Distinct Activities: Shackled by a Common Name?

[goboatingnow]

DockHead

I must say your analysis feat is substantial

I have one issue ( other then the average CTS calculation tidal vector )

If you are only given the tidal effects for the passage of the boat following a constat heading and hence spending 4 hours on the water. How can you deduce what the tides are for any boat following a slower water track. Its like you have the Answer but cant find the question.

Hence how can you compare different time transits' with each other ( in other words , what does a slower boat experience as tides, since you arnt given that information in the first place.)

but proceed

dave

[goboatingnow]

CTS CALCULATION USING THE RYA METHOD

A = departure point
B = destination
C = sum of current vectors for the number of hours taken to get close to B
D = arc off of boat distance travelled from end of sum of current vectors for however many hours the current applied to the course line (= "rhumb line cut position")

NO this isnt correct , This is only ONE way of getting the estimate of how many tides you will pass through , its not a good way for complex tides

[goboatingnow]

Seaworthy for that method you have drawn , the chart is correct , its a one hour plot, and it seems completely correct . ( ps you could just review my previous post out lining the full method) , saves you drawing graphs

PS you can call me Dave , I dont mind , its shorter then the moniker , or use GBN as some do.

Dave

[Seaworthy Lass]

Originally Posted by goboatingnow

Seaworthy , No graph paper, and currenty with the snow my Internet is wobbly

So Ive read back and I think there is some confusion lets review each step and you can tell me what you agree and disagree with , In the process we will go through the whole RYA method

For completeness the A = Start B = Destination, C= tidal vector and D is the rhumb line cut position

YES

These are the statements for agreement or other wise

(1) We all agree the hourly plot of a tide and the boat speed over that time results in a point D that may lie before or the ahead of the destination B, ( whether one hour or multi hour), D is a one hour vector addition point but also the boat is physically there after the 1,2,3, hour plot time

YES

(2) If D its lies before B it indicates that the journey over the water is slower then estimated, if it lies after it is faster then estimated

YES

(3) For a multi hour crossing the RYA adopts the following

* Estimate the journey , That estimate of time is the journey time ALLOWING for the anticipated tides , Given tidal data is in general specified in hourly increments , then the intended passage is estimated in quantums of hours

YES, this is what the RYA method does

* It doesnt matter if you over or under estimate the time, the first graph will show the errors

NO, it does matter if you over or underestimate the time. For this reason, you need to examine BOTH cases. You need data for the last current influence and you need to plot D for the beginning AND the end of that full hour (I call that D1 and D2 in my method, which I will explain properly in a couple of posts).


(4) The tidal data is valid for 30 minutes +- ( this is by definition) , in other words you have no other data upto 30 minutes before of after the hour. HENCE it is acceptable to inflate the vector triangle or deflate the triangle pro rata for 30 minutes each side of point D. - have a look at this

YES, but not beyond it. A one hour block centres on one figure. If you go half an hour past the time this figure is valid, you are subject to the next figure for current.

(5) so simple case we etimate a one hour crossing at a given time, we look up the tide vector and do a one hour plot, equally we plot the one hour boat speed through the water . we cut the rhumb line with the boat vector, gving us a one hour plot, WE NEVER jooin the end of the tidal vectors to the destination

YES for the RYA method. NO for my method. If you want to finally arrive at B you must produce a method of calculating CTS that has you arriving at B, not arriving at D then extrapolating the result. This extrapolation is the source of the inaccuracy in the RYA method.

(6) The bearing of the line CD is read , this our CTS to point D

YES

(7) NOW, lets say point D is in front of B(dest), by inspection we can see that D is within the 30 minute time from B. HENCE given the validity of the tidal vector ( we have no < 30 other tide data available) , HENCE we " inflate " prorate the one hour plot ( as I did previously) so that D coincides with B, Our CTS remains the same, all thats different is that we <30 more at sea getting to the destination

D could be much further from B. In the hovercraft example it was 3.51 miles away for a 16 mile journey. In this example you calculated the hovercraft had another hour and seven minutes to go. this is an entirely new hour for current and the current could do anything in that time, even reverse in direction. You need a method that is accurate for ALL circumstances, not just when D is very close to B. Extrapolations may be very inaccurate and need to be avoided.

(8) Lets say D falls behind B , again by inspection we see that D is within 30 minutes of B, So based on the assumptions behind tidal data, WE deflate the triangle so that D co-incides with B, hence our time at sea is proportionally less then 60 minutes ( in a one hour plot), Our CTS bearing for the voyage always remains constant the bearing of CD.

NO, this will only give an approximation. It may work some/most of the time, but not all the time. I made this mistake this morning thinking my 'parallel' method would work which is similar to this. It does not, so I have reverted to using the method I came up with at 3am.

(9) OK , you say , and this is your bit, the hourly ( 1,2,3, hour it matters not) , plot point D falls more then 30 minutes either behind or ahead of B.

What thats means is your original journey estimate was incorrect and you need to either add the next hours tide or remove ( ignore ) one , Hence if your 2 hour plot results in a >30 "error" of DB time, You rework the plot adding increasing the number of hourly plots until point D is within the 30 minute margin allowed by the tidal data.

Then you inflate of deflate the resulting vector plot as before ( ie with the 30 minute window)

This would help, but it is not what you did in the hovercraft method, you used the exact procedure given in the second RYA video link you gave. You didn't add on another hour, so I presume it is not what the RYA are stressing is necessary. Even if you added on another hour the method would still be approximate, not accurate. You would still be extrapolating the last bit.

Thats the RYA method in full. it takes into account all possible tides that may apply for the full journey within the limitations of the tidal data

Dave

Glad we are getting some common ground Dave
My biggest point during all this discussion is that the RYA method gives you the CTS to get you on the rhumb line hopefully close to B, then you must decide what to do when you get there (even if all predictions are spot on, the RYA method does not give you the CTS to arrive at B unless the current for the last hour is EXACTLY the average of the sums for the journey up to that point (and why on earth should they be? That would be a very unique case, not the general rule).

[Seaworthy Lass]

Originally Posted by goboatingnow

NO this isnt correct , This is only ONE way of getting the estimate of how many tides you will pass through , its not a good way for complex tides

Not sure what you mean. How does the RYA teach that you should estimate the number of hours a journey will take?

[Andrew Troup]

Originally Posted by HappySeagull

boy, this is difficult... but the ants are gone, anyways
... .

To those who dread any return to the status quo ante, I can only remind them of what Bob Dylan famously had to say on the topic.

[Andrew Troup]

Originally Posted by Dockhead

Oh, so you just used it to graph the data . Excel can do that too.

Dockhead

thanks, but I did know that, witness the opening sentence of my post of the CAD plot

http://www.cruisersforum.com/forums/...ml#post1133411

where I acknowledged that, but mentioned a benefit of doing it using CAD.

[Seaworthy Lass]

CTS CALCULATION USING MY METHOD

A = departure point
B = destination
C = sum of current for the entire journey
D1 = arc off of boat distance vector from the beginning of the last current vector (bisecting just before B)
D2 = arc off of boat distance vector from the end of the last current vector (bisecting just after B)

AB = AB = 12 nm (OOPS wrote it incorrectly as 16 in the RYA example, it was correctly marked on the chart photograph)
B is due east of A
Boat speed is constant at 4 knots
Current:
1st hour: 2.5 knots from N to S
2nd hour: 1.5 knots from N to S
3rd hour: 1 knot from N to S
4th hour: 1 knot from S to N

Plot of the vectors on a chart is shown below.

Calculations:
D1B / D1D2 = 1.1 / 4.6 = 0.24
Last current displacement vector = 1 nm S to N (note reversal of direction).
So mark C at 4.72 ( = 2.5 + 1.5 + 1 - 0.24)

What is the time taken to travel between A and B?
= CB / 4 = 12.9 / 4 = 3.2 hours
What is the constant CTS? 66.6 degrees (angle of CB)

Well, as D was so very close to B in this example, my method and the RYA method give similar for this example (but not identical results).
It is not as dramatic as the 7 degree difference in the hovercraft example! I should have picked better figures to demonstrate my method, as there would be lots of instances the difference would be marked, as in the hovercraft case, but getting tired (nearly midnight and I am desperate for some sleep and too tired to play with data). Just wanted to demonstrate the technique for my method.

I will try an example with more current tomorrow and see how the two methods compare. I think they will often produce similar results (the extraploation the RYA is using wont have a big effect if D is close to B) but very significantly not always. This is the thrust of the message.

The RYA method to find the CTS is a great approximation in many cases, but it has its limitations.


Latest details of my method for determining CTS (based on the method in post #564, not using the later post #569, as #569 makes assumptions that won't always apply):

Proceed as you would normally in the existing RYA method and mark this point D1 if it is close to B. (If not, add on another hour's worth of displacement, arc the distance travelled vector off that and call that D1 instead).

Extend the course line past B.
Mark on the next lot of current displacement from the end of the last current vector. Repeat the arcing of the distance travelled vector for yet another hour. Hopefully this will lie past B. Mark this point D2.

Work out the proportion of the final hour it would have taken to get to B instead of past it (= D1B / D1D2) and mark off the same proportion on the final current vector. This point on the last current vector is now C. Join C to B. This is your true course to steer.

The length of CB divided by your boat speed is the time taken for the entire journey.

Make your correction for magnetic compass deviation and voila, you have an exact course to steer to get you exactly to your destination and you can also calculate the time taken for the journey exactly.

Of course this is limited by all sorts of factors like your speed varying, the current not being as predicted, leeway, etc, but at least it is now an excellent starting point.


In our hovercraft/sailing example D1 is at 12.49 nm from A. D2 is 17.32 nm from A. As the current for this final hour is zero, there is no need to work out what proportion of time we are subjected to it and the end of the first four current vectors is therefore C so we simply connect this with B.

So for the hovercraft example the length of AC is 10 miles for the journey. It happens to be perpendicular to AB, but the beauty of the RYA method is that it easily allows for the summation of complex currents from any direction.
The length of CB in this case is 18.87 miles.
The time taken is 18.87/4 = 4.72 hours.
The course to steer is 58 degrees true if you measure it with the protractor (the original heading was 90 and allowance for current is 32 degrees). This gives the correct course to steer for the journey (instead of being nearly 7 degrees off as it is with the RYA method).



That's all for tonight folks.
Sorry I was annoyed at you this morning Dave. I was just tired of being repeatedly told how confused and wrong I was . My equilibrium is now restored .

Night all.

[goboatingnow]

Originally Posted by Seaworthy Lass

Not sure what you mean. How does the RYA teach that you should estimate the number of hours a journey will take?

by quesstimation, for example you couldnt solve DockHeads channel crossing correctly without a reasonable guess at the total time, otherwise you have the wrong number of tides.

Note however that the doesnt matter once you abide by the 30 min rule

dave

[goboatingnow]

Seaworthy , I understand what you are doing, but your examples simply cannot cope with the real world, equally you could not get the tidal data like you are saying. There is a danger of fiting the tides to the result here

Firstly the RYA method does not always result in D being in front of the destination. any tides or set of tides in that pushs you towards the destination will result in D ahead of B

Are you saying I cant proportionally reduce the triangle ( by 30 minutes )

The RYA method DOES correctly give you course to steer.

Are you saying

[goboatingnow]

Extend the course line past B.
Mark on the next lot of current displacement from the end of the last current vector. Repeat the arcing of the distance travelled vector for yet another hour. Hopefully this will lie past B. Mark this point D2.

Work out the proportion of the final hour it would have taken to get to B instead of past it (= D1B / D1D2) and mark off the same proportion on the final current vector. This point on the last current vector is now C. Join C to B. This is your true course to steer.

The length of CB divided by your boat speed is the time taken for the entire journey.

YesThe mistake I made in Hovercraft Seaworthy that is in essence the RYA method except, You remove the time from the last current

The RYA removes the time averages across all the currents, because in a say 6 hour plot averaging all the tidal vectors up or down by 30/6 ie 5 minutes is more accurate then reducing linearly the last current by say 40 minutes.

in practice there is no need simply compute the advance or retard at the average speed of the all the tidal vectors. The tidal data is valid enough to sustain that the change over 30 minutes is simply minute. In practice with your example No tide reverses from forward to back in effect in an instance.

IN the real world there is simply no need to go to your effort.

(a) compute point D either behind or in front of Destination B. calculate the SOG as a result of that . Apply that SOG to the remaining (<30) min journey. The CTS remains the bearing CD at all times.

The results are accurate , in fact more accurate then what happens in real life ,

The danger here is you are attempting more accuracy then the underlying data represents and therefore such calculations are a waste of time.

If you look at tides you will see that the first several minutes of each time are actually virtually identical, hence whether you use say +10 minutes from the earlier one or -10 minutes from the later one is irrelevant.

Hence assuming the tide runs on for a certain time carries little error in real life.

Remember , you have laid out a table of four tides , in practice from tide tables you would getting nothing of the sort as you would have to interpolate the set and drift from a fixed HW time point , incremented by a hour,

example was that I should have stated that the RYA technique would recalculate with 5 tides , which would move D within the "error" circle of the tidal information


The RYA method is as good as any given the data involved and easier to apply


Dave

[goboatingnow]

In our hovercraft/sailing example D1 is at 12.49 nm from A. D2 is 17.32 nm from A. As the current for this final hour is zero, there is no need to work out what proportion of time we are subjected to it and the end of the first four current vectors is therefore C so we simply connect this with B.

who said the current was zero in the last hour.?

The correct answer to the hovercraft example, is (a) We only have 4 hours of tidal data and hence it isnt possible to accurately compute the result.

therefore its either (a) based on what I gave , just inflate the triangle or (b) declare the the result void because there is no data.

Your method simply isnt right you have made an assumption of slack tide.

[goboatingnow]

Glad we are getting some common ground Dave
My biggest point during all this discussion is that the RYA method gives you the CTS to get you on the rhumb line hopefully close to B, then you must decide what to do when you get there (even if all predictions are spot on, the RYA method does not give you the CTS to arrive at B unless the current for the last hour is EXACTLY the average of the sums for the journey up to that point (and why on earth should they be? That would be a very unique case, not the general rule)

So explain the RYA method when D ends up behind B????

again you miss the point I made a few posts ago, You get D within 30 minutes of B for a hourly plot. Anymore and you add or subtract a whole tidal hour.

The tidal data DOES remain valid if you proportionally move the triangle up or down in order to make D coincide with B, This is a basic fact of tidal data, the error one encounters in doing that is less then the basic tidal interpolation to get the vectors in the first place. its certainly valid for the couple of minutes thats typically needed spread over a multi hour tide, any more and you add or subtract a whole tide

You persist in using the hovercraft example when I have already said repeatedly that I should have rerun the plot with five hours of data as, my original planning estimate was wrong , YOU DO not average up or down the tides by an hour. The max is 30 minutes and the data is valid for this .

The RYA for example also shows that you can use 30 minute tidal plots for areas subject to precise tides and where you have detailed set and drift data.

Just remember the cardinal error is to join the end of hourly tidal plots to the destination, You always set your dividers to the boat speed time number of hours of potting and arc that through the rhumb ( to do the vector addition). Thats the correct course to steer in all cases.

To finish your method is merely the RYA case where you always plot enough tides to make D go behind B.

Whats more accurate Seaworthy D 15 minutes away from B, inflated proportionally or your method that could put D 45 minutes behind B and your suggesting the last tide is still valid shrinking it back by 45 minutes .

You are aware that tide data is only actually valid for a point in time , The assumption that the tide is precisely constant over the whole hour is a mathematical abstraction and in practice nonsense.

The RYA state that inflating each tide proportionally over the whole plot is more accurate that any other method , once such method does not extend beyond 30 minutes, otherwise you add or subtract a tide.

Your method is based on the fact that 10% of a final tide in your view is more accurate that me taking say a 6 hour plot and merely increasing each tidal vector by 5mins. PLease I know from real life which is more reliable.!!!

peace and love

Dave

[Jammer Six]

After a brief stumble through the thread, I'm not sure why we switched from boats and current to hovercraft and wind.

[Andrew Troup]

I let the idea quoted below swim quietly past, at the outset of this thread.

At that time there was much thrashing about by a few other people, who insisted on retaliating with frantic splashing and verbal waterbombs to ideas nobody had actually advanced.

It seemed to me that in that climate, any sort of rational discussion was effectively drowned.

Originally Posted by DeepFrz

Wotname wrote:

Ahh... my friend, I would class you as a true navigator as you are using some of the finest navigational tools available; the Mk 1 eyeball, the active brain and the sense to use them when and as required

DeepFrz replied:

That would be called Pilotage rather than Navigation would it not?

Quote from Wikipedia:

Pilotage is the use of fixed visual references on the ground or sea by means of sight or radar to guide oneself to a destination, sometimes with the help of a map or nautical chart ....

When visual references are not available, it is necessary to use an alternative method of navigation such as dead reckoning (typically with a compass), radio navigation, and satellite navigation (such as GPS).

end of quote from Wikipedia

So it seems that other forms of navigation have different names already.

When you say "That would be called Pilotage rather than Navigation", isn't that a false dichotomy?

Surely pilotage is a sub-branch of navigation (as the last part of the wikipedia quote implies - "When visual references are not available, it is necessary to use an alternative method of navigation"

- I'd be the last to claim that wikipedia is any sort of authoritative source, but it's what you chose to post (AT wrote mildly)

Question 2:
Are you suggesting, in your last sentence, that, just as pilotage is considered a sub-branch of navigation, satellite navigation could be so considered?

Note to our delegates from the Ministry of Defence: I'm asking a question here, not stating a position