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Old 20-11-2013, 02:34   #661
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

Quote:
Originally Posted by Paul Elliott View Post
WHAT current-related hydrodynamic forces? There are none, unless you are merely talking about there being a down-current vector to your speed/course over ground, or something about the True Wind being affected by the current. The boat is sailing through the water. There is no additional current-related force on the hull. Please explain what you are talking about.
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What a daft statement
why would we work in current wind effect if there was not
the problem is how I have seen if from the start
your are turning a current into wind and that is inaccurate because of the true wind being an equation of hydrodymaic and aerodynamic forces which unless referenced to gps are not real.
the real affect is the current moving past your hull with a seabed fixed point and a ground fixed a wind point, after all this is the true affect the hull reacts to, we are so ingrained at working courses to steer and VMG that sailers look at it the other way because the variables are so complex
As I have said a CTS or actually DR which what you are all doing (are inherently inaccurate as we all know) and you have to be amazingly accurate to work of the effects of current on the lee bow
I think we are getting there !
Hoof, Paul is correct. His statements are not daft. Quite the contrary.

Imagining the current flowing past your hull is your great stumbling point.
In a constant current the boat is moving with the current, at the same speed as the current. There is no flow over the foils created by the current. The log will read zero unless there are other forces acting on the boat.

Another analogy would be a balloon pilot who feels no wind in a steady wind even if the wind is propelling him rapidly across the earth (think of world record attempts in the jet stream)

It is a difficult concept to grasp because our normal frame of reference is the earth's surface. It is difficult to let go of this frame of reference and and make the mental leap to a water frame of reference.

If you think there is flow of water over the hull in steady current, then logically there will be some action on the log. If you have no sails up and you are drifting with the current what do you think the log reads? ZERO. There is no flow of water over the hull.

BUT, your wind instruments would register the apparent wind. If the ground wind was zero, the wind instruments would show a wind of the same amount as the current and opposite in direction to the way the current was moving.

Dave (Goboatingnow), it would really help if you thought about these last few paragraphs as well. You are writing a lot of nonsense (apart from misinterpreting much of what I have said). This is said kindly, not harshly .
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Old 20-11-2013, 03:52   #662
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Quote:
Originally Posted by Paul Elliott View Post

Dave, you've said this many times before (the Ground Wind being generated in part by the current). It was wrong the first time, and it's always going to be wrong. Ground Wind is the wind measured by a device attached to the ground. Current has absolutely no effect on Ground Wind.

Combine the Ground Wind and the current vector and you get True Wind (water-referenced). These are the *only* factors in True Wind. Boat motion or heading have no affect on True Wind.

On your boat you directly measure Apparent Wind. This is True Wind combined with the boat motion vector. If you like, you can also work from Apparent Wind to Ground Wind (or the other direction) by factoring in current and boat motion.

Can we agree on these terms? I'm having a very difficult time understanding your diagrams, and this might be one of the reasons.
I referenced all vectors to ground. As you cannot draw vectors referenced to different reference points.

Firstly

I have shown the wind generated by current pushing the boat through the air as a ground referenced vector opposite to the current direction.

Secondly

Forget True wind as mariners call it. There is only a set of ground winds and boat movement winds

Now lets restate.

a boat sails in apparent wind that's it nothing else. It does not experience true wind ground wind my wind or your wind. It experiences apparent wind , the sailing angle of that wind relative to the boat is the determining factor which is related to the direction of ground wind and boat heading.

Thirdly

Dock head continuously mixes up an issue.

Current can be represented by a " ground wind vector " and that is what I said when I mentioned current wind.

So when sailing in a current you have ground wind that represents the sum of current and earth ( ground wind ) this wind is fixed in direction for the duration of that current

You will notice if my diagram that even though the boat heading vector changed the current wind and ground wind do not change they remain constant on the vector diagram.

What changes when you change heading is the apparent wind. What I show is that the current wind cause an increase in apparent wind angle against the situation where no current exists.

This improvement continues to exist when the helm comes up to weather over the situation where he cones up to weather without the current . This is maximised when current is on the bow. The effect of current wind also continues round the compass

So yes the effect should be called bow current effect , rather then the lee bow effect , but the effect is still evident in the particular case of the current ending up fine on the lee bow, hence the moniker

It has nothing to do CTS. It has to do with apparent wind angles , apparent wind is made up of all ground wind effects , ( current as represented by a wind in opposition ) , boating heading vector as expressed by vector wind in opposition to boat heading

Just forget the STW true wind. It plays no part in anything. it's merely an intermediate vector sum of " certain " vectors. The boat never ever experiences it.

You cannot look at the bow effect of current unless you resolve everything apparent , because its a sailing angle issue.


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Old 20-11-2013, 04:02   #663
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BUT, your wind instruments would register the apparent wind. If the ground wind was zero, the wind instruments would show a wind of the same amount as the current and opposite in direction to the way the current was moving.

Dave (Goboatingnow), it would really help if you thought about these last few paragraphs as well. You are writing a lot of nonsense (apart from misinterpreting much of what I have said). This is said kindly, not harshly .
Seaworthy you have just shown the existence of a current generated wind referenced to ground

Do this excercise

Take a ground wind direction , take a current set and drift vector , assume a given boat speed and direction

Now please compute the wind vector the crew actually feel and the one the boat actually sails in

Please draw this for me and we can proceed step by step from there.


( by the way do you accept The fact that current on the bow improves apparent wind angle )

( said in a nice way )

The problem with this discussion is people switching from water references to ground references willy nilly

And Dockhead. The issue I'm discussing has nothing to do with CTS or navigation.

Ts merely about improvements in apparent wind angle. You are confusing attempting to resolve progress to a mark. ( I don't care why the Helmans wanted to come up to weather , maybe he wanted a closer look at the babe sailing the boat upwind from him ). You are mixing in navigation ( lee bowing the tide ) with the " bow current effect " which is purely about sailing wind angles.

Please even if seaworthy doesn't draw the vectors making up apparent wind. Someone do it. No point me doing it.

PS. We have to draw diagrams. I have, anyone on this post needs to draw not write, because English is a bad substitute maths

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Old 20-11-2013, 06:43   #664
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

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Originally Posted by Seaworthy Lass View Post
Hoof, Paul is correct. His statements are not daft. Quite the contrary.

Imagining the current flowing past your hull is your great stumbling point.
In a constant current the boat is moving with the current, at the same speed as the current. There is no flow over the foils created by the current. The log will read zero unless there are other forces acting on the boat.

Another analogy would be a balloon pilot who feels no wind in a steady wind even if the wind is propelling him rapidly across the earth (think of world record attempts in the jet stream)

It is a difficult concept to grasp because our normal frame of reference is the earth's surface. It is difficult to let go of this frame of reference and and make the mental leap to a water frame of reference.

If you think there is flow of water over the hull in steady current, then logically there will be some action on the log. If you have no sails up and you are drifting with the current what do you think the log reads? ZERO. There is no flow of water over the hull.

BUT, your wind instruments would register the apparent wind. If the ground wind was zero, the wind instruments would show a wind of the same amount as the current and opposite in direction to the way the current was moving.

Dave (Goboatingnow), it would really help if you thought about these last few paragraphs as well. You are writing a lot of nonsense (apart from misinterpreting much of what I have said). This is said kindly, not harshly .
Your examples show you do not apply physics

simply put ..... your sails are not filled when drifting we are talking about movement ..... linear velocity

I do not know how to change your perception of the real world away from a the maths involved for a DR or Cts
As much as you wish to change mine
I agree that in reality CTS or DR is the best method to anylis how to get from A to B
but this cannot be used in working out such fine changes in direction or speed
Linear Velocity Dependence

For objects moving at relatively low speeds through a liquid, where turbulence is not a significant factor, then the viscous resistance to the object's motion is approximately proportional to its velocity. Even in gases there are circumstances where the frictional resistance is approximately proportional to the velocity - such as the motion of tiny dist particles through the air. More commonly, air friction has terms proportional to the square or even higher powers of the velocity.
For linear velocity dependence the resistance force can be written
fresistance = -bv . If this resistance is the only force acting, then Newton's second law becomes
Newtons second law is the resistance we are interested in

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Old 20-11-2013, 08:45   #665
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

Hoofsmit and Dave
I will try and reply later tonight. We are hauling out tomorrow and needed to service two engines and pickle the water maker and I haven't finished with chores yet . Long day and I still have to get dinner on the table.

Hoof, just a quick response to what your posted. if you had no ground wind and a sufficient amount of current (and very lightweight sails) of course your sails would fill if you put them up. You can sail to this wind generated by current.

Also you still have not answered what your log would read if you were drifting in current with no ground wind. For example you have a 4 knot current and no ground wind. If your log impellor was clean, what do you think your instruments would display for boat speed? (The answer is zero, as the water is not moving over the hull).
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Old 20-11-2013, 09:28   #666
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

Here's something that I had started two days ago, but ran out of steam (and time). I have salvaged some (I hope) germane points. I am using a rather strong current just to make the numbers easy to work with, but the principles hold for any current. Please work with me here, and tell me what you think:

Quote:
Originally Posted by goboatingnow View Post
Current moves a boat with reference to the ground hence it's produces a ground referenced wind.

Dave
No. I'm going to describe a scenario that ties together (I hope) several of the issues we've been discussing. I'm *not* going to mention Lee Bow, but just the wind and the current. If we can't agree on how these work, there's not much point in arguing about the rest. Here we go:

Assume there is zero ground wind. This is our ground wind. Assume there is a ten knot current flowing south to north. Leeway is not being considered -- that comes later. Assume that your boat has no windage, for now we are sailing an "ideal" boat.

Your boat is tied to a mooring in this ten knots of current flowing south to north. Ground Wind is still zero. You measure an Apparent Wind of zero, but for what it's worth you do measure boat speed (through the water) of ten knots, and your rudder has some effect.

Now keep your sails stowed, but cast off from the mooring. You are now drifting north with the current. The Ground Wind as measured by the anemometer on shore is still zero. The True Wind (water-ref) is ten knots from the north, as is your Apparent Wind. Your speed through the water has now dropped to zero. Your rudder has no effect at all.

With me so far? Have I made any mistakes?

The Ground Wind is still, by definition and on-shore measurement, zero, but we have this 10kt True Wind. So, let's hoist the sails and go! Of course our destination had better be north of us!

We are now sailing. Say we've got a high-performance boat and can sail 75% of the True Wind Speed, except when the bow is within 45 degrees of the True Wind where our speed quickly drops to zero.

Point the boat due south. Remember, the True Wind is 10kts from the north. Our boat is sailing through the water, pointing south, at 7.5 kts. Our Apparent Wind is 2.5 kts on the bow. The current is still flowing north at 10 kts. Our SOG is 2.5 kts, and our COG is due north. Our boat is moving backwards over the ground, but we knew that was going to happen.

Now point the boat due east. True Wind is 10 kts from the north, our boat speed is 7.5 kts through the water. The Apparent Wind is 12.5 kts, and has moved forward (True Wind is at 90 degrees Port, and Apparent Wind is at 53.2 degrees Port). Our SOG is now 12.5 kts (which makes perfect sense since the Ground Wind remains at zero). Our COG is 036.8 degrees.

Do the math. Check my work. Can we agree that these are the facts? If so, we can then start analyzing the other "effects", on other points of sail, under different conditions.
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Old 20-11-2013, 10:34   #667
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When you head south the Apparent wind is 2.5 from the stern not the bow as it would be impressive to sail in an apparent wind dead on the bow !

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Old 20-11-2013, 10:57   #668
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Furthermore your use of the term true wind does not comply with this definition

"If you are motoring directly against a 5 knot current, however, then you have Apparent Wind of 10, True Wind of 0, and Ground Wind of 5. To calculate Ground Wind, your instruments take out the vector for SOG, rather than STW. Ground Wind is of no relevance to sailors unless we are sailing in a strong tidal current and are trying to anticipate what the wind will be like when the tide changes."[dockhead 22-12-11]

( caveat of course Dockhead could be wrong )

Since current wind is ground referenced , its a ground wind. Ie it modifies the real ground wind.


Remember all the vectors must remain referenced To the same plane.

I think water referenced vectors and ground referenced are being mixed up here

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Old 20-11-2013, 11:41   #669
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Hoofsmit and Dave I will try and reply later tonight. We are hauling out tomorrow and needed to service two engines and pickle the water maker and I haven't finished with chores yet . Long day and I still have to get dinner on the table. Hoof, just a quick response to what your posted. if you had no ground wind and a sufficient amount of current (and very lightweight sails) of course your sails would fill if you put them up. You can sail to this wind generated by current. Also you still have not answered what your log would read if you were drifting in current with no ground wind. For example you have a 4 knot current and no ground wind. If your log impellor was clean, what do you think your instruments would display for boat speed? (The answer is zero, as the water is not moving over the hull).
O dear ....... You need to do this when you have time : flowers x

Read your paragraphs
1 st point made
Ground wind .......tidal induced wind = 6 knots ( in theory 6 knots in practice nothing or at least not enough to keep you boom on one side, wonder why that is? Because you consider the current is inducing real wind, this might answer why your applied vectors are incorrect )
2.
Ground wind........ 4 knots of wind in my sails apparently !.. forward motion ..... In this example my log would read motion since "you can sail in that wind induced current" in reality nothing

I think this is what Dave is saying .
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Old 20-11-2013, 13:30   #670
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

the problem is when you start referencing ground ( i.e. compass) headings with water referenced headings

This is how complex it gets


To show how it can go wrong , Paul , put your resulting SOG and AWS into US sailing true ( i.e. ground wind calculator )


True and Apparent Wind Calculator

Note their idea of true is ground wind. but if your right you should derive your current influenced wind

i.e. using a SOG of 12.5, AWS of 12.5 and AWD of 53.2


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Old 20-11-2013, 20:15   #671
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

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When you head south the Apparent wind is 2.5 from the stern not the bow as it would be impressive to sail in an apparent wind dead on the bow !

Dave
Yes, my mistake. I had originally had the current flowing in the opposite direction, and I obviously missed that in the editing.
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Old 20-11-2013, 21:17   #672
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

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Furthermore your use of the term true wind does not comply with this definition

"If you are motoring directly against a 5 knot current, however, then you have Apparent Wind of 10, True Wind of 0, and Ground Wind of 5. To calculate Ground Wind, your instruments take out the vector for SOG, rather than STW. Ground Wind is of no relevance to sailors unless we are sailing in a strong tidal current and are trying to anticipate what the wind will be like when the tide changes."[dockhead 22-12-11]

( caveat of course Dockhead could be wrong )
Dockhead is correct, and so am I. In his example the ground Wind is 5kts (say it's from the north). The current is 5kts from north to south.

True Wind (water-referenced) is 0 kts, since the water is moving the same direction and speed as the Ground Wind.

If you then motor up-current at 10kts (through the water) your Apparent Wind is 10kts.

You can check this by looking at SOG (5 kts north) and Ground Wind (5 kts from the north). These add up to that 10 kts.

Quote:
Since current wind is ground referenced , its a ground wind. Ie it modifies the real ground wind.
Your repeating this does not make it true. It's not.


Quote:
Remember all the vectors must remain referenced To the same plane.

I think water referenced vectors and ground referenced are being mixed up here
I agree, but I think you're the one who is mixing them up.

I will check the rest of my math later tonight. Perhaps I made some more mistakes. If you like, I would appreciate a critique, with your results for the two scenarios I proposed.
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Old 20-11-2013, 23:07   #673
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

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the problem is when you start referencing ground ( i.e. compass) headings with water referenced headings

This is how complex it gets


To show how it can go wrong , Paul , put your resulting SOG and AWS into US sailing true ( i.e. ground wind calculator )


True and Apparent Wind Calculator

Note their idea of true is ground wind. but if your right you should derive your current influenced wind

i.e. using a SOG of 12.5, AWS of 12.5 and AWD of 53.2


DAve
Dave, where in the world did that diagram come from? Are any of those numbers or angles related to any of the examples we have been discussing? Certainly none of mine!

That U.S. Sailing calculator does not allow for current. If adjust the conditions to reflect zero current and 10 kts of Ground Wind then the calculator spits out the same numbers I provided.

You are confusing Ground Wind, and current-induced wind. They are different things.

Why don't you actually try calculating it my way, and see if you agree with the results. Or using my conditions calculate it your way (whatever that is) and present your results. That calculator isn't going to do the job for you, it's not designed for a case where there is current.

And of course they use Ground Wind. They are using SOG and ignoring current!
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Old 21-11-2013, 05:22   #674
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

Paul I put up that diagram to illustrate the complete ground based referenced vector system that shows all the relationships between every relevant vector

The answers once you add in ground wind, current wind, etc , all these vectors interfere with each other.

for example compute true wind , from STW while under the influence of current, is incorrectly computed on all instruments, because the drift issue is not handled.


But proceed. to the next step.

It would be better to draw vectors so we could see the thinking. The key to a robust solution is the process should be reversible. so if you plug the items computed on the boat , AWS, COG.SOG, STW, AWA, you should be able to recreate inputs.

PS ground wind and current wind, in effect combine to provide a composite Ground wind. since current wind is a ground referenced valued as is ground wind, the resulting vector is a ground referenced vector. ( which makes adding a water referenced vector,such as STW interesting) The results are a series of interrelated non right angle triangles. which only in certain cases reduce to simple right angle triangles.

Part of the problem is the mariners definition of "True Wind", can you state what you regard as "True wind" and how does that resolve to Dockheads definition

But continue to the next step.



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Old 21-11-2013, 05:28   #675
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Re: Let's Get to the Bottom of the Lee-Bow Effect Once and For All

Quote:
Dockhead is correct, and so am I. In his example the ground Wind is 5kts (say it's from the north). The current is 5kts from north to south.

True Wind (water-referenced) is 0 kts, since the water is moving the same direction and speed as the Ground Wind.

If you then motor up-current at 10kts (through the water) your Apparent Wind is 10kts.

You can check this by looking at SOG (5 kts north) and Ground Wind (5 kts from the north). These add up to that 10 kts.
No he is motoring at 5kts against the current, not 10kts ( he never mentions that ) so rerun your logic above

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