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Re: a go at Silicon Sea:Leg 81
From: Michael Wescott
Date: 2001 Oct 19, 1:40 PM

```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{at}EMC.COM
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