Distinct Activities: Shackled by a Common Name?

[Andrew Troup]

Originally Posted by Jammer Six

After a brief stumble through the thread, I'm not sure why we switched from boats and current to hovercraft and wind.

It did occur to me to wonder the same thing.

I'm all for abstractions

either as a way of reducing a problem to the minimal set of behaviours which bear on the solution

or to move the thinking of experienced people into new territory, in the hope they'll leave preconceptions behind

but I'm not sure what the value of the hovercraft abstraction is: it seems to me to carry approximately the same number of behaviours, including a few which are unusual and unhelpful (such as having no preference for travelling in the direction they're pointing).

It is new territory, though, so maybe that's helpful for some people. For me, it's too confusing and unfamiliar.

Ants, for me, are very predictable and familiar. I'm trying to eradicate discourage a supercolony as we speak, one ant at a time ....

Perhaps the analogy was chosen by one of our resident antiphobes:

If you keep the main fan running, it's very difficult for ants to find their way onto a hovercraft

[Wotname]

Originally Posted by Andrew Troup

I let the idea quoted below swim quietly past, at the outset of this thread.

..................
When you say "That would be called Pilotage rather than Navigation", isn't that a false dichotomy?

Surely pilotage is a sub-branch of navigation (as the last part of the wikipedia quote implies - "When visual references are not available, it is necessary to use an alternative method of navigation"

- I'd be the last to claim that wikipedia is any sort of authoritative source, but it's what you chose to post (AT wrote mildly)

Question 2:
Are you suggesting, in your last sentence, that, just as pilotage is considered a sub-branch of navigation, satellite navigation could be so considered?

Note to our delegates from the Ministry of Defence: I'm asking a question here, not stating a position

I also let this float past. I think it is self-evident that pilotage is a sub-set of navigation or perhaps even its founding father! As for satellite positions and subsequent electronic charting methods, these are all sub-sets (or tools) of the science art or craft of navigation - IM(somewhat)HO.

Originally Posted by Andrew Troup

It did occur to me to wonder the same thing.

I'm all for abstractions

either as a way of reducing a problem to the minimal set of behaviours which bear on the solution

or to move the thinking of experienced people into new territory, in the hope they'll leave preconceptions behind

but I'm not sure what the value of the hovercraft abstraction is: it seems to me to carry approximately the same number of behaviours, including a few which are unusual and unhelpful (such as having no preference for travelling in the direction they're pointing).

It is new territory, though, so maybe that's helpful for some people. For me, it's too confusing and unfamiliar.

Ants, for me, are very predictable and familiar. I'm trying to eradicate discourage a supercolony as we speak, one ant at a time ....

Perhaps the analogy was chosen by one of our resident antiphobes:

If you keep the main fan running, it's very difficult for ants to find their way onto a hovercraft

Likewise I have been wondering just how helpful the hovercraft example has been. It seems to be a constuct to help "simply" an understanding of perhaps a complex concept of tidal (current?) influcence on passage planning. However I just see it a simple example of a everyday aviation navigational excercise (VFR) albeit with some interesting forecast winds. As such, a vector summation is all that is needed to determine the required aircraft heading (or in maritime language CTS). This is why Seaworthy's approach seems intuitively correct to me.

But is it not great example of the effects tidal currents, winds don't normally increase and decrease in same manner as tidal currents and don't reverse at 6 hours (or so) intervals so it does not (IMO) help understand the problems and thus solutions of passage planning. On the other hand, winds affect the track of and aircraft in the same mathematical way that currents affect the track of boats. So the maths principles are the same for both.

[Seaworthy Lass]

'Morning all.
Beautiful morning here, the water is like a mirror and with a southerly blowing straight from Libya I don't even need my hot water bottle this morning. Unlike yesterday when the gas ran out, I even have cuppa in my hand not long after awakening .

I just had a quick look at the overnight posts. Folks, don't stress whether it is a hovercraft or a boat. I have redrawn the data for a boat and it is equally wrong - CST 51.3 degrees using the RYA method when the correct answer is 58 degrees (which my method computes correctly. Dave, please don't say the data is unrealistic. A method should be able to cope with any data at all thrown at it and the RYA method can't - it fails in some circumstances like the example above. This means the RYA is just an approximation. I have a better method that cover ALL circumstances and I would like to put it up for consideration. It is not much more difficult to do and it gives you the correct answer every time .

Dave, you say in the above example you would have actually drawn on another current vector and swung another arc had the data for the 5th hour been available. I am sorry, but you would NOT have according to the RYA method as then D would have been beyond B (be honest here , the RYA does not teach you to arc off the distance beyond the destination unless there has been current with you for the journey).

Many would say this is a totally unrealistic lot of current data and would never happen, but what if it did? The RYA method simply does not give you a good result. Furthermore if this had been an RYA exam example I would have failed miserably saying the CTS was 58 degrees (which it is) instead of 51.3 degrees, as the limitations of the RYA method are not recognised. The RYA teach that their method of determining CTS will have you arriving exactly at B, not that it is just an approximation and that it may possibly be quite wrong in some circumstances. This is the key message I would like to impart to the RYA and readers of CF.

Now that I think I have my method sorted out I would like to take it for a test run.

I will no longer explain what I am doing. The RYA method is well known, my final method is given in post # 623 and I will follow this exactly. I will just give the data and the results and take photos of the chart plots using the RYA method and my method for each case. I will try and use realistic data for my examples. Let's see how the RYA method compares with mine .

Ready to fasten your seatbelts folks?

[Dockhead]

Originally Posted by goboatingnow

DockHead

I must say your analysis feat is substantial

I have one issue ( other then the average CTS calculation tidal vector )

If you are only given the tidal effects for the passage of the boat following a constat heading and hence spending 4 hours on the water. How can you deduce what the tides are for any boat following a slower water track. Its like you have the Answer but cant find the question.

Hence how can you compare different time transits' with each other ( in other words , what does a slower boat experience as tides, since you arnt given that information in the first place.)

but proceed

dave

I am indebted to Andrew for the solution of the transit time problem for the GPS track boat. I sliced the GPS track boat's passage into distance slices. The problem had currents varying only by position (so not tidal currents), so I think my model works perfectly to account for the different amount of current the boat is exposed to with different passage times, for this particular problem. Tidal currents varying by both time and position will be more complicated.

What problem did you have with my CTS boat calculation? That one is dead simple. All you need is the average current, which forms the short side of the overall vector triangle. Or?

[Seaworthy Lass]

COMPARING 'THE RYA COURSE TO STEER METHOD' TO MY 'NEW IMPROVED MODEL'

You have a crossing you need to make. It is foggy, you can't see the other shore until you are very close. Your chartplotter is not functioning and you need to work out an accurate course to steer to get from A to B.

There is no wind, so you are motoring. It is an emergency, so you don't have the luxury of picking a time close to slack water to go and you need to get there ASAP.

Boat speed is constant at 4 knots.
Destination is 10nm due east.
Currrent is always from the north:
1st hour 3 knots
2nd hour: 1.5 knots
3rd hour: 0.5 knots

RYA method:
Calculated CTS: 55.8 degrees
Calculated time taken: 3 hours 2 minutes.

My method:
Calculated CTS: 63.9 degrees
Calculated time taken: 2 hours 47 minutes.

Note: D1B / D1D2 = 3.4 / 4.3 = 0.79
So the amount of current that applies to get to B = 0.5 x 0.79 = 0.4
Hence total current for the journey = 4.9 knots
If you didn't have a calculator, just eyeball the proportion of the last current vector that will apply by estimating the proportion of the distance between D1 and D2 that will apply. Mark that proportion on the last current vector and call that C. Join C and B. that is your CTS and of course the the length of CB is your total distance travelled. It still gives you a very close result, certainly dramatically better than the RYA computation.

If the tide data is correct and the RYA method is used for calculations and the CTS followed, the skipper is going to be SERIOUSLY unhappy when he hits the shore 2.3 nm north of his destination. He would be even more seriously unhappy and his life may even be at risk if there was an obstruction he hit so far north of the course line, as the RYA method would have predicted he would never be north of the course.

If the tide data is correct and my method is used for calculations and the CTS followed, the skipper it thrilled to find that he arrives exactly at his destination. He also makes it in the shortest possible time and is able to evert the emergency that was the cause for his passage. So my method results in a very happy ending.

The plot using my method is in the next post.

So folks. That's it for examples for now. I took delivery of a new Sailrite sewing machine two days ago and I am just dying to play with it. Lots of boat chores to catch with as well, this has been very time consuming over the last two days. I will check posts now and again as I am sewing. .

[Seaworthy Lass]

This is the computation for the above example using my method:

[Seaworthy Lass]

Explanation of why you need to mark D2 on the chart as well as D1 (D1 is usually the same point as D using the RYA method).

Note: You need to work relative to the rhumb line as B is on the rhumb line.

Firstly estimate (an educated guess will usually do) the whole number of hours to get you nearly to B. If this was actually where B was, this would give you your CTS. Mark this point D1.

D1 is however unlike to coincide with B, and you don't know how long the next hour of tide will apply as your CTS depends on the length of time you are subjected to the last bit of tide and the amount you are displaced by the last bit of tide depends on the CTS. So at a glance it seems unsolvable (except to a mathematician), so I have come up with an easy solution.

See where you end up on the rhumb line if the journey went for another whole hour (so you know you are displaced by the whole amount of tide). If B was actually there, that gives you the CTS. Mark that as D2.

Now you know that to reach B you must steer somewhere between these 2 CTS figures.
But how much?
Its nice and easy. If D1 is really close to B, the CTS is really close to the one for D2.
If D2 is really close to B, the course to steer is really close to the one for D2.
So measure (or simply eyeball the proportion) and mark that same proportion off the last current vector.
Label that C.
Join C to B. that is your CTS.

If the current is zero for the last hour you still need to plot D1 and D2 (as they may not fall either side of B and if not, you need to allow for another hour of current).
If D1 and D2 do fall either side of B, then there is nothing more to do (you don't need a proportion of zero LOL), so just mark the end of the end of the last current displacement vector as C and join that to B and that is your CTS.

Hope I have explained that so everyone can follow

[Seaworthy Lass]

Ahoy there!

Nearly eleven hours later and there are no replies at all?

So much is hotly debated on CF: anchors, anchoring, what boats are bluewater capable and what building materials are best. Occasionally even discussions on other topics sneak in LOL. A wonderful exchange of ideas often occurs (and often smiles are raised) and everyone benefits.

I have posted a realistic scenario where my method of obtaining CTS varies eight, in fact more than eight degrees from that obtained by the RYA method and there is no comment all day .

I would prefer so that I can work on refining my method, to no repsonse LOL!

[Wotname]

Originally Posted by Seaworthy Lass

Ahoy there!

Nearly eleven hours later and there are no replies at all?

So much is hotly debated on CF: anchors, anchoring, what boats are bluewater capable and what building materials are best. Occasionally even discussions on other topics sneak in LOL. A wonderful exchange of ideas often occurs (and often smiles are raised) and everyone benefits.

I have posted a realistic scenario where my method of obtaining CTS varies eight, in fact more than eight degrees from that obtained by the RYA method and there is no comment all day .

I would prefer so that I can work on refining my method, to no repsonse LOL!

No one likes someone who is always right .
Those who agree has nothing to say and the nay-sayers are either re-grouping for a new offensive or just licking their wounds

Prolly just snow, life and the lack of graph paper is delaying a proper and reasoned response.

Meanwhile I'll start saving for the limeburners .

[Seaworthy Lass]

Originally Posted by Wotname

No one likes someone who is always right .
Those who agree has nothing to say and the nay-sayers are either re-grouping for a new offensive or just licking their wounds

Prolly just snow, life and the lack of graph paper is delaying a proper and reasoned response.

Meanwhile I'll start saving for the limeburners .

Hi Wotname
Do you think the RYA will maybe send me a box or two of some fine single malt in appreciation of discovering a better technique for determining CTS? If they do I will be happy to share it with you . I know the RYA will probably be very grateful being able to improve their teaching methods .

All the old textbooks will become collectors pieces illustrating what they did in the 'old' days before my method was adopted.

Thanks for your moral support xxxxxxx

[conachair]

Originally Posted by Seaworthy Lass

Ahoy there!

Nearly eleven hours later and there are no replies at all?

Still here

just a bit mildly obsessed with trying to graph it, without too much success but fun all the same

[Seaworthy Lass]

Originally Posted by conachair

Still here
just a bit mildly obsessed with trying to graph it, without too much success but fun all the same

Hi Conachair
Many thanks for trying.
I have plotted the chartplotter track and I do arrive there as expected, and the RYA method gives a way off result, but it would be really good to get some independent confirmation.

Also good to get some confirmation that the computations I have made for the RYA method are actually correct according to what they teach.

Lots of CF members and lurkers must have done RYA navigation courses. Could someone PLEASE tell me if what I have done for the RYA method is correct according to what they teach? It would be really appreciated.

[goboatingnow]

Hi seaworthy.

The RYA method says the following ( and please you have to except that I know the method , I can't realistically copy the whole instructors note section )

Ill look, at your chart work later , its quite difficult from the pictures

The RYA says estimate in hours the journey time ( or whatever tidal quantum you have available the , in some places you have 30 minute tides ). That estimate is the total expected passage time allowing for any tides. typically where tides are 25% or less then boat speed or the journey is short then the rhumb line can be used. But in long runs with many tides ( like across the Irish Sea ) you have to guesstimate.

Derive the tidal data from the appropriate sources. Manually interpolating if your passage point do not coincide with the HW ( high water marks )

Draw a line through from A ( start ) to B the destination and carry that line beyond B

Plot as many hourly ( or whatever tidal time quantum ) as feel is neccessary to cover the intended JOURNEY time. For multi hour tides, add the one hour tidal vectors together call this Point C

Draw the boat speed vector in quantum's of the tidal vector quantum , typically one hour. , so if the boat speed is 4 kn , open the dividers to 4nm and arc of D on the rhumb line. , if its a 2 hour plot then its a 8nm arc and so forth call this intersection Point D

despite what you persist in saying , D depending on how many hours you estimate , and the effects of the tidal vectors may lie in front , at or behind B

Perform a check to ensure that D lies approx less then 30 minutes away from B , some instructors omit this bit. , ie to ensure the tidal data remains valid. This can be done by simplest calculating the rate of advance , ie the average SOG to D and by inspection determine how far B is away. If its greater then 30 minutes ( or less) redraw the plot using one more or less tide ,( note this is rarely done in examples as the test questions are typically picked to ensure it isn't needed.

If D lies in front of B , calculate the time of arrival by applying the rate of advance to D to the whole journey B, equally if D is behind reduce the time by using the SOG to B accordingly ( ie inflate or deflate the whole tidal triangle

CTS is the bearing of line CD at all times.

That's it

As I mentioned inflating or deflating the total tidal triangle is as accurate as any other method , given the interpolation of tidal vectors in the first place. Ie in a six hour plot inflating each vector by say 7 minutes , is much as accurate as deflating the last vector by 42, in fact its actually more accurate

Dave

[goboatingnow]

Originally Posted by Seaworthy Lass

Explanation of why you need to mark D2 on the chart as well as D1 (D1 is usually the same point as D using the RYA method).

Note: You need to work relative to the rhumb line as B is on the rhumb line.

Firstly estimate (an educated guess will usually do) the whole number of hours to get you nearly to B. If this was actually where B was, this would give you your CTS. Mark this point D1.

D1 is however unlike to coincide with B, and you don't know how long the next hour of tide will apply as your CTS depends on the length of time you are subjected to the last bit of tide and the amount you are displaced by the last bit of tide depends on the CTS. So at a glance it seems unsolvable (except to a mathematician), so I have come up with an easy solution.

See where you end up on the rhumb line if the journey went for another whole hour (so you know you are displaced by the whole amount of tide). If B was actually there, that gives you the CTS. Mark that as D2.

Now you know that to reach B you must steer somewhere between these 2 CTS figures.
But how much?
Its nice and easy. If D1 is really close to B, the CTS is really close to the one for D2.
If D2 is really close to B, the course to steer is really close to the one for D2.
So measure (or simply eyeball the proportion) and mark that same proportion off the last current vector.
Label that C.
Join C to B. that is your CTS.

If the current is zero for the last hour you still need to plot D1 and D2 (as they may not fall either side of B and if not, you need to allow for another hour of current).
If D1 and D2 do fall either side of B, then there is nothing more to do (you don't need a proportion of zero LOL), so just mark the end of the end of the last current displacement vector as C and join that to B and that is your CTS.

Hope I have explained that so everyone can follow

As I read it this is exactly the same process as the RYA , except you only apply the proportion to the last tide.

Where you to assume like you did , that the tides are actually exactly constant for each hour AND the hourly, tidal boundaries are immutable your method would be accurate.

In practice , as I have repeated over several posts , proportioning , say 4 minutes of additional vector to each hour say over a 6 hour plot in order to " advance " the whole triangle is MORE accurate then deflating the last triangle by 60-24= 38.

The fact is that by the nature of determining tidal vectors your method results in a less accurate result where you proportion the last tide by a large percentage of an hour.

Accepting that mis - assumption in your method , then all you have done is the RYA method.

( note computing numbers based on wrong assumptions , may mean your maths are right, but the data is simply flawed to begin with )
See my next post as a summary

I must say you have a bee in your bonnet about this. You must have done lots of real CTS calculations in real life.


Dave

[goboatingnow]

To summarise the errors in your method Seaworthy


(a) you apply all the adjustments only in the last hour. So for example if D1 and D2 end up equidistant , you would add or deduct 30 minutes of tide vector to the last tidal vector.

Yet the RYA would proportion ( in a 2 hour plot , 15 minutes to the two preceding tidal vectors , whereas you would assume all the correction applies to the last one. Yet all tide data gives you is a speed and direction at a particular point in time. , in a 6 hour plot , the RYA would apply only 5 minutes of proportional time to each vector as against you applying it all to one tide.

That in itself makes your method more inaccurate in real life. You are assuming accuracy that isn't there and abrupt transitions from one tide to another after 60 minutes , accuracy which isn't there at all

(B) you admit that in plotting hourly D1 and D2 , that CD1 and CD2 would represent the CTS to get to those points ( because you said if they happen to lie on or near B that's the CTS)

Yet by inflating or deflating only the last triangle , you are in effect creating two CTS, the one to get to D1 and the one from D1 to B , or if you state that the last triangles CTS is the CTS for the whole journey, that that's clearly an error , you are in effect applying a partial tidal vector to all previous one hour vectors. In vector addition with a time quantum all the vectors must be a proportion of the same time , what you are doing is mathematically incorrect in that regard

(C) think about this. , tides are SOG based. , a boat with a water track is slowed by a tide from say 6 kn to 4kn SOG , therefore its spends more then one hour being affected by that tide. Ie lets say it was stopped by the tide. Your assumptions say the boat must actually get out of the tide in one hour ( because you apply all the corrections at the end ) but in reality if I spend 6/4 x 60 minutes in a 2kn tide I am affected by it for that time. , ie the tidal quantum is simply not abrupt., the tide does not disappear after 60 minutes. hence average back each tide for the appropriate additional time is more correct then applying all at the end.

The RYA method has withstood the test of 50 years of application, I suspect it will withstand the next 50.

Dave