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Re: Interpolation of Meridional Part Table
From: George Huxtable
Date: 2009 Mar 24, 22:17 -0000
From: George Huxtable
Date: 2009 Mar 24, 22:17 -0000
Brad wrote- responding to a question about meridional parts- "= LOG(TAN(RADIANS(Latitude/2+45)))*7915.704468" ================= Comment from George. Well, not quite. Not if you're working to high precision, as a constant such as 7915.704468 implies. That would be right for a spherical Earth, but needs correcting for its actual non-spherical shape. Taking the flattening to be 1 part in 298, a more accurate answer is given by MP= log(tan(radians(lat/2 +45)))*7915.7 - 23*sin L, which is good to about a tenth of a mile. A more precise expression can be found in Bowditch ((1981) vol 2, page 4 (table 5 explanation) Slightly different answers, in different sets of tables, may reflect nothing more than updating the model for the Earth's shape, which isn't a precise spheroid anyway. In that edition of Bowditch the meridional parts is found in table 5, not table 6 as John Parsons quotes from his 1972 edition, but the interpolated value at 48� 23.5' is exactly the same as his, as 3309.5, interpolating for the half-minute. There's no problem at all with making such a linear interpolation. Most likely, that example in Dutton's was formulated many years earlier, when a slightly different model had been adopted for the Earth's flattening. What intrigues me is why John is worrying about such small discrepancies. Navigation isn't such an exact science as that. George. contact George Huxtable, at george@hux.me.uk or at +44 1865 820222 (from UK, 01865 820222) or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. -----Original Message----- From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of NavList@aol.com Sent: Tuesday, March 24, 2009 3:42 PM To: NavList@fer3.com Subject: [NavList 7754] Interpolation of Meridional Part Table Anyone, I am looking at a sample problem in section 505 of the 1972 Dutton's. It says that the latitude of Cape Flattery Light is 48d23.5' and says that the meridional parts (m) for that latitude are (is?) 3309.2. Using an on-line versuion of Bowditch Table 6, I located 48 deg latitude and went down the minutes column to 23' to find m=3308.8, and to 24' to find m=3310.3. A linear interpolation would put 48d23.5' at 1/2 the difference, or 3309.55 (round up or down to taste), and one would then have m=3309.5 or 3309.6. I recognize that a linear interpolation doesn't really fit the facts as far as the way meridional parts change; so is there a technique I shold know about that would have given me the result Dutton's got, 3309.2? That is, how did they do that? -John Parsons --------------------------------------- [Sent from archive by: jkp-AT-obec.com] "Confidentiality and Privilege Notice The information transmitted by this electronic mail (and any attachments) is being sent by or on behalf of Tactronics; it is intended for the exclusive use of the addressee named above and may constitute information that is privileged or confidential or otherwise legally exempt from disclosure. If you are not the addressee or an employee or agent responsible for delivering this message to same, you are not authorized to retain, read, copy or disseminate this electronic mail (or any attachments) or any part thereof. If you have received this electronic mail (and any attachments) in error, please call us immediately and send written confirmation that same has been deleted from your system. Thank you." --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---