Local Apparent Noon

[HenryC]

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.

[Craig Weis]

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.

[HenryC]

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. 

[HenryC]

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.

[curtisv]

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

[HenryC]

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.

[curtisv]

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

[Salty19]

Just a quick comment on this topic that may be of interest to those studying the motion of the sun.

I have been working on an Excel/Visual Basic application that provides the exact local apparent noon, azimuth, declination, elevation, rt ascension, sunset/sunrise and many other useful bits of information on solar motion that takes into account your specific Lat/Long coordinates and your local time, but not altitude which is assumed to be sea level.  There is no need in the application to know anything except the DMS GPS coordinates, and to make sure the time on your computer is as accurate as possible.   It will provide up the minute (every 60 seconds) data on each calculation so there is no need to wait to take fixes at certain hours, perform ANY manual calculations or use almanacs.   

The solar motion portion of the application is completed, save some minor visual cleanup work.  While I have not sea tested the solar motion calcs with a sextant to validate it, it appears accurate based on "front porch" observations over the last few weeks with more primitive angle and directional measuring tools.

The application actually does much more than tracking the sun, and am planning a fully features, cruising oriented application.  I am integrating NMEA data from a Sirf chipset GPS receiver and building a full-on navigation and anchoring program, with waypoint tools, anchoring monitor, and much more.  I doubt I will integrate actual maps into it but rather am designing it as a supplement to paper maps-however the possibility of outputting the data to Google Earth might be a reality in the future.     Some of these functions are completed, and am working on other tools such as logs, moon motion, and integrating with other NMEA outputs (such as a depthfinder) but for now I will only share the Solar information since it's more or less ready for use.  If the project grows wings, I'll compile it and sell as a standalone application, without the need for Excel-but that's further down the road. 

Henry, I am especially interested if you would like to take a test drive of the solar motion data, mainly to help validate accuracy using your advanced knowledge of this subject and use of a sextant, and to provide feedback.

Would anyone like to try out the Solar motion portion of this program, especially those with a sextant and the knowledge to use it?  You will need Microsoft Excel (any version) to use it. I just need a week or so to clean up some code and lock it down so you won't inadvertently make critical changes.

Henry, if you do not have Excel, I may be able to email you snapshots of the output for future days with coordinates you choose to test drive accuracy.  That way you can print the output and compare the data to some future date/time/location when you may be motivated to provide feedback.     

[HenryC]

http://aa.usno.navy.mil/data/docs/celnavtable.php

I'd be delighted to do anything I can to help with your project, but I doubt that there's much I can do that you can't easily accomplish yourself just by using the web link above: the Almanac Office's online Almanac Calculator.  I have used it extensively and it is extremely precise and dependable.  You can trust it.  It will allow you to check your software and do multiple tests and QA.  It covers the whole navigational sky for hundreds of years into the past and future.

The dip (height above MSL) corrections can be taken care of by the Nautical Almanac, either in tabular form, or by equations in the Almanac text itself which can easily be programmed into your code.

Keep us up to date on your progress, and please feel free to contact me if you have any questions.  I'll do what I can to help, although I suspect that the link I provided above will be able to give you everything you need.

Good Luck, and keep in touch.

HC

PS: Keep in mind, the Earth rotates one revolution every 24 hours,
that's 15 degrees/hr, 1/4 deg/min, or a 1/4 of a nautical mile per SECOND of time at the equator.  For accurate navigation, you're going to need to provide solutions more frequently than once per minute.  The clocks on the platform your app runs on will have to be accurate to the nearest second.

[Salty19]

Thank you, Henry!   It appears as if my calculations are spot on.  Actually, there was a minor discrepancy (less than .5*), but I am attributing this to
rounding error.  The site you linked allows you to enter degrees and minutes with one decimal point in the minutes, whereas my program allows you to extend it to 5 decimal points.  When I forced the program to use one decimal point, it was exactly the same elevation and azimuth at the minute that was used.

However, my program provides much more data, and provides data for each minute of the day should the user choose to click update each minute, whereas the link only provides information on a 10 minute basis. Whether that data is more useful is the determination of the user.  Here is a screenshot from a little earlier today at my location in Ohio (ignore the GMT offset and time, it's incorrect as shown but the calculations are based on the correct time).   Also ignore the horizon line--that was placed manually, I will plan to fix that to self-adjust. 





I did find it interesting that my GPS, a cheap Garmin E-trex showed different sunset and sunrise times, off by a considerable amount.
I suspect the GPS is not calculating to the level of detail that it should, but do not know.

Thank you for the link!   I will look into building in elevation (altitude) into the equations, but admittedly my priority is to work on the GPS aspects of the program.

As you mentioned above, the solar noon is not necessarily local noon. Today at my location the maximum height of the sun occurred at 1:16:30PM--quite a difference from the commonly accepted 12:00:00!!!  The elevation at solar noon was 36.98*.  This is the time of year when the sun is always in your eye here at 40*N , 82* W. 

So with the solar declination at -9.98*, one would need to be near 10* south of the equator for the sun to be directly overhead. Following the sun indeed!

Another detail that I choose to leave out is the refraction of the sunlight through the atmosphere that changes the observed location from the perspective of being on the ground.   I have the calculations for it (lifted from a NOAA spreadsheet), but the differences where pretty minor.  It appears as if the sunset as seen by me would be a few seconds after what is indicated due to the refraction, but that's really not too important for me to speak of. 

[HenryC]

I don't understand what you mean by "the link only provides information on a 10 minute basis".  The Almanac calculator allows you to enter time to the exact second and lat/lon to the nearest tenth of a second of arc..  Am I missing something?

Also, you mention a "minor discrepancy (less than .5*)".  I presume ' * ' stands for 'degree', but a half of a degree (30') is an enormous error (the sun and moon discs are both about a half a degree in diameter!).  A skilled sextant operator under ideal conditions can measure an angle to about a half MINUTE of arc.

A few other comments about nomenclature:

"Sun degrees from North" should be labeled " Sun Azimuth".
"Sunrise Angle" should likewise be labeled  "Sunrise Azimuth"
"Right Ascension" is always measured in time units (h m s) east from the Vernal Equinox (called "Aries" by navigators) to the celestial body.  It is always a positive number.  If you want to express the angular measure from Aries to the celestial body in degrees, navigators use the term "Sidereal Hour Angle" or SHA.  This is the quantity listed in the Nautical Almanac and it is always positive.  "Aries" is just the imaginary point in the sky where the Sun crosses the equator on the first day of spring.  All this quibbling is important because users may not be navigators or astronomers and it is important to use consistent terminology to avoid confusion.

Also, be very careful when dealing with Civil Time, especially when daylight savings is in effect.  That can really be confusing, especially when its a different date in Greenwich than where you are.  The standard time in your entire time zone is arbitrarily set to the Local Standard Time at the central meridian of your time zone. (That's how the entire time zone clocks are set to the same time).  Local Time varies within your time zone depending on your longitude. It is only used in calculations. The sun operates on Local Mean Time (LMT) which is the Local Time referenced to the Mean Sun, not the real sun,  but your watch reads civil time, which is always Local Standard Time (possibly corrected for daylight savings).  Got that?

Just a review.  The Almanac data is listed in Universal Time,  which is the hour angle of the Mean Sun at the Prime Meridian.  It is, by definition, the Greenwich Mean Time (GMT) at Her Majesty's Naval Observatory in Greenwich, UK. They paint a stripe on the sidewalk there to let you know exactly where it is.  We use the Mean Sun (its like an average) because the real Sun doesn't travel at a uniform speed due to the Earth's elliptical orbit around the Sun.  Your Local Mean Time (LMT) is the hour angle of the Mean Sun as seen from YOUR longitude.  Someone standing just a few feet E or W of you has a different LMT.  Every time zone is assigned a Standard Time, which is the LMT at the central meridian of that time zone. This way everybody's clocks in each time zone agree. 

However, the Standard time is often messed around with by local authorities to line up with political boundaries, or for daylight savings purposes.  The civil time on your watch is only distantly related to what the real Sun is doing.  And that is not the same as what the Mean Sun is doing.  Unfortunately, the Mean Sun is invisible, your sextant or sundial can only measure the real one.  Local Apparent Noon (LAN) does not occur at 12:00 LMT because the real Sun may be a few minutes ahead or behind the Mean Sun. But the difference between the two is never more than about 17 minutes.  Incidentally, this difference (between Mean and Solar time is called "The Equation of Time" and is printed for every day in the Nautical Almanac.

Your "NOAA Spreadsheet" is probably the same algorithm that's on the website I pointed you to, so they will always agree.  Your major source of potential error is making a mistake in getting the time sorted out right.  That's why it is so critical to get these definitions straight.

HC

[Salty19]

Thank you for the great feedback, Henry!

I'm glad you brought up the 1 minute increment...I did not notice this capability at first and just saw the default 10 minutes.

I am still seeing an error to the tune of appx .31 degrees in the elevation figure, but the azimuth angle is correct as far as I can tell.  The website only gives 1 decimal point and the program gives as many as I want.  I do see higher or lower azimuth figures that round up or down to the websites' figures.   Not the case with elevation and I need to look into that.

The labels were just to help me keep all the azimuths straight during the learning process.  With less than 3 weeks of experience with diving into this subject, it was helpful for me to separate the meaning of all these values with plain english.  It's been 25 years since I've had an astronomy course!

The program based off NOAA calculations does appear to take into account longitude when determining the solar figures.   For a given minute, if I alter the longitude--even a little (fraction of a minute), the results do change, albeit slightly. So I believe it's considering actual solar time, not a mean.  Time and experimentation will tell me for sure.

From the reading I've done, the program will not benefit a user within the polar regions.  That's OK, I have zero plans to ever step foot there, let alone be in a boat at these locations.

I also did figure out how to determine distance between two waypoints based on radians which is pretty exciting.  It was surprising to see how much the distance between lines of longitude changes with each degree, which obviously is important to calculate. The ancient mariners must of been off by dozens of miles before elliptic geometry was invented.

I'm going to leave the solar items alone for now and focus on importing and processing the GPS signals based on the several NMEA sentence types available with the Sirf GPS chipset.  With some time I should be able to produce all the information available to any GPS, and crunch my own figures as needed for derived results.

I'm having fun "messing around", and that is what is keeping the motivation going.

Once again, thank you!

[HenryC]

A .31 degree error is enormous, which means its probably pretty easy to fix.  It's those tiny errors that are hard to diagnose and cause the real problems.  .31 degrees is about 18' of arc, or about equal to the semi-diameter of the sun or moon disc.  Remember, the almanac gives you the value of the center of the orb, but what you measure with the sextant is the upper or lower limb (edge).  And are you correcting for refraction and parallax?  Or are you correcting twice without realizing it?

The NOAA website calculator gives you the almanac values and it also gives you the elevation corrections separately along the right hand side of the table. I suspect that's where the problem lies.

Refraction is a function of altitude, the higher the elevation the less the correction.  Directly overhead it is zero, for objects near the horizon it can be quite large. That's why when you see the sun rise clear of the horizon, it is actually still below it.  The light bends in the atmosphere so you have to correct for that.

I'm not surprised the azimuth is OK, the earth is so big your estimated position could be a hundred miles off and the azimuth probably wouldn't be affected. But every minute of arc the elevation is off is 1 nautical mile of error transferred to the earth's surface.

Parallax is especially a problem with the moon, because its so close to earth that where you are on earth affects its position in the sky, so you have to correct for it.  It consists of a horizontal (HP) and elevation component (PA) so that PA =HP cos H, where H is the raw sextant altitude. 

The other corrections, dip and index error, cannot be calculated ahead of time. They depend on high your sextant is above MSL, and what its non-adjustable mechanical error is. These have to be determined at the time of observation by the navigator, for his ship and his instrument, respectively.

PS: Just to make sure your problem is with the software, not your check procedure, let me see if I can duplicate your figures.

Give me the exact lat and long you are using, and the precise time and date using your local time to the second (which I presume is EDT) .  I will try and determine the elevation and azimuth at my end for that moment in time at that spot.