Topic: Flat Earth - page 66. (Read 1095196 times)

^^^ You get the same response as the moon thread... https://bitcointalksearch.org/topic/m.51577028.

Besides, you don't show us how you get the distance. Eyeballing it doesn't work, especially when the distances are great.

^^^ You get the same response as in the Moon thread:


You have to have a distance and I provide a distance. You provide semantics and pipul, go curl up inside a gas oven.

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 non-accurate, hazy, limited observations of the eye? Show us if you do.

   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!

@notbatman

You 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.

90°? Where does that come from?

The distance to the horizon A is divided into 90° because there are 90° between the observer and the horizon vanishing point at 0°.

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.

... except where it goes over the horizon.     

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

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.

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I found this, it's made out of cardboard.

   How To Build A Homemade Sextant -- https://youtu.be/hOLjEj8OxJM

The angles that you have marked as 90° are not 90°. As a result, the calculations for P, E, and C are not correct.

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

You remember the bottomless pit in Revelation, don't you? Revelation 9:2 (KJV):
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, globe-gravity 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.

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?

Also in Italy they see too far and supposed curvature of a spinning ball earth is nowhere to be found. Earth is flat and stationary, covered by a dome-firmament of God.

https://www.youtube.com/watch?v=scgrpmQa0J8

Reality:

@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:

^^^ But the ONLY thing we know from direct, simple sextant measurements is the angle... not the distance or the size. Distance and size measurements come from other sources, which might include some complex sextant measurements, as in parallax measurements.

@odolvlobo,

   The image I posted is for proving that the angle defined in red is the same for every pole and, that the distance of the pole from the observer doesn't affect this angle. The image is a photograph not a human eye+sextant and, it shows the angle from a bit of a side perspective so you can see multiple poles and, that they all fit perfectly within the angle.

oh shit...

Look, I'm sorry if I'm not completely there yet either but some you are defiantly disinformation agents and cause me to fuck up and act fucked up.

I don't actually own a sextant I just know it measures angles between objects and that objects including the Sun can be measured based on the horizon and the angular limits of the human eye. All sources of information are corrupt and I'm forced to figure out everything with no help. My point is still valid even if the sextant usage is a bit more complex and possibly non-standard than I first realized. Your point about the red angle being different form what the sextant is measuring is also valid, sorry i fucked up there.

The Sun doesn't change apparent size (except under certain conditions) so measurement with the sextant is just the standard elevation above the horizon from the bottom and top of the Sun. The poles however do change apprent size and this has to be accounted for first to get the size measurement.

So what values do we know?

1. & 2. The distance to the horizon is ~3 miles for a 6 foot person.

3. The angle from the horizon to the top of the object.

4. the angle from the horizon to the bottom of the object.

5. 1 nautical mile = 1 minute.

6. Google says the angular resolution limit of the eye is 1 minute.

This is enough to calculate the 3rd side of the triangle, the height of the object. Right?