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    Re: Practice problem
    From: Joe Shields
    Date: 1999 Dec 07, 11:18 EST

    Does the answer in the back of the book show what they used as AP?
    Obviously intercept will be relative to AP used, although both  40.9 Nm away
    at 212.9 deg, and 10 nm towards at 211.1 deg should wind up on the same LOP.
    If you then used your DR position (employing a method like H.O. 211 that
    allows it) you would still wind up with the same LOP.  That's the whole
    point, no matter what your assumed position is, or your DR position is (or
    how accurate it is), you wind up with a line of position (LOP) that you know
    you are on (somewhere).   Catch a later shot of the sun, or use another
    celestial body to get another LOP, and where they cross that's where you are
    (or were in the case of a sun-run-sun shot where you retarded the later LOP
    by the distance and direction of the run).
    > ----------
    > From:         Chan, Sammy W[SMTP:Sammy.Chan@XXX.XXX]
    > Reply To:     Navigation Mailing List
    > Sent:         Tuesday, December 07, 1999 9:57 AM
    > Subject:      Practice problem
    > I am working on the book "Practical Celestial Navigation" by Susan P.
    > Howell
    > and I have a question about one of the exercises in the book.
    > Specifically, in exercise 7 problem 5, the altitude and azimuth calculated
    > for the Sun LL sight given by the answer in the back of the book
    > disagrees with my calculation. However, if I recalculate it with an
    > assumed latitude one degree higher, I get the answer in the book.
    > What am I doing wrong?  Here is the problem:
    > The 1500 DR is at 36 deg 20' N and 156 deg 16' E.
    > W.E. on Z.T. is 47 sec fast. Height of eye is 6 feet, I.C. is 0.
    > The Sun LL sight at W.T. of 12-16-13 is Hs=73 deg 39.7'
    > Local date is 6-Jul-76
    > A summary of my calculations follows:
    > GMT 02-15-26
    > Date 6-Jul-76
    > GP of the Sun is GHA 212 deg 41.7, Dec = 22 deg 41.3 N
    > AP used is 36 deg N, 156 deg 18.3' E.
    > LHA is 9 deg
    > Ho is 73 deg 53.0'
    > The intercept from HO 229 is 40.9 Nm away at 212.9 deg.
    > If I use an assumed latitude of 37 deg N, then I get
    > a = 10 nm T at 211.1 deg. which is the answer given in the book.
    > Any comments?
    > Sam Chan

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