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    Re: accuracy of sights and averaging
    From: Peter Fogg
    Date: 2010 Dec 6, 16:14 +1100
     Alan 202 you wrote:

    Re graphing, the appropriate paper I can buy. Would you please point me toward the formula you referenced.

    With lots of nav topics, there is a simple answer and, if you are so inclined (and many here are) it can get much more complex.  A simple example is the statement that the earth is a sphere.  And so it is, more or less (mostly more) although it tends to bulge out a little at the equator and be a trifle squashed at each pole.  Actually this is just the beginning of the story of how uneven the earth's surface is.  Such matters are endlessly discussed here.  But for many practical purposes you can just go with the sphere.

    The reason for that long intro is twofold: 

    *  Firstly, don't be disheartened when every foundation notion (eg; the earth is a sphere) gets contradicted, especially by the nitpickers.  Keep the contradictions in their correct proportion within your own mind.

    *  Secondly, because when it comes to this formula there is, to begin with, a simple answer and here it is:

    Delta H = 15 x cosine latitude x sine azimuth
    Where Delta H = rate of change in arc minute per minute of time

    Since that answer is for a minute of time it needs to be multiplied by 5 if you are using a 5-minute period.  My favourite scientific calculator allows me to enter that formula in narrative form, just as set out, without needing to use any memory functions.  Way to go.

    The more complex answer is what has been discussed here recently.  My conclusion is that none of it matters much unless you have a great need for precision and/or are on a VERY fast vessel at the time.  Just be aware that the slope of apparent rise or fall (apparent since this movement is caused by the rotation of the earth) is not really a straight line but an arc, but over a short period like 5-minutes you can mostly not worry about this, either.

       
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