Distinct Activities: Shackled by a Common Name?

[Dockhead]

Originally Posted by conachair

Moving forward with Desmos graph app in chrome..

Still a way to go with sliders for easy manipulation of the data but moving in the right direction

From Capt example but with a single tide max flow of 4Kn, as a sine wave.

https://www.desmos.com/calculator/6quwldikyl



The curve from 0 - 8 is (hopefully) total distance the tide has moved (north in this example) calced by Y=(-cos(x)+1)*4. Suspect this might be wrong though, the "-cos(x)" is straight out of google but can't see why I had to add 1 to get it all positive. Looks about right though.

The next curve from 0 - 4 back to 0 should be the tidal rate. y=sin(x) x 4

Straight line should be the boat heading through the water.

And the final line is the tidal distance and boat track through water added together, ie boat track over the ground.

If you are using chrome the the graph app is a simple add on -

https://chrome.google.com/webstore/d...ojekgdko?hl=en


EDIT. I've just divided the Y axis by 5 to make the whole thing look a bit better. Dunno how to zoom just one axis yet.

Brilliant!! Perfect graphical illustration of the Capt Force scenario! A little bit of trig is just what was needed, I see!

[goboatingnow]

Originally Posted by Seaworthy Lass

Please don't be pedantic. Your exact words were:

"You know, because some nerd has worked out the winds speeds and direction across the 16 mile lake. He tells you that the winds are all blow North to south as follows as you know from experience that it has this effect on your hovercraft
first hour 3 miles/hour ( ie it will blow your hovercraft 3 miles south in an hour if it was just sitting there)
2nd hour 2 miles/hour
3rd hour 4 miles/hour
4th hour 1 mile/hour"

So why can't the nerd in this example tell me what the wind is for the 5th hour?

Please don't start picking the English , I gave you what I gave you. I was trying to simply the problem away from the ground track / water track issue. So I used terms that typically are used to extract tides.


Look here's what you get in a tide table.

Lets say your crossing the channel , you'll get something like this

At a tidal diamond positioned at a particular point on the chart ( which could be some ways from your ground track )

Something like this( though often represented in several ways )

hW Dover 2kn 90t
+1 2.1 100t
+2 etc etc


So yes you'll have 24 hours of data

Now you don't know in advance what your journey will take. You can only estimate. If its subject to complex tides the only assumption you can make is that the rhumb line is X and the boat can do Y , using that you extract the data.

Yes this bits imperfect.

But now that have the vectors you inflate or deflate the plot as required. In practice you spend some minutes less then 60 in the tide when you are bring slowed and some minutes faster then 60 in every tide if bring pushed. The tide vector is deemed to,be valid +- 15 minutes from the hour.

Hence you can't determine what other tide might apply because you have no data as to where to apply it and when ( ie your solution suggests the addition is at the end , but in practice it could be added in anywhere.

What you are doing is computing a tide that in effects suits the answer.

YES. You could do a recursive iterative method, ie guesstimate the effect of the tide over the whole " guesstimated " journey time. Yes in simple cases you could extract all the correct tides.

In complex cases with non right angled tides you could use your method. In complex cases it would require a lot of further recursive computation


To summarise the assumptions and to reiterate

(a) I said the RYA method is mathematically correct to apply tidal rate vectors and boat rate vectors. I will show that by graphs, ( or do the dont know how far the destination is example )

(B) it does not work well where the boat speed is similar to the tide as the final passage time may extensively affected by the passage of so many tides. It is therefore difficult to work out exactly what tide will affect what and would require recursive iteration , that hopefully converges. In most cases over a journey the boat is significantly faster then the tide over the sane journey. ( otherwise wed never get anywhere )

And yes it doesn't therefor work well when D turns out to be way short of the destination. Ie as I said where the tide is a significant proportion of boat speed, this has to do with the limitations of computing the course and time in advance and then attempting to determine the actual tides affecting the boat. ( this is actually very difficult in practice)

(c) in practice all combined CTS expose you to non intuitive ground tracks. as I pointed out to DockHead , you cannot easily determine safe clearance. , a point he did not agree , had he postulated the ground track of the varying non reversing tide he would only originally checked one side of the lines!! for that reason in real life I don't recommend multi hour complex CTS


Both of us are mathematically correct. I am constructing a known ( in advance ) vector addition and assuming it can be expanded ( for a limited time) ie the math is correct , ( note that the tidal information assumes i can )

but the real world may be slightly different , you are attempting to construct a vector based on attempting to predict all possible vectors that may apply to the REAL journey time.

Dave

[goboatingnow]

Further more Seaworthy. The method you mention where you continue adding vectors until past the destination is just the RYA method where the tides are pushing you , ie D ends up slightly beyond the destination

Dave

[goboatingnow]

Originally Posted by conachair

Moving forward with Desmos graph app in chrome..

Still a way to go with sliders for easy manipulation of the data but moving in the right direction

From Capt example but with a single tide max flow of 4Kn, as a sine wave.

https://www.desmos.com/calculator/6quwldikyl

The curve from 0 - 8 is (hopefully) total distance the tide has moved (north in this example) calced by Y=(-cos(x)+1)*4. Suspect this might be wrong though, the "-cos(x)" is straight out of google but can't see why I had to add 1 to get it all positive. Looks about right though.

The next curve from 0 - 4 back to 0 should be the tidal rate. y=sin(x) x 4

Straight line should be the boat heading through the water.

And the final line is the tidal distance and boat track through water added together, ie boat track over the ground.

If you are using chrome the the graph app is a simple add on -

https://chrome.google.com/webstore/d...ojekgdko?hl=en

EDIT. I've just divided the Y axis by 5 to make the whole thing look a bit better. Dunno how to zoom just one axis yet.

Where does the assumption 0-8 come from. I don't understand that graph would suggest the " average " vector to compute CTS is 1.25 or there abouts.

Ie how is the straight line computed. I know the slope of that one isn't germane to the overall pictorial representation but still

Dave

[conachair]

Originally Posted by goboatingnow

Where does the assumption 0-8 come from. I don't understand that graph would suggest the " average " vector to compute CTS is 1.25 or there abouts.

Ie how is the straight line computed. I know the slope of that one isn't germane to the overall pictorial representation but still

Dave

The 0-8 comes from the Y=(-cos(x)+1)*4. *4 is a multiplier from the max current being 4Kn.

Which is the area under the sine curve, assumed to be also the total distance the tide has moved. Having very grave doubts about that now (see edit), hindered by the fact I really don't know what I'm doing and this is all coming from trying to put equations together from visualization. Intuition suspects the total distance might be an RMS value of the area under the speed curve, but that's just a guess.

to get the area, from here

So you want the area under f(x) = sin(x), from 0 to pi.

Your first step is to determine whether sin(x) crosses the x-axis at the given interval. I'll save steps and say that it doesn't (if it did, we would be splitting this into two areas).

Since f(x) = sin(x) falls above the x-axis from 0 to pi, it is directly taking the integral.

Integral (0 to pi, sin(x) dx )

Take the integral. The integral of sin(x) is -cos(x).

[ -cos(x) ] {evaluated from 0 to pi}
[ -cos(pi) ] - [ -cos(0) ]

Solve each.

[ -(-1) ] - [ -1 ]
1 - (-1)
2

The area is two square units.

The straight line comes from Y=x*cos(pi*a)*b
Clumsy and no units but I haven't figured out how to do a polar line yet , A and B are just variables which can be assigned to sliders for easy manipulation. Bear in mind the y axis is divided by 8 to make it look a bit better. This is quite good fun and learning lots so i might have a hunt for some better graphing software.

Edit. Maybe I was right after all ....
http://physics.info/motion-graphs/

On a velocity-time graph …
the area under the curve equals the change in displacement.

I got to the area= displacement by imagining breaking down the curve into lots of little rectangles and placing them on top of each other to find the distance traveled. Luckily Newton had invented calculus already to figure out what goes on as as the rectangles got towards zero width.

Edit 2 But the displacement graph still doesn't look right . Something's wrong.

[Dockhead]

Originally Posted by goboatingnow

(c) in practice all combined CTS expose you to non intuitive ground tracks. as I pointed out to DockHead , you cannot easily determine safe clearance. , a point he did not agree , had he postulated the ground track of the varying non reversing tide he would only originally checked one side of the lines!! for that reason in real life I don't recommend multi hour complex CTS



Dave

I don't know how you would cross the English Channel without a multi-hour combined CTS. With 6 knots and more of tide on the French side, you run a real risk of not arriving at all, if you don't do your CTS right, especially in a slower boat.

I think the ground tracks are already no longer counterintuitive for me, after this educational thread . In my home waters, I like other sailors know very well what the ground track will look like - we've done the vectors and know exactly what the XTE is supposed to be every hour (that's how we monitor our progress). Of course you have to watch the chart, but I don't why this is a reason not to steer correctly.

[conachair]

Originally Posted by conachair

Edit 2 But the displacement graph still doesn't look right . Something's wrong.

Or maybe not. I was trying to visualize it using decimal x scale from 0 to 2, but it is actually 0 to Pi. Makes more sense now, looks right after all.

[Seaworthy Lass]

Hi folks, I am back . BBQ ashore was wonderful!

I see there are lots of posts I need to reply to. Will take me ages , I want to work on the new method I am proposing and I have a brad new Sailrite I want to play with. Not enough time in one day.

I sat at the BBQ and my mind wasn't there, it was thinking about my model. The technique has to be the one I described at 3 am last night. The change I made to it this morning to make it an even simpler method won't work in extreme conditions.

I really want to try and test this out by plotting a few examples.

I am getting tired of disagreeing with GoBoatingNow and I am feeling a little sorry for him as if his technique is the technique the RYA are teaching (and it most probably is if he is an RYA instructor) then he is only following what they are telling him to teach.

He is teaching that the RYA method gives the CTS to B. I am saying it gives the CTS to D and you will arrive at D. He is saying you never necessarily arrive at D. He (or the RYA if they are saying this) is incorrect.

GoBoating is vehemently saying the RYA method gives a perfectly accurate CTS if the data it is using is perfectly accurate. I am saying it is not as it is not taking the final angle and amount of tide into account. In strong tides it may be particularly wrong.

Forget about the mistakes he has made, will all make mistakes. It is the principle of what he (RYA?) is teaching I am challenging.

There are a few issues here:

1. The accuracy of the data (will the current be what is expected for a whole hour hour under consideration even if you are in the exact spot indicated on the map for that current? - this is unlikely)

2. Which bit of current do you pick? It depends on where your CTS takes you and the previous current you have encountered and is very tricky to estimate well and figures may need to be repeated when your ground track is not what you expected when you start checking your result.

3. With absolutely perfect data for the current you will be in, will the RYA method give you a CTS to get you exactly to your destination?

I recognise all the difficulties associated with points 1 and 2, but if you need to have an accurate method for when the data is correct. Points 1 and 2 introduce enough inaccuracies without an inaccurate method to determine CTS.

It is the method of determining CTS when the data is correct, that I am working on.
The RYA method is only an approximation. It is a very useful approximation, but it is not accurate as GoBoating keeps repeating. It will get you to D, hopefully not too far from B (but no guarantees, you could be way short depending on the current. You then need to decide what to do. This is the limitation of the method.

I am sure whoever thought of the RYA CTS method knew it was just an approximation, this bit of info just got lost over the years and it seems it is now being taught as being an accurate to get to B if the data is accurate. It is NOT.

I am working on a more precise method. Maybe I cant find one that is absolutely perfect, but it will be a big improvement on the RYA technique. Ideas are still being tossed around in my head and it is tantalisingly close.

Let me work on a few examples I make up and see if the method I described at 3 am works in all instances.

I will leave responding to the last posts. What I have have said here covers everything I want to say well I think.

[HappySeagull]

boy, this is difficult... but the ants are gone, anyways, so I think I understand where the drift has taken this thread.

I use the rule of Twelfths for current. Maybe I missed it in all this. ( but, Yes, I saw the word sinusoidal and I see the word sin popping up more)

if I were crossing a tide,I just use the curve in the tide book to see somewhat what my track will look like in the time elapsed. A vector sum from it will give me a CTS that will agree roughly with the rule of twelfths sums.
Calculating elapsed Time is tricky, beforehand, so I add allowance to arrive a little upstream.
More than that, would be too easy to make mistakes. Estimated Margin of Safety is easier.
But,
I don't cross the tide that much. I tack against it .

[HappySeagull]

... Forgot to mention Simpson's Rule...ie: place your time of departure and time of arrival on the tide curve.....but really, the twelfths rule is easier.

[Andrew Troup]

Originally Posted by HappySeagull

boy, this is difficult... but the ants are gone, anyways, so I think I understand where the drift has taken this thread.

Just when you thought it was safe to get back in the thread,
watch this space for (taa - daaaaa !)

Return of the Killer Ants ! ! !

[Seaworthy Lass]

Originally Posted by HappySeagull

... Forgot to mention Simpson's Rule...ie: place your time of departure and time of arrival on the tide curve.....but really, the twelfths rule is easier.

Thanks, I use the twelfths rule.

[Seaworthy Lass]

I have just finished phototographing two plots for an journey using the RYA method to determine CTS and my method. My method is looking very promising.
I think it gives me an accurate CTS. Fingers crossed!

Will write up the details in a sec and post them and the photos .

[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

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

(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

(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

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

(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 , as I think this is your argument , ie you cannot inflate the triangle for ever . I dont think I made this provision clear.

(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

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

(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, if i allowed that to exceed 30 minutes i was wrong BTW) 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

(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, in both bases actual transist time can be deduced along the X axis.

(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)

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

PS you seem to be saying the RYA method always put D in front of B , it doesnt it depends on how many hourly plots you run.

Dave

[Seaworthy Lass]

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")

AB = 16 nm
B is due east of A
Boat speed is constant at 4 knots (motoring, no wind)
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

What is the quickest time taken to travel between A and B?
What is the constant CTS?

RYA method result:
Time taken = 3.3 hours
CTS = 65.4 degrees

Please assume current is correct. I am not here to argue how hard it is to obtain this data well (I know it is difficult), I am just disputing the accuracy of the RYA method if the data IS accurate.

GoBoating, we should work together on this .
You are obviously really familiar with the RYA method. I am only following the method I saw yesterday on the RYA video link you gave me. Could you confirm what I have done is correct according to the RYA for this example and that you arrive at the same figures?

Conachair, could you please the vector diagrams for this example if you can, showing where this heading will take you. Your diagrams are lovely. I am getting tired of drawing up graph paper by hand and haven't the time .