Distinct Activities: Shackled by a Common Name?

[goboatingnow]

Originally Posted by Seaworthy Lass

Dave, could you please type out what is the CTS and time taken?
Also any other comments you have written.
The photo is too indistinct for me to read it.

Jeepers I can easily read it back from my ipad. , the picture is high res , it should zoom up easily.

I dont have a Breton plotter handy. A can't easily measure the the actual CTS angle. But the time to the destination is either 145 minutes at an average SOG of 3.5 kn , using the two hour plot. , in the 3 hours its 159 minutes at an average SOG 3.2 kn

Given the D3 plot is more accurate , I d used that , but you could just interpolate between the two numbers


Dave

[Seaworthy Lass]

Originally Posted by Jammer Six

It would also help, for the benefit of us newbies, if someone could be kind enough to post a blow-by-blow description of the RYA method.

I sail here in Puget Sound, in the United States. I'm trained in the U.S. Sailing program, and had never heard of the RYA until I read this thread.

My training in these matters consists, therefore, of the U.S. Sailing Piloting & Navigation course, and that is the first and, so far, only method I know of plotting courses and currents.

I've been keeping my mouth shut, because it seemed clear to me that everyone in the discussion understood all facets of the RYA method, but at this late date, I'd like to point out that I don't, and that means there are, perhaps, many folks following this thread who don't.

As you are willing to accept egg as makeup, (an extremely worthy stance!) so am I much too old to care about how I look or sound. So I will speak for all the clueless, curious newbies reading this thread, and give voice to all the stupid questions that only a newbie would ask.

(Those who think there is no such thing as a stupid question haven't met my apprentice. But I digress.)

So as a prelude to comparison between methods, I would like to hear both methods described as Lass has described hers.

One of the most valuable threads I've ever read on the internet. And that's saying a lot.

Hi Jammer
Welcome to CF! Hang around, it can be a rough ride, but it is very worthwhile .

The RYA method is actually an excellent one for in many cases computing a reasonable estimate of the CTS (course to steer for the entire journey if a constant speed is assumed). You then need to make allowances for variation and allow for leeway.
The method allows easy summation of complex tides and the beauty of the system is that computations just need a parallel ruler, dividers, pen, eraser and plastic plotter protractor.

This RYA video shows how it is done:
YouTube
It much simpler watching this than trying to work out written instructions .

You will note in this example that D (the point where the arcing off of the distance travels intersects the course line) is a fair way from B in this case. What does the tide do during the third hour of travel where you need to get from B to D?
No idea. The RYA method ignores this and instead extrapolates what has happened the first two hours.

Anyone following this thread, you will gain a lot from watching this RYA video if you have not done an RYA course:
YouTube
.

[Seaworthy Lass]

Originally Posted by goboatingnow

Jeepers I can easily read it back from my ipad. , the picture is high res , it should zoom up easily.

I dont have a Breton plotter handy. A can't easily measure the the actual CTS angle. But the time to the destination is either 145 minutes at an average SOG of 3.5 kn , using the two hour plot. , in the 3 hours its 159 minutes at an average SOG 3.2 kn

Given the D3 plot is more accurate , I d used that , but you could just interpolate between the two numbers
Dave

So you are selecting D determined by the three hour calculations are you?
Even though you think it is slower journey doing that than using two hours of tide data?
Interesting!

I will wait for you to give us the CTS and time taken exactly so that I can plot a vector diagram showing the chartplotter track if a boat follows the CTS you determine

PS And what is all this about interpolating between two numbers?
You have repeatedly said you determine the CTS by selecting the most appropriate point for D, then extrapolating the data, or as you put it "applying the same proportion to the whole journey".

I don't think the RYA teach you to work on two lots of D and 'interpolate' between the two different CTS computations in any way do they?

[Wotname]

Originally Posted by Seaworthy Lass

[B][COLOR="Blue"]Huh?
"TIME AND TIDE WAIT FOR NO MAN".

I believe the expression is better stated as "Time and tide wait for no man and precious few women"

Originally Posted by Seaworthy Lass

Boat speed is constant at 4 knots throughout the journey.
Destination is 8.5 nm due east.
Current is always from the north:
1st hour 3 knots
2nd hour: 2 knots
3rd hour: 0.5 knots

What is the CTS and the time taken as estimated by the RYA method?

Please give me the values you come up with and we can settle the accuracy of the RYA method once and for all!

OK, I'm not going to use the RYA method; rather just treat it as a pure navigational exercise based on the given data.

Rightly or wrongly, I would depart on a CTS of 058.3 degrees and I would expect to arrive at the destination 2.5 hours later. Well actually I would just steer a little higher than 060 as I couldn't hold a heading of 058.3

[Seaworthy Lass]

Originally Posted by Wotname

I believe the expression is better stated as "Time and tide wait for no man and precious few women"

OK, I'm not going to use the RYA method; rather just treat it as a pure navigational exercise based on the given data.

Rightly or wrongly, I would depart on a CTS of 058.3 degrees and I would expect to arrive at the destination 2.5 hours later. Well actually I would just steer a little higher than 060 as I couldn't hold a heading of 058.3

LOL. I am not one if those women .

Great to have some response on CTS estimates. Hope more people will chime in

I will not give my result until I have heard what Dave's answer is.

I will then plot the vectors giving the charplotter track for the two boats for the journey (one following Dave's estimated CTS and one following mine) .

[Wotname]

Originally Posted by Seaworthy Lass

LOL. I am not one if those women .

Great to have some response on CTS estimates. Hope more people will chime in

I will not give my result until I have heard what Dave's answer is.

I will then plot the vectors giving the charplotter track for the two boats for the journey (one following Dave's estimated CTS and one following mine) .

Well, it was either break out pencil, paper and grey matter or clean the shower. Unfortunately the shower still needs cleaning

[goboatingnow]

Ill print out a protractor , now that I got down the mountain in the snow.
BUt this statement

You will note in this example that D (the point where the arcing off of the distance travels intersects the course line) is a fair way from B in this case. What does the tide do during the third hour of travel where you need to get from B to D?
No idea. The RYA method ignores this and instead extrapolates what has happened the first two hours.

THIS is factually wrong, The RYA method as clearly outlined in text books, say that based on your estimate of the tides that affect you D, may lie infront of or behind B, WHY do you insist on telling me a person that teaches the method, the opposite, The VIdeo only takes ONE case.

please this is unbecoming of you Seaworthy.

Dave

[goboatingnow]

You have repeatedly said you determine the CTS by selecting the most appropriate point for D, then extrapolating the data, or as you put it "applying the same proportion to the whole journey".

I don't think the RYA teach you to work on two lots of D and 'interpolate' between the two different CTS computations in any way do they?

I give up , Im being told by someone , who is learning the RYA method by looking at a you tube video ( one that I sent her) and then TELLING me what or what not the RYA does.!!!.

The whole process of determining the tides in the first place its based on interpolation ( ie you have to interpolate the tidal date to arrive at a value for your passage through that tide) Thats process in itself adds enough error to render any further accuracy moot.

The RYA in a class room ( The theory class is 16 night classes of about 4 hours each) explains the limitations of the tidal data, gets students to look at CTS graph and evaluate what seems appropriate given the quality of the tide data, the effective CTS that a helmmans can steer and the vagaries of leeway and boat speed

All these errors introduce far greater errors that the precision you are attempting to find.

You are turning this into a angels on the head of a pin argument.

[goboatingnow]

So you are selecting D determined by the three hour calculations are you?
Even though you think it is slower journey doing that than using two hours of tide data?
Interesting!

Correct it is a slightly better fit as D3 is closer. In practise the CTS and the time using either method are well within the level of accuracy that can be expected in real life and thats all that matters

Dave

[Dockhead]

Originally Posted by Wotname

I believe the expression is better stated as "Time and tide wait for no man and precious few women"


OK, I'm not going to use the RYA method; rather just treat it as a pure navigational exercise based on the given data.

Rightly or wrongly, I would depart on a CTS of 058.3 degrees and I would expect to arrive at the destination 2.5 hours later. Well actually I would just steer a little higher than 060 as I couldn't hold a heading of 058.3

OK, here's how I would solve this problem, according to the method I was taught admittedly decades ago.

It can be done the simple way since the tides are perpendicular.

I start by guessing it's going to take three hours. That will give me an aggregate set of 5.5 miles. I solve a triangle with c=8.5, b=5.5, and I get a water track of, oops, 10.5 miles, which ain't no 3 hours at 4 knots. So I guess 2.5 hours. That gives an aggregate set of 5.25 miles. I solve that triangle and, bingo, I get 9.99 miles which is my 2.5 hours. There it is!

So, it takes 2.5 hours and my steering correction is 31.7 degrees. Course to steer is 90 minus 31.7 or 58.3. I will want to err on the side of coming out uptide, so so I'll knock a couple of degrees off that -- call it 56.

Looking down I see that this is exactly Wotname's solution, too.

I will scroll around and see how Dave and Seaworthy came out.

But as far as I know, this one is a no-brainer, and this is a mathematically perfect solution. We don't stretch the vectors here, and it would be incorrect to do so, because the tides are measured by hours. If it takes us 2.5 hours to get across, we will get all of the first two and half of the last one. The tides will be expressed as averages during the given hour, and that is also mathematically perfect. The set during a given hour will be a pure function of the average current (and direction if it weren't perpendicular) during that hour.

As I was taught, the vectors need to be stretched if the "rate of advance", as Dave calls it, is affected by a non-perpendicular current. I don't actually remember how we solved those cases because all my real cases have involved currents which were close enough to perpendicular to ignore that. I seem to remember we used a marine slide rule.

All right, tear me apart. The learning is 30 years old and might be mis-remembered here or there, but it's been working really well for me for a couple dozen English Channel crossings.

[goboatingnow]

The course to steer is 60 degrees for D2 model and 64 degrees for D3, either are well within the accuracy of the tidal data and CTS error in real life are the differences are irrelevant,. Use which ever you want. of average it or whatever it matters not.

Most people would do like Wotname , Most students would merely ignore the quality of the result analysis and use D2. Agai it really doesnt matter,

The answer is steer about 60 degrees and expect to arrive in about 2 hours ans half hours.

This discussion is useless in advancing anything new.

[goboatingnow]

start by guessing it's going to take three hours. That will give me an aggregate set of 5.5 miles. I solve a triangle with c=8.5, b=5.5, and I get a water track of, oops, 10.5 miles, which ain't no 3 hours at 4 knots. So I guess 2.5 hours. That gives an aggregate set of 5.25 miles. I solve that triangle and, bingo, I get 9.99 miles which is my 2.5 hours. There it is!

So, it takes 2.5 hours and my steering correction is 31.7 degrees. Course to steer is 90 minus 31.7 or 58.3. I will want to err on the side of coming out uptide, so so I'll knock a couple of degrees off that -- call it 56.

Looking down I see that this is exactly Wotname's solution, too.

I will scroll around and see how Dave and Seaworthy came out.

But as far as I know, this one is a no-brainer, and this is a mathematically perfect solution. We don't stretch the vectors here, and it would be incorrect to do so, because the tides are measured by hours. If it takes us 2.5 hours to get across, we will get all of the first two and half of the last one. The tides will be expressed as averages during the given hour, and that is also mathematically perfect. The set during a given hour will be a pure function of the average current (and direction if it weren't perpendicular) during that hour.

As I was taught, the vectors need to be stretched if the "rate of advance", as Dave calls it, is affected by a non-perpendicular current. I don't actually remember how we solved those cases because all my real cases have involved currents which were close enough to perpendicular to ignore that. I seem to remember we used a marine slide rule.

The method you outline is impossible to apply in any real life is non perpendicular tides., The result must be derived graphically from the chart. I dont know who taught it to you , but its of no real use in real world applications.

Dave

[goboatingnow]

But as far as I know, this one is a no-brainer, and this is a mathematically perfect solution. We don't stretch the vectors here, and it would be incorrect to do so, because the tides are measured by hours. If it takes us 2.5 hours to get across, we will get all of the first two and half of the last one. The tides will be expressed as averages during the given hour, and that is also mathematically perfect. The set during a given hour will be a pure function of the average current (and direction if it weren't perpendicular) during that hour.

This isnt really true Dockhead, tides are stated as a set and drift at a point in time usually in increments of 1 Hour referenced back to the nearest standard port , or a reference port.

You extract the actual set and drift you experience at the time you pass through the area, by interpolating the results from the tide table, You then assume that the 60 minutes rule applies, In practice it ,may or may not apply , particulary if the underlying tide is non sinusoidal. Hence a 2kn tide can be valid for 60 minutes, 70 minutes 80- minutes , or invalid after 45 minutes, you have no way of knowing

What we do know is that tide A at + 45 minutes is inreality very little different from trhe next hour B, at 15 minutes , ie in reality, not what the tide tables say.

Hence it is perfectly acceptable to advance or retard tides appropriately within a reasonable time limit. ,ie whether you take 2 tide and spend 2.5 hours in them or spend 60,60 and 30 in the next tide has very little difference in real life as the tide is a constantly changing event and never remains still for 60 minutes , or whatever.

[Wotname]

Originally Posted by goboatingnow

Please STOP telling me what you think the RYA method is. I have provided a detailed description of the complete method. I've met and talk to one of the people who designed the course You estimate the passage time using any available means..........

Hmm... I am not sure that you have; at least I can't find it even are scanning back over the many pages of this thread.

Originally Posted by goboatingnow

......THIS is factually wrong, The RYA method as clearly outlined in text books, say that based on your estimate of the tides that affect you D, may lie infront of or behind B, WHY do you insist on telling me a person that teaches the method, the opposite, The VIdeo only takes ONE case. .........

Well, I was fooled, I never picked up on the video that is just one case of perhaps many cases. I took the video as the method.

I may be overly suspicious here but why do I have a feeling the goal posts are moving

[goboatingnow]

So, it takes 2.5 hours and my steering correction is 31.7 degrees. Course to steer is 90 minus 31.7 or 58.3. I will want to err on the side of coming out uptide, so so I'll knock a couple of degrees off that -- call it 56.

again false logic, in complex tides it can can be very difficult to actually determine at the outside what way to adjust the CTS to end uptide. ( Consider how many saw the constaly varying one way current caused a rhumb line transition).

If you want to be uptide, you would typically adjust the CTS as you progress across and it is clear what is really happening.