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Subject: Re: leg 60
From: Dan Hogan (dhhogan@XXX.XXX)
Date: Thu Mar 23 2000 - 16:00:03 EST
R.:
># 2. Distance 88.5 nm
># 3. 190d t
>
>But I cheated and I am somewhat confused by my answers. I used a
>program entered into my TI 60 calculator for determining Hc and Z. I
>computed lha by the difference between my lat and the long. of the
>waypoint. I entered my lat and substituting the lat for the waypoint for
>dec in the program. Then I checked my calculations using the Nav20 program
>in the archive.
I haven't used either Nav20 or a TI. But to get the correct answer you
need to use one of the Sailings for a solution. Using Sight Reduction
provides a Great Circle answer.
>My question is this. The program calculates Hc which is in degrees
>between the assumed position and the gp of the body (in this case it
>would be the waypoint. Since 1 degree equals 60 nm the calculated
>distance should have been 60 X the degrees arrived at from the
>calculation or 5310 nm. How did Nav 20 come up with 88.5 nm?
Real quick to use an SR program for GC use:
Latitude of Departure
Latitude of Destination
LHA= diff in longitude between departure point and destination
Az= TC
GC distance = 90 - Hc x 60
Distances will vary depending on the formula used for the SR.
Dan Hogan WA6PBY
C27 "Gacha"
dhhogan@XXX.XXX
Navigation-L: http://nav.cnchost.com
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