Inaccurate RYA Teaching : CTS - Quest For a New Method

[Dockhead]

Originally Posted by 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!

Whoops! I didn't see the hour of slack water in the middle. Rerun! Thanks a lot!

[Dockhead]

OK, since channel tides vary by time as well as location (so not like the Capt Force example), we don't have the hour of the 2.4 knot current at the end.

In real life we would look at the tidal atlas to check the velocity of the current at the particular predicted location at each hour, but we don't have that much information here so we have to assume that the first ten hours of tides correspond to what we will experience on passage.

This problem is not a very good one because the corrections are very small and so the vector triangle is very skinny. The net tidal set now is only 1.1 miles E so steering correction will be very small -- 1.1 degrees by our rule of thumb plus whatever margin of error we want, so 2 or 3 degrees. So CTS is 181.1 plus margin or 182 or 183 (you also need to add deviation to get a magnetic CTS from true chart directions but no need to get into this). The water line track is 60.01 miles so passage time is 10.001 hours. The method is described in my previous post.

That was the rule of thumb calculation. The exact calculation is 1.05 degrees steering correction and water track 60.01 miles, passage time 10.001, showing again that the rule of thumb method gives us a result well within any usable precision.

The only error in these calculations is ignoring 0.001 hours of the 2.4 knot final hour tide, which is vanishingly small.

[LJH]

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).

Nice when the 1:60 rule is easy to apply.
offset 1.1 miles west = 181.1 course
time taken will be [60/cos(1.1)]/6=10 hr 6 mins
monitor your XTE diverging out to 6.5 miles west of the rhumb line before you start converging..

[Dockhead]

Originally Posted by LJH

Nice when the 1:60 rule is easy to apply.
offset 1.1 miles west = 181.1 course
time taken will be [60/cos(1.1)]/6=10 hr 6 mins
monitor your XTE diverging out to 6.5 miles west of the rhumb line before you start converging..

It's always easy to apply --

1 mile at 60 miles is 1 degree
1 mile at 120 miles is 1/2 degree
1 mile at 180 miles is 1/3 degree
etc.

It's a linear relationship so just divide by 60 and invert.

It's superb because it is so close and so easy. The real relationship is 0.9458 degrees per mile at 60 miles. The error is less than a tenth of a degree, a useless degree of precision. The relationship is almost linear for angles less than 30 degrees.

So there's actually no need to ever do the exact calculation.


And you've exactly got the easy method of checking your progress . Your XTE should = net cumulative tides you have been through at any given moment. Another beauty of perpendicular tides -- you always know at a glance whether your passage it going to plan or not.

[Dockhead]

Originally Posted by LJH

Nice when the 1:60 rule is easy to apply.
offset 1.1 miles west = 181.1 course
time taken will be [60/cos(1.1)]/6=10 hr 6 mins
monitor your XTE diverging out to 6.5 miles west of the rhumb line before you start converging..

I think your time is 10 hours six seconds, not minutes, no? Should be 10.00168 hours, I think?

[LJH]

Originally Posted by Dockhead

I think your time is 10 hours six seconds, not minutes, no? Should be 10.00168 hours, I think?

Yes, it is typo as I don't normally use seconds!

[Seaworthy Lass]

Hi all!
Just having lunch ashore in between doing sail repair work.

Use the SWL method and you will know exactly how much of the last hour is needed . If the journey was 63 nm instead of 60 nm and the current was stronger in the 11th hour, the method would be useful, as it is the proportion is too small.

By the way, Dockhead, what is the strongest current you are likely to encounter before the breakwater?

[Zonker]

Just passed the YM theory exam last week ... Seaworthy Lass ... did u ever hear of something like a "computer" ?

The RYA CTS method is just an interpolation - even if my teacher did disagree on that - but who cares. If you want it really correct, then you need a) computer an b) have a numerical calculation and c) have it all the time measured (control system with feedback) by sensors. Everything else is splitting hairs.

[Dockhead]

Originally Posted by Seaworthy Lass

Hi all!
Just having lunch ashore in between doing sail repair work.

Use the SWL method and you will know exactly how much of the last hour is needed . If the journey was 63 nm instead of 60 nm and the current was stronger in the 11th hour, the method would be useful, as it is the proportion is too small.

By the way, Dockhead, what is the strongest current you are likely to encounter before the breakwater?

The SWL method will give us a proportionate share of the average current of the last hour.

If we suppose that the current develops like a sine wave, the inherent error in this approach will reach its maximum in a situation like this where we have an extremely small part of the last hour. In this case it's six seconds. This is a case where the RYA method will give almost exactly the same answer as SWL's method (and the difference will be far too small to plot). That is because the problem is extremely close (within six seconds) to a passage of 10 whole hours. RYA calculates whole hours passages with mathematical perfection (and incidentally, it is only whole hours passages where SWL method is mathematically perfect).
Using SWL method, we will get 6/3600 * 2.4 = 0.004 miles (that is, 4/100 of a cable or about 7.5 meters of tidal set to the E. She will not be able to plot this; it will be far beyond the accuracy of her protractor. She would need a protractor about 30 meters long to get meaningful accuracy for such a problem. But we can easily determine digitally what her method will give, it's 0.004 miles.

A much more precise method for this particular case (orders of magnitude more precise than SWL) will be to assume that at 0 minute of the 11th hour, the tide will run at the average of the 10th and 11th hour rates, which it will exactly do if it is developing as a sine wave. So it is a very robust assumption to take the tidal stream during the first six seconds of the 11th hour to be 2.85 knots. At that rate, we will be moved 0.00475 miles during the six seconds we sail in the 11th hour. This solution very closely approaches mathematical perfection. The course correction for this is practically unmeasurable so might as well stop there.

As to currents off Cherbourg -- at springs at HW Dover +3 the nearest tidal stream diamond reads 6.3. That's an average over an hour and over a certain area. The peak local rate will be around 8 knots.

That's why you're really stuffed if you arrive downtide at springs and peak rate in a small boat. It's actually very dangerous, because there is no refuge to the West -- the widow-making Alderney Race awaits you around Cap Hague. Standard procedure for this case is to get an anchor down if you can -- if the sea is not too rough -- and wait for the rate to subside to less than your cruising speed. If it's too rough to anchor, you motor against the stream to reduce lost miles until you can start to make headway. Or you turn around and go back to England -- not such an unusual case.

[Dockhead]

Originally Posted by Zonker

Just passed the YM theory exam last week ... Seaworthy Lass ... did u ever hear of something like a "computer" ?

The RYA CTS method is just an interpolation - even if my teacher did disagree on that - but who cares. If you want it really correct, then you need a) computer an b) have a numerical calculation and c) have it all the time measured (control system with feedback) by sensors. Everything else is splitting hairs.

You should read the thread from the beginning.

SWL has come up with an ingenious method of doing this calculation which with practically zero extra effort will give you a much more precise result than RYA. It is really worth learning -- it is really totally superior to RYA.

The problem with RYA is that it uses the average of the whole passage and applies this to the last partial hour, which it does not analyze at all. This can lead to very large errors. There has been some debate about whether these errors are inside or outside the useful amount of precision -- in other words whether we are splitting hairs or not. We will soon figure that our empirically), but in my opinion it's pretty clear there are many cases where RYA will leave us pretty stuffed. If we use it uncritically -- of course, using it intelligently we would know how to fudge it. But SWL method obviates the need to do any fudging, because it uses the average of the last hour, instead of the whole passage, which is a very large increase in precision. And with no extra effort, almost!

The other ingenious thing about SWL method is that it greatly simplifies getting the passage time. It's really worth learning!

If SWL agrees, I'm going to send it to Dick Durham at Yaching Monthly and see if he will write an article about it. It's that good.

[Zonker]

Originally Posted by Seaworthy Lass

In other words it isn't taught by the RYA and students aren't alerted to the possibility of problems if D is a long way from B (and it could be several nm)

Wrong - there was a exercise where that was addressed.

Seriously ... I think you are making a big thing of nothing. If you are so worried about the precision then you gotta setup A->B that way so it fits the tidal stream hours. So for B->B(port) a separate calculation is needed if at all.

Further if you want it super exact you gotta split up A->B in many many pieces and do the calculation based on these little tracks assuming you have the exact tidal data but most likely you wont.

@Dockhead - I've read that ... and see my answer above. The main issue is that folks use the "hourly" base. If you would subdivide according your needs then RYA "method" is sufficient.

[Andrew Troup]

Dockhead is exactly right to say that the rhumb line (aka "Course Line") forms an essential element of the vector solution for CTS, in both the RYA and Seaworthy methods.

I apologise to him for my stubborn inability to see that, and to anyone who might have been misled by my forceful sophistry in putting the contrary view.

If anyone was misled, I'm happy to explain in more detail what I'm retracting and why.

In brief, I was forgetting that vectors do not have position.
They ONLY have magnitude and direction.

(Which is why we can freely 'drag them around' to add them head to tail)

Consequently a vector diagram must form a closed loop, because leaving it open can only be done by bringing in positional information from the "outside world" about the two endpoints of the open diagram.

Another way of saying this:
The only information we can put into (or extract from) a vector diagram is vectors.
Smuggling fixed or isolated points in, or out, is not valid.
　
The vector which closes the CTS diagram, whether or not we choose to draw it or measure it, is identical in magnitude and direction to a line joining the departure and arrival points, as Dockhead correctly and repeatedly pointed out.

While the diagram can remain open in practice, if we are working directly on the chart, pretending the diagram can remain open in theory, as I was somehow managing to do, was a nonsense.

[Dockhead]

Originally Posted by Andrew Troup

Dockhead is exactly right to say that the rhumb line (aka "Course Line") forms an essential element of the vector solution for CTS, in both the RYA and Seaworthy methods.

I apologise to him for my stubborn inability to see that, and to anyone who might have been misled by my forceful sophistry in putting the contrary view.

If anyone was misled, I'm happy to explain in more detail what I'm retracting and why.

In brief, I was forgetting that vectors do not have position.
They ONLY have magnitude and direction.

(Which is why we can freely 'drag them around' to add them head to tail)

Consequently a vector diagram must form a closed loop, because leaving it open can only be done by bringing in positional information from the "outside world" about the two endpoints of the open diagram.

Another way of saying this:
The only information we can put into (or extract from) a vector diagram is vectors.
Smuggling fixed or isolated points in, or out, is not valid.
　
The vector which closes the CTS diagram, whether or not we choose to draw it or measure it, is identical in magnitude and direction to a line joining the departure and arrival points, as Dockhead correctly and repeatedly pointed out.

While the diagram can remain open in practice, if we are working directly on the chart, pretending the diagram can remain open in theory, as I wa
s somehow managing to do, was a nonsense.

Welcome to club of all of the rest of us who have taken off in a wrong direction, and after much argument ended up learning a lot from it! What's cool is that we all learned as much from the argument as you did.

It's a great method we have here, learning and teaching at the same time, saying one stupid thing for every 1.5 right things (that's about my ratio, anyway), getting called on it, arguing about it, finally figuring it out. I've learned more than I could have learned in a year of nav classes.

[Zonker]

It's time folks are getting faster boats ... :-)

[Lodesman]

Originally Posted by Andrew Troup

　
The vector which closes the CTS diagram, whether or not we choose to draw it or measure it, is identical in magnitude and direction to a line joining the departure and arrival points, as Dockhead correctly and repeatedly pointed out.

While the diagram can remain open in practice, if we are working directly on the chart, pretending the diagram can remain open in theory, as I was somehow managing to do, was a nonsense.

Hallelujah! Y'know Dockhead wasn't the only one telling you this. My posts #175 and 180:

Originally Posted by Lodesman

Of course SWL's method still has the 'rhumb line'. She starts at point A and goes to point B - and measures the distance between them. The only thing she doesn't do is draw the pencil line between them. It is still part of the tidal triangle that is the basis of her method. I really can't believe this argument.

Originally Posted by Lodesman

The combined set vectors = a single vector that is one side of the triangle. The line between start point and destination, whether drawn or not, is a side of the triangle. And your CTS is the third side of the triangle.