Distinct Activities: Shackled by a Common Name?

[goboatingnow]

I may be overly suspicious here but why do I have a feeling the goal posts are moving

Nope , becuase to see the method , you have to apply the process from the start and derive the tides, apply the information, access the relability, ensure the ground track is safe , factor in real life CTS, and boat speed errors, apply an destimate of leeway, deviation and variation etc. All this is covered in the classes, All these introduce errors well beyond any precise mathematical derivation.

Seaworthy is seeking an accuracy that simply isnt there in real life and hence any such method has little point. ( even if it was right and precise)

[Wotname]

Originally Posted by goboatingnow

The method you outline is impossible to apply in any real life is non perpendicular tides., The result must be derived graphically from the chart. I dont know who taught it to you , but its of no real use in real world applications.

Dave

Well this thread is expanding faster than I can read.
Please correct me if I am wrong but wouldn't non-perpendicular tides just be charted accordingly and the result vector give the CTS in a similar manner as an aircraft calculates for winds from any angle. I might have to break out my EB6 soon. I seem to remember that is wasn't hard to do at the time.

[goboatingnow]

Well, I was fooled, I never picked up on the video that is just one case of perhaps many cases. I took the video as the method.

The RYA method overall say "estimate the journey time and the number of tides that you will pass through". Thats the method, Most people are not dealing with perpendicular tides, hence D can lie either side of B.

Rework Seaworthy example with all these tides at say 135, 130 and 125 degrees T to see what I mean.

Dave

[goboatingnow]

Originally Posted by Wotname

Well this thread is expanding faster than I can read.
Please correct me if I am wrong but wouldn't non-perpendicular tides just be charted accordingly and the result vector give the CTS in a similar manner as an aircraft calculates for winds from any angle. I might have to break out my EB6 soon. I seem to remember that is wasn't hard to do at the time.

Dont know have no idea of the method aircraft use, I suspect the forward speed is so much greater then the winds that the errors are different.


IN non perpendicular, you again estimate the journey time , lets say in this case you still felt it was a 3 hour plot.

Youd plot the hourly tides, join C to the 3 hour boat speed track and very quickly realise that that D was way to far behind B, you recompute based on 2 tides and deflate proportionaly accordingly ( or recompute to one hour if the estimate was grossly out.)

The exercises in the RYA YM handbooks tends to cover far more difficult cases, and the actual instructor packs contain solution accetates to these more complex ones, so that students can see how to interpret the method in the lite of the graphs they produce.

Equally in most cases , there are reall significant errors in deducing the tidal data in the first place. Tidal diamonds may be mile away from where you actually cross, the whole thing is a estimate, There are too m,any real variables that make computing exact precise figures nonsense.

[barnakiel]

None of our GPS units shows RYA though.

Here we are limited to XTE, VMG, COG, SOG and WPT.

b.

[Wotname]

Originally Posted by goboatingnow

Nope , becuase to see the method , you have to apply the process from the start and derive the tides, apply the information, access the relability, ensure the ground track is safe , factor in real life CTS, and boat speed errors, apply an destimate of leeway, deviation and variation etc. All this is covered in the classes, All these introduce errors well beyond any precise mathematical derivation.

Seaworthy is seeking an accuracy that simply isnt there in real life and hence any such method has little point. ( even if it was right and precise)

Fair enough for the emphasised section but I have always (and perhaps wrongly) tried to be as accurate as possible during the planning phase (including the maths) and then update as the trip progresses with best possible fixes and revamping the DR accordingly. Why start off with imprecise data when more precise data is available.

I do accept that a heavy sea, late night departure party and lack of sleep throws this internet passage planning into disarray

[Seaworthy Lass]

Originally Posted by goboatingnow

I give up , Im being told by someone , who is learning the RYA method by looking at a you tube video ( one that I sent her) and then TELLING me what or what not the RYA does.!!!.

The whole process of determining the tides in the first place its based on interpolation ( ie you have to interpolate the tidal date to arrive at a value for your passage through that tide) Thats process in itself adds enough error to render any further accuracy moot.

The RYA in a class room ( The theory class is 16 night classes of about 4 hours each) explains the limitations of the tidal data, gets students to look at CTS graph and evaluate what seems appropriate given the quality of the tide data, the effective CTS that a helmmans can steer and the vagaries of leeway and boat speed

All these errors introduce far greater errors that the precision you are attempting to find.
You are turning this into a angels on the head of a pin argument.

Dave, I am not about to enter into a discussion of the accuracy of the tidal data. I know it is defective.

But for pages of posts you have kept hotly insisting insisting the RYA method itself was precise mathematically.

Originally Posted by goboatingnow

No again you are confused.
.....
However once you exact ( ie whatever data you decide to apply ) the tidal vectors the RYA method is mathematically correct
.....
And if you want me to question THE WORLD greatest training organisation , that certifies sailors in 6 countries and all over the globe , you've another thing coming. The method is right mathematically.
Dave

I have been saying all along it is only an approximation. Sometimes very good, sometimes poor.

This is what we are debating here.

Whether tidal data is correct or not, you don't want to be adding extra errors eg ten degrees of inaccuracy in some examples I have looked at if you use the RYA incorrectly computed CTS. This is 3.5 nm over a 20 nm journey. Not insubstantial!

The RYA does not warn students that their method in some case will produce this type of error. I do not think you are the only one teaching that the "method is right mathematically".

Rather than worrying about being shown to be wrong, why don't you work with me in solving this. The method I last described is a lot better than the RYA method, but it is still very flawed and needs a complete revision, but even as it is, it performs a lot better that the RYA method, as I will show you once you give the CTS and time taken for the example I set.

PS It shows the value of Youtube videos doesn't it!

[Dockhead]

Originally Posted by goboatingnow

The method you outline is impossible to apply in any real life is non perpendicular tides., The result must be derived graphically from the chart. I dont know who taught it to you , but its of no real use in real world applications.

Dave

It was taught to me by a U.S. Navy WWII veteran navigator who spent the whole war fighting the Japanese. What I have written above is only a small part of what he taught me; the fun part of course is when the tides are not perpendicular. So for a situation with rotary tides like in the Channel Islands, I'm screwed and don't know how to solve them, as I admitted a few dozen posts ago.

But I must disagree that this method is "no real use in real world applications". That is because tides which are with or against you don't pose a CTS problem. Tides which are within 10 or 15 degrees of perpendicular also don't pose a problem, especially if they tend to cancel each other out. So that covers most real world situations -- Channel Islands rotary tides are not that common. You can fudge those hours with an element of tide against you -- add some set because your rate of advance slows down. The opposite if the tide is with you a bit. These analogue fudges tend to work very well, because you only need to get the average right, in order to get the right CTS. So if you're off by a knot under here and a knot over there, you're still going to come out ok. The main thing is that you know it's a little more, and not a little less, and you still come out far ahead in the game.

But for the problem as given by Seaworthy, I submit that this method gives a mathematically perfect solution. In 2.5 hours, your set is exactly 5.25 miles. Seaworthy's problem is an artificial situation, and we accept the data as given and so perfect, which is not like the real world. For one thing, the tide doesn't vary by position like it does in the real world. But based on that data and that scenario, the correct CTS is not, mathematically, 60 degrees, which if you steered, you would end up downtide, and be sorry about it, if you're sailing in Seaworthy's constructed world.


Using this method in the more complex real world, I get across, under sail (so less predictable speed), sailing sometimes 70 - 75 miles through the water over 8 or 9 hours, and often arrive within 1 mile of my calculated waypoint. One mile out of 75! Without any mid-Channel correction. When I'm off it's always because my average speed did not go to plan, and a mid-Channel correction always does it. I don't think I've ever been as much as 3 miles off. So clearly it works, somehow.

This method is also used in English Channel pilot books such as Tom Cunliffe's. Except that it is simplified even further -- you are supposed to assume that one mile = one degree at 60 miles, which happens to be the distance between Needles and Cherbourg. So you don't even need to do any square roots.

[Wotname]

Originally Posted by goboatingnow

Dont know have no idea of the method aircraft use, I suspect the forward speed is so much greater then the winds that the errors are different.
Yes and no, a 25 kt wind can play havoc when you have an airspeed of 80 kts, not much different than 2.5 kt tide on say a boat speed of 6 or 8 kts but we digress

Rework Seaworthy example with all these tides at say 135, 130 and 125 degrees T to see what I mean.
.......

Tomorrow perhaps, the shower really does need cleaning - nite all!

[goboatingnow]

Originally Posted by Wotname

Fair enough for the emphasised section but I have always (and perhaps wrongly) tried to be as accurate as possible during the planning phase (including the maths) and then update as the trip progresses with best possible fixes and revamping the DR accordingly. Why start off with imprecise data when more precise data is available.

I do accept that a heavy sea, late night departure party and lack of sleep throws this internet passage planning into disarray

Thats the key point , no more precise data is available anyway. Tide is an estimate ( an estimate YOU make), hence the vectors you plot have an inherent error in them, Just add that to the fact that few can steer on average better then +- 5 degrees, and you have errors that completely outweigh the precision of teh maths.

The underlying data simply isnt precise. Hence merely attempting to be precise has no effect ( because you dont in practice know how the errors add up),


The only difference is that Seaworthy disputes that 2 tide inflated by 15 minutes is less accurate then three tide with teh last tide deflated by 30.

You can then take matematical examples to prove this or not, IN real life none of these vectors are that accurate to begin with. Hence I can argue that 2 trides inflated is simply as valid as any other method, becuase specifically the underlying data has greater errors then MY assumption.

Your guestimate of your CTS and time of arrival is exactly how it would be determined inreal life and applied in practice.
dave

[goboatingnow]

But for pages of posts you have kept hotly insisting insisting the RYA method itself was precise mathematically.

The vector additon is precise, thats what I argued. I was drawing attention that in your original drawings you connected C to B, which is not correct , unless by coincidence D is at B. I notice now you dont do that anymore, thats All I meant by the precision of the maths.

[goboatingnow]

Whether tidal data is correct or not, you don't want to be adding extra errors eg ten degrees of inaccuracy in some examples I have looked at if you use the RYA incorrectly computed CTS. This is 3.5 nm over a 20 nm journey. Not insubstantial!

The RYA does not warn students that their method in some case will produce this type of error. I do not think you are the only one teaching that the "method is right mathematically".

Rather than worrying about being shown to be wrong, why don't you work with me in solving this. The method I last described is a lot better than the RYA method, but it is still very flawed and needs a complete revision, but even as it is, it performs a lot better that the RYA method, as I will show you once you give the CTS and time taken for the example I set.

The RYA method of inflating the triangle or deflating it in proportion is as valid as any other method and easy to do. Hence the CTS is as valid as any other method. You can take "make up tides" to try and disprove the point or seriously argue that tide cut off at 60 minutes . Are you seriously suggesting your method really reflects whats happening, no it doesnt.


Hence there isnt the CTS error you suggest , nor have you ever established there is.

Basically as wotname suggested in the example steer 60 and get there after 2 and half hours, that about as precise as you get and you will get that level of accuracy from the 2 hour plot alone.

Youve admitted yourself that your method has flaws and cant cope with more complex non perpendicular tides, whats the point of inventing another method , righty or wrongly when the RYA method is proven in real life to work.

Telling me CTS errors of 10 degrees result in error of X just make me laugh , IN reality there are far greater inaccuracies that dwarf this particular "claimed inaccuracy",and since you cant know what way the errors add up, seeking greater precision is nonsense.

Your quest is a lab experiment Seaworthy , amusing and takes up pages of discussion , when fundementally the underlying data in the real world offers no such precision.

[goboatingnow]

The RYA does not warn students that their method in some case will produce this type of error. I do not think you are the only one teaching that the "method is right mathematically".

Youve obviously sat through the 16 night theory class then , OK good

The RYA method given the assumptions of the underlying real tides is as good as any method and easy to apply in complex situations

[goboatingnow]

Originally Posted by barnakiel

None of our GPS units shows RYA though.

Here we are limited to XTE, VMG, COG, SOG and WPT.

b.

Try crossing from Poole to Cherbourg using that , Dockhead, Seaworthy and Me , despite all our differences will have several rounds of Pernod drunk before you in Cafe de PAris in Cherbourgh.,

Seaworthy is arguing that she will have the waiter to the table first. Im arguing I know the waiter and hence Ill will be served first

[Dockhead]

Originally Posted by goboatingnow

This isnt really true Dockhead, tides are stated as a set and drift at a point in time usually in increments of 1 Hour referenced back to the nearest standard port , or a reference port.

You extract the actual set and drift you experience at the time you pass through the area, by interpolating the results from the tide table, You then assume that the 60 minutes rule applies, In practice it ,may or may not apply , particulary if the underlying tide is non sinusoidal. Hence a 2kn tide can be valid for 60 minutes, 70 minutes 80- minutes , or invalid after 45 minutes, you have no way of knowing

What we do know is that tide A at + 45 minutes is inreality very little different from trhe next hour B, at 15 minutes , ie in reality, not what the tide tables say.

Hence it is perfectly acceptable to advance or retard tides appropriately within a reasonable time limit. ,ie whether you take 2 tide and spend 2.5 hours in them or spend 60,60 and 30 in the next tide has very little difference in real life as the tide is a constantly changing event and never remains still for 60 minutes , or whatever.

OK, thanks, I get that -- and that's a new insight for me.

In Seaworthy's case we have half of the last tide, and I assumed half of the set -- 0.25 miles. That corresponds to our assumptions and so is the right answer to the given problem. But in a real world situation the first 30 minutes of that tide of course will not be the same as the last 30 minutes. But it will also not be the average of the previous hour's tide. It will be somewhere between, so both approaches are imprecise, and an analogue fudge -- a guess with taste and feeling -- will give the best results.

But still it seems to me, I humbly submit, that you are missing the point that the tidal data is expressed as an average over that hour, and the average is exactly what we need. That's the beauty of steering a constant heading -- how you come out is purely a function of how close your calculation came to the average tide -- you don't need to get any particular hour exactly right. So analogue fudging really works extremely well in practice -- I think you are dissing the quality of the data issue, and in my experience a degree or two in CTS over 60 or 70 miles is significant, and is within our ability to calculate meaningfully. The tides on the typical x-channel passage are mostly perpendicular-ish, and if they get to 10 degrees or more away from perpendicular, I just apply an analogue fudge. I have sailed many passages with no mid-Channel correction and arrived within a mile or so of plan, and that's with the huge tides of this region, with XTE over 12 miles at times. Just luck?