Inaccurate RYA Teaching : CTS - Quest For a New Method

[Seaworthy Lass]

Example 4
RYA method results:

In the last example I was criticised by the RYA instructor for selecting an unrealistic situation. He said the current was too high (8,6,2 then 2 knots) and this would never be encountered and the only reason the boat could get there was because the tide turned at the end.

So in this example the current is as follows:
Ist hour: 3 knots 135 T
2nd hour: 1.5 knots 135 T
3rd hour: 0.5 knots 135 T

So nothing weird about this is there?

Boat speed = 1.8 knots
AB = 4.24 nm due east

You have a short journey 4.24 nm due east to get to B, which is a spot you consider safe to be before you proceed into the harbour.

This is an emergency situation and you must get there ASAP. Conditions are calm, there is no wind and the water is flat.
Unfortunately your prop is damaged and your maximum boat speed is 1.8 knots (your know this from the last trip).

Can you get to your destination to deal with this emergency?
It depends on the current for the next few hours (when you check it will actually reverse and act at 315 T, increasing in strength), so I doubt it.

What is the CTS?
That is a real dilemma. You arc off your distance after 3 hours and find it it cuts the rhumb line twice, equidistant from B.
Oh! What to do? Is it 39 degrees or is it 321? I guess if I had to pick I would choose 39 if you thought you could get there, but perhaps an RYA instructor could answer that.

What is the estimated time taken?
Time taken = AD divided by the boat speed, so it depends on which D you choose.

Anyway, you get totally wacky results using the RYA method!

[Paul Elliott]

Originally Posted by Seaworthy Lass

Close

Oops. I now see that I was using a distance AB of 4.2 miles, and not the specified 4.24 miles. You know what they say: GIGO (Garbage In, Garbage Out).

[jackdale]

I am sorry. I would not take a boat capable of 1.8 knots into a 3 knot current. Especially one with a damaged propeller, or any other equipment, on which I cannot rely. I would be inviting an emergency.

BTW I am NOT an RYA instructor. Just about everything but.

[Paul Elliott]

Originally Posted by jackdale

I am sorry. I would not take a boat capable of 1.8 knots into a 3 knot current. Especially one with a damaged propeller, or any other equipment, on which I cannot rely. I would be inviting an emergency

No doubt true, and prudent. But let's say that a deranged terrorist was forcing you to do it, and was pointing a gun at my head, threatening to shoot if you didn't make the trip as specified. I'm a nice guy. You wouldn't want anything bad to happen to me, right?

[jackdale]

Originally Posted by Paul Elliott

No doubt true, and prudent. But let's say that a deranged terrorist was forcing you to do it, and was pointing a gun at my head, threatening to shoot if you didn't make the trip as specified. I'm a nice guy. You wouldn't want anything bad to happen to me, right?

I assess the terrorists I invite on board. I exclude deranged ones.

[Paul Elliott]

Originally Posted by Paul Elliott

Oops. I now see that I was using a distance AB of 4.2 miles, and not the specified 4.24 miles. You know what they say: GIGO (Garbage In, Garbage Out).

And, I was using 1st hour current of 3.5 kt, not 3kt. Totally screwed up that one!

[TeddyDiver]

Originally Posted by Paul Elliott

And, I was using 1st hour current of 3.5 kt, not 3kt. Totally screwed up that one!

No you didn't

"So in this example the current is as follows:
Ist hour: 3.5 knots 135 T
2nd hour: 1.5 knots 135 T
3rd hour: 0.5 knots 135 T

So nothing weird about this is there?

B is due east of A, 4.24 nm away
Boat speed = 1.8 knots (see explanation below as to why it is low)"

That was the data as presented originally..

[Seaworthy Lass]

Originally Posted by Paul Elliott

And, I was using 1st hour current of 3.5 kt, not 3kt. Totally screwed up that one!

Oooooh, it was 3.5 and all the calculations on the chart were for 3.5
And the original write up was 3.5 knots in post # 354

The . 5 just mysteriously disappeared when I copy and pasted it in post # 361

Can't edit the text now

SORRY TO CONFUSE ANY OF YOU

[Seaworthy Lass]

Example 4
SWL method results:

In the last example I was criticised by the RYA instructor for selecting an unrealistic situation. He said the current was too high (8,6,2 then 2 knots) and this would never be encountered and the only reason the boat could get there was because the tide turned at the end.

So in this example the current is as follows:
Ist hour: 3.5 knots 135 T
2nd hour: 1.5 knots 135 T
3rd hour: 0.5 knots 135 T

So nothing weird about this is there?

You have a short journey due east of roughly 4 nm to get to B, which is a spot you consider safe to be before you proceed into the harbour.
(I will tell you this distance is actually 4.24 nm so that you can mark it on the chart for the purposes of this calculation, but on the chart it is just a spot you want to go to, the exact distance is irrelevant unless you want to calculate your SMG for some reason).

This is an emergency situation and you must get there ASAP. Conditions are calm, there is no wind and the water is flat.
Unfortunately your prop is damaged and your maximum boat speed is 1.8 knots (your know this from the last trip). It is a feathering prop jammed at the wrong pitch so that it is quite safe to travel.

Can you get to your destination to deal with this emergency?
What is the CTS?
What is the time taken?

You estimate the journey will take 2-3 hours, so you look up that tide data.

The speed vector off the 2 hour current displacement vector does not quite reach B, but it is close so draw a line to B and mark it off as S.
The speed vector off the 3 hour current displacement vector is well beyond B so draw a line and mark that off as L. You can see immediately it will be closer to 2 hours than 3!

Measure the proportion of time accurately
= SB / SB + BL = 0.3 / 0.3 + 1.2 = 0.2
So the last bit of the journey is 0.2 hours

To find the position of K on the last current displacement vector:
= 0.2 x the length of the vector
= 0.2 x 0.5 = 0.1

Mark that as K

The angle of KB is the CTS
CTS = 4 degrees (true)

Time taken = 2 + 0.2 hours = 2.2 hours (2 hours and 12 minutes)

Distance travelled through water = length of KB = 4 nm

SMG if you need to know it:
Now is the time to accurately measure the distance on the chart between A and B. It is 4.24 nm
SMG = distance between A and B divided by the time taken
So SMG = 4.24 / 2.2 = 1.9 knots (note that it is actually faster than your speed through the water of 1.8 knots as a component of the current is with you)

Well done Paul Elliott for being close in your estimate, whatever method you used. Not exactly right but very close!

This is a perfectly possible journey in an emergency and safe in my opinion as:
- it is short so it is easy to pick the correct tide data
- you arrive close to slack water, so you can easily make any small adjustments when you get close to B

The RYA result will either tell you your CTS (true) is 39 degrees (35 degrees off) or it will make you think the journey is not possible. Both these conclusions are incorrect!

[Paul Elliott]

SWL, using my cellphone calculator, and solving the triangle using the Law of Cosines, I think that you are overshooting the target. I can only make your numbers work with a boat speed of 1.66 Kts.

With your current vectors as stated (3.5, 1.5, 0.5 @ 135 deg), and the AB distance of 2.4 miles, I find that a 2-hour passage (@1.8 Kts) on a CTS of 11.2 degrees puts me within 300 ft of the goal.

My triangle has the sides:
4.240 (AB)
5.000 (Current, 2 hours)
3.605 (course through the water, actually takes 2.0028 hours @ 1.8 kts)

I could extend the current vector for two minutes, but it's just not worth it.

(ridiculous precision just so someone can check my math -- please!)

I think the differences between our results exceed the "pencil-width" issue -- one of us is probably wrong. I'm not suggesting that my method is better than your pencil and paper method (I need a calculator, or table of cosines), I'm just trying to check the results.

[Seaworthy Lass]

Originally Posted by Paul Elliott

SWL, using my cellphone calculator, and solving the triangle using the Law of Cosines, I think that you are overshooting the target. I can only make your numbers work with a boat speed of 1.66 Kts.

With your current vectors as stated (3.5, 1.5, 0.5 @ 135 deg), and the AB distance of 2.4 miles, I find that a 2-hour passage (@1.8 Kts) on a CTS of 11.2 degrees puts me within 300 ft of the goal.

My triangle has the sides:
4.240 (AB)
5.000 (Current, 2 hours)
3.605 (course through the water, actually takes 2.0028 hours @ 1.8 kts)

I could extend the current vector for two minutes, but it's just not worth it.

(ridiculous precision just so someone can check my math -- please!)

I think the differences between our results exceed the "pencil-width" issue -- one of us is probably wrong. I'm not suggesting that my method is better than your pencil and paper method (I need a calculator, or table of cosines), I'm just trying to check the results.

Hang on, I have just finished dinner, so I will draw in the vectors to mark the ground track for you on my plot and photograph it.

Give me a few minutes .

[nigel1]

While freezing my bits off today trying to run the solar cables through the dink davits, I was thinking about SWL method of finding a CTS with variable tidal streams, and came up with a hypothesis.
BTW, it needs a bit of imagination to work around the method.

Working on example 3, with the strong tides, and a boat speed of 4 hours.
Now imagine for the first 3 hours, the boat just drifts, but while its drifting, its building up kinetic energy.
At the end of 3 hours, the boat reaches the end of the 3rd tidal vector.
It now needs to get to point B.
Connect the end of the third tide vector to B. Lets call the end of the 3rd tide vector X
Now need to consider the 4th tidal vector, so draw in for 1 hour.
Now, from the end of the 4th tide vector, (call this Y) strike an arc of distance 16 miles to cut the extension of the line X - B, and call this Z
Why 16 miles?, the boat built up 12 miles of kinetic energy while drifting, and to that we add the 4 miles for the 4th hour vector.
The line Y - Z is the CTS line, and the distance through the water over the last 4 hours.
This method has a virtual Course to make good line (the rhumb line), and a triangle of vectors which all meet at the right places.
OK, so its a bit Star Trek, but it seems to look right, and it uses the RYA method.
My fingers have not thawed out so not able to draw it out at the moment.

[LJH]

Originally Posted by Dockhead

You are fundamentally not understanding the method or even the reason why Seaworthy is plotting A to K.

The purpose of her exercise is to get an angle -- the course to steer. She's working it with a protractor because it's easier and quicker than math, but the mathematical principle is the same. In order to solve the problem, she is building up a triangle with vectors equal in time.

Strictly speaking, trig needs lines, but it's ok to talk about points if it is easier to understand. If you have three points, you have enough information to do trig and get a CTS. But you need three if you don't know any of the angles; two is not enough.

Two of the three points are A and B. The third point is K. Any two points are a line, in math, which is why it doesn't really matter whether you talk about points or lines.

This is the principle which she, and everyone else, is using to derive that angle which is CTS:

"The law of cosines relates the lengths of the sides of a plane triangle to the cosine of one of its angles. Using notation as in Fig. 1, the law of cosines says
where γ denotes the angle contained between sides of lengths a and b and opposite the side of length c. . . . The law of cosines is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known. "

Attachment 53456

Law of cosines - Wikipedia, the free encyclopedia

The word "trigonometry" means "triangle-measuring" in Greek, and in fact that is literally what Seaworthy is doing when she builds up her triangle, and then literally measures the angle between two of the sides with her protractor. In any case, whenever you do trig, you are putting a problem into the form of a triangle so you can figure out an angle or the length of a line, or both. There are three lines and three angles in a triangle -- six things, altogether; the Law of Cosines allows you to know any three unknown things if you know the other three things.


When you sum the vectors of the tides, you are finding the position of the end point -- K in this case -- of the last unknown line. The starting point of that line is A, so you must know A to do that one. Once you know the positions of A, B and K then you can get the angle you need by plotting it and measuring the angle with a protractor. It doesn't make any difference whether you think of the points, or of the lines between them; mathematically it's all the same. In any case, you're doing trig -- three angles, and three lines (or if you like, three points), and the relations between them.


We are not getting anywhere near Great Circle paths. No one has mentioned this at all. For the sake of simplicity we are doing the problems in two dimensions (which is plenty close enough at these distances). People are having enough trouble with regular trig; I'm not even going to think about spherical trigonometry -- let's not even go there (it would be useless anyway, since I am clueless about spherical trig ). But joking aside, spherical trig IS used in navigation over large distances.

Yes the solution is made using triangles, however I have made a simple example to show that the line from A-B is not one of the lines used.
Point A: N 50 30.00/W002 00.00
Tide Set : 180 degrees 2 Kts
Boat speed 6.4 Kts
Point C: N50 28.00/W002 00.00 (2 miles south of point A)

Now lets say I want to go to
Point B: N50 28.00/W001 50.0

Point C to B: 090/6.4

What I have here are two rhumb lines, AC and BC. I do not need to plot AB in order to find out how I will get from A to B in this case.

Using trig
from point A to C
Delta Lat = cos (tide set) x tide strength = cos(180)*2 = 2 minutes
Delta Long = sin (tide set) x tide strength = sin(180)*2 = 0
Point C = N50 28.0/W002 00.00

From point C to point B
Delta Lat = 0
Delta Long = 10 minutes (6.4 miles at N50 28 Lat)
Heading = arc tan (delta Lat/delta long) +90 = 090 (I added 90 since the formula is based on the x axis as 0)
using Pythagorus distance = sqrt(6.4*6.4+0*0)=6.4

If you change the tide set to 210/2.3 and the boat speed to 7.3 and try it you should get the same answer as my example.

I will stay away from the spherical stuff since over these short distances the rhumb line and the great circle route for all intensive purposes are the same. Sorry I mentioned it, a bit of my past life coming through.

[Seaworthy Lass]

The ground track = the dotted red line

For each hour it is simply the sum of the boat displacement due to current vector (blue dotted line which is 3.5 nm long for the first hour at 135 T, 1.5 nm long for the second hour 135 T and 0.2x0.5 = 0.1 nm for the last 0.2 hour) plus the boat displacement due to speed vector (black line which is constant at 1.8 nm long at 4 degrees true for the first two hours, then 0.2x1.8 = 0.36 nm still at 9 degrees for the last 0.2 hours)

The last 0.2 hour plot is so tiny that I have reproduced it in finer pen at the top to illustrate what it looks like.

All the computations were done in extra sharp pencil (I kept sharpening frequently), but I have marked them in black pen and then coloured texta so that they stand out for the photos

You can see the ground track is never actually any further than about 0.7 nm from the rhumb line.

Note that this CTS gets you exactly to your nominated point B (it is sensible not to make B the actual point you want to finish (eg the spot you plan to anchor etc), but a point a bit up current (for the current when you arrive) and a bit short for safety. The amount depends on the expected current when you arrive.

Travelling at one constant correct CTS (in this case 9 degrees) gets you there in the quickest time using minimal fuel

Plotting your course (the ground track) is essential to check that there are no obstructions along the way.
It only takes a few minutes to pencil the sum of the lines in for each hour, indicating where you expect to be.

Paul, the ground track exactly reaches point B, so my computed CTS etc must be very very close.
I will look at vector analysis mathematically for you in a sec to see how we compare.
edited to add : Anyone with a version of CPN that can plot the result?

[Andrew Troup]

Originally Posted by jackdale

I am sorry. I would not take a boat capable of 1.8 knots into a 3 knot current. Especially one with a damaged propeller, or any other equipment, on which I cannot rely. I would be inviting an emergency.

You seem to rule out the possibility of force majeure

There are a wide variety of possibilities which could force your hand, of which an obvious couple are:

1) Your boat can to 6 knots, at the time you make your passage plan.

Your propeller gets damaged en route.

In order to get to safety, or back to your point of origin, you have to traverse the currents as described.

These would have been no problem at normal speed.

2) Due to stresses of weather or temporary problems with steering or rig, you end up in a location you did not plan to end up in.



Faced with the choice of two simple methods to calculate CTS, why would you want to compound the difficulties and dangers you correctly identify, by choosing the method which fails over the one which copes effortlessly ?