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Author Topic: Local Apparent Noon  (Read 2337 times)
HenryC
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« on: February 02, 2010, 12:17:29 PM »

As the earth rotates on its axis from W to E, all the celestial bodies appear to move in the opposite direction from E to W.  This means that the way the sky looks like to you now is exactly the way it looked to someone E of you some time ago, and exactly the way it will look like to someone W of you some time from now.  In other words, you can't tell how far E or W you are (Longitude) without an accurate timepiece to correct for the earth's motion.  On the other hand, how far north or south of the Equator a star is in the sky changes only very gradually, it is written down in the Almanac where you can look it up.  The sun does move N or S throughout the year, but this is also published in the Almanac for each day, you don't need to know the precise time.  So it is possible to determine your Latitude without a chronometer, even if you can't determine your Longitude. This is called Latitude Sailing.

The Arabs and probably the Chinese knew about Latitude Sailing back in medieval times, and Columbus knew about the technique (although he wasn't very good at it).  I'm sure even the Ancient Greeks were familiar with it, their astronomers were, although merchant sailors weren't too eager in those days to share their trade secrets with potential competitors so history doesn't tell us for sure.  Until the invention of chronometers in the middle of the 18th century, Latitude Sailing was the only form of celestial navigation available.  You simply sailed N or S to the Latitude of your destination, then E or W until you hit it.  Of course, if the winds didn't cooperate it could be quite inconvenient, and if you missed your destination, you could keep right on going past it!   

The stars and sun all rise in the east, more or less, and roll around the sky until they cross the meridian, that imaginary line that goes from the N to S pole, directly over your head.  At meridian crossing, the star is due S of you if its declination is S of your Latitude, due N if its declination is N of your Latitude.  If it's declination is the same as your Latitude, then it will be directly overhead. In either case, this is as high as the star (or Sun) can get from the horizon as seen from your position.  Declination is just another way of saying the Latitude of the point directly beneath the star.

Since the celestial equator is directly above the earth's equator, it can be said to have a declination of zero degrees.  Now you can't see the equator to measure its altitude above the horizon, after all, it is an imaginary line; but if you could, your Lat would be 90 minus the equatorial altitude.  So for example, if you saw the equator at it's highest point to be 10 deg above the horizon then your Lat would be 80 degrees.   If the equator arced to 75 deg above the horizon, your Latitude would be 15 deg.  Whether those would be N or S Latitudes would depend on whether that high point where equator crossed meridian to be N or S of you.

Unfortunately, the equator's altitude can't be measured with a sextant, but a star's altitude can, and we know exactly how far each star is from the equator (the declination) from the Almanac. So a simple addition or subtraction tells us where the equator is!  Of course, the devil is in the details, and exactly how to do the sums depends on whether the star is N or S of you, N or S of the equator, and whether you are N or S of the equator.  I'll work the easiest case, and leave the others to you as "an exercise for the student".

At LAN (Local Apparent Noon, the moment the Sun is as high above the horizon as it can get) you observe it with your sextant to be 47 deg 14.7 min above your S horizon. You look up the declination of the sun in the almanac for that day and make a rough correction for the time of day
(remember, you don't have a chronometer but you have a rough idea of your DR Longitude and the time correction for it in the almanac).   Incidentally, LAN doesn't necessarily happen at twelve o'clock, but that's a topic for a different lecture. You find that according to the Almanac, the sun's declination is -18 deg 05 min. So the sun is 18 deg 05 min S of the equator,  therefore, adding the two,  the equator must be 65 deg 19.7 min above the S horizon.  Consequently, (subtracting from 90) you must be at Lat 24 deg 40.3 min N.

Since you don't know the exact time (we're assuming you don't have a chronometer or other access to Greenwich Mean Time), you can't can't calculate exactly when LAN is going to happen.  So you have to start making sextant observations sometime before , and writing down your altitudes, watching them get higher and higher until they start shrinking again (this means you've passed the exact moment of LAN). Unfortunately, the sun moves very slowly at the top of its arc and it is very difficult to exactly determine or predict the exact moment it is at its peak. Still, the LAN sight (navigators do a "noon sun" every day just to keep in shape because this is such an important backup procedure)  will give you a fairly good latitude, so once you have reached the latitude of your destination, you simply sail east or west until you hit it. 

When I lived in San Francisco I read about a sailor who was knocked down in a Pacific storm, severely damaging his rig, and taking out all his electronics.  Without electricity, he lost GPS and time ticks, so even his celestial navigation was compromised.  He was able to erect a jury mast and fly a spare jib as a lateener, and he used Latitude sailing to find the Golden Gate and safety.  So this is not just an academic exercise.
« Last Edit: February 02, 2010, 03:12:40 PM by HenryC » Logged
skip
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Logan and skip at dock in Sister Bay, Wisconsin.


« Reply #1 on: February 03, 2010, 08:45:14 AM »

It's very difficult for me to grasp these concepts. It's a little easier having read this book. I though it would be boring but instead it was great! Your explanation all through this topic have sharpened my understanding. And it's very cool that I live just three miles away from the 45th parallel line. The sun marches no further North in it's yearly trek than my house. It's on its way here now. Some 40 days till the first day of spring when the angle of the dangle will be heating my part of the earth.

So did all of you here the cyclical cycle of the ozone hole is letting out the earth's heat? There is no such thing as global warming, rather a normal cooling off and heating up. But not a constant ratcheting up in temperature.
Say the earth is a 100 foot dia blown up beach ball. Take a one ten thousandth of an inch flat feeler gauge and slap it down on the ball. That represents the earth's 85,000 or so height of atmosphere. Give or take a bit.

skip.

« Last Edit: February 03, 2010, 11:25:57 AM by skip » Logged

HenryC
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« Reply #2 on: February 03, 2010, 11:21:05 AM »

Actually, the sun never gets to your house, or even mine, and I live in S Florida!  At Lat 45N, halfway between equator and pole, the celestial equator is always 45 degrees above the Southern horizon.  At the sun's highest point around June 21st it is 23.5 deg north of the equator, so the highest the sun can be from your location is 68.5 degrees above the S horizon--certainly not overhead.  The distance of the equator from your overhead point is equal to your latitude, so the distance of the equator from your horizon is 90 minus the Lat.

I live at Lat 26N, so the equator is always 64 degrees above my Southern horizon. At its northernmost point, the sun is 87.5 degrees above my S horizon, and still not quite directly overhead.  The only places on earth where the sun can possibly be directly overhead is between the tropics of Cancer and Capricorn, 23.5 degress N and S, respectively, of the equator.

That 23.5 degree magic number that keeps on popping up in these calculations is the amount the earth is tilted in relation to the plane of its orbit around the sun. If the earth's axis of rotation was perpendicular to its orbital plane, there would be no seasons, and the sun would just travel along the equator all year round. 
« Last Edit: February 03, 2010, 11:32:21 AM by HenryC » Logged
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« Reply #3 on: February 03, 2010, 11:24:24 AM »

So were just talking about the angle of the sun reaching my house before heading South. Not over head.
Being a dummy I didn't know that. And all this time...

skip.
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HenryC
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« Reply #4 on: February 03, 2010, 12:47:45 PM »

It's really noticeable where you live, in the mid-latitudes.  On the first day of winter, the sun's lowest point around 21 December, it is 23.5 degrees South of the equator, so it would appear to you to be 45 - 23.5 = 21.5 degrees above your southern horizon at Local Apparent Noon.  On that same day, the sun would be 64 - 23.5 = 40.5 degrees above my southern horizon at LAN.  Which is why it is warmer here than it is in Yankeeland; on the first day of winter my sun is almost twice as high in the sky as yours.  Now you can see how navigators can exploit this phenomenon to determine their latitude.

It gets really bizarre in the polar regions; north of the Arctic Circle or south of the Antarctic Circle (Latitude 66.5 deg) the sun never sets on the first day of summer, and it never rises on the first day of winter.

The ancient Greeks had a myth about the country of Cimmeria, far to the north and inhabited by savages, where "the sun never set in summer, or rose in winter".  They must have sent explorers to, or heard rumors about, those distant, frozen  lands.
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curtisv
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« Reply #5 on: February 15, 2010, 06:41:29 PM »

Henry,

Interesting posting.  I haven't looked into this topic in quite a while but at some point before accurate time, it was observed that if both the sun and moon were visible at local noon, you could tell the lattitude by declination of the sun and get an approximate longitude with the declination of the moon.  Of course, you couldn't see stars or planets at local noon so all you had to go by was the sun and moon.  You needed to know what day it was and have some tables handy.

I think Celestaire has a few good books on the history of naviagation and the math behind it.  I bought only one book on astonomical math but it wasn't a good choice.  This is where thumbing through the pages at a boat show before buying would be helpful.

Curtis
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HenryC
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« Reply #6 on: February 15, 2010, 08:19:53 PM »

You are referring to the technigue of "lunars" which essentially means using the moon as a clock, since it's position is noted accurately in the almanac.  Without a chronometer, it's about the only way you can tell time at sea without a chronometer.  (The astronomer Maskelyne suggested using eclipses and occultations of Jupiter's moons, but this was impossible unless you put in ashore and set up a telescope on stable ground.

Lunars were used as a backup for chronometer failure, but they are difficult in practice.  Youy have to lay the sextant on its side and bring the moon and some other celestial body of known position  together, and then do the spherical trig to infer the moon's position.  There are specialized tables that allow you to do this, but I've never taken the trouble to try to learn the technique. That qualifies as advanced navigation!

Another problem is that the moon's orbit is not known very accurately (it is subject to all sorts of gravitational perturbations from the earth, sun, and planets) and it can only be predicted with confidence for several years into the future.
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curtisv
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« Reply #7 on: February 15, 2010, 08:28:25 PM »

Henry,

Thanks for the info.

I think the technique was to use the sun to determine time (determine local apprarent noon) and take a sighting on the moon.  This only worked when the moon was out during daylight hours.  In the days when there was no means to determine longitude and no direct way to measure time, an approximation of longitude would be the best information they could hope for.

It sounds like you've studied this quite a bit.  I appreciate the enlightenment.  Keep it up.

Curtis
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