Determining Latitude and Longitude from the Sun

Basic principles for determining latitude and longitude from the Sun. Background music is by Bach and Haydn, two composers who lived during the age of exploration.
YES there are actually TWO math errors in this video !!! Ahhhh - it's a math based video and there are TWO MATH ERRORS!!! Well, some day I will fix this!! Both example 3 and example 4 contain minor errors of multiplication.

[@csvaughen]

Sorry!! There are two errors in this video and I haven't found the time yet to re-do the video. Both example 3 and 4 have errors of multiplication. Example 3: 7 hrs x 15 deg/hr = 105 degrees (so final answer is 115.75 E) Example 4: 6 min x 0.25 deg/min = 1.5 min (so final answer is longitude 76.5 W)

[@nathanbruneau1822]

You were testing us ;) That's a good teacher :P

[@jayh9529]

It's better understood with the astrolabe device before they transferred it to a globe ,globalists lol horizons curve apparently

[@jima4656]

Example 4 - 5:06 after noon - 6 minutes is 1/10 of an hour, so 5:06 is equal to 5.1 hours. If longitude is 15⁰ per hour, then 5.1 x 15⁰ = 46.5⁰ or 46⁰ 30'. Is this not correct?

[@jima4656]

Never mind. I fat fingered the calculator. It's 76.5

[@tcz1757]

I was going to point it out, but you got it man.

[@timfronimos459]

This topic is one of my favorite. I was trained at US Navy OCS in celestial navigation. We had an amazing MasterChief as our instructor. Thirty years later, I can still perform most of the calculations due to his masterful teaching.

[@charlesklimko492]

I was Kilo Company, Class 83006. I reported on 22 JUL 83. My dumb, immature ass disenrolled.

[@Bazzawombat]

at 20 mins into your video, time is 5.06 (1706) London-GMT. That makes 5 hrs ahead of GMT= 15*5 = 75. ).25deg per min, add the 6 mins is 1.5 deg, total 76.5 degrees west, or did I get mixed up somewhere?

[@helenfu77]

I finally understood how latitude and longitude are calculated. Thank you for the teaching!

[@ninjarider2941]

I like this Solar meridian transit method of celestial navigation because it is the simplest. For demonstration purposes, you can use a modified protractor in place of an expensive sextant to measure the Sun's declination (see David Colarusso's quadrant). However, this method only works during local noon when the Sun is on the prime meridian. You need to know the exact time that the Sun culminates, which can be obtained with the equation of time. It is also shown at the 'Heavens Above' web site under the 'Astronomy: Sun' section for any given day (so is the Sun's declination). The Catch-22 is that you have to first specify your location to the web site. For maritime and aviation navigation, the alternative 'intercept method' is commonly used (there is a Wikipedia article about it). This method is more complicated, but it works with other celestial objects, and measurements can be made at any time day or night under a clear sky. It involves spherical trigonometry, but the data is tabulated in various ephemerides such as the Astronomical Almanac, the Nautical Almanac, and the Air Almanac published by the US Naval Observatory. The latter is currently available as a free PDF file at the US Govt Printing Office web site. A sight reduction table such as HO-249 is also required. Another free annual govt publication is called Astronomical Phenomena, which predicts planetary configurations, eclipses, etc. Regardless of the method used, accuracy is important. The declination measurement must be corrected for dip angle, atmospheric refraction, and instrument calibration error. An error of one arc minute corresponds to a linear measure of one nautical mile. A timing error of one second corresponds to 15 arc seconds which corresponds to 1/4 nautical mile. Note that an ordinary watch measures civil time rather than true solar time, so there is a difference due to time zones, daylight saving time, and the equation of time. The USNO or NIST web sites can be used to synchronize a watch to UTC/GMT.

[@condor5635]

Excellent overview, bravo. Thanks for sharing!

[@henrywilson5204]

Simple but to the point, outstanding class. Thank you.

[@hari20001]

Hi Chris, Really enjoyed this educational talk, Thank you, great fun.

[@joecastanza9822]

Okay!!! I'm a theoretical physicist, a former airline pilot and an expert navigator!!! I dabble in celestial navigation for fun and, sometimes, professionally. I just want to say, this was very well put together. Despite some mathematical errors, it was a wonderful presentation and, yes... the longitude problem was quite a big deal back in the early days, ESPECIALLY when the trade routes opened up... and, of course, when people were beginning to sail to the "New World". Bravo, Christopher!!! Nice work.

[@csvaughen]

Thank you!! I know - I can't believe after all the time putting that together it has a couple of basic arithmetic errors. Eventually I have to redo this and repost it. Anyway, if you are interested, as a physicist and airline pilot, please check out this newer video on great circle navigation kzbin.info/www/bejne/fInFoK16h8uoirs. Absolutely fascinating mathematics, and there's more at sites.google.com/view/kspmath. Thanks again for the kind comments!

[@dgmansmile3051]

great overview. really helped me to understand the concepts before learning celestial navigation. thank you

[@GrahamDodd8]

One of the best presentations of navigating by sun.

[@r.firerunner7017]

Thank yo Christopher. I have been wanting to find a way to determine both latitude and longitude by the sun. Great explanation and thanks for making the post on the two multiplication errors in examples 3 and 4. I am glad you did. Admitting the errors proves we are all human! Keep up the fantastic content.

[@csvaughen]

thank you - yes, old video, if I did it now, I think I could do much better, some day I'll do a new version

[@Gismotronics]

Good info, well presented. Thank you. I am trying to find out ways to counter flat Earthers and it seems to me that the navigational aspect would be impossible to explain with regard to a flat Earth model. I just can't figure how how to explain this in a definitive way! As I understand it, the flat Earth model requires that the north pole is in the center and the lines of latitude surround the center like onion rings until they reach the 'ice wall'. The lines of longitude start at the north pole and go out in straight lines - until they also hit the ice wall. I know, I know... crazy stuff but please try to think of undeniable facts/logic where the flat Earth model breaks down, given fact we know about global navigation. For example, would it be even theoretically possible to calibrate a sun's trajectory above a flat Earth (they say 3000 miles above Earth) in a way that replicates the navigational techniques for finding latitude and longitude outlined in this video? I would also wonder if 'east to west' travel, according to the flat earth model, would necessitate going around a large circle along or in parallel with a line of latitude... Whereas, if we were on flat earth, traveling in a straight line would by logical reasoning always result in hitting their ice wall? Oh boy.

[@csvaughen]

I think most flat Earther's are actually just joking, to get a reaction, but if not, I'd ask, is Mars flat? What about all the other planets? The moon? Why do we have night and day? How do long airlines plan long distance flights? Probably best to engage on things that make a person curious and stick with that, encourage curiosity, without insults, and conclude that a spherical model of Earth does a very good job providing useful accurate predictions and if any other model was more useful then I would choose the more useful model.

[@free2contemplate]

Extremest helpful. You have a gift for clearly, concisely explaining concepts.

[@AceNinja2112]

Excellent tutorial! Thank you for making it!!!

[@josephlai9759]

Thank you for your clear explanation which has been made meticulous plain and simple by the workings of the examples. A well-planned video indeed.

[@luminatuber]

Excellent video, its a thing of past but must know. Now they use gps, speed of light, lasers and atomic clocks for high accuracy using advanced matrixes.

[@johnlbales2773]

Christopher Vaughen Christopher, this video is outstanding! You helped me soooooo much! Thank you. Great video. I subbed. You really advanced my understanding. Clear, timely, & efficiently. You even managed to keep it light hearted. What a delightful fellow. Thanks.

[@csvaughen]

So nice to hear, thank you so much for writing that!

[@greauxpete]

Good stuff.. going on a cruise going to try it with my kids..

[@breezybeats6315]

When people with no respect for J.S. Bach are complaining about your video, you've done well. Very informative 👍

[@poly_hexamethyl]

15:46 This is an approximation that would work if the earth's orbit was circular. But since it isn't, the result you get from this will have an error that varies throughout the year (I think by about +/- 15 minutes). Since the error is known, it can be corrected for if you have the appropriate graph/table (analemma).

[@leo29july]

Great video. I'll be looking for the book and movie you mentioned as well. Thanks so much. :)

[@588158]

Well done! You managed to put the essentials of navigation in a 21 minute video. With GPS we navigation buffs may seem like dinosaurs but knowing celestial navigation is a statement of resourcefulness and independence. I dream that some day someone will have a Celestial navigation computer simulation where the player could take sites and navigate around the world.

[@robinj.9329]

What is not realized is that the incredibly weak signals from our GPS satellites is very prone to "Jamming" or scrambling! As the Communist Chinese have demonstrated numerous times over huge areas of the Western Pacific ocean! This is one of the main reasons that our US Naval academy has had to reintroduce Celestial Navigation into it's curriculum! The LORAN signals, their (much more powerful) transmitters being land based were/are much harder to jam!

[@QuantumMechanic_88]

You might be surprised how easy it is to make a clinometer - basic sextant and accurate compass . Making an hour glass is an entirely different thing .

[@paulscousedownie]

just goes to show how easy it is to make simple mistakes, thats okay if your way out to sea not so good if you're making a landfall. But the explanation was clear and concise.

[@theyoungmarshjones]

Thank you so much for the informative video! It was very helpful. And "Longitude" is a great book, highly recommend it. I have a puzzle I'm working on that you might be able to help me with. My survivor wakes up stranded on an island with no tools to measure the sun's declination and no radio. He does not know which hemisphere he is in, but the sun seems to pass directly overhead, and the climate is tropical. By standing on a mountaintop, he can observe the sun rising and setting upon the ocean. He can even time the sun's journey. Does he have enough information to find his coordinates, and what steps should he take to do so?

[@mikepuckerburg1568]

Sun directly overhead is elevation is 90' Only happens at the Equator so you already have latitude is approx ZERO ,! Next longitude well complex math you can take a tangentially revolving ellipsoid and compare the Exact daytime to find where on the solenoid you'd be Another much easier way is brute deduction as there are very few remote islands right on the Equator IE Maldives or Kiribati or Sao or Indonesian Archipelago www.thoughtco.com/countries-that-lie-on-the-equator-1435319. Lastly you can estimate elevation by your hand FFS so he does have a tool for estimating elevation it's so easy

[@Andrew-md4rp]

Very clear and concise explanation

[@doglao1236]

Thank you very very very veeeery very much!!!!!

[@williamthethespian]

You have provided the best navigation education I have yet experienced. Thank you.

[@mariemorris7031]

Thank you best tutorial yet! went thru several Thanks!

[@Gismotronics]

About how to know the longitude using a watch set to the time in the UK (GMT) and the local time... That's fine if travelling over land when you can ask a local what time it is but how did the early navigators find out what the local time was when crossing a large ocean like the Atlantic or Pacific? I'm guessing that they would have had to have kept a record of how far they had sailed but I don't think that's correct as that's kind of estimation proved to be inaccurate and, sometimes, disastrous.

[@csvaughen]

they needed a very accurate clock, check out www.rmg.co.uk/stories/topics/harrisons-clocks-longitude-problem or kzbin.info/www/bejne/il7KY2qBiJWsn7s or www.imdb.com/title/tt0192263/ or www.amazon.com/Longitude-Genius-Greatest-Scientific-Problem/dp/080271529X

[@jaakkooksa5374]

How does this work on the flat earth model? :-)

[@neoXXquick]

Isnt there an error in example 3: 7 * 15 is never 75?? but its 105...

[@csvaughen]

yes, thank you, you are right - a goof up there, such a chore to fix... some day.... thanks for checking out the video and for the comment

[@kepspark3362]

@@csvaughen & also in the last example.... 6 x .25 = 1.5. You took 1.75. Anyways, nice explanation! Thanks!

[@csvaughen]

@@kepspark3362 yeah, I can't believe how many errors I made in that video. I REALLY want to redo it, but I'm working on another project right now, I'll get to fixing this sometime

[@voidisyinyangvoidisyinyang885]

@@csvaughen errors probably due to dehydration. Drink more water. thanks - still very concise and lucid.

[@jimegnor4442]

Yea...a few mathematical errors but I figured you were just wanting to see that we were paying attention and not falling asleep in class!! Nice tutorial btw...will have to check out that 1707 British naval calamity you mentioned for additional readings.

[@denisrogers4543]

Thank you for the video. Can one compare the local apparent noon sun at the observer's position to the mean sun at Greenwich, without having a longitude that is constantly shifting position throughout the year? Does the equation of time not come into the calculations?

[@csvaughen]

yes, I think it would, my video just skips over those complicating details, I've been wanting to redo the video, with that level of detail, (and to fix the obvious errors in this one) but it's taking me forever to get around to it... thanks for asking, I'm just hoping maybe this summer I can return to this topic for a redo...

[@veronicalchukwu8780]

Thanks for reviving my lost memory

[@ibraheemtalash5094]

finally someone who explains it !

[@markberry7193]

Great video!

[@nasirabbasj]

Great video, very helpful

[@gilligansail]

Thanks that was a great explanation.

[@ebenv1]

Christopher, nice video but I suggest you also factor in the Equation of Time along with your calculation of Longitude. This changes with I think just more that 31 minutes during the year which is more than 7.5 degrees and can put someone off by quite a bit. Overall a very good explanation. Thanks.

[@csvaughen]

thank you for the suggestion... sounds complicated!

[@ebenv1]

Christopher Vaughen Hi Christopher, no not really complicated at all. You can simplify it by looking at the graph of the equation of time and get an estimate of what that is for the given time of the year. If the equation of time is let's say -4 then it means that solar noon will occur 4 minutes after local solar mean time. Local solar mean time is determined by your longitude.

[@johnlbales2773]

Eben Venter Eben, thanks. Suggestions for source material on your comment? I would like to understand your comment, please.

[@johnlbales2773]

Eben Venter ??? I still didn't get it. I'll look up "time equation." It's been 2 years since your comment, so don't worry about digging this video up. I understand. I can't remember things from 2 days ago, let alone 2 years, lol!

[@t147han]

This is great.Even though you got sums wrong.I understand how to do this now.Thank you.Ive watch a few of these types of video now.And just give up.

[@pukhrajmansion8445]

GREAT 👍

[@veronicalchukwu8780]

Very knowledgeable

[@veronicalchukwu8780]

Very fantanstic

[@MrMike77471]

MATH ERROR: On the longitude equation, 6 minutes * .25 degrees/minute = 1.5 [NOT 1.75]. Therefore, the correct longitude is 76.5 degrees and NOT 76.75.

[@markmorrow3199]

Very well explained

[@SaveStandard]

Please tell lawmakers to restore permanent Standard Time to preserve our connection to latitude, longitude, and solar time. Daylight Saving Time needlessly obscures these.

[@ajepjep6167]

The guitar music on the bacground gets distracting while the speaking happens at the same time. The video otherwise is very good and informative.

[@doctorae724]

Question. How did Magellan circumnavigate the globe if he didn't know where Prime Meridian was?

[@jkbatna]

Very interesting indeed

[@jesse6468]

But for longitude, the sun does not travel straight from east to west all year long due to seasons, so isn't it that the latitude you are on also is of effect on this method?

[@csvaughen]

A rotation of the earth is always 360 degrees any time of year, and a calendar day is always 24 hours, all year long. So every hour is always 15 degrees of longitude and 1 minute is always .25 degrees of longitude. This doesn't change. The method I'm describing in the video is simplified and approximate from what one might really do in practice, but it's the basic idea: and in that method longitude calculations do not depend at all on the time of year. Now latitude does depend directly on the time of year because the maximum elevation of the sun changes during the year.

[@softjet6293]

Still a good video…I got the errors.. but I understood what you were talking about.

[@penttisydanmaa4025]

are you sure that declaration is 23.45 at tropic of cancer? what about 23.5?

[@P6009D]

How to translate Latitude and Longitude coordinates, to MGRS or UTM coordinates without a computer?

[@csvaughen]

Yes interesting question- I’m sorry I don’t have any experience with that

[@philipteyssier]

Would someone be able to explain why when we take a sighting of Polaris, it is a direct read and we don’t have to subtract 90 degrees.

[@csvaughen]

Sorry, I wish I could help with that, I don't know

[@karhukivi]

Because it is almost exactly overhead the North Pole. For precise position there are a few small corrections to be made.

[@samuelgoldenbaum9569]

Content seems great, but i find the background music is pretty distracting.

[@csvaughen]

thanks, I know, sometime I really want to redo this video - so many things can be improved, including the obvious math error

[@dan75401]

Well done video but your math is wrong in example 4 also on longitude. 6x .25 is 1.5

[@hans1066]

Nice baroque music in the background... I think good video but these topics are better conquered with a book in a quiet place...

[@melda9148]

Thanks for trying to help me.. But I'm Dutch and don't understand it in Dutch, let alone in English, oh well.. inadequate here I come!

[@csvaughen]

I could email you the powerpoint and try to answer questions via email. I also have practice questions with work and solutions that could help that I use with my classes. I love this topic, and would enjoy the discussion. I really want to redo that video someday, maybe your questions what help me make it better. I'm so busy with other things right now, it will be a while before I can redo this video.

[@aaronroach5153]

How you got 75 degrees for 7x15 degree hours?

[@csvaughen]

yes, I know, sorry, and that is only one of the two errors in the video ... I have been meaning to redo this video but keep procrastinating..

[@alabamamothman2986]

I don't understand how using a clock can compute longitude. Not taking into account currents and the winds how could it be computed like that. it seems like it could be grossly inaccurate.

[@csvaughen]

I recommend this book: www.amazon.com/Longitude-Genius-Greatest-Scientific-Problem/dp/080271529X there's a movie version also www.imdb.com/title/tt0192263/ it's a fantastic story

[@marcg1686]

Perhaps an example would suffice. You're in the middle of the Atlantic. Your watch is set to Greenwich Mean Time. You note the time the shadows are at their shortest. 15:00. That's 3 hours after the shadows were at their shortest in Greenwich. The Earth rotates at a rate of 15°per hour. Hence you are 45° to the west of Greenwich.

[@prakashbhatt4085]

you saved me!!!!!!!!!!(tomorrow is mah exam)

[@tobiaspiechowiak5432]

Puhh.... I first thought you were lost at sea ;-)

[@MrKlaudiosable]

Tobias XD

[@drdsouza5285]

@@tobiaspiechowiak5432 Me also🤣🤣🤣

[@ryananthony6416]

Thanks for putting that together, exactly what I was looking for. Now just need to figure out how to use a sextant haha...

[@boanthun]

Go to a Nautical Collage. I did in the 1990s and haven't regret it. 6 months paid vacation every year and a good salary. However I have to warn you though. Celestial Navigation is not much used nowadays. We use GPS and Glonass, but still....

[@resonant_theories]

Very good work, i will add a link to your video in two videos of mine regarding ancient solar geometry. I think someone should see your video also to better understand what i mean.

[@csvaughen]

thank you! Too bad I have two arithmetic errors at the end... I've been wanting to do a new version, and physically measuring shadows but waiting for a free day with good weather -- one of these days!!

[@resonant_theories]

@@csvaughen personally did not care at all for the minor mistake!

[@resonant_theories]

@@csvaughen judging from what you know regarding solar geometry and how to use the sun to navigate, have a look at these two kzbin.info/www/bejne/ooraiKCgg9Vqnqs, kzbin.info/www/bejne/Y57KqatmirxjkLc and think how old those methods are!

[@roomofidiots]

Gets more exciting when finding long and lat when it’s not noon

[@csvaughen]

LOL! Sure!

[@sulimanemad9261]

❤

[@justpassingwind]

Why that bloody music this is not a music channel ? otherwise good content

[@freedom_aint_free]

But...but...How does one find the time IN THE SUN?!!

[@bingosunnoon9341]

Who is the target audience for this video?

[@deirdrerobertbean6064]

Yes Both example 3 and 4 have errors

[@QNFee]

the dutch navigate the seas long before all those navigation tools like clocks , between 1600-1700 the VOC lost 335 ships and between 1700 and 1800 they lost 360 ships , that is a loss of 3.5 ships a year on average , comparing those numbers with todays numbers the people in the 1600 were beter navigators than the people today

[@markjensen9453]

You need to refigure your math. 6 minutes X .25 degrees /minute = 1.5 degrees, not 1.75 degrees.

[@Bazzawombat]

Thanks Mark, I just made a comment, thought I'd lost the plot, yep, he made a slight miscalculation.

[@oldbloke135]

I'm not sure the latitude calculation is as straightforward as you make it seem. If you don't know your latitude then you don't know whether you are in the northern or southern hemisphere! What changes the calculation (+ or - d) is not which hemisphere you are in, it is which direction you are facing when you measure the elevation, north or south. It is that which changes the sign of e in your diagrams, not being north or south of the equator. Using your rules, moving from one mile north of the equator to one mile south, changes your latitude calculation by up to 47 degrees!

[@csvaughen]

that is a really interesting question.. I've been thinking to delete this video and redoing it and I'm waiting until I have free time on a sunny day to actually do some real examples... now I'm glad I didn't delete the video yet -exactly because of interesting questions like this - now I have something to think about how to answer when I redo this one... I'm thinking during Christmas break I'll get a chance, hopefully on a sunny day, I'll do an example with real measurements, and now I have another interesting thing to address... what if you are really close to the equator ... I think it's really just going to be whether the sun is in the southern sky (you would then be in northern hemisphere) or if the sun is in the northern sky (you would be south of the equator) and if you can't tell that much, you know you are very near the equator which is basically knowing your latitude... anyway, something that might be interesting to add into a redo of this video once I get the chance... thank you!

[@robinj.9329]

I wouldn't say that the choice of the Prime meridian was arbitrary. It's the location of the Royal Observatory of Great Britain. The Nation that figured out how to navigate by the stars in our modern age. The 1700's

[@csvaughen]

yes, that's right, I meant arbitrary in a mathematical sense, that is, no reason to choose any particular line of longitude when looking at the Earth just physically as a sphere spinning on one axis

[@seedaworld2406]

6×.25=1.5

[@longpinkytoes]

70 people disliked that half of the video title is incorrect there is no way to determine longitude by the sun in this video

[@csvaughen]

Yes, well not only the sun, you also need a clock - check out this video, it's better than mine - kzbin.info/www/bejne/eImynYWJfJh2nac --- I've been wanting to redo mine and do an example like in this other video, need a sunny day and lots of time, maybe during the upcoming holiday break

[@vsplit141]

Background music is annoying as hell. Hard to concentrate and listen to the speaker.

[@mosessalazar5484]

🤔 I'm a Filipino, so I'll make Philippines my reference not England...😂😂😂🤣🤣🤣😎

[@alandonly]

Seems a good video. I had to stop watching because of the completely irrelevant music.

[@auraveenley.8743]

The earth is a flat motionless plane.

[@csvaughen]

or it isn't... check out my new video: kzbin.info/www/bejne/paWYiqCenbWGqbM

[@EmersumBiggins]

Your brain is a smooth non-functional lump.

[@veronicalchukwu8780]

Thanks for reviving my lost memory

[@veronicalchukwu8780]

Very fantanstic

[@veronicalchukwu8780]

Very knowledgeable

[@BertTotzke]

are you sure that declaration is 23.45 at tropic of cancer? what about 23.5?

[@csvaughen]

Interesting question: From wikipedia: The Tropic of Cancer's position is not fixed, but constantly changes because of a slight wobble in the Earth's longitudinal alignment relative to the ecliptic, the plane in which the Earth orbits around the Sun. Earth's axial tilt varies over a 41,000-year period from about 22.1 to 24.5 degrees, and as of 2000 is about 23.4 degrees, which will continue to remain valid for about a millennium. This wobble means that the Tropic of Cancer is currently drifting southward at a rate of almost half an arcsecond (0.468″) of latitude, or 15 m (49 ft), per year. The circle's position was at exactly 23° 27′N in 1917 and will be at 23° 26'N in 2045. en.wikipedia.org/wiki/Tropic_of_Cancer