Distinct Activities: Shackled by a Common Name?

[Seaworthy Lass]

Originally Posted by goboatingnow

The graph is traditionally labelled A ( start) , B ( destination) , C ( sum of hourly tidal vectors ) and D the vector addition of the tidal vectors and boat speed ( hourly vectors ) through the water vector.

In the vast majority of cases D is before or after B on the rhumb line. D is merely a vector addition vertices. You are not physically there. ( which is why it confuses people when they use a chart to graph speed vectors )

D is always the intersection of the rhumb line and the boat vector because you are engaging in vector addition ( the closing of the triangle ). You are not engaged in determining where the boat physically is at the time of D that's a different story all together.

Dave

Dave, AD is the vector addition of AC (distance pushed by current over however many hours you are considering) and CD (your distance travelled through water during this time, deliberately positioned so it hits the rhumb line). The sum gives you AD, where D is where you end up on the rhumb line.

If AC is 3 nm perpendicular from the rhumb line and CD is five nm, put at a heading designed to hit the rhumb line, you would end up at D, 4 nm up the rhumb line and on it.

Where do you think you would end up?

[conachair]

Originally Posted by goboatingnow

You are both mixing up vectors that represent speed and direction with distance.

Opposite I think. You plot the vectors for each hour. Speed = distance/time and you are putting a constant in for time , 1 hour, so you are plotting the change in position after an hour, ie distance with a direction.

[Seaworthy Lass]

Originally Posted by goboatingnow

....D is always the intersection of the rhumb line and the boat vector because you are engaging in vector addition ( the closing of the triangle ). You are not engaged in determining where the boat physically is at the time of D that's a different story all together.
Dave

Conachair commented on the above:

Originally Posted by conachair

Now that makes no sense again
To take an extreme example, boat at A with a destination due north, boat s speed 5Kn, CTS due north, tide sets due south at 1Kn, so after doing vector analysis for an hour (velocity vectors convert to distance) boat position D is 4Nm due north of A, you seem to be saying somehow that isn't the case??

Yes it makes no sense at all. I am glad you agree with me Conachair. This is in fact not an isolated exception, if you add two or more distance vectors together (which is what the RYA are doing) you end up at the sum of these (and since in this case you deliberately made the boat heading vector meet the rhumb line, that is where you will always end up (if the figures for the current were accurate to start with - that is a totally seperate issue).

Because I am disputing what GOBoating says is the mathematical perfection of the method used by the RYA, everyone is being strangely silent. Yours is the only voice of reason piping up. Thankyou.

The RYA method is an excellent approximation allowing rough computation of very complex currents, but it is NOT mathematically perfect.

Two examples of its limitations:
- If the cross current is partly with you, you will end up at some other spot other than the destination unless you realise the error and change course.

- If the current is partly against you, once you have arrived at the rhumb line the current is very different to the AVERAGE you have had in speed and direction up to that point, you will also arrive elsewhere unless you realise the error and change course.

[Andrew Troup]

Dockhead:

I don't think the answer to your specific question (how far did my GPS boat get in four hours) is very useful, because with my 20 mile wide current, it was only part way across. In case it does make some sense to you, the answer is about 16.8 nm
(provisionally - the spreadsheet is not set up to answer such questions easily. Distance-based questions, I can answer more easily)

At the four hour mark, the constant heading boat has covered 18.35 nm of Northing, and I reckon the Easting is -0.57nm, in other words it's still that far downstream of the rhumb line.

To give a more meaningful answer which might be comparable with your analysis, I quickly massaged the current width separately for each boat, to get each boat arriving at the far side of its private current scenario (linear variation with distance, av 2 knots, max 4kn midchannel) at the four hour mark, but I haven't currently got the luxury of time to ensure the results are robust.

For what it's worth, here they are (provisionally!)

adjusting current width for rhumb line boat to take four hours:
Current width 17.18nm

adjusting current width for constant heading boat to take four hours:
Current width 18.26nm

[goboatingnow]

Originally Posted by conachair

Okay, we are talking about the same thing then. It was the use of "destination" instead of "d" in an earlier post which confused things.

Now that makes no sense again
To take an extreme example, boat at A with a destination due north, boat s speed 5Kn, CTS due north, tide sets due south at 1Kn, so after doing vector analysis for an hour (velocity vectors convert to distance) boat position D is 4Nm due north of A, you seem to be saying somehow that isn't the case??

No in a simple one hour plot the boat runs down the rhumb. In a case where there are multiple vectors of differing angles the boat does not run down the rhumb. ( which is a safety issue. ) all the vectors in line is a simple case.


There are three threads here all debating different things. My fingers cant keep up.

Irrespective of where D is. The key is that you ARE plotting speed vectors. That is you lay down distance per unit time. That's speed Seaworthy

Ill come back later tomorrow. The battery thread is getting interesting


seaworthy the method ( ie the intersection ) does not end up ALWAYS in front of the destination. Equally the method of retrieving the tidal vectors assumes they don't stop. Ie its a 2kn tide not a 2nM tide.



Dave

[Seaworthy Lass]

Originally Posted by goboatingnow

No in a simple one hour plot the boat runs down the rhumb. In a case where there are multiple vectors of differing angles the boat does not run down the rhumb. ( which is a safety issue. ) all the vectors in line is a simple case.

Dave

I agree that if there is any cross current of varying degree the boat does not run down the rhumb line. But it does end up on the rhumb line at the end of the specified number of hours the current was examined for using the RYA method, as the vector for the distance the boat travelled was deliberately positioned to hit the rhumb line to compute what course it need to follow to hit the rhumb line.

Two am here. Time to say goodnight to all.

GoBoating, I am looking forward to hearing what your compass heading and time to arrive at the destination are for the hovercraft example.

This was your final plot to illustrate the course to steer. If you would like to look at that when you have a chance and tell me what they are, it would be much appreciated

[goboatingnow]

Originally Posted by Seaworthy Lass

Conachair commented on the above:

Yes it makes no sense at all. I am glad you agree with me Conachair. This is in fact not an isolated exception, if you add two or more distance vectors together (which is what the RYA are doing) you end up at the sum of these (and since in this case you deliberately made the boat heading vector meet the rhumb line, that is where you will always end up (if the figures for the current were accurate to start with - that is a totally seperate issue).

Because I am disputing what GOBoating says is the mathematical perfection of the method used by the RYA, everyone is being strangely silent. Yours is the only voice of reason piping up. Thankyou.

The RYA method is an excellent approximation allowing rough computation of very complex currents, but it is NOT mathematically perfect.

Two examples of its limitations:
- If the cross current is partly with you, you will end up at some other spot other than the destination unless you realise the error and change course.

- If the current is partly against you, once you have arrived at the rhumb line the current is very different to the AVERAGE you have had in speed and direction up to that point, you will also arrive elsewhere unless you realise the error and change course.

I couldn't resist looking at this last line because yes its shows the limits of tidal stream atlas. Not the vector method in case limitations b above this is a limitation of hourly tidal vectors not the method. If you have 30 minutes vectors information then the accuracy , ie the correspondance with the real world gets better. In the absence of such additional information the tidal runs as per the vector for infinite time. ( you simply have no other information one way or the other. )

So yes of course in the real world you could be hit with a tide you don't know about.

That's why at the very opening of this thread I said that I don't like large multi vector tidal computations , its too open to challenge by real world situations.

More graphs over the next few days. But they key is the tidal vectors are assumed to be constant +- 30 minutes from the centre point. That's nothing to do with vectors its to do with the assumptions of tidal stream atlas hence as you say Seaworthy it does assume that the rate is constant within that period. , outside of that you fall into the next period.

I will demonstrate that your limitation A above does not hold

Night night

Dave

[Dockhead]

Originally Posted by Andrew Troup

Dockhead:

I don't think the answer to your specific question (how far did my GPS boat get in four hours) is very useful, because with my 20 mile wide current, it was only part way across. In case it does make some sense to you, the answer is about 16.8 nm
(provisionally - the spreadsheet is not set up to answer such questions easily. Distance-based questions, I can answer more easily)

At the four hour mark, the constant heading boat has covered 18.35 nm of Northing, and I reckon the Easting is -0.57nm, in other words it's still that far downstream of the rhumb line.

To give a more meaningful answer which might be comparable with your analysis, I quickly massaged the current width separately for each boat, to get each boat arriving at the far side of its private current scenario (linear variation with distance, av 2 knots, max 4kn midchannel) at the four hour mark, but I haven't currently got the luxury of time to ensure the results are robust.

For what it's worth, here they are (provisionally!)

adjusting current width for rhumb line boat to take four hours:
Current width 17.18nm

adjusting current width for constant heading boat to take four hours:
Current width 18.26nm

I wasn't asking about the constant heading boat. Why would you break your head doing a ground referenced calculation of the constant heading boat when you have a simple, mathematically perfect water based method?

I was interested in the GPS track boat.

[Seaworthy Lass]

Originally Posted by goboatingnow

Try This for a plotting exercise

(a) You on a frozen lake, hence no water movement in a hovercraft.
(B) Your destination is directly East, 16 miles away
(c) your hovercraft does 4 miles an hour.

(c) 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

You can cross the lake in two ways,

(a) Using Track , you in effect stem each hours wind as you meet it.
(b) compute a single heading for the whole journey

Now tell me the time it takes for the first ,method , the second method and compare it the time without wind, ie 4 hours.
If you can do this you will remove all the confusion.
Dave

Originally Posted by goboatingnow

Seaworthy heres the actual graphical solution.
The track method time is 6 hours 24
The CTS method is 5 hours 7 ( approaching the destination at a rate of 12.5/16)

Hi GoBoating
I am bright eyed and bushy tailed here after six hours of deep sleep.
It is amazing what inspirations can come after a good night's rest .

The penny has dropped for me why you think it will take five hours and seven minutes to reach the destination if you steer a constant heading and travel at a constant speed for the duration of the journey - see post #407 of yours which I have reproduced above. The diagram you gave in this post is attached below.

It is because if you have a shore perpendicular to the bearing of A to B that is exactly how long it will take you to reach the shore upcurrent of B (if B is on the edge of land in the hovercraft example). You will in fact never reach B if you follow the heading deviation derived from the RYA method.

I have carefully rewatched your second video link and there is nothing wrong with your diagram (given below) according to the RYA method.

You have plotted exactly what they tell you to do:
- at the departure point mark on the sum of the distances pushed by the current (in this case 10 miles perpendicularly)
- as the trip is 16 miles and your speed is 4 miles/hour, the time taken is very roughly four hours, so mark off 16 miles from the tip of the final current vector and swing it around until it hits the rhumb line - this point in our case is 12.49 miles from the departure point.
- the RYA says this is the compass deviation required from whatever the rhumb line heading is in order to reach the destination (in this case 38.7 degrees, so you steer this much north of east ie 51.3 degrees).
They are incorrect.

You will hit the rhumb line 12.49 miles from A (3.51 miles from B). At this point you are still heading at 51.3 degrees and the bearing of B from there is 90 degrees, so given the current has dropped to zero at this point you will never reach the destination if you keep going on this heading.

At the 5 hr 7 minute mark you will reach a point upcurrent of B and perpendicular to it.

The reason is that the correct deviation required is 32 degrees (sorry, I called this the compass heading before, I meant the compass deviation, I had forgotten you said the destination was due east). If you deviate 32 degrees (compass heading 58 degrees) you will get there in 4:72 hours (4 hours 43 minutes).

I have taken the current to be zero in the fifth hour as you did not respond when I asked you what it was (if it is 'known' for the first 4 hours, it should also be 'known' for the fifth hour). It is not unreasonable to call it zero, it was one mile/hour the previous hour. The RYA does not allow for the correct current in the lt portion of the journey. It assumes it is exactly the average you have encountered along the journey up to that point. This is the fault in the RYA method.

Look at your diagram and you will see this. You have actually dotted on an average amount of extra current all the way from the point you hit the rhumb line (D) to B. Why do you think the current is this amount during the fifth hour? Why would it be?

This example you set should above all else show you the RYA method is mathematically incorrect.

It is not an exception because the current in the last part of the journey is zero. It will always be incorrect unless the current encountered in the last portion of the journey is EXACTLY the average of the current encountered for the journey up to then (that is the exception where the RYA method will work but only if part of the current is not pushing you towards the destination - it will only work if part of the current is against you).

Plot out where your position is every hour of the journey and see. Please don't just say "this is not the case, the RYA would never make a mistake like this after 50 years of teaching this", please actually do it (I learned this the hard way in one of my responses to CaptForce earlier).

The RYA method is incorrect not because the current information is not accurate, and not because the wind is pushing you off course or because the calculations may push you into a different tidal area or because the current does not change exactly once an hour and stay constant for the hour (although in reality these are all factors that will affect you).

It is simply because the RYA method is only an approximation, it is not a mathematically correct method, unless you have the unique example of the current encountered in the last portion of the journey being EXACTLY the average of the current encountered for the journey up to then (and why would it be?).

Does someone want to tell the RYA not to tell people that this method gives the precise heading to the destination of all the current data is correct and there is no leeway? If they have truly been teaching this for 50 years I am a little concerned they will shoot the messenger and I am simply a coward at heart .

I repeat what I have said before: the RYA method is an excellent tool as it allows you to easily calculate your approximate compass heading for very complex situations where the current is varying in strength AND direction for the journey. It is not actually accurate mathematically though and this limitation needs to be recognised.

We have had over eight thousand views of this thread now. Will someone with some mathematical knowledge please work though this hovercraft example using the RYA method and confirm what I am saying is correct (or incorrect if this is the case).

I throw down the gauntlet (not dropping any more towels LOL).

The link to the RYA method is:
YouTube

If I am incorrect (not impossible, I have occasionally been known to be wrong ), I will be embarrassed, but I would have learned a tremendous amount from this thread and will be forever grateful .

This was your diagram:

[Seaworthy Lass]

PS please see post #403 for my calculations for the hovercraft example and post #449 for my vector diagram of distances each hour, including the dotted line giving the chartplotter track for the journey.

The diagram shows that by following a compass deviation of 32 degrees (the actual compass heading is 58 degrees if B is magnetically east of A), you will arrive at B at 4.72 hours (4 hrs 43 min). You have travelled at a constant heading and speed for the entire journey.

[Andrew Troup]

Originally Posted by Dockhead

I wasn't asking about the constant heading boat.

I was interested in the GPS track boat.

And I answered your question, for the GPS track boat.

I then answered a question you didn't answer, because your question didn't seem to me to provide any basis for comparison with your four hour scenario.
And I answered for both boats, because I thought the entire point of the exercise was to compare the two tactics.

That's a long answer: the short answer is that I was trying to be helpful.

Originally Posted by Dockhead

Why would you break your head doing a ground referenced calculation of the constant heading boat when you have a simple, mathematically perfect water based method?.

Ummmm... two reasons:

1) I was trying to meet CaptForce, to the extent possible, on his own territory. He, unlike you, was not convinced that the water based method was mathematically perfect - or, at least, that it was any better than the rhumb line. Presumably, he remains unconvinced, but that's his prerogative. I've done all I'm going to do.

2) Have you already forgotten that we all made prize dicks of ourselves through having not the faintest idea what the ground track looked like for the constant heading transit?

Also: What if there are places en route we do not want to go ?

I like to know whereabouts on the chart I'm going, thanks very much. My spreadsheet provides a prediction of where I can expect to be at any stage of a constant heading transit; I can compare this against the GPS to find out how accurate the current data are proving to be.

In any case, I wanted to be sure I was correct to change sides, in favour of CF's 'S' curve

(you need to remember that at the time when I put the spreadsheet together, nobody else had come on board with his provocative idea. In abstract theory, I had pretty much convinced myself, but I wanted a second opinion, from a practical worked example)

3) OK, so I lied about 2 reasons:

So that the relative progress of the two competing boats could be meaningfully compared at any point.

4) It also provided a cross check that the constant heading boat ends up back on the rhumb line (and crosses it midway, and runs parallel to it at the quarter and three-quarter marks).

As for "breaking my head":

If there's one thing this exercise has (re)taught me, it's the perils of seeking the easiest answer.

In any case, it was not difficult with distance slices, especially in comparison with trying to reconcile time slices with a current which varies as a function of distance, which presumably you somehow found a way to do.

[Seaworthy Lass]

Hi Dockhead and Andrew
We have two lots of discussions going on concurrently here. I have given up trying to convince CaptForce, but full marks to you two for persisting .

When you are through with CaptForce's example, could you please go through GoBoating's hovercraft example? It would be very appreciated. All the details are in post #519. Unlike your case, the maths are extremely simple and shouldn't take long.

Other than Conachair briefy joining me out on the limb, I seem to be entirely on my own putting forward my case and I am getting a little lonely .

[Seaworthy Lass]

In response to Dockhead Andrew posted this:

Originally Posted by Andrew Troup

......Have you already forgotten that we all made prize dicks of ourselves through having not the faintest idea what the ground track looked like for the constant heading transit?

You are being very diplomatic here .

It was only me and Dockhead and GoBoatingNow who made prize dicks of ourselves LOL. You were the only one who realised CaptForce was correct with his symmetrical sinusoidal shape for the chartplotter track .

[Dockhead]

Originally Posted by Seaworthy Lass

In response to Dockhead Andrew posted this:


You are being very diplomatic here .

It was only me and Dockhead and GoBoatingNow who made prize dicks of ourselves LOL. You were the only one who realised CaptForce was correct with his symmetrical sinusoidal shape for the chartplotter track .

Ha, ha. There were plenty of brain farts all around on here. I think I probably get the prize for the most!

But my darling -- Capt Force was not correct about the symmetrical sinusoid shape for the chart plotter track for a constant heading boat in a unidirectional current. It's symmetrical only in a symmetrical changing current. For a unidirectional current, the vast majority of the passage is spent on the upstream side of the rhumb line. You fall off downstream of the rhumb only just at the end -- if the current is unidirectional like in Capt. Force's scenario.

I think that was one of the least egregious of the brain farts on here, certainly I have much worse ones When you imagine a constant heading passage, you just don't need to think about the ground track -- it's irrelevant to the mathematical problem (NOT of course to avoiding obstacles, as Dave pointed out ).

[Seaworthy Lass]

Originally Posted by Dockhead

Ha, ha. There were plenty of brain farts all around on here. I think I probably get the prize for the most!

But my darling -- Capt Force was not correct about the symmetrical sinusoid shape for the chart plotter track for a constant heading boat in a unidirectional current. It's symmetrical only in a symmetrical changing current. For a unidirectional current, the vast majority of the passage is spent on the upstream side of the rhumb line. You fall off downstream of the rhumb only just at the end -- if the current is unidirectional like in Capt. Force's scenario.

I think that was one of the least egregious of the brain farts on here, certainly I have much worse ones When you imagine a constant heading passage, you just don't need to think about the ground track -- it's irrelevant to the mathematical problem (NOT of course to avoiding obstacles, as Dave pointed out ).

LOL Dockhead. Dont make the same mistake I did. Plot it out!
If the current is unidirectional but a mirror image in strength, your chartplotter track WILL actually be a mirror image track (crossing the rhumb line at the halfway point - cross my heart LOL)

Check it out. I have learned a lot from this thread on what your ground track will be like following a constant heading to compensate for variable current .