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    Re: Predicting the time of a desired azmuth
    From: Joe Shields
    Date: 1998 Oct 29, 08:35 EST

    Of course the point of celestial navigation is to do things without
    relying on electronics, but ...
    If you have a computer at your disposal and a copy of MICA,  I would
    just print out the altitude and azimuth of the sun for the whole day
    for a given DR pos.  Takes just a few seconds.  Here is a sample I
    used last weekend when I was up at the lake doing some short dip
    sights.  It is configured to print Alt./Az for 5 min increments,
    however it could be 1 min increments and still fit on a 8.5 x 11 page
    (with a small but still readable font).  Let that beast do the work if
    you have it!
    [The number between Hc and Az is the Hc increment between each 5 min
    readout -- for use in interpolating for any time in between.  I have a
    corrections chart.]
    1998 Oct 24 altitude and azimuth of sun at 40d 57.7'N 80d 05.9W in 5
    min. increments
    time  Hc        Az
    hhmm ddmm.m    ddd.d  hhmm ddmm.m    ddd.d  hhmm ddmm.m     ddd.d
    hhmm ddmm.m    ddd.d  hhmm ddmm.m    ddd.d
    0645-00146 546 1055 | 0855 21287 441 1293 | 1105 35260 181 1620 | 1315
    34432 198 2011 | 1525 19480 449 2329 |
    0650 00398 544 1063 | 0900 22122 435 1303 | 1110 35427 167 1635 | 1320
    34221 211 2025 | 1530 19025 456 2339 |
    0655 01340 542 1071 | 0905 22549 428 1314 | 1115 35581 153 1649 | 1325
    33597 224 2039 | 1535 18163 461 2350 |
    0700 02280 540 1079 | 0910 23370 421 1325 | 1120 36120 139 1664 | 1330
    33360 237 2053 | 1540 17296 467 2359 |
    0705 03217 537 1088 | 0915 24184 413 1336 | 1125 36245 125 1679 | 1335
    33111 249 2067 | 1545 16424 472 2369 |
    0710 04152 534 1096 | 0920 24589 406 1347 | 1130 36355 110 1694 | 1340
    32450 261 2081 | 1550 15546 478 2379 |
    0715 05083 532 1105 | 0925 25387 398 1358 | 1135 36451 096 1709 | 1345
    32177 273 2094 | 1555 15063 483 2388 |
    0720 06012 529 1113 | 0930 26177 390 1370 | 1140 36532 081 1725 | 1350
    31492 285 2107 | 1600 14176 488 2398 |
    0725 06537 526 1122 | 0935 26559 381 1382 | 1145 36598 066 1740 | 1355
    31196 296 2121 | 1605 13283 492 2407 |
    0730 07460 522 1130 | 0940 27331 373 1394 | 1150 37049 051 1755 | 1400
    30490 307 2134 | 1610 12386 497 2417 |
    0735 08379 519 1139 | 0945 28095 364 1406 | 1155 37085 036 1771 | 1405
    30172 317 2146 | 1615 11485 501 2426 |
    0740 09294 515 1148 | 0950 28450 355 1418 | 1200 37106 021 1786 | 1410
    29445 328 2159 | 1620 10580 505 2435 |
    0745 10206 512 1157 | 0955 29195 345 1430 | 1205 37112 006 1801 | 1415
    29107 338 2172 | 1625 10071 509 2444 |
    0750 11114 508 1166 | 1000 29530 335 1443 | 1210 37102 010 1817 | 1420
    28359 348 2184 | 1630 09158 513 2453 |
    0755 12018 504 1175 | 1005 30255 325 1455 | 1215 37077 025 1832 | 1425
    28002 357 2196 | 1635 08241 517 2461 |
    0800 12518 500 1184 | 1010 30569 315 1468 | 1220 37038 040 1847 | 1430
    27236 366 2208 | 1640 07321 520 2470 |
    0805 13413 495 1193 | 1015 31273 304 1481 | 1225 36583 055 1863 | 1435
    26461 375 2220 | 1645 06397 524 2479 |
    0810 14304 491 1203 | 1020 31566 293 1494 | 1230 36513 070 1878 | 1440
    26077 384 2231 | 1650 05470 527 2487 |
    0815 15190 486 1212 | 1025 32248 282 1508 | 1235 36428 085 1893 | 1445
    25285 392 2243 | 1655 04541 530 2496 |
    0820 16072 481 1222 | 1030 32518 270 1521 | 1240 36329 100 1908 | 1450
    24485 400 2254 | 1700 04008 533 2504 |
    0825 16548 476 1232 | 1035 33176 258 1535 | 1245 36214 114 1923 | 1455
    24077 408 2265 | 1705 03072 536 2513 |
    0830 17419 471 1241 | 1040 33422 246 1549 | 1250 36086 129 1938 | 1500
    23262 415 2276 | 1710 02134 538 2521 |
    0835 18284 465 1251 | 1045 34055 234 1563 | 1255 35543 143 1953 | 1505
    22439 423 2287 | 1715 01194 541 2529 |
    0840 19144 460 1261 | 1050 34276 221 1577 | 1300 35386 157 1968 | 1510
    22009 430 2298 | 1720 00250 543 2537 |
    0845 19598 454 1272 | 1055 34484 208 1591 | 1305 35215 171 1982 | 1515
    21173 437 2309 | 1725 00295 545 2546 |
    0850 20446 448 1282 | 1100 35079 195 1606 | 1310 35030 185 1997 | 1520
    20330 443 2319 |
    ---"Daniel K. Allen (Visual C++)" <danallen{at}XXX.XXX> wrote:
    >
    > I have solved this problem using 3 programs that I have written for
    my HP-48
    > calculator.
    >
    > Program A determines the position of the sun (altitude and azimuth)
    given
    > date, time, time zone offset, latitude, and longitude.  It is based
    on a
    > paper written by the U.S. Naval Observatory back in 1981 and was in
    the
    > Astrophysical Journal Supplement series.  It is accurate to within a
    minute
    > of arc, usually to within a tenth of a minute of arc.
    >
    > Program B allows you to input two points: a source position's
    lat/lon, and a
    > destination position's lat/lon, and a constant speed to travel
    between those
    > points.  Using the built-in clock of the HP-48 it calculates in real
    time
    > your position as you move in a straight line from A to B.  This does
    not
    > take into effect currents, winds, and changes in course and speed,
    but over
    > short distances it works well in practice.
    >
    > Program C solves your problem by iteratively solving for a
    particular sun
    > altitude (zero gives sunrise/sunset), or azimuth (180 degrees gives
    local
    > noon), etc.  It extrapolates forward in time your intended position
    using
    > Program B and feeds that position into Program A and then sees if the
    > altitude or azimuth is the desired value.  I use the built-in root
    finder of
    > the HP-48 for this, which works very well.
    >
    > The full solution takes 20 seconds or so on the 2 MHz 4-bit CPU that
    runs
    > for most of a year on 3 AAA batteries.
    >
    > Dan
    >
    > -----Original Message-----
    > From: psmith{at}XXX.XXX]
    > Sent: Wednesday, October 28, 1998 2:11 PM
    > To: navigation{at}XXX.XXX
    > Subject: [Nml] Predicting the time of a desired azmuth
    >
    >
    > During Dan's well-earned vacation from the Silicon Sea cruise, we've
    > not had any practical exercises.  Here is an example of a problem
    > that I've worked at sea many times, but have always wondered if
    > there is a better solution.
    >
    > How do you predict when the Sun's azmuth will be a particular value?
    > The simplest case is predicting meridian passage -- local apparent
    > noon -- and that is a fairly straightforward calculation involving
    > only the almanac.  A more complex case is predicting when the Sun's
    > azmuth will be perpendicular or parallel to your course (or more
    > likely, your Intended Track).
    >
    >  -- A sight taken directly abeam will yield a Line Of Position (LOP)
    >     parallel to your track, giving you a measure of course error.
    >     It can be useful in confirming that you are keeping clear
    >     of some lateral hazard (reef or shoal) in the same way a danger
    >     bearing is used in piloting.
    >
    >  -- A sight taken directly ahead or astern gives you an LOP perpen-
    >     dicular to your course/track and thus distance since your last
    >     fix and/or distance to your destination.
    >
    >  -- The two sights combined give a particularly useful running fix.
    >
    >
    > Example:
    >
    > I am off the US east coast. All times are GMT-4 (Eastern Daylight
    Time)
    > and all courses and azmuths are in degrees TRUE. The date is
    16-Jun-1998.
    > My destination is Buzzards Bay entrance tower at 41d 23.8'N  71d
    02.0'W.
    > My morning star sights gave a 04:39:00 fix at 39d 48'N  72d 14'W. My
    > course and speed made good are estimated as 030d at 5 knots.  Since I
    > will be approaching this often-foggy destination at night, I would
    like
    > a good distance-off measure during the day in case I can't get a fix
    by
    > evening star sights.  When will the Sun's azmuth be 210d, parallel to
    > my intended track?
    >
    >
    > Solution by Ho-229 and Nautical Almanac:
    >
    > Step 1: Rough estimate of time and DR position
    >
    >     An azmuth of 210d will occur sometime not too long after Local
    >     Apparent Noon (LAN), so as a first approximation, I project my
    >     position forward to 13:00, getting 40d 24'N, 71d 47'W. I figure
    >     the time of the Sun's meridian passage at this longitude:
    >
    >         71d 47'  Estimated longitude
    >        -60d 00'  Time zone meridian (GMT-4)
    >         ------
    >         11d 47'  Angular distance from time meridian to local meridian
    >
    >         00h47m   Meridian angle converted to time
    >         12:01    Local Apparent Time of meridian passage (from
    almanac)
    >         -----
    >         12:48    Zone Time of local noon
    >
    >     So far, so good.  Had LAN been at or before the time I selected,
    >     (13:00) I would advance the DR/EP another hour and use that in
    >     the subsequent calculations.
    >
    >
    > Step 2: Use HO-229 to find LHA that approximates the desired azmuth
    >
    >      40d 24'  N  Estimated latitude
    >      71d 47'  W  Estimated longitude
    >      23d 21.3'N  Sun's declination at 13:00
    >     150d         Desired azmuth of Sun
    >
    >     Our latitude and the Sun's declination are both North, so we will
    >     use the "latitude SAME name as declination" pages and look in the
    >     40d latitude column. Our desired normalized azmuth (Zn) is 210d,
    >     corresponding to a tabulated (i.e., unnormalized) azmuth of 150
    >     (N210E = N150W; put another way, we know the Sun must be West of
    >     us to give the desired azmuth, so the LHA measured West will be a
    >     small number, and the rule at the top of the page for North
    >     latitudes and LHA<180 is Zn=360-Z). Since the Sun's declination
    >     is between 23d and 24d, we look for entries where Z for 23d and
    >     24d declination straddle 150d. On the page for LHA 10d we see:
    >
    >         LHA 10d, 350d
    >         Dec    ...     40d
    >          .              Z
    >          .
    >          .
    >          23    ...    150.6
    >          24    ...    149.3
    >
    >     So, an LHA of 10d will give Z~150d and thus Zn~210d (at 40dN and
    >     declination between 23d and 24d).
    >
    >
    > Step 3: Figure out when the Sun's LHA will be 10d at our position
    >
    >     There are probably several ways of doing this. Here, I add the
    time
    >     it takes the Sun to move 10d West to the previously computed value
    >     for the Sun's meridian passage at my estimated position.
    >
    >     00h40m  10d LHA converted to time
    >     12:48   Zone time of meridian passage (from Step 1)
    >     -----
    >     13:28   Approximate Zone Time when Sun's LHA will be 10d
    >
    >
    > Inaccuracies in this method:
    >
    >  -- The estimated position may be off, but an updated DR & EP as the
    >     time approaches should indicate if a re-apraisal is necessary
    >
    >  -- The latitude and declination we enter the tables with are rounded
    >     to the nearest 1d
    >
    >  -- The LHA we get out of the table is only given to 1d
    >
    > Still, experience has shown this method to be accurate to within five
    > minutes if the estimated position is reasonably good.
    >
    >
    > Questions:
    >
    >  -- Is there a better way of doing this using HO-229?
    >
    >  -- What are some other methods?  Has anyone written a calculator or
    >     computer program to solve this?  Can navigational calculators like
    >     the Celsticomp spit this right out?  Is there an efficient
    solution
    >     for non-programmable calculators?
    >
    >     My gut feeling is that since your estimated latitude and longitude
    >     are changing linearly, and the first derivative of any of the
    azmuth
    >     formulae will give a linear approximation of the trend of Zn, one
    >     could work a solution for a time near the expected time, then
    solve
    >     for the exact time. (Of course, I was shakey in calculus 25 years
    >     ago, and now wouldn't even attempt to differentiate something like
    >
    >                                cos d * sin LHA
    > 	Z = arctan ---------------------------------------
    >                    cos L * cos d - sin L * cos d * cos LHA
    >
    >     so this is only a suppostion on my part.)
    > --
    > Peter Smith -- psmith{at}XXX.XXX
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