Distinct Activities: Shackled by a Common Name?

[Hudson Force]

Let's look at the following representaion of a crossing where we assume that two vessles are traveling at indentical speeds of five knots if slack water. The passage will be calculated for a period of four hours while crossing a current at 90 degrees to the passage with a maxium current at the center and zero current at the start and finish with an average current of 2 kts. One vessel takes the constant heading approach an sums the vestors that results in the sigmoid curve on the following diagram and the other vessel remains on the GPS track while changing heading throughout the passage in order to stay on this shortest distance path.



You will notice that the pathway for the constant heading path has been assigned some approximations of straight lines to accomodate the math. If we solve for time, being the distance divided by the speed, you will see that both vessels reach their destination withing the same elapsed time......


The constant heading vector summation increases the speed and the distance proportionally resulting in no change in time. Following the GPS track decreases the speed and distance proportionally resulting in the same time for the passage. In reality the two processes are the same process with one taking on the current vector in one step and the other taking on the current vector in many steps.

[Andrew Troup]

CaptForce:

What is the geographical distance from departure to destination?

Also, please clarify your current scenario.

1) Is this a current which varies with location, rather than with time?

2) Do you mean that the current increases in a linear way from zero at the start to 4 knots at the centre, then decreases linearly to zero at the finish, giving the average you refer to of 2 knots, flowing throughout the crossing from the same direction(say right to left)?

If not, can you please spell out what you do mean, in a way which enables quantifying the current at any location (or, if it's time-variable, at any time)?


I take it from the diagram that when you say "GPS track" you mean this boat follows a straight (rhumb) line over the bottom?

Once you've answered these, I may still have some questions, because I am struggling to understand your solution, under any scenario I can conceive of.

[Seaworthy Lass]

Hi CaptForce
Glad you are still with us.

In your example there is no difference between the two methods in time taken as there is no change in direction of the current. We both agree on that

The curve you have drawn for Dockhead's boat is not correct though. Halfway across both boats are in the strongest current and Dockhead would actually be the furthest away from the straight line you are travelling. His path from A to B would be a big loop out in one direction, and to complicate matters it would not be a semicircular curve if the current in your example behaves like currents usually do (you have not given us any info about this).

The vectors you are using are not correct if there is an average tide of zero though (sweeps one way then the other as you have drawn).

Dockhead''s log speed is not 5 knots diagonally then. His log speed is 5 knots directly towards the other shore as that is his compass heading. The diagonal speed (speed the chartplotter shows he is doing) is higher as the current is pushing him over. See my diagram below.

Try repeating your calculation and using the correct vectors for both boats and assuming a change in current as you have drawn, not the current you have described .

It will be really great when the penny drops for you. It makes discussions like this very worthwhile!

[Dockhead]

Originally Posted by CaptForce

Let's look at the following representaion of a crossing where we assume that two vessles are traveling at indentical speeds of five knots if slack water. The passage will be calculated for a period of four hours while crossing a current at 90 degrees to the passage with a maxium current at the center and zero current at the start and finish with an average current of 2 kts. One vessel takes the constant heading approach an sums the vestors that results in the sigmoid curve on the following diagram and the other vessel remains on the GPS track while changing heading throughout the passage in order to stay on this shortest distance path.



You will notice that the pathway for the constant heading path has been assigned some approximations of straight lines to accomodate the math. If we solve for time, being the distance divided by the speed, you will see that both vessels reach their destination withing the same elapsed time......


The constant heading vector summation increases the speed and the distance proportionally resulting in no change in time. Following the GPS track decreases the speed and distance proportionally resulting in the same time for the passage. In reality the two processes are the same process with one taking on the current vector in one step and the other taking on the current vector in many steps.

The track you drew for the constant heading boat is obviously derived by backing into it based on your theory that vector sums will always be the same. It is not the track which a constant heading boat would take, as a result of which you have not calculated the distance over ground sailed by the constant heading boat. If the current does not change direction, the constant heading boat track will not even be an S curve. That boat will never appear downcurrent from the rhumb line.

Besides that, you have a false assumption: "Avg Speed S-Curve = Boat Speed". This is false. Speed over ground for a boat following a constant heading path will vary constantly.

So all of your calculations about speed and distance of the constant heading boat are false.

In order to do this exercise correctly, you first need to know how to calculate a CTS for the constant heading approach and you need to make that calculation. Without doing that, you are nowhere. In order to calculate CTS, you need to know more than the assumptions in your problem. You need to know or at least be able to guess at the set of the current every hour (or in this example with only 4 hours sailed, better every 30 minutes). You add up the total set over the passage and then calculate the difference in course required to offset that set. That's your CTS. The constant heading boat will be heading more upcurrent at the start of the passage, then will be sailing with the current towards the destination later. The distance over ground for the constant heading boat in your example will be much less than what you calculated. Incidentally the average speed over ground of the constant heading boat will be less than boat speed, but it will be much higher than the average speed over ground of the GPS track boat, more than making up for the difference in distance over ground sailed. It is in fact axiomatic, and you will see that if you do the calculations correctly.

[goboatingnow]

Phew
CaptForce

IN situations where the tidal vector remains constant over the journey, then it matter not how many triangles you break it up into , its the same as you said

to abstract this a little
Ie, two 4,3,5 triangles, results in a 8,6,10 one. , ie 2 smaller hypotenuse= larger hypotenuse

However where the triangles vary, the larger composite one always has the smaller hypotenuse ( because the progression of square roots is not linear)

Hence say a 4,3,5 triangle and a 4,2,4.472 , the large triangle is 8,5,9.433 , hence the 5+4.472 = 9.472 which is greater then 9.433 , the hypotenuse of the larger composite triangle, in this case a 1.5% percent improvement

As you can see the advantage will increase given the number of triangles that differ.

a journey of 20 miles with , a 90 degree current of 2,6,4,1 over the four hours, would result in a 3.6% advantage to the CTS route.

In practice its actually worse then that as the GPS track does not sail in "hourly intervals", but creates in effect a series of small triangles, depending on the control algorithm. Since in fact each triangle is of varying ( the current is not actually constant over a hour just deemed so.

What I was interested in BTW, Dockhead is the progression of the "advantage", ie a 1.5% increase mightnt have many getting out their vectors!. but perhaps a 10% over a long journey would.

And Seaworthy Lass, where the tidal components are with or against you , ie parallel, results in no CTS gain,( by which I mean there isnt a better course then track) The track and the CTS align. Of course there is the issue of calculation of arrival time, which the vector will give you, whereas the GPS cant see into the future. Time of arrival calculation is useful in some cases, ie arrivals at ports and cills. But of course while we can go slower, we usually cant go faster!.

Dave

[conachair]

I'm just glad I started off learning in the Thames estuary. It all seems fairly natural instead of having to jump through a load of intellectual hoops every time you need to factor in tidal set to get a course to steer.

[Dockhead]

Originally Posted by goboatingnow

Phew
CaptForce

IN situations where the tidal vector remains constant over the journey, then it matter not how many triangles you break it up into , its the same as you said

to abstract this a little
Ie, two 4,3,5 triangles, results in a 8,6,10 one. , ie 2 smaller hypotenuse= larger hypotenuse

However where the triangles vary, the larger composite one always has the smaller hypotenuse ( because the progression of square roots is not linear)

Hence say a 4,3,5 triangle and a 4,2,4.472 , the large triangle is 8,5,9.433 , hence the 5+4.472 = 9.472 which is greater then 9.433 , the hypotenuse of the larger composite triangle, in this case a 1.5% percent improvement

As you can see the advantage will increase given the number of triangles that differ.

a journey of 20 miles with , a 90 degree current of 2,6,4,1 over the four hours, would result in a 3.6% advantage to the CTS route.

In practice its actually worse then that as the GPS track does not sail in "hourly intervals", but creates in effect a series of small triangles, depending on the control algorithm. Since in fact each triangle is of varying ( the current is not actually constant over a hour just deemed so.

What I was interested in BTW, Dockhead is the progression of the "advantage", ie a 1.5% increase mightnt have many getting out their vectors!. but perhaps a 10% over a long journey would.

And Seaworthy Lass, where the tidal components are with or against you , ie parallel, results in no CTS gain,( by which I mean there isnt a better course then track) The track and the CTS align. Of course there is the issue of calculation of arrival time, which the vector will give you, whereas the GPS cant see into the future. Time of arrival calculation is useful in some cases, ie arrivals at ports and cills. But of course while we can go slower, we usually cant go faster!.

Dave

I am rushing around now getting out of Cowes to make the ferry, but I'll run some numbers on the plane tomorrow.

In Capt Force's example, the current has varying strength, so the constant bearing approach will give a different track and will provide an advantage. How big it is depends on the variation, which we don't know since Capt Force only told us slack at the ends and an average of 2. I will hypothesize some half-hourly streams and run the numbers off that. The track over ground will be very different from the one which CF drew, it will be a single curve with XTE always upcurrent, not an "S" curve, and the distance over ground will be much less than what CF calculated.

When the two approaches coincide, then the track over ground will also be identical. This will take place only in slack water, and in a current which is constant in both speed and direction for the entire passage. All other conditions, the track over ground will vary between the two approaches. Distance over ground will always be greater for constant bearing approach; distance through the water will always be less. So average SOG will always be higher for constant bearing approach (except for the two cases above), and enough higher to, in every case, make up for the increased distance over ground. This is axiomatic and could be proven mathematically.

[goboatingnow]

Originally Posted by Dockhead

This is axiomatic and could be proven mathematically.

The additions of two unequal triangles as I have shown demonstrates that in all cases except in equal triangles ( ie constant current) the CTS method is an advantage.

Demonstrably the advantage can be small depending on the precise set and drift that ones encounters. hence there is a trade off .

So average SOG will always be higher for constant bearing approach (except for the two cases above), and enough higher to, in every case, make up for the increased distance over ground. This is axiomatic and could be proven mathematically.

I believe you mean constant heading.
Dave

[goboatingnow]

It's important to remember here that we are comparing vectors. Neither the GPS method or the CTS have anything to go with the ground. Because in effect all the GPS does is sail small CtS triangles.

So there's no advantage in attempting to compare actual ground tracks.

The fact is that the sum of the hypotenuse of a number of smaller vector triangles is always equal of greater them the hypotenuse of the larger combined vector triangle its has nothing to do with symmetrical tides ( that's merely the poster child )

It doesn't matter whether you use time or distance or speed since they are linear functions of each other.

The actual ground track in each method is utterly irrelevant. This is a mistake many people make when drawing vectors on a physical chart. The actual ground track especially with vectors at non right angles can be quite difficult to work put. The maximum xte are not at least

[Dockhead]

Originally Posted by goboatingnow

It's important to remember here that we are comparing vectors. Neither the GPS method or the CTS have anything to go with the ground. Because in effect all the GPS does is sail small CtS triangles.

So there's no advantage in attempting to compare actual ground tracks.

The fact is that the sum of the hypotenuse of a number of smaller vector triangles is always equal of greater them the hypotenuse of the larger combined vector triangle its has nothing to do with symmetrical tides ( that's merely the poster child )

It doesn't matter whether you use time or distance or speed since they are linear functions of each other.

The actual ground track in each method is utterly irrelevant. This is a mistake many people make when drawing vectors on a physical chart. The actual ground track especially with vectors at non right angles can be quite difficult to work put. The maximum xte are not at least

Yes! The ground tacks are irrelevant indeed, and thinking about with a fixation on the ground will lead to errors.

The constant in the problem is speed through the water. Ergo, that approach which produces the shortest distance through the water will get you there faster. The shortest distance between two points is a straight line. A straight line through the water can only be sailed on a constant heading. Ergo, the fastest way between any two points across moving water will always be on a constant heading. All of that is irrefutable.

A GPS track approach will require frequent changes of heading. Thus you are not sailing a straight line through the water and thus you will always be sailing a greater distance through the water. At a constant speed through water you will thus always get there later, except the two cases where the two approaches coincide (slack water, and constant speed and direction of current). In these two cases, the GPS track approach will not require any changes of heading; thus you are sailing a straight line through the water exactly as if you had figured it out using a constant heading approach.


What is really essential is an understanding of the concept of distance through the water. Anyone hung up on distance over ground and ground tracks will find it very hard to visualize the problem.


In practice, of course, as you say, the difference will be small if the difference in current is small, so in many cases not worth the hassle of calculating. The other problem is information -- the less information you have about the currents and about your own speed on passage, the less accurate your CTS calculation will be and the less valuable it will be.

[Seaworthy Lass]

OK, CaptForce one final attempt from me to convince you using the example you gave. I am throwing in the towel then LOL.

I have made up some figures for the current during the four hours as you gave none, apart from saying current in 4 knots at the two hour mark and averages 2 knots during the trip.
A constantly changing current is an infinite number of vectors, so I have made the current constant for eight half hour blocks (as per your example the direction of the current is constant and from 90 degrees to the course). It varies as follows:
1st half hour: 1 knot
2nd half hour: 1 knot
3rd half hour: 2 knots
4th half hour: 4 knots
5th half hour: 4 knots
6th half hour: 2 knots
7th half hour: 1 knot
8th half hour: 1 knot
It averages 2 knots over the four hours as you specified.
For the rest of the journey however long it is, there is zero current

Dockhead wanting to follow a constant heading calculates there is an average of 2 knots current. If he aims to end up on the "rhumb" line (the initial destination point) after the four hours of current, his distance reached on the line is the square root of (5x4 squared minus 2x4 squared) which equals the square root of 336 which is 18.33 nm.

Meanwhile you change course ever half hour for the four hours and your progress along the straight line to the destination is as follows:
1st half hour: square root of (5 squared minus 1 squared)/2 = 2.45 nm
2nd half hour: square root of (5 squared minus 1 squared)/2 = 2.45 nm
3rd half hour: square root of (5 squared minus 2 squared)/2 = 2.29 nm
4th half hour: square root of (5 squared minus 4 squared)/2 = 1.5 nm
5th half hour: square root of (5 squared minus 4 squared)/2 = 1.5 nm
6th half hour: square root of (5 squared minus 2 squared)/2 = 2.29 nm
7th half hour: square root of (5 squared minus 1 squared)/2 = 2.45 nm
8th half hour: square root of (5 squared minus 1 squared)/2 = 2.45 nm

total distance travelled along the straight chartplotter line over four hours = 17.38 nm

So CaptForce, you are 0.95 nm behind after 4 hours if you change course every half hour to follow the straight line on the chartplotter.

That's me done unless you would like me to draw the vector diagrams for the above. Happy to do so if it helps you

[Hudson Force]

Originally Posted by Andrew Troup

CaptForce:

What is the geographical distance from departure to destination?

Also, please clarify your current scenario.

1) Is this a current which varies with location, rather than with time?

2) Do you mean that the current increases in a linear way from zero at the start to 4 knots at the centre, then decreases linearly to zero at the finish, giving the average you refer to of 2 knots, flowing throughout the crossing from the same direction(say right to left)?

If not, can you please spell out what you do mean, in a way which enables quantifying the current at any location (or, if it's time-variable, at any time)?


I take it from the diagram that when you say "GPS track" you mean this boat follows a straight (rhumb) line over the bottom?

Once you've answered these, I may still have some questions, because I am struggling to understand your solution, under any scenario I can conceive of.

Hello all! I've been refitting my rubrail today and only now back to this site,- much response!

First, Andrew, yes to both your questions and with such a tide as described, with the current stronger at the center and proportionally decreasing at the margins, we should all agree that the vessel on the constant heading would not remain entirely up current thoughout the crossing, but meet the rumbline in the center and then approach the final destination at that same initial heading from a down current position resuting in this sigmoid curve.

I notice that Seaworthy Lass points out in their most recent post that the vessel following the GPS track would, by their scenario, adjust their course every half hour. This, of course, would not be the case and would result in an inefficient series of zig-zags. Following the GPS track would result in a continual and gradual change of heading to allow for the shortest distance.

Any repeated discription of "distance through the water" being different from the distance between two fixed points continues to blur the distinct definitions of distance and speed.

My thought is that most of those that feel the vessel following the curve that would remain up current are accustomed to thinking of the average current in drawing their vectors and considering this average to represent a constant current from start to finish. True, if the current were to remain consistant during the entire crosssing, then the constant heading vessel would remain upcurrent. During any crossing where the current is uniform at the same set and drift for a passage, then the course followed by both plans would be identical.

I appreciate the enthusiasm of those that continue to support the idea that traditional navigation techniques trump the data supplied by the GPS and I honor the thought that many hold when considering that employing the GPS allows those with less understanding of navigation to be at sea. I'm sure there was a captain with an RDF that was suspicious of LORAN.

Nothing presented here suggests to me that there is a difference in the time elapsed during a crossing when the relationship between speed and distance remain proportional.

I never heard the idiom before, but I like it. "The penny dropped for me" when I accepted that the GPS track IS a summation of vectors.

[Andrew Troup]

Capt Force

Your statement

"we should all agree that the vessel on the constant heading would not remain entirely up current thoughout the crossing, but meet the rumbline in the center and then approach the final destination at that same initial heading from a down current position resuting in this sigmoid curve."

is a bald assertion, and bald assertions are, in general, not particularly helpful. However I have to make a confession:

I was going to ask you to justify it, but in trying to imagine what that might be, it occurs to me that (I think) I could do so myself.

This is going to require more thought.

I'm warming up the humble pie warmer as we speak ... early days, but if you are correct, I will be offering my warmest thanks, because you've exposed a gaping deficiency in my understanding.

There's nothing better than filling such gaps...

To clarify what I'm expecting to concede: It's restricted, at present, to the specific situation you describe here, and the specific question of the shape of the track of a constant heading vessel over the ground, in the given scenario. (Note that's not a scenario, as far as I can recall, which anyone else has yet analysed on this thread)

I still disagree with many of your other contentions, particularly in the case where the current reverses direction:

In such cases (unless the set cancels out) don't see how a GPS can be equipped to position the boat optimally, as doing so relies on being able to predict the future vagaries of the current, in order to borrow the correct amount of "free" distance sideways to 'cash in' against future tideflow.

[Andrew Troup]

Capt Force

I will post my analysis of the scenario you've clarified, but it may be a while before I can make time, as on the face of it, calculus is required to solve it (because the current is continually changing rate, instant by instant.)

It's an intriguing scenario - have you given any thought to how it would play out with five knots of current midstream, rather than four? If you work out how long your rhumb line guy takes at five knots through the water, (I already think I know, without doing any calculations) I'll work out how long the constant heading guy takes.

PS: Reminder: You still haven't told us the distance in nautical miles from start point to finish point

[Seaworthy Lass]

Originally Posted by CaptForce

......
I never heard the idiom before, but I like it. "The penny dropped for me" when I accepted that the GPS track IS a summation of vectors.

OH, the penny has just dropped for me CaptForce! I know now why you are going wrong.
Vectors DON'T sum up as you think they do.

Take a very simple example:
- Track is north magnetic
- Boat speed in still water is 5 knots for both your boat and Dockhead's
- Perpendicular current from the east, 2 knots first hour all the hour, then 4 knots from the same direction for the entire next hour.
- You follows a straight line on the chartplotter
- Dockhead works out the angle needed for a total of 6 knots of current over two hours and follows that as his compass heading.

You allows for 2 knots of current in the first hour and travel 4.58 nm along the straight chartplotter line (north)
In the second hour with 4 knots of current you travel 3 nm.
Total progress = 7.58 nm

Dockhead travels allows for 3 knots of current in the first hour and his shift to the north is 4 nm
This is repeated in the second hour.
Total progress north = 8 nm and he end up on your line at this point.

I repeat again, vectors don't sum up as you have assumed! This is where you are going wrong.

This is the diagram for the speed vectors for each of you (the chartplotter diagrams are shown in the next post, I can't get them to attach together:

PS it is a woman's prerogative to pick up the towel again at any time . I will give a big cheer at this end when you realise why constant heading is always the quickest way of travelling between two points.