Distinct Activities: Shackled by a Common Name?

[goboatingnow]

n 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.

I dont disagree, with the paragraph above all these methods are to some extent fudges because the underlying data is inherently fudged. In some cases the next hours tide will exhibit virtually no change in some it will be more and often say around the channel islands you have to plot 30 minute tides ( and the tables give that) to up the accuracy.

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?

Now I beleive you are contradicting yourself. CTS errors due to leeway, helmans error, wave action ,and the fact that the underlying data is only an approimation ( and a roiugh approximation in some cases ) means that saying a CTS error of +- 2 degree is significant is nonsense , its only meaningful in theory in practice its irrelevant .

You apply an "analogue fudge" of course thats what everyone does, we inherently review the progress, we rework the model as we get closer and our progress differs from theory and lots of the time we just adjust intutitively based on previous experience.

for example have you heard of the 1 in 60 rule of thumb of course correction, often it give completely acceptable result , have a look at that in Bowditch.

In most cases Dockhead, leaving aside the poster child x-channel situation , Working up a complex multi hour CTS is fraught with error ( not to mention determining the ground track). hence its really not a good ideas anyway. Better to work up hourly or maybe two hourly plots and recompute as you go. Thats's what I do in practice and thats where I came in at the beginning.

ITs notable that very few textbooks , especially American ones bother going beyond the one hour plot. ( and inflating or deflating the triangle ) , Its not covered in Annapolis quide to navigation, Bowditch, Rousmarine, `even in the UK the method is glossed over, ( akak Tom cunliffe) in m,any books and in one or two there are howling errors.

Simply in practice very few people bother with multi hour ( leaving aside x-xhannel) becuase its so error prone.

[Seaworthy Lass]

Sorry for the delay folks. It was time for cockpit showers and for a drink to be poured. It is a celebratory one here .

Originally Posted by goboatingnow

The course to steer is 60 degrees for D2 model and 64 degrees for D3, either are well within the accuracy of the tidal data and CTS error in real life are the differences are irrelevant,. Use which ever you want. of average it or whatever it matters not.

The answer is steer about 60 degrees and expect to arrive in about 2 hours ans half hours.

This discussion is useless in advancing anything new.

The CTS is actually 51 degrees if you use the two hour method to pinpoint D.

The CTS is actually 63 degrees if you use the three hour method (the method you have chosen).
You would arrive on the opposite side in a bit under two hours actually, but you would be over a nautical mile south of your destination (your chartplotter is broken and it a bit foggy like the last example) so you would have to make your way north when you realised where you were.

The correct CTS is 58 degrees (58.3 degrees, but you would not compute it this accurately from the chart method, even if your pencil is super sharp and the lines super thin as the RYA insists on for accuracy (by the way is it true that you fail if you use a thick line for computations because it is less accurate?)
Time taken is 2.5 hours.

I arrive at these figures using my method, but as I have said a few times today even this method will fail. It is based on rocky ground. It needs to be modified and I need help with that. We need to eliminate D and just concentrate on B in computations.

So Dave, you are five degrees off when calculating the CTS using the RYA method (63-58, NOT 60-58 as you have given).

This is not a useless argument. Having insisted the RYA method was mathematically accurate, it has been shown in this simple exercise that you arrive at an answer that is 5 degrees off.

This information about the accuracy of the RYA method for determining CTS is IMPORTANT.

Originally Posted by goboatingnow

again false logic, in complex tides it can can be very difficult to actually determine at the outside what way to adjust the CTS to end uptide. ( Consider how many saw the constaly varying one way current caused a rhumb line transition).
If you want to be uptide, you would typically adjust the CTS as you progress across and it is clear what is really happening.

This discussion is not about accuracy of the tide data. It is about the accuracy of the RYA method .

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)

Are you now saying the RYA method is not precise mathematically as you have been repeating endlessly up to now?
Dave, I know I should be a little more gracious in victory, but I find it hard when you are now shrugging and saying there is "little point" in using an accurate method so why bother having one!

Originally Posted by goboatingnow

The RYA method overall say "estimate the journey
Rework Seaworthy example with all these tides at say 135, 130 and 125 degrees T to see what I mean.
Dave

Dave, it doesn't matter what the angles are. The only difference is that you need to do it on a chart/graph paper if the angles are not all 90 degree ones. The RYA method does not become more accurate when angles change.

[Seaworthy Lass]

Originally Posted by goboatingnow

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.

You have been saying the RYA METHOD is precise. Don't deny that now. There are numerous posts where you say this.

And I certainly am connecting my C to B (my C is different to yours, in this case it is at 5.25). The technique can only be accurate if you work with B, not D which may or may not be close to it. My technique needs revising to focus purely on B somehow, rather than the rhumb line.

[Seaworthy Lass]

Originally Posted by goboatingnow

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.

Dave, I just can't believe what you are saying. The RYA method is not as precise as my one that still needs to be worked on.

There ARE better methods!

Originally Posted by goboatingnow

Basicallyt 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.

How did you arrive at 60 degrees? The RYA method gave 63 degrees as the CTS if you are going heading to D. You were FIVE degrees off.

Originally Posted by goboatingnow

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.

Sorry, in all circumstances that I have tried the RYA method gives an even worse result when my method is failing. So yes, there is a better method possible.
And yes, I am humble enough to admit my method has flaws and I announced that it did as soon as I realised. I want the method to be improved, not to cover up my errors.

Originally Posted by goboatingnow

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.

This is more than an amusing discussion. I am challenging the accuracy of the RYA method that you have said repeatedly is mathematically perfect and that you would get precisely to your destination if the data was correct and now you are saying it doesn't get you there, but it doesn't matter?

And an 3.5 nm error over 20 nm is certainly significant! Or it is when I am navigating.

If anyone would like to help me iron out the flaws in my method it would be very welcome.

I am about to have a second glass of some wonderful red wine from Paros and to watch a movie. Many many thanks to everyone for chiming in at the end. It has been a really hard slog battling on my own against Dave.

Have a lovely day/ evening everyone .

[Lodesman]

Originally Posted by Seaworthy Lass

If anyone would like to help me iron out the flaws in my method it would be very welcome.
.

For someone joining late in the discussion, could you please describe your method or point me to the post where you describe it? Thanks.

[Dockhead]

Originally Posted by goboatingnow

again false logic, in complex tides it can can be very difficult to actually determine at the outside what way to adjust the CTS to end uptide. ( Consider how many saw the constaly varying one way current caused a rhumb line transition).

If you want to be uptide, you would typically adjust the CTS as you progress across and it is clear what is really happening.

This seems to me, simply incoherent. If you know at least during what tide you will arrive in (and they last 6 hours, duh!), you know whether nor not you need to add or subtract degrees to give an uptide margin of error!! You don't need to think even one millisecond about your - for these purposes, utterly irrelevant - ground track, to know that!

If you know what you're doing, and if you have a bit of luck with estimating your average speed, you don't even need a mid-Channel correction! You are far too pessimistic!

[Jammer Six]

Looking at the videos, they appear to be exactly what the U.S. Sailing courses teach here. With an accent.

[HappySeagull]

Originally Posted by Lodesman

For someone joining late in the discussion, could you please describe your method or point me to the post where you describe it? Thanks.

here? post #635
http://www.cruisersforum.com/forums/...ml#post1134450

saw it flying over...
but , hey, I thought it should agreed on the twelfths rule Post #612
1st hour 3 knots.....................=MAX? I am guessing so = 6 hour tide.
2nd hour: 1.5 knots..............I figger 2.25
3rd hour: 0.5 knots............. I figger .75
There's been ants and hovercraft, so this seagull is going back to preening

[Lodesman]

Thanks HSeagull

Originally Posted by Seaworthy Lass

Explanation of why you need to mark D2 on the chart as well as D1 (D1 is usually the same point as D using the RYA method).

Note: You need to work relative to the rhumb line as B is on the rhumb line.

Firstly estimate (an educated guess will usually do) the whole number of hours to get you nearly to B. If this was actually where B was, this would give you your CTS. Mark this point D1.

D1 is however unlike to coincide with B, and you don't know how long the next hour of tide will apply as your CTS depends on the length of time you are subjected to the last bit of tide and the amount you are displaced by the last bit of tide depends on the CTS. So at a glance it seems unsolvable (except to a mathematician), so I have come up with an easy solution.

See where you end up on the rhumb line if the journey went for another whole hour (so you know you are displaced by the whole amount of tide). If B was actually there, that gives you the CTS. Mark that as D2.

Now you know that to reach B you must steer somewhere between these 2 CTS figures.
But how much?
Its nice and easy. If D1 is really close to B, the CTS is really close to the one for D2.
If D2 is really close to B, the course to steer is really close to the one for D2.
So measure (or simply eyeball the proportion) and mark that same proportion off the last current vector.
Label that C.
Join C to B. that is your CTS.

If the current is zero for the last hour you still need to plot D1 and D2 (as they may not fall either side of B and if not, you need to allow for another hour of current).
If D1 and D2 do fall either side of B, then there is nothing more to do (you don't need a proportion of zero LOL), so just mark the end of the end of the last current displacement vector as C and join that to B and that is your CTS.

Hope I have explained that so everyone can follow

I may have misinterpreted your method from the example, but assuming I understood your method, it seems accurate but a bit labourious. Where your ownship vector (CTS/SRO) crosses the CMG vector (rhumb line) you can calculate speed made good - in the example 11 nm over 3 hours for SMG of 3.67 kts - at that SMG 10 miles is coverered in 2h44m. Shorten the final tidal vector (0.5 kts for 44 mins = 0.367NM). The resultant vector from the end of the truncated tidal vector to the destination should be the course to steer and the length should equal S*T (4 kts * 2h44m = 10.93). I suppose it's possible (particularly with strong currents and/or slow boat speeds) it may not intercept exactly, but would refine the SMG based on where it crosses the rhumb line; in this case you would just do another correction as above to get the final accurate solution.

[Seaworthy Lass]

Originally Posted by Lodesman

I may have misinterpreted your method from the example, but assuming I understood your method, it seems accurate but a bit labourious. Where your ownship vector (CTS/SRO) crosses the CMG vector (rhumb line) you can calculate speed made good - in the example 11 nm over 3 hours for SMG of 3.67 kts - at that SMG 10 miles is coverered in 2h44m. Shorten the final tidal vector (0.5 kts for 44 mins = 0.367NM). The resultant vector from the end of the truncated tidal vector to the destination should be the course to steer and the length should equal S*T (4 kts * 2h44m = 10.93). I suppose it's possible (particularly with strong currents and/or slow boat speeds) it may not intercept exactly, but would refine the SMG based on where it crosses the rhumb line; in this case you would just do another correction as above to get the final accurate solution.

Hi Lodesman
Yes, it did seem accurate to me at the time, that is why I proposed it LOL. It seemed to work for all the cases I could think of including wildly high and low figures for the current at the beginning and end of a journey (and gave the correct answer) but this is not a good basis for proposing a new method. It needs to be based on good underlying principles.

The problem with this method is that is uses the rhumb line as a reference. If we look at this problem freshly without the distraction of the RYA method, why on earth should we use the rhumb line at all? Think about it. It has nothing to do with our course over ground (your actual position) and sum of current displacement vectors could end up just about anywhere. Why reference the rhumb line at all? We need to work with B, not some line that just looks convenient.

I did not see this at all until yesterday. I was smuggly thinking I had come up with a good method (pride comes before fall LOL), when a CF member sent me a PM shooting holes in the method. In fact he took a great deal of trouble to hammer that point home (very kindly, but hammer he did in case I merrily kept advocating this method). I thank him deeply for this, even though he made me feel incredibly stupid LOL. He told me to go back to the drawing board.

I have just woken up after about 10 hours sleep, and as usual inspiration hit me on opening my eyes again. I have come up with yet another technique, this one totally new as it ditches the rhumb line completely. There is no need to draw it on at all using my method.

I am just typing it up now and will then draw up a diagram to illustrate this.
I will start a new thread, as I don't want this issue lost in a very long thread about all sorts of other navigational matters.

Watch out for the next instalment coming up shortly .

[Dockhead]

Originally Posted by Seaworthy Lass

Sorry for the delay folks. It was time for cockpit showers and for a drink to be poured. It is a celebratory one here .



The CTS is actually 51 degrees if you use the two hour method to pinpoint D.

The CTS is actually 63 degrees if you use the three hour method (the method you have chosen).
You would arrive on the opposite side in a bit under two hours actually, but you would be over a nautical mile south of your destination (your chartplotter is broken and it a bit foggy like the last example) so you would have to make your way north when you realised where you were.

The correct CTS is 58 degrees (58.3 degrees, but you would not compute it this accurately from the chart method, even if your pencil is super sharp and the lines super thin as the RYA insists on for accuracy (by the way is it true that you fail if you use a thick line for computations because it is less accurate?)
Time taken is 2.5 hours.

I arrive at these figures using my method, but as I have said a few times today even this method will fail. It is based on rocky ground. It needs to be modified and I need help with that. We need to eliminate D and just concentrate on B in computations.

So Dave, you are five degrees off when calculating the CTS using the RYA method (63-58, NOT 60-58 as you have given).

This is not a useless argument. Having insisted the RYA method was mathematically accurate, it has been shown in this simple exercise that you arrive at an answer that is 5 degrees off.

This information about the accuracy of the RYA method for determining CTS is IMPORTANT.



This discussion is not about accuracy of the tide data. It is about the accuracy of the RYA method .



Are you now saying the RYA method is not precise mathematically as you have been repeating endlessly up to now?
Dave, I know I should be a little more gracious in victory, but I find it hard when you are now shrugging and saying there is "little point" in using an accurate method so why bother having one!



Dave, it doesn't matter what the angles are. The only difference is that you need to do it on a chart/graph paper if the angles are not all 90 degree ones. The RYA method does not become more accurate when angles change.

So, that's three independent calculations, using possibly three different methods, giving precisely the same result -- CTS 58.3, and time 2.5 hours.

I have to agree with Seaworthy on the point about accuracy -- the correct and logical path to truth is to test methods with idealized perfect data first. Will a given method produce the correct results with idealized data, or not? You can't test the method with corrupt data.

Once you see that you have a method which is internally consistent and is based on sound principles, only THEN do you start to worry about what to do with the imprecise data you have.

And in any case -- do you want to add imprecision of method on top of imprecise data? This is just wrong.

Concerning imprecise tide data, I say again -- you are looking for an average. You do not need to nail every hour, you just need a reasonable approximation of the average effect of the tides. I don't think that this is as futile as Dave makes it out to be.

I don't know about Annapolis or Bowditch or altogether what they teach in the U.S. these days, but I was taught 30 years ago with pencil, paper, protractor and slide rule (most of which, to my shame, I have forgotten) was to always calculate a unified course to steer especially when it is a multi-hour passage. If you are affected by tides or currents which are not substantially in your direction of travel or against it, you are really screwed if you don't calculate a CTS. It does not need to be precise. Even a very rough approximation is going to be better than dumbly crabbing along the rhumb line -- in fact if the only thing you get right is to correct your heading in the right direction, this will in most cases will be better than just crabbing along the rhumb line. You can't just do it as you go along, or you just end up crabbing along. That's not what I meant by an "analogue fudge" -- I meant making intelligent approximations in an analogue manner -- for example, if you are 2/3 of the way between neaps and springs, you make up a number between the springs stream and neat spring which you feel is the right distance between.


I have been trolling the net trying to get a detailed explanation of the RYA method of calculating CTS, which I admit I still don't understand (and therefore can't critique). One thing I found was this:

Sailtrain: Navigation and Chartwork,

Imagine my amazement, and delight, when I found that in these professionally written training materials, I found the SAME MISTAKE (!!) we all made about the ground track!! I just about fell on the floor laughing! What it means is that all these concepts are really bloody hard to get your mind around! Even professionals, even teachers screw it up! This mistake is even worse, because the problem studied was in a changing tide!!



Note the ground track shown!!! These are official training materials! I was flabbergasted!!!

[Andrew Troup]

Originally Posted by Dockhead

Sailtrain: Navigation and Chartwork,

Imagine my amazement, and delight, when I found that in these professionally written training materials, I found the SAME MISTAKE (!!) we all made about the ground track!! I just about fell on the floor laughing! What it means is that all these concepts are really bloody hard to get your mind around! Even professionals, even teachers screw it up! This mistake is even worse, because the problem studied was in a changing tide!!

Attachment 53300

Note the ground track shown!!! These are official training materials! I was flabbergasted!!!

Hmmm - isn't that track correct, though, for the situation given?

(Assuming the tidal current is at a constant rate in each direction except around slack water - but even if sinusoidal, it would still be very roughly correct, no?)

[Dockhead]

Originally Posted by Andrew Troup

Hmmm - isn't that track correct, though, for the situation given?

(Assuming the tidal current is at a constant rate in each direction except around slack water - but even if sinusoidal, it would still be very roughly correct, no?)

Seaworthy pointed out the same thing to me in a PM.

I throw up my hands -- I confess once again that although I have a good grasp of water tracks and how to compute CTS in most situations, I am clueless about ground tracks. I seem to remember on real passages being on both sides of the rhumb line. Maybe you can run that situation through your CAD program and see?

Otherwise, where is that sheepish smiley?

[Wotname]

Originally Posted by Seaworthy Lass

Hi Wotname
Do you think the RYA will maybe send me a box or two of some fine single malt in appreciation of discovering a better technique for determining CTS? If they do I will be happy to share it with you . I know the RYA will probably be very grateful being able to improve their teaching methods .

All the old textbooks will become collectors pieces illustrating what they did in the 'old' days before my method was adopted.

Thanks for your moral support xxxxxxx

Cworthy, you either have very dry sense of humour or very little understanding of the working of the upper echelons of any "Royal Association".

I suspect the former and guess you are not standing by the mailbox with a developing SM thirst.

Keep the buggas honest

[Wotname]

Originally Posted by Dockhead

Seaworthy pointed out the same thing to me in a PM.

I throw up my hands -- I confess once again that although I have a good grasp of water tracks and how to compute CTS in most situations, I am clueless about ground tracks. I seem to remember on real passages being on both sides of the rhumb line. Maybe you can run that situation through your CAD program and see?

Otherwise, where is that sheepish smiley?

You were comparing apples with oranges. One example was simply a current all going in one direction (but at varying rates) for the duration of the passage while the other is a current that reverses direction about midway during the passage.

They can't both give the same ground track effect.