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    Re: A slope example
    From: Peter Fogg
    Date: 2010 Dec 1, 19:58 +1100

     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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