't think you're in a position to deem your method to be the superior approximation.

An easy way to check for flaws in the method is to throw some extreme data at it. I cannot find any situation that my method will not work for, but would love some feedback if you have found anything to the contrary. This would send me back to the drawing board, but that is a help, not a hindrance

Again without knowing how your method works mathematically, I'm not sure I would trust it to be infallible. Sure it may work for some randomly sampled extremes; that is not a guarantee it will work for every normal situation. If you're happy using your system then that's all that matters

[HappySeagull]

1st guess. few steps.
practise makes perfect...but no no foolin' 'round is really required beyond a good estimate in a scale drawing..
green dots are vectors and track and is proof of arrival within the limitations of the data....
I might say,gee, instead of 10T, I'll steer 10.5T...