Distinct Activities: Shackled by a Common Name?

[Seaworthy Lass]

Originally Posted by four winds

......I have a math background but didn't feel qualified to participate with such knowledgeable people debating. About four pages back I finally chose sides and should apologize to Seaworthy for not stating so publicly. Her graphs and text were the easiest to digest for me and helped me get a handle on the issues.

CaptForce's last post was perfect.

Thanks for the lively, courteous, enlightening debate. (not saying the debate is over, never is)

Thanks Four Winds. Better late than never .

I was so out on a limb on my own . And not just disagreeing with GoBoating but the whole accuracy of the RYA method if it was supposed to be an accurate mathematical formula when all the data was correct (as GoBoating kept insisting it was).

GoBoating just about had a fit when I told him the RYA was not infallible.

Yes, CaptForce's last post was brilliant! I loved it. Wittiest post I have ever read on CF .

[noelex 77]

From post 1

Originally Posted by Andrew Troup

I have been puzzling over the absence of discussion of navigation on this forum. Even the navigation forum is largely a discussion of the purchase and troubleshooting of electronic equipment

Andrew I think you got your wish

This has been a great thread.

I think everyone has learned something. While I feel we have achieved agreement over many of the issues, there is still some debate, particularly over the mathematical accuracy of the methods taught by the RYA.

I have sent the RYA the following email and hope we can look forward to their participation in our discussion.

I am a moderator on Cruising Forum. There has been a recent long (it has had over 500 posts and 10000 views already), and very fruitful thread about the finer points of navigation.
Some of the detailed discussion has focused on the techniques taught by the RYA.
I think it would be helpful if we could have some comments from your organisation.

The discussion is here:
http://www.cruisersforum.com/forums/...ml#post1133059

The relevant discussion is mostly from post 395

It a would be great if a senior RYA instructor, experienced in navigation could join in the discussion, the forum is easy to join.
Alternatively if you would like to email your thoughts to me I will pass these on to our members.

Despite the above, I am still hopeful we can come to a consensus. We have some very experienced members that are confronting these problems every day.
Keep the posts coming.

[conachair]

Originally Posted by Seaworthy Lass

Thanks Four Winds. Better late than never .

I was so out on a limb on my own . And not just disagreeing with GoBoating but the whole accuracy of the RYA method if it was supposed to be an accurate mathematical formula when all the data was correct (as GoBoating kept insisting it was).

GoBoating just about had a fit when I told him the RYA was not infallible.

Yes, CaptForce's last post was brilliant! I loved it. Wittiest post I have ever read on CF .

Well, I'll continue to back up the maths side . (we call it maths over here )
Lets say you're heading up the english channel from A to Destination. On the charts there are tidal diamonds in various places which give the average tidal speed and direction at that point for each hour for 6 hours before and after high water somewhere defined.
You can plot a , say, 3 hour passage using the data from the tidal diamond which is closest to your position at each hour.

With the vectors for the boat speed and direction then you will be exactly where the plot puts you, assuming the tides from the tidal diamonds were exactly what you were experiencing.

Maybe not exactly where you want to go but from the data you are stuck with averages over an hour. The maths, I think, cannot be wrong. It's just when you get into the real world it starts drifting out because there isn't actually that much data.

I think

[conachair]

Originally Posted by four winds

I have a math background but didn't feel qualified to participate with such knowledgeable people debating.

Join in

I need help with math(s)

Been thinking about equations for graphing a boat going across a tide.

Assuming that the tidal velocity is a sin wave, it looks like it's important to know the actual distance that the water has moved at any given moment, not just the speed at any given time, somehow you need to add all the speed X time. Been a long time since school but it looks that maybe this is proportional to the area under the sin wave graph.

So how do you do that? Suspect I'm trying to reinvent calculus here, a pointer might speed things up. I haven't tried google yet, thats next

[four winds]

Well unfortunately, I'm a couple or three decades passed my last class on differential equations. But yes, it is solving for an integral. It would be a complex formula for the shape of the path over ground.

Once done, any elapsed time or distance through water input could be solved for the position on the curve.

I'm just happy to be able to follow the discussion using vector analysis.

[wolfenzee]

It a appears as though this forum has evolved into a very good demonstration of how to navigate "by hand" in ways that the gadgets and gizmos just are not capable of.
If something is idiot proof...what about for those of us that are not idiots
If navigation gizmos are to designed to do the thinking for those of us that can't do the thinking themselves and/or don't have the knowledge....what about the rest of us.
Electronic navigation tools are just that, tools, every navigator should be familiar with everything in there navigation "toolbox" and have as many options as possible

[Andrew Troup]

Here's the CAD plot of the constant heading passage – one advantage of doing it in CAD as opposed to simply graphing in Excel is the ability to attach dimensions.

Distances are in nautical miles, angles in degrees.

I did have to fudge the headings slightly: the CAD program fitted a smooth curve to the points, but it's an approximation: the tangency values are out by a few tenths of a degree from the true values tabulated in the spreadsheet.

The even balance of the upstream vs downstream curves, E vs W, might surprise many.

A even simpler explanation for this aspect, which might work for those who haven't the patience for my previous efforts, goes like this:

A constant heading passage means that we angle upstream to the minimum degree sufficient to offset the average tide. It stands to reason that, half the time, the tide will have the upper hand.

If we fantasise a track which keeps us mostly above the rhumb line, we're approaching Garrison Keiler's mythical community of Lake Wobegon, where
"all the children are above average."
　
I've explained earlier in more detail how I managed to infer the probable shape, but it was still quite a shock to see how perfectly balanced it is about the horizontal axis, as the midchannel heading angle shows.

I remind you that the current is twice as strong midchannel as at the quarter points, and I half expected the heading midstream to be steeper than at start and finish points, as a result.

I guess the key is that the current INCREMENT from quarter to half way across is the same as the increment during the first quarter of the trip.

The only parallel I can think of is another very surprising mathematical truth: if you cut the top off a cone, the cut shape, viewed square on, will be a circle, provided you cut it "squarely" (no surprises yet)

However if you tilt the saw, the cut shape will be a perfect ellipse, not (as I would expect) a lopsided egg-shape, bigger at one end.

This result still has the capacity to surprise me every time I think of it. I'm in awe of the ancient Greek geometer who worked out that this would be true, before he ever tried it.
(Many Greeks of his standing hated practical real-world approaches to problem solving; even making scale drawings was generally looked down on – it may even be true that he never felt the need to try it.)

- - - - -

CaptForce's post showed me something else I never knew: I always thought that saving face was incompatible with admitting a mistake: this post was an eyeopener, a very classy way of achieving both ends, with panache and humour. Kudos, and thanks again for the initial revelation about track.

One final thought: it seems to me that reliance on one intellectual tool (eg shortest line through the water) as a proxy for reality, has similar dangers to relying on one technological tool (eg GPS)

[Seaworthy Lass]

Originally Posted by Andrew Troup

Here's the CAD plot of the constant heading passage – one advantage of doing it in CAD as opposed to simply graphing in Excel is the ability to attach dimensions.

Distances are in nautical miles, angles in degrees.

I did have to fudge the headings slightly: the CAD program fitted a smooth curve to the points, but it's an approximation: the tangency values are out by a few tenths of a degree from the true values tabulated in the spreadsheet.

The even balance of the upstream vs downstream curves, E vs W, might surprise many.

A even simpler explanation for this aspect, which might work for those who haven't the patience for my previous efforts, goes like this:

A constant heading passage means that we angle upstream to the minimum degree sufficient to offset the average tide. It stands to reason that, half the time, the tide will have the upper hand.

If we fantasise a track which keeps us mostly above the rhumb line, we're approaching Garrison Keiler's mythical community of Lake Wobegon, where
"all the children are above average."
　
I've explained earlier in more detail how I managed to infer the probable shape, but it was still quite a shock to see how perfectly balanced it is about the horizontal axis, as the midchannel heading angle shows.

I remind you that the current is twice as strong midchannel as at the quarter points, and I half expected the heading midstream to be steeper than at start and finish points, as a result.

I guess the key is that the current INCREMENT from quarter to half way across is the same as the increment during the first quarter of the trip.

The only parallel I can think of is another very surprising mathematical truth: if you cut the top off a cone, the cut shape, viewed square on, will be a circle, provided you cut it "squarely" (no surprises yet)

However if you tilt the saw, the cut shape will be a perfect ellipse, not (as I would expect) a lopsided egg-shape, bigger at one end.

This result still has the capacity to surprise me every time I think of it. I'm in awe of the ancient Greek geometer who worked out that this would be true, before he ever tried it.
(Many Greeks of his standing hated practical real-world approaches to problem solving; even making scale drawings was generally looked down on – it may even be true that he never felt the need to try it.)

Absoltely brilliant Andrew! Well done .

Originally Posted by Andrew Troup

CaptForce's post showed me something else I never knew: I always thought that saving face was incompatible with admitting a mistake: this post was an eyeopener, a very classy way of achieving both ends, with panache and humour. Kudos, and thanks again for the initial revelation about track.

+1 . It was a real eye opener!

Originally Posted by Andrew Troup

One final thought: it seems to me that reliance on one intellectual tool (eg shortest line through the water) as a proxy for reality, has similar dangers to relying on one technological tool (eg GPS)

Yes, I don't know if the RYA teach that you should plot your position roughly for every hour of a journey if there is any current and you are using a constant course to steer to get there. If not, they should be, as it is essential to know if there are any dangers in the way. The actual track on the chart that the boat will take may be totally counter intuitive as we have demonstrated so well here.

[Seaworthy Lass]

Our brains must really work overtime when sleeping LOL. I woke up after a lovely three hour nap on the settee and an inspiration hit me regarding a new method for the RYA to teach for "course to steer".

More pennies dropped when the light went on . Hope the noise did not wake my better half! I have just scrambled on my hands and knees frantically trying to find where all of them rolled as it is not good in an aluminium boat if any of them escape and sneak into the bilge LOL.

OK, so here goes:

NEW METHOD FOR CALCULATING "COURSE TO STEER":

The whole problem with the current (excuse the pun) RYA method is that it does not arc off the 'distance travelled' vector to the arrival point B. In fact it stresses not to do this (and rightly so if you are only considering a fraction of the time for the total journey).

So proceed as you would normally in the existing RYA method, but repeat the process for yet another hour, extending the course line past B. Hopefully you are now arcing off past B, but if the current is very adverse, you may need to repeat again for yet another hour. It is not impossible you may need to repeat this again (if so reconsider your departure time and go when the current is more favourable).

Finally when you can arc off a distance vector past B, you must work out the proportion of the final hour it would have taken to get to B instead of past it and mark off the same proportion on the final current vector. This point is now C. Join C to B. This is your true course to steer. The length of CB divided by your boat speed is the time taken for the entire journey. Or alternatively you can also work out the time easily by adding together the number of whole hours plus the proportion of the final hour it takes to arrive at B (if the current for the last bit of the journey is not zero, this may be a better way of working out time taken, as less measurement errors are introduced).

Make your correction for magnetic compass deviation and voila, you have an exact course to steer to get you exactly to your destination and you can also calculate the time taken for the journey exactly. Of course this is limited by all sorts of factors like your speed varying, the current not being as predicted, leeway, etc, but at least it is now an excellent starting point.

In our hovercraft example the last current vector was zero, so the next whole hour would have taken D past B. As the current for this final hour is zero, there is no need to work out what proportion of time we are subjected to it and the end of the first four current vectors is therefore C so we simply connect this with B.

So for the hovercraft example the length of AC is 10 miles for the journey. It happens to be perpendicular to AB, but the beauty of the RYA method is that it easily allows for the summation of complex currents from any direction.
The length of CB in this case is 18.87 miles.
The time taken is 18.87/4 = 4.72 hours.
The course to steer is 58 degrees true if you measure it with the protractor (the original heading was 90 and allowance for current is 32 degrees). This gives the correct course to steer for the journey (instead of being nearly 7 degrees off as it is with the current method the RYA are teaching).

The brilliant thing is that if a component of the average current is with you for the first bit of the journey, D is already past B, so you modify the last hour as above. By the way, how does the RYA cope with this situation at the moment? It seems to me their method will not work if the current has a component significantly with you for the journey. This problem is taken care of with the new method

Any problems anyone can see with my method? I would love any feedback.

If there are no problems anyone can see (I have been known to have really stupid brain waves past midnight LOL) the RYA should consider teaching this new "Cruising Forum Method" instead of the old one .
Modestly prevents me from suggesting they call it the "Seaworthy Lass Method" .

[Jammer Six]

So far, after reading and following, I've read one more thing to take away: the reason I would want to plot my course out over the ground is to avoid those pieces of ground that may wish to interfere with flirting with the bow babe.

Avoiding obstinate pieces of rock and errant, misinformed ground is an excellent reason, in my (new) opinion, and worth lots and lots of work.

[Seaworthy Lass]

Originally Posted by Jammer Six

So far, after reading and following, I've read one more thing to take away: the reason I would want to plot my course out over the ground is to avoid those pieces of ground that may wish to interfere with flirting with the bow babe.

Avoiding obstinate pieces of rock and errant, misinformed ground is an excellent reason, in my (new) opinion, and worth lots and lots of work.

100% correct. This is a huge thing that I have gained from this thread, as I instinctively imagined the course over ground to be totally diiferent in the "current going all one way, but mirror imaged in strength over distance" example that CaptForce gave.

It was a real shock for me realising the actual 'course over ground' is a curve which crosses the midline at the halfway mark, so that you end up spending half your time downcurrent. That is so totally counterintuitive to me (and obviously to several other people who have been posting on this thread).

So learning how to plot your course over ground (I suggest you mark it hourly) if you are planning to use the "single course to steer" route for a journey (which could shave hours off a long passage) is very important if you dont want to run into stuff.

[Seaworthy Lass]

OK folks, it is now an hour since I posted my
NEW METHOD FOR CALCULATING "COURSE TO STEER"

No replies about it yet and since it is nearly 3:30 am I am off to bed.

Hope to have lots of replies to read when I wake up .

Night all.

[Dockhead]

Originally Posted by Andrew Troup

Here's the CAD plot of the constant heading passage – one advantage of doing it in CAD as opposed to simply graphing in Excel is the ability to attach dimensions.

Distances are in nautical miles, angles in degrees.

I did have to fudge the headings slightly: the CAD program fitted a smooth curve to the points, but it's an approximation: the tangency values are out by a few tenths of a degree from the true values tabulated in the spreadsheet.

The even balance of the upstream vs downstream curves, E vs W, might surprise many.

A even simpler explanation for this aspect, which might work for those who haven't the patience for my previous efforts, goes like this:

A constant heading passage means that we angle upstream to the minimum degree sufficient to offset the average tide. It stands to reason that, half the time, the tide will have the upper hand.

If we fantasise a track which keeps us mostly above the rhumb line, we're approaching Garrison Keiler's mythical community of Lake Wobegon, where
"all the children are above average."
　
I've explained earlier in more detail how I managed to infer the probable shape, but it was still quite a shock to see how perfectly balanced it is about the horizontal axis, as the midchannel heading angle shows.

I remind you that the current is twice as strong midchannel as at the quarter points, and I half expected the heading midstream to be steeper than at start and finish points, as a result.

I guess the key is that the current INCREMENT from quarter to half way across is the same as the increment during the first quarter of the trip.

The only parallel I can think of is another very surprising mathematical truth: if you cut the top off a cone, the cut shape, viewed square on, will be a circle, provided you cut it "squarely" (no surprises yet)

However if you tilt the saw, the cut shape will be a perfect ellipse, not (as I would expect) a lopsided egg-shape, bigger at one end.

This result still has the capacity to surprise me every time I think of it. I'm in awe of the ancient Greek geometer who worked out that this would be true, before he ever tried it.
(Many Greeks of his standing hated practical real-world approaches to problem solving; even making scale drawings was generally looked down on – it may even be true that he never felt the need to try it.)

- - - - -

CaptForce's post showed me something else I never knew: I always thought that saving face was incompatible with admitting a mistake: this post was an eyeopener, a very classy way of achieving both ends, with panache and humour. Kudos, and thanks again for the initial revelation about track.

One final thought: it seems to me that reliance on one intellectual tool (eg shortest line through the water) as a proxy for reality, has similar dangers to relying on one technological tool (eg GPS)

Fantastic, Andrew!

That is really useful, and something completely new I've learned, thanks to you and Capt Force.

Concerning straight lines through water - I think this is the key insight for understanding the problem as originally presented. It's not just an explanation, or method of visualization, but the reason why constant heading paths are faster.

It's not enough, however, as we now know, to understand all aspects of the problem. It tells us nothing whatsoever about the ground track!

Andrew, maybe you could check my last spreadsheet with your CAD program? Maybe everyone else is bored with it, but I for one would really like to know that we've done all the maths correctly.

[Seaworthy Lass]

Originally Posted by Seaworthy Lass

NEW METHOD FOR CALCULATING "COURSE TO STEER":

The whole problem with the current (excuse the pun) RYA method is that it does not arc off the 'distance travelled' vector to the arrival point B. In fact it stresses not to do this (and rightly so if you are only considering a fraction of the time for the total journey).

So proceed as you would normally in the existing RYA method, but repeat the process for yet another hour, extending the course line past B. Hopefully you are now arcing off past B, but if the current is very adverse, you may need to repeat again for yet another hour. It is not impossible you may need to repeat this again (if so reconsider your departure time and go when the current is more favourable).

Finally when you can arc off a distance vector past B, you must work out the proportion of the final hour it would have taken to get to B instead of past it and mark off the same proportion on the final current vector. This point is now C. Join C to B. This is your true course to steer. The length of CB divided by your boat speed is the time taken for the entire journey. Or alternatively you can also work out the time easily by adding together the number of whole hours plus the proportion of the final hour it takes to arrive at B (if the current for the last bit of the journey is not zero, this may be a better way of working out time taken, as less measurement errors are introduced). Ignore this last paragraph, see my explanation below for a quicker easier technique.

Make your correction for magnetic compass deviation and voila, you have an exact course to steer to get you exactly to your destination and you can also calculate the time taken for the journey exactly. Of course this is limited by all sorts of factors like your speed varying, the current not being as predicted, leeway, etc, but at least it is now an excellent starting point.

...............

The brilliant thing is that if a component of the average current is with you for the first bit of the journey, D is already past B, so you modify the last hour as above.

By the way, how does the RYA cope with this situation at the moment? It seems to me their method will not work if the current has a component significantly with you for the journey. This problem is taken care of with the new method Ignore this last paragraph, I will explain below why this situation works.

What? Did I get up for nothing?
Five hours later and one one person has chimed in LOL.
Is my solution so stunning it has left you all speechless

Morning here, more sleep more brain waves, oops, wait need to pick up a few more pennies . No I haven't dreamed of vectors LOL, but the minute my eyes popped open the thought hit me (as it did after my nap last night).

There is no need to actually calculate the proportion of time subjected to the last lot of current at all (so no need for any calculators). That last vector of distance travelled connecting the end of the current displacement vectors to D (where D is past B on the course line) already gives you the course to steer, but D must be past B (or at B) for this to work.

If you want time taken for the journey, just draw line parallel to this last vector of distance travelled which passes through B and the line for the distance displaced by the last lot of current. The point where it hits that line is where you label C.

This is the key to the new method: (OK, I admit I got a bit carried away describing it as 'new', it is only a modification to the old method, but is is an extremely important modification)
You must keep drawing hourly current displacement lines and trying to arc off the distance travelled vector on the course line until for the very first time you can get it to arc PAST B or exactly AT B.

The problem with the current RYA method is that they stop the process before D is longer than AB (or at least equal to it).

This is of course why the RYA method currently works if the current sum actually pushes the boat towards B and D ends up behind B on the AB line (sorry RYA - I thought your method would not work in this case, I asked twice about that here hoping for a reply, but no one explained why I was wrong).

I will make myself a cuppa now, then I will draw out my explanation as I suspect I am not explaining it well and that is why no one is responding.

[Seaworthy Lass]

Gas ran out so no tea or coffee. At least we have another gas bottle on board, but I need to wait for it to get connected (my hubbie's job) .

Come on folks, please respond. Don't be worried about making a mistake and looking foolish. If you drop that fear we can bounce ideas back and forth - brainstorming produces wonderful results .

I will draw out the hovercraft example on a chart to show you what I mean and then maybe I will make up another example and plot it out for you too. It may make sense then.