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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Time of meridian passage accuracy
From: Andrés Ruiz
Date: 2009 Oct 5, 12:32 +0200

Last friday early in the morning I went to sail. Maximum VOG = 8 kt, and average of 3, 4 kt. This is the real world on a sloop.

Now back in business again, a lot of mails about the subject in navlist. A good test bench.

With the data provided by Gary LaPook in Navlist 10041, using general SR methods, the fix at 17:00:00 is:

Fix(24.161666623101908, -79.441666294194576)           ( 24º  9.7' -79º 26.5')      LopLSSR

Fix(24.161666624470591, -79.441666332809163)           ( 24º  9.7' -79º 26.5')      KaplanSR_Bm = 0

Solving for course and speed:

Fix(24.161665279196630, -79.441682214842487)           ( 24º  9.7' -79º 26.5' 180.0 300.0)            KaplanSR_Bm_BLRV = 0

If anybody wants to use the standard intercept method:

 t B2 L2 gha dec HC Z HO LHA LHA p 17 24.1616667 -79.4416667 75.3093102 -23.4350226 42.2346691 174.876739 42.2347 -4.13235651 355.8676 0 17.0833333 23.745 -79.4416667 76.5588778 -23.4349961 42.7370395 176.39795 42.737 -2.88278887 357.1172 0 17.1666667 23.3283333 -79.4416667 77.8084456 -23.4349696 43.2097856 177.943863 43.2098 -1.63322105 358.3668 0 17.25 22.9116667 -79.4416667 79.0580133 -23.4349429 43.6518901 179.513496 43.6519 -0.3836534 359.6163 0 17.3333333 22.495 -79.4416667 80.3075809 -23.4349161 44.0623645 181.105656 44.0624 0.86591425 0.8659 0 17.4166667 22.0783333 -79.4416667 81.5571487 -23.4348893 44.4402573 182.718937 44.4403 2.11548207 2.1155 0 17.5 21.6616667 -79.4416667 82.8067164 -23.4348623 44.7846626 184.351715 44.7847 3.36504972 3.365 0 17.5833333 21.245 -79.4416667 84.0562842 -23.4348352 45.0947284 186.002154 45.0947 4.61461754 4.6146 0 17.6666667 20.8283333 -79.4416667 85.3058519 -23.4348081 45.369665 187.668209 45.3697 5.86418519 5.8642 0 17.75 20.4116667 -79.4416667 86.5554197 -23.4347809 45.6087536 189.347636 45.6088 7.11375301 7.1138 0 17.8333333 19.995 -79.4416667 87.8049873 -23.4347535 45.811354 191.038003 45.8114 8.36332067 8.3633 0 17.9166667 19.5783333 -79.4416667 89.0545552 -23.4347261 45.9769119 192.736715 45.9769 9.61288849 9.6129 0 18 19.1616667 -79.4416667 90.3041228 -23.4346986 46.1049657 194.441025 46.105 10.8624562 10.8625 0 18.0833333 18.745 -79.4416667 91.5536905 -23.434671 46.1951522 196.148068 46.1952 12.1120238 12.112 0 18.1666667 18.3283333 -79.4416667 92.8032583 -23.4346432 46.2472108 197.854883 46.2472 13.3615916 13.3616 0 18.25 17.9116667 -79.4416667 94.052826 -23.4346154 46.2609872 199.558445 46.261 14.6111593 14.6112 0 18.3333333 17.495 -79.4416667 95.3023938 -23.4345875 46.236435 201.255697 46.2364 15.8607271 15.8607 0 18.4166667 17.0783333 -79.4416667 96.5519615 -23.4345595 46.173617 202.943584 46.1736 17.1102948 17.1103 0 18.5 16.6616667 -79.4416667 97.8015293 -23.4345314 46.0727033 204.619083 46.0727 18.3598626 18.3599 0 18.5833333 16.245 -79.4416667 99.0510969 -23.4345033 45.9339704 206.279233 45.934 19.6094303 19.6094 0 18.6666667 15.8283333 -79.4416667 100.300665 -23.434475 45.7577962 207.921172 45.7578 20.8589981 20.859 0 18.75 15.4116667 -79.4416667 101.550232 -23.4344466 45.5446567 209.542153 45.5447 22.1085658 22.1086 0 18.8333333 14.995 -79.4416667 102.7998 -23.4344181 45.2951193 211.139581 45.2951 23.3581334 23.3581 0 18.9166667 14.5783333 -79.4416667 104.049368 -23.4343896 45.0098361 212.711022 45.0098 24.6077013 24.6077 0 19 14.1616667 -79.4416667 105.298936 -23.4343609 44.6895371 214.254228 44.6895 25.8572689 25.8573 0

Simulating with the data provided by Gary:

 t B2 L2 gha dec HC Z LHA LAN 17.27583333 22.7825 -79.44166667 79.44537934 -23.43493459 43.78256526 180.0047182 0.003712671 H max 18.23833333 17.97 -79.44166667 93.87788659 -23.43461934 46.26136592 199.3202558 14.43621992

LAN: 17:16:33

Time of Hmax: 18:14:18

Conclusion:

• The generality and power of the least squares SR method based in the Marcq Saint Hilaire, and the one by Mr. Kaplan is checked for cases of high speed.
• For obtain the Hmax, the Least squares fitting H = a0+a1*t+a2*t2 is OK. The maximum gives:  t max = -a1/(2a2) and  Hmax = a0-a2t2. It retains the asymmetrical nature of the curve.
• For shoots made aboard a vessel the equation for time of meridian transit proposed by Jim Wilson in his paper “Position from observation of a single body” work fine.
• For high speed, in aircrafts, the equation doesn’t work, and iteration doesn’t improve the result. I think the problem is in the assumption made in the appendix I: the error in time is little, and differentiating is possible.

Is Jim Wilson agree with these last points? Or I am doing some thing wrong?

Regards,

Andrés Ruiz

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