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    Re: Non-graphical way to solve a wind triangle
    From: Frank Reed
    Date: 2020 May 11, 20:39 -0700

    Equations as posted by Greg:
    "CORRECTION ANGLE = ASIN(((DRIFT X SIN(SET-TC))/STW)
    SOG=SQRT((STW2 + DRIFT2) - (((2 X STW) X DRIFT) X COS(TC - SET + CORRECTION ANGLE)))"

    I was hoping someone else would deal with this, but no luck. I'll take a stab at it.

    First, I'm going to make one change in the internal terminology here and replace "CORRECTION ANGLE" by plain old 'A' (for angle, if you like). And just for notational aesthetics, I'll replace "big fat X" for multiply with a mid-dot: A X B becomes A · B. Next it's clear that Greg or his source have gone crazy with parentheses. Almost none of the parentheses pairs are necessary thanks to simple order of operations rules and commutativity of multiplication. Changes so far:

    A = asin[drift·sin(set - tc) / stw],
    sog = sqrt[stw2 + drift2 - 2·stw·drift·cos(tc - set + A)].

    Now even without getting into the underlying geometry of this, it's clear that the second equation is a representation of the plane trigonometry law of cosines:
       c2 = a2 + b2 - 2·a·b·cos θ.
    See it?

    And therefore those 2's following stw and drift and really exponents -- squares. So finally:

    A = asin[drift·sin(set - tc) / stw],
    sog = sqrt[stw2 + drift2 - 2·stw·drift·cos(tc - set + A)].

    Does that make sense? I'm just distilling equations and decoding symbols here and not deriving anything from principles, so it could all be wrong. As for the inputs and variable names, set and drift are obvious enough terms. And I suppose tc is "true course", stw is "spreed true wind" (or true wind speed), and sog is "speed over ground". Yes?
    Frank Reed
    Clockwork Mapping / ReedNavigation.com
    Conanicut Island USA
       
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