Inaccurate RYA Teaching : CTS - Quest For a New Method

[Dockhead]

Originally Posted by nigel1

When I was studying coastal nav. we were taught that if your point D was a significant distance from your destination, and if tidal variations were significant, then to re work the problem using additional vectors to get closer to your destination.
I dont really know what the RYA teach as I have never done any of there courses, but the solution to working out CTS is based on the same methods used by the RYA.
Navigation is not just about knowing calculations etc, its also about using a bit of common sense

That seems eminently sensible!

[Seaworthy Lass]

Originally Posted by nigel1

When I was studying coastal nav. we were taught that if your point D was a significant distance from your destination, and if tidal variations were significant, then to re work the problem using additional vectors to get closer to your destination.
I dont really know what the RYA teach as I have never done any of there courses, but the solution to working out CTS is based on the same methods used by the RYA.
Navigation is not just about knowing calculations etc, its also about using a bit of common sense

Sounds like a good course .

The RYA teach a very standard formula for determining CTS. Common sense doesn't seem to come into it .

Even their instructors follow the procedure to the letter and some preach that its accuracy is only limited by the accuracy of the data .

[jackdale]

Originally Posted by nigel1

When I was studying coastal nav. we were taught that if your point D was a significant distance from your destination, and if tidal variations were significant, then to re work the problem using additional vectors to get closer to your destination.
I dont really know what the RYA teach as I have never done any of there courses, but the solution to working out CTS is based on the same methods used by the RYA.
Navigation is not just about knowing calculations etc, its also about using a bit of common sense

Absolutely!

As a a bit of an aside instructors are expected to teach the standards of their respective organizations. As an ISPA instructor evaluator, I expect instructor candidates and those who are advancing their status to teach the ISPA way of doing things. When I teach CYA or IYT course I follow those standards. There are differences in line handling, knots, MOBs, plotting symbols, and docking to name a few.

If you are going to take a test to determine if you meet a particular standard, follow that standard.

This is true in other areas as well. There are different protocols and standards for first aid.

[cal40john]

Originally Posted by Seaworthy Lass

The trip is not unfeasible by any means.
The speed made good is 3.4 knots (11 nm / 3.2 hours).

Using the RYA method would have shown the trip was not possible.
This is CERTAINLY not the case, although it seems I am not going to be enjoying any single malt anytime soon. Pity as it is rare as hen's teeth on the Greek islands .

I didn't say it was unfeasible, I said I wouldn't waste my time in that kind of current. Using the RYA method you travel for 4 hours. I plotted the ground track and found that you have to motor another hour to reach your actual destination.

Your method says 3.2 hours, I still would have stayed home for 2 hours, motored for less than 3 and gotten there 2 hours later than you. I'm not saying your method doesn't work, though I intend to look at it to see if I can figure out the math behind it.

JOhn

[Seaworthy Lass]

Originally Posted by Dockhead

I still don't understand your method but now have time to study it -- did you write a concise but comprehensive instruction? Can you point me to it? I'm lost in the hundreds of posts. I don't have access to a protractor until next week but will try to understand it abstractly.

It was explained in post # 2

[jackdale]

Originally Posted by Seaworthy Lass

The SMG using my method is the distance between A and B divided by the total time taken to get from A to B.
SMG = 11 / 3.2 = 3.4 knots, not an unfeasible journey at all for a boat only doing 4 knots.

This is huge flaw.

Use some common sense. The current is giving a boost throughout the entire voyage. You SMG cannot possibly be slower that your S.

Some basics

When constructing a vector the following are paired

• Set and drift
• Course and speed
• SMG and CMG

[Seaworthy Lass]

Originally Posted by cal40john

I didn't say it was unfeasible, I said I wouldn't waste my time in that kind of current. Using the RYA method you travel for 4 hours. I plotted the ground track and found that you have to motor another hour to reach your actual destination.

Well I wouldn't waste my time in that sort of current either if I had to rely on the RYA method for a CTS .

Originally Posted by cal40john

Your method says 3.2 hours, I still would have stayed home for 2 hours, motored for less than 3 and gotten there 2 hours later than you. I'm not saying your method doesn't work, though I intend to look at it to see if I can figure out the math behind it.
JOhn

I was motoring too. If you stayed home for two hours before departing the cross current would have been just as bad in the opposite direction and worst still would have been against you rather than with you!

So you would have literally gone backwards initially!

At the time I went although there was a fair amount of cross current, there was just as much pushing me to my destination.

And when you finally decided to go what method for determining CTS would you have selected?

[Seaworthy Lass]

Originally Posted by jackdale

This is huge flaw.

Use some common sense. The current is giving a boost throughout the entire voyage. You SMG cannot possibly be slower that your S.

Sorry, my data is actually correct.

The reason why my SMG is less than my boat speed is that although a component of the current is with me, there is also a component pushing me perpendicularly from the rhumb line.

Originally Posted by jackdale

Some basics
When constructing a vector the following are paired

• Set and drift
• Course and speed
• SMG and CMG

I am drawing 'distance displaced over one hour' vectors for both the boat and current, displaying the correct amount displaced and the direction of displacement.

The sum of these vectors for each hour shows the course over ground if the CTS is followed.

[Lodesman]

Originally Posted by Seaworthy Lass

Example 3
I have computed the CST for the example above using the SWL method.

Here is the data again.
Boat speed is constant at 4 knots throughout the journey.
You are motoring in flat water.
Destination is 11 nm due east.
Current:
1st hour: 8 knots 135 T
2nd hour: 6 knots 150 T
3rd hour: 2 knots 170 T
4th hour: 2 knots 10 T

What is the CTS and time taken?

Using the SWL method
Computations are plotted in the diagram below (note, these are usually made on a chart).
This was done according to the instructions in my second post.

The position of K on the last vector is determined by the proportion of SB relative to SB + BL.
Measuring this is comes to 1.1 / 1.1 + 4.9 = 0.2
As the last current vector displacement is 2 nm, it means you are subjected to it for 2 x 0.2 nm = 0.4 nm. Mark K at that spot on the vector.

Join K to B.

Measure the angle of the line KB. This is 9 degrees
CTS = 9 degrees true

Time taken is simply 3 hours plus 0.2 hours = 3.2 hours
(No need to even measure KB and divide by the boat speed as I said before to determine that, but if you want to double check by doing it:
Time taken = KB / boat speed = 12.8 / 4 = 3.2 hours)

Not to burst your bubble, but plotting the ground track shows your course of 009º overshooting your destination by 2 cables; and you actually pass it at 3h8m, vice 3h12.

Calculating using my method required replotting the CMG, but I had my course of 008º (and some fractions, I rounded down) with just one refinement. The rough first course was 010º.

It was a rather tortured example to showcase your method - a prudent sailor probably would have waited 2 hours then drove straight East for 2h33m to arrive (you didn't give the 5th hour's tidal set - I would like to see what sort of set you would envision that could not be mitigated in the first two hours where you can actually go faster than the stream?).

[Seaworthy Lass]

The red dotted line on this plot gives the ground track.

The result is computed by adding
1. the boat displacement due to boat speed over one hour - constant at 4 nm at a constant direction of 9 degrees (indicated by a thick black line)
TO
2. the boat displacement due to current over one hour a variable: 8 nm, then 6 nm, then 2 nm, then 2 nm in the direction shown (indicated by a dotted blue line).

In the last 0.2 hours of the journey the current direction almost coincides with the speed of the boat direction. This last distant displaced is short as it is only for 0.2 hours.

[jackdale]

A realistic scenario, not unusual for us.

Depart at 1200

Set and drift

1200 1 knot 000 T
1300 0.50 knot 320 T
1400 0.0
1500 0.0

Boat Speed 4.0
CMG = 238 T

Distance 16.46 miles

[Dockhead]

Originally Posted by Seaworthy Lass

It was explained in post # 2

I'm still not entirely sure I understood it - still working on it. But if it consists of what I think it does, then it is truly brilliant. And if MISUNDERSTOOd it, then I have the makings of a truly brilliant system of my own .

It exactly uses the analogue method which has existed for hundreds of years to overcome the lack of number crunching power in the pre-digital age . Brilliant simplicity. Fantastic; truly inspired. Unbelievable, really. I'm in awe (pending final confirmation of my understanding of it ).

What it does is simply finds the proportionate average set and drift of the full hour of which the final partial hour is a part, and uses that as the final jink of the vector sum leg.

It is - again, if I understood it correctly - mathematically perfect IF we assume that any part of a whole hour is average. This is definitely not true in reality, but it's as good as we can get without assuming something about the shape of the current development through time.

In any case, it is simple to apply, and the built-in error is approximately an order of magnitude smaller than what is produced by the RYA method. This is really good stuff

I tank we might do even better with a sine curve model of tidal currents, without any hourly analysis at all. But that still has to be developed. . .

Bravo, Seaworthy

[Seaworthy Lass]

Originally Posted by Lodesman

Not to burst your bubble, but plotting the ground track shows your course of 009º overshooting your destination by 2 cables; and you actually pass it at 3h8m, vice 3h12.

Well I would not have overshot it LOL, I would have stopped when I got there. If your calculations are correct, being 4 minutes out with my time estimate is insignificant. I have rounded figures off to the closest 0.1 nm as this was a limitation of my measurements, so that is how the small discrepancy occured.

Had my boat not been violently knocked about by 30+ knot gusts (which it still is) maybe I would have had a better plotting accuracy.

Originally Posted by Lodesman

Calculating using my method required replotting the CMG, but I had my course of 008º (and some fractions, I rounded down) with just one refinement. The rough first course was 010º.

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?

All contributions are welcome .
I am on a quest to find a better method than the RYA's and to have it considered, not to promote my method.

Originally Posted by Lodesman

It was a rather tortured example to showcase your method - a prudent sailor probably would have waited 2 hours then drove straight East for 2h33m to arrive (you didn't give the 5th hour's tidal set - I would like to see what sort of set you would envision that could not be mitigated in the first two hours where you can actually go faster than the stream?).

Had the 'prudent' sailor waited two hours I think he would not have made any progress against the current (in fact he would have gone backwards).

[Seaworthy Lass]

Originally Posted by Dockhead

I'm still not entirely sure I understood it - still working on it. But if it consists of what I think it does, then it is truly brilliant. And if MISUNDERSTOOd it, then I have the makings of a truly brilliant system of my own .

It exactly uses the analogue method which has existed for hundreds of years to overcome the lack of number crunching power in the pre-digital age . Brilliant simplicity. Fantastic; truly inspired. Unbelievable, really. I'm in awe (pending final confirmation of my understanding of it ).

THANKS Dockhead
It sounds like you have understood it!

[goboatingnow]

I'm back .
Chores are finished for the day and armed with a cup of coffee rather than a glass of wine, I have drawn up what would be drawn on a chart for the example I set the RYA instructor in the other thread where all this discussion started.

Details:
Boat speed is constant at 4 knots throughout the journey.
Destination is 8.5 nm due east.
Current is always from the north:
1st hour 3 knots
2nd hour: 2 knots
3rd hour: 0.5 knots

MY METHOD
Step 1: Mark A and B on the chart

Step 2: Distance from A to B = 8.5 nm (specified in this case, but you would usually measure it)

Step 3: estimate time taken by dividing distance by your speed (8.5 / 4)
It is more than 2 hours, so you know at least three hours of current vectors need to be drawn.

Step 4: determine current (specified in this example, see above)

Step 5: Mark the current vectors starting from A, adding each one to the tip of the previous. In this case nice and easy, the are all running due south)

Step 5A: Check at the 3 hour mark you have gone past B (measure 3x4= 12 nm from the end of the 3 hour current vector)
- Check at the 2 hour mark you haven't reached B (measure 2x4= 8nm from the end of the 2 hour vector)

Step 6: Draw a line from the end of the 2 hour vector and mark off 8 nm and label that S

Step 7: Draw a line from the end of the 3 hour vector and mark off 12 nm and label that L

Step 8: Calculation to determine the proportion of the final current:
Distance from S to B divided by the distance from S to L = 0.5
Position of K = 0.5 x (length of last current vector) = 0.5 x 0.5 = 0.25
Mark K at this point

Step 9: Join K to B.
CTS as measured on chart: 58 degrees
Time taken = length KB/ boat speed = 10/4 = 2.5 hours

SO heres a summary of SWL method, Its the RYA method up to the Step 8.

Its worth noting that either by design or luck , this example puts B the destination almost midway between the two 2 and 3 tidal plots

Note the only difference between SWL methods and the RYA, is that the RYA would take a 3 hour plot and "deflate" all tides by the same proportion of minutes, where as SWL applies this to the last tide only.

IN practice and in the classroom, I have pointed out the issue to students where B ends up almost in the middle of the two plots, and have suggested that the answer just be averaged out forexample , my dodgy protractor gives me a CTS from D2(S) as 53 and a CTS from D3(L) of 64, A quick visual inspection shows B in approx the middle, so add 5.5 to D2(s) gives us a CTS of 58.5 - Voila.

In real life none of this matters, as thats not what happens, as D(S or L) comes either reasonably close ( as a proportion of the last tide as hence can in effect be ignored, or is very far away and more tides have to be plotted.

For example , its clearly nonsense to take a one hour tidal plot and expand it to 3 hours proportionally , ( Though in practice in many places the tides don't change that much over three hours.)

Computing CTS of course to this level of accuracy is of course ridiculous anyway.

Again, I state that the RYA method is mathematically correct, That means its "triangles add up". Thats statement holds. WHat we are arguing here is the correlation between a mathematical model and the real world.

so lets have that discussion ( I said this in the other thread)

In reality SWL method provides no greater accuracy then the RYA method if that . like all methods is intelligently applied, The real world conditions make a mockery of any such accuracy anyway

In the classroom, there is always a spirited discussion on CTS, As I said many times , I don't like or recommend multi hour CTS computations anyway. and in reality with changing boat speeds, leeway etc, One never does what SWL does, You end up recomputing often several times and in effect virtually arrive at a one hour plot anyway.

The multihour CTS , simply provides too many opportunities to get the ground track wrong with disastrous conclusions