Distinct Activities: Shackled by a Common Name?

[Hudson Force]

Friends, I've taken a one minute glance at the added posts since I last viewed, but I've little time now. I'm trying to finish my rub rail refit before the rain hits tomorrow morning. 'much for me to review, - 'thanks for all the insights.

[goboatingnow]

Originally Posted by Seaworthy Lass

GoBoating, where is the boat speed vector of 5 knots in your second diagram?
I am not sure what you are demonstrating here .
Specifically what is the big triangle in your second diagram????

Dont get het up about whether the vector is speed, time or distance, since all these are merely derivatives of each other.


The key to the CTS method, is you sum all the vectors in advance.

The right hand vertical vector represents the sum of all the intermediate vectors ( which is how you calculate CTS).

so in the first case this demonstrates that there is no CTS advantage where the current remains the same. sailing an infinite amount of small triangles is equivalent to working out the vectors in advance ( ie the large hypotenuse)

In the second case , however where the tide vectors varying , there is always and advantage to precomputing a CTS in advance and sailing that, Here the right hand vector represent the vector sum of the individual vectors, The CTS is computed in advance as the hypotenuse and as you can see the sum is less then the individual hypotenuse. ( The actual tidal vectors are in fact opposite to the ones drawn, but this is often the technique used to minimise chart space. )

You can transfer the number into anything you like, speed time or distance, for example. If speed it just means you will have to travel faster in the small triangles, if distance, you will travel a shorter distance in the larger hypotenuse etc.

Nothing to do with symmetrical tides , that as I said is the poster child of the usefullness of the system.
Dave

[Seaworthy Lass]

GoBoating, in the example given, your vectors are unfortunately not correct .
What exactly is the hypotenuse representing in each little triangle - lengths labelled 5, 4.47, 5.66, 4.12? It cannot be faster than the log speed unless you are heading down current, in which case you are not allowing for the current, but exacerbating its effect


The constant heading of the boat needs to be represented along the hypotenuse (the boat is heading off at a tangent to the 'x axis' to compensate for the tide.

I have redrawn the diagrammatic representation of the vectors for the boat following the constant heading in your example.

The big triangle this is equivalent to is 16 along the hypotenuse (4 hours of 4 knots) and the current component is 10 (3 + 2 + 4 + 1).

Note that the black dotted line is the actual track on the chartplotter. There is nothing to work out if you just draw the vectors correctly. In the complex other example you gave where the current was coming from varying directions and strengths, absolutely everything (including and most importantly the compass course to steer) would be a nightmare to work out accurately.

[Dockhead]

Whew, you guys are beating your heads against the wall trying to calculate the constant heading passage in a ground-referenced manner. In order to do it this way, you have to calculate exactly where the boat will be every hour (or better, more often than that). In order to do that, you will have to calculate SOG and COG derived from the tidal vector, STW, and heading, and work out every position. Whew! Capt Force already went wrong by assuming that SOG = STW for the constant heading boat -- not so.

Am I the only person here who, rather than inventing some method based on rusty college trig, actually learned how to do it? It's very easy to do it with a water-based reference, which is the way they teach you to do it.

All you do is add up all the vectors to get a total net tidal set. Say it's 8 miles. Then you draw a right triangle with the other short side having a length equal to the distance to your waypoint over ground. Et voila, you have the length of the hypotenuse, which is the distance through water you will sail, which will tell you the amount of time it will take at a constant speed through water. And you have the angle of the apex of the triangle, which is the correction factor for your course.

So you now have exact course to steer and exact time for your passage, exact, that is, to the same extent that the input information is exact, but my point is that there are no mathematical approximations involved in this method.


You can get the same results with a ground referenced method, but it is 20 times the amount of work, and it is slightly less precise, because you have a limited number of data points. No one here has done it correctly so far.

I worked up Capt. Force's example. I will post the results next.

[goboatingnow]

Originally Posted by Seaworthy Lass

GoBoating, in the example given, your vectors are unfortunately not correct .
What exactly is the hypotenuse representing in each little triangle - lengths labelled 5, 4.47, 5.66, 4.12? It cannot be faster than the log speed unless you are heading down current, in which case you are not allowing for the current, but exacerbating its effect


The constant heading of the boat needs to be represented along the hypotenuse (the boat is heading off at a tangent to the 'x axis' to compensate for the tide.

I have redrawn the diagrammatic representation of the vectors for the boat following the constant heading in your example.

The big triangle this is equivalent to is 16 along the hypotenuse (4 hours of 4 knots) and the current component is 10 (3 + 2 + 4 + 1).

Note that the black dotted line is the actual track on the chartplotter. There is nothing to work out if you just draw the vectors correctly. In the complex other example you gave where the current was coming from varying directions and strengths, absolutely everything (including and most importantly the compass course to steer) would be a nightmare to work out accurately.

Sorry Lass, you are wrong here. what you have represented is incorrect, you are confusing vectors with actual travelling over teh ground

The third example I gave is actually typically of a proper CTS exercise given to any Yachtmaster candidate. The CTS course is as Ive plotted and the required angle is read using a normal Breton plotter.

remember CTS computations in varying tide, NEVER show the ground track. All the vectors are in effect water based.


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

[goboatingnow]

All you do is add up all the vectors to get a total net tidal set. Say it's 8 miles. Then you draw a right triangle with the other short side having a length equal to the distance to your waypoint over ground. Et voila, you have the length of the hypotenuse, which is the distance through water you will sail, which will tell you the amount of time it will take at a constant speed through water. And you have the angle of the apex of the triangle, which is the correction factor for your course.

So you now have exact course to steer and exact time for your passage, exact, that is, to the same extent that the input information is exact, but my point is that there are no mathematical approximations involved in this method.

my illustrated examples are just that, but what in effect you are doing Dockhead is vector addition. what confuses people is that think you actually sail along the vectors

[Seaworthy Lass]

Originally Posted by goboatingnow

Sorry Lass, you are wrong here. what you have represented is incorrect, you are confusing vectors with actual travelling over teh ground

The third example I gave is actually typically of a proper CTS exercise given to any Yachtmaster candidate. The CTS course is as Ive plotted and the required angle is read using a normal Breton plotter.

remember CTS computations in varying tide, NEVER show the ground track. All the vectors are in effect water based.


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

One question: what is the cross wind after 4 hours? The journey will take longer than four hours if there is a cross wind.

[Dockhead]

OK, so here’s the deal.


Capt Force’s scenario is:
1.

* Current stronger in the middle, slack at the beginning and end.
* Average force 2 knots
3. * 4 hours passage.
4. * Speed through the water of constant 5 knots

That’s not actually enough information to calculate any passage times. We need to know the distance, and we need to know the streams hour by hour. Calculations made without that data are meaningless -- in fact, the missing data points to corruption in the logic -- conclusions are being presumed ("the vectors must add up, because that's the only way the two approaches can give the same result").



I’m going to make up the essential missing data. I will calculate every half hour, and I’m going to use the following streams:


0 hour - 0:30 0 knots
0:30 – 1:00 2 knots
1:00 – 1:30 2 knots
1:30 – 2:00 4 knots
2:00 – 2:30 4 knots
2:30 - 3:00 2 knots
3:00 – 3:30 2 knots
3:30 – 4:00 0 knots


OK, average 2 knots, as per the assumptions. The current is perpendicular to the rhumb line. Let’s say BTW is 180 and current runs E to W.

First, let’s calculate a valid CTS. Without doing this correctly the whole exercise is useless. In this case, it’s a simple calculation – we simply add up the set of the current, which amounts to 8 miles to the W. So we need a CTS which will put us 8 miles further E.


Next, we will need to know the distance, which Capt Force did not give us. The distance over ground is not 20 miles (sorry Andrew), because we cannot make 5 knots made good towards the waypoint with a unidirectional current across our path. Average VMG towards the waypoint must be less than boat speed. I can calculate how far I can get in 4 hours by solving a right triangle with hypotenuse of 20 miles, and C side of 8 miles – simples. So the rhumb line has a length of 18.33 miles, and we have a distance to waypoint which can be sailed in the presumed 4 hours at the presumed 5 knots, provided we sail the correct path.



We also have my course correction 23.58 degrees, and angle of the apex of the triangle. Subtract that number from 180, the bearing to waypoint, and we have our CTS.



So now we have a complete picture of the constant heading passage:


* CTS constant heading of 156.42
* Distance made good towards waypoint 18.33 miles
* Distance covered through water 20 miles
* Time on passage 4 hours.



Now let’s calculate the GPS track passage. Since this vessel is moving in a straight line over the ground, this calculation is much easier if we do it in a ground-referenced manner.


0 hour – 0:30 HDG 180 SOG 5.0 Distance covered over ground 2.5
0:30 – 1:00 HDG 156.42 SOG 4.58 Distance covered over ground 2.29
1:00 - 1:30 HDG 156.42 SOG 4.58 Distance covered over ground 2.29
1:30 – 2:00 HDG 126.87 SOG 3.0 Distance covered over ground 1.5
2:00 – 2:30 HDG 126.87 SOG 3.0 Distance covered over ground 1.5
0:30 – 1:00 HDG 156.42 SOG 4.58 Distance covered over ground 2.29
1:00 - 1:30 HDG 156.42 SOG 4.58 Distance covered over ground 2.29
0 hour – 0:30 HDG 180 SOG 5.0 Distance covered over ground 2.5

So in four hours the GPS track boat has sailed the same 20 miles through the water as the constant heading boat (obviously -- 4 hours * 5 knots), but has only covered 17.16 miles over ground along the rhumb line, which is its ground track. So after four hours, it has not arrived yet – it has another 1.17 miles to go. Because of the longer passage time, it will be exposed to that much more set of the current, so we can approximate with some accuracy that it will make 4.58 knots SOG along the rhumb line during this extra time (that is your SOG staying on the rhumb line with a perpendicular current of 2 knots, which is the average). So the GPS track boat will need 0.26 hours more to get there. So now we have a complete picture of the GPS track boat's passage.



Results:


* Constant heading boat gets there in 4 hours
* GPS track boat gets there in 4.26 hours
* Both boats cover the same distance made good – 18.33 miles
* The constant heading boat sails 20 miles through water
* The GPS track boat sails 21.3 miles through water.
* The constant heading boat sails 20 miles through water.

* The constant heading boat has an advantage of 6.5%

QED.

The difference reflects the crooked path through the water produced by the GPS track approach. It matters not whether the changing heading is a smooth progression or a series of instant changes. The varying headings themselves represent an indirect path through water. I know Capt Force doesn’t like the idea of distance through water, but not liking it doesn’t mean that it doesn't exist, just like repeating over and over again that the world is flat doesn’t make that so. We sail in water – our speed is speed through water, the constant in all of our equations, so the fastest way to get anywhere is that way which puts the smallest number of miles of water between you and your destination, which optimizes your distance through water. Exactly water – your ground path, as Dave keeps reminding us, is totally irrelevant. You cannot get away from that fact. A straight path is shorter than a crooked one, a curved one, or whatever shape, any shape at all, which varies from a straight line. Any change in heading, any single isolated change, any set of changes of any number, any constantly changing heading, whether smoothly or in jumps, any change whatsoever, means that you are deviating from the most efficient passage, that is, the shortest and fastest way to get where you are going.


The GPS and modern chart plotter are wonderful devices, but they simply do not calculate the most efficient path over a body of moving water. The navigator has to do that (the original theme of this thred). Electronic devices cannot make such calculations without current data over the whole passage.


The advantage of the constant heading approach increases with increasing current differential, and especially with reversing currents. On an English Channel crossing, the difference is huge, truly dramatic, which is why these principles are second nature to us English Channel sailors As Dave said.


For anyone who STILL doesn't get it, then just picture this:


The same scenario, the same 18.33 miles, the same 4 hours and 5 knots, BUT -- the current is slack the whole way except for the last hour, when it runs at 8 knots (not absurd -- we have such currents in the Channel Islands).


So the constant bearing boat will sail exactly the same CTS and will arrive in exactly the same about of time -- 4 hours.


How about the GPS track boat?


Well, it will NEVER arrive. Never! Because at the beginning of the last hour, the GPS track boat will be on the rhumb line 5 miles away from the destination, which bears 180. But now the GPS track boat is making 5 knots through the water in an 8 knots current. Its maximum VMG towards the waypoint will be on course 090, stemming the current, but its VMG towards the waypoint will be -- negative.


NOW can anyone still hang on to the belief that the two approaches bring the same results? And there is no, zero, difference between this scenario and Capt Force's scenario, in terms of the total vectors, as witnessed by the identical strategy of the constant heading boat, bringing identical results.


Let's say we make it a little milder -- two hours of slack water followed by two hours of four knots.


The constant heading boat will still use the same CTS and will still arrive in four hours.


The GPS track boat will make 5 knots SOG for the first two hours, but then will only make 3 knots SOG at a heading of 132.8. With 9.17 miles to go, he's got more than 3 hours to go before arrival. His total journey time will be more than five hours, and he will arrive more than an hour later than the constant heading boat.

[goboatingnow]

One of the issues here is that people quote numbers for a graphical issue. If everyone interested solved the crossing the lake issue, I presented, graph it and post it, we could easily see whats wrong ( as one of the solutions will be the right one) . Mine is a simpler example then dockheads, but it needs to be done graphically

dave

[goboatingnow]

can I say OPPS to some people, I just realised I transposed my hypotenuses, ( The classic CTS error). sheesh , I repost my graphs

[Seaworthy Lass]

Originally Posted by goboatingnow

my illustrated examples are just that, but what in effect you are doing Dockhead is vector addition. what confuses people is that think you actually sail along the vectors

You do actually sail along the vector which sums up: vectors of "boat speed in the water" vector plus "current" vector. This vector of the sum will be constantly changing though if either your boat speed of the current speed is constantly changing (thus creating a curved path). Any period of time where current and speed are steady, you sail a stright line along the vector summing the two.

So, yes, the dotted line on my plot is the charplotter track .

PS Just read your other post. It is an easy thing to do, but you can see now why your graphs didn't make any sense to me.

I am just doing the other calculations for your hovercraft, but my better half is clamouring for dinner.
Calculations win LOL. Will post in a sec.

[goboatingnow]

this is the correct computation of CTS

[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.

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

As no info is given for the cross wind after 4 hours I will assume it is zero.

a) First method where you are making corrections as you hit them and travelling a straight line:
Progress along track:
1st hour: square root of (4 squared minus 3 squared) = 2.65 miles
2nd hour: square root of (4 squared minus 2 squared) = 3.46 miles
3rd hour: no progress (hovers on the spot)
4th hour: square root of (4 squared minus 1 squared) = 3.87 miles

So over 4 hours he travels 9.98 miles. There is 6.02 miles to go, and he travels there at 4 miles/hr so it takes 1.5 hrs.

Total time for the journey for a) is 5.5 hours (= 5 hrs 30 minutes)

b) Second hovercraft calculates the distance he has to travel through the air is:
The square root of (16 squared plus 10 squared) = 18.87 nm
At 4 knots this takes 4.72 hours

Total time for the journey of b) is 4.72 hours (= 4 hrs 43 min)

[Seaworthy Lass]

Originally Posted by goboatingnow

this is the correct computation of CTS

Sorry Dave, you still have the boat speed along one of the short axes, not the hypotenuse .

Try again?
This is your second diagram I am referring to:

[goboatingnow]

forget the little triangles , the large one is what I reworked, That is a 4 hour CTS plot
Dave