Distinct Activities: Shackled by a Common Name?

[Dockhead]

I'll play with it next time I'm on the boat where my charts and Portland plotters live.

But in general I agree with you, if I am correctly understanding the problem (maybe Dave will jump in to clear up some misunderstanding). I want to know the whole steering error for the whole passage, as accurate as possible given the data. Then I will add an uptide margin of error myself.

One thing worth repeating for anyone who hasn't done this kind of navigation - you don't need perfect data at all, in order for this to be an extremely valuable technique. Even pretty wild guesses will do. The errors seem to cancel out (I don't know what is the mathematical principle). You just need to know direction of the current and a rough idea of speed and you already start getting good results. Obviously better and better, the better the data is, but already very good, from rough guesses. The main variable is passage speed, which is why I always do three sets of calculations, so I know the range of steering corrections. So if my base case is 8, and I see I'm doing more like 8.3, I look at my calcs for 9 and immediately see what I need to do.

[Dockhead]

Originally Posted by Seaworthy Lass

Gas ran out so no tea or coffee. At least we have another gas bottle on board, but I need to wait for it to get connected (my hubbie's job) .

Come on folks, please respond. Don't be worried about making a mistake and looking foolish. If you drop that fear we can bounce ideas back and forth - brainstorming produces wonderful results .

I will draw out the hovercraft example on a chart to show you what I mean and then maybe I will make up another example and plot it out for you too. It may make sense then.

Don't you have an electric kettle?

[Seaworthy Lass]

Originally Posted by Dockhead

Don't you have an electric kettle?

We run purely off solar, so I have no electrical appliances in the kitchen at all.

PS just photographing the chart I have drawn the hovercraft example on.

[Andrew Troup]

Originally Posted by Dockhead

....
Andrew, maybe you could check my last spreadsheet with your CAD program? Maybe everyone else is bored with it, but I for one would really like to know that we've done all the maths correctly.

Dockhead

Maybe you thought I used the CAD package to solve the triangles. I didn't, Excel did all that, with a bit of coaxing, and gave me the sideways displacement from the rhumbline at each constant heading slice.

So the CAD package just fitted a curve through the xy coordinates I extracted from the spreadsheet.

I might be wrong, but I don't see any sideways displacement from the rhumbline in your spreadsheet, so I'm not sure what you would like me to plot ?

[goboatingnow]

Originally Posted by conachair

Agree with that, of course. It was the "D is merely a vector addition vertices. You are not physically there." bit which was confusing.

I can't see how you come up with that. You start with a velocity vector, speed (in Kn or Nm/H) of tide and direction, then constrain the time to 1 (hour) leaving you with a unit of distance, Nm/1. This is what you plot, distance and direction.

Otherwise the scale of the vectors wouldn't matter as long as they were all the same scale but not necessary the same scale as the chart, all you would get would be a CTS, no idea where you would be after the 3 hours or however many you plotted.

Sorry about delays in replying , bit of Snow got in the way and a power failure !!

You are correct at least you accept the CTS is right.

The Fundementally mistake being made by seaworthy is for a vector plot over time ( forget the real tides for a minute) , if you inflate the water track ( to take accout of the actual time of the journey. ) then you must inflate all time based vectors.

The current flowing at 2knots does not stop after 1 hour. , if your out there in reality for 2"5 hours the tidal vector have a 2.3 hour effect.

There is no forgotten tide in the RYA method. The maths simply deal with the given vectors.

If I can get some graph paper in my local shop ill do a series of vector graphs to show what happens.


Dave

[Seaworthy Lass]

EXAMPLE USING THE 'CF MODIFICATION' TO THE RYA METHOD OF DETERMINING THE COURSE TO STEER:

Here I have illustrated GoBoatingNow's hovercraft example we have been discussing with (changing units to nm not miles).

These are the details of the journey:
Your destination is directly east, 16 miles away.
Your boat speed is a constant 4 knots (motoring with no wind so no leeway).

You have estimated the time of the journey as 4+ hours and thefore found the current for five hours (very important to do this for at least an hour longer than the time taken to go straight from A to B with no current).
Current is from north to south as follows:
first hour 3 knots
2nd hour 2 knots
3rd hour 4 knots
4th hour 1 knot
5th hour slack water (no current)

A = departure point
B = destination
C = displacement of the total current for the entire journey (it will fall somewhere along the vector for the displacement due to current for the final hour of the journey, not usually at the end of it, but in this case it is at the end, as the current in the last hour is zero)
D = the point where the distance travelled for a whole number of hours first extends past B when the distance is arced off along a line extended through AB.

What is the course to steer and how long does it take to arrive?

Solution:
- Plot five lots of the current south going from A (10 nm). You do 5 not 4 hours worth, as D must extend past B and it there is any current at all during any part of the journey, you need to work on the next hour as well.
- Use dividers to arc off 20 nm from the end of the current displacements (5 hours times 4 knots). As it extends past B, we can label that D.
- we can see that it extends past B so we know the journey will take between 4 and 5 hours. If it fell short if B we would have to add on another hour of current displacement and arc off 24 nm this time and again check if it extended past B.
- as the current is zero for the last hour we cannot take a proportion for that and instead of drawing a line to D as we usually would do first for this method, we need to take it straight to B.
- we can now label the end of the sum of the current displacements as C

Measure the length of CB. It is 18.9 nm. Divide this by 4 knots to give the time taken.
The time taken is 4.72 hours equalling 4 hours 43 minutes (agrees exactly with my calculated time).

The course to steer is the angle of CB which you measure to be 58 degrees (agrees with my previous calculations of 90 minus 32).

Don't forget to then need to work out the magnetic compass deviation for the area you are in .

[goboatingnow]

Originally Posted by Dockhead

ROTFLMAO!

The most hilarious, wittiest, and best concession speech I ever read! Love it!

I have to say, Capt Force, that we all owe you a huge debt of gratitude!

You stubbornly challenged something all the rest of us took for granted, and forced us to think it through to a level of detail we had never ever imagined even existed. In the process of pompously pontificating on various aspects of the problem, every one of us, including Dave (!) had brain farts, made mistakes, and learned the hard way that none of us understood it at all as well as we thought we did, including even Dave, who teaches this ****!!!

I might have to rate this as the single best thread ever on CF, at least in terms of how much people learned from it.

You may make a concession speech. , I maintain the way the RYA does it is correct given the vectors concerned, there is no missing tide. I will show it is correct

I confess I haven't followed Captforce thread.

What I can't understand is , of you are given the tide for your particular journey. , how can you extrapolate the data for longer journeys. , all you can do s make an assumption that irrespective of the time taken , the current vectors remain the same irrespective of the crossing time. I still don't understand how this can be resolved

Dave

[goboatingnow]

Originally Posted by Seaworthy Lass

The hovercraft scenario and all the data was repeated in post #519, so you don't have to go back far.

I would love it if someone would work it through from first principles as I did, then work it out following the RYA method in the youtube that GoBoating was using (and that I also repeated) and compare the two results.

I am convinced the RYA method is not a precise formula as GOBoating keeps insisting, but a rough approximation to get you on the rhumb line hopefully not too far from the destination. The heading needs to changed well before if there is any component of the current running with you, as that places you off to the side aiming to a point behind the destination.

The heading also needs to be altered significantly if the current in the final hour of your journey is predicted to be significantly different than the average current over the part of the journey before you hit the rhumb line. I am not talking about unexpected current, but the predicted one.

In the hovercraft example if GoBoating persisted in following the course to steer that the RYA method determines, he would pass up current of the destination and never reach it!

The RYA method does NOT give you the quickest route to the destination and it also does not give you a course to steer to get you to the destination (as GoBoating keeps repeating it is), it just gets you close. It is, however, a GREAT way of getting close if you have very complex current vectors that would otherwise need to be resolved.

The method is very useful, but its limitations need to be recognised particularly if you are teaching the subject (and I am talking limitations even when you have perfectly accurate data to work with - if that was only possible in the real world!)

no no no

You still are one applying the vectors incorrectly. What you are doing in all your graphs is applying the rules backwards. You are computing the start from knowing the end.

Later today ill have graph paper.

Think of it like this.

You do not know the amount of time it takes to actually cross , so you don't know in advance what the length of the tide vectors will be. The more time you spend the greater the effect of the tidal vector. In other words if you. Spend 90 minutes in a 2kn tide , you are not subject to 60 minutes a d then 30 minutes of nothing. You are subject to 90 minutes of the 2kn tide

I'm sorry Seaworthy I will show you that the graph maths you use are wrong. You are constantly in effect computing Captforces scenario , ie you are computing the result from the end of the journey

To say it again , try doing time plots rather then distance to try and see it.

Look at it this way. You are given speed vectors. Yet you persist in plotting distance vectors

Dave

[Seaworthy Lass]

Originally Posted by goboatingnow

Sorry about delays in replying , bit of Snow got in the way and a power failure !!

You are correct at least you accept the CTS is right.

The Fundementally mistake being made by seaworthy is for a vector plot over time ( forget the real tides for a minute) , if you inflate the water track ( to take accout of the actual time of the journey. ) then you must inflate all time based vectors.

The current flowing at 2knots does not stop after 1 hour. , if your out there in reality for 2"5 hours the tidal vector have a 2.3 hour effect.

There is no forgotten tide in the RYA method. The maths simply deal with the given vectors.
Dave

This is PRECISELY why the RYA 'course to steer' method is inaccurate.
Why on earth would you inflate the vectors for the previous tide . You would simply use the predicted value of the tide for the following hour for the calculations.

And please don't tell me you don't know what that is. It you can find what the predicted tide is for the first four hours, you can find out what it is for the fifth!

Great errors come about if the tide is not the same as the average for the previous part of the journey (as they did for you in the above example, where using the RYA method gave a course to steer of 51 degrees not 58).

The RYA method is to inflate the total size of the current vector and use this for the last part of the journey - you have just admitted so yourself!!! This is what I have been saying all along .

What if the predicted current is vastly different for this final period of time? What if it is zero as it was in the hovercraft case? What if it reverses in direction?
Why use an "inflated figure" as you call it and not the predicted figure?

I repeat, this is PRECISELY why the RYA method is not accurate .

[goboatingnow]

Originally Posted by Wotname

The RYA method seems to be in a similar vein as say using 22/7 as the value of pi. That is, a good approximation and quite useful, just not exact by any means.

No its is mathematically rigerious given the information you know. It cannot compensate for tides you don't know about.

Seaworthy triangles in effect compute the tide to suit the crossing ( like capt force ) , but we don't usually need to work out what we have done ( past time)

Again think of it like this if you are in a 2kn tide and as a result you go slower , do you still only spend 1 hour in its influence ! Of you accept that you the resultant vector of the tide looked at in a time plot , still be a unit of 2 !!

Dave

[goboatingnow]

Originally Posted by Seaworthy Lass

Yep . I agree, we must have just rounded off the figures differently, I had 4.72 hours.

The RYA method calculated five hour seven minutes and what GoBoating refuses to recognise is that that is the time take to arrive at a point upcurrent of the destination (he will never pass through the destination point). He insists I am the one who will not arrive following a steady course of an offset of 32 degrees (B is due east of A magnetically, this makes my compass heading 58 degrees, GoBoatings offset is 38.7 degrees making his compass heading 51.3. This is not an insignificant difference).

It is not the we "don't know about it" it is that the RYA method completely ignores the predicted current in the final bit of the journey (and in this case it was not an insignificant bit, it was 3.51 miles out of a 16 mile journey!).
If the data is available for the first four hours of this journey, then it is also available for the 5th hour. The RYA is simply ignoring it.
This is precisely why the RYA method is not precise. It does NOT have us arriving at the destination as GoBoating also insists it does.

In addition GoBoating is also insisting he will not arrive on the the rhumb line at the four hour mark. This is incorrect. He certainly does! This is the whole thrust of the RYA exercise, to get you on the rhumb line close to the destination every single time.
Unfortunately the method is not always successful at getting you close to the destination when you arrive at the rhumb line (the hovercraft case illustrates one such example).

Yep, demo graphs are there, including a dotted line indicating the chartplotter track.

Wrong , all tidal plots are plots in unit time , ie speed vectors not distance vectors

Dave

[Seaworthy Lass]

Originally Posted by goboatingnow

Wrong , all tidal plots are plots in unit time , ie speed vectors not distance vectors

Dave

Dave they are not!
You say you teach this, but you are making huge errors and not only with this point.

I was mislead by you when I was watching the first video link you gave, as you told me they were speed vectors. The example went for only one hour so I did not realise the error then. Watching the second video link you gave it is very clear they are NOT speed but distance vectors as the duration was 2+ hours for the journey and distance not speed was marked.

Go watch the second RYA video you posted a link to. The instructor actually says he is "marking how far the tidal flow will have taken us". He marks this in nm. Later he says "so in two hours this boat will have travelled 12 nm, set dividers for 12, we put one end on the end of the second tidal vector and we arc off until we meet the course line". So the 12 he is marking is nm not knots.

The reason this works is that he is doing this in hour lots. One hour at 'x' knots equals 'x' nm.

This may seem pedantic, but it is in fact critical as in the RYA method the point D is where you will end up on the rhumb line, NOT at the destination as you keep repeating.

Could you please tell me for the hovercraft example how far from the rhumb line you think you are at 4 hours? You have said before you are somewhere south of it which is totally incorrect. You are AT it at the four hour mark .

[goboatingnow]

Originally Posted by conachair

Well, I'll continue to back up the maths side . (we call it maths over here )
Lets say you're heading up the english channel from A to Destination. On the charts there are tidal diamonds in various places which give the average tidal speed and direction at that point for each hour for 6 hours before and after high water somewhere defined.
You can plot a , say, 3 hour passage using the data from the tidal diamond which is closest to your position at each hour.

With the vectors for the boat speed and direction then you will be exactly where the plot puts you, assuming the tides from the tidal diamonds were exactly what you were experiencing.

Maybe not exactly where you want to go but from the data you are stuck with averages over an hour. The maths, I think, cannot be wrong. It's just when you get into the real world it starts drifting out because there isn't actually that much data.

I think

You are right this is the RYA method , when you lift whatever tidal vectors and you apply them your result is only as good as that data. BUT the RYA method is mathematically correct

All you know at the start is the tidal speeds and directions and the boat speed.

Tide speeds are the same tide simply running at different directions and speeds over TIME.

What seaworthy is doing is retrospectively fitting the tides into the actual journey. , that would be correct if you travelled across and somehow guessed the right angle and as you went and measured the distances the affected you over actual passage time. Ie the Captforce scenario. This would result in the triangles Seaworthy computes.

Of course this is impossible to do in advance to use that method.

So what we have is a boat speed and tidal vectors by the hour , the tidal vectors are spaced across your journey in Time not in distance.

Simply if you water track speed of the boat is 4kn and lets say your not sure how far away the other side is. , but you know there is a current of 1 knot. directly perpendicular , compute the course to steer , after 1 hour how far away are you from the destination, at 4:20 how far are you away from the destination , at 4:40 how far away are you from the destination.

This will demonstrate a plot in TIME , which is what you have , NOT a plot in distance, which you do not have.

seaworthy continues to mx up time based plots and distance ones

Dave

[Seaworthy Lass]

Originally Posted by goboatingnow

..... the RYA method is mathematically correct

It is NOT always correct, it is only correct if the tide in the last portion of the journey is EXACTLY the average in speed and direction to the tide for the previous portion of the journey. Why on earth would it be?

Originally Posted by goboatingnow

What seaworthy is doing is retrospectively fitting the tides into the actual journey. , that would be correct if you travelled across and somehow guessed the right angle and as you went and measured the distances the affected you over actual passage time.

I most certainly am not LOL. I am not doing anything at all retrospectively. I am just taking into account the predicted tide for journey in the 5th hour, not some extrapolated amount that you (and the RYA) for some reason want to do.

Originally Posted by goboatingnow

seaworthy continues to mx up time based plots and distance ones
Dave

I do not. You are the one mixing it up. Please go and watch the link to the second video that you posted .

[goboatingnow]

Originally Posted by Seaworthy Lass

This is PRECISELY why the RYA 'course to steer' method is inaccurate.
Why on earth would you inflate the vectors for the previous tide . You would simply use the predicted value of the tide for the following hour for the calculations.

And please don't tell me you don't know what that is. It you can find what the predicted tide is for the first four hours, you can find out what it is for the fifth!

Great errors come about if the tide is not the same as the average for the previous part of the journey (as they did for you in the above example, where using the RYA method gave a course to steer of 51 degrees not 58).

The RYA method is to inflate the total size of the current vector and use this for the last part of the journey - you have just admitted so yourself!!! This is what I have been saying all along .

What if the predicted current is vastly different for this final period of time? What if it is zero as it was in the hovercraft case? What if it reverses in direction?
Why use an "inflated figure" as you call it and not the predicted figure?

I repeat, this is PRECISELY why the RYA method is not accurate .

Yes the RYA method applies the the Rate of known tides over the course of the actual journey. You cannot of course find all the tides in advance , because you cannot know in advance how long the journey will take. You can only estimate the tides

HOWEVER

Once you have lifted whatever tidal information you determine in advance , the vector plot is a plot in TIME , not distance. ( because you only know distance by hour not in total.

I'm not arguing that there are issues in determining what actual vectors apply, ie what you should lift from the tidal stream atlas. , these are all approximations.

Again the key issue is. Plot a CTS knowing boat speed and and estimation of the tides that apply

Your idea is in effect back computing tides vectors based on completing the journey then re examining the journey to get more tidal vectors. Ie a recursive and hopefully converge series. In practice you can't do that as you can't determine it with any accuracy in any sort of complex case.

Dave