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    Re: Silicon Sea: Leg 77 Rev 08/13/2001
    From: Michael Wescott
    Date: 2001 Aug 19, 2:02 PM

    Does no one else wish to publish their results?
    
    > Promptly at Sunrise on 24/06/2001 we raise anchor at Isl Flamenco
    > (08d 55.2'N  79d 31.6'W) and head for our departure point at Isla
    > San Jose Light(08d 10.5'N  79d 15.0'W) motoring.
    
    > Speed = SQRT LWL x 1.34
    
    > 1)  What is the Zone Time (ZT) of our Estimated Time of Arrival(ETA)
    >     at the Isla San Jose Light?
    
    >     What is the UT/GMT ETA?
    > --  ----------------------------------------------------------------
    
    Answer:
    ETA = 1141 GMT-5 = 1641 GMT
    
    Discussion:
    We need starting time, and time en route. For time en route we need the
    distance and speed. Speed is given as 1.34*sqrt(LWL) a.k.a. "hull speed".
    This is easy with a calculator and S=8.4.
    
    The distance I'd get directly from the chart with dividers. Without the
    chart, any of the sailings will do. Simple traverse with Mid Latitude
    of 8d 30' gives us D=47.6 (so does a Great Circle calculation, Meridional
    Parts is just enough more so as to round up to 47.7).
    
    Time en route is T=D/S=5.67hrs=5h 40m.
    
    Sunrise is calculated by taking the almanac entry (0559), interpolating
    for latitude (-16m) and interpolating for longitude differnce from central
    meridian of the zone (+18m). Dawn is at 0601 (give or take a little).
    
    
    > On our ETA at Isla San Jose we take a departure FIX off of the light that
    > is (08d 10.5'N  079d 18.0'W). We set our course for the Mid-Ocean Point
    > (MOP) at (03d 00.0'N  110d 00.0'W).
    
    > Variation 2.0dW, Dev 0.5dE. Drift 0.6 Kts Set 285.0d. Speed 10.1 Kts.
    
    > 2)  What is the TC and Distance from the Isla San Jose Light departure
    >     point (08d 10.5'N  079d 18.0'W) to the MOP(03d 00.0'N  110d 00.0'W)?
    > --  --------------------------------------------------------------------
    
    Answer:
    C = 260.4T D = 1858.0
    
    Discussion;
    I used plane sailing with mid lat of 6d (easy with a calculator). Mercator
    Sailing calculation (Meridional Parts) gives same course and a slightly
    longer distance (1871.0), but still only a difference of 0.7%.
    
    
    > We plot our track on INT Chart 811 and see that we can not head direct for
    > the MOP(03d 00.0'N  110d 00.0'W) without crossing land above Punta Mala in
    > Panama. So we alter course to MOP2 at (05d 00.0'N  80d 00.0'W).
    
    > Variation 2.0dW, Dev 0.5dE. Drift 0.6 Kts Set 285.0d. Speed 10.1 Kts.
    
    > 3)  What is the True Course and distance to MOP2 from our Departure point?
    >     What is the Compass Course(CC)/Course-to-Steer?
    > --  ----------------------------------------------------------------------
    
    Answer:
            C = 192.3T
            D = 195.0 nmi
            CC/CTC = 188.9T = 190.9M = 190.4C
    
    Discussion:
    I used plane sailing calculation (mid lat = 7) for Course and Distance.
    Comparison with Mercator calculations show a difference of .1d and .1nmi.
    For me, it is easier to use a plotting sheet to adjust for current than
    to use the law of sines, since I have to draw it out either way. The
    difference should be less than a half a degree, and in this case I
    lucked out and got it within 0.1d
    
    > The weather has high Cirrus clouds building, wind from the Southwest.
    
    Hmm. A tropical Low to the west, perhaps?
    
    > At 14:45:30UT the next morning we take a SUN LL sight. Other Data as above.
    
    > |    Hs 47d 27.2'
    
    > 4)  What is the DR position?
    > --  ------------------------
    
    Answer:
            DR = 4d 32.6N  80d 05.8W
    
    Discussion:
    Time difference is 22h05m, speed is 10.1, D = 223.0.
    Plane sailing calculation with Mid lat = 6.5d. C = 192.3T
    dLat = 3d 37.9' dep = 47.5, dLo = 47.8'.
    
    > At 22:45:23UT we take another sight of the SUN LL
    
    >     Hs 10d 07.2'
    
    > 5)  What is the DR position?
    > --  ------------------------
    
    Answer:
            DR = 3d 13.8N   80d 22.9'W
    
    Discussion:
    Using last fix as starting point we have T = 30h 04m. S = 10.1, so
    D = 303.7. dLat = 296.7' = 4d 56.7', dep = 64.9, dLo = 64.9' = 1d 04.9'
    (with mid lat of 6d).
    
    > 6)  What is the Running FIX(RFIX) position at the 22:45:23UT sight?
    > --  ---------------------------------------------------------------
    
    Answer:
            RFIX = 3d 14.7'N  80d 23.5'W
    
    Discussion:
    For a running fix, there is a distinct advantage to using a sight
    reduction method that uses AP = DR. It means that we can translate
    the derived LOP to the next DR directly.
    
    The first sight reduces to Ho = 47d 36,9' Hc = 47d 36.9'  Z = 59.8
    The second one reduces to  Ho = 10d 12.5' Hc = 10d 11.7'  Z = 293.2
    
    The first is spot on. The second is .8T. Plotting gives the RFIX.
    
    > Checking our Running FIX against the position of MOP2(05d 00.0'N
    > 080d 00.0'W) it seems we have passed MOP2. So we alter course to our
    > original Mid-Ocean Point which we re-name MOP1(03d 00.0'N  110d 00.0'W).
    
    > Speed 10.5 Kts. Current Set 285, Drift 0.6 Kts. Var 1dW Dev 2dE.
    
    > 7)  What is the TC and Dist to the MOP1(03d 00.0'N  110d 00.0'W)?
    >     What is the Compass Course(CC)/Course-to-Steer?
    > --  -------------------------------------------------------------
    
    Answer:
            C = 269.5T
            D = 1777.4
            CC/CTC = 267.8
    
    Discussion:
    Plane sailing calculation used, but in a non-trafitional way. In
    this part of the globe, a degree of Latitude is smaller (by 0.5%)
    than 60 nmi. and a degree of Longitude is actually slightly more
    (by 0.05%) than 60 nmi. For that reason Mercator sailing will tend
    to overstate the distance by 0.5%. By calculating the departure
    directly from the length of a degree of Longitude and likewise
    the N/S distance travelled we get D = 1777.4. Courses calculated
    by either method should be about the same within 0.1d.
    
    The CTC was obtained by plot. Calculation by law of sines puts
    the course 0.2d more southerly.
    --
            Mike Wescott
            Wescott_Mike{at}EMC.COM
    

       
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