Distinct Activities: Shackled by a Common Name?

[goboatingnow]

Originally Posted by Seaworthy Lass

GoBoating, here is a really simple but dramatic example of best course to steer for "current then no current" journey that will hopefully help convince you.

Same 16 mile journey as the last example.
Same boat/hovercraft speed of 4 miles/hour.
Again, four hours of current, but this time a steady four knots during this time.

Boat that wants to get to the rhumb line at the end of the current sits in the same spot for 4 hours as he can make no progress against the current.
Then travels 16 miles at 4 knots, taking 4 hours.
Total time = 8 hours

Boat 2 works out that the 4 knots of current over 4 hours will move him 16 miles over a journey of 16 miles, so he heads off at 45 degrees.
Total time = (square root of 16 squared plus 16 squared) divide by 4 = 5.66 hours

The boat following the constant heading for the entire journey beats the boat who is following RYA instructions by 2.34 hours.

If the RYA is teaching that you should aim to be on the rhumb line at the end of the current, but before the end of the journey, they are incorrect.

Actually the red quote is not the case. I graphed it right but wrote a set of words down incorrectly. You end up at the destination directly using my graph even though the vector falls short. And it deals with the discrepancies I believe because of the underlying assumptions in tidal streams you do end up n the rhumb line where I said ( what you don't do is then change course you keep heading on you CTS and you will get home

In my original case , you arrive directly at the destination, merely it takes longer. There is no course heading change because of the underlying assumptions, this also deals with Dockheads conundrum that the end up downtide ( which he would if he connect the water track to the destination. In summary here's why the RYA method is actually right

( a) tidal vectors are rate based , ie say 2kn. Ie two nautical miles in an hour. It does not mean that all you experience is that though. The vectors don't stop after 2 knots , no more then a car travelling at 30mph , stops at 30 miles.

Since the plot is In hourly vector plot. ( why because we have no other information)

Implicit in using a rate vector is the assumption that it maintains a constant rate In effect for ever , or more correctly for all the time being considered under question.

This is the key. If I have an tidal vector specified as a rate of 2 kn , if I subject myself to it for 70 minutes instead of 60 minutes , its not that I travel 60 minutes and then 10 minutes of nothing. I am instead subject to the vector for 70 minutes.

So in you case seaworthy. With the hovercraft if you put 4 hours of time on the hypotenuse , ie 16 and. , in the same time that would equate to 10 on the x-axis. , this would indicate for 4 hours of water travel an 4 hours of perturbation vectors. You would arrive at a point somewhere short of the destination.

What you do is Continue on the the same course as indicated by the 4 hour plot ( ie mine not yours ) AND the rate vectors continue as you extend your time to reach the destination ( sorry about my tongue tied here )

Hence I will arrive at my destination in the time I calculated ( I will in effect continue on south of the rhumb line )

You would never actually arrive , because you would actually pass north of the objective. ( remember you are out there longer so you must continue to allow for the rate vectors ( ie the tide or the wind ). In effect as I steer My course to steer the hole triangle grows proportionally to allow the vertice to meet the intended destination.

Dockhead you were right and I was right. , yes irrespective of slack or not , you reach the destination on a single CTS.

However you must plot in unit time rate vectors as this is how tides vectors are specified. Hence the water track is never connected to the destination, its a hourly vector plot. If the destination is further away you continue on and the rate vectors continue on an you are brought to the destination.

All over my lecture notes ( I dug them out , I have written DO NOT connect the water track to the destination, its the most common mistake made buy students. It took me time to put words on it.

Hope everyone can see it ill draw later today when I'm in the office ( where the graph paper is )

My maths are right my English was wrong , your answer Seaworthy is wrong not because your maths are wrong, but your application of rate vectors is incorrect. , if you spend for example 4.5 hours out on a 4hour crossing , you must " inflate " the tidal vectors accordingly ( if you see what I mean) your maths is right , its just you never reach the destination on the CTS you computed.

Hopefully I ve clarified the situation and answered Dockhead conundrum as well

[Dockhead]

Originally Posted by goboatingnow

Actually the red quote is not the case. I graphed it right but wrote a set of words down incorrectly. You end up at the destination directly using my graph even though the vector falls short. And it deals with the discrepancies I believe because of the underlying assumptions in tidal streams

In my original case , you arrive directly at the destination, merely it takes longer. There is no course heading change because of the underlying assumptions, this also deals with Dockheads conundrum that the end up downtide ( which he would if he connect the water track to the destination. In summary here's why the RYA method is actually right

( a) tidal vectors are rate based , ie say 2kn. Ie two nautical miles in an hour. It does not mean that all you experience is that though. The vectors don't stop after 2 knots , no more then a car travelling at 30mph , stops at 30 miles.

Since the plot is In hourly vector plot. ( why because we have no other information)

Implicit in using a rate vector is the assumption that it maintains a constant rate In effect for ever , or more correctly for all the time being considered under question.

This is the key. If I have an tidal vector specified as a rate of 2 kn , if I subject myself to it for 70 minutes instead of 60 minutes , its not that I travel 60 minutes and then 10 minutes of nothing. I am instead subject to the vector for 70 minutes.

So in you case seaworthy. With the hovercraft if you put 4 hours of time on the hypotenuse , ie 16 and. , in the same time that would equate to 10 on the x-axis. , this would indicate for 4 hours of water travel an 4 hours of perturbation vectors. You would arrive at a point somewhere short of the destination.

What you do is Continue on the the same course as indicated by the 4 hour plot ( ie mine not yours ) AND the rate vectors continue as you extend your time to reach the destination ( sorry about my tongue tied here )

Hence I will arrive at my destination in the time I calculated ( I will in effect continue on south of the rhumb line )

You would never actually arrive , because you would actually pass north of the objective. ( remember you are out there longer so you must continue to allow for the rate vectors ( ie the tide or the wind ). In effect as I steer My course to steer the hole triangle grows proportionally to allow the vertice to meet the intended destination.

Dockhead you were right and I was right. , yes irrespective of slack or not , you reach the destination on a single CTS.

However you must plot in unit time rate vectors as this is how tides vectors are specified. Hence the water track is never connected to the destination, its a hourly vector plot. If the destination is further away you continue on and the rate vectors continue on an you are brought to the destination.

All over my lecture notes ( I dug them out , I have written DO NOT connect the water track to the destination, its the most common mistake made buy students. It took me time to put words on it.

Hope everyone can see it ill draw later today when I'm in the office ( where the graph paper is )

My maths are right my English was wrong , your answer Seaworthy is wrong not because your maths are wrong, but your application of rate vectors is incorrect. , if you spend for example 4.5 hours out on a 4hour crossing , you must " inflate " the tidal vectors accordingly ( if you see what I mean) your maths is right , its just you never reach the destination on the CTS you computed.

Hopefully I ve clarified the situation and answered Dockhead conundrum as well

Extremely interesting.

Might it not be that the RYA instructions are also directed at the practical, realistic situation, rather than the theoretical ideal?

The theoretical ideal CTS requires infallible knowledge about currents and own boat speed. By the way, we have also been ignoring leeway, which must be considered in CTS as well.

In reality we will be off so we have to build in some margin for error, and naturally, we want to err on the side of being a bit uptide of the destination.

Furthermore, at some point we will stop steering our CTS and just line up COG with BTW. It is not realistically possible to steer a constant heading the last cables and meters to the destination -- only in theoretical exercises like these. So in practice surely it's right that we will try to navigate not to the very quay we intend to tie up to, but to a point a mile or so away, using an excellently calculated CTS, and then set the pilot on "track" after we've achieved that goal.

That's why in real Channel crossings, I calculate a CTS for a point a mile or two uptide of Cherbourg (say). If everything works out really well I will end up a little uptide and a mile off, and will steer by hand or put the pilot on "track" from there. If I'm less close, that point will occur further out, maybe up to an hour off. It is not the theoretical ideal track but it is a practical and realistic one.

[Seaworthy Lass]

Originally Posted by goboatingnow

Actually the red quote is not the case. I graphed it right but wrote a set of words down incorrectly.

What should you have written?

Originally Posted by goboatingnow

You end up at the destination directly using my graph even though the vector falls short. And it deals with the discrepancies I believe because of the underlying assumptions in tidal stream vectors. So we have to leave the hovercraft for now as wind vectors have different
assumptions.

Vector analysis actually works for any mechanics - wind if flying, current if sailing.

Originally Posted by goboatingnow

In my original case , you arrive directly at the destination, merely it takes longer. There is no course heading change because of the underlying assumptions, this also deals with Dockheads conundrum that the end up downtide ( which he would if he connect the water track to the destination. In summary here's why the RYA method is actually right

Of course you have a heading change if you aim to be on the rhumb line at the end of the current. During the first four hours where there is current if you aim to be on the rhumb at the end of this your heading has to be 38.7 degrees from the rhumb line (if the rhumb line is due east as in your diagram, then you are heading 90-38.7= 51.3. When you have reached the rhumb line your heading is then 90 degrees. So you have 2 headings for the one journey.

Originally Posted by goboatingnow

( a) tidal vectors are rate based , ie say 2kn. Ie two nautical miles in an hour. It does not mean that all you experience is that though. The vectors don't stop after 2 knots , no more then a car travelling at 30mph , stops at 30 miles.

It does if they are both only travelling for one hour.

Originally Posted by goboatingnow

Since the plot is In hourly vector plot. ( why because we have no other information)
Implicit in using a rate vector is the assumption that it maintains a constant rate In effect for ever , or more correctly for all the time being considered under question.

Yes, what is wrong with that? we are using constant rates so we have figures to work with given no one has volunteered to do any calculus.

Originally Posted by goboatingnow

This is the key. If I have an tidal vector specified as a rate of 2 kn , if I subject myself to it for 70 minutes instead of 60 minutes , its not that I travel 60 minutes and then 10 minutes of nothing. I am instead subject to the vector for 70 minutes.

Which is why the hourly plots consider what is happening during 60 minutes. The next 10 minutes are dealt with in the next vector.

Originally Posted by goboatingnow

So in you case seaworthy. With the hovercraft if you put 4 hours of time on the hypotenuse , ie 16 and. , in the same time that would equate to 10 on the x-axis. , this would indicate for 4 hours of water travel an 4 hours of perturbation vectors. You would arrive at a point somewhere short of the destination.

What you do is Continue on the the same course as indicated by the 4 hour plot ( ie mine not yours ) AND the rate vectors continue as you extend your time to reach the destination ( sorry about my tongue tied here )
Hence I will arrive at my destination in the time I calculated ( I will in effect continue on south of the rhumb line )

What time did you calculate? I arrived in 4.72 hours (4:43).

Originally Posted by goboatingnow

You would never actually arrive , because you would actually pass north of the objective. ( remember you are out there longer so you must continue to allow for the rate vectors ( ie the tide or the wind ). In effect as I steer My course to steer the hole triangle grows proportionally to allow the vertice to meet the intended destination.

I did arrive - look at my plot! I would only end up ahead of the destination and north of it if I kept going after I had reached it!

Originally Posted by goboatingnow

Dockhead you were right and I was right. , yes irrespective of slack or not , you reach the destination on a single CTS.

GoBoating you are NOT travelling a single CTS you alter your CTS by 38.7 degrees when you hit the rhumb line.

Originally Posted by goboatingnow

However you must plot in unit time rate vectors as this is how tides vectors are specified. Hence the water track is never connected to the destination, its a hourly vector plot. If the destination is further away you continue on and the rate vectors continue on an you are brought to the destination.

My vectors DID show hourly plots.
The water track may not connect to the destination (it will if there is no current for the last part of the journey) but the sum of the water track and current track for the last part of the journey will always connect to the destination. How else do you arrive .

Originally Posted by goboatingnow

All over my lecture notes ( I dug them out , I have written DO NOT connect the water track to the destination, its the most common mistake made buy students. It took me time to put words on it.

I have connected the TOTAL track to the destination. It is only that there is no current for the last bit that means the last water track vector is connected to the destination (here the water track is the total track for that bit of the journey as there is no current).

Originally Posted by goboatingnow

Hope everyone can see it ill draw later today when I'm in the office ( where the graph paper is )

Would be interested in seeing your graph.

[goboatingnow]

Originally Posted by Dockhead

Extremely interesting.

Might it not be that the RYA instructions are also directed at the practical, realistic situation, rather than the theoretical ideal?

The theoretical ideal CTS requires infallible knowledge about currents and own boat speed. By the way, we have also been ignoring leeway, which must be considered in CTS as well.

In reality we will be off so we have to build in some margin for error, and naturally, we want to err on the side of being a bit uptide of the destination.

Furthermore, at some point we will stop steering our CTS and just line up COG with BTW. It is not realistically possible to steer a constant heading the last cables and meters to the destination -- only in theoretical exercises like these. So in practice surely it's right that we will try to navigate not to the very quay we intend to tie up to, but to a point a mile or so away, using an excellently calculated CTS, and then set the pilot on "track" after we've achieved that goal.

That's why in real Channel crossings, I calculate a CTS for a point a mile or two uptide of Cherbourg (say). If everything works out really well I will end up a little uptide and a mile off, and will steer by hand or put the pilot on "track" from there. If I'm less close, that point will occur further out, maybe up to an hour off. It is not the theoretical ideal track but it is a practical and realistic one.

No Dockhead the approach is the mathematical correct application of rate vector addition , the tidal vectors are not fixed length over any random time extension., just like the boat is a rate over total time the the tidal vectors as they are specified in speed are also.

Joining the end of the tidal vectors to the destination as people have being coming is factually wrong and that CTS never reaches the destination if one accepts that both the boat vector and tidal vectors are rate( ie time based) since both are specified in kn there inherently are.

Also applied my graph correctly applied shows that you (a) CTS gives you the quickest time ( we all except that ) and you don't inherently end up down tide , ( unless you want to reposition the destination)


To summarise , in these examples( right angles triangles) the X-axis is time constant , but the Y -Axis expands with total journey time as does the hypotenuse , as both are rate based.

For an example of three side time dependant vectors, try calculating the case where the destination moves in time ( ie another ship )


Dave

[Seaworthy Lass]

Originally Posted by Dockhead

Extremely interesting.

Might it not be that the RYA instructions are also directed at the practical, realistic situation, rather than the theoretical ideal?

The theoretical ideal CTS requires infallible knowledge about currents and own boat speed. By the way, we have also been ignoring leeway, which must be considered in CTS as well.

In reality we will be off so we have to build in some margin for error, and naturally, we want to err on the side of being a bit uptide of the destination.

Furthermore, at some point we will stop steering our CTS and just line up COG with BTW. It is not realistically possible to steer a constant heading the last cables and meters to the destination -- only in theoretical exercises like these. So in practice surely it's right that we will try to navigate not to the very quay we intend to tie up to, but to a point a mile or so away, using an excellently calculated CTS, and then set the pilot on "track" after we've achieved that goal.

That's why in real Channel crossings, I calculate a CTS for a point a mile or two uptide of Cherbourg (say). If everything works out really well I will end up a little uptide and a mile off, and will steer by hand or put the pilot on "track" from there. If I'm less close, that point will occur further out, maybe up to an hour off. It is not the theoretical ideal track but it is a practical and realistic one.

+1
I agree with all that. The chart GoBoating attached showed had very shallow water just before arrival. I would want to go the safest way through that. I would also want to hedge my bets and try and stay up current a bit more than than calculations indicated I should.

BUT, this is not what GoBoating is saying. He is saying I would not get there following a constant heading and constant speed for the entire journey. He says the quickest way of actually arriving is to hit the rhumb line after the current stopped, then change heading and head for the destination and that the RYA is teaching this.

[goboatingnow]

BUT, this is not what GoBoating is saying. He is saying I would not get there following a constant heading and constant speed for the entire journey. He says the quickest way of actually arriving is to hit the rhumb line after the current stopped, then change heading and head for the destination and that the RYA is teaching this.

No seaworthy I corrected that. What I said was my english was wrong but my maths and graph was correct. You maintain a constant heading as per the calculated CTS, In the hovercraft case you never arrive at the rhumb or cross it until you reach the destination, in sum vectors summations you will of course cross the rhumb line

I agree with all that. The chart GoBoating attached showed had very shallow water just before arrival. I would want to go the safest way through that. I would also want to hedge my bets and try and stay up current a bit more than than calculations indicated I should.

The rya example i showed is a pure maths workup and takes no account of any chart obstacles, The chart is merely being used as graph paper


You hover craft triangle is still wrong. as your are misinterpreting vectors over time. As is Dockheads conundrum that he arrives downtide mathematically,

[goboatingnow]

Here's what actually happens

In practice you dont compute that larger triangle, you compute the mean SOG and apply the extra distance ( its all the same )

Your CTS would never actually reach the destination.


The exact same approach is applied for tidal vectors , But you do have to cognisant of how the underlying tidal vector was got in the first place. The exactness of these maths, is not reflected in what really happens.

Dave

[Dockhead]

Originally Posted by goboatingnow

No Dockhead the approach is the mathematical correct application of rate vector addition , the tidal vectors are not fixed length over any random time extension., just like the boat is a rate over total time the the tidal vectors as they are specified in speed are also.

Joining the end of the tidal vectors to the destination as people have being coming is factually wrong and that CTS never reaches the destination if one accepts that both the boat vector and tidal vectors are rate( ie time based) since both are specified in kn there inherently are.

Also applied my graph correctly applied shows that you (a) CTS gives you the quickest time ( we all except that ) and you don't inherently end up down tide , ( unless you want to reposition the destination)


To summarise , in these examples( right angles triangles) the X-axis is time constant , but the Y -Axis expands with total journey time as does the hypotenuse , as both are rate based.

For an example of three side time dependant vectors, try calculating the case where the destination moves in time ( ie another ship )


Dave

Oooooooh, that's subtle -- new territory for me -- now I'm learning something totally new!

My own analysis of Capt Force's scenario cannot deal with that issue.

I am so far not able to deal with the greater influence of current due to longer passage time of the GPS track boat. So I am only able -- so far -- to analyze how far the GPS track boat gets in 4 hours, which is the time it takes for the constant heading boat to get across.

Now you have me doubting whether I have got the constant heading boat right -- do you mind having a look?

[Jammer Six]

Originally Posted by Seaworthy Lass

We need to lighten up this thread a little .

Drop your towel again.

That, out of all this thread, I understood.

[goboatingnow]

Originally Posted by Dockhead

Oooooooh, that's subtle -- new territory for me -- now I'm learning something totally new!

My own analysis of Capt Force's scenario cannot deal with that issue.

I am so far not able to deal with the greater influence of current due to longer passage time of the GPS track boat. So I am only able -- so far -- to analyze how far the GPS track boat gets in 4 hours, which is the time it takes for the constant heading boat to get across.

Now you have me doubting whether I have got the constant heading boat right -- do you mind having a look?

Im sorry , Im an engineer, Im a terrible writer of explanations as I put down wrong things, which I teach this stuff I have a blackboard and in a few swipes of a chalk , its easy to see, even if I babble the explanation badly. ( in reality Im a poor teacher, I often know it from first principles , but I still get it accros badly),

whats important is that when applying tide vectors to a CTS plot you do it increments of time, if the tides are in hours then the plots is in increments of hours. the destination vertices will be ahead, in front or at the destination. You then "scale" the result to get the time to the destination as a result of the tidal vectors OVER the whole passage time.

IN all cases

(a) SIngle CTS computation is the most efficient method where there is varying tidal vectors

(b) You arrive at the destination without changing heading

(c) You arrive precisely at the destination, ( mathematically) in real life the tidal vectors are constantly varying with both direction and speed and your progress may not be constant over the water anyway.


The method ( and the maths) have quite severe limitations where the tide approaches the speed of the craft , as it is difficult in advance to ascertain what tidal vectors may actually affect the boat.

IN my view , with the exception of some "poster child" examples, like sinusoidal symmetrical tides, It is better to compute as you go in hourly increments , or multiple of hourly increments to avoid large excursions from the rhumb line that cant be easily determined., ie avoid large multi hour in advance CTS computations.

You will observe that pre-computing your exact or even approximate ground track is quite involved, as the hourly incremental plot is not what happens on the ground.

dave

[Dockhead]

So here is my workup of Capt Force's scenario using constantly changing current which starts at 0 increasing to 4 then decreasing back to 0, with average of 2.

The current increases and decreases in a linear manner and is at 0 for both first periods and at 4 for both middle periods. This is rather artificial but I think it works ok (if anyone objects, please let me know and suggest a better approach.

captforcescenario.xls

The main formula is for SOG for a given period:

=SQRT(SUM(POWER5,2)-(POWER"X",2) where "X" is the set of the current for the given period.

I have run checks on average current and linearity of current change, which are shown in the spreadsheet.

I ran three scenarios for the GPS track boat:

1. hourly
2. five minutes (48 periods analyzed)
3. one minute (240 periods analyzed)

SOG was calculated for each period from which I derived distance made good towards the waypoint.

The results of course confirm that sailing a GPS track, however you analyze it, is considerably slower than sailing the constant heading, something I think everyone here except Capt Force accepts.

Unfortunately, the results did not correspond to my hypothesis that the smaller the analysis period (the smaller the vector triangles we calculate), the slower the GPS track boat will go. I am perplexed by this and will welcome any insight into why. The only thing I can imagine is that my current scenario, while perfectly linear at the given level of granularity, is not really -- because the beginning, ending, and two middle periods are all at the maximum and minimum values, not at the average value which would be achieved is the "line" went all the way to instantaneous ends and middle of the graph. So perhaps the slightly faster passage with one-minute analysis reflects the shorter period of time at maximum current -- two minutes instead of 10 minutes -- as between the one-minute and five-minute analyses. I don't know if this is enough to explain that; insights welcome.

The results are as follows:

Constant heading 18.33 miles made good in four hours
GPS track, hourly analysis, 17.7980 miles made good in four hours
GPS track, five minutes analysis, 17.5217 miles made good in four hours
GPS track, one minute analysis, 17.5778 miles made good in four hours

The most serious failing of my model is that the currents to which the GPS track boat is exposed to under various scenarios is the same as the currents the CTS boat is exposed to. And so I could only run 4 hours of the GPS track boat's passage, which does not get the boat into port.

This is flattering to the GPS track boat, as in reality, going more slowly (more slow VMG towards the waypoint) it will be explosed to more current, slowing it down further. So my analysis exaggerates the speed of the GPS track boat by that whatever that bit.

If anyone can suggest how I can overcome that -- "stretch" the current graph to match the pace of the boat in each scenario, I will be grateful. Anyone can download and hack on the model, just please rename it every time so that we don't get versions mixed up.

[Seaworthy Lass]

Originally Posted by goboatingnow

Here's what actually happens

In practice you dont compute that larger triangle, you compute the mean SOG and apply the extra distance ( its all the same )

Your CTS would never actually reach the destination.

The exact same approach is applied for tidal vectors , But you do have to cognisant of how the underlying tidal vector was got in the first place. The exactness of these maths, is not reflected in what really happens.

Dave

I am sorry but your diagram is wrong

[goboatingnow]

Dockhead unless I reading this The whole Captforce scenario is weird, its a backwards computation, ie the tide is assumed to have occured in whatever passage time is allowed.

Tides runs at a rate per hour, The do not run at a rate per passage, Hence the tide cannot varying as a function of the passage time.

The CTS calculations actually make no sense

The energy in a triangle , is not the Average its the Root mean square, ie .577 peak. in this case approx 9.23, I it can also be deduced graphically, In electrics, if your triangle was heating your one bar fire, this figure is the energy produced.

Heres the Science, as Jenifier would say

Triangle Wave Voltages - Vpk, Vpk-pk, Vavg, Vrms - RF Cafe

Ill look at the spreadsheet next

[goboatingnow]

Originally Posted by Seaworthy Lass

I am sorry, but your diagram is simply wrong.
The current is NOT perpendicular to the log speed of the boat (4 knots) - it is perpendicular to the rhumb line.
The boat speed/distance travelled vector should be on the hypotenuse, not on one of the short sides.

Easy There Seaworthy, lets not start shouting, the 4 is a 4 hour plot

huh, the Y-axis is the tidal current vectors!( in this case of perpendicular tides)

Im in the position of having clarity of thought here, as this is the correct way, so hit me with it.

[Seaworthy Lass]

Originally Posted by goboatingnow

huh, the Y-axis is the current vectors!( in this case of perpendicular tides)

Sorry, just saw your horizontal axis was marked in 4's

So what is your course to steer?