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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@EMC.COM

   

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