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    Re: Spherical triangle split by right triangles
    From: Bill B
    Date: 2016 Jan 22, 15:01 -0500

    If I understand your quest, Ageton's H.O. 211 tables are based on
    dividing the triangle into two right triangles. Construction line "R" is
    through the sun's GP and perpendicular to the observer's meridian.
    Angles and legs of the two right triangles can be calculated using GP,
    LHA, latitude/etc.from the original triangle.
    On 1/22/2016 2:02 PM, Mark Coady wrote:
    > Ok, I’m feeling stupid again.
    > While I’ve pretty much used spherical law of cosines happily enough for
    > my purposes, my introduction to HO208 and other methods of Nav triangle
    > calculation had me attempt the split of  an obliquesphericaltriangle
    > into two Right spherical triangles for an experimental calculation  on a
    > circle route.So I sketched out a simple great circle route with uneven
    > legs.Then debated how I would know where to split it. I know the
    > included angle for the difference in Longitude,I know the
    > Latitude/co-latitude of my start (last fix) and of my destination. Now
    > if it were isosceles...ok..midpoint of base/ split the vertex and away
    > we go....but oblique.....I kept googling up the question various ways
    > with everything coming back that you just draw a perpendicular to the
    > line opposite my vertex.   Yup...ok..makes perfect sense.....thats
    > exactly what last math site said...BUT HOW ??
    > Its apparently so easy..nobody tells you ...so I feel dumb, but I just
    > don’t see ...
    >   Geometric construction...ok...but.....What if I don’t want to guess
    > drawing a circle course curved line perpendicular intersection on a flat
    > chart....or don't have a plot and just  want math. Mathematically how do
    > I proportion the values of the Difference  in longitude angle and course
    > line?
    > So I can then solve for the other unknowns with uncle napier?
    > I feel like I am missing one more number I need to calculate using
    > napier’s normal cos & tan the rules.
    > I have the colatitude side from the N or S pole,  opposite the 90 degree
    > angle for each of the two split triangles, but I don’t know the vertex
    > split, don’t know intersection distance from the start or end point of
    > the perpendicular?.
    > is there a good explanation somewhere?   I looked up Napier's agian, but
    > the relationships do not prescribe solutions with the 90 degree angle
    > and only one side.
    > What am I missing...... ? besides the obvious....
    > View and reply to this message

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