International 14 Mailing List
Re: [I14] leeway angles = AOA

14

From: Ted Rogers (twrogers@earthlink.net)
Date: Mon 21 Apr 2003 - 05:03:55 BST

  • Next message: Grant Bourke: "Re: [I14] leeway angles = AOA"

    Guess that explains why trapeezing in light air doesn't work, especially if
    you're big.....but seriously, this looks like it assumes full load on the
    rig, even at low speeds through the water. Maybe a reasonable description
    of what happens when coming out of a tack in a good breeze, but not quite
    representative of straight line sailing in light air?

    Looks like the spreadsheet is using Lift = 1/2 * density * velocity^2 *
    Area * Lift Coefficient.

    If you approximate Lift Coefficient as = 0.1 * AOA in degrees, and ask "For
    a given lift, and velocity, what lift coefficient, and therefore what AOA,
    do I need?", I think you get the results shown.

    The flaw in the ointment (the fly in the logic?) is the assumption that
    lift needs to be constant for any velocity. The rig loads also go as
    velocity squared, so the lift required from the board goes up as the
    velocity increases. The only caveat is that it's possible to come out of a
    tack and jump out on the wire with very little boat speed, loading up the
    rig, which has already assumed some angle of attack, and is seeing some
    finite true wind speed, before the board is moving forward at any
    reasonable velocity. In that case, you slip sideways at a significant angle
    until the boat accelerates enough for the board to work.

    The fact that the rig is experiencing (true wind + boat speed) while the
    board only sees (boat speed) for flow velocity is also why we need less
    board when we're powered up and going fast, and why fast boats can get away
    with smaller foils generally.

    Ted Rogers
    US 1140

    At 10:35 AM 4/16/03 -0700, PETER MOHLER wrote:
    >leeway angles
    >
    >An aerodynamics pal of mine worked up a spreadsheet to
    >calculate leeway angles for a 14. (Leeway = Angle of
    >Attack, AOA)
    >
    >It has lots of variables, but if you assume the
    >daggerboard is symmetrical and has a chord of 12
    >inches and is 40 inches deep (30 X 100 cm) and that
    >two heavy dudes are on the trap (400 lbs total, (181
    >kg)); then the following AOA will result for the
    >indicated hull speed.
    >
    >13 kts n 1.3 deg
    >10 kts n 2.4 deg
    >7.5 kts n 4.3 deg
    >5 kts n 9.6 deg
    >3 kts n 26.6 deg
    >
    >
    >
    >Pete Mohler
    >USA 1150
    >Almost Lunchtime
    >
    >Thanks to Brooke Smith for the spread sheet
    >
    >
    >
    >
    >
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