Distinct Activities: Shackled by a Common Name?

[Seaworthy Lass]

Originally Posted by goboatingnow

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

Sheesh.

Dave

Dave could you please (pretty please with sugar on top) tell me your compass heading and time taken for the hovercraft journey?

[conachair]

Originally Posted by goboatingnow

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.

Yes, but you know s the rhumb line distance and hence the time it would take to cross at 6Kn with no tides. Using that and the tidal data you can calculate, in this example, where you will be after a couple of hours. Which is about 2 miles short of where you want to be from the original rhumb line.

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.

I disagree with that in the example on the video. The tide has a component which is pushing you away from your destination. If you do the vectors with the destination 12Nm away then after 2 hours you won't reach it, you will be on the rhum line but short of the destination. If the tide doesn't change much then you will hit the bouy, but that hasn't been calculated so isn't known for sure.
I know how to do it, on paper and in real life. We must be missing something here, it's not that hard.

[goboatingnow]

Originally Posted by Seaworthy Lass

Dave could you please (pretty please with sugar on top) tell me your compass heading and time taken for the hovercraft journey?

I can but I don't have my graph in front of me , its at the office.

[Seaworthy Lass]

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

+ 1
I agree with all that.

If the currents can be perfectly predicted though, the RYA method is not perfect mathematically, it will only be so if the current for the final bit of the journey is the average of the currents you have been subjected to along the way.

[Seaworthy Lass]

Originally Posted by goboatingnow

I can but I don't have my graph in front of me , its at the office.

I await it eagerly .

[goboatingnow]

Originally Posted by conachair

Yes, but you know s the rhumb line distance and hence the time it would take to cross at 6Kn with no tides. Using that and the tidal data you can calculate, in this example, where you will be after a couple of hours. Which is about 2 miles short of where you want to be from the original rhumb line.

I disagree with that in the example on the video. The tide has a component which is pushing you away from your destination. If you do the vectors with the destination 12Nm away then after 2 hours you won't reach it, you will be on the rhum line but short of the destination. If the tide doesn't change much then you will hit the bouy, but that hasn't been calculated so isn't known for sure.
I know how to do it, on paper and in real life. We must be missing something here, it's not that hard.

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

The RYA method is the correct addition of speed vectors. , no more then you'd do the same to compute true wind from apparent and boat speed.

You do not , except in very simple cases ever end up at the rhumb at D , after 4 hours with the RYA you are 12.5 ( if I remember ) towards your destination. Thats all you know, you actually cannot determine where you are physically Obviously due to 4 hours of vectors you haven't made the destination, ( sorry my hover craft example)

In the two plot RYA one YOU are not on the rhumb line after two hours. Yes your short of your destination because you are progressing slower towards it due to the tidal vectors. Mathematically if you expand the graph to include the desired destination ALL time related vectors increase in length.

YOU DO NOT CHANGE COURSE , you proceed steering the same CTS. Hence the time line extends. With that time line extending ALL rate vectors increase in length appropriately. If you were to back compute , you would find each tidal vectors were proportionately longer then originally plotted ( ie my inflated triangle )

To illustrate that a two hour plot that place the vector position D BEHIND the destination. Are use suggesting you have to double back !!!!!!!


I've given you 4 links all are mathematically correct. I can prove multiple book references if so so desire

The RYA method simply computes the CTS and time to destination correctly given the tidal vectors that you can determine in advance. In real life a large complex multi vector multi hourly plot can cause trouble in real life as your actual ground track can bring you into a different tidal area
Dave

[conachair]

Originally Posted by Seaworthy Lass

+ 1
I agree with all that.

If the currents can be perfectly predicted though, the RYA method is not perfect mathematically, it will only be so if the current for the final bit of the journey is the average of the currents you have been subjected to along the way.

Well, it is mathematically perfect but you are limited to hourly tidal data so anything you do will be rounded up to the nearest hour. In the video example most people would be looking at the third hour to keep an eye on what was going on anyway,

[goboatingnow]

Originally Posted by conachair

Well, it is mathematically perfect but you are limited to hourly tidal data so anything you do will be rounded up to the nearest hour. In the video example most people would be looking at the third hour to keep an eye on what was going on anyway,

In real life you are recomputing based on actual boat speed, actual tidal vectors , wind , leeway etc.

But within the data you have starting out , that method is the only one you can bring to bear and it is mathematically correct for the vectors plotted. You can add in other vectors if you know they exist , but that doesn't change the method


Remember you are not at D , physically after 2 hours. You are actually south of the rhumb as you're red over the remains time to the destination the tidal vectors continue to work bringing north you to your destination.
Dave

[conachair]

Originally Posted by goboatingnow

You do not , except in very simple cases ever end up at the rhumb at D , after 4 hours with the RYA you are 12.5 ( if I remember ) towards your destination. Thats all you know, you actually cannot determine where you are physically Obviously due to 4 hours of vectors you haven't made the destination,

That what I just tried to say as you appeared to say the exact opposite

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.

[goboatingnow]

Originally Posted by conachair

That what I just tried to say as you appeared to say the exact opposite

Where did I say the opposite. Except about 8 pages ago when I put the wrong wording to my correct hovercraft graph I corrected that subsequently and to my knowledge haven't maintained that since a d if I have I'm wrong.

You do not hit the rhumb at D its just a vector addition vertice, since we seemingly all agree on that let's move on to the next point

And Seaworthy I will do the hovercraft graph with the wind signs the right way and show you I am correct.

Dave

[conachair]

Originally Posted by goboatingnow

Where did I say the opposite. Except about 8 pages ago when I put the wrong wording to my correct hovercraft graph I corrected that subsequently and to my knowledge haven't maintained that since a d if I have I'm wrong.

You do not hit the rhumb at D its just a vector addition vertice

Dave

as far as the maths go you arrive directly at the destination

[goboatingnow]

Originally Posted by conachair

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

[Seaworthy Lass]

Originally Posted by goboatingnow

You are both mixing up vectors that represent speed and direction with distance.
The RYA method is the correct addition of speed vectors, no more then you'd do the same to compute true wind from apparent and boat speed.

Goboating, no they are not speed vectors . You are very confused.
The RYA is plotting distance vectors. They are taking units of distance off the longitude and plotting nm. Rewatch the second video you posted. The demonstrator actually says he is plotting "how far the tide has taken us".
He later tells us to "mark on the the distance we would have travelled". He sets the dividers to 12 nm as that is the distance.

Originally Posted by goboatingnow

You do not , except in very simple cases ever end up at the rhumb at D , after 4 hours with the RYA you are 12.5 ( if I remember ) towards your destination. Obviously due to 4 hours of vectors you haven't made the destination,

You ALWAYS end up on the rhumb line short of the destination with the RYA method if the estimates for current have been accurate (or if the current is with you will arrive north or south of the destination as the rhumb line intersection may not be reached as it may be on land). Your position is the sum of the distances travelled (due to current plus boat speed), so if you position your distance travelled through water vector to end up on the rhumb line, that is where you will end up.

Originally Posted by goboatingnow

YOU DO NOT CHANGE COURSE , you proceed steering the same CTS.

Well, the RYA tell you to keep steering at this heading. you will then only reach your destination if the current you then encounter is EXACTLY an average of the previous currents you have encountered.

Originally Posted by goboatingnow

To illustrate that a two hour plot that place the vector position D BEHIND the destination. Are use suggesting you have to double back !!!!!!!

No LOL, if D is behind the destination it means that you arrive somewhere close to your destination under two hours as the current is with you, but you do not arrive AT the destination (unless you realise the error and change course) because you have drawn a track for 2 hours and you arrive under two hours.
This is another illustration of why the RYA method is not mathematically correct.

[Andrew Troup]

Originally Posted by Dockhead

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.

Dockhead

Firstly, it seemed to me that taking time slices (which CaptForce's scenario virtually mandates) does not permit a meaningful comparison between the rhumb line and constant heading strategies.

In the constant heading case, the slices will be constant in geographical width. No problem.

However in the rhumb line (straight GPS track) instance, uniform time slices will become progressively slimmer in distance as you progress into stronger current.

This must surely make analysis difficult, it seems to me, because we are told that the change in tidal strength is geographic. The current strength does not change with time, but with distance.

This is an interesting feature of the scenario, characteristic of 'brand name' ocean currents*, which I did not want to discard.

Consequently the variation per slice in current, in the scenario you are solving, will not be linear, in the case of the rhumb line boat – which (without puzzling it too deeply) seems to be to add a recursive aspect to the analysis.

(We don't know the current until we know the heading; we don't know the heading until we know the current.)

If this difficulty alone was not enough to put me off, the overwhelming reason why I discarded the constant time aspect of CaptForce's otherwise excellent scenario, was because he doesn't say which boat takes four hours.

Given that they travel at the same speed through the water, they cannot arrive at the same time, so the scenario cannot be solved as posed. Unless we toss a coin, we cannot solve it at all, which seems to me a nonsense.

(You have implied the same thing, indirectly, when you talk of 'scaling' one of the results in order to compare – but if you can see a way of doing this which preserves the validity of the conclusion well enough to persuade a sceptic, you're a better analyst than I will ever be)
So my assertion that your results will not be comparable with mine (particularly in the rhumb line case) was based on our solving different scenarios.

When I get a moment, I'll adjust the width of the current separately for each boat to bring the time to four hours (which incidentally disobeys the clause in CF's scenario which states that the current varies geographically, rather than on a 'per boat' basis) and post that.

If you PM me with your email address, I'll send you my spreadsheet, and would be interested to look over yours.

*sure, ocean currents change over time, particularly as eddies form and shed, but generally much slower than the regular change of tidal currents. I don't expect they would move very far (or change in intensity at a given location) during a single transit.

It seems realistic to hope that we will soon access accurate realtime data on ocean currents from satellite, if we cannot do so already. This could be a big deal for slow vessels, like sailboats with limited range under power, or no engine, in a light wind scenario.

[conachair]

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.

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.

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

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??