Inaccurate RYA Teaching : CTS - Quest For a New Method

[Andrew Troup]

Never mind, Lodesman.

I don't know where to start.

There's nothing about your posts about other people's understanding of the relationship between vectors and triangles that I can relate to anything anyone has posted.

Some of your posts make so much sense, this must surely be just an isolated misunderstanding.

If you care about it, you can clear it up; just re-read any posts containing the word "Triangle", starting with the first, and you'll see how we got to here.


I frankly don't know that it's worth it: here's a potted synopsis:

I'm trying to prise Dockhead's fingers off his fixed notion that Seaworthy's method cannot possibly have arisen from her realising the following:

The reliance of the RYA method on the rhumb line is the root cause
of the two distinct problems it experiences with certain scenarios.

(one of which problems - the serious one - we have not yet got to ....
I'm starting to appreciate that SwL's appetite for generating scenarios is limitless,
so I hope I am still alive when we do. < rueful grin > )

Dockhead said, some time ago, that her method could not possibly relinquish the RYA reliance on the rhumb line, because it forms one leg of a triangle which is indispensable to a solution.

My contention is that the triangle he refers to, in fact ANY triangle, is only indispensable, in the SwL method, to those for whom triangles are inseperably bound up with vectors. (To be completely clear: I am NOT such a person)

And no one has yet shown any flaw in my whimsical proof of that, involving a small dinghy puttering across the chart equipped only with a tape measure, a watch, a protractor and their eyes, exactly duplicating the indispensable elements of Seaworthy's method, with zero recourse to triangles. Or the rhumb line.

The participants do not at any stage KNOW where the rhumb line is, or how long it is, and yet they solve the CTS, the elapsed time for the passage, and the distance through the water.

[Lodesman]

My method. Call it the RYA method corrected for the final part hour.

Originally Posted by Seaworthy Lass

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; draw line from A through B

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 6: Set compass to measure CTS vector (speed 4kts x 3 hours = 12nm); put the point on the end of the 3-hour composite tide vector and mark the line AB (in this case beyond B); call this point C.

Step 7: measure AC (10.7nm); divide by 3 hours to get SMG of 3.6kts. Calculate time to cover 8.5 nm at 3.6kts = 2h23m

Step 8: refine tidal set vector - ie. third hour vector shortened to 23 mins (23m at 0.5kts = 0.2nm)

Step 9: refine CTS vector (speed 4kts x 2h23m = 9.5nm); set the compass to that length, reposition it on the previously shortened set vector and mark AB - call this point D. In this case D falls short of B, so we further refine:

Step 10: restarting at step 7, measure AD (8nm); divide by 2h23m to get SMG of 3.4kts. 8.5nm at 3.4kts SMG = 2h32m

Step 11: refine tidal set vector - third hour of tide 0.5kts x 32mins = 0.3nm

Step 12: refine CTS vector (4kts x 2h32m = 10.1nm); replot with compass.

It falls a tenth of a mile past B on the line; close enough for this purpose. To get closer, you'll need to measure distances to hundreths or thousandths and angles to fractions of degrees. The CTS when measured is 058º and it takes 2h32m.

[boatman61]

$hyte.... glad I'm not a Yachtie...

[Dockhead]

Originally Posted by Lodesman

You've said it twice:





Near as I can tell Dockhead mentioned "vector triangle" which is clear from the context means "triangle using 3 vectors as sides".

Is that a structural problem or a weak link?

I did say vectors are triangles, and I was wrong - brain fart. I have to claim responsibility for that particular stupidity. Momentarily confused vector triangles with the vectors themselves :sheepishgrin:.

It was part of a conversation where SWL and Andrew were claiming that the course line plays no role in SWL's calculation, and that the course line is "the whole problem" with the RYA method. They claimed that there are "no triangles in sight" in SWL's method.

I finally proved to SWL's satisfaction that she uses the same vector triangles as RYA and all the rest of us use. Andrew still thinks, apparently, that some kind of magic is involved

It's a big mistake to think that the course line has anything to do with the shortcomings of the RYA method. sWL uses the course line in exactly the same way (as she now sees, I think). It's the same method, but with a brilliant (IMHO) refinement which greatly reduces the inherent error with no additional effort. Dave doesn't admit this last part, but it really is a mathematical fact which can be shown fairly easily.

[Paul Elliott]

Originally Posted by Andrew Troup

Dockhead

The position of B is not determined by distance.

It's determined by being a point we are trying to reach.

The positions of all three points (A, B, and the end of the combined current vector) are determined by distance and angle. You can't have one of these without all the others. The current vectors start at "A", so you have to know where "A" is. You have to know the distance between "A" and "B", otherwise how do you know where to start the current vector? Claiming anything else is pretty silly. I don't care if you draw the line AB or not, you still need to know the distance.

All this should be quite obvious, and no doubt to many of us it is, but we have some people arguing about the underlying mathematic/geometric principles, and others arguing about the implementation, or construction. They are arguing apples and oranges, without knowing it.

The discussion of "inflation" of the entire passage vs interpolation or extrapolation of the last hour seems to be the root issue here (as well as the effects of the underlying accuracy of the available data). This can be considered without getting into the mechanical details of the solution. The actual pencil and paper technique for arriving at the solution, and appropriate simplifications, approximations, and shortcuts, should be isolated from the "higher level" discussion.

Just a suggestion,

[Seaworthy Lass]

Originally Posted by boatman61

$hyte.... glad I'm not a Yachtie...

Hiiiiiii Boaty
Welcome to the current discussion. Great to see you here
My initiation to CF has been trial by fire LOL.

Bet you always plan passages well, so I assume like me you have never used the RYA method for determining CTS?

[Seaworthy Lass]

Originally Posted by Dockhead

I finally proved to SWL's satisfaction that she uses the same vector triangles as RYA and all the rest of us use. Andrew still thinks, apparently, that some kind of magic is involved

NOOOOOOOOOOOO. I do not use the same vector triangles as the RYA.
And I admitted that you could think of vectors as triangles for the purposes of mathematical calculations. But you could also think of them as rectangles or rhomboids or whatever you fancied (not that this would help with mathematically solving anything LOL).

The only time I work with triangles is right at the end after I have computed the CTS and I want to plot my ground track .

Originally Posted by Dockhead

It's a big mistake to think that the course line has anything to do with the shortcomings of the RYA method. sWL uses the course line in exactly the same way (as she now sees, I think). It's the same method, but with a brilliant (IMHO) refinement which greatly reduces the inherent error with no additional effort. Dave doesn't admit this last part, but it really is a mathematical fact which can be shown fairly easily.

The rhumb line has EVERYTHING to do with the shortcomings of the RYA method. Try working through example 3 using the rhumb line as a reference - you are stuffed if you do.

My lines from the current are directed at B, the destination. There is no D in my method at all. I ignore the rhumb line totally.

I think what you are doing when you use my method, is the procedure I followed in the first thread where I was looking at the proportion of the under or overshoot when the distance vector was arced off at the rhumb line. This method will fail. You need to join the current vector to B and then arc off the distance travelled and then compare the under and overshoot distances.

THERE IS NO RHUMB LINE IN MY METHOD!

[Seaworthy Lass]

Originally Posted by Paul Elliott

The discussion of "inflation" of the entire passage vs interpolation or extrapolation of the last hour seems to be the root issue here (as well as the effects of the underlying accuracy of the available data). This can be considered without getting into the mechanical details of the solution. The actual pencil and paper technique for arriving at the solution, and appropriate simplifications, approximations, and shortcuts, should be isolated from the "higher level" discussion.
Just a suggestion,

Paul, you are right (the voice of reason LOL). The waters are being muddied by whether or not triangles are involved.

The main point to consider is that reference to the rhumb line, not the destination is the main source of the problem with the RYA method.

We need to focus on this (see example 3 where the RYA method was 37 degrees out with the CTS it computed).

This is not because the average current at the four hour mark was significantly different to the average at the 3 hour mark. It was because the RYA method insisits you NEVER NEVER NEVER arc off the distanced travelled vector towards B. they say you MUST do it toward the rhumb line, giving truly ridiculous results in some circumstances as I have shown in example 3 .

[Seaworthy Lass]

Originally Posted by Lodesman

My method. Call it the RYA method corrected for the final part hour.


Step 6: Set compass to measure CTS vector (speed 4kts x 3 hours = 12nm); put the point on the end of the 3-hour composite tide vector and mark the line AB (in this case beyond B); call this point C.

Step 7: measure AC (10.7nm); divide by 3 hours to get SMG of 3.6kts. Calculate time to cover 8.5 nm at 3.6kts = 2h23m

Step 8: refine tidal set vector - ie. third hour vector shortened to 23 mins (23m at 0.5kts = 0.2nm)

Step 9: refine CTS vector (speed 4kts x 2h23m = 9.5nm); set the compass to that length, reposition it on the previously shortened set vector and mark AB - call this point D. In this case D falls short of B, so we further refine:

Step 10: restarting at step 7, measure AD (8nm); divide by 2h23m to get SMG of 3.4kts. 8.5nm at 3.4kts SMG = 2h32m

Step 11: refine tidal set vector - third hour of tide 0.5kts x 32mins = 0.3nm

Step 12: refine CTS vector (4kts x 2h32m = 10.1nm); replot with compass.

It falls a tenth of a mile past B on the line; close enough for this purpose. To get closer, you'll need to measure distances to hundreths or thousandths and angles to fractions of degrees. The CTS when measured is 058º and it takes 2h32m.

Lodesman, could you please draw a few step by step diagram using the data in example 3, not example 1 to show us how this works? This would be very helpful.

And why mark the distance off the rhumb line, why not directly toward B and see how much it overshoots? Then you could try half that amount, then half again etc, until you get close.

Either way, that would take AGES!!!!!

How many times do you have to keep stabbing guesses before you arrive at the correct answer of the amount of time you a subjected to the last lot of current?

No need to stab any guesses with my method .
I just just do it twice, once for the beginning and once for the end of the last current vector, then I look at the proportion of the undershoot compared to the overshoot. If they are equal for example, then you use half the hour. Simple as that, no trial and error until you can eventually get close as you are suggesting.

[Wotname]

@SWL,

I admit that I haven't worked out all your examples but from the ones I have, your method as it appears to me is a simple navigation exercise that one would do to solve any nav problem where there is an outside influence that would take the moving object (boat in this case) away from it's intended ground track and where you wish to steer a single CTS.

Whether that outside influence is a fixed or varying current, a single tide or mutli tides, leeway, windage, sea state influence or whatever is immaterial; providing it can be predicted, it can be accounted for vectorially.

In this sense you have re-invented the wheel but what stands out for me is that the RYA and it's method (at least as described here on CF) never knew the wheel ever existed.

You have shown them your wheel and they don't like it . However others have been using this wheel for centuries (or more). You have done well to work out your wheel independently without knowing that such a thing existed.

If I have confused your method, then forgot what I just posted .

[LJH]

Good Morning, SWL asked me to show her method used on CPN to find the CTS. (note the A-B course line is not drawn, measured or used)

Steps:
1. Plot A and B (triangles on attachments)
2. Use the measure feature to plot the tide set for each period (magenta on attachments) (A-C1, C1-C2, C2-C3, C3-C4)
3. Use the measure feature from point C4 through B to 16.0 miles and drop a mark there (this is L)
4. Use the measure feature to measure 12 miles from C3 towards B and drop a mark there (this is S)
5. Use the measure feature to measure from L to B to S. This will give you the ration SB/SL = 1.1/6.1)
6. Apply that ratio to the last current to measure .36 miles from point C3 toward C4. Drop a mark there. This is K, the blue diamond on the attachments.
7. Create a route from K to B and view route properties. Enter your boat speed and it will show 010 degrees, 12.8 nm and 3 hrs 12 mins.
8. To draw ground track use the measure function to measure from the tide points parallel to the K-B course (010) for the distance travelled (e.g. From C1 4nm, from C2 8nm, etc). These are the red “X”s.
9. Make a route from A, to B by joining the red Xs in between and you get the ground track (red line).
10. If you wish to you can transfer this route to your chart plotter and monitor you progress hourly relative to the red track, or just monitor you XTE.

[mcarling]

Originally Posted by Paul Elliott

The discussion of "inflation" of the entire passage vs interpolation or extrapolation of the last hour seems to be the root issue here (as well as the effects of the underlying accuracy of the available data). This can be considered without getting into the mechanical details of the solution. The actual pencil and paper technique for arriving at the solution, and appropriate simplifications, approximations, and shortcuts, should be isolated from the "higher level" discussion.

Yes. The essential difference between the RYA method and the Seaworthy Lass method is that the former is based on extrapolation while the latter is based on interpolation. Interpolation tends to be more accurate than extrapolation if done correctly, at the cost of being slightly more prone to the introduction of human error in implementation.

Anyway, excellent work by Seaworthy Lass. If the Chief Instructor position opens up at RYA ....

[Seaworthy Lass]

Originally Posted by Wotname

@SWL,

I admit that I haven't worked out all your examples but from the ones I have, your method as it appears to me is a simple navigation exercise that one would do to solve any nav problem where there is an outside influence that would take the moving object (boat in this case) away from it's intended ground track and where you wish to steer a single CTS.

Whether that outside influence is a fixed or varying current, a single tide or mutli tides, leeway, windage, sea state influence or whatever is immaterial; providing it can be predicted, it can be accounted for vectorially.

In this sense you have re-invented the wheel but what stands out for me is that the RYA and it's method (at least as described here on CF) never knew the wheel ever existed.

You have shown them your wheel and they don't like it . However others have been using this wheel for centuries (or more). You have done well to work out your wheel independently without knowing that such a thing existed.

If I have confused your method, then forgot what I just posted .

Wotname, this is EXACTLY what I am doing, no confusion. Very simple vector addition.
The only dilemma is how much of the last hour to consider. The amount you are displaced by the current in this time is depended on your CTS and your CTS is dependent on the amount of time subjected to the current. This is solvable for a minimum KB length using differential equations, but I have a simple way of determining it.

Just compare the amount of undershoot towards B at the beginning of that hour of current, compared to the overshoot at the end. Look at the size of the undershoot compared to the total under and overshoot.
If it is a half, mark a half of the last current vector and label that K.
If it is a quarter, mark a quarter etc etc. you could just eyeball it, but you would have greater accuracy measuring it.

Draw a line between K and B. That is your CTS
The time taken is the number of current hours used plus the proportion of the last one. Oh so easy

[Seaworthy Lass]

Originally Posted by mcarling

Yes. The essential difference between the RYA method and the Seaworthy Lass method is that the former is based on extrapolation while the latter is based on interpolation. Interpolation tends to be more accurate than extrapolation if done correctly, at the cost of being slightly more prone to the introduction of human error in implementation.

Anyway, excellent work by Seaworthy Lass. If the Chief Instructor position opens up at RYA ....

They couldn't afford me LOL.

They are very lucky to be getting a markedly superior method of determining the CTS at no charge whatsoever (although I am still hopeful that a case or at least a bottle of Scotch may come my way ).

[Wotname]

Originally Posted by Seaworthy Lass

They couldn't afford me LOL.

They are very lucky to be getting a markedly superior method of determining the CTS at no charge whatsoever (although I am still hopeful that a case or at least a bottle of Scotch may come my way ).

There is that dry SOH again