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    Re: Formulas to Compute LHA
    From: Andrew Corl
    Date: 2009 Jun 27, 17:27 -0700
    I am interested in this diagram,  where can I find it?
     
    Andrew


    From: chempro <snav-coxn---.net>
    To: NavList <NavList@fer3.com>
    Sent: Saturday, June 27, 2009 11:00:31 AM
    Subject: [NavList 8849] Re: Formulas to Compute LHA


    Why dont you learn to use a TIME DIAGRAM? It simplfies ALL of the
    concepts.It can universally used for all reduction methods from Ageton
    to NASR. Chempro-Dr Will

    On Jun 24, 8:07 am, Gary LaPook <glap...---.net> wrote:
    > To see that you can use hour angle and LHA equally well, you just have
    > to look at H.O. 249. Look at the LHA columns at each edge of the page
    > and you will see that you get the same Hc for two values of LHA. Look at
    > LHA 10 and LHA 350 on this example page:
    >
    > http://www.fer3.com/arc/img/106576.2007-page%20152.pdf
    >
    > LHA 10 is the same as hour angle 10 west., LHA 350 is the same as hour
    > angle 10 east.
    >
    > This relationship is not so evident if you just look at H.O 229 due to
    > the arrangement of the tables.
    > gl
    >
    >
    >
    > Gary LaPook wrote:
    > > Remember, LHA is in the range of 0º to 360º and is always measured to
    > > the west from the assumed longitude (ALon). It is the angle measured
    > > westward from the meridian of the ALon to the meridian containing the
    > > body's grographic position (GP). There is no such thing as easterly LHA.
    >
    > > In the past, various computations methods and tables (e.g H.O. 214) used
    > > "hour angle," (abbreviated "H.A." or "H" or "t"), which is the angle
    > > measured between the meridian of the ALon and the meridian containing
    > > the body's GP measured the shortest way, either west or east. Using this
    > > notation, hour angle ends up in the range of 0º to 180º only and is
    > > denoted "east" or "west." Because of the the way the trig formulas work,
    > > using either method computes the same Hc and the same "azimuth angle"
    > > ("Az" or "Z"). The only thing affected by choice of notation is the
    > > method used for the final determination of Zn,(azimuth used for plotting
    > > the LOP.)
    >
    > > The original Bygrave used hour angle, not LHA, and the scales were
    > > marked from 0º to 90º and then back the other way, 90º to 180º. My
    > > implementation of the Bygrave eliminated the second set of markings on
    > > the scale, 90º-180º, to eliminate clutter so I added an extra step to
    > > bring hour angle into the range of 0º to 90º only and provided the
    > > necessary rules for the final computation of Zn. This is what is
    > > happening on the top of the form I provided. If LHA is less than 90º, H
    > > = LHA; if LHA is greater than 90º but less than 180º, H = 180º - LHA; if
    > > LHA is greater than 180º but less than 270º then H = LHA - 180º; and if
    > > LHA is greater than 270º but less than 360º, H = 360 º- LHA.
    > > Conceptually, this is the smallest angle measured from either the upper
    > > branch or from the lower branch of the observer's meridian to the
    > > meridian containing the body's GP.
    >
    > > See the revised form at:
    >
    > >http://www.fer3.com/arc/img/108719.revised%20form%206-18-09.pdf
    >
    > > The formulas for calculating LHA are:
    >
    > > If your AP is in west longitude: LHA = GHA - ALon. (If necessary, add
    > > 360 ºto GHA prior to subtracting ALon.)
    >
    > > If your AP is in east longitude: LHA = GHA + ALan. (if LHA then exceeds
    > > 360º, subtract 360º from the result.)
    >
    > > Using the first formula for your first two examples, GHA (55º + 360º) -
    > > 77º = LHA = 338º.
    >
    > > GHA 95º - 77º = LHA = 18º
    >
    > > The third example you bring up makes no sense since GHA is never
    > > measured to the east, it is always measured west from Greenwich.
    >
    > > gl
    >
    > > Andrew Corl wrote:
    >
    > >> All,
    >
    > >> I need some help.  I am attempting to work the problem in Ocean
    > >> Navigator using the Lapook-Bygrave Slide Rule.  I am uncertain how to
    > >> compute the Local Hour Angle (LHA).
    >
    > >> In the following cases I am assuming my longitude to be 77 degrees
    > >> west of Greenwich.
    >
    > >> 1. The GHA of the Sun is 55 degrees west of Greenwich
    > >> 2. The GHA of the Sun is 95 degrees west of Greenwich
    >
    > >> In the following cases I am assuming my longitude to be 120 degrees
    > >> west of Greenwich
    >
    > >> 1. The GHA of the sun is 170 degrees east of Greenwich.
    >
    > >> Also what is the formula if my position is east of Greenwich and the
    > >> sun has a GHA of more than 180 degrees?
    >
    > >> Thanks
    > >> Andrew- Hide quoted text -
    >
    > - Show quoted text -




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