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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

    Navigational Algorithms

    http://sites.google.com/site/navigationalalgorithms/

     


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