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^^^ You think it's fucking reasonable to believe that we're stuck to the surface of a spinning ball in a vacuum orbiting a million mile wide thermonuclear bomb in an endless virtually empty universe created by nothing exploding for no reason?
The only way anybody could possibly believe any of that shit is if they were indoctrinated and brainwashed from birth.
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Keep in mind that I'm combining angles from both physical space and optical space and the diagram is less than clear in this regard. R & T also need to be moved over to the left side.
First point, the distance to the sun can't be calculated from the sun's apparent size alone. As you can see I didn't use "I" or "U" at all in my calculation at all, only "Q" and "A".
You can if you know the effect of refraction. Do you know how to compute that? ...
Finally the sun is always 32 minutes wide due to refraction and maintains the same diameter as if it was at 90° because of it. As you can see the calculations for the angular size (X) have not been completed yet. In the case of the sun and moon here X will equal U and refraction plays a role such that a refracted optical space is created in addition to the physical and optical spaces.
I see. The sun appears to have the same size because of refraction. That raises a question for me. Why are the sun and moon affected by refraction in this way, but other objects are not? No, no, no and no! You can not calculate the distance to the sun or moon from just apparent size you have to have another object, in this case I used the horizon. Atmospheric refraction is not another object. Refraction can be determined by calculating the sun or moon's apparent size and position then comparing them with measured values. The difference will be the effect of refraction. The sun and moon are affected by atmospheric refraction because there are layers of different density gases below them and they causes the light to change direction. Other objects, what other objects? Stars are points of light and remain points of light after being refracted. @BADecker, Accuracy is irrelevant for determining approximate values. The fact you throw the baby out with the bathwater showcases your intellectual dishonesty. |
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....How accurate are you when you eyeball the distance across your room in the funny farm? If you are anywhere close to the sun, you should have your room down to the thousandth of an inch.
The famous aviator Wiley Post, 1920ish, had lost one eye when young. He trained to estimate distances very accurately with one eye. Hence, with no triangulation. He would often win wagers in bars this way. It enabled him to pass the Airman's exam, which required good vision. So I'll let that issue pass. However, you are correct in that NotBat has not shown the total of the triangulation issue. There are several issues. (A) How much does the Moon change in diameter, from the Supermoon to the minima, and why? I'll need to read up on supermoons; I'm sure somebody has documented and published those values. As for the why I'll say it's atmospheric refraction unless there's other greater factors I'm not considering.What does it mean, if an object in the sky periodically, and predictably, changes in diameter? I'd say it means it has a cycle.(B) From where might we stand, say in a couple of places, to measure the Sun and Moon over the course of a year? This would be to find their distance from the Earth. Anywhere they're visible in your field of view, preferably with a high angle to minimize atmospheric refraction.(C) We commonly measure the height of mountains on the Moon, and the depth of craters, using a form of simple triangulation. Measuring the spreading of the shadows as the Moon moves to a crescent. Notbat, care to comment? Who is "we"? Your "team" (an anagram for meat) is probably making assumptions about the distances involved.(D) Given that the terminator line moves precisely across the lunar surface over the course of a lunar day (27+ Earth days) and we measure heights on the Moon of objects as they approach that terminator, geometrically this is only possible if the Moon is a sphere like object. NotBat claims the Moon is just a bright light? ... projected off of a concave mirror.(E) A lunar eclipse is when the Earth occludes the Sun, as viewed from the Moon. You're making an assumption, what if the eclipse is a filter passing in between the Moon's projection? We will consider the trig involved in this, along with a quite interesting topic, after NotBat answers the above questions and issues. This topic is how we can predict lunar eclipses. I'll have to check and confirm this but I've read that modern astronomers still use the old geocentric flat earth equations to make lunar eclipse predictions; it aint broke so they never fixed it.Any scientific theory is affirmed if it predicts correctly. We have a model of elliptical orbiting masses in space, which we can use to accurately predict many things. All on the basis of a synergistic development of math, and orbital mechanics, over three thousand years. But you reject all of that. No problem. We can step through the fundamentals. There are no heavy balls floating above us being held up by a magical force a butterfly can overcome by flapping its wings. The lights in the sky are just that, lights in sky. |
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Keep in mind that I'm combining angles from both physical space and optical space and the diagram is less than clear in this regard. R & T also need to be moved over to the left side. let me go over the possible inconsistencies, I was overcome with fatigue for no particular reason and passed out for 5 hours after posting that... edit: There is an inconsistency. If Q is 59.73 degrees and E is 1.99 nm and the angle marked 90 degrees is 90 degrees, then Z can't be 3068 nm.
Z = E x tan Q = 1.99 x 1.73 = 3.4 nm
I think you are making it overly complex. If the apparent size of the sun is 32 minutes and it is 32 nm wide, then the distance to the sun can be calculated with this:
distance = width / 2 / tan(angle / 2) = 32 / 2 / tan(0.533 / 2) = 3438 nm
Now that you know the distance to the sun, you can calculate its height:
height = distance * sin(Q) = 3438 * sin(59.73) = 2969 nm
There is a problem. If the sun is always 32 nm wide, then it must always be 3438 nm away, and if it is always 3438 nm then its height changes as Q changes. In other words, all three - width, height, and distance, cannot be constant.
First point, the distance to the sun can't be calculated from the sun's apparent size alone. As you can see I didn't use "I" or "U" at all in my calculation at all, only "Q" and "A".
Second point, like I mentioned there are two spaces combined into a single diagram/calculation. The left side is mostly optical space for calculating physical distance (X=angular size) and the right is mostly physical space for calculating optical angles (U=apparent size).
I hope my second point makes sense.
The square angle triangle formed by E-Z-O is apparent and Z can't be calculated on this side. The square angle triangle formed by D-Z-[unlabelled leg between sun and horizon] is physical and the distance is calculated on this side. The square angle triangle [unlabelled leg between horizon and the ground below the sun]-V-D is where the physical angle to the sun is calculated. So the left side is actually two triangles with P = T.
If that makes sense.
Finally the sun is always 32 minutes wide due to refraction and maintains the same diameter as if it was at 90° because of it. As you can see the calculations for the angular size (X) have not been completed yet. In the case of the sun and moon here X will equal U and refraction plays a role such that a refracted optical space is created in addition to the physical and optical spaces.
If that makes sense you get an award!
Overly complex, no, no not exactly...
@BADecker,
I could be wrong but I think the aspect ratio is maintained when changing the resolution limit.
edit:
I updated the image to include the calculation for Z, I didn't realize it was missing. I also made a few other adjustments. |
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This is what I've got so far:
[image]
edit:
Removed image/calculation due to horrific errors, I must have mixed up my notes.
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My first attempt at calculating the distance to the Sun resulted in a height of 3068.4 nautical miles. I'm still going over the calculations...
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^^^ I'm upset because you're being intellectually dishonest.
@odolvlobo,
I'll do up a diagram for measuring the Sun and Moon (this may take me a while), the 3100 mile distance calc was not done by me so I'll do it and see what I get. BTW don't confuse angle U and angle X, only with angle X does 1 minute = 1 nautical mile. With the Sun and Moon however atmospheric refraction keeps them the same size all the way to the horizon so X may in fact equal U for these two objects. Their distance makes an accurate measurement in this manner difficult, as BAD keeps pointing out however, an error in measurement of few miles doesn't save the globe.
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@BADecker,
Measuring with a sextant isn't fucking "eyeballing it", you're throwing out direct measurement of the Sun and Moon, pretending it's not possible then giving a lame answer from NASA that's absolute bullshit. If the they're both 32 miles wide and both 3100 miles high then a sextant and the human eye work just fine for making an approximate measurement. You have to beg the question about the distance to the Sun and Moon being millions and hundreds of thousands miles away to table your parallax bullshit.
Go crawl in the oven and shut the door behind you.
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No, no you can use for example a Nikon P1000 with an angular resolution limit in the seconds, at ~6 feet an eye with a 1 second resolution limit can see a ~1 foot object up to ~180 nautical miles away (if the atmosphere was removed). My calculations allow for any eye to be used, I just chose the human eye because there's 1 nautical mile per 1 minute, it's the standard.
The Sun and Moon are measuring ~3,100 nautical miles away, not millions. You claim they're millions of miles away and thus a 100 million dollar telescope is needed for accuracy but you're begging the question (petitio principii).
Stop assuming the Copernican model is correct, it's cut from the same cloth as evolution...
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^^^ That's why they make the Nikon P1000, while I don't have the maximum resolution limit handy I'm confident it's measured in seconds. BTW I do show how I get the the distance, it's based on 1 minute and I show my source for that including the page number.    This means that on a plain with an eye height of 0 a ~1 foot tall object will not be visible beyond ~1/2 nautical miles thus, with an eye height of ~1 foot the distance to the horizon vanishing point will be ~1/2 nautical miles. With a ~2 foot eye height the distance to the horizon is ~1 nautical miles. At ~6 feet the distance is 3 nautical miles. I used (eye_height + eye_maximum_resolution_angle) / 2 to calculate this. Is there a better way? edit: I was trying to avoid a divide by zero error for a zero input and was fatigued and constantly disrupted. It looks like the output is off by 1/2 miles and 0.5 needs to be subtracted from the answer. Help me me fix this, I shouldn't have posted this without double checking but I did... edit 2: I'll change it to ((eye_height + eye_maximum_resolution_angle) / 2) - (eye_maximum_resolution_angle / 2) as a temporary fix. edit 3: You can see how I've obtained the distance to the horizon, even if the way I've calculated it is not optimal you can see where the values come from... |
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@Spendulus, the Moon is 32 nautical miles wide and I can prove it in a court of law with a sextant. You would have to debunk/discredit "Yanoff, Myron; Duker, Jay S. (2009). Ophthalmology 3rd Edition. MOSBY Elsevier. p. 54. ISBN 978-0444511416" to win.
Be my guest, go right ahead. Show your work. It's all right here ( https://i.imgur.com/zS0G3hs.jpg edit: version 2.1: https://i.imgur.com/15e5eZH.jpg), a pole of known height and distance then just add a Moon that measures 32 minuets: 1 minuet = 1 mile for the human eye. Size and distance are then known; 32 nautical miles wide at an altitude of ~3,100 nautical miles. Source: "Yanoff, Myron; Duker, Jay S. (2009). Ophthalmology 3rd Edition. MOSBY Elsevier. p. 54. ISBN 978-0444511416" edit: To simplify it, we have a known distance (observer to horizon), and three known angles (90°, horizon > base of Moon and horizon > top of Moon).  |
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@Spendulus, the Moon is 32 nautical miles wide and I can prove it in a court of law with a sextant. You would have to debunk/discredit "Yanoff, Myron; Duker, Jay S. (2009). Ophthalmology 3rd Edition. MOSBY Elsevier. p. 54. ISBN 978-0444511416" to win.
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^^^ You get the same response as in the Moon thread: ^^^ You're a clown using pipul to win an argument with intellectual dishonestly. Every eye has an angular resolution limit or distance and I'm using that distance to measure other objects. You response (you fucking clown) is to conflate the maximum distance the eye (any eye or camera) can see with with the concept that the human eye has limited abilities thus can't be used.
This like telling somebody you're having guests for dinner then killing and eating them when they arrive for supper. You had "guests for dinner" and everybody you told about having "guests for dinner" believes that you met with some people and shared a meal together.
You have to have a distance and I provide a distance. You provide semantics and pipul, go curl up inside a gas oven.  |
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^^^ You're a clown using pipul to win an argument with intellectual dishonestly. Every eye has an angular resolution limit or distance and I'm using that distance to measure other objects. You response (you fucking clown) is to conflate the maximum distance the eye (any eye or camera) can see with with the concept that the human eye has limited abilities thus can't be used.
This like telling somebody you're having guests for dinner then killing and eating them when they arrive for supper. You had "guests for dinner" and everybody you told about having "guests for dinner" believes that you met with some people and shared a meal together.
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@BADecker,
The maximum distance the eye can see (angular resolution limit) can be used to measure objects, so I'm not measuring an objects size and distance using only degrees like you claim.
@Spendulus,
NASA states in their documentation on their space sextant that an astronaut has to re-calibrate it every day by spotting a star. However when asked at the press conference after they returned, the Apollo 11 astronauts claimed they didn't recall seeing any stars on their journey to the Moon.
How do you reconcile this discrepancy?
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@notbatmanYou can end most of the disagreement and misunderstanding by simply showing us the math/trig that allows us to find the distance and size using only the angle. If you do this, 90% of all the rest of the talk will disappear. Just spit it out right here. You know, like 1+1=2. Or whatever it is. And if you use unconventional math, show us why your math stands over standard math.  The angular resolution limit of the human eye determines how far the human eye can see (source: Ophthalmology 3rd Edition, ISBN 978-0444511416). Tell us why you think the angular resolution limit of the human eye isn't a factor in determining the distance the human eye can see? Why are you asking me to provide a formula for calculating the distance to an object without including the angular resolution limit of the eye? If you don't include the angular resolution limit of the eye in your calculation, then the distance to an object can not be calculated. If there's an error in my formulas for calculating size and distance then show us! |
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I see. The issue then is that you have drawn the line to the horizon incorrectly because it is parallel to the ground and should never intersect it. If it is parallel, then the angles are 90° and P, E, and C are correct.
Next, why do you divide A by 90 when computing O? I don't understand that equation.
The distance to the horizon A is divided into 90° because there are 90° between the observer and the horizon vanishing point at 0°. 90°? Where does that come from?  |
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The angles that you have marked as 90° are not 90°. As a result, the calculations for P, E, and C are not correct. P, E and C are not calculated from the diagram markings, they are calculated using measured and known values thus they are correct. The marked angles can't be drawn at 90° due to optical convergence, hence the 90° angle notation to indicate their actual values. The reason for this is that parallel lines optically converge to a point at the horizon (see photograph with angle marked in red) but, physically parallel lines never converge. The 90° angle can be confirmed empirically by observing that the horizon line is at eye level. I see. The issue then is that you have drawn the line to the horizon incorrectly because it is parallel to the ground and should never intersect it. If it is parallel, then the angles are 90° and P, E, and C are correct. Next, why do you divide A by 90 when computing O? I don't understand that equation. The distance to the horizon A is divided into 90° because there are 90° between the observer and the horizon vanishing point at 0°. |
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^^^ I found this, it's made out of cardboard. How To Build A Homemade Sextant -- https://youtu.be/hOLjEj8OxJM edit: Even easier, clip a bic mechanical pencil to the protractor and hold it with two fingers by the eraser. |
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^^^ The preponderance of evidence that indicates it's a hoax makes the likelihood it's made of cardboard almost a certainty. Just like the preponderance of evidence that the globe is a lie makes flat earth almost a certainty; what little doubt remains is put to rest by direct measurement.
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