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    Re: calculated slope questions
    From: Gary LaPook
    Date: 2010 Dec 13, 09:59 -0800

    Below is a compilation of previous posts on this topic from November 27 to December 1.

    My updated table only goes up to 15 knots but you can calculate you own table or you
    can use the MOO table in H.O. 249 by using the 900 knot line, dividing by 100 and then multiplying by 2
    to get the correction for 18 knots. If you don't have that table, I posted it before, see below.

    ------------------------------------------------------

    I am attaching an updated motion of the observer table that now


    includes data for up to 15 knots.

    http://fer3.com/arc/img/114591.moo-rev.pdf

    ---------------------------------------------------------------------------

    There has been some discussions lately about using the technique of
    graphing a number of sextant observations on the same body and comparing
    them with a line having a slope representing the change in altitude of
    the body over time to allow discarding erroneous observations so as to
    keep them from distorting an average of the good observations. If you
    are going to do this then, if you are not taking the sights from the
    beach but from a moving vessel, you must adjust the slope of the line to
    account for the movement of the vessel. Back in February I posted a
    table I computed for adjusting sextant observations for the movement of
    the vessel and this same table can be used for adjusting the slope of
    the line for "manual slope plotting" technique. The table is here:

    http://fer3.com/arc/img/111994.moo.pdf

    For even more precision, and for speeds in the range of .5 to 9, knots
    you can use the Motion of the Observer (MOO) table found in H.O. 249
    simply by dividing the tabulated speeds, 50 to 900 knots, by 100 and at
    the same time dividing the tabulated values for the correction by the
    same factor of 100. There are two tables, one for a one minute period
    and one for a four minute period. I posted excerpts of H.O. 249
    including the MOO tables here:

    http://fer3.com/arc/img/105924.ho%20249%20extracts%20.pdf

    You can review my previous posts on how to use these tables at these links:

    http://www.fer3.com/arc/m2.aspx?i=111885&y=201002

    http://www.fer3.com/arc/m2.aspx?i=111994&y=201002

    http://www.fer3.com/arc/m2.aspx?i=105937&y=200807

    http://www.fer3.com/arc/m2.aspx?i=105938&y=200807


    You can also use the "Motion of the Body" (MOB) tables in the excerpts
    of H.O. 249 for actually plotting the sloping line representing the
    change in altitude of the body over time. The two MOB tables show this
    for a one minute period and for a four minute period and are equally
    valid standing on the beach, sailing at 6 knots or flying at 900 knots.
    When you study this table you will see that the slope of the line is
    determined solely for the azimuth of the body and the latitude of the
    observer. You can simply refer to this table and then plot the sloping line.


    Think these tables through and you can figure the sign to use for
    plotting the line, (although this should be obvious, bodies to the east
    are rising and those to the west are descending.)

    Give em a try.

    http://www.fer3.com/arc/m2.aspx?i=114523&y=201011

    --------------------------------------------------------------------------------

    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. Completing the computation of the slope, -58.35 + 1.1 makes the slope -57.25' in 5 minutes of time which is different than your result of -52'.


    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.)

    I should have put an explanation on my table for how to use it but I did explain it in the original posts. You find the difference between the TR (track or course) and the ZN to find relative bearing. In this case it is approximately 20°. Use the first column which has a plus sign. The first column is for objects ahead of the ship and the second column, with a minus sign, for objects behind the ship. (I am going to change the table and place "or" in the place of the "-" between the ZN-TR columns which might make it less confusing.)

    gl

    http://www.fer3.com/arc/m2.aspx?i=114568&y=201011

    And
    one more refinement (which I know is not really needed but I mention it
    for completeness sake) is that if the altitude of the body changes
    significantly during the period of the graph then the refraction
    correction might also have changed so should be included in drawing the
    line. Above about 24° the altitude would have to change by more than a
    whole degree for the refraction correction to fall into the next bracket
    thus changing the refraction correction by 0.1', not much to worry


    about.


    gl

    -----------------------------------------------------------------

    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


    --------------------------------------------------------------------------


    http://www.fer3.com/arc/m2.aspx?i=114582&y=201012


    gl

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