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    Re: A slope example
    From: Gary LaPook
    Date: 2010 Dec 1, 02:08 -0800


    --- On Wed, 12/1/10, Peter Fogg <piterr11@gmail.com> wrote:

    From: Peter Fogg <piterr11@gmail.com>
    Subject: [NavList] Re: A slope example
    To: NavList@fer3.com
    Date: Wednesday, December 1, 2010, 12:58 AM

    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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