# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**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@EMC.COM