Inaccurate RYA Teaching : CTS - Quest For a New Method

[Paul Elliott]

Here are two vector triangles, one for 2.2 hours (SWL's solution), and one for 2.0 hours (my solution). Note that the triangles are *not* drawn to scale, and do not represent the actual dimensions or angles. They are simply used to illustrate where the corners and lines are. Do *not* try to take any measurements from them!


I couldn't get your numbers to add up exactly (not surprising, since you are drawing and measuring), so I used the 5.1 mile / 2.2 hour current vector and solved for the resulting CTS vector. As you will see, the speed needed to complete the triangle is 1.66 kts, not 1.8 kts.

So (assuming my math is correct), is this degree of (im)precision to be expected with mechanical construction of the triangle? I don't know, but it seems excessive to me. I would love to have someone critique this for me.

[Seaworthy Lass]

Paul, 0.2 hrs (with 1.8 knots) fits well.
See photo.

Where d= 'total distance travelled through water' vector

Working on 4 degrees, d = 4.05

Working on 0.2 hours, d= (2 + 0.2)x 1.8 = 3.96 (round off to 4.0)

So 0.2 hours for the last bit with a speed of 1.8 at 4 degrees seems to fit well

[Seaworthy Lass]

Paul, just looking at your figures for the SWL method, you have plugged in 10 degrees as your angle, not the 4 degree CTS that I found

I think that is why you think my data won't fit!

Try it with 4 degrees (making the full angle 49 not 55)

[jackdale]

Originally Posted by Andrew Troup

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 ?

Andrew

I am probably the last person to rule out force majeur.

In 2000 on a race boat delivery from Honolulu to Vancouver, we hit a fishing net under power and took out our transmission in the middle of the Pacific High - no wind. After about 3 days of drifting we managed to find some wind and get to the entrance of San de Juan where the wind died and we spend another 1.5 days floating around the start of the TSS. We were able to dock under sail in to the fixed dock Port San Juan, where we were met by the owners wife who had the tender and outboard.

A few years ago the intake and outlets hoses on either side of the raw pump split and the engine overheated. We sailed into Pender Harbour (not easy) and docked a Beneteau 50 under sail. We did some ersatz fixes and powered the boat out of Pender Harbour and sailed to Nanaimo and picked up a mooring ball using the engine for a short period of time. Some mechanics from the charter company / sailing school came out with spare hoses and fixed the problem.

In July of 2011 our alternator seized after we got sails up just after we had crossed the Nahwitti Bar at the end of Vancouver Island. We turned around to sail back over the bar (in a less than opportune current). I noticed that we were losing distance between our selves and the shore. So I contacted the CG (just a notification, not a Pan Pan) who sent out a cutter to escort us across the bar (I have been across it more than a dozen times.) After we were across I sailed the 22 miles back to Port Hardy during which time I arranged for the spare to be sent up from Sidney. I now longer make that trip without a spare alternator.

I also effected an AMVER medical evacuation off my boat to an 1100 foot container ship 1040 miles north of Hawaii last summer.

My first responsibility is the safety of my crew. If I found that I could not keep my crew safe I have no problems issuing a Pan Pan or Mayday. I have only issued a Pan Pan once and that was for another vessel. Even in the case of medical evacuation, I conducted CG Honolulu via satellite phone and my crew member spoke with their Flight Surgeon who recommended the evacuation.

BTW - no terrorists - yet.

What I am asking of the OP is a realistic, credible scenario; not the extremes that keep getting thrown out. Of the OP cannot do it, can someone else present a real planning scenario that involves crossing the Channel using real current numbers?

Since I am refereeing at a squash tournament I have not had as much time as would have liked to follow this thread. The tournament has me tied up for tonight (Z+7) and all day Saturday and Sunday.

[Paul Elliott]

Originally Posted by Seaworthy Lass

Paul, just looking at your figures for the SWL method, you have plugged in 10 degrees as your angle, not the 4 degree CTS that I found

I think that is why you think my data won't fit!

Try it with 4 degrees (making the full angle 49 not 55)

If I use a CTS 4 degrees, then the current vector has to be extended to about 5.5 miles in order to make the triangle come together. This gives a time of over three hours, a distance of 3.97 miles, and a speed of about 1.32 kts.

We can't arbitrarily assign angles and lengths to the triangle. They all have to fit. Change one angle, or one side, and other lengths and angles have to change too (or you no longer have a triangle).

You know, it's possible that my math is wrong. I don't see it though. I am using the Law of Cosines.

[conachair]

Originally Posted by Paul Elliott

Here are two vector triangles, one for 2.2 hours (SWL's solution), and one for 2.0 hours (my solution). Note that the triangles are *not* drawn to scale, and do not represent the actual dimensions or angles. They are simply used to illustrate where the corners and lines are. Do *not* try to take any measurements from them!


I couldn't get your numbers to add up exactly (not surprising, since you are drawing and measuring), so I used the 5.1 mile / 2.2 hour current vector and solved for the resulting CTS vector. As you will see, the speed needed to complete the triangle is 1.66 kts, not 1.8 kts.

So (assuming my math is correct), is this degree of (im)precision to be expected with mechanical construction of the triangle? I don't know, but it seems excessive to me. I would love to have someone critique this for me.

On CAD I got close to you, Paul.



Though in the real world I wouldn't be out there, too close to the edge. Makes me nervous just thinking about it.



edit:PS no whiskey for me then

[Seaworthy Lass]

Originally Posted by conachair

On CAD I got close to you, Paul.
Attachment 53482

Though in the real world I wouldn't be out there, too close to the edge. Makes me nervous just thinking about it.

Oh, yes, I realised my error when my data didn't line up with Paul's when I looked at his closely.

All my figures were done (both for the RYA method and my method) with 6 knots of current after 3 hours, not 5.5 as I specified. I wanted to specifically make it a right angles triangle to have 2 lots of D produced, so I altered all the second lot of current to 2, not 1.5. (See my plots). I didn't correct the data.

So Paul will be correct for 1.5 knots of current for the second hour as I wrote. Apologies Paul. I put it down to fatigue.

Conachair, could you please check the CTS with 2 knots as the second lot of current? I think that will come out as 4 degrees.

[jackdale]

Anyone want to try this. It looks more realistic.



Boat speed 6.0 knots. No wind flat seas.

An east-west ebb and flood.

Destination 60 miles due south (180T).

Source - Crossing Channel

The English Channel rarely gets a 6 knot current, so why are numbers like that being thrown around.



I will have to wait until Monday at the earliest to complete the numbers. But I will try to follow the discussion as much as possible.

[Seaworthy Lass]

Apologies to all.

Getting so tired here.
I was checking Paul Elliott's calculations and found they just didn't make sense. I had made an error in the text. Thanks Paul for pointing it out .

I wrote the text before tackling the 'chart' work for example 4 and for the results to be striking I had to alter the current in the second hour from 1.5 to 2 knots (that little change can mean the difference between the RYA technique giving a good result and an appalling one).

All my diagrams and ground plot were worked out with 2 knots of current, but I didn't alter the text and didn't notice the error . I put it down to exhaustion.

So here is that data again (diagrams unaltered):

Example 4 TAKE 2
RYA method results:

In the last example I was criticised for selecting an unrealistic situation. The comment was that 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: 2.0 knots 135 T (was written as 1.5 in error last time)
3rd hour: 0.5 knots 135 T

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 feathering prop is jammed at the wrong pitch and your maximum boat speed is 1.8 knots (your know this from the last trip out).

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 if the RYA result is used.

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.

What is the estimated time taken?
Time taken = AD divided by the boat speed (depends on which D you choose)

Anyway, you get totally wacky results!

[Seaworthy Lass]

Example 4 TAKE 2
SWL method results:

The current is as follows:
Ist hour: 3.5 knots 135 T
2nd hour: 2.0 knots 135 T (was written as 1.5 in error last time)
3rd hour: 0.5 knots 135 T

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).

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)

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!

[Seaworthy Lass]

Text altered simply to reflect the 2 knots of current on the second hour:

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, 2.0 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 4 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 4 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.

[Seaworthy Lass]

Originally Posted by jackdale

Anyone want to try this. It looks more realistic.

Boat speed 6.0 knots. No wind flat seas.

An east-west ebb and flood.

Destination 60 miles due south (180T).

The English Channel rarely gets a 6 knot current, so why are numbers like that being thrown around.

I will have to wait until Monday at the earliest to complete the numbers. But I will try to follow the discussion as much as possible.

I will check it out tomorrow after sail repair work (boat chores a getting in the way of CF LOL).
Have a lovely weekend

[Paul Elliott]

Originally Posted by conachair

On CAD I got close to you, Paul.
Attachment 53482


Though in the real world I wouldn't be out there, too close to the edge. Makes me nervous just thinking about it.



edit:PS no whiskey for me then

This morning I re-worked the numbers using vector addition (instead of directly solving the triangle) and verified my results. Of course this makes sense now, since SWL has found her error.

This strong-current stuff reminds me of the San Francisco "Three Bridge Fiasco" race, where we often have strong currents and light winds. One year I crept ahead of many competitors by casually slipping my anchor overboard and relaxing in the cockpit eating lunch. The other guys were fighting their way upstream, all the while drifting slowly backwards.

Here are my "3BF" blog posts: Three Bridge Fiasco Report « VALIS, Three Bridge Fiasco, 2010 « VALIS, Three Bridge Fiasco 2012 « VALIS. We did absolutely no course plotting, but we did look at the current predictions to see which was the favored route for the day. Of course the wind usually determines the route, and all my current projections go out the window.

[Dockhead]

Originally Posted by jackdale

Anyone want to try this. It looks more realistic.



Boat speed 6.0 knots. No wind flat seas.

An east-west ebb and flood.

Destination 60 miles due south (180T).

Source - Crossing Channel

The English Channel rarely gets a 6 knot current, so why are numbers like that being thrown around.



I will have to wait until Monday at the earliest to complete the numbers. But I will try to follow the discussion as much as possible.

That's a real easy one, which you can do in your head.

Looks like neaps.

You will swept West a total of 6.5 miles, then East 10 miles, total error 3.5 miles E, assuming 10 hours passage time. In reality it will take you a bit longer, as your water track will be a bit longer than 60 miles. So you will get a bit more of the East going tide. You are arriving nearly at slack water so this correction is small - maybe a few cables.

So your CTS is 180 + 3.5 + a smidge. We would add at least a mile uptide margin of error, so we would steer no less than 185. Less margin motoring in flat water as no leeway and no uncertainty about time. We are using the rule of thumb that 1 mile at 60 miles = 1 degree.

Now, the exact calculation. Current vector is 3.5 miles long, making a right triangle. So we add the squares of 60 and 3.5, which gives us 3612.25, and take the square root, which is 60.102, which is our water track. To get the steering correction, we take arcsin (3.5/60) and convert to degrees (divide by (pi/180)). We get 3.34, so steer 183.34, obviously unrealistically precise, so we see we got good results from the rule of thumb calculation, within our usable precision.

Time on passage is 10.02 hours.


Now, you all understand WHY this is an easy problem, right? Because we have perpendicular tides. That means our vector triangle is a RIGHT triangle.

That means we know one of the angles. Since you only need to know any three of six things to solve any triangle, we don't need all three sides (anyway we get the third side easily using the Pythagorean Theorem).

So that means you don't need to plot your vector triangle, per RYA or SWL or whatever method. The purpose of plotting is to get the length of the sum-of-tide-vectors line, which we don't know if the tides are not perpendicular. So in such cases (non=perpendicular tides) we are able by plotting to build up the vector triangle in analogue, and simply measure the angle we need. The digital way to do the same thing is to run the sum of the tide vectors and get the length. Once you have the length of the tide vectors line, then you have all three sides of the vector triangle, so it doesn't matter that you don't know any of the angles. You just solve using the Law of Cosines. By running the sum of the tide vectors, you also get the angle between the tide vector line and course line of our vector triangle, so we already know more than we need to know to solve it.

[Paul Elliott]

Dockhead, you are using the 11-hour current vector (3.5 miles), but your passage as calculated only takes 10.02 hours. You are going to have to interpolate within the last hour's current, and adjust your transit time, before it all comes together exactly.

I'm on the road again, so I will leave the fun part to you!