Inaccurate RYA Teaching : CTS - Quest For a New Method

[Seaworthy Lass]

Originally Posted by HappySeagull

...then what are the displacements you want to use? and I will send Hapless out again. I don't expect him to beat anybody, but I am curious as to what advantage can be displayed graphically.

Hapless, there was no significant advantage of the SWL method for this example LOL. All I was doing was giving step by step instructions with step by step diagrams for those that had not followed my written instructions previously. The RYA method does produce similar results to mine if D happens to fall very close to B (in some cases the closest D can be is several nm away from B and here results can be way out).

It is 1am here now and I am off to bed, but I will post an example for you to work through tomorrow with clearly specified data in whole hours at the beginning so that you can compare the results of the two methods .

[jackdale]

Originally Posted by Seaworthy Lass

Hapless, are you still with us?

I won this race and have been in the bar sipping celebratory drinks for at least 10 minutes before you and Jackdale got in .

Your 10 step model (maybe it should the 12 Step) takes way longer to compute. I was half way to the destination before you even left the dock. BTW I even stretched mine out somewhat. Check the PPS that I posted earlier, way fewer steps.

[jackdale]

Originally Posted by Lodesman

At 4kts, you cover a distance of 4 cables in 6 minutes.

OK

[Lodesman]

If we assume your tidal figures of 1.3nm the first half-hour, 2nm the next hour and 0.4 for the last half-hour, for a total 2-hour vector 3.7nm long; and an 8nm boat vector, we get the mark on the "rhumb line" 7.1nm along. Time required to get to 6.8nm is 6.8/7.1*120= 114.9 mins.

Recalculate tidal vector, reducing the final 0.4 to 24.9 mins worth 0.4*24.9/30=0.3, therefore tidal vector is 1.3+2+0.3=3.6nm and;
boatsp vector is 8*114.9/120=7.7nm.

Confirm plot - sqrt((7.7)² - (3.6)²) = 6.8nm
Solve for angle: asin(3.6/7.7) = 27.9º. Course 090-27.9= 062.1º

Course 062º for 1h54.9m. No drawing needed.

[jackdale]

SWL - I using the numbers that you provided.

Had I been using tide and current tables, I would have used them to ascertain the strength and time of the maximum as well as to the time of the turn (slack) and use either Happy's formulas or my own worksheet for intermediate currents.

This is what we get

http://tides-marees.gc.ca/Content/ti...100_public.pdf

Can we include that information next time?

[HappySeagull]

Hapless doesn't read instructions. He (and I!!)just wants the drift as a sum because"Calculating displacement" might be a hundred page thread. This is why I covered his *ss with :

"destination bears 90T 6.8mi
displacement hrs 1,2,3 all 180T: 2.8mi, 2mi, 0.8mi
Note that Hapless is just tooling along by considering 1 hour increments of tide displacement...
after 2 hours, Hapless is a little behind Jackdale who has a longer view...
(Hapless bears 222T@150meters from Jackdale) after 2 hours"
so, I had Jackdale to race, "hourly" versus a sum.
Those displacements aren't unreasonable?,compared to some I could dredge up.
IF a "rule" is used to calculate displacement, the rule must be argued etc etc. It's too much, although I think half-hour "drifts" would make better plots.

"Hapless" is supposed to lose. It's the plot, and by how much that's interesting (to me).

noted:
If the SWL method has "no significant advantage" over the RYA (and whatever Hapless does) where the current is decreasing, that's half the time.
That is not encouraging.

However,
I do not recall a single example where the current is INCREASING (despite what I thought was a tempting tale of a vessel being sucked back into Race Passage for want of an application... )
I am hoping for a much better "race" and significant result.

[Seaworthy Lass]

Originally Posted by HappySeagull

If the SWL method has "no significant advantage" over the RYA (and whatever Hapless does) where the current is decreasing, that's half the time.
That is not encouraging.

It is the combination of numbers, not whether or not the current is increasing or decreasing that produce errors in the RYA method

[Seaworthy Lass]

Originally Posted by HappySeagull

I do not recall a single example where the current is INCREASING (despite what I thought was a tempting tale of a vessel being sucked back into Race Passage for want of an application... )

In example 3 of mine in this thread the current had changed direction. I can make this increase at this stage and still produce a result for the RYA that is 30+ degrees in error. I will do this later just to demonstrate this .

I never got back to finishing the "sucked back" example for Bewitched. I will do this at some stage too. Come spring and with winter chores behind me, I will have a lot more time on my hands .

[Seaworthy Lass]

Originally Posted by HappySeagull

Hapless doesn't read instructions. He (and I!!)just wants the drift as a sum because"Calculating displacement" might be a hundred page thread.

I will give hourly averages for the next few examples so that there is no confusion, but in the real world Hapless does need to consider what to do with the tidal stream data in the way it is presented, particularly as Hapless will not always leave exactly at the beginning of one lot of average data

Originally Posted by HappySeagull

"Hapless" is supposed to lose. It's the plot, and by how much that's interesting (to me).

We can't have Hapless losing all the time, he will get far too discouraged and give up racing or worst still he will start crabbing along the rhumb line to get anywhere LOL.

So here is an example where Hapless is right there at the finish line with SWL (maybe even a whisker ahead since I am limited by the sharpness of my now stubby pencil).

To avoid criticism of making up unrealistic tidal streams, I have selected data from the example Jackdale gave in post #383 of this thread.

I have made it very simple so that anyone trying to follow my method for the first time after my lengthy explanation a few posts back can do so easily.

It is also very simple to check this one mathematically as the last hour is slack water, so no proportions of tidal displacement need to be calculated for that (mathematical calculations get more complex with tidal streams that are not conveniently perpendicular and are varying in direction and if proportions of the last hour need to be determined).

Example 6 (I think, losing track now):

Heading due south 6.7 nm
Conditions are calm, no wind, flat seas, and you are motoring in a small boat at 3 knots to conserve fuel.
Tidal stream:
Average for the first hour: 2.2 knots west
Average for the second hour: 1.3 knots west
Average for the third hour: zero (slack water)

What is the CTS (true) and time taken:
Using the RYA method?
Using the SWL method?
Calculating it mathematically? (this only a minute or two for this example and will give Hapless the opportunity to be in the lead).

Sunrise this morning distracted me from posting this earlier and my numb fingers could not grasp a protractor very easily until they had warmed up on a couple of cups of coffee

I am sharing the view (unphotoshopped) for all of you to enjoy:

[Wotname]

Originally Posted by Seaworthy Lass

...........
Sunrise this morning distracted me from posting this earlier and my numb fingers could not grasp a protractor very easily until they had warmed up on a couple of cups of coffee .......

Thread drift alert - 'bout time anyway

Just what are the am temps where you are - I ask as we are currently getting mins of 25 and maxes of 40 (all C); numb morning fingers would be nice for a change.

[Seaworthy Lass]

Originally Posted by Wotname

Thread drift alert - 'bout time anyway

Just what are the am temps where you are - I ask as we are currently getting mins of 25 and maxes of 40 (all C); numb morning fingers would be nice for a change.

It was a chilly ten degrees outside this morning. Temps had dropped overnight as a slight north westerly has temporarily kicked in. It felt a lot worse as I didn't have a lot on LOL and the sun took FOREVER to peek over the horizon and I wanted to snap a few shots.

It has been reaching 16, 17, and even 18 or 19 here lately (61-66 F). Best winter we have had yet in the Med! I wouldn't mind a taste of 40 again though.

It is 10:30 and about 16 on board. May be time to get up .

[Wotname]

[LJH]

Originally Posted by Seaworthy Lass

I will give hourly averages for the next few examples so that there is no confusion, but in the real world Hapless does need to consider what to do with the tidal stream data in the way it is presented, particularly as Hapless will not always leave exactly at the beginning of one lot of average data



We can't have Hapless losing all the time, he will get far too discouraged and give up racing or worst still he will start crabbing along the rhumb line to get anywhere LOL.

So here is an example where Hapless is right there at the finish line with SWL (maybe even a whisker ahead since I am limited by the sharpness of my now stubby pencil).

To avoid criticism of making up unrealistic tidal streams, I have selected data from the example Jackdale gave in post #383 of this thread.

I have made it very simple so that anyone trying to follow my method for the first time after my lengthy explanation a few posts back can do so easily.

It is also very simple to check this one mathematically as the last hour is slack water, so no proportions of tidal displacement need to be calculated for that (mathematical calculations get more complex with tidal streams that are not conveniently perpendicular and are varying in direction and if proportions of the last hour need to be determined).

Example 6 (I think, losing track now):

Heading due south 6.7 nm
Conditions are calm, no wind, flat seas, and you are motoring in a small boat at 3 knots to conserve fuel.
Tidal stream:
Average for the first hour: 2.2 knots west
Average for the second hour: 1.3 knots west
Average for the third hour: zero (slack water)

What is the CTS (true) and time taken:
Using the RYA method?
Using the SWL method?
Calculating it mathematically? (this only a minute or two for this example and will give Hapless the opportunity to be in the lead).

Sunrise this morning distracted me from posting this earlier and my numb fingers could not grasp a protractor very easily until they had warmed up on a couple of cups of coffee

I am sharing the view (unphotoshopped) for all of you to enjoy:

Great view, thanks. 11 degrees and rain here
Using SWL on OpenCPN 152 for 2:31
Maths:
Crse = 90= (tan-1(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

[Seaworthy Lass]

Originally Posted by LJH

Great view, thanks. 11 degrees and rain here
Using SWL on OpenCPN 152 for 2:31
Maths:
Crse = 90= (tan-1(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

10/10 LHJ!

Does anyone want me to post diagrams or are you all getting the hang of it?

The SWL method gives the same result as calculating it mathematically (with the limitation of the care you take).

The RYA does not give the correct result regardless of which hour is selected. I doubt very much that the 3rd hour would have been chosen by the RYA, as if I understand correctly from how the method is taught, overshooting deliberately is not the favoured thing to do. Anyway, although the CTS is closer using three hours of data rather than two (157 instead of 144 when it should have been 152), the predicted time is out.

As one instructor pointed out, checking which hour to use is not consistently taught and certainly not given in examples or test questions:

Originally Posted by goboatingnow

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.

Either way, following the RYA CTS result of either 144 degrees or 157 degrees true would not actually see you ever arriving at B.

[HappySeagull]

Heading due south 6.7 nm
Conditions are calm, no wind, flat seas, and you are motoring in a small boat at 3 knots to conserve fuel.
Displacement(Drift):
first hour: 2.2 miles west
second hour: 1.3 miles west
third hour: zero (slack water)

(it's getting to be a peeve, but don't worry, I'll get over it.
Hapless is doin' pretty good just stayin' inside his attention span.
If I could get his attention away from feeding seagulls, I daresay he could do better.)
2 hours 27 minutes