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    Re: Wind & Current Navigation
    From: George Huxtable
    Date: 2003 Apr 17, 15:16 +0100

    Dan Allen has pointed us away from astronomy, and refreshed us with an
    interesting question about wind-forces and waterflow forces on a craft.
    Taking a simple situation: consider the horizontal drag on a spherical
    buoy, anchored in the water such as to be completely submerged, in a
    tideway. Compare it with the windforce in that same buoy if it is dangled
    in the air on a string from the vessel's rigging, in a breeze.
    At the same rate of flow, the water force would be about 830 times greater,
    because water is denser than air by that factor. The force will vary as the
    square of the velocity, so the force due to a 1 knot tide on the buoy would
    be equivalent to the force of a breeze of nearly 30 knots (28 being the
    square root of 830, roughly). That gives an idea of the order-of-magnitude
    of differences between water forces and wind forces.
    It's more complex than that, of course. The air drag on Dan's boat will be
    roughly proportional to the above-water cross-sectional area of its hull.
    This will vary a bit (but not a lot, presuming sails aren't hoisted) with
    the way the boat is facing with respect to the wind.
    The waterflow forces will be to some extent proportional to the underwater
    area of the boat, but this  will be GREATLY dependent on the way the vessel
    is facing with respect to the tidal current. It's obvious, isn't it: the
    hull is deliberately shaped to allow it to slip through the water in a
    fore-and-aft direction, but to minimise leeway by offering a large
    resistance sideways. It's surprising how great the tidal forces on a hull
    can become, even at quite feeble currents.
    Manoevring a vessel in a restricted area, where cross-tide flows through
    obstacles, is always a tricky business. There's a marina in the beautiful
    Breton town of Treguier which is particularly bad in this respect, because
    the pontoons, with their finger-berths, poke out into the tidal River
    Trieux. It's always fun to watch the performance of the locals compared to
    the visitors. Particularly, those that know the river well are aware of the
    times when it's wisest to stay put and wait an hour or two for slacker
    It's all very well being wise after the event, but my strategy, in Dan's
    situation, would be to power the vessel slowly astern, without turning
    across tide, for as far as possible (space allowing) to get clear of
    down-tide obstructions, such as his dock and other craft: then, helm hard
    over, applying sufficient power to get enough way on quickly, to turn
    sharply across-tide.  The tide will immediately grab the keel and try to
    force the craft to the South, so that tendency must be counteracted by
    crabbing a bit to the North of East. This can only be done with some
    boat-velocity in hand: if you are caught facing across-tide and stalled in
    the water, big trouble will inevitably ensue! I am speaking without
    knowledge of the details of Dan's berth geometry, and how much clear space
    lies astern as he exits his berth, so my suggestions are somewhat
    tentative. The trouble is that if the boat fails to turn to port quickly
    enough (perhaps because its prop-walk is in the other direction) an
    unpleasant splintering noise may result, and the more momentum has
    accumulated, the worse that will be.
    Perhaps, with luck, there's another pontoon, or another craft, astern, as
    Dan reverses out, with a bollard on which he can drop a loop of line from
    his stern. If so, he can then relax, and slowly "walk" the attachment point
    of that doubled line from the stern of his craft to somewhere amidships on
    the port side. This will allow his boat to pivot around to port, before the
    line is slipped and he can set off, facing more-or-less in the right
    direction. The problem here is that as the boat pivots to a cross-tide
    position, the tidal forces on its keel can then become very great.
    Many craft, mine included, have a strong tendency, when not moving through
    the water, to "weathercock", and in my case the bow wishes to point
    downwind. This effect may have hindered Bill's efforts to turn.
    Finally, I think Dan's attempt to create a formula specifying the magnitude
    and direction of the net forces on his boat will fail, unless he can find a
    way to account for the big effect on the current forces of the orientation
    of the keel with respect to the waterflow. I doubt if ships' masters rely
    on such calculations when berthing, but more on instinct and experience.
    George Huxtable.
    Dan Allen said-
    >Today I went out on my boat, but leaving the slip became dicey.  This
    >experience leads to a couple of questions about determining the
    >cumulative effects of winds and currents.
    >My slip faces due south.  The 15 knot wind was coming from the SE.  I
    >released my lines and engaged reverse and pulled out of the slip, so
    >far so good.  As I entered the fairway and the boat began turning so
    >that the bow began to face East, I stayed relatively close to my slip,
    >reasoning that the wind would push me out more to the middle of the
    >fairway, and when I had turned enough, I would go forward and that
    >would be that.  This plan has worked fine in no wind, or in light
    >winds, or even winds this strong that I have encountered before.
    >However, the incoming invisible tidal current was coming from the
    >North.  It pushed me back towards the dock when I was starboard side to
    >and I narrowly missed hitting docks and boats.
    >The wind seemed to be controlling the situation.  The waves were all
    >heading NW due to the wind from the SE.
    >The current seemed negligible.  I knew that low-tide was just occurred
    >about 30 minutes earlier; the flood was beginning again, but I did not
    >think it had much strength compared to the wind, as I saw no sign of a
    >current pushing me back into my slip.
    >I was wrong!
    >The current was darn near invisible to me. The winds appeared to be all
    >powerful, but the quiet, silent, invisible currents got me again.
    >Let us say that we knew that the wind was 15 knots from SE and that the
    >current from the incoming tide was 2 knots from the North; are they
    >factored equally in how they combine to move the boat?  If they were
    >weighted equally the result would be a 13.66 knot force coming from 129
    >degrees, but this would not have blown me against the dock.  It still
    >would have blown me away from the dock, but this is not what happened.
    >A current from the North of 11 knots combined with a 15 knot wind from
    >the SE would give a total force of 10.6 knots coming 2 degrees North of
    >East.  This would have just barely begun to nudge me back toward the
    >dock, but the tidal current certainly was nowhere near 11 knots!  It
    >was more like 2 knots, perhaps 3 or 4 at the most.
    >This leads me to believe that the current needs to be weighted much
    >more than the wind, at least in my craft, but is this always the case?
    >How can you determine the relative strength of a current compared to
    >the wind and determine the outcome of the two?  I know how to
    >mathematically add two vectors, so it is not the actual math that I am
    >asking about, but rather, how empirically does one determine the
    >relative forces involved?
    >How does one weight the effects of wind and weight the effects of
    >How do the size, shape, and mass of the boat and the hull, and the
    >superstructure (masts, flying bridges, etc.) effect these relative
    >Let the vector "W" be the wind, and the vector "C" be the current.  Let
    >"s" be a factor for the amount of superstructure on a boat, which the
    >winds effect.  Let "h" be a variable representing the hull under the
    >waterline, which the currents effect.  It seems to me that there should
    >be some formula to determine the final speed and direction of the boat
    >under these two forces of Wind and Current, the result being the vector
    >B (for boat).  The formula would look something like:
    >       B = s * W + h * C
    >s and h obviously are determined by a variety of measures and need to
    >be further broken down.
    >Are there any rules of thumb that help one determine the sum of the
    >wind and current accurately?  How do master mariners figure this out?
    >What observations can be made to help predict the cumulative effects of
    >wind and current accurately?  In other words, how could I prevent such
    >near disasters in the future?
    contact George Huxtable by email at george@huxtable.u-net.com, by phone at
    01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
    Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.

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