

17012013, 11:10

#481

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Re: Distinct Activities: Shackled by a Common Name?
Quote:
Originally Posted by goboatingnow
Nope by the action of the tide you dont cross the tide in an hour, hence you are effected by it longer.
dont beleive me, Ive searched the net see here

Sigh. GoBoating, you have totally misinterpreted what figures to apply where.
In the video AC represents average current (2.2 knots in the case of the video)
You have made AC represent total current which is 10 knots.
The point D is plotted only as a way of working out the course to steer, it is not the point your track will necessarily cross at any stage.
For the hovercraft example, as the total distance travelled in the water is 18.87 miles, the time taken is 4.72 hours. If there is a total displacement of 10 miles from the current over this time, the average current is 2.12 knots. If you look at my vector diagram 2.12 is the average tide for every hour in my four triangles (the ACTUAL tide is the thick line).
So if you were to apply the method the video uses, the length of AC is 2.12 NOT 10.
The length of AD is your speed (in the case of the video 6 knots, in our case 4 knots).
You mark where a line of 4 knots crosses the rhumb line.
This angle is then the course to steer.
See my diagram below.
Sin (angle to steer) = 2.12 / 4
Therefore 'angle to steer' is 32 degrees.
Note this is EXACTLY the heading I said I needed to steer, and I steer it for the ENTIRE journey and I DO arrive at the destination.
PS the diagram below is not to scale, but all the lengths are noted.
PPS you have never actually told us what course you think you need to steer, but from your diagram it is NOT 32 degrees.
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17012013, 11:19

#482

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Re: Distinct Activities: Shackled by a Common Name?
Quote:
Originally Posted by goboatingnow
no different from a multihour tide computation. Thats exactly what you get on a passage in one varying tide. ( its actually the same tide, just flowing at different rates in different places).

Yes, I understand that  and for the base case of the CTS getting across in four hours, its simples.
The problem is with the GPS track boat which does not get across in four hours. This boat will be subject to the same streams but over a longer period of time, so the vectors will be different. I need to figure out how to "stretch" the current scale to fit the passage times of the GPS track boat.
Here it does make a difference that the currents vary by place only, not by place + time like tides. Capt Force's scenario is easier to analyze, in this respect.
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17012013, 11:34

#483

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Re: Distinct Activities: Shackled by a Common Name?
GoBoating, this is a to scale diagram of what would be drawn on your map if you were using the RYA video technique.
Note the purpose of the AB line is simply so that a constant angle to steer can be worked out relative to the this line (the rhumb line).
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17012013, 11:50

#484

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Re: Distinct Activities: Shackled by a Common Name?
Quote:
Originally Posted by Dockhead
Yes, I understand that  and for the base case of the CTS getting across in four hours, its simples.
The problem is with the GPS track boat which does not get across in four hours. This boat will be subject to the same streams but over a longer period of time, so the vectors will be different. I need to figure out how to "stretch" the current scale to fit the passage times of the GPS track boat.
Here it does make a difference that the currents vary by place only, not by place + time like tides. Capt Force's scenario is easier to analyze, in this respect.

Dockhead
I regret that I wasn't able to talk you into discarding the one aspect of CaptForce's scenario which made it completely unrealistic.
Namely the duration of the crossing being preordained.
I'm guessing he did this because he already "knew" the answer he was setting out to prove: namely, that both boats would arrive at the same time.
I can imagine a current being geographically fixed, and if it's a strong ocean current like the Gulf Stream or Agulhas, it will flow like a river through the ocean, strongest in the middle and weaker at the margins (due to friction).
So although it won't be a linear rise and fall as you sail across it (more sinusoidal, I guess) it was easy enough to write a single spreadsheet which could cope with either linear or sinusoidal scenario. In fact my spreadsheet lets me adjust the current at every 'slice' across the current, not necessarily symmetrically about midway.
Having failed to get CaptForce to specify a distance across the current, I wrote my analysis so that I could specify a distance. I don't expect ever to encounter a current where I know how long it will take to cross it, but I don't know the distance.
I couldn't get interested in writing a spreadsheet where you had to know the transit time in advance; I can't see any way I would ever get to reuse that in the real world.
So rather than breaking the transit into time slices as you were forced to do, I broke it into distance slices.
My spreadsheet works out the time to cross each slice; then sums those times to tell me how long each boat takes to arrive on the far side of the current.
Hence I can't see any way we can compare results, which seems a shame.



17012013, 11:59

#485

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Re: Distinct Activities: Shackled by a Common Name?
Quote:
Originally Posted by Andrew Troup
Dockhead
I regret that I wasn't able to talk you into discarding the one aspect of CaptForce's scenario which made it completely unrealistic.
Namely the duration of the crossing being preordained.
I'm guessing he did this because he already "knew" the answer he was setting out to prove: namely, that both boats would arrive at the same time.
I can imagine a current being geographically fixed, and if it's a strong ocean current like the Gulf Stream or Agulhas, it will flow like a river through the ocean, strongest in the middle and weaker at the margins (due to friction).
So although it won't be a linear rise and fall as you sail across it (more sinusoidal, I guess) it was easy enough to write a single spreadsheet which could cope with either linear or sinusoidal scenario. In fact my spreadsheet lets me adjust the current at every 'slice' across the current, not necessarily symmetrically about midway.
Having failed to get CaptForce to specify a distance across the current, I wrote my analysis so that I could specify a distance. I don't expect ever to encounter a current where I know how long it will take to cross it, but I don't know the distance.
I couldn't get interested in writing a spreadsheet where you had to know the transit time in advance; I can't see any way I would ever get to reuse that in the real world.
So rather than breaking the transit into time slices as you were forced to do, I broke it into distance slices. Hence I can't see any way we can compare results, which seems a shame.

Distance slices  hmm, that's very good! That would solve the current stretching problem. I like it! What formulae did you use? I might try running it the same way.
Why can't we compare? Time and distance are in a simple relationship. How far does your GPS track boat get (in distance made good to waypoint) in 4 hours? Or, if you like, how long does it take your GPS track boat to go 17.53 miles? It's simples. Did you come out the same as me?
I didn't throw out CFs four hours because we are solving his problem. It's fairly trivial to fill in the one blank, and it's no big deal how you choose to fill it in.



17012013, 12:05

#486

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Re: Distinct Activities: Shackled by a Common Name?
Quote:
Originally Posted by Andrew Troup
Dockhead
I don't expect ever to encounter a current where I know how long it will take to cross it, but I don't know the distance.
.

But it's a simple problem. If you know your speed through water and the average set of the current, you know your distance made good in any given amount of time if you're sailing constant heading.
It's much easier to calculate because you are sailing a straight line through water at a constant speed.
Why don't you upload your spreadsheet and let us play with it? If you have a recent version of Excel, you have to save as Excel972003, because Cruisers Forum doesn't recognize and won't upload xlsx files, only xls.



17012013, 12:40

#487

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Re: Distinct Activities: Shackled by a Common Name?
Quote:
Originally Posted by goboatingnow

Very interesting video!
The RYA method is only an approximation for the journey. It assumes the journey is roughly two hours and only gives the current for those two hours.
The journey will run longer. It is a simple technique that will only work when the current in the third hour is very similar in direction and strength to the average of the currents during the first 2 hours.
The RYA technique will not work if the current is very different in the third hour of their example in this video.. This by the way is not an unrealistic scenario, the current could increase or decrease rapidly in the third hour of a journey compared to the first two.
Do you remember as soon as you posted the hovercraft problem I asked you what the current was in the fifth hour? You had specified the current only affected the boat for four hours (3,2 4 then 1 knot).
I said as I had no reply from you I would take it as zero. Not unreasonable, as the current for the last hour had decreased to only one knot.
This is why your calculations are wrong. You are following the RYA method and it will not work in this example as the current in the fifth hour is NOT roughly the average of the currents during the first four hours!
If you apply the true average current over the hovercraft journey (2.12 knots) and the average boat speed (in this case it is a constant 4 knots) you will come up with the correct solution of the compass course required.
Maybe someone should let the RYA know the problem with their technique .
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17012013, 12:46

#488

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Quote:
Originally Posted by Seaworthy Lass
GoBoating, this is a to scale diagram of what would be drawn on your map if you were using the RYA video technique.
Note the purpose of the AB line is simply so that a constant angle to steer can be worked out relative to the this line (the rhumb line).

Seaworthy. Have a look at the two hour plot. Version, you do not average out the current ( would you care to do that on a non penpendicupar current
At least you'll agree that at no point did any one in the videos connect the end of the vector to the destination , maybe that's a start
The steps to plot a multi stage tide CTS. All the video show this technique
(A) determine the rhumb line distance. Draw a line through and beyond the destination ( as the vector could land beyond the destination ) , conventionally this is labelled AB
(B) determine how many tides. Set and distance will be encountered on that rhumb line. ( you cannot determine anything else) its in effect one tide that's changing course and speed over tie, but that's not important.
(C) look up the tidal data, you are given a speed ( the drift ) and the direction ( the set) note that speed is a distance in an hour. ( not the total distance the tide runs ) in this case 4 tides
(D) plot the hourly vectors including their direction and length for one hour. These combined vectors are called AC
(E) since you plotted 4 hours of tide , you then determine the distance the boat at its fixed speed through the water can travel in the same time. , ie 4 hours. Again all you know at this time are these facts.
(F) graphically or mathematically, you perform the vector addition to determine the effects of the 4 x hourly tides on the boat. Ie where the 4 hours boat plot cuts the rhumb , this is point D it may be behind the destination
(G) where the boat vector cuts the rhumb does not represent anything other then the distance covered towards the destination, equally the course to steer can be lifted from the chart ( or graph paper ) Only in the perpendicular case can you use the angle with the tide. Your ground track never hits the rhumb line
(H) you can now deduce the effect the accumulated tides are having on the boat. By using the rhumb line vector. You can work out the average SOG .
(G) advance at the average mean SOG the rest of the distance to compute the total time , there a few ways to do this.
(I) the CTS angle is the same bearing as the 4 hour water track is. Its NOT the angle between the tidal vectors and the water track. That's only the case in perpendicular tides.
The key you are missing is that let's say the expected 4 hour journey takes 5 hours , because of the tide , does not mean that is 4 hours of tide and a slack tide ( or something you don't know about)
what is means is that you were subject to a nM per hour tide for longer then an hour , hence the real ( after arrival destination triangle ) would have larger tidal vectors then the ones plotted. This is intuitive because if you think about it you end up actually crabbing across the tide at an angle so unlike going straight across you are exposed to it for a longer time and hence its bigger ( sorry rather wordy )
You made a Fundemental mistake of looking at an hourly plot video and mis constructing it advice into a multi hour plot , you certainly don't average vectors because that would only work in one case where they were all aligned , in this case it would.
Note you are right in one respect , I plotted the sign of the tidal vectors the other way. It's makes no odds as you can compensate by flipping the CTS . This is my method , it doesn't invalidate the method
Look specifically at the two hour plot video to see the technique. Nowhere will you ever see the water vector connected to the destination directly.
Dave
How can you compute water track in advance before applying the effect of the vectors.
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17012013, 13:01

#489

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Quote:
Originally Posted by Seaworthy Lass
Very interesting video!
The RYA method is only an approximation for the journey. It assumes the journey is roughly two hours and only gives the current for those two hours.
The journey will run longer. It is a simple technique that will only work when the current in the third hour is very similar in direction and strength to the average of the currents during the first 2 hours.
The RYA technique will not work if the current is very different in the third hour of their example in this video.. This by the way is not an unrealistic scenario, the current could increase or decrease rapidly in the third hour of a journey compared to the first two.
Do you remember as soon as you posted the hovercraft problem I asked you what the current was in the fifth hour? You had specified the current only affected the boat for four hours (3,2 4 then 1 knot).
I said as I had no reply from you I would take it as zero. Not unreasonable, as the current for the last hour had decreased to only one knot.
This is why your calculations are wrong. You are following the RYA method and it will not work in this example as the current in the fifth hour is NOT roughly the average of the currents during the first four hours!
If you apply the true average current over the hovercraft journey (2.12 knots) and the average boat speed (in this case it is a constant 4 knots) you will come up with the correct solution of the compass course required.
Maybe someone should let the RYA know the problem with their technique .

No again you are confused. What you are describing is a problem with determining what actual tides apply for the journey. Since you cannot tell in advance exactly what tides apply. For example your actual ground track could take you into an area with different tides.
However since all you can is by inspection of your rhumb line determine the set and drift.
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
It's as simple as this , if you know you have a tide of 2 kn , and you find due to the effect of that tide that you do spend longer under its influence, does its effect stop after 2 nM , no of course not . Your journey method would require inventing imsginary tidal vectors to justify how you got to you destination.
Quote:
again , try a simple test . 5 knot boat ,destination rhumb line due east distance 5 knots , by inspection you determine there is a tide of 2 nM per hour direction ( ie 2kn) direction 135T compute the CTS only. A simple graph on rough squared will do. Lets just do that.

Dave
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17012013, 13:11

#490

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Quote:
Originally Posted by Dockhead
But it's a simple problem. If you know your speed through water and the average set of the current, you know your distance made good in any given amount of time if you're sailing constant heading.
It's much easier to calculate because you are sailing a straight line through water at a constant speed.
Why don't you upload your spreadsheet and let us play with it? If you have a recent version of Excel, you have to save as Excel972003, because Cruisers Forum doesn't recognize and won't upload xlsx files, only xls.

Dockhead you've not addressed my point that you cannot use the a average of a triangle to determine the vector. You have to use the RMS the CTS angle is not correct as the X axis is larger then your one.
Dave
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17012013, 13:22

#491

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Dockhead. Captforces problem can not be solved , because as you discovered , you where not given the actual tidal set and drift , ie the hourly rate. , what you have been given is the tidal vectors that boat in that crossing at that speed experienced. Hence you have no data to apply to a boat following a different course, at a different speed. In order words you have the resulting tide the first boat experienced but you cannot compute the underlying vectors because you don't have that data. Captforce fitted the tide to the time of the first boat. You have no data how its fits to any other.
I suspect its not thought out.
Hence the apples and oranges problem
Dave
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17012013, 13:25

#492

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Re: Distinct Activities: Shackled by a Common Name?
Quote:
Originally Posted by Seaworthy Lass
The RYA technique will not work if the current is very different in the third hour of their example in this video.. This by the way is not an unrealistic scenario, the current could increase or decrease rapidly in the third hour of a journey compared to the first two.

In a paper world yes, but by the time the 2 hours are up you are only 20 minutes or so away from the destination so into pilotage and bearings/transits if it's that crucial. The tidal data usually covers quite a big area so could well be out for those 2 hours anyway. From memory they teach to take the closest hour of data so to plot a cts for an extra half hour at the end in this example you would might well use the 2nd hour data anyway.
Edit. that last bit is probably wrong, think you would be into the 3rd hour data



17012013, 13:40

#493

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Quote:
Originally Posted by conachair
In a paper world yes, but by the time the 2 hours are up you are only 20 minutes or so away from the destination so into pilotage and bearings/transits if it's that crucial. The tidal data usually covers quite a big area so could well be out for those 2 hours anyway. From memory they teach to take the closest hour of data so to plot a cts for an extra half hour at the end in this example you would might well use the 2nd hour data anyway.
Edit. that last bit is probably wrong, think you would be into the 3rd hour data

No this isn't the underlying method.
All you know is the rhumb line distance , the time that would take to cross that rhumb and hence the hence tides that affect you. It nothing to do with the RYA method. It's all the data you have. You cannot reasonably recursively compute the real effects base on the actual voyage. There not much use determining the past. .
You extract the set and drift tidal streams from the atlas.
That's all you know folks , that and boat speed.
From now on in , you are plotting vectors. These vectors APPLY for the WHOLE time. Ie if the tide is 2kn , it doesn't stop after 1hour. If you are in it as a result of it for 1.5 hours its still flowing ( you have no other vectors to contradict that )
Using the vector addition , provides you with a CTS directly to the destination, you DO NOT arrive outside or near the destination , as far as the maths go you arrive directly at the destination without changing course.
So on your graph when you determine D it does not mean the distance DB is in some unknown tide.
You can relate the whole lot triangle back to real life , but in that case all the vectors can be out.
There is a still massive confusion of how to actually compute a CTS from the data available ie a chart known boat speed and a tidal Stream atlas , in this thread. Possibly because few do it.
Dave
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17012013, 13:51

#494

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Re: Distinct Activities: Shackled by a Common Name?
Quote:
Originally Posted by goboatingnow
No again you are confused.

I am not confused at all.
Quote:
Originally Posted by goboatingnow
What you are describing is a problem with determining what actual tides apply for the journey. Since you cannot tell in advance exactly what tides apply. For example your actual ground track could take you into an area with different tides.

That is a different problem. We can only work with what we know and work out an accurate course to steer for that.
Quote:
Originally Posted by goboatingnow
However once you exact the tidal vectors the RYA method is mathematically correct.

It is only correct to get you back on the rhumb line at the end of the time subjected to the current for the time taken to "nearly" get there (in the hovercraft example you still had 43 minutes to go). At that point you must steer whatever compass heading you need to to get to the destination. If there is no current you then need to steer straight to the destination point. If the current is the average you have had over the journey so far, you continue on the same heading.
This is NOT the quickest route to destination if once you reach the rhumb line the current is not the average you have been subjected to so far, as you will need to alter heading.
The technique does however get you nearly there and it does allows you to compute complex situations where the current is not perpendicular to the course and varies not only in speed, but in direction also.
So the technique has huge advantages.
However, it is NOT calculating the exact compass direction to get to the destination, only to the rhumb line close to the destination.
The limitations of the technique need to be recognised. You cannot blindly say it is correct mathematically when it is not.
In your hovercraft case I have given you perfect vectors illustrating a constant heading and constant speed will get you to the destination the quickest. You can not fault these.
You get there slower using the RYA technique, so how can you say the RYA technique is mathematically correct? It will only work exactly if the current over the last part of the journey (in the hovercraft case 43 minutes) is the average speed and direction of the current you have previously been subjected to. On the hovercraft example it isn't, so it doesnt work.
GOBoating, what is your compass course to steer and what is your time taken for the journey for the hovercraft example?
Mine is 32 degrees off the rhumb line and it takes me 4.72 hours (4:43). I arrive exactly at the destination at that time following a constant heading and speed for the entire journey.
Quote:
Originally Posted by goboatingnow
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.

The RYA method is a wonderful technique, but it is not "right" mathematically, it is an approximation unless the current in the final time is exactly the average in speed and direction to what you have applied for the rest of the course.
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17012013, 13:57

#495

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Quote:
It is only correct to get you back on the rhumb line at the end of the time subjected to the current for the time taken to "nearly" get there (in the hovercraft example you still had 43 minutes to go). At that point you must steer whatever compass heading you need to to get to the destination. If there is no current you then need to steer straight to the destination point. If the current is the average you have had over the journey so far, you continue on the same heading.
This is NOT the quickest route to destination if once you reach the rhumb line the current is not the average you have been subjected to so far, as you will need to alter heading.

No no no. , at no point in the hovercraft do you end up on the rhumb line, with the RYA method , ie the ONLY correct method , you DO NOT change change course at the rhumb. You hit the rhumb at the destination directly
In proper chart work of course you don't put the destination at the end of the finger berth you put the destination a little always from the entrance. Etc etc.
Your computation of CTS, by joining up the end of the vector to the destination would put you downtide of the destination.
You have Fundementally not understood that a speed vector is not the same length irrespective of the time.
The RYA is mathematically the correct addition of SPEED vectors.
Sheesh.
Dave
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