Inaccurate RYA Teaching : CTS - Quest For a New Method

[LJH]

The Neptune Planner looks like a great piece of kit. Enjoy it!

Originally Posted by Dockhead

Attachment 53652

So how the instructor teaches it is obviously key!

It must be, given how little is in the example!

[Dockhead]

Originally Posted by LJH

The Neptune Planner looks like a great piece of kit. Enjoy it!



It must be, given how little is in the example!

That was just the example - I've added the actual text now.

[LJH]

Originally Posted by Dockhead

That was just the example - I've added the actual text now.

Yes they do leave a lot for the instructor to fill in!

[Lodesman]

Originally Posted by Seaworthy Lass

Sounds like it has produced a good result in this case. Would you like to write up a step by step description with diagrams so that we can follow what you do?

Long post warning!

As I said it would take awhile (spotty wifi amongst other issues), here is my take on problem #3. I realize we've moved onto problems 4, 5, and more, so this is a step back, but I think it does reasonably illustrate that with modification the RYA method (or the basic navigation principles) can be utilized to resolve even the most extreme problems. While I've put it down in step-by-step, I don't advocate rote application of such - it's better to understand how the tidal triangle works and apply to the individual situation. This particular problem required a few departures from the normal 'tidal triangle' model.

Steps:

1. Lay off course 090º, distance 11nm. Mark start A, end B;

2. 11nm at 4kts = rough time required 2h45m, so mark off 3 hours of tidal vectors;

3. Due to high current and low boat speed it is impossible to close the vector triangle - the best that we can hope to achieve is minimal loss in the first two hours. Draw an initial CTS - 009º;

4. Using initial CTS, plot CMG for first two hours; mark endpoint C; draw a line from C through B - this for all intents and purposes, is the "rhumbline";

5. Plot third hour set vector from C; mark end D and from D measure 4nm and draw the arc where it crosses CB - mark this E;

6. Measure CE (2.2nm) and CB (3.2nm) - CE is SMG, and based on it, it requires 1h27m to make good CB; plot 27 mins worth of the fourth hour's set vector (0.9nm); measure CTS vector (1h27m at 4kts= 5.8nm) and plotting from end of set vector, mark an arc on the rhumbline (call this F);

7. Measuring EF (2.8nm), you should realize immediately that SMG has changed dramatically in this last hour and is now 6.2 kts(2.8nm over 27m); calculate time required at 6.2kts to cover EB (1.0nm) = 10 mins;

8. Refine final set vector (10 mins at 2kts = .3nm) mark it G; calculate CTS vector (1h10min @ 4kts = 4.7nm); connect G and B, measure it as just a smidge over 4.7nm, now refined as close as it will likely get; measure it, the course is 008º.

At this point, you would then plot the CMG based on CTS of 008º - I haven't done that here (see first image) as it would require erasing the previously drawn lines; the second picture shows the refined solution plotted. In doing so it will be clearer to you that the actual course is fractions of a degree above 008º (it's greater than 008, but less than 009) and the actual total time can be further refined to 3h09m.
(see second image)

You will note of course that my method is quite complicated and would not appear to provide any substantial benefit over the SWL method. I would have to agree - as I said it was a rather extreme and improbable scenario that was designed specifically to flatter the SWL method. While the SWL method could go into a sailor's bag of tricks, I think it is rather limited in its utility. Knowing the standard tidal triangle solution and being able to manipulate it provides for all your navigation needs as I will demonstrate.

After going through all the calculations above (RYA, SWL or other method) a smart sailor would realize that they are only spending 9 mins (or 12) with the set that helps them get to destination, and would realize it would make more sense to maximize time running with a helpful current and less time fighting a counter current. With my method it is simple to work backwards:

1. Start from the basic course line (rhumbline); start A, destination B and plot the final two hours' of set vectors back from B; mark the start of the vectors C;

2. Calculate CTS speed component (2 hours at 4kts = 8nm); measure it on a compass, point on C, draw an arc;

3. Although we'd make better way against the first hour's current, it would be awkward going to anchor in the middle, so I'll plot the second hour's current (150º @ 6kts) from A; mark the end D;

4. Measure 4nm on a compass to mark the CTS vector - with the compass point at D scribe the arc; this crosses the arc drawn at (2), so it describes a range of course options - basically anywhere between 352º and 118º will provide a relative vector (CMG) that intersects the arc drawn at (2). This is a bit of a conundrum determining the optimum course - balancing relative speed versus distance to cover. I settled on a course of 085º - I don't know if it's the best, so if one of the mathematicians on the forum could determine an easy way to accurately determine the optimum course, please chime in;

5. Draw on the 085º CTS, complete the triangle - here I've bolded the section from A to E, where it intercepts the 8nm arc; for those who do such things, calculate SMG (8.5kts) and distance of AE (3.0nm) to give a time required of 21 mins;

6. Connect EC - this is CTS (078º); to complete the chartwork I connect EB for CMG and I'm ready to go.
Now all I have to do is stay at dock or anchor for the first 1 hour 39 mins, get underway and steer 085º for the first 21 mins, and 078º for the rest to arrive in a total time of 2h21m. Yes I have to alter course, but it's supposed to be about efficiency and it's a small price to pay to knock 50 minutes of motoring off the trip. (see third image)

Can you do this with the SWL method?

[jackdale]

Originally Posted by Dockhead

Sine waves would work to give a superior tidal rate prediction. You need the limits. And -- drum roll please -- you can get them. They are reported by the Admiralty H.O., and you can get them out of programs like the Neptune Planner:

Attachment 53636

However, without stream direction, this is useless -- a dead end.

Current direction is shown at the bottom of each page of the current tables for reference stations and are on the correction table for secondary stations. Flood and ebb directions are also shown on the chart.

BUT -- if you have a computer on board, why bother? I just bought Neptune, and ironically after all this work on CTS theory and practice, I may never do one by hand again. I ran some Channel crossings and found out that the non-perpendicular element of the tide, which we are trained to ignore, has a BIG influence on your passage.

And Neptune will do the rotary tides in the Channel Island, which NO hand method can deal with. I'm in lurrrvvvv -- for the first time all day!

For the three standards that I teach we use traditional, non-electronic, navigation using official publications. Once we understand the basics we get into chartplotters, computers, etc.. Most of the courses I teach are at advanced levels so we do teach electronic charting.

In Canada, vessels are required to have official publications unless they have ECDIS with ENC charts. No recreational vessel is equipped with ECDIS.

And if your vessel is struck by lightning....

[jackdale]

Originally Posted by jackdale

Shall do. I am not sure that level of significant figures is realistic

Happy Seagull

Just used your method for intermediate times. This is an example that I use for teaching intermediate times for current strength.

We may wish to know the state of current at a particular time. For this we can use a sine wave.
In the example below we want to know the currents in Active Pass at 1130. The turn before 1130
occurs at 1000, the maximum ebb of –3.3 knots is at 1300. These values are placed on the sine
wave. The time of 1130 is interpolated. Where the time crosses the sine wave a line is drawn to
the vertical axis. The current at 1130 is 7/10 of the maximum –3.3 ebb or –2.3 knots.

Same answer - not surprising, sines are used in both cases.

Another arrow in the quiver. I will add this to my more advanced courses and compare results.

[Seaworthy Lass]

Originally Posted by Lodesman

Long post warning!

As I said it would take awhile (spotty wifi amongst other issues), here is my take on problem #3. I realize we've moved onto problems 4, 5, and more, so this is a step back, but I think it does reasonably illustrate that with modification the RYA method (or the basic navigation principles) can be utilized to resolve even the most extreme problems. While I've put it down in step-by-step, I don't advocate rote application of such - it's better to understand how the tidal triangle works and apply to the individual situation. This particular problem required a few departures from the normal 'tidal triangle' model.

Steps:

1. Lay off course 090º, distance 11nm. Mark start A, end B;

2. 11nm at 4kts = rough time required 2h45m, so mark off 3 hours of tidal vectors;

3. Due to high current and low boat speed it is impossible to close the vector triangle - the best that we can hope to achieve is minimal loss in the first two hours. Draw an initial CTS - 009º;

4. Using initial CTS, plot CMG for first two hours; mark endpoint C; draw a line from C through B - this for all intents and purposes, is the "rhumbline";

5. Plot third hour set vector from C; mark end D and from D measure 4nm and draw the arc where it crosses CB - mark this E;

6. Measure CE (2.2nm) and CB (3.2nm) - CE is SMG, and based on it, it requires 1h27m to make good CB; plot 27 mins worth of the fourth hour's set vector (0.9nm); measure CTS vector (1h27m at 4kts= 5.8nm) and plotting from end of set vector, mark an arc on the rhumbline (call this F);

7. Measuring EF (2.8nm), you should realize immediately that SMG has changed dramatically in this last hour and is now 6.2 kts(2.8nm over 27m); calculate time required at 6.2kts to cover EB (1.0nm) = 10 mins;

8. Refine final set vector (10 mins at 2kts = .3nm) mark it G; calculate CTS vector (1h10min @ 4kts = 4.7nm); connect G and B, measure it as just a smidge over 4.7nm, now refined as close as it will likely get; measure it, the course is 008º.

At this point, you would then plot the CMG based on CTS of 008º - I haven't done that here (see first image) as it would require erasing the previously drawn lines; the second picture shows the refined solution plotted. In doing so it will be clearer to you that the actual course is fractions of a degree above 008º (it's greater than 008, but less than 009) and the actual total time can be further refined to 3h09m.
(see second image)

You will note of course that my method is quite complicated and would not appear to provide any substantial benefit over the SWL method. I would have to agree - as I said it was a rather extreme and improbable scenario that was designed specifically to flatter the SWL method. While the SWL method could go into a sailor's bag of tricks, I think it is rather limited in its utility. Knowing the standard tidal triangle solution and being able to manipulate it provides for all your navigation needs as I will demonstrate.

Lodesman, well done for having a go at finding a better alternative to the RYA method. The more people that put their heads to this, the more likely we will have the best alternative.
I wish more people would have a go!

My only comment is that I have no idea why you are drawing a rhumb line from your position at the end of two hours. You are being subjected to more than another hour of current, and possibly these next currents may be severly adverse, so why draw a line to B from the end of the second hour as if there was no current after that?
It is late here now and maybe I am missing something, but I just don't get it.

Originally Posted by Lodesman

After going through all the calculations above (RYA, SWL or other method) a smart sailor would realize that they are only spending 9 mins (or 12) with the set that helps them get to destination, and would realize it would make more sense to maximize time running with a helpful current and less time fighting a counter current. With my method it is simple to work backwards:

1. Start from the basic course line (rhumbline); start A, destination B and plot the final two hours' of set vectors back from B; mark the start of the vectors C;

2. Calculate CTS speed component (2 hours at 4kts = 8nm); measure it on a compass, point on C, draw an arc;

3. Although we'd make better way against the first hour's current, it would be awkward going to anchor in the middle, so I'll plot the second hour's current (150º @ 6kts) from A; mark the end D;

4. Measure 4nm on a compass to mark the CTS vector - with the compass point at D scribe the arc; this crosses the arc drawn at (2), so it describes a range of course options - basically anywhere between 352º and 118º will provide a relative vector (CMG) that intersects the arc drawn at (2). This is a bit of a conundrum determining the optimum course - balancing relative speed versus distance to cover. I settled on a course of 085º - I don't know if it's the best, so if one of the mathematicians on the forum could determine an easy way to accurately determine the optimum course, please chime in;

5. Draw on the 085º CTS, complete the triangle - here I've bolded the section from A to E, where it intercepts the 8nm arc; for those who do such things, calculate SMG (8.5kts) and distance of AE (3.0nm) to give a time required of 21 mins;

6. Connect EC - this is CTS (078º); to complete the chartwork I connect EB for CMG and I'm ready to go.
Now all I have to do is stay at dock or anchor for the first 1 hour 39 mins, get underway and steer 085º for the first 21 mins, and 078º for the rest to arrive in a total time of 2h21m. Yes I have to alter course, but it's supposed to be about efficiency and it's a small price to pay to knock 50 minutes of motoring off the trip. (see third image)

Can you do this with the SWL method?

Picking the best time to travel is a different matter. I did not give the current data for the full tidal cycle. Some other portion may well have been a better time to travel.
If the passage takes advantage of the full hour of the optimum time, the total time taken can probably be minmimised by working backwards to determine the time to leave. This can be done with my technique also I think. I will look at that tomorrow (it is 1 am here now and my mind is getting a bit fuzzy).
Thanks for that great thought, it may work.
Anyone see any flaws with doing this?
The more we put our heads to it, the better an alternative method will be.

[HappySeagull]

Exactly, Jack. I just made it up as being easy. You could have saved me the trouble, and I will blame you forever (I apologize for the thread drift.) But don't forget it's for slack-to max or max-to-slack. The moon really moves quick and wild.

Did I say it works( sort of ) for tides too?

Added:While I'm at it, Thanks to Dockhead for posting those pages. I'm a little behind here as to bandwidth and without a Royal Yacht, it's been hard to Associate just what the RYA is in the flurry.

[Seaworthy Lass]

Originally Posted by Dockhead

The lake bit was a joke. I have done a fair bit of sailing in the Med and it can be challenging. The Meltemi in the Aegean can be wild, especially in the acceleration zones between islands!

For sheer sh*tty weather, the Black Sea is hard to beat -- the sea state can be horrendous with the long fetch and shallow depth . . .


Before we finish our conversation about the RYA method of calculating CTS, I though I would actually put up the pages!

It would be an exaggeration to say that I sleep with the RYA navigation handbook, but I always have it in my briefcase, even here in Helsinki in my hotel room:

Attachment 53651

Here's all it says about multi-hour CTS "course shaping":

Attachment 53652

Attachment 53654

Attachment 53655

So how the instructor teaches it is obviously key!

Dockhead, thanks so much for showing us these pages.

I cannot read the text clearly on my iPad, either by clicking on the image or by selecting 'save image'. It seems to be fuzzy even if I try loading the images I personally put on originally that I know were clear. Any suggestions anyone?

Anyway the diagrams in the book indicate the RYA arc off the 'distance displaced by boat speed vector' on the rhumb line just as the RYA YouTube videos showed (either before or after B). Does the text say something different?

This is the method I used for all the RYA calculations I did and Jackdale confirmed he had the same results using the RYA method as I did.

[LJH]

Originally Posted by Seaworthy Lass

Anyway the diagrams in the book indicate the RYA arc off the 'distance displaced by boat speed vector' on the rhumb line just as the RYA YouTube videos showed (either before or after B). Does the text say something different?

No, but it does not say anything more than the steps required to do the chart work. Nothing about adjusting for a partial hour! They seem to leave that for the instructor to talk about. If there is an instructor handbook, I hope it has some more guidance.

[Seaworthy Lass]

Originally Posted by LJH

No, but it does not say anything more than the steps required to do the chart work. Nothing about adjusting for a partial hour! They seem to leave that for the instructor to talk about. If there is an instructor handbook, I hope it has some more guidance.

This was the explanation given in the 'Distinct Activities Shacked by a Common Name' by an RYA instructor:

Originally Posted by goboatingnow

Hi seaworthy.

The RYA method says the following ( and please you have to except that I know the method , I can't realistically copy the whole instructors note section )

Ill look, at your chart work later , its quite difficult from the pictures

The RYA says estimate in hours the journey time ( or whatever tidal quantum you have available the , in some places you have 30 minute tides ). That estimate is the total expected passage time allowing for any tides. typically where tides are 25% or less then boat speed or the journey is short then the rhumb line can be used. But in long runs with many tides ( like across the Irish Sea ) you have to guesstimate.

Derive the tidal data from the appropriate sources. Manually interpolating if your passage point do not coincide with the HW ( high water marks )

Draw a line through from A ( start ) to B the destination and carry that line beyond B

Plot as many hourly ( or whatever tidal time quantum ) as feel is neccessary to cover the intended JOURNEY time. For multi hour tides, add the one hour tidal vectors together call this Point C

Draw the boat speed vector in quantum's of the tidal vector quantum , typically one hour. , so if the boat speed is 4 kn , open the dividers to 4nm and arc of D on the rhumb line. , if its a 2 hour plot then its a 8nm arc and so forth call this intersection Point D

despite what you persist in saying , D depending on how many hours you estimate , and the effects of the tidal vectors may lie in front , at or behind B

Perform a check to ensure that D lies approx less then 30 minutes away from B , some instructors omit this bit. , ie to ensure the tidal data remains valid. This can be done by simplest calculating the rate of advance , ie the average SOG to D and by inspection determine how far B is away. If its greater then 30 minutes ( or less) redraw the plot using one more or less tide ,( note this is rarely done in examples as the test questions are typically picked to ensure it isn't needed.

If D lies in front of B , calculate the time of arrival by applying the rate of advance to D to the whole journey B, equally if D is behind reduce the time by using the SOG to B accordingly ( ie inflate or deflate the whole tidal triangle

CTS is the bearing of line CD at all times.

That's it

As I mentioned inflating or deflating the total tidal triangle is as accurate as any other method , given the interpolation of tidal vectors in the first place. Ie in a six hour plot inflating each vector by say 7 minutes , is much as accurate as deflating the last vector by 42, in fact its actually more accurate

Dave

Note that if D is not close to B then the instructions are to "redraw the plot using one more or less tide" although apparently "some instructors omit this bit".
I shudder to think how inaccurate those computations become if you don't find D at it's closest to B. The method is bad enough in some cases even when the closest possible D is selected!

Results are interpolated by "inflating or deflating the total tidal triangle" rather than by examining the actual current that will occur in the last portion of the journey.

[Lodesman]

Originally Posted by Seaworthy Lass

My only comment is that I have no idea why you are drawing a rhumb line from your position at the end of two hours. You are being subjected to more than another hour of current, and possibly these next currents may be severly adverse, so why draw a line to B from the end of the second hour as if there was no current after that?
It is late here now and maybe I am missing something, but I just don't get it.

As I said before, the tidal triangle principle doesn't work unless you can make a triangle - taken by themselves the sets during the first and second hours could not be solved as you could not overcome the current to get to destination. In my solution, I made best effort against the set, but was still off the original rhumbline. In the third hour the set could be overcome and a CMG to destination was possible. While we've been calling it a rhumbline, let's not forget that in the tidal triangle it is Course Made Good. Although in the RYA/other multiple set scenarios the actual CMG does not follow the rhumbline, the resultant vector does.

[HappySeagull]

Lodesman, SWL, et al.
I got interested in that nasty one too.Anathema here, but I just steered for B ...boat merely changes course to aim for B every 15 minutes. Result is "more than 3 .25 hours and less than 3.5hours" Sorry for decimal hours.



Compares to SWL model 3.2 hours. About 20 minutes improvement. Not bad, but the problem is in the model being just too close such that improvements are difficult to express and all jammed into the end.SWL caught up to Hapless Seagull at about 2.75 hours.
As to whether Hapless could have simply chosen to steer SWLs course when ahead- aye there's the rub and Lodesman is "on it" too apparently. Apologies if sketch is wrong. False alarms are constant in the Colony.
http://www.cruisersforum.com/forums/...0&d=1359039346

[HappySeagull]

Ha! excellent Lodesman. It works, per my scribble.

added: it does assume there is somewhere to anchor though

[Seaworthy Lass]

Originally Posted by Lodesman

Steps:

1. Lay off course 090º, distance 11nm. Mark start A, end B;

2. 11nm at 4kts = rough time required 2h45m, so mark off 3 hours of tidal vectors;

3. Due to high current and low boat speed it is impossible to close the vector triangle - the best that we can hope to achieve is minimal loss in the first two hours. Draw an initial CTS - 009º;

4. Using initial CTS, plot CMG for first two hours; mark endpoint C; draw a line from C through B - this for all intents and purposes, is the "rhumbline";

Where do you get an "initial CTS - 009º"
You need to determine a CTS before you can start drawing any ground tracks. You cant draw a ground track and then estimate CTS.
You are missing some steps between step 2 and step 3