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    Re: A slope example
    From: Gary LaPook
    Date: 2010 Nov 30, 15:15 -0800
    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'.


    --- On Tue, 11/30/10, Gary LaPook <glapook@pacbell.net> wrote:

    From: Gary LaPook <glapook@pacbell.net>
    Subject: [NavList] Re: A slope example
    To: NavList@fer3.com
    Date: Tuesday, November 30, 2010, 2:17 PM

    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.

    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 subtract the TR (track or course) from 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.)


    --- On Tue, 11/30/10, Peter Fogg <piterr11@gmail.com> wrote:

    From: Peter Fogg <piterr11@gmail.com>
    Subject: [NavList] A slope example
    To: NavList@fer3.com
    Date: Tuesday, November 30, 2010, 11:50 AM

    Slope doc
    Antoine, you provided this wealth of information below (what is GAST?) but not the azimuth of Vega at that time and place which I've calculated as 289d (expressed to the nearest whole degree).

    Antoine quote:
    28 Oct 1984, TT-UT = 54.8s, FN Aircraft Carrier FOCH (now property of the Brasilian Navy), on an evening where I was not night flying (maybe because the Moon was out that night - since I also shot it on that evening ... - , so these "moonly" nights were devoted to "nuggests" training ... :-)).
    Loran C Position at 16H30M00.0s UT : N34d30.3m/E019d59.1m, and
    Course/speed made good 270d/14.0kts
    Height of Eye 28 meters, P=1016 Mb, T=23�C
    Vega observations are given under the following format : UT / Sextant value corrected for instrument error (cfie) :
    16h28m24.0s / 69d31.9m
    16h29m21.3s / 69d21.4m
    16h30m17.3s / 69d11.2m
    16h31m05.7s / 69d01.8m
    16h31m47.0s / 68d55.1m
    I am getting the following results (FAR too many digits given ....):
    1 - Averaged observation time : 16h30m11.060s, and Averaged Observed geocentric height 69d02.558m, and
    2 - Vega apparent geocentric coordinates at averaged observation time :
    Apparent RA = 18h36m23.984s, Apparent Dec = 38d46m18.355s , GAST = 284d49m39.404s, and
    3 - DR Position at the time of the averaged observation time : N34d30.300m/E019d59.048m, and
    4 - Geocentric height and azimuth derived from DR at the averaged observation time : 69d04.213m, which by substraction from true observed height yields an intercept of 1.655 NM rounded to 1.7 NM
    ******* (Antoine quote ends)

    The apparent rise or fall of any celestial body is a function of latitude and azimuth, and on this occasion I've calculated it as minus (ie; 'falling') 52m.

    Gary LaPook: Thanks for posting that table for adjusting slope for the vessel's movement. It went to 12 knots, while Antoine tells us he was steaming at 14, but a bigger problem what that I couldn't figure out which row of azimuth range this 289d would fit into? In any case looking at the adjustments needed for 12 knots and 5 minutes of time it didn't seem likely to be more than about 1 minute of arc. As you can see by perusal of the linked image, the line of best fit is drawn parallel with the calculated line of slope, by eye in my case, but even if parallel rulers or a roller device was used an error of a single minute of arc could easily creep in somewhere in the process. While the procedure should increase the accuracy of a round of sights by reducing random error its not particularly precise. Could be another example of the difference between accuracy and precision.

    The linked file should show: the calculated line of slope, being the bottom line, the observations plotted according to altitude and time, and the 'line of best fit'; the top line which attempts to reconcile those sights. You can see that in this case I've given less weight to the first sight, although it is different by far too little to call it an outlier. Choosing the 'line of best fit' is a judgement call.

    Having done that, any point along that 'line of best fit' can be adopted for sight reduction. I have chosen two at random:
    16h 29m 00s 69d 24m
    16h 32m 00s 68d 52m

    Alternatively, a whole degree of altitude could be adopted, matched to the appropriate time.

    Antoine, you tell us "Averaged observation time : 16h30m11.060s, and Averaged Observed geocentric height 69d02.558m" but that time is quite close to one of your observations: "16h30m17.3s / 69d11.2m" with about 10m greater altitude. Could there be an error here somewhere?

    Your observations, judged from the pattern so close to the line of slope, are very good and this case either averaging or use of this slope technique should yield a similar result. From the deck of a small sailing boat my observations tend to be considerably more erratic, and use of the slope becomes vital to derive a good or even reasonable result.

    However, it seems to me that adding up all those figures and dividing by the number of observations (have I got this right?) is more tedious and error prone than graphing the slope, and you don't get a picture of your sights to contemplate as reward for the small effort...

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