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Peter Fogg wrote:

>>  3) What is the DR position at the end of this time?
>  N10d52'.6 W131d 59'.5

>> 5) What is the TC & Distance from the DR to our Hawaii
> landfall point at 19d 30.0'N 154d 45.0'W

> Approx. 1400nm at 292d TC (I'll have to find a way to work this out
> accurately - any clues?)

Here are 3 ways.

1) Use a Great Circle calculation. Treat the destination as star.
   LHA is difference in Longitude. Declination == Destination Latitude
   Distance is (90d - Hc) * 60.  TC = Z. The problem with this method
   is that's not how you'll sail. The TC changes as you progress and
   must be continually recalculated to sail a Great Circle course.

2) Plane Sailing. Treat the problem as a simple triangle. The North-South
   distance is simply the latitude difference (in degrees) times 60 to
   give nautical miles. The East-West difference is the difference in
   Longitude (times 60) and times the cosine of the Mid-Latitude.
   Unfortunately the mid-Latitude is not so easily calculated. But a
   reasonable approximation is to split the difference, and maybe shade
   it a little closer to the pole. Pythagorean theorem gives us the
   distance and simple trigonometry for the course. This is the easiest
   to calculate with a simple calculator (with sin, cos, sqrt and inverse
   functions).

3) Meridional parts is a more accurate variation on the theme of Plane
   Sailing. Instead of factoring the change of scale into the Longitude
   difference, here it is done to the Latitude. Each latitude is converted
   into meridional parts, the difference of the MP's is taken as the
   N-S side of the triangle and the Longitude difference (aka depature)
   is the E-W side. The course is computed from this triangle. The distance
   is computed by dividing the Latitude difference (in minutes) by the
   cosine of TC (ignoring signs). This will tend to be a bit more accurate
   except where the courses are almost due East or West. The calculations
   are bit more complex, but can be easily programed or approximated by
   a simple series.

The Meridional parts method usually gives the best answer with respect
to the Silicon Sea series.

For much better and more detailed explanation of all of this see chapter
24 of Bowditch:

        http://pollux.nss.nima.mil/NAV_PUBS/APN/Chapt-24.pdf

and for details on Meridional parts calculation, see the explanation of
table 6.

        http://pollux.nss.nima.mil/NAV_PUBS/APN/Tables/TblExpl.pdf



Using Meridional Parts (#3) on this problem:

d.Lat = 19d 30.0'N - 10d52.6'N = 8d 37.4' = 517.4 nm.
d.lLo = 154d 45.0'W - 131d 59'.5W =  22d 45.5' = 1365.5'
d.MP = MP(19d 30.0'N) - MP(10d52.6'N) = 1185.58 - 652.21 = 533.37

TC = 270 + arctan(533.37/1365.5) = 270 + 21.3 = 291.3
Dist = 517.4/sin(21.3) = 1422.1




Using Plane Sailing (#2):

D.Lat =  517.4 nm
departure = d.Lo * cos((19.5+10.8)/2) = 1365.5 * 0.965245 = 1318.0
dist = sqrt(1318.0^2 + 517.4^2) = 1416.0
TC = 270 + arctan(517.4/1318.0) = 291.4




For me, I have trouble holding course within 2.5d for any length of time
let alone worry about a .1d difference. And the distance calculation
difference is 6 miles out of 1400 or 0.4%.




--
        Mike Wescott
        Wescott_Mike@EMC.COM

   

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