Inaccurate RYA Teaching : CTS - Quest For a New Method

[HappySeagull]

...heyyy This is an irrelevant post, but I'm doing it anyway to show my interest.
That's a brute,jackdale. very cruel. My pencil is too dull to plot the wee tail and I gave up where the reward don't suit the punishment...

so for fun, I looked at messy paper and did as (cos235*3.7)+(cos295*.2)=-2.04miles Southing .( now,I know I'm going to get there in less than 4 because current helps in this southerly direction16-2=14..so can drop the .2 kn in 5th hour.)

then (sin235*3.7)+(sin295*.2)=-3.2121212..for displacement West

3.2/14=tan12.87 180-12.87=CTS =167.13T and is 1/cos167.13*14 in length =14.36 miles/4=3.6 hrs= 3hrs 36 minutes

CTS 167 for 3 hours and 36 minutes. Hope it's not foggy when I get CLOSE to B...I haven't actually plotted it hour by hour or even to the 1/2s because it'll be slightly wrong without being able to see the error well.
when you post, I'll be able to see gross mistakes anyways.

[jackdale]

RYA solutions for my example.

I did two versions; a 4 hour vector and a 3 hour vector both on graph paper. These are big images. I also used the chart and got the same answers. If you want to see that let me know.

Since I am only 2.1 cables from my destination in version 1 I would simply alter course using my Mark 1 eyeball.





Direct access to images

http://www3.telus.net/jackdale/pics/RYAv1.gif
http://www3.telus.net/jackdale/pics/RYAv2.gif

Course 180 T 16.0 Miles

Depart at 1000

Boat speed (knotmeter) – 4 knots

Current
1000 1.1 knots 235T
1100 1.3 knots 235
1200 1.3 knots 235
1300 0.20 knots 295
1400 0.20 knots 080

CTS (aka C)
-170 (4 hour vector )
-164 (3 hour vector) + 184 for last 2.65 miles

SMG – 4.4 knots (both vectors) 4.1 knot on last leg of version 2

ETA - 1339 (both cases)

[jackdale]

Happy

Very similar results. The blue green grid did not show in the scan.

I am a rather visual learner, which is why I like to plot them on the chart. I am impressed by the trig. My trig class was in 1964.

[HappySeagull]

Ok, I plot as "Hapless Seagull" with no method to fine up that last bit... just figure displacement and fudge the end.
My Plot is hour by hour. Note that I failed somewhere in the trig post above on elapsed time!
here it is 3 hours and 21 minutes per the hapless plot


Now I will look at jackdale's I see above.

Editadded first thought given Jack's-oh. well, maybe my plot is (very possibly)wrong per time error. aargh.coffee.
Glad I am in the ballpark anyways.

[HappySeagull]

Originally Posted by jackdale

Happy

Very similar results. The blue green grid did not show in the scan.

I am a rather visual learner, which is why I like to plot them on the chart. I am impressed by the trig. My trig class was in 1964.

Yes, it'd be possible to build a system of geometry alright- likely been done for this somewhere .
For me, geometry is no good if I don't understand it, so I often need to dissect or at least check it as a drawing too. Hand me a formula or a book of tables and I don't really like it until I can figure out where it "came from"

[Seaworthy Lass]

After reading the text in Tim Bartlett's Navigation handbook, it is actually even worse than I thought!

The text book is not just one for people doing an RYA course. The forward states "this book provides the knowledge needed by the modern small boat navigator".

The instructions for finding CTS are straightforward and as follows:

The section is titled: "Finding course to steer by chart work"

It is described as: "Using the chart itself as a worksheet"
In summary (quotes have been taken directly from the text):

- "draw the intended track onto the chart, as a straight line from the point of departure to the intended destination and beyond"

- "assess the tidal stream that the boat will experience at each hour of the intended passage" (this is somehow estimated)

- "the tidal stream vectors can then be drawn onto the chart, starting from the intended starting point, but joining onto each other, nose to tail"

- "it does not matter what scale you use: most navigators prefer to stick to the scale of the chart. There is no reason, however, why you shouldn't use centimetres, quarter inches, or any other unit you like, so long as you are absolutely consistent throughout the calculation"

Note: the diagram shows all this being drawn on the chart, there is no talk of doing it on graph paper at any stage, or of adjusting the scale of the distance, which is fine as using this method you are simply finding a CTS, not SMG or time taken.

- "draw an arc with a radius equivalent to the distance the boat is expected to travel in the time, and its centre on the end of the tidal vectors. This is why it may be useful to use something other than the chart scale for the tidal vectors"

- "finally, join the end of the tidal vector to the point at which the arc cuts the intended track. The direction of this line represents the course to steer"

There is absolutely no suggestion of checking if you have roughly the correct number of hours of boat displacement due to current. You could easily be an hour or two out if the current is strong and adverse. And if you are using a scale that is different to that of the chart (recommended as "useful") there is actually no way of checking this. And certainly no scope for adding more hours of current or subtracting some if you have got it wrong.

Originally Posted by Dockhead

Obviously you have to lay the vector triangle to scale on the chart to see whether you get there or not

No, Tim is saying any scale can be used on the chart as long as you keep the current and boat speed vectors to the same scale. If all you are seeking is a CTS, the distance from A to B does not need to be to the same scale. There is no suggestion or implication of using graph paper.

There is nothing to suggest you look at the arc relative to the destination, just to what they call the "intended track" (a VERY misleading term anyway, as it is actually the rhumb line, not a track you intend being on).

This focus on the "intended track" is catastrophic in some situations, producing results that can be wildly off (30+ degrees) as I have shown in some of my examples if you are approaching the destination obliquely.

The RYA method will work well for small amounts of current relative to boat speed, particularly if the journey is very close to a whole number of hours, or if the average amount of current for the trip is roughly the same whether or not the final amount of current encountered in the journey is included in the average.

This is a VERY limited set of conditions for a method to work. there is no warning at all in the text about errors in the computed CTS if these conditions are not met.

From people I have spoken to that have recently undergone RYA courses, there is also no warning about this during the course. As an RYA instructor here on CF has repeatedly said, the technique is a mathematically precise way of getting to the destination, limited only by the data input. He does not seem to be unique in teaching this.

Does this not bother anyone else?

[jackdale]

SWL - I have provided a real life example. I would like to see your solution please.

Also, Can you provide a real life example to which I can provide solutions? By real life I mean using actual current rates and directions along a real route.

[HappySeagull]

well, SWL, I'd like to nail down my speed versus drift for the best CTS...this text don't do it for me, I'd agree. But, "assess the tidal stream that the boat will experience at each hour of the intended passage" is pretty similar to your first instruction here:
Step 4: You know the passage will take longer than that if there is total current against you (and shorter if it it with you), so make an educated guess how much longer by looking at the current along the way (this may actually be difficult, but that is a whole different topic). Which made me fall back on my own ignorant charms and incantations to ward off the squalls in the thread....
I expect you have a revision for this? It seems to be the crux of your method?
I apologise if I missed it but there's a ton of stuff here and you will know best.
I did get " the averaging bit" by the way but got distracted ...
(There is nothing to suggest you look at the arc relative to the destination, just to what they call the "intended track" (a VERY misleading term anyway, as it is actually the rhumb line, not a track you intend being on).Not the track you will be on, that's for sure. ....My Peeve:Is it mentioned as a caution to plot where I will be, hour to hour (or half-hour) , if I blithely follow this CTS? That bugs me. You could easily sail into danger or a shipping lane , because in these diagrams you are only known to be at Start + Finish. It could be elsewhere in this book- I -don't- have.

[Dockhead]

Originally Posted by jackdale

Happy

Very similar results. The blue green grid did not show in the scan.

I am a rather visual learner, which is why I like to plot them on the chart. I am impressed by the trig. My trig class was in 1964.

Me too! The trig is dmaned good!

[conachair]

Originally Posted by Seaworthy Lass

Does this not bother anyone else?

Not really Though the wording's a bit ambiguous. But day to day in the real world I don't want an exact cts to the destination, just something to get within a few miles preferably uptide a little then have another look and take it from there, usually into pilotage with transits or whatever suits the conditions. Maybe there is a little too much emphasis on a high level of accuracy to the destination , there is a slight danger that you might start to think the sea will behave the same as the numbers om a piece of paper.
So I've never had a problem with the RYA method of calculating tidal vectors, they get you near enough.

[Dockhead]

Originally Posted by Seaworthy Lass

From people I have spoken to that have recently undergone RYA courses, there is also no warning about this during the course. As an RYA instructor here on CF has repeatedly said, the technique is a mathematically precise way of getting to the destination, limited only by the data input. He does not seem to be unique in teaching this.

Does this not bother anyone else?

I've just arrived in the UK. I'll try to talk to the author about it, if he's not at sea.

But I think that Dave already backed away from the proposition that it's mathematically precise, which it obviously is not. I rather doubt that he teaches that -- would be quite silly. I think he just had a little brain fart like all the rest of us have had from time to time.

The matter of scale is a matter of common sense. No one would use a different scale from the chart unless they were not able to measure the angle because the vector triangle is too small. You certain are taught (it's in there somewhere) to evaluate how long the passage will be -- that's where he talks about using dividers or a drawing compass and marking off arcs for the hours on the chart. You draw those arcs from the end of the tidal vectors line and so you know quite exactly how far you will get towards the destination in a given number of hours at a given boat speed. Obviously that bit has to be done at the chart scale. That's just common sense and I do not believe that either Tim Bartlett, or Dave, or anyone else using the method is so extremely stupid as to not get that.

[HappySeagull]

jackdale, not to whine, but I was hoping for a grosser model.( SWL, you will need a very sharp pencil!) See how everybody is slacking off? they believe the difference won't pay 'em. We need some meat wae can eat!
I'm considering an example that is "all about me" except it's predictable-something like :
It's crucial to know the answer before I set sail.
"beating with a fading current to round the point to harbour before current changes to overstrong flood.." Hereabouts, the 5-6 knot examples are in deep water channels. I could drop the "channel" aspect but would rather keep it in.
because Tacks could then be counted and just when to make the last tack! might be as interesting as the actual arrival time. In the example, it would be a blood sport-no tacks for safety. Just clear the point !
Fail, and you fall back into the increasingly savage maw of the pass!

Wind is relative to boat ie: fail and you lose it quick! but a boatspeed 4 to start ,I'd include a table of boat speed versus wind but must stick to the slant. No "lifts" or it'll get out of hand as an example.Tacks -do they need a penalty? yes. I think so. Chop has a penalty in life but not in this example.

[Seaworthy Lass]

Originally Posted by jackdale

Happy et al

These are real numbers.

Course 180 T 16.0 Miles
Depart at 1000
Boat speed (knotmeter) – 4 knots
Current
1000 1.1 knots 235T
1100 1.3 knots 235
1200 1.3 knots 235
1300 0.20 knots 295
1400 0.20 knots 080

For those interested. It is from the Botchie Ledge light off Victoria, BC to the Ediz Hook light off Port Angeles, WA across the eastern portion of The Straits of Juan de Fuca.

I used two current stations, first three hours are from 5.2 miles SSW of Trial Island, the next two hours are from 1.2 miles north of the Ediz Hook Light. The data is from OpenCPN. I used data from today.

I used a NOAA chart 18400 which is available for free as a download at Pacific Coast NOAA Nautical Charts

If you could please provide an answer to Andrew Troup's example in post #485, I will provide an answer to yours tomorrow when I am back on board after a day's work replacing the sacrificial strips on our headsails.

[jackdale]

Happy

Those beats in channels will either have the current on the nose or astern. And in those situations you just have make do. They are very hard to predict. Or you try to avoid them. I go through Whirlpool Rapids rather than getting caught in Current Passage.

I will try to get something stronger. I will look for a strong current off Ediz Hook with a strong current SSW of Trial. Will that work?

We need a longish to get to multiple currents.

[LJH]

Below is an OpenCPN solution using SWL's method. 168 degrees for 3 hr 34 minutes. (Blue line is k to B, Red line is ground track)

Also done by spreadsheet I got 167 degrees 3 hr 34 mins.
(https://docs.google.com/file/d/0Bz8m...dGeG80ZTA/edit)