Can You Tell Me Where I Am By Using These Two Sextant Readings? -- A Great Math Exercise
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- Опубліковано 30 вер 2024
- Can you calculate my latitude and longitude? I took two sun shot measurements using a sextant through an artificial horizon --one shot in the morning and one shot in the afternoon with a time signal.
Wolfram MathWorld Spherical Trigonometry mathworld.wolfr...
Science Math Celestial Navigation Artificial Horizon latitude longitude sextant location sun spherical trigonomentry
This challenge was so interesting that I now have bought a second hand and unused Freiberger drum sextant. I also orderd a Davis artificial horizon, same as you have.
Nice job to Rob Howell. He says he solved it by using Nories tables 1983 edition.
Nice job Mehmet Guzey. You are within 0.25 miles.
Soon I will delete your answer so it won't be a spoiler.
Reply
Mehmet Guzey
Mehmet Guzey7 hours agoHighlighted reply
Thank you so much for this great exercise. I learned a lot and had fun.
Congratulations to jrinconr. Jrinconr's solution is within 2 miles.
This is the video I was looking for to understand how to use an artificial horizon with the sextant. Thank you so much!! Greetings from Spain!
Yes, they are very easy to use. Also, there is no need to consider height of eye (dip angle) as with the sea horizon.
Congratulations to andypdq. His answer was within 0.25 mi. He says he worked with pencil, paper and a calculator, then plotted using the altitude intercept with assumed position method.
björn: Your solution is about 1/2 miles from the actual position. Not bad. I will delete your answer soon so that it won't be a spoiler.
Congratulations to yaronh hnoray. Yaronhy hnoray's solution is within a mile.
what is the date and the time?
iviewthetube
1 year ago (edited)
+Michael Jones Congratulations Mr. Jones. Your calculation is within 3 miles from the actual position. Yes, you are correct about the artificial horizon. Limb shots are taken by joining the reflected edge of the Sun with the direct edge of the Sun.
Congratulations to Axel Woters. His answer was within 1/2 miles. Says he used a Self written program version 0.1 :-)
Congratulations to +Ryan Hardy. Using a Python script he wrote with the PyEphem library, his answer was within 0.6 miles.
UA-cam user 19Yves84 submitted his answer and got within about 0.3 nautical miles from actual.
19Yves84 performed his calculation using paper, a Texas Instrument TI-83 and great perseverance.
49°36'34"N
121°36'05"W
I just got into celestial navigation so I am very curious how I did. When making altitude corrections, how does the artificial horizon effect it?
The artificial horizon produces a reflection off the water. The actual horizon is precisely half way between the direct image and its reflection. You must divide the measured angle by 2.
@@iviewthetube I think there is still a small altitude correction required which is caused by refraction. The almanac has tables for that divided into Oct-Mar and Apr-Sept and for angles less than 10º and angles greater than 10º. It is also divided into sun, stars and planets, on one table and the moon on another.
Yes, there is no dip correction with the AH. But there is still a small refraction in the atmosphere when looking up at high angles and it is not cancelled out by the short path of the same light reflected from the liquid in the tray. There is only one "ray" coming down from the body being measured. The general altitude correction accounts for parallax and this refraction. @@mikefochtman7164
@@karhukiviIf you use the 'Altitude Correction" table for the sun, based on whether it is a lower limb, or upper limb sighting (be careful here, ivewthetube performed one of each in this video), that correction has refraction, parallax, and semi-diameter all combined. You can, of course, look up the refraction for the measured elevation (often found on the same page of the almanac) and the sun's semidiameter (found at the bottom of each GHA column in the 'daily pages' of the almanac) and apply these corrections separately.
Yes, I think most navigators use the general altitude correction for the Sun. @@mikefochtman7164
@iviewthetube I certainly will email you but be patient. Celestair is just processing my order for the artificial horizon and I think I need some excercise before I get as good mesurement as you have. They are remarkably good for a plastic sextant.
Congratulations Greg Rudzinski, your answer is correct. Your Pilot Navigator Version 6.9 seems to be working correctly. The mirrors were adjusted to zero out the index error prior to use. Perhaps I should have demonstrated that in the video.
IFRINST is 71 years old. His solution was about 3.5 miles SE of actual location.
Congratulations to björn johansson. His solution was 0.5 miles from the actual location.
Congratulations to IFRINST. His solution was 3.5 miles from the actual location.
bjorn: Unfortunately I ended up in a small wood but
there are still some houses near by.
I used Google Earth to plot the results from
the equations in spherical trigonometry.
I had some doubts on howe to deal with the upper
and lower limb with respect to the artificial horizon
If I'm wrong perhaps this is the reason
Congratulations to Stan Klein. His solution was within 1/2 mile.
Nice job, brohen. Very similar to others' results within 1/4 nautical miles away.
HE ID OBVIOUSLY IN THE UNITED STATES!!!!!! JEEZ LOOK AT THE FLAGS.....LOL
Congratulations to Giorgos Raptis. His answer was 1/2 Nm from actual position.
Congratulations to Roland Dutton. His answer was 1.4 miles from the actual position.
Congratulations to +rtbertz Nice job rtbertz. Your calculation is within 1/3 mile of true position. Nice job. I will erase your answer as to not be spoiler.
Congratulations to sm2iuf who submitted the correct answer using Henning Umlands Javascripts as an almanac.
Congratulations to Cori Corgi. His solution was within 2/3 miles.
mikefochtman7164 Calulated the position to within 1 mile using the spherical law of cosines done in Excel spreadsheet of his own making.
Dip does not need to be considered when using this type of artificial horizon.
pmh099 submitted his answer and got it within a 1/2 mile.
pmh099 performed the calculations with Microsoft Excel using Navigation Spreadsheets.
@iviewthetube I can't figure out the correct UTC hour for the afternoon shot. Minutes and seconds is clear. The audio from WWV is strange when he says the hour is 2 .....
By guessing the hour to 21, I get the position 48 N 59.6' and 92 W 16.6 ' This is in a lake in Canada but you are obviously on dry land.
@iviewthetube Be sure to note that Excel calculates in radians rather than degrees.
@bodybager Congratulations. You made a very precise calculation.
I've used the T-2 for sunshots also for establishing a basis of bearing for surveying; you are lucky to have such a great gun.
Theoretically these sun shots could have two mathematical solutions -- one in the Northern Hemisphere and one in the Southern Hemisphere. The flag is a nice hint, huh?
I place more stock in seeing you rotate your chair right for the PM shot. Flags cam be false.
Thinking back to my Bowditch, I initially thought noooo, you need to shoot a second celestial body so as to get two circles. But wait, being stationary, the 2nd sun shot gives a different circle on the same longitude. Whether the 2nd circle is bigger or smaller than the first, they will intersect at only one point, your latitude. I solved it in concept, I have no need to wade through the math. :D
You can plot those two circles on Google Earth using the circle measuring tool. For greater accuracy and resolution, you can write your own kml file to draw those circles.
How cool is this? Thanks! I was searching 'Davis Instruments Artificial Horizon' and found this. I live inland in Oregon and want to practice with Mark 25. Visits to the ocean beaches are often frustrated by clouds or fog. After seeing this I will be purchasing the artificial Horizon. Looks like it works well.
Yes, the Artificial Horizon works well. Just remember to divide the measured angle by two. Also with the Sun, limb shots are easier than trying to overlay one image with the other.
Another advantage to the Artificial Horizon is that dip can be ignored. With an Artificial Horizon, going up in elevation does not change the perceived horizon.
@iviewthetube Thanks for the time info. I now end up just outside the US west coast in the Pacific Ocean at 35 N 56.6' and 122 N 45.6' this must also be in error. I have checked and double checked and noticed that the first shot is upper limb and second shot is lower limb. However I think I am much closer now.
I will work some more on this problem to find errors in my calculations. I like this kind of problems. Please make some more sun shots when you are visiting relatives in other places.
I also ended up there until I realized he says February and not January.
Just drive to the beach. It’s easier and more enjoyable. Of course it doesn’t hurt that the Pacific Ocean is less than an hour away
This is what inspired me to do this: lewis-clark.org/sciences/geography/celestial-data/
Correct within 1/2 mile. Good job.
iviewthetube 1 month ago (edited)
+Rob Howell Congratulations Rob Howell. Your second pen, paper and calculator calculation put you to within about 3 miles. I removed your second reply with the more accurate calculation so it would not be a spoiler.
+Rob Howell Congratulations Rob Howell. Your second pen, paper and calculator calculation put you to within about 3 miles. I removed your second reply with the more accurate calculation so it would not be a spoiler.
iviewthetube 1 month ago
Rob, your first calc is about 80 miles north.
Rob Howell 1 month ago
+iviewthetube Thanks iviewthetube for the challenge, I had fun putting all the information together to get the final answer. Is 3 miles good or bad in this kind of work???. May go out and buy a sextant, almanac and artificial horizon. Cheers
This was a fun exercise, thank you! I didn't realize it was possible to find lat+lon from just two readings throughout the day. I had also never heard of spherical trigonometry until now.
My answer, using a Nautical Almanac PDF and a SUN HEMMI slide rule:
56°13.15'N, 116°4.5'W
Hello K. I'm not sure what went wrong. Your calculation is 21°, 580Nmi from my position. For this artificial horizon, to obtain the angle above the horizon, be sure to divide the measured angle by two. (1/2 of the reflected angle) Also, to get your angular distance from the Sun, use 90°- the angle above the horizon.
@@iviewthetube Thanks so much for the notes, 14 years later. :)
I made both of those adjustments in my original calculations. I'll review my work!
I tried this method using a home-made sextant-like device, making my own sightings, and found myself off by only 1,000km! I assumed it was an instrumentation issue, but maybe there's a math error somewhere in here. Thanks so much!
Checking: the times are 1726h Feb 20 and 0016h Feb 21 UTC respectively, right? I found it hard to hear.
@@K-vp5sq I don't have my notes handy so I just watched the video again and came up with 2010, Feb 20, 17:26:26 UTC (upper limb) for the morning shot and 2010, Feb 21, 00:16:02 UTC (lower limb) for the afternoon shot.
I may give it a shot. However I hear What I think are Stellers Jays squawking in the background which gives me a clue of very rough location
A buddy was Asst Navigator and plank owner on the Conny. RIP Chuck. His best was using three stars and laying out a 1 mile triangle on the chart.
What is the position of the sun should in this box of water? I see the sun in the box, what is a line of horizon?
The horizon is precisely 1/2 between the Sun and its reflection.
Is this the only way to find your GPS on land, while using this tool?
Hello M R. The sextant is very good at measuring the angle between two objects. For example, if you turn the sextant on its side you can measure the angle between two towers. If you know the distance between those two towers, with trigonometry you can determine an arc of your possible positions. Regarding Artificial Horizons, there are other types other than the one I have shown.
Ha, ha, are you trying for centimeter accuracy?
Temparature that day was about 70 degrees Fahrenheit. This sort of weather the barametric pressure is usually around 30.00 in Hg. The elevation won't affect the dip, but it might affect the amount of refraction. Elevation here is 200 feet above sea level.
I must have deleted my comment when I tried to X out my answer.
Thank you for the most entertaining video I have seen.
@sm2iuf
That is not correct. You are about 1400 miles east.
Regarding the time: The afternoon shot is the next day UTC (Feb. 21, 2010). Try 00 hrs., 16 min., 02 seconds.
@iviewthetube 19Yves84 informed me that most of his calcs were done on Microsoft Excel and that a scientific calculator is not even needed.
Ok, so since I'm not so good at math, I tried to go the easy way and use almanac tables, but I couldn't find conclusive results with it. So, a bit frustrated, I resorted to the best tool I possess : my sense of logical deduction (and internet)!
First I used a bunch of clues from your video :
1. American flag and American accent : You are in USA.
2. No snow on the ground, wearing only a shirt in late February : You are not in the NE nor the Midwest states.
3. The vegetation around you looks like a temperate climate zone (non-arid) : You are in the NW states.
4. You refer to 1726 UTC as "morning" and your "afternoon" shot is 7 hours later. The sun is (obviously) up (and not setting) in the second shot and the sun sets before 18:00 local time on the west coast (UTC-8) at that time of year : The only time zone where you can be is Pacific Standard (UTC-8).
Then, I opened two suncalc.org tabs and entered the morning time in the first and the afternoon time in the second. All I had to do now was to patiently move each location until the sun altitude of both had the correct reading (half your corrected measurement, 19.87 at 09:26 PT and 12.23 at 16:16 PT) when at the same place.
*You are here : 47°30'13.08'N 122°6'45.69''W*
This method as a precision of about +/- 1 mile from what I can see, so I hope you can respond and tell me how close I got!
If you don't know math, you can string two circles on a globe or Google Earth. Your answer is within six miles.
How do I get you my answer without divulging it on the youtube screen? Can't seem to find an email address, but hopefully I have found your position.
You can provide the lat, long on this thread. I will congratulate you if you are correct and delete the correct answer so as not to give it away to others.
Nick Evans got it to within 1/2 mile.
47.32N 121.90W, sorry I can't understand all words in the video. Can you write the time and the date in the comments?
The morning shot is an upper limb shot taken on February 20, 2010 UTC. The first beep is 1726 UTC and the second beep it 1727.
The afternoon shot is a lower limb shot taken on February 21, 2010 UTC. The beep is at 0016 UTC.
Could you make another vid that goes into more detail about the artificial horizon (AH)? Please explain the concepts, demonstrate in detail how it works, and the adjustments you have with your calculations. Have plenty of illustrations, drawings, and of course demonstrations. Can you use the moon, and the brighter stars with the AH?
The concept of this type of artificial horizon is very simple: When you see the sun's reflection in a calm lake, the horizon is precisely half way between the sun and its reflection.
I don't like to use the Moon for celestial navigation because there is too much parallax to consider.
Stars will work on the very brightest stars with a very dark, clear sky; but it is very difficult.
Oh man... really? Why not just write out the problem data and take a picture for us? Be happy to give you the answer. Jiggle-jiggle and have hearted information about some book you think you want to buy? Try harder next time.
Crazy, send my your sun measurements via video and/or comment and I'll make a video on how it is solved.
Hi! I have a question about this problem. Combining 2 equations, I obtain 1 equation with the unknown being the latitude. Though the equation contains Lat and sqrt(1-C*cos(Lat)^2). Now before I continue solving this via iteration methods, is this how it is meant to be solved? I hope my question is clear.
+David VDC This video shows the order I solve the angles and sides of the spherical triangle: ua-cam.com/video/uZk7W0i3jd8/v-deo.html
User DO1VX got the correct answer within 1/2 mile.
n205mk submitted an answer which was about 5 miles SSW.
When I was learning, 30 years ago, I used to practice by taking moon shots reflected in my backyard birdbath 😉. I’d check my results against my known position to see how close I got.
It is difficult to use the Moon for navigation because of its parallax.
There are more corrections for the Moon, including parallax (as mentioned in the other reply) due to its closeness to Earth. Also, depending on the segment of the Moon visible, you might have to use the upper limb which can be confusing when using the artificial horizon! But the stars and planets are very difficult to see in the AH reflection so basically it is the Sun by day and the Moon by night when using the AH.
@sm2iuf
Thanks for the update. I am looking forward to hearing how it goes.
Hi iveiwthetube. I want to solve this, but I need help. Do I need to also have some sort of Assumed Position (long. and lat.) that is relatively close to where you actually are, which it seems is what celestial navigation books say you will need in addition to a time and sun sighting/angle?
I am referring to this: en.wikipedia.org/wiki/Intercept_method
thanks
Since many calculators today will solve different variables in an equation, I prefer to use spherical trig over dead reckoning.
I solve the spherical trig angles and distances in this order: ua-cam.com/video/uZk7W0i3jd8/v-deo.html
If you want a guestimate then use a globe.
1 Make a tape measure that is calibrated from 0 to 360 degrees for the circumference of the globe.
2 Locate where the Sun was on the two measurements.
3 Using your special tape measure draw a circle from the each Sun location a distance of (90-sextant reading)
4 There will be two points where the circles intersect; one of them will be the correct solution.
@sm2iuf
Now you are about 800 miles south.
Will you please explain the difference between a lower limb shot and an upper limb shot using an artificial horizon. I was led to believe that the two images were superimposed when using an artificial horizon.
I suppose you could superimpose but that would be very difficult; whereas, limb shots are very definite -- just touch the edge of the direct with the edge of the reflected image. In January the Earth is closest to the Sun and has an apparent diameter of 32′32″. In July the apparent diameter is 31′27″. The chart on the first page of The Nautical Almanac corrects for both refraction and the radius of the Sun for both an upper and lower limb shots depending on the time of year and the altitude of ones shot.
@@iviewthetube This guy seemed so sure of himself at 13:08 of this video: ua-cam.com/video/i9gUs3cLgx0/v-deo.html
Thanks for the tip.
@sm2iuf
Hey, how are your Sun shots going?
I have no answer, rather I have a question. Suppose you were David Thompson the explorer . How did he know the DIP when he did not know the ELEVATION at any given location? Charting the direction of a river would seemingly be impossible without reasonably accurate DIP correction.
The angle you want to measure is the angle between the celestial object and level. Dip corrections are not needed when using an artificial horizon such as a reflection off of water or a mirror leveled with a spirit level because they are level, no matter what elevation one is at. Conversely, dip corrections are only needed when you are measuring off of the physical horizon which lowers as you increase in elevation -- the horizon becomes no longer level as you rise in elevation.
iviewthetube Thank you ever so much for replying to my question so quickly. Now I have something to do, to make the time pass. I need a diversion to occupy my mind during this Flu virus. Taking Sun Shots should help the time pass. I guess Star Shots are out of the question on land when reflection off an Artificial Horizon is impossible.
No dip correction for the artificial horizon
I would very much like do this calculation. My problem is to figure the date and the time.
The noicy sound in combination with my poor hearing and also being a forigner
makes it difficult for me to get this data correct.
Björn Johansson
Morning shot: 2010 FEB 20, 17:26:26 UTC (Upper Limb shot)
Afternoon shot: 2010 FEB 21, 00:16:02 UTC (Lower Limb shot)
Good luck!
@@iviewthetube I was using lower limb shots for both of them. Also misheard the first time to be 36 min. Rewatching it, you made it clear.
Would i do a correction for refraction or is that only over the ocean?
@@yottawatt I announced in the video whether I used upper-limb or lower-limb. Also, there is refraction on land. However, you can ignore the dip since the artificial horizon does eliminate that error.
48.060, -122.186
Why didn't you step through the entire process? The tutorial on the artificial horizon was excellent. Turning the sextant and time readings into an actual position seems to have been left out.
david
This video illustrates the steps I use to calculate position using Spherical Trigonometry. ua-cam.com/video/uZk7W0i3jd8/v-deo.html
www.starpath.com/online/celestial/vernier.pdf
Sean, If you provide me a video and/or info with your measurements & time of your position then I will provide you with a video on how to do the calculations. If you do not have a sextant, then I can give you a rough estimate if you use a string and protractor. Keep in mind that I need a very good UTC time mark to get accurate longitude.
How is the assumed latitude estimated without a DR position or a noonday sunshot? As your shots are approx 5 hours apart, I could guess that the noonday sun might be 40 deg and adding the declination that gives me about 50 deg N (US flag) but is there another way to obtain Ap LAT?
I use spherical trigonometry which does not require dead reckoning. Here is the order I solve the angles and sides: ua-cam.com/video/uZk7W0i3jd8/v-deo.html
@@iviewthetube Thanks I got that. In effect your methods needs two measurements for calculation as opposed to the intercept method which needs (at least) two position lines calculated from two measurements. The "assumed position" required for the latter is just a convenient point nearby (can be DR or something else) so that the position lines are not too long (they are circles, in fact). In effect your "assumed latitude" is replaced by the zenith angles from the Pole, and the "assumed longitude" is replaced by the GHA from Greenwich.
My interest in astro nav is also from reading the Lewis&Clark expedition and Sir Francis Chichester's round the world trip where he was relying on an ageing sextant with the silvering falling off and making mistakes due to fatigue. I'll get back to you with my results - thanks again!
@@karhukivi Admittedly, computers & scientific calculators make the spherical trig method a lot more possible.
@@iviewthetube The intercept method also uses spherical trig to calculate the Hc or calculated altitude and the azimuth. Then it reverts to a plotting sheet to locate the position. The spherical trig calculations were summarised in the sight reductions books before calculators were available. however not every position of latitude was compiled (the books would have been enormous) so only integral latitude values were shown. This accounts for the navigator having to round up or down the assumed latitude by shifting the assumed position (or DR) to an appropriate value that would cancel out the minutes and decimal minutes. I sometimes get schoolkids to make a rough sunsight at local noon using a simple clinometer and a compass and watch, and they are astonished how it can place them on a globe to within 60 nm. The clever ones are then shown the trig and the almanac, and if they are still interested, we have a go with a sextant!
Visually you can use Google earth using the circle measuring tool set for units of nautical miles. Center the mouse on the Sun's position and draw a circle which represents your angular distance from the sun (1 degree = 60 nautical miles)
Do this twice and one of the intersections of the two circles represents your position. If you want it to be accurate then use Excel to write a kml file with more resolution.
What liquid do you use in your AH now? You used water in the video but I found water leaves condensation on the coloured windows. Cooking oil is messy and, if left too long, leaves a sticky deposit. Now I'm trying propylene glycol which is sold to people who make their own e-cigarette vaping solutions for about $6 a litre. It's viscous, clear, dissolves in water and is non-toxic. Did Lewis & Clark use water or mercury?
Water would have frozen on the days that Lewis took his measurements in the middle of January 1805 at Fort Mandan. If I remember correctly, Lewis used a mirror leveled with a spirit level -- sort of like a surveyor's tribrach. I may be totally wrong so I would appreciate hearing back from you if you find out differently.
Regarding condensation, I've rarely have a problem with that. Have you tried very cold water? Seems to me that a non toxic anti-boil/anti-freeze would work fine too.
@@iviewthetube On the summer days when I was working with my sextant, the condensation made a "halo" around the Sun's image and it was difficult to see when the limbs touched. The PG seems to be an ideal liquid - I heard about it from another video on how to remove bubbles in a marine compass.
@@iviewthetube Good to hear from you - I read some historic accounts online and they seem to agree that it was water or a bubble level as you say. One book stated that they only made observations which had to be computed on their return, but I'm inclined to disagree and think they had some ability to compute their positions on the go, as they had a table of lunars in their kit. Many thanks for your insights!
I've used black coffee lol. I get a stronger reflection from the black liquid and it's not too hard to clean up. I let it cool thoroughly before hand to minimize condensation. Other bit of advice, don't put the glass shades on the artificial horizon until just before taking the sighting.
Hello iviewthetube,
would be nice to point me in the right direction. The methods I have read about so far in several webpages are always need more or less accurate position that get corrected by 2 or more "shots".
best regards,
Thomas
First step: What date, times, and measurements did you come up with on the two Sun shots? Were they upper limb shots or lower limb shots?
Hello, thanks for the quick response. Here are the measurements I have got from the Video:
20.02.2010
Morning: 17:26:26
Upper Limpshot Sun
40°18.2' /2 = 20°9.1'-18.3' = 19°50,8'
21.02.2010
Evening: 00:16:02
Lover Limpshot Sun
24°4.4' / 2 = 12°2.2'+11.7' = 12°13,9'
regards,
Thomas
Thomas Baumann OK good, close enough. Now determine the Sun's GHA and declination at the time of measurement for those two shots. Watch this vid to determine the order in which you will solve the spherical triangle's sides and angles: Celestial Navigation of Lewis and Clark
Thomas Baumann For some reason you are about 20 miles SSW
iviewthetube I found my mistake, the formula to calculate the declination of the sun was not that accurate. But thanks a lot for your very helpful hints. Greetings from
31°38,2' - 09:13:35 (UTC lower limb SUN)
36°1,1' - 11:33:15 (UTC lower limb SUN)
What were you listening to that was giving time?
WWV transmits time signals on 2.5, 5, 10, 15, and 20 MHz.
iviewthetube 10-4 Good Buddy, I ended up spending the morning finding that out which lead me to another hour or so figuring out how they have a W call letter. I'm a nerd like that. It's great how it all goes together in some way shape or for.
@@TempoDrift1480 When I was a ham radio operator I had the call letters KE7I. I let my license lapse and those call letters have been snatched up by someone else.
Congratulations to Jacob brown. His solution was within a mile and a quarter.