
NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Figuring Course given Lat/Long of destination
From: Roger M. Derby
Date: 2000 Feb 27, 7:24 PM
From: Roger M. Derby
Date: 2000 Feb 27, 7:24 PM
Note that for great circle sailing, you only compute your "departure heading". The heading you sail changes continuously (or as often as you refigure it.) Roger ----- Original Message ----- From- Luis Soltero To: NAVIGATION-L@LISTSERV.WEBKAHUNA.COM Sent: Sunday, February 27, 2000 8:42 PM Subject: Re: Figuring Course given Lat/Long of destination There are several ways to solve this problem. 3 come to mind a) Using plane sailing formulations which are very inaccurate due to reasons you mention below. b) rhumbline computations using meridionals which compute distances based on Mercator projections. c) great circle sailing which will compute the shortest distance between both points very accurately. Alton B. Moody in navigation afloat also discusses something called mid latitude sailings which if memory serves me right uses the mean of the two given longs in the calculation. Anyway, all these sailings are discussed in great detail with formulas and examples in Bowditch. Cheers, --luis ----- Original Message ----- From- Ed Kitchin To: NAVIGATION-L@LISTSERV.WEBKAHUNA.COM Sent: Sunday, February 27, 2000 5:49 PM Subject: Figuring Course given Lat/Long of destination An interesting problem appears in the latest issue of "Ocean Navigator" Which asks that you figure the course to a destination given origination and destination. It would seem easy to determine the difference in lat. (The destination was over several degrees of lat.), but deg. of long. differ in length as you change lat. One could simply take the mean of the two given long. and use that, but that bothers me as not being all that accurate. There is the error of the Macerator thing. You could use universal plotting sheets and construct using a vertical representing diff./lat., then draw a horizontal from the top of the lat. fig., representing the long. at the destination, and draw a hypotenuse as the course line. (???) Are there any mathematicians out there to give me a good formula to learn for this task? Thank you. Ed Kitchin