In this two part episode, Dr. Bob looks at the basic operation of a Sextant for Navigation. For more science and technology discussions in understandable videos make sure you like and subscribe to our channel.
I just bought a Davis mark 15 sextant and I can’t wait to test it out. The video is a tremendous help. Btw, Im very close to replicating the lake curvature experiment. It took time to calibrate the auto level so now I’ll get my brother to hold the metric surveying rod for me. I’ll do as you say and make a video. Sorry for such a long delay.
This is so funny. I actually came here for a good video to give as an example to a Flate earth, how the earth can't be flat... lol - love the precise earth model at: 0:35 :)
This is why I'm a failure. I would get lost ten minutes into the lesson then just keep nodding sagely till I could escape. It is crippling being so utterly stupid, but I suppose somebody has to mow the grass.
Or by doing the same thing with a 2nd or even better a 3rd celestial body. But during the day, my understanding is the same as the other commenter. It requires some time to have passed for you to get that 2nd position. I am no sailor - I can’t even swim! My understanding is that with one celestial body that will put you somewhere in a circle, but you don’t really know where. With a 2nd celestial body it will give you another circle. The two points where those circles touch means you are one of those two points. And a 3rd would put you at the intersection of those three circles. Majorly oversimplified and if anyone thinks this isn’t a good kinda explanation, please set me straight. I think this stuff is fascinating.
Awesome! An actual mathematician that can help me understand something! Whenever I drive a pole into the ground perfectly plumb, the sun will cast a shadow on the ground. I can calculate the angle of the sun with basic trigonometry using the length of the pole, length of the shadow and the 90 degree angle. So take that information and calculate the distance of the zenith of the sun on earth at that exact moment to the end of the shadow. The angle is the same. So the distance of the hypotenuse and legs of this triangle should be proportional to the triangle of the pole and shadow triangle. The distance cannot be more than 24,000 miles (the approximate circumference of the globe). It’s honestly around 3000 miles if you use the multitude of apps/websites showing you the zenith. Question, how is the sun 93,000,000 miles away using the empirical evidence of the shadow,angles and known distances? Oh, I get it😅 we don’t live on a plane!! Well here’s the next question, say the earth is a globe, using the Law of intersecting tangent lines of a circle/sphere, using cities viewing sun rise sun set simultaneously, it still doesn’t compute the sun being 93,000,000 miles away with the given distance of half the globe. In other words, you have 2 cities on the same latitude, one viewing a sunrise the other viewing a sunset, the line of sight from the city is a tangent to the sun. The two tangents intersect at the sun. It is impossible for the sun to be 93,000,000 miles away when the radius from the cities to the center of the globe is approximately 3,963 miles. And if the sun isn’t 93,000,000 miles away using basic geometry, then we have a huge problem with the heliocentric model. And don’t give me the garbage, cop out parallax measurements. Start with known irrefutable distances and measurements on earth and go from there. Show me how you compute the distance to the sun using the shadows cast at different cities at a given moment. It should be a very basic computation that even a junior high student can walk outside and prove the distance.(approximately)
Wow you are pretty deep in that rabbit hole. That isn't how you do it. The transit of Venus is better and yes, involves parallax. If you want to know the mathematics, here is the first of the 2 videos that show you just how to do it. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-99PjF6HxD2A.html
@@sliderulesandmathematics9232 nah man. That tells me enough about you. Parallax is a cop out lie to fool the masses. If Eratosthenes used geometry to “prove” the circumference of the globe, which is also a lie, then you can use geometry to prove the distance to the sun.
@@sliderulesandmathematics9232 to find common ground here, could you please tell us how much of a 500 foot tall building should be hidden by the curvature of the earth at the following distances with the given viewer height and assuming the exact same elevation: 6 foot viewer height 500 foot building 3 miles away 6 miles away 10 miles away 15 miles away 20 miles away 30 miles away 50 miles away If you could, tell which formula you use to compute as well. Thanks in advance.
walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator That is the link to the Advanced Earth Curve Calculator. Input your observer height, distance and click standard refraction. That will let you fill out your table.
You bet story of guys in gippesland in Australia, old fisherman used sextant to measure angles on lighthouses ,during a rally bad electrical storm old fisherman able to get to harbour others using gps had to heave too. Electrical storms will knock out gps.
So you seem to understand. HOWEVER, the earth is a globe, and sextants work, so you need to dismiss your false belief. Now rewatch the video and see why it is REQUIRED for earth to be a globe and the stars to be distant for the sextant to give an accurate fix.
Several centuries of sextant use for celestial navigation prove you wrong. Two questions: 1. Why, after taking your sight, do you have to use spherical trigonometry to find your position on Earth? B. Why has no FlAt EaRtH JeEnYuS ever shown how a sextant and Euclidean geometry can give your position on Earth? Inquiring minds want to know. Elucidate us.
@@sliderulesandmathematics92322:02 “ANGLE OF ELEVATION: The angle formed by the line of sight and the ‘HORIZONTAL PLANE‘ for an object above “THE HORIZONTAL“ There are no horizontal planes on an oblate spheroid in a vacuum.
can you explain the lack of a measurable curve? if you have the size of the earth given by nasa, use spherical geometry and you can see exactly how much curve should occur at a given distance. There is no measurable curve and the surface of the earth is not a ball lmao. there is not people living upside down under your feet dumbass we are all on the same plane of existance. the earth is flat just as it looks to the naked eye. the higher up you go the more you can see, the horizan always rises to meet your eye level. this is a property of a flat surface not a curved surface.@@sliderulesandmathematics9232
