Distinct Activities: Shackled by a Common Name?

[Jammer Six]

Dockhead, if all this is so, then why did they teach me all about set, drift and laying out a course to compensate for all manner of water movement?

[Andrew Troup]

Jammer Six

Presumably they stopped when it was just starting to get interesting.

Many teachers do that...

(There's probably nothing wrong with what you were taught, but presumably they didn't cover the situation of sailing across an appreciable current which reverses direction during the transit - possibly something that doesn't happen much, except around and between major islands with a decent north/south extent)

[Hudson Force]

Dockhead, I was intrigued by your 237 post indicating that the route you would take would, as you admit, not be the shortest route (speaking of coordinate to coordinate), but that it would be the faster route due to spending less time working against the current. This opens more agreement in our comparisons. You appear to accept that your route would result in the longer sigmoid curve in georaphical measurement and I can accept that throughout your longer route you would likely make better speed. Now I'm interested in applying the math to Noelex 77's scenario, but I have to admit that, at my coodinates I've reached my bedtime! You have difinitely given me cause to recognize merit in your one compass heading plan, but I'm not ready to give you the win in time at the finish line! Thanks for your thoughts,- I'll be back!

[goboatingnow]

CaptForce. CTS computation properly done ALWAYS results in the best sped to the destination. It's taught extensively in all RYA courses.

Anyway
What is important however is to recognise the limitations. Where you have well documented symmetrical tides. ( like in the channel and in the Irish Sea. ) totalling all the tidal vectors over the complete passage resulting in one course to steer makes sense, because if your off at least you have the benefit of a set in the opposite direction. This was the root of my " theoretical" comment to Dockhead.

The alternative approach is like that laid out by Noelex. Here you compute multiple course to steers based on defined increments often 1 hour traditionally. This is not as efficient as the first method, but has the advantage of ( a) reduced XTE. And ( b) inherently recalculated as the passage progresses. In fact in some autopilots merely widening the allowed XTE to the desired amount of drift and dropping waypoints ahead will approximate the same method.

Furthermore the single combined CTS has the disadvantage that the vector addition doesn't show the ground track, so checking the route for hazards is a little more difficult, furthermore while certain areas have very good tidal stream data, many places do not, hence an intuitive approach has to be taken.

In classic applications like fairly symmetrical tides, the classroom theory works well and is often applied intuitively by sailors familiar with the passage.

As to terminology, I was always taught that currents apply to one directional flows, ie current in a river. Tides have tidal streams which have set ( direction) and drift ( speed ) . So no such thing as tidal currents.

I did my YM ( 20 years ago now) on the western approaches to GIb, here we had to compute tidal stream vectors AND current vectors for the trip, lots of arrows on be chart.

I can't follow the ant on be belt at all , its seems to be complicating a simple idea.

Dave

[Jammer Six]

Can't both cases be modeled with sufficient theoretical accuracy to determine which is a) faster and b) more efficient? (I suspect they're not the same answer, and so economic and other goals will pay a part in the decision.)

[goboatingnow]

Originally Posted by Jammer Six

Can't both cases be modeled with sufficient theoretical accuracy to determine which is a) faster and b) more efficient? (I suspect they're not the same answer, and so economic and other goals will pay a part in the decision.)

Which two cases

[Jammer Six]

You know, if the treadmill doesn't reverse, the ant better be headin' upstream, or changing the address on his driver's license, 'cause he ain't gettin' home headed straight across.

[Seaworthy Lass]

Originally Posted by Jammer Six

You know, if the treadmill doesn't reverse, the ant better be headin' upstream, or changing the address on his driver's license, 'cause he ain't gettin' home headed straight across.

Jammer, he is getting home straight across the water, not straight across the ground if he takes a constant heading which is the quickest route .

Good discussion everyone. Heading off the boat for the day but will look at the posts as soon as I come back .

[Dockhead]

Originally Posted by Jammer Six

Dockhead, if all this is so, then why did they teach me all about set, drift and laying out a course to compensate for all manner of water movement?

They taught you right - with one set of vectors for set and drift in one direction, that is, with a current with constant speed and direction, then you have one simple calculation to get the right course to steer -- CTS.

It gets interesting when the current varies in speed and/or direction, which is what we're talking about. You still make that same calculation which was taught to you, only you use the SUM of the vectors over the time of your passage, not the single vector of a single current.

People, I didn't make this up myself, as much as I would like to take credit for it , it is the correct and mathematically only answer to the problem.

[Dockhead]

Originally Posted by Jammer Six

You know, if the treadmill doesn't reverse, the ant better be headin' upstream, or changing the address on his driver's license, 'cause he ain't gettin' home headed straight across.

It's true. If the current doesn't behave as predicted, you will be off course.

But if you stay on the rhumb line, you are off course to begin with and for sure.

[Dockhead]

Originally Posted by Andrew Troup

The problem is that unlike the ant, we sailors do not have an aiming spot. We are relying on a compass bearing, and while that's a perfect proxy for an aiming spot if the belt is always oriented (say) North/South, it's not if the belt slews to a new axis. If our ant continued crawling (say) NE, he would no longer be crawling towards the aiming spot, and consequently his trail would no longer be a straight line.

So I respectfully disagree with Seaworthy Lass in post #227, for the moment at least.

She's right.

Unlike ants, we have compasses

My example had vectors cancelling each other out, so we were steering 180, which is the same CTS we would have in still water. That was intentionally for simplicity, to make it easier to understand.

If the vectors don't cancel each other out, and the sum of the vector puts us 5 miles East of Cherbourg, then we must calculate a CTS which gets us 5 miles West of there. At 60 miles it's easy - one degree equals one mile, so we add 5 degrees, so our CTS is 185.

For the ant -- the shortest way across is still a straight line in relation to the treadmill -- it's a perfect metaphor. But not necesssarily perpendicularly across. If the sum of the movements of the treadmill mean that over the ant's walk the treadmill will have moved 1 foot to the right, then the ant has to steer to a point on the edge of the treadmill one foot to the left of the point which is directly across from him. And to get there fastest (which is the same as most efficiently), he needs to walk towards that point in a straight line.

[carstenb]

Let's try another analogy in the real world. BAck when I used to do these things, I worked a bit with a customer who manufactured plate glass, including window panes etc.

For those who don't know, glass is made by extrusion, meaning a machine produces an endlessly long sheet of glass continuously. This sheet is say 3 meters wide and moves continuously on a roller belt.

Now it is no problem to imagine cutting the sheet into ribbons lengthwise. You set a couple of glass cutters and as the sheet rolls under them they cut and you now have 3 ribbons 1 meter wide and endlessly long (it is still being extruded at the other end.). So how do we cut to get 1 meter X 1 meter panes?

We have to cut while the glass is still moving, since we can't stop it (this is = to current). The solution is to set a glass cutter on a roller going ACROSS the sheet of glass (= sailboat crossing channel).

Now if the cutter is set to cut at right angles then in the time it takes for the glass cutter to move from one side to the other, the angle will cease to be 90 degrees. (= current offset) This is because the sheet of glass underneath it is moving.

To get square wind panes, the cutter is set at an offset angle towards the direction the glass is coming from (= offset against the current) If the math has been done correctly, the glass cutter now moves across the ribbons of glass at an angle which in conjunction with the movement of the glass produces a right angle. (which it does and that is why you can buy square window panes at home depot)

Now just to make things a bit more interesting, the extrusion process can have varying speeds (= more or less current), to compensate, the roller arrangement's angle across the glass can be adjusted by the computer, thereby resulting in always having square panes.

The point of all the above is to show that even though the glass cutter is SET at an angle which is offset against the current, at any given point on its travel across the ribbons (water) it actually is traveling at a 90 degrees angle to the glass (water.)

Phew. Anyone understand the analogy?

[noelex 77]

Lets work out the maths hour by hour.
A) steering a constant due North compass course (swinging east and then back on course)
1st hour our boat is 5Nm north and 1Nm east
2nd hour our boat is 10Nm north and 3Nm east
3rd hour our boat is 15Nm north and 6Nm east
4th hour our boat is 20Nm north and 9Nm east
5th hour our boat is 25Nm north and 11Nm east
6th hour our boat is 30Nm north and 12Nm east

Then the current changes

7th hour our boat is 35Nm north and 11Nm east
8th hour our boat is 40 Nm north and 9Nm east
9th hour our boat is 45Nm north and 6Nm east
10th hour our boat is 50Nm north and 3Nm east
11th hour our boat is 55Nm north and 1Nm east
12th hour our boat has arrived at its destination

Journey time 12 hours.

B) steering a constant COG ( with zero cross track error)
1st hour our boat is 4.9Nm on track due north
2nd hour our boat is 9.5Nm on track due north
3rd hour our boat is 13.5Nm on track due north
4th hour our boat is 17.5Nm on track due north
5th hour our boat is 22.1Nm on track due north
6th hour our boat is 27Nm on track due north

Then the current changes

7th hour our boat is 31.9Nm on track due north
8th hour our boat is 36.5Nm on track due north
9th hour our boat is 40.5Nm on track due north
10th hour our boat is 44.5Nm on track due north
11th hour our boat is 49.1Nm on track due north
12 hour our boat is 54 Nm on track due north

There is still 6 Nm to go when the boat that has followed Dockheads plan has arrived and is enjoying a beer
If we assume no current from now on. (any current at right angles would be slower)

13th hour our boat is 59Nm on track due north

Journey time 13hours and 12mins

You have lost 1hour and 12 mins of beer (or even worse French wine) drinking time

( the actual Vmg can be calculated using Pythagoras. Vmg = the square root of 5squared -the current squared.
For example if the current is 2 knots the Vmg will be the square root of 25-4=4.6)
Normally the current would not be exactly at right angles and a simple vector diagram is easier, quicker and accurate enough)

Note the above example is for the simplest case, but same principal applies if the currents do not cancel. Just add add up the currents and make allowance depending on the journey time ( refer to Dockheads post on how to do this)

In practice a lot of other factors may determine the best course, especially when sailing. Factors like expected wind shifts can make a big difference.

A good navigator will put all this information together to determine the best course to steer at any one time, but you must correctly understand the effects of currents and how to optimise you course to take advantage of them, or to minimise there adverse effects.

This illustrates the point that knowing your position is only a small part of navigation, or pilotage if you prefer.

[goboatingnow]

I think everyone excepts in a basically symmetrical tide, that calculating overall CTS is the best method. I can't see from a purely speed perspective why anyone would argue otherwise.

However on a unidirectional tide is there any difference. Once an autopilot essentially sails a series of miniature course triangles , its something I've never done the maths on.


Dave

[Dockhead]

Originally Posted by goboatingnow

I think everyone excepts in a basically symmetrical tide, that calculating overall CTS is the best method. I can't see from a purely speed perspective why anyone would argue otherwise.

However on a unidirectional tide is there any difference. Once an autopilot essentially sails a series of miniature course triangles , its something I've never done the maths on.


Dave

Constant compass heading = straight water track. So that's always the shortest and fastest way to get anywhere, assuming of course you know the correct heading. That's in the RYA training materials, even.

If the tide is unidirectional with a constant speed, your CTS = COG to your waypoint -- the two approaches converge. So your ideal constant heading happens to take you over the rhumb line.

If the tide is unidirectional with varying speeds, like the Gulf Stream for example, then you will go wrong with steering to COG -- that is, staying on the rhumb line. Here you get back to the classical CTS calculation to get your ideal way across, which is a straight line through the water, which is a constant compass heading.

And the approach is exactly the same -- you just add up the set and drift for every hour and calculate a CTS to get you to the right spot at the edge of the treadmill, to use the ant analogy, to get to the right spot by the shortest possible path.

If you change headings every hour, then you are sailing an erratic course through the water and losing miles. However straight the line may appear over ground. The key thing is really to just forget about the ground, which is not relevant when you are crossing a moving body of water, even if it is moving in only one direction.