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Old 11-02-2013, 10:14   #661
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Re: Inaccurate RYA Teaching : CTS - Quest For a New Method

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Originally Posted by HappySeagull View Post
[/B](it's getting to be a peeve, but don't worry, I'll get over it.
Hapless is doin' pretty good just stayin' inside his attention span.
If I could get his attention away from feeding seagulls, I daresay he could do better.)
2 hours 27 minutes

Attachment 54476
Hapless, I think you have plotted your position at the end of the second hour incorrectly .
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Old 11-02-2013, 10:17   #662
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Re: Inaccurate RYA Teaching : CTS - Quest For a New Method

you are right! 1.3 needed west
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Old 11-02-2013, 10:19   #663
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Re: Inaccurate RYA Teaching : CTS - Quest For a New Method

I really hate to do this, but I personally would not take a boat "capable of three knots" into a 2.2 knot current. I would argue that we have another extreme example. If you have to conserve fuel on a 6.7 mile voyage you need to re-think the voyage..

EDIT

If I wait one hour.

I have SMG of 2.9, a CTS of 168 and an ETA 139 minutes. My CTS line intersects my CMG 9 cables from the destination.

BTW are we still waiting for a scenario based on real data?
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Old 11-02-2013, 10:32   #664
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Re: Inaccurate RYA Teaching : CTS - Quest For a New Method

there that's better. 2 hours 34 minutes. Loses.as he should.
I can't fit the rya in here right now. attention span batteries dead.

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Old 11-02-2013, 10:43   #665
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Re: Inaccurate RYA Teaching : CTS - Quest For a New Method

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Originally Posted by HappySeagull View Post
there that's better. 2 hours 34 minutes. Loses.as he should.
I can't fit the rya in here right now. attention span batteries dead.

Attachment 54483
I will pass over a little hint. Tell Hapless the quickest route is one using just one CTS .

His problem was that he wanted to change course a couple of times along the way - he initially set off at 162, then altered course to 146, then finished off at 147 .
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Old 11-02-2013, 11:27   #666
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Method for plotting the ground track

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Originally Posted by Seaworthy Lass View Post
Step by step explanation of the SWL method of determining the CTS with variable cross current.

Boat speed is 4 knots (log reading with freshly cleaned log)
Calm conditions, no wind, you are motoring at a steady throttle setting
Departure time 09:00
Destination due east (6.8 nm)
Tidal stream data:
09:00 2.8 knots 180T (= rough average 08:30-09.30)
10:00 2.0 knots 180T (= rough average 09:30-10.30)
11:00 0.8 knots 180T (= rough average 10:30-11.30)

What is the expected CTS?
What is the expected time taken?
What is the expected ground track?
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Originally Posted by Seaworthy Lass View Post
[B]
Tomorrow I will complete the step by step process for plotting your expected ground track (I feel it is important to do this). Getting late here, so I will bid everyone goodnight .
So here is the last chapter in the process of finding a CTS with variable cross current:

STEP 11
Plotting the expected ground track.


The position at the end of each time segment (it may be a partial hour at the beginning and end of the trip) is plotted simply by adding the 'boat displacement due to boat speed' onto the 'boat displacement due to current' for each time segment.

Mark all these positions at the end of each time segment and join them to produce your ground track.

I think it is important to plot the expected ground track for three reasons:
- to determine what hazards are present on or close to the track
- to check the accuracy of your CTS computation (if your vectors don't finish exactly at the destination, you have made an error)
- to be able to easily see along the way if you are deviating significantly from your expected track (this will occur for example if your actual boat speed is significantly different to that predicted initially or the current encountered is not what is expected). The CTS needs to be recomputed if this is the case.

For this example:
Start plotting the expected ground track by looking at the first time segment (half an hour in this case).

Add the 'boat displacement due to boat speed' (4 knots for half an hour = 2 nm @ 62 degrees) to the tip of the first 'boat displacement due to current' (2.65 knots for half an hour = 1.3 nm @ 180 degrees).

Mark this spot.
This is your position after the first half hour.
Draw a line between the start and this point. This is your 'expected' track for the first half hour. I have marked this with a dotted red line in the diagram.

Note, because the current will not actually be steady for this period, your actual ground track will be curved.
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Old 11-02-2013, 11:42   #667
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Method for plotting the ground track

STEP 11 continued

From your position at the end of the first time segment, add the next lot of 'boat displacement due to current' (2 knots for 1 hours = 2 nm @ 180 degrees) then add the amount of 'boat displacement due to boat speed' (4 knots for 1 hour = 4 nm @ 62 degrees) to the tip of that.

Mark this point.
This is your position after the first hour and a half hour.
Draw a line between the position after the first time segment and this point. This is your 'expected' track after the first hour and a half.
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Old 11-02-2013, 11:53   #668
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Method for plotting the ground track

STEP 11 continued

The last time segment is a partial hour in this case.

From the position at an hour and a half, add the next lot of 'boat displacement due to current' (0.8 knots for 0.43 hours = 0.34 nm @ 180 degrees) then add the amount of 'boat displacement due to boat speed' (4 knots for 0.43 hours = 1.72 nm @ 62 degrees) to the tip of that.

This coincides with B, so you know your computation of 62 degrees as the CTS is correct for the given data. Finish drawing in your ground track from the position after the first two time segments to B.

This track can also be easily plotted on OpenCPN.

That's it, task complete .
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Old 11-02-2013, 12:07   #669
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Re: Example 6

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Originally Posted by Seaworthy Lass View Post
I will give hourly averages for the next few examples so that there is no confusion, but in the real world Hapless does need to consider what to do with the tidal stream data in the way it is presented, particularly as Hapless will not always leave exactly at the beginning of one lot of average data

Tidal stream:
Average for the first hour: 2.2 knots west
Average for the second hour: 1.3 knots west
Average for the third hour: zero (slack water)
I have to call poppycock on this one. Where in the world do you have slack that lasts an hour? Or is your example assuming the slack in the middle of the third hour, with the previous 30 mins flooding and the final 30 mins ebbing? If that's the case then your solution will fail due to the last half-hour's missing set.

For the sake of interest, I took a look at Jackdale's excerpt from CHS's current data for SJDF, and found one with the max flood of 2.2 kts, and a little over 3 hours later, slack water; I drew the graph and measured the expected set at 15-minute intervals. The first hour average was 2.1kts, the second hour 1.68kts and the third 0.74kts. If you were to be affected by all three hours the set vector would be 4.5nm long, but if you were to arrive at destination after 2h30m, then the tidal vector would only be 4.3nm long.
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Old 11-02-2013, 12:31   #670
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Re: Example 6

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Originally Posted by Lodesman View Post
I have to call poppycock on this one. Where in the world do you have slack that lasts an hour? Or is your example assuming the slack in the middle of the third hour, with the previous 30 mins flooding and the final 30 mins ebbing? If that's the case then your solution will fail due to the last half-hour's missing set.

For the sake of interest, I took a look at Jackdale's excerpt from CHS's current data for SJDF, and found one with the max flood of 2.2 kts, and a little over 3 hours later, slack water; I drew the graph and measured the expected set at 15-minute intervals. The first hour average was 2.1kts, the second hour 1.68kts and the third 0.74kts. If you were to be affected by all three hours the set vector would be 4.5nm long, but if you were to arrive at destination after 2h30m, then the tidal vector would only be 4.3nm long.
This why I am looking for real scenarios of real voyages in real places with realistic boat speeds.

My earlier Port Angeles - Victoria scenarios were based on real passages with real numbers.

For anyone with Navionics or OpenCPN look at the current data or use your hydrographic service data.
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Old 11-02-2013, 12:33   #671
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Re: Example 6

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Originally Posted by Lodesman View Post
I have to call poppycock on this one. Where in the world do you have slack that lasts an hour? Or is your example assuming the slack in the middle of the third hour, with the previous 30 mins flooding and the final 30 mins ebbing? If that's the case then your solution will fail due to the last half-hour's missing set.

For the sake of interest, I took a look at Jackdale's excerpt from CHS's current data for SJDF, and found one with the max flood of 2.2 kts, and a little over 3 hours later, slack water; I drew the graph and measured the expected set at 15-minute intervals. The first hour average was 2.1kts, the second hour 1.68kts and the third 0.74kts. If you were to be affected by all three hours the set vector would be 4.5nm long, but if you were to arrive at destination after 2h30m, then the tidal vector would only be 4.3nm long.
I can't win LOL.
The previous example I gave was for tidal stream information on the hour. I extrapolated the average and was told this was too complex a way to present an example and that people wanted the simplicity of hourly averages.

I give hourly averages in this example and I am told that this is not suitable either. The hourly averages I selected were the ones given by Jackdale in his English channel example.

I addressed the issue of inaccuracy creeping in due to the estimate of the average current for the last bit of the journey in post #629:

Quote:
Originally Posted by Seaworthy Lass View Post
Inaccuracy due to the estimate of the average current for the last portion of the journey:

An error will occur using a partial portion of the last amount of current as the average will be different for the partial hour compared to the whole hour. It will be higher initially if the current is decreasing and lower if it is increasing.

I addressed this issue a couple of weeks ago:

Originally Posted by Seaworthy Lass
And yes, it does assume the current is reasonably constant during that hour, but once you get good with the method you could tweak that proportion yourself according to whether the current was increasing or decreasing.


In very rough terms, if the average amount of cross current is about a quarter of the average boat speed and roughly half of the last hour of current data was used, then I would take off a degree or two if the current was decreasing, or add a degree or two if it was increasing.

I would add a proviso that the CTS determined using the SWL method would consistently give you an accuracy within about 5 degrees given the limitations of using hourly tidal data.

The RYA method can produce a result that is close in lots of circumstances, but it may sometimes also be 10 or 20 or even 30+ degrees in error. These large inaccuracies will not occur using the SWL method .
If examples are chosen where the amount of average current for a journey is tiny in comparison to boat speed, then problems with a method are hidden.

I am deliberately picking average values that may occur for tidal streams during spring tides. If anyone feels the values I am selecting are incorrect, then could you please give me some values for tidal stream over a 12 hour period (in areas with a high tidal range) that you are satisfied with and I will work examples around these .
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Old 11-02-2013, 14:54   #672
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Re: Example 6

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Originally Posted by Seaworthy Lass View Post

I am deliberately picking average values that may occur for tidal streams during spring tides. If anyone feels the values I am selecting are incorrect, then could you please give me some values for tidal stream over a 12 hour period (in areas with a high tidal range) that you are satisfied with and I will work examples around these .
The numbers that I posted were ones I found on another web site. I would not trust a skipper who tried to may a Channel crossing in a vessel that needed to maintain 3 knots to conserve fuel. I know about conserving fuel. I ran Turicum at 1700 rpm to maximize efficiency from Maui to Vancouver. When made about 5 knots at that rpm.

Will this work? This is real

Here are the currents for Juan de Fuca Entrance for Feb 13. They are fairly strong.

Currents at Strait of Juan de Fuca Entrance, Washington

You will have to interpolate the strength, but should not be that far off.

I used the numbers from Open CPN:

0900 1.63 290
1000 1.85 290
1100 1.75 290

CHS Table

Turns Maximum
Day Time Time Knots

13 1:18 AM 3:49 AM +1.0
13 6:20 AM 10:11 AM -1.9
13 2:18 PM 4:21 PM +0.6
13 6:27 PM 10:18 PM -1.6

+ Flood 115
- Ebb 290 (caught the error)

If you want them they are here Juan De Fuca-West - Tides, Currents, and Water Levels

The Navionics numbers are somewhat discrepant.


Flood direction 115 T
Ebb direction 290 T

A to B - 10.3 miles
Bearing A to B 023T

Depart 0900
Speed (knotmeter) 5.0 knots

(for those with the chart the voyage from the Port Hand Bell Buoy "G1" at the entrance to Neah Bay to the Fairway (Mid Channel) Buoy "YK" at the entrance to Port San Juan)

SWL I will give you a 24 hour cycle for this current station. I need to key it in as I cannot cut and paste the numbers.
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Old 11-02-2013, 15:05   #673
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Quote:
Originally Posted by Seaworthy Lass

Yes, I agree the average amount of current in any one hour period is not necessarily 50%. Either plot out the tide of the entire cycle and extrapolate the average from there, or make some simple assumptions as I have or find data presented more frequently than hourly.

Deal with the current data however you like. I am not here to argue about that.

But once you have decided on what data to select, the critical thing is which method do you use to compute the CTS?

The instructors here have basically been saying the data is so inaccurate that it doesn't matter how inaccurate the method is (although one instructor still feels that the CTS determined by the RYA method will consistently get you to B).

I think working with an inaccurate method only compounds the errors. Although the RYA method frequently gives a good 'rough and ready' approximation of the CTS, sometimes it fails miserably. My method can handle any data you care to throw at it .
This is not a correct summation SWL, your method gives more precise answers. We have no idea if it is more accurate , we await Dockheads deliberations

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Old 11-02-2013, 15:17   #674
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Re: Inaccurate RYA Teaching : CTS - Quest For a New Method

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This is not a correct summation SWL, your method gives more precise answers. We have no idea if it is more accurate , we await Dockheads deliberations

Dave
Thanks Dave

Those are my thoughts as well. That is why real world examples are needed. SWL's method has some apparent precision in extreme situations. I have expressed several times that I would not venture out in to the conditions described.

Jack
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Old 11-02-2013, 16:37   #675
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Re: Inaccurate RYA Teaching : CTS - Quest For a New Method

Solution to post 672



SMG - 4.6
CTS - 044
ETA 1114

EP based on DR and current.

Almost perpendicular to TSS.
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