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Old 17-01-2013, 05:18   #436
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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.
I don't beleive it is. Nor is it the way taught , what you are doing is connecting the water track vector to the destination. You can't its a constant speed vector.

Dave
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Old 17-01-2013, 05:25   #437
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Re: Distinct Activities: Shackled by a Common Name?

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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
Sorry, if RYA a teaching that, they are wrong.

I dont understand at all what you are saying about vector size changing. In your hovercraft example I did not alter speed at all (I am keeping a constant 4 knots and a constant heading of 32 degrees for the entire journey). It takes me 4:43 hours to get there.
I am getting there quicker than you. Getting there quicker is the bottom line .
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Old 17-01-2013, 05:27   #438
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We do need put to bed the actual CTS angle calculation


Lets take a simple one hour plot. The destination is approx one hour away. The destination is due east. , the tide over that approx travel time to the destination is 2kn at 135T , the boat speed is constant at 5 kn through the water



The correct technique is to plot a one hour water track vector. This will end up " past" the destination. , ie this actually where we would end up if we kept going for one hour. The CTS is read off the chart appropriately.

If you join the water track vector to the destination ( as some would suggest. ) what you are doing is adjusting the speed of the vessel. ( since the vector is shorter ) so now you will arrive in one hour , put at a slower speed ( which is intuitive as the tide is pushing you to your destination)

It's a cardinal rule of CTS plotting that you DO NOT join the end of the tidal vectors to the destination. ( because you have set a constant water speed )

Dave
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Old 17-01-2013, 05:28   #439
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Originally Posted by Seaworthy Lass

Sorry, if RYA a teaching that, they are wrong.

I dont understand at all what you are saying about vector size changing. In your hovercraft example I did not alter speed at all (I am keeping a constant 4 knots and a constant heading of 32 degrees for the entire journey). It takes me 4:43 hours to get there.
I am getting there quicker than you. Getting there quicker is the bottom line .

The vector length of the water track represents the vessels speed. That vector By Definition must be a fixed length. To compare it to the perturbation vectors.

Please plot it and show me, that's the only way it can be demonstrated. ( measure the length of the water track vector )

Thanks

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Old 17-01-2013, 05:32   #440
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Re: Distinct Activities: Shackled by a Common Name?

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I don't beleive it is. Nor is it the way taught , what you are doing is connecting the water track vector to the destination. You can't its a constant speed vector.

Dave
Dave, would you like my vector diagram for the journey?

Edited to add: posted this at the same time you did. Will do this in a sec xxx
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Old 17-01-2013, 05:35   #441
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Dave, would you like my vector diagram for the journey?

Edited to add: posted this at the same time you did. Will do this in a sec xxx
Please I can then see exactly what you did. The problem is the lowest number doesn't mean its correct

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Old 17-01-2013, 05:43   #442
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Seaworthy. As an example to demonstrate. You should plot my 135 degree tidal example. This demonstrates exactly why the water track cannot be connect to the destination

Dave
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Old 17-01-2013, 05:47   #443
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Re: Distinct Activities: Shackled by a Common Name?

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Please I can then see exactly what you did. The problem is the lowest number doesn't mean its correct

Dave
The time taken was a very simple calculation - no error.

Distance is 16 miles.
Tide offset is 10 miles for the journey.
Distance travelled through water = the square root of (16 squared plus 10 squared) = 18.87 nm
Speed is constant at 4 knots through the water.
Time taken is therefore 4.72 hours (4:43)

Working on vector diagram now. Taking forever as I have no graph paper and have to rule lines .
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Old 17-01-2013, 05:58   #444
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Quote:
Originally Posted by Seaworthy Lass

The time taken was a very simple calculation - no error.

Distance is 16 miles.
Tide offset is 10 miles for the journey.
Distance travelled through water = the square root of (16 squared plus 10 squared) = 18.87 nm
Speed is constant at 4 knots through the water.
Time taken is therefore 4.72 hours (4:43)

Working on vector diagram now. Taking forever as I have no graph paper and have to rule lines .
Loads of free graph paper on the Internet if you have a printer

No offence
By the way its not that your maths arnt right. , its your understanding of unit vectors that's misleading you. ,

That's why plotting will show it.

Dave
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Old 17-01-2013, 06:01   #445
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Or just plot the 135 current example this demonstrates the issue


Dave
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Old 17-01-2013, 06:12   #446
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Re: Distinct Activities: Shackled by a Common Name?

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Originally Posted by goboatingnow View Post
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
I have enormous respect for your knowledge, which is much greater than mine in this area, obviously. But I don't see how that could be right. I will devise a test case after I finish with Capt Force's scenario.

Can you maybe give me a hypothetical case where this principle applies? And I will run the numbers.
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Old 17-01-2013, 06:19   #447
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I have enormous respect for your knowledge, which is much greater than mine in this area, obviously. But I don't see how that could be right. I will devise a test case after I finish with Capt Force's scenario.

Can you maybe give me a hypothetical case where this principle applies? And I will run the numbers.
Usually people flatter me when I'm about wrong.!! I'm just graphing two scenarios with a slack tide at the end.

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Old 17-01-2013, 06:29   #448
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Re: Distinct Activities: Shackled by a Common Name?

https://docs.google.com/folder/d/0Bx...RqTTg0ZzQ/edit

Google drive folder we can share stuff in if you want. - does it work??

My spreadsheet with the 360deg increments is in there. Don't expect anything tidy though

Still not completely convinced calcs are right but off to boat show to drink some of the worlds most expensive guiness now so will have a look later.

Also , I should have a cad prog somewhere which makes doing plots really fast and easy. Will dig it out later.
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Old 17-01-2013, 06:39   #449
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Re: Distinct Activities: Shackled by a Common Name?

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Originally Posted by goboatingnow View Post
The vector length of the water track represents the vessels speed. That vector By Definition must be a fixed length. To compare it to the perturbation vectors.

Please plot it and show me, that's the only way it can be demonstrated. ( measure the length of the water track vector )

Thanks
Dave
Here you go:
Constant speed of 4 knots, constant heading of 32 degrees, quickest time.
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Old 17-01-2013, 07:04   #450
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Re: Distinct Activities: Shackled by a Common Name?

GoBoating, here is a really simple but dramatic example of best course to steer for "current then no current" journey that will hopefully help convince you.

Same 16 mile journey as the last example.
Same boat/hovercraft speed of 4 miles/hour.
Again, four hours of current, but this time a steady four knots during this time.

Boat that wants to get to the rhumb line at the end of the current sits in the same spot for 4 hours as he can make no progress against the current.
Then travels 16 miles at 4 knots, taking 4 hours.
Total time = 8 hours

Boat 2 works out that the 4 knots of current over 4 hours will move him 16 miles over a journey of 16 miles, so he heads off at 45 degrees.
Total time = (square root of 16 squared plus 16 squared) divide by 4 = 5.66 hours

The boat following the constant heading for the entire journey beats the boat who is following RYA instructions by 2.34 hours.

If the RYA is teaching that you should aim to be on the rhumb line at the end of the current, but before the end of the journey, they are incorrect.
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