notbatman
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June 18, 2019, 07:46:17 PM Last edit: June 18, 2019, 09:04:38 PM by notbatman 

@BADecker, the back of the napkin trigonometric illustration I made look like a game of hangman, or perhaps somebody being crucified. I'm going with a known distance of 3 minutes based on the angular resolution limit of 1 minute with an elevation above the plain of 3/50 seconds and two measured angles. edit: So here's what the problem looks like after MS paint:





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odolvlobo
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June 18, 2019, 11:06:58 PM Last edit: June 18, 2019, 11:24:29 PM by odolvlobo 

That's a great diagram. Here is where my confusion comes from: First, replacing the pole with the sun gives you X + Z = 0.533 degrees or 32 nm. But if you make the same measurements with the pole (and you are close enough to it), you will get an angle more than 0.533 degrees and the pole certainly isn't more than 32 nm tall. Another way to look at it is this: suppose the sun is setting directly behind the pole and the top of the sun lines up with the top of the pole while the bottom of the sun is exactly at the horizon. Now, since part of the pole is below the horizon, its angular size is bigger than the sun, but the pole isn't more than 32 nm tall. So, 1 minute can't always be 1 nm. (X+Z) depends on the distance K, assuming P is constant, or P depends on K, assuming (X+Z) is constant. That's what it looks like to me. Do you understand my confusion now?

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BADecker
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June 18, 2019, 11:25:15 PM 

You remember the bottomless pit in Revelation, don't you? Revelation 9:2 (KJV): And he opened the bottomless pit; and there arose a smoke out of the pit, as the smoke of a great furnace; and the sun and the air were darkened by reason of the smoke of the pit. So, on a flat earth, how deep is the bottomless pit? And if the dragon was cast into the bottomless pit, how far did he fall? Does density explain how the attraction/repulsion that is normally called "gravity" works in an endless, bottomless pit? In a globe earth, the bottomless pit would go down through the center of the earth. Why? Because at the center of the earth, globegravity would be attracting everything equally in every direction. This makes a pit, but there doesn't have to be any bottom. By the time you reach the center of the earth, you stop falling, because you have become weightless. You never splat on the bottom. In fact, the attraction of gravity as you get closer to the center of the earth does funny things. For example, if you went 1 mile in any direction away from the center, there would be a slight gravitational effect. But it would still be less than it would be on a small asteroid... almost unnoticeable. What this means is that other earthly forces may have made the earth hollow, because of the small amount of gravitational effect at the center. The highest pressure effect on a hollow earth would be somewhere in between the center and the surface. The earth is hollow, and people are finding it out. That's why the powers that be are using you jokers to promote a flat earth theory. The powers are using the center of the earth for military experimentation. And their leader is the devil, who was cast into the bottomless pit.




notbatman
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June 21, 2019, 08:23:11 PM Last edit: June 23, 2019, 09:43:18 PM by notbatman 

I created this to show how to measure the angular size of objects. If you see any errors let me know. edit: version 2.1 corrected calculation for A




odolvlobo
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June 22, 2019, 05:17:41 AM Last edit: June 22, 2019, 05:29:21 AM by odolvlobo 

I created this to show how to measure the angular size of objects. If you see any errors let me know.
The angles that you have marked as 90° are not 90°. As a result, the calculations for P, E, and C are not correct.

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notbatman
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June 22, 2019, 06:27:24 AM Last edit: June 23, 2019, 09:43:29 PM by notbatman 

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 found this, it's made out of cardboard. How To Build A Homemade Sextant  https://youtu.be/hOLjEj8OxJM




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June 22, 2019, 08:36:41 PM 

Everyone sees the world in one’s own way..




BADecker
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June 22, 2019, 09:58:45 PM 

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. ... except where it goes over the horizon.




odolvlobo
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June 23, 2019, 03:05:37 AM 

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.

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notbatman
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June 23, 2019, 04:46:20 AM Last edit: June 23, 2019, 09:43:40 PM by notbatman 

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°.




odolvlobo
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June 23, 2019, 04:55:29 AM 

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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notbatman
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June 23, 2019, 05:19:21 AM 

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?




BADecker
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June 23, 2019, 12:37:28 PM 

@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.




notbatman
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June 23, 2019, 02:14:47 PM 

@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 9780444511416). 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!




BADecker
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June 23, 2019, 03:24:44 PM 

@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 9780444511416). 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! I ask for this formula because of what you said.If we get rid of the limited human eye in the equation altogether, we also get rid of eye limits. Trigonometry doesn't use human eye limitations. Rather, trig calculates the answer accurately no matter what the eye thinks it sees... and especially if we use calculus along with the trig. If there is an error in your calc, the error is using your formulas at all, because there is trig and calculus that will give the answer easily and accurately, and (as you said) the eye has limitations.What is the basic answer that trig and calculus gives? It gives the answer that says that you can't find distance and size with only the angle. You need another measurement along with the angle to show distance or size. Let me say it another way. Go inside a building with no windows, let someone give you the angle (32 degrees), and calculate the size/distance of anything accurately without using trig or calculus. You can do it accurately if you have the distance or size (and use trig), but not without one of these in addition to the angle (and trig). Or do you have a way without nonaccurate, hazy, limited observations of the eye? Show us if you do.




notbatman
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June 23, 2019, 04:11:45 PM Last edit: June 23, 2019, 04:28:23 PM by notbatman 

^^^ 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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June 23, 2019, 08:43:02 PM Last edit: June 23, 2019, 10:51:27 PM by notbatman 

^^^ 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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June 23, 2019, 10:14:48 PM Last edit: June 23, 2019, 11:09:24 PM by BADecker 

^^^ What you are showing is that the eye has limits. The camera has greater ability than the eye. The simple telescope has greater ability than the camera. And an extremely powerful telescope has greater ability of sight than the simple telescope. Now be patient with me for a moment, and don't just lay into me because of the distances I am using. What if the object is a parsec away? Even the best telescope would see it way less than a parsec. The eye, camera, and simple telescope would see it essentially the same distance away, way less than the best telescope. But if you use trig triangulation, you can find a more accurate distance than any of them... if you have two measurements. If the object is 20 parsecs away, all standard eyeballing through any instrument we have in existence would show a tremendously nearer object than the 20 parsecs. And the eyeballing would be wrong. Triangulation has been done on distant objects using the double earthtosun distance, at opposite times of the year, as the measuring method. While the distances are not absolutely accurate, they are a "million" times more accurate than eyeballing with one of the instruments... which is way, way more accurate than eyeballing directly with the eye. Why trust triangulation based on a distance of 186 million miles (the distance of the earth in it's orbit on the exact opposite side of the sun)? Because all kinds of other measurements have been done that show that 186 million miles is near accurate. These measurements include combining several parallaxes of several planets. Your measurements are fine and good at earthsurface distances (much of the time). But they fail at interplanetary, interstellar, and intergalactic distances. And you admit it by stating the limits that eyeballing has. Your mistake lies in trying to shrink the universe so that it will match earthsurface eyeballing distances.




