

11022013, 19:15

#676

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
SWL  Promised you currents in hourly increments.
Currents Juan de Fuca West Entrance February 13, 2013
0000 00.87 290
0100 00.20 290
0200 00.42 115
0300 00.84 115
0400 00.95 115
0500 00.72 115
0600 00.21 115
0700 00.46 290
0800 01.13 290
0900 01.63 290
1000 01.85 290
1100 01.75 290
1200 01.35 290
1300 00.78 290
1400 00.16 290
1500 00.33 115
1600 00.58 115
1700 00.54 115
1800 00.21 115
1900 00.30 290
2000 00.86 290
2100 01.31 290
2200 01.54 290
2300 01.48 290
0000 01.15 290
0100 00.62 290
All times PST (Z+8)
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11022013, 19:45

#677

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by goboatingnow
This is not a correct summation SWL, your method gives more precise answers. We have no idea if it is more accurate , we await Dockheads deliberations
Dave

Sorry, I've been very busy with work. I have a delegation of investors today.
I did learn to use the Admiralty tide program. I will post a scenario I later today.
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11022013, 20:22

#678

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
I can't see any problem in the last example. Not extreme. Perhaps you want to arrive somewhere at a particular time. Or you want to fish along the way...Perhaps the motor is overheating after 15 minutes of running. ... 2+ knots that's no problem . I can row my skiff in that.
I can't see where the SWL method relies on a one hour 0 drift either.
An average of displacement for 1/2 hour each side of slack might just sum as that.
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11022013, 22:39

#679

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by goboatingnow
This is not a correct summation SWL, your method gives more precise answers. We have no idea if it is more accurate , we await Dockheads deliberations
Dave

Quote:
Originally Posted by jackdale
Thanks Dave
Those are my thoughts as well. That is why real world examples are needed. SWL's method has some apparent precision in extreme situations. I have expressed several times that I would not venture out in to the conditions described.
Jack

Come on guys! How difficult is it for two instructors to admit that the RYA method is flawed?
The SWL results are not more precise (the results have only ever been specified one degree, exactly as you have specified the RYA results), they are more ACCURATE. The RYA method fares well only if the amount of average current is very small in relation to boat speed, or if the journey is a whole number of hours or very close (so that D coincides with B or is very close) or if the the average current for the journey is nearly the same whether or not the last hour of data is included in the average. These are a very limited set of conditions for a method to work. My method works well regardless of the circumstances.
LJH correctly posted the results using all 3 methods for the last example:
Quote:
Originally Posted by LJH
Great view, thanks.
Using SWL on OpenCPN 152 for 2:31
Maths:Crse = 90= (tan1(6.7/3.5)= 152 degrees
Time = Sqr(3.5*3.5+6.7*6.7)/3 = 2.51 Hr = 2:30:36
Now doing the RYA method one can choose either the 2 or 3 hr triangle
2hr = 144 degrees 2.75 hrs
3hr = 157 degrees 2.4 hrs

The answer here was 152 degrees true, limited only by the fact that only a partial hour of current was used at the end.
The SWL method gave this result.
The RYA method was in error by 8 degrees!
The RYA method is just inherently inaccurate .
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12022013, 00:35

#680

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Join Date: Jun 2009
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Sorry seaworthy , the RYA method is " different " by 5 degrees. Ie 157 v 152. It's different not inaccurate. The correct RYA method is always to use the closest tidal hour. ( 5 degrees well within the resulting real life error)
The only way you can prove " accuracy" is to compare each method to reality. Your method is only " accurate" if you " assume" the tides are actually in real life like you state them and apply rigid time quantums. the RyA gives a different answer because it is based on different assumptions on the tide data , that's all
You have made no attempt to critique your own methods , you assume your 100% " accurate" and hence all the error is in the RYA method. Again this is simply wrong.
Why you can't see this I don't know, you fail to see that the methods can only be compared to real life not to each other as they make different assumptions about the data . I've set out the limitations and assumptions of the RYA method , you've performed no critical analysis of your method , just a shrill repeated call telling us repeatedly how " bad" the RYA method is
shheesh this is getting tiring
Precision has nothing to do with the number of digits in the calculation, in your method your precision is that you apply increasing " precise" interpretations of tidal data. ( your assumptions )
accuracy is comparing both methods to real life. You will find the error distributes between both the RYA method and Yours. , since you accept 0 error, you will find that in reality your error grows and the RYA diminishes.
Please stop telling us its more accurate , you have no idea and seeking far fetched examples( like consistently picking solutions that arrive at near the 30 minute boundary ) does little to advance your methods applicability to real life usage. ( nor in fact " prove" anything )
The proof of the pudding is the comparison to real life, even them the resulting error difference has to substantially more then the real life error to actually have any meaningful result.
Dave
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12022013, 06:26

#681

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
SWL  I have I have explained repeatedly, your scenarios are extreme ans unrealistic; some verge on poor seamanship. I pointed this out again at Inaccurate RYA Teaching : CTS  Quest For a New Method. You have not responded.
Any realistic scenario that I have posted has not resulted in any advantage to your method. It involves more steps to get similar answers. Occam's razor at work.
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12022013, 06:56

#682

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Re: Example 6
Quote:
Originally Posted by Seaworthy Lass
I can't win LOL.
The previous example I gave was for tidal stream information on the hour. I extrapolated the average and was told this was too complex a way to present an example and that people wanted the simplicity of hourly averages.

This issue I had there, was that you interpolated a figure without giving a methology; in fact it looked like a WAG. You then pronounced that your solution was better than the RYA and mathematical methods which used different data  that's comparing apples to oranges. I would like to see a real comparison of the methods, with resultant plots to see if the accuracy/precision of your method offers any real advantage. I worked to same level of precision in my method, but I don't think it makes any difference in the end  we'll all be rejigging our calculations to make it to the destination.
Quote:
I give hourly averages in this example and I am told that this is not suitable either. The hourly averages I selected were the ones given by Jackdale in his English channel example.

The channel crossing link posted by Jackdale, did not make it clear that the graph showed average speeds. I would interpret those numbers as the current speed at particular points on the graph. I don't know what tidal program you use, but would suggest that any one that gives you an hourly average of 0 kts does not do you any favours. Most tidal current tables that I'm aware of give the time and strength of maximums and slacks  you (and current programs) interpret from that.
Quote:
I addressed the issue of inaccuracy creeping in due to the estimate of the average current for the last bit of the journey in post #629:

This wasn't so much a case of "the inaccuracy of the average speed over a partial hour, compared to the average speed of the whole hour"; this was a case that the entire hour was missing. If in fact, you knew that the final hour had no current  you would be out of the stream in a bay or something  then the correct method of drawing the current triangle would be from the end of the 3.5nm current vector to the destination, to give you a CTS of 152º
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12022013, 08:33

#683

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Use this one ,as an example and this interminable and separate argument can be put to rest.
The method of calculating the displacement due to current is a separate subject unless a method relies on it .
BECAUSE
In the above example,
0845 266 0.08kn
0945 263 1.3kn
1045 254 2.0kn
How many miles did the vessel drift in the hour 0845>0945?
not .08 miles.not 1.3 miles.
How do you calculate average velocity or the total "drift" in miles when the change of velocity is not linear?
Beyond the scope of either method that's for sure.Obviously beyond the scope of the book too.It plots these as though the boat would drift these distances.
In a complete current cycle from slack to max, a boat will drift roughly according to the average 60% of the max current velocity.
"The method of calculating the displacement due to current is a separate subject unless a method relies on it ."...
Added:::if the RYA method demands the data presented as above, to be accurate, then someone who knows the method should prove it, share it, so everybody can be on the same page.
Boat speed is linear, current data is not. They do not mesh.
Therefore, present a HYPOTHETICAL sum of drift (for whatever period but half hour is best) that is within reason, and I for one, will look the other way.
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12022013, 08:36

#684

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Deficiencies with RYA teaching:
Quote:
Originally Posted by Lodesman
If in fact, you knew that the final hour had no current  you would be out of the stream in a bay or something  then the correct method of drawing the current triangle would be from the end of the 3.5nm current vector to the destination, to give you a CTS of 152º

PRECISELY!
That is exactly what my method allows you to do. That is why you end up with the same result using my method as is calculated mathematically. If there is zero current at the end of the journey you simply connect the tip of the last 'boat displacement due to current' vector to B when you use the SWL method.
A big flaw with the RYA method is that they don't allow you to do this:
Quote:
Originally Posted by goboatingnow
However you must plot in unit time rate vectors as this is how tides vectors are specified. Hence the water track is never connected to the destination, its a hourly vector plot. If the destination is further away you continue on and the rate vectors continue on an you are brought to the destination.
All over my lecture notes ( I dug them out , I have written DO NOT connect the water track to the destination, its the most common mistake made buy students.

And again:
Quote:
Originally Posted by goboatingnow
You cannot connect the water track to the destination. The length of the water track vector is set by the assumed vessel speed , since we are not allowing it to change speed. If there perturbation vectors are in hour increments , then the vessel water track vector is in hour increments. , hence a 4 hour tide vector plot has the water track at 4x vessel water speed. Extending the vector to fit the destination in essence is increasing the vessel speed. Which is why you get there earlier.
The RYA is infallible here, this method has been taught for 50 years.
Dave

RYA instructors even amazingly say that "you can't have a constant heading if you have a period of slack at the end":
Quote:
Originally Posted by goboatingnow
You can't have a constant heading if you have a period of slack at the end. There simply no perturbation vector. The CTS method only works with a vector causing the vessel to leave its ground track. There is no point otherwise.
What I did was straight from the RYA course notes.

Then there were all the comments from this RYA instructor about how mathematically precise the RYA method is and how you arrived at B (not D) without changing heading of you followed the determined CTS:
Quote:
Originally Posted by goboatingnow
In the vast majority of cases D is before or after B on the rhumb line. D is merely a vector addition vertices. You are not physically there. ( which is why it confuses people when they use a chart to graph speed vectors )
D is always the intersection of the rhumb line and the boat vector because you are engaging in vector addition ( the closing of the triangle ). You are not engaged in determining where the boat physically is at the time of D that's a different story all together.
Dave

Not only are you never at D according to how some instructors teach the method, but it is claimed the method will get you directly to the destination itself without changing heading:
Quote:
Originally Posted by goboatingnow
IN all cases
(a) SIngle CTS computation is the most efficient method where there is varying tidal vectors
(b) You arrive at the destination without changing heading
(c) You arrive precisely at the destination, ( mathematically) in real life the tidal vectors are constantly varying with both direction and speed and your progress may not be constant over the water anyway.

It was again repeated that you never arrive at D using the RYA method, you "hit the rhumb at the destination directly" (D was several miles before B in this example):
Quote:
Originally Posted by goboatingnow
No no no. , at no point in the hovercraft do you end up on the rhumb line, with the RYA method , ie the ONLY correct method , you DO NOT change change course at the rhumb. You hit the rhumb at the destination directly

The bottom line is that the RYA method is NOT a mathematically precise way of getting to the destination (B), yet it is taught that this is the case.
It is stressed that you do not arrive at D, but at B. This is plain incorrect.
I have said several times in previous threads that the RYA method is a good "rough and ready" method to get you close to the destination in most cases. The RYA does, however, not teach this. It attributes accuracies to this method that it simply does not possess.
Apart from presenting what I believe to be a better method for determining CTS, I have wanted to highlight that the RYA is not teaching students about the limitations of the actual RYA method itself (only of the data used for calculations).
Computing a CTS with variable cross current is very poorly taught by the RYA, yet all the instructors participating on this thread can say is that I am selecting unrealistic data for my computations .
The accuracy of the RYA method is not the only problem. How it is presented to students is the greater issue I would like to draw attention to!
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12022013, 09:29

#685

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Re: Deficiencies with RYA teaching:
Quote:
Originally Posted by Seaworthy Lass
PRECISELY!
That is exactly what my method allows you to do. That is why you end up with the same result using my method as is calculated mathematically. If there is zero current at the end of the journey you simply connect the tip of the last 'boat displacement due to current' vector to B when you use the SWL method.
A big flaw with the RYA method is that they don't allow you to do this:

I don't know what the RYA party line is, but it seems to me that they are using standard navigational vector calculations; so do not prevent anyone from connecting the C to the B, if B and D are coincidental. In the case that the tidal vector (AC) is the same length at every time from 2h0m to 3h0m, means that any boatspeed vector (CD) for the time period 2h0m3h0m can be drawn from point C, and it will intercept the rhumbline anywhere between D(2hr) and D(3hr). Common mathematical sense tells you that a vector of 2 hrs x min in length can be drawn from C to B  dividing it by boatsp solves x.
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12022013, 09:35

#686

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Re: Deficiencies with RYA teaching:
Quote:
Originally Posted by Lodesman
I don't know what the RYA party line is, but it seems to me that they are using standard navigational vector calculations; so do not prevent anyone from connecting the C to the B, if B and D are coincidental. In the case that the tidal vector (AC) is the same length at every time from 2h0m to 3h0m, means that any boatspeed vector (CD) for the time period 2h0m3h0m can be drawn from point C, and it will intercept the rhumbline anywhere between D(2hr) and D(3hr). Common mathematical sense tells you that a vector of 2 hrs x min in length can be drawn from C to B  dividing it by boatsp solves x.

Yes, in the rare example that the journey takes precisely a whole number of hours, and D therefore coincides with B, a line can be connected between C and B using the RYA method. Otherwise it is taught that C cannot be connected with B, even if there no current at the end of a journey.
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12022013, 09:50

#687

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
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12022013, 13:21

#688

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Anyone familiar with
TidePlan2
Quote:
The program uses a model of the tidal flows occuring in the English Channel. It relates this model to the magnitude and state of the tide at short intervals during the planned passage and calculates the tidal vector that is likely to affect the vessel at each interval. It integrates this with the vessel's speed and heading for the same interval and recalculates the vessel's course accordingly.
It repeats the process above and adjusts the heading at each iteration until a heading is found that takes the vessel to within approximately 100 yards of the destination position. It uses this last iteration to calculate the waypoints and ETA.. It then displays the ground track on the minichart.
Accuracy and reliability
The program is tested repeatedly each year and the results compared with those obtained by careful manual plotting.

I get lots of warnings if I try to install it.
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14022013, 10:22

#689

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
ok. I mada a scenario from the first scanned page Dockhead had up. as above in this thread too, where I posted it.
Here's the gpx...RENAME the downloaded ".doc" by removing the .doc from the end. It's a gpx file from OpenCPN's newer beta. unknown if it will display in your nav app.
RYAexample.gpx.doc
....so,
Start>Destination (A>B)=11 miles
Boat speed =4 knots
RYA calculation from the page I posted is 3 hours =12 miles and I got CTS =355.
current
0845 263 0.8kn
0945 266 1.3kn
1045 254 2.0kn
Departure Time is 0815
this is important. ...a half hour of 0.8 current=0.4 knots and boat speed vector is therefore 2 knots
blue diamond is RYA POS at 1045
the green lines are the vectors speed and drift for the times 845,0945,1045
The triangles represent where I think the boat will actually be nearer to given easy "sum/2" velocity averages ..except .the first half hour  .35 versus half hour current.4 so it's of no consequence anyways.
..You might say the RYA takes into account that the posted currents are not actual displacement. It is Possible, looking at this...
...You might also say, "THIS EXAMPLE IS FUDGED" by the author to make the example ( nearly!)work IS the halfhour before 0845 is suspicious? I'd say not because current is small.
However, I'd say the current is increasing on a three hour curve ( it does so hereabouts) and therefore the actual arrival at Destination is a bit more motoring against current than I'd like, but you could argue that it's near to shore, close counts, yada yada
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14022013, 11:31

#690

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
at 1045, (2 1/2 hours from start) most "methods" are going to be ~0 .85 miles from destination...? From this example, where current is increasing, it's the last mile  the last 15 minutes that is , aah,"important".
It's not clear to me in this example and other "reasonable" ones why the example boat won't just tweak her way home when she spys a mark.
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