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    Re: A slope example
    From: Gary LaPook
    Date: 2010 Dec 01, 02:00 -0800

    Peter Fogg wrote:
    
    "Yep, you're right.  There is another way to calculate it:
    Delta H = 15 x cos latitude x sin azimuth
    Where Delta H = rate of change in arc minute per minute of time"
    
    ________________________________________________________________________
    
    Yep, that's the formula I posted on July 24, 2008, see:
    
    http://www.fer3.com/arc/m2.aspx?i=105929&y=200807
    
    But, if you really wanted to be persnickety, when doing stars you should start 
    with 15.042 as that is the rate
    for Aries, 15 is the rate for the Sun and planets but for practical navigation 
    and short time periods no significant error is introduced by using 15 for 
    both stars and objects in the solar system.
    
    gl
    
    
    
    
    
    On 12/1/2010 12:58 AM, Peter Fogg wrote:
    >   Gary LaPook wrote:
    >> I get the calculated slope for Vega for the movement of the body (change in 
    LHA) for the five minute period from 1628 to 1633 to be -58.6' doing the 
    sight reduction and -58.35' by using the MOB table from H.O. 249. This table 
    show -11.7' per minute of time at latitude 35� and azimuth 288� and -11.5' at 
    latitude 35� and azimuth 290�. For latitude 30� the values are -12.4' and 
    -12.2' respectively. Doing some simple interpolation makes the slope for the 
    movement of the body -11.67' per minute of time and -58.35' for the 5 minute 
    time period.
    > Yep, you're right.  There is another way to calculate it:
    > Delta H = 15 x cos latitude x sin azimuth
    > Where Delta H = rate of change in arc minute per minute of time
    >
    > Using that formula (x5) gives - 58.3
    >
    >> The adjustment for the movement of the ship is plus 1.1' minutes. If you 
    use my table you can take the value for 7 knots and relative bearing of 20� 
    (ZN-TR, 289-270 = 19, ~ 20) which is .5' and double it to find the correction 
    to be + 1.0'. You can also do the same thing using the MOO table from H.O. 
    249 which shows + 11.05' (interpolating) for 700 knots and relative bearing 
    of 19�. Divide by 100 to get the adjustment for 7 knots, multiply by 2 to get 
    it for 14 knots and then multiply by 5 minutes to get the adjustment of + 
    1.1'.  (You get the same value using a calculator, 14 knots divided by 60 
    minutes gives a ship movement of .23 NM per minute, times 5 minutes makes 
    1.17 NM times the cosine of the relative bearing, 19�, makes the correction 
    of + 1.1' also.)
    > Thanks for this explanation.  So applying this correction results in a
    > corrected slope of - 57.2 ?
    >
    >
    >
    >
    
    
    
    
    

       
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