Distinct Activities: Shackled by a Common Name?

[Seaworthy Lass]

The chartplotter diagrams for the two boats over two hours:

[Andrew Troup]

I've done a quick and dirty mathematical analysis of CaptForce's own scenario, assuming a distance across of twenty nautical miles.

I broke the crossing into multiple strips and applied simple trig. There was no preconception built into the analysis as to what the track across the ground would be for the Dockhead constant heading transit, but it turned out to be exactly as CaptForce has predicted.

I was not surprised because I had already convinced myself he was right on this point.

I thought of several different ways of explaining it to myself, because it was a counterintuitive proposition at first.

One way of working out what heading to adopt, for a Constant Heading transit, is to steam at the correct angle to exactly cancel the average tidal current, in other words, to make our track across the ground exactly at right angles to the average current, when we encounter it.

In this case, that's 2 knots, and we'll encounter that much current one quarter of the way across.

Until then, we'll arc upstream. But at the quarter transit mark, we'll be making way parallel to the rhumbline. It cannot be otherwise, because we've calculated our heading to make that exact thing come true.

For the next quarter of the crossing, the tide increases, and just as an additional two knots of tide was enough to bring our detour upstream to an end in the first sector, a further two knots is just enough to cancel the excursion and bring us back to the rhumb line. Because the current is a maximum at this point, when we cross the line we are now heading more steeply downstream than at any other time before or after.

(The track across the ground for the remainder of the passage is a mirror image, with an instant double flip about E-W and then N-S.)

Initially we are losing badly and get swept downstream, but we pluckily maintain the same heading as before, and sure enough by 3/4 of the way across, the crabbing due to our angled heading is once again sufficient to cancel the 2 knot current at this location and we're back parallel to the rhumb line, as far downstream as we were upstream at the maximum extent.

From now on, we're winning against the decreasing current. 7/8 across, we're parallel to our track 1/8 across, angling steeply up, almost making good our heading.
By the time we touch the far side, our track has angled up even further, exactly on our heading.

When you think about it carefully, it does make perfect sense.

The question of which tactic wins the race is a different question, but I would prefer to leave that until there's substantial agreement on the question of the track (whether CF and I are right or wrong) - because if I'm wrong, the rest of my analysis must also be wrong.

The analysis is not a trivial issue, if you want to be able to plot the ground track accurately, because the current in this scenario is increasing or decreasing every inch of the way.

My calculus was never great and it's virtually non existent now, so I resorted to brute force and used an iterative method with a spreadsheet. It was not difficult in a quiet office on land, but I wouldn't be quite so keen on tackling it when the 'office' was leaning over and leaping off waves.

[Jammer Six]

Wait, wait, WAIT!

Seaworthy Lass dropped her towel?

[Seaworthy Lass]

Originally Posted by Andrew Troup

I've done a quick and dirty mathematical analysis of CaptForce's own scenario, assuming a distance across of twenty nautical miles.

I broke the crossing into multiple strips and applied simple trig. There was no preconception built into the analysis as to what the track across the ground would be for the Dockhead constant heading transit, but it turned out to be exactly as CaptForce has predicted.

I was not surprised because I had already convinced myself he was right on this point.

I thought of several different ways of explaining it to myself, because it was a counterintuitive proposition at first.

One way of working out what heading to adopt, for a Constant Heading transit, is to steam at the correct angle to exactly cancel the average tidal current, in other words, to make our track across the ground exactly at right angles to the average current, when we encounter it.

In this case, that's 2 knots, and we'll encounter that much current one quarter of the way across.

Until then, we'll arc upstream. But at the quarter transit mark, we'll be making way parallel to the rhumbline. It cannot be otherwise, because we've calculated our heading to make that exact thing come true.

For the next quarter of the crossing, the tide increases, and just as an additional two knots of tide was enough to bring our detour upstream to an end in the first sector, a further two knots is just enough to cancel the excursion and bring us back to the rhumb line. Because the current is a maximum at this point, when we cross the line we are now heading more steeply downstream than at any other time before or after.

(The track across the ground for the remainder of the passage is a mirror image, with an instant double flip about E-W and then N-S.)

Initially we are losing badly and get swept downstream, but we pluckily maintain the same heading as before, and sure enough by 3/4 of the way across, the crabbing due to our angled heading is once again sufficient to cancel the 2 knot current at this location and we're back parallel to the rhumb line, as far downstream as we were upstream at the maximum extent.

From now on, we're winning against the decreasing current. 7/8 across, we're parallel to our track 1/8 across, angling steeply up, almost making good our heading.
By the time we touch the far side, our track has angled up even further, exactly on our heading.

When you think about it carefully, it does make perfect sense.

The question of which tactic wins the race is a different question, but I would prefer to leave that until there's substantial agreement on the question of the track (whether CF and I are right or wrong) - because if I'm wrong, the rest of my analysis must also be wrong.

The analysis is not a trivial issue, if you want to be able to plot the ground track accurately, because the current in this scenario is increasing or decreasing every inch of the way.

My calculus was never great and it's virtually non existent now, so I resorted to brute force and used an iterative method with a spreadsheet. It was not difficult in a quiet office on land, but I wouldn't be quite so keen on tackling it when the 'office' was leaning over and leaping off waves.

OH, you are right about the track being sygmoidal for current coming constantly from one direction but varying in strength.

It is soooooooooooo very counterintuitive.
Well done for working this out.

I have realised my speed vectors are displayed a bit wrongly so will post a new diagram in a sec (end result is still the same though ).

[Seaworthy Lass]

Originally Posted by Jammer Six

Wait, wait, WAIT!

Seaworthy Lass dropped her towel?

Gosh, it was so quick I thought no one noticed .
Were you peeking LOL?

[Seaworthy Lass]

Sorry folks, my last speed diagram was a little incorrect. Net result is the same, I have just shown the vectors correctly here:

[Sir Rondo Normal]

Originally Posted by goboatingnow

pS the drum kit is a ridiculous analogy.

Actually, from a drummer's perspective, the analogy is perfectly sound. Drummers can and do use drum machines along with any other tools to achieve the desired sound/effect they're looking for.

So too should a sailor when it comes to navigation.

[goboatingnow]

Originally Posted by Seaworthy Lass

Sorry folks, my last speed diagram was a little incorrect. Net result is the same, I have just shown the vectors correctly here:

to my understanding that is not what happens in Dockheads case, your diagram suggest a heading change. Dockhead does not change heading. Your

All these attempts to resolve the matter to a ground track completely miss the point, its irrelevant.

dave

[Seaworthy Lass]

Originally Posted by goboatingnow

to my understanding that is not what happens in Dockheads case, your diagram suggest a heading change. Dockhead does not change heading.

The speed vector diagram below shows absolutely no change in Dockhead's heading (the thick red lines 5 long are parallel). Only the ground track direction changes after an hour.

Originally Posted by goboatingnow

All these attempts to resolve the matter to a ground track completely miss the point, its irrelevant.
dave

It is critical to be able to calculate and plot your ground track - how else do you work out if you are going to hit any obstacles along the way

[Dockhead]

Originally Posted by Seaworthy Lass

It is critical to be able to calculate and plot your ground track - how else do you work out if you are going to hit any obstacles along the way

Peace, y'all!

Dave just means irrelevant for purposes of calculating an efficient passage. Naturally, we also need to have an idea of where we are going in relation to obstacles, in order to avoid them. No one would disagree with that. You two are not disagreeing with each other.

I must confess to having had a brain f*art about the shape of the ground path in a unidirectional current -- YES, it is also sinusoidal, and yes, you end up downcurrent from the rhumb line for a little bit at the end of your passage. Sorry, my mistake -- I take back my statement. A momentary lapse in visualization.

I'm getting on the plane to Helsinki and will have some numbers to post when I land!

[Seaworthy Lass]

Originally Posted by Dockhead

Peace, y'all!

Dave just means irrelevant for purposes of calculating an efficient passage. Naturally, we also need to have an idea of where we are going in relation to obstacles, in order to avoid them. No one would disagree with that. You two are not disagreeing with each other.

I must confess to having had a brain f*art about the shape of the ground path in a unidirectional current -- YES, it is also sinusoidal, and yes, you end up downcurrent from the rhumb line for a little bit at the end of your passage. Sorry, my mistake -- I take back my statement. A momentary lapse in visualization.

I'm getting on the plane to Helsinki and will have some numbers to post when I land!

Yes, I had a brain freeze too LOL, apologies to CaptForce for my error. But the shape of the curve does not alter the fact that following the constant heading is the quicker route!

Look forward to some figures on time to get across the channel. Yours on a straight heading are easy (once we know the width of the channel, we already know average current and you log speed), but I find CaptForce's speed is impossible to calculate without knowing exactly how the current varies across the channel he describes.

Have a good trip. Make sure the pilot examines his likely crosswind and calculates the constant heading needed for the quickest flight LOL .

PS You don't end up downstream for a little bit of the journey, but for half if the current is symmetrical

[goboatingnow]

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.

Captforce, despite multiple mathematical profs, you persist in seeing the wrong thing

Firstly it has nothing to do with GPS versus "older technology". GPS cannot see into the future, using a tidal stream atlas you CAN. Hence GPS sails only for the conditions it detects right now. In effect , you sail a series of very small triangles.

as you point out if the tide remains exactly the same through the voyage there is no difference

However in any situation where the tide ( in the future) varies, The GPS/autopilot method is not as efficient.

Case (a) Constant tide case (b) Varying tide, as you clearly see summing the triangles in advance produces a advantage, it matters not what the actual units are.

[Seaworthy Lass]

Originally Posted by goboatingnow

Captforce, despite multiple mathematical profs, you persist in seeing the wrong thing

Firstly it has nothing to do with GPS versus "older technology". GPS cannot see into the future, using a tidal stream atlas you CAN. Hence GPS sails only for the conditions it detects right now. In effect , you sail a series of very small triangles.

as you point out if the tide remains exactly the same through the voyage there is no difference

However in any situation where the tide ( in the future) varies, The GPS/autopilot method is not as efficient.

Case (a) Constant tide case (b) Varying tide, as you clearly see summing the triangles in advance produces a advantage, it matters not what the actual units are.

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????

[goboatingnow]

Originally Posted by Seaworthy Lass

It is critical to be able to calculate and plot your ground track - how else do you work out if you are going to hit any obstacles along the way

Actually only in simple cases it is possible to do so with any degree of accuracy. In complex cases with tidal vectors not at right angles etc its very difficult , even in simple cases you again only approximate , because the tide is never actually constant.

remember the ground track cannot be deduced by breaking the composite CTS triangle into say 1 hour smaller triangles , thats not whats actually happening.

for example compute the ground track on this more typical complex tidal stream example

[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. )

And the fundemental reason is based on the progression of squares ( or square roots if you will), is squareroot (x) + squareroot(x)+square root(x) is always greater then square root(3x)

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.

Dave