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[ksurf] Re: KitePhysics 102

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  • Ken Winner
    From: ... heading Kite ... ...used as a sail to ... I guess their ... In that case, this statement from kitephysics 102:
    Message 1 of 5 , Sep 30 8:12 AM
      From: <KiteBoard@...>
      > The . . . quote was from about the 5th paragraph under the
      heading "Kite
      > Sailing". The second line under that heading states
      "...used as a sail to
      > propel a small land-sailing or ice-sailing vehicle..." so
      I guess their
      > assumptions are based on those vehicles speed polars.


      In that case, this statement from kitephysics 102:

      << "When kite-sailing a beam reach -- which happens to be
      the course where
      apparent-wind speed and pull are at their highest possible
      levels while
      sailing -- you must careful about where on the flight
      envelope you dare fly."

      is simply wrong. If you look at the speed polar for an
      iceboat, you'll see that increases in boat speed below a
      beam reach give higher apparent wind and thus more pull than
      the boat experiences on a beam reach.

      Having found such a big error on this page, I would have to
      read it carefully before taking anything else there
      seriously.

      KW
    • KiteBoard@aol.com
      In a message dated 99-09-30 11:10:06 EDT, kw@kenwinner.com writes:
      Message 2 of 5 , Sep 30 9:53 AM
        In a message dated 99-09-30 11:10:06 EDT, kw@... writes:

        << In that case, this statement from kitephysics 102:

        "...small land-sailing or ice-sailing vehicle...<snip> When kite-sailing a
        beam reach -- which happens to be the course where apparent-wind speed and
        pull are at their highest possible levels while sailing ...<snip>"

        is simply wrong. If you look at the speed polar for an iceboat, you'll see
        that increases in boat speed below a beam reach give higher apparent wind and
        thus more pull than the boat experiences on a beam reach.

        Having found such a big error on this page, I would have to read it carefully
        before taking anything else there seriously.

        KW >>

        I would also tend to not take them seriously if that is in fact wrong, but if
        it IS wrong, it isn't so simple: When travelling below a beam reach at
        multiple wind speed, VMG downwind is pretty high relative to true wind speed.
        This at least somewhat reduces the effect of true wind on the apparent wind,
        possibly as much as the increase caused by the increased boat speed. In
        other words, just because boatspeed is higher, it doesn't automatically
        increase apparent wind speed if your course is downwind to any degree.

        Tom
      • Ken Winner
        From: ... fact wrong, but if ... beam reach at ... to true wind speed. ... the apparent wind, ... boat speed. In ... automatically ... any
        Message 3 of 5 , Sep 30 10:27 AM
          From: <KiteBoard@...>
          > I would also tend to not take them seriously if that is in
          fact wrong, but if
          > it IS wrong, it isn't so simple: When travelling below a
          beam reach at
          > multiple wind speed, VMG downwind is pretty high relative
          to true wind speed.
          > This at least somewhat reduces the effect of true wind on
          the apparent wind,
          > possibly as much as the increase caused by the increased
          boat speed. In
          > other words, just because boatspeed is higher, it doesn't
          automatically
          > increase apparent wind speed if your course is downwind to
          any degree.

          It's true that the angle can subtract from apparent wind
          velocity, but most often and for fast craft, that doesn't
          happen until the craft is sailing pretty broad -- at least
          below 100 degrees.

          The attached sketch doesn't reflect the speed polar of some
          craft in particular, and is actually a bit of an
          exageration, but it illustrates the idea. By the way, it's
          not there for your edification, as I'm sure you know what
          I'm talking about, but for anyone who hasn't been following
          this carefully (and I wouldn't have attached it were it not
          quite small).

          KW
        • Dave Culp
          ... I know you know what you are talking about, Ken, but your drawing is inaccurate. If one draws the remaining two bounding lines, from the ends of the
          Message 4 of 5 , Sep 30 11:28 PM
            >The attached sketch doesn't reflect the speed polar of some
            >craft in particular, and is actually a bit of an
            >exageration, but it illustrates the idea. By the way, it's
            >not there for your edification, as I'm sure you know what
            >I'm talking about, but for anyone who hasn't been following
            >this carefully (and I wouldn't have attached it were it not
            >quite small).

            I know you know what you are talking about, Ken, but your drawing is inaccurate. If one draws the remaining two "bounding lines," from the ends of the vectors shown, they must define a parallelogram, but your drawing doesn't. As drawn, the vector geometry won't work.

            While it's true that apparent wind is greatest, for very high Vb/Vt sailcraft, on broad reaching courses, this is only due to the great craft speed, as you mention. However, the apparent wind will almost always be *less* than craft speed, on any course much over 100 degrees true, and certainly on the craft's fastest course, which will be nearer 115-120 degrees. This is true of all sailcraft, fast or slow.

            FWIW, for a given, fixed craft speed, apparent wind will be highest on close reaching (windward) courses, not broad reaching.

            Dave
          • Ken Winner
            From: Dave Culp ... drawing is inaccurate. If one draws the remaining two bounding lines, from the ends of the vectors shown, they must
            Message 5 of 5 , Oct 1, 1999
              From: Dave Culp <dave@...>
              > I know you know what you are talking about, Ken, but your
              drawing is inaccurate. If one draws the remaining two
              "bounding lines," from the ends of the vectors shown, they
              must define a parallelogram, but your drawing doesn't.

              Yeh, you're right about the sketch.

              KW
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