NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: leg 60
From: Dan Hogan
Date: 2000 Mar 23, 2:00 PM
From: Dan Hogan
Date: 2000 Mar 23, 2:00 PM
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@nav.cnchost.com Navigation-L: http://nav.cnchost.com
