

31012013, 17:23

#616

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Location: B.C.,Canada
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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
SWL:When I have some free time I will go through my method (as per instructions in Post # 2 of this thread) step by step with diagrams so that it is easy for everyone to follow.
thanks SWL, until then, I actually did some rooting around (41 pages!) and found a pretty good explanation here... Inaccurate RYA Teaching : CTS  Quest For a New Method
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01022013, 03:02

#617

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Posts: 3,888

Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Can you good folks please take a look at my posting in the "Single CTS..." thread ( here)? The SOG vs BSP (through the water) question seems to be a good one.
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01022013, 03:26

#618

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Join Date: Aug 2010
Location: Arctic Ocean
Boat: Under construction 35' ketch
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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
If you are speaking of the SOG on a given moment it's just to vectoring up log speed and the current speed (or just peek from the plotter) Anyway one should allways be aware of that, you are quite right
about the importance. Thou the problem is more accurate when you are excecuting the CTS plan not when planning it.
BR Teddy
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01022013, 04:18

#619

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
I may be wrong. Given the triangle ABV (A = starting point, B = destination, V = endpoint of the current vector AV), is the segment VB equal to the distance sailed through the water?
Foe a moment I thought it wasn't, but right now I believe it is indeed the distance through the water. If so, then BSP is the parameter we should be using (and we have been using), and the calculations are correct.
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01022013, 04:58

#620

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Location: Cowes (Winter), Baltic (Summer) (the boat!); somewhere in the air (me!)
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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Paul Elliott
I may be wrong. Given the triangle ABV (A = starting point, B = destination, V = endpoint of the current vector AV), is the segment VB equal to the distance sailed through the water?
Foe a moment I thought it wasn't, but right now I believe it is indeed the distance through the water. If so, then BSP is the parameter we should be using (and we have been using), and the calculations are correct.

VB is absolutely the distance through the water if you sail an ideal CTS passage. The end of the vector line AV is the aggregate of all of the current effects during the passage. That's why you get passage time by dividing the length of VB by speed through water.
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09022013, 14:55

#621

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Join Date: Oct 2008
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SWL method for CTS: step by step instructions
Step by step explanation of the SWL method of determining the CTS with variable cross current.
More than a week has passed since I promised to post step by step instructions with diagrams for my method of determining CTS with cross current (chores on board have been keeping this lassie occupied and temporarily out of trouble), but here they finally are .
The example I will work through:
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 08:3009.30)
10:00 2.0 knots 180T (= rough average 09:3010.30)
11:00 0.8 knots 180T (= rough average 10:3011.30)
What is the expected CTS?
What is the expected time taken?
What is the expected ground track?
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09022013, 15:10

#622

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SWL method for CTS: step by step instructions
STEPS 15
Step 1: Mark the departure and destination points on a chart. I will refer to these as A and B respectively.
Sharpen your pencil. You will need it for the accuracy
Step 2: Measure AB (= the course distance) to enable you to start gathering data on the currents. It need not be measured very accurately at all unless you need to know your predicted SMG for some reason (the time taken is computed separately and does not require this distance to be accurately known).
Step 3: Divide the course distance by the expected boat speed. This gives time of the journey if there is no current or leeway.
= 6.8 / 4 = 1.7 hours for this example
Step 4: You know the passage will take longer than that if there is total current against you (and shorter if it it with you), so make an educated guess how much longer by looking at the current along the way (this may actually be difficult, but that is a whole different topic).
This gives you a starting point for how many hour lots of current you need to consider (or half hour lots if this current info is provided). For simplicity I will describe hourly data here, but the technique works equally well for any segment you care to look at.
Divide the course into an equal number of sections as per the number of hours you estimate and determine the current in each segment for the specific hour you will be there.
This is only an approximation as the current will not be constant over the hour.
For this example you can guess that the journey will take over 1.7 hours, but probably less than 2.5. As the first hour is partial, you will need a couple of extra hours of tidal stream data.
Step 5: For this next step, it is important to draw all distance displacement vectors to the same scale as the chart.
Determine what proportion of the first hour the first lot of tidal stream data will apply and work out how much the boat would be displaced by this.
In this case departure is at 09:00 so the average of the last half an hour of tidal data applies (estimate 2.65 knots) for half an hour.
So the boat displacement due to current during this time is 1.3 nm.
Draw 1.3 nm starting from A heading south, the direction the current will take you.
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09022013, 15:34

#623

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SWL method for CTS: step by step instructions
STEP 6
Step 6: Add the next lot of boat displacement due to current (2.0 nm south over the next hour) onto the tip of the 1.3 nm vector.
The aim is to continue adding on hourly boat displacement amounts due to current from the tip of the last current displacement vector until you can arc off a speed displacement vector towards B that is getting close to reaching B. they must be drawn at the correct angle of the current (in this case 180 degrees).
This particular journey is expected to take between 1.7 and 2.5 hours, so check how close you would be after 1.5 hours, as you want a position at the last time you would be undershooting.
Boat speed is 4 knots, so at 1.5 hours (total time to now) you would have been displaced 1.5 x 4 = 6 nm due to boat speed.
If you measure 6nm from the end of the current displacement vector after 1.5 hours in the direction of B, you can see you are short of B.
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09022013, 15:46

#624

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SWL method for CTS: step by step instructions
STEP 6 continued
Mark on the next lot of boat displacement due to current (0.8 nm over the next hour).
Total time to this point = 2.5 hours.
Check where you would be at the end of 2.5 hours of travel by measuring the 'boat displacement due to boat speed' from the tip of the last 'boat displacement due to current' vector.
2.5 x (boat speed)= 2.5 hours x 4 knots = 10 nm
Mark this off on the direction of B.
This takes you beyond B.
Almost all the hard work has been done. You know you will arrive sometime during this last hour of current.
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"To me the simple act of tying a knot is an adventure in unlimited space." Clifford Ashley



09022013, 16:00

#625

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Posts: 12,819

SWL method for CTS: step by step instructions
STEP 7
Step 7: To determine the proportion of the last hour which will be needed to get to B, draw a line between the start of this final 'boat displacement due to current' vector and B.
Measure off the 'boat displacement due to boat speed' for this time (1.5 hours travelling 4 knots = 6 nm) from the start of this line, and mark this point S (for short).
This is the closest you could get to B after 1.5 full hours of travel travelling on one compass heading.
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09022013, 16:16

#626

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SWL method for CTS: step by step instructions
STEP 8
Step 8: Draw a line between the end of the final 'boat displacement due to current' vector and B.
Measure off the 'boat displacement due to boat speed' for this time (2.5 hours travelling 4 knots = 10 nm) and mark this point L (for long).
This overshoots B and is the closest you could get to B after 2.5 full hours travelling on one compass heading.
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09022013, 16:23

#627

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Posts: 12,819

SWL method for CTS: step by step instructions
STEP 9
Step 9: Look at the proportion between SB and BL. This will immediately give you an idea how much of the final current needs to be applied.
In this case it looks a bit under a half, so you know roughly 0.4 of the last hour will apply.
Determine this proportion accurately by measuring the distance from S to B (=SB) and also from B to L (=BL) and calculating:
SB / SB+BL
= 1.6 / 1.6+2.1 = 0.43
Multiply this by the strength of the current for this last hour and you now know the boat displacement due to current for this period.
= 0.8 x 0.43 = 0.3 nm
Mark 0.3 nm on the final current vector and label it K
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09022013, 16:30

#628

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SWL method for CTS: step by step instructions
STEP 10
Step 10: Join K to B
CTS = the angle of the line KB before you have made allowances for compass variation and for leeway = 62 degrees true
Distance travelled through water = the length of KB = 7.7 nm
Time taken = the length of KB divided by the boat speed
= 7.7 / 4 = 1.93 hours = 1 hour 56 minutes
(It can also be determined by adding up the hours of 'boat displacment due to current' = 0.5 + 1 + 0.43 = 1.93 hours)
SMG (if you need to know this for some reason) = the length of AB divided by the time taken
= 6.8 / 1.93 = 3.5 knots
Note: calculating this is the only reason you would need to measure the distance from A to B accurately.
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09022013, 17:48

#629

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Boat: Aluminium cutter rigged sloop
Posts: 12,819

Estimate of last amount of current
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:
Quote:
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 .
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 .
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09022013, 21:35

#630

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Join Date: Mar 2008
Location: Calgary, AB, Canada
Posts: 5,048

Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
My Solution
Step 1 – Draw CMG AB
Step 2  Estimate time on route  approx 2 hours
Step 3  Draw Set and Drift Vector  4.8 nm
Step 4  Arc line with hours boat speed 8 knots
Step 5  Measure distance (AD) 6.34 miles
Step 6  Measure CTS = 053 T (CD)
Step 7  SMG = 6.34 / 2 = 3.17
Step 8  ETA = 128 minutes 1108 ( was using a circular slide rule to avoid any electronics, other than plots using Open CPN)
The distance DB is approx. .4 miles which at 4 knots is less than 4 minutes.
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