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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Pub.229 GC-Calculation distance and course.
From: Guus Dekker
Date: 2009 Nov 28, 12:48 -0800
From: Guus Dekker
Date: 2009 Nov 28, 12:48 -0800
jm7857 Guus Dekker. I am exercising all kinds of nav problems. Can somebody explain me about 1 tiny problem I can not solve by my self. It deals about a sailing calculation problem finding GC-Distance and GC-Course from Fremantle to Durban, with nonintegral degrees and using a graphical method to simplify the DSD (Double-Second Difference) calculations. It's about the supplement of pub. 229 (vol.3 all the supplements are more or less the same). CHAPTER: D. Other applications 2. Great-Circle Sailing page xx. Case 1 example. point (6). (6) The Azimuth angle interpolated for declination, LHA, and latitude increments is S66.3W; the initial greatcircle course from the point of departure is 246.3* How is this calcutalted ?! What is the 0.2* correction from 246.1*(S66.1*W) The Z part is calc. like this: Z=66.9* Zdiff.= 66.9-66.0 =0.9 corr.= -0.8* distance = S66.3W - corr. = S66.1W Next: I think the 0.2* is the correction angle between - AP to destination - and - Departure Point to destination. - Pub.229 says on page 172 LHA 85* lat. 32* Decl. 29* that Hc=18* 45.4' d= +27.0' Z=66.9* d (from decl. 28* just above) = +27.2'. Is this the 0.2* diff. between the outcome S66.3*W and S66.1W !? Please... somebody !!?? -- NavList message boards: www.fer3.com/arc Or post by email to: NavList@fer3.com To , email NavList+@fer3.com
