Distinct Activities: Shackled by a Common Name?

[Seaworthy Lass]

Hi Conachair
Welcome The more the merrier .

I think your formula is correct. I haven't checked the figures (no spreadsheet on an iPad) but they look reasonable.

The only thing is that you are introducing a new situation - speeding up whenever in current, constantly maintaining a VMG of five knots. Your formula is limited by the hull speed of your boat. It fails when the current is equal to or greater than it.

It is an unusual situation too, normally we would not alter the throttle constantly (and of course we would not keep deliberately increasing and decreasing speed while sailing, we would just go as fast as we could ).

Someone here on CF must remember their calculus and come up with the calculations for a constantly changing current (linear or sinusoidal). I have not done calculations like this for well over 30 years now and my brain is very rusty).

Conchair and Astrid, I rolled my eyes LOL, but I smiled. Keep the puns coming .
We need to lighten up this thread a little .

[Dockhead]

Originally Posted by Seaworthy Lass

Someone here on CF must remember their calculus and come up with the calculations for a constantly changing current (linear or sinusoidal). I have not done calculations like this for well over 30 years now and my brain is very rusty).

A constantly changing current can be linear, too. Doesn't have to be a sine wave.

Now I'm back in my office with a full sized computer and screen, so I can work easily in Excel. I'm going to run Capt Force's scenario with continuously changing current.

I predict the results will be pretty close to what I got with half-hourly changing currents.

Now I have another prediction, which will be another proof of the constant heading principle --

I will run the scenario with granularity of one minute -- so minute-by-minute analysis of the path of the GPS track boat.

Then, I will run the scenario with granularity of five minutes. Then with one hour.

What will we get? The GPS track boat will be slightly slower, the finer the granularity we use. WHY? That is because every period of analysis is a micro passage. The GPS track boat will run better, the less often it changes heading. So if it changes heading only once per hour, it will run a whole hour on an ideal constant heading course. If it changes heading every minute, it will run slower. If it changes heading continuously (infinitely fine granularity), it will run slowest of all. This is the exact opposite of what Capt Force thinks -- he thinks we slow down the boat by making less frequent changes of heading.

That is my hypothesis. If it is confirmed by the numbers, it is a powerful additional proof of the constant heading principle. We shall see.

[Dockhead]

By the way, one knotty challenge to this analysis is the fact that the GPS track boat will run slower and thus get out of synch with the 4 hours of current, which Andrew correctly says if a function of distance rather than time. I am trying to think how to deal with this. One way to do it is recursively revising the distribution of the current based on derived passage time. I'll see if I can think of a more elegant way to do it.

[conachair]

Originally Posted by Seaworthy Lass

The only thing is that you are introducing a new situation - speeding up whenever in current, constantly maintaining a VMG of five knots. Your formula is limited by the hull speed of your boat. It fails when the current is equal to or greater than it.

It is an unusual situation too, normally we would not alter the throttle constantly (and of course we would not keep deliberately increasing and decreasing speed while sailing, we would just go as fast as we could ).

Agree, not the real world, the only way my creaky brain could figure out to get the boat covering 60Nm made good over the ground in 12h, same as the constant bearing boat, was to stick that constant in there. Tides don't cycle over 12h either.
I've never come across linear tides this neck of the woods, would make the calcs a little easier. Some places round here there are 2 high waters each cycle, so calcing anything is next to impossible, drawing is the only option. Or like many people, do a quick calc in your head and enjoy the trip trying not to hit anything on the way.

All good fun though, nice to stretch the gray matter now and again.
Doubt if I'll get too far with this, forgotten it all from school and the first session made my head hurt a bit, but it's fun to try.

[Dockhead]

OK, here are the first numbers. I analyzed Capt Force's scenario based on these parameters:

5 knots constant STW

2 knots average current, perpendicular to BTW, varying linearly from 0 to 5 knots to 0

18.33 miles over ground to waypoint (longest possible distance which can be reached given the previous parameters).

Here is what the GPS track boat would do if we use granularity one hour -- that is, we analyze one hour at a time based on the average current and average heading during that hour:

Constant speed through water:5,00Distance over ground to waypoint:18.33Average current:2,00period:av current during that periodSOG (VMG to waypoint)distance made goodhourly11,004,904,9023,004,004,0033,004,004,0041,004,904,9017,80

OK, so the GPS track boat will travel 17.8 miles during four hours. I have not yet solved the problem of current varying by distance rather than by time. Varying by distance I will have to stretch out the array of current speeds (which will further slow down the GPS track boat). Right now we are seeing the scenario as if the current varies by time regardless of how far the boat has gotten.

Remember that the constant heading boat will travel 18.33 miles made good towards the waypoint, and will arrive, in any combination whatsoever of currents which add up to an average of 2 knots.


Now here is what the GPS track boat looks like if we analyze every 5 minutes. That gives us 48 periods of 5 minutes each. I have analyzed the SOG and therefore distance made good towards the waypoint for each period, using a spreadsheet:

period:av current during that periodSOG (VMG to waypoint)distance made good10,005,000,4220,175,000,4230,354,990,4240,524,970,4150,704,950,4160,874,920,4171,044,890,4181,224,850,4091,394,800,40101,574,750,40111,744,690,39121,914,620,38132,094,540,38142,264,460,37152,434,370,36162,614,270,36172,784,150,35182,964,030,34193,133,900,32203,303,750,31213,483,590,30223,653,410,28233,833,220,27244,003,000,25254,003,000,25263,833,220,27273,653,410,28283,483,590,30293,303,750,31303,133,900,32312,964,030,34322,784,150,35332,614,270,36342,434,370,36352,264,460,37362,094,540,38371,914,620,38381,744,690,39391,574,750,40401,394,800,40411,224,850,40421,044,890,41430,874,920,41440,704,950,41450,524,970,41460,354,990,42470,175,000,42480,005,000,4217,52Checksum average current:2,00

Et voila!! Eureka! Now the GPS track boat travels only 17.52 miles towards the waypoint, compared to 18.33 for the constant heading boat.

This is an additional proof of the constant heading principle -- the smaller the triangles and more frequent the changes of heading, the slower the GPS track boat will go over ground. This trend will continue until the heading changes are continuous, at which point there will be maximum disadvantage of the GPS track compared to constant heading.

The reason is that every "grain" of our analysis, in this case one hour, and then five minutes, is a micro passage which is sailed on an ideal constant heading course. The larger these grains are, the more the GPS track boat will deviate from the rhumb line, and the faster the GPS track boat goes over ground.

I have to run off to a meeting now, but I will do another set of numbers analyzing every minute (240 periods). We will see that the distance achieved by the GPS track boat will fall further.

Next task will be to correct the current-over-time verus current-over-distance problem.

If anyone wants to check my formulae, I can email the spreadsheet.

[goboatingnow]

Originally Posted by Seaworthy Lass

Constant heading to steer means just that, a constant heading from departure to arrival, not until the current/crosswind runs out. It is slower to have more than one compass heading over a journey, I thought we had agreed on that.
My calculations were for a constant heading over the entire course you set. The time taken for a constant compass heading all the way was 4 hours and 43 minutes.

You can't have a constant heading if you have a period of slack at the end. There simply no perturbation vector. T he. CTS method only works with a vector causing the vessel to leave its ground track. There is no point otherwise.

If you are subject to a 4 hour tide , you must plan keep all the vectors to 4 hours.

What I did was straight from the RYA course notes.

[goboatingnow]

Just to show a worked up example. ( From RYA YM ) Here is a two tide plot , ie a two hour vector plot

Seaworthy


The background is a yacht having arrived at the waypoint ( near the rose) , wants to plot approaching Cherbourg. It will experience two tides ( from approximating the time ) in its journey
The tides are 1.6 kn 089T first hour and 1.6kn 098T for the 2nd hour boat speed is 6.5 knots. Through the water

See attached solution, note the water track vector never is connected to the destination point. that would not give the course to steer. The fact that the track may cut the ground track behind or ahead of the destination merely indicates that you arrive after or before you two hours. If the vector was incorrectly joined to the destination, you would actually arrive downside of the destination.

The further assumption made is remaining distance closed is made at the average SOG. ( not the water track speed)

This technique can easily be extended to,multiple hourly vectors
Dave

[Dockhead]

Originally Posted by goboatingnow

You can't have a constant heading if you have a period of slack at the end. There simply no perturbation vector. T he. CTS method only works with a vector causing the vessel to leave its ground track. There is no point otherwise.

If you are subject to a 4 hour tide , you must plan keep all the vectors to 4 hours.

What I did was straight from the RYA course notes.

I'm not arguing with you, just curious -- why not?

If you have one hour of slack at the end of the passage, your constant heading will simply put you the same bearing from your waypoint as your constant heading, at the beginning of that hour. What's the problem with that? And that's got to be the most efficient path considering the whole passage. No? Am I missing something?

It seems to me that a constant bearing must always be the fastest way over any bit of water, moving, or still, or any combination, axiomatically as the constant value is STW, and so optimum time must result from the shortest distance through water, which can only be a constant heading. If the entire body of water is still, then speed, distance, and track through water is identical to speed, distance, and track over ground. If the body of water moves constantly at the same speed and direction, then speed, distance, and track through water will be different from speed, distance, and track through ground, but the following the straight path through water will take you right over the rhumb line.

Is this not all true? If I am not getting something, I will be grateful for enlightenment.

[goboatingnow]

Originally Posted by Dockhead

I'm not arguing with you, just curious -- why not?

If you have one hour of slack at the end of the passage, your constant heading will simply put you the same bearing from your waypoint as your constant heading, at the beginning of that hour. What's the problem with that? And that's got to be the most efficient path considering the whole passage. No? Am I missing something?

It seems to me that a constant bearing must always be the fastest way over any bit of water, moving, or still, or any combination, axiomatically as the constant value is STW, and so optimum time must result from the shortest distance through water, which can only be a constant heading. If the entire body of water is still, then speed, distance, and track through water is identical to speed, distance, and track over ground. If the body of water moves constantly at the same speed and direction, then speed, distance, and track through water will be different from speed, distance, and track through ground, but the following the straight path through water will take you right over the rhumb line.

Is this not all true? If I am not getting something, I will be grateful for enlightenment.

Well thinking on my feet. I would subscribe that the quickest way from A to B , where you are subject to tidal currents is a single CTS. ( we're all agreed on that ). Equally the quickest time from B to C , where there is no tide ( or other forces) is a a straight line. Hence I would have thought that's the quickest way. ( a+ b , c+d) At the end of the tidal vector time ( and yes this is stylised as that's how the data is presented)

What I was arguing with seaworthy that the actual computed CTS water is not drawn to the destination to do so would put you down tide.

Dave

[Seaworthy Lass]

Originally Posted by goboatingnow

Just to show a worked up example. ( From RYA YM ) Here is a two tide plot , ie a two hour vector plot

Seaworthy

The background is a yacht having arrived at the waypoint ( near the rose) , wants to plot approaching Cherbourg. It will experience two tides ( from approximating the time ) in its journey
The tides are 1.6 kn 089T first hour and 1.6kn 098T for the 2nd hour

See attached solution, note the water track vector never is connected to the destination point. that would not give the course to steer. The fact that the track may cut the ground track behind or ahead of the destination merely indicates that you arrive after or before you two hours. If the vector was incorrectly joined to the destination, you would actually arrive downside of the destination.

The further assumption made is remaining distance closed is made at the average SOG. ( not the water track speed)

This technique can easily be extended to,multiple hourly vectors
Dave

Dave in your hovercraft example the time taken for a constant compass heading from start to finish was 4.72 hours (= 4:43).
My constant heading for the entire journey was QUICKER than your method of arriving on the rhumb line at the end of the current then going straight.

If the RYA is telling you it is best to arrive on the rhumb line at the end of the current, they are wrong (they are not infallible).

If you are only going by the track they have drawn in the example it does not really say anything. It may have been straight here to avoid the shallows or obstructions (the map is unclear).

[Dockhead]

I made the same mistake, but realized (with help from some on here) that indeed you will always, in any case, be downtide for at least a short bit at the end of any CTS passage. It's still the most efficient way.

Think about an 8 hour passage with 6 knots of tide for just the first hour, and slack tide for the rest. On the ideal passage, which is as always a constant heading, you will only make up 1/8 of the 6 miles you are knocked off during that first hour when the current flows. The other 7 hours you will stay on the same heading and make up the rest of your XTE as you go, and you will approach from "downtide", based on the sum of tides you were overcoming.


This thread has been enormously valuable because no matter how well you think you know this stuff, more and more complex implications and nuances which are often counterintuitive keep coming out of the woodwork. I have learned a tremendous lot; hope others have too.

[Seaworthy Lass]

Originally Posted by goboatingnow

Well thinking on my feet. I would subscribe that the quickest way from A to B , where you are subject to tidal currents is a single CTS. ( we're all agreed on that ). Equally the quickest time from B to C , where there is no tide ( or other forces) is a a straight line. Hence I would have thought that's the quickest way. ( a+ b , c+d) At the end of the tidal vector time ( and yes this is stylised as that's how the data is presented)

What I was arguing with seaworthy that the actual computed CTS water is not drawn to the destination to do so would put you down tide.

Dave

Dave, stop and think - you are saying to use two different headings for the hovercraft journey you described. You have agreed before that ONE heading for a journey was the quickest. I dont understand why you want to use two in this example (particularly since two is slower). .

[goboatingnow]

You cannot connect the water track to the destination. The length of the water track vector is set by the assumed vessel speed , since we are not allowing it to change speed. If there perturbation vectors are in hour increments , then the vessel water track vector is in hour increments. , hence a 4 hour tide vector plot has the water track at 4x vessel water speed. Extending the vector to fit the destination in essence is increasing the vessel speed. Which is why you get there earlier.

The RYA is infallible here, this method has been taught for 50 years. But it makes sense from a vector perspective. The lengths must be proportional

Dave

[goboatingnow]

Originally Posted by Seaworthy Lass

Dave, stop and think - you are saying to use two different headings for the hovercraft journey you described. You have agreed before that ONE heading for a journey was the quickest. I dont understand why you want to use two in this example (particularly since two is slower). .

CTS Only makes sense where one is subject to a perturbation vector.

If my journey from A to B is subject to perturbation vectors. Then a computed single vector gets me to B quickest , if I then extend my journey another 50 mins to C ( say) but I am subject to no perturbation. Then that is a straight line. Hence my journey consists of a Heading to B and a heading to C. Hence my journey A, C consists of two headings.

Remember this is a stylised case. The whole method is actually inaccurate. Since by the fact that we are delayed by one tidal vector( or advanced) we may not actually experience the correct tidal vectors at all. ( this can easily be seen if you plot large tides against small vessel speeds.

Dave

[Seaworthy Lass]

Originally Posted by Dockhead

I made the same mistake, but realized (with help from some on here) that indeed you will always, in any case, be downtide for at least a short bit at the end of any CTS passage. It's still the most efficient way.

Think about an 8 hour passage with 6 knots of tide for just the first hour, and slack tide for the rest. On the ideal passage, which is as always a constant heading, you will only make up 1/8 of the 6 miles you are knocked off during that first hour when the current flows. The other 7 hours you will stay on the same heading and make up the rest of your XTE as you go, and you will approach from "downtide", based on the sum of tides you were overcoming.


This thread has been enormously valuable because no matter how well you think you know this stuff, more and more complex implications and nuances which are often counterintuitive keep coming out of the woodwork. I have learned a tremendous lot; hope others have too.

Enormously valuable for me too! I aways knew one compass heading was most efficient , but I never thought too much about the ground track. It is obviously really important to do so as there may be obstructions in the way.