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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@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@XXX.XXX]
> Sent: Wednesday, October 28, 1998 2:11 PM
> To: navigation@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@XXX.XXX
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