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Subject: RE: [Nml] Predicting the time of a desired azmuth
From: Daniel K. Allen (Visual C++) (danallen@XXX.XXX)
Date: Wed Oct 28 1998 - 19:00:35 EST
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 =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=--=-= =-= TO UNSUBSCRIBE, send this message to majordomo@XXX.XXX: =-= =-= unsubscribe navigation =-= =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=--=-= =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=--=-= =-= TO UNSUBSCRIBE, send this message to majordomo@XXX.XXX: =-= =-= unsubscribe navigation =-= =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=--=-=
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