Please prove why on a flat Earth the horizon would be a blurry haze.Because light light disperses as it travels through the atmoplane, making very far away things (like the horizon) a blurry haze.
I can stand on a tall mountain, and on a clear day, the outline of distant mountains will be pretty crisp - especially when looking through binoculars.
Most importantly in this case, do you know how close the photographer was to the water when taking the photo?
Judging by the photo itself, I'd say he was pretty darn near water level. That close to the water your view distance would be severely limited by the water itself - as it's well known that you can see farther at greater height; and you can not see far at all when close to ground-level (or water-level).
Therefor if the person is close to water level, it makes sense that he would't be able to see very far at all, and hence the waterline would be crisp.
Now go get a photo of a large sea vessel taken at a great distance, and then explain why on a ball earth the horizon is so incredibly blurry. Now your argument is working against you. . . lol.
After you do that, go the the ocean with a telescope and watch a ship "go over the horizon." Once the ship is totally gone "over the horizon" zoom in with your telescope and try to explain why the ship is still there, and not over the horizon. Our view distance is only limited by a) how far we can see, and b) atmospheric weather conditions (haze).
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Why should you be able to see further the higher you are in FE?
Pick a point in the distance that you can see standing on the ground in FE. Now scale an imaginary ladder that extends as far as you like into the sky. The distance to that point is greater from the tall ladder than it is from the surface. You are viewing along the hypotenuse of a triangle. The length of your elevated viewing distance is the square root of the surface distance squared plus the height above the surface squared which is greater than the surface distance.
So altitude should cause you to lose sight of things in the distance in FE. The higher you climb the less far you should be able to see if FE were true. The fact you get to see farther the higher you go falsifies FE.
But these kinds of arguments are not what real FEers (if there are any) need. Hypothetical real FEers would need interaction with people from a different profession.
Why should you be able to see further the higher you are in FE?
Pick a point in the distance that you can see standing on the ground in FE. Now scale an imaginary ladder that extends as far as you like into the sky. The distance to that point is greater from the tall ladder than it is from the surface. You are viewing along the hypotenuse of a triangle. The length of your elevated viewing distance is the square root of the surface distance squared plus the height above the surface squared which is greater than the surface distance.
So altitude should cause you to lose sight of things in the distance in FE. The higher you climb the less far you should be able to see if FE were true. The fact you get to see farther the higher you go falsifies FE.
But these kinds of arguments are not what real FEers (if there are any) need. Hypothetical real FEers would need interaction with people from a different profession.
Untrue, untrue, untrue.
a) Go to a parking lot, lay right on the tarmac with your stomach on the ground, take a picture of how far you can see. Not very far.
b) Next, stand up and take a picture and see how far you can see. You will be able to see farther.
c) Stand on a ladder or another high object, you will see even farther—you will even be able to see over objects that were in your way while standing.
It's true that your maximum view distance never changes; but your chances of seeing the maximum distance increases with height.
One reason you can see farther is simply because your perspective has changed. Another reason is that while you're on the ground, objects on the ground can obstruct your view. In the case of ocean water you have WAVES rising into your field of view. When looking out into the ocean the water could even be relatively calm where you're standing, but it could be windy with high waves in the distance. Those waves will obstruct your view as the vanishing point converges.
As to your claim that "So altitude should cause you to lose sight of things in the distance in FE. The higher you climb the less far you should be able to see if FE were true. The fact you get to see farther the higher you go falsifies FE."
You're acting as though being on a flat earth changes the laws of perspective and view distance. If what you say is true, then the same would hold true with RE.
In addition, because on a round earth you would have to look DOWN at the curve, your distance would actually be less on a ball than on a plane where you don't have the earth getting in the way of your view. On a round earth you would get to a height where it's impossible to see any farther, because you would be looking at the edge of the curve—whereas on a plane there would be nothing getting in the way of you seeing farther.
You really like to pray on ignorance don't you. Unfortunately for you there aren't many ignorant FErs here—which leaves only you. ;)
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No RE-er would expect to see curvature on such a small FOV, sorry to bust your bubble.
The horizon is always a distinct line where earth - or sea - meet the sky. It is never blurred except in rare occasions when there might be haze, mist, fog, darkness or some other condition to cause the horizon to not be distinct. The horizon is most distinctly defined on open seas in the middle of the ocean on a clear day.
So the bottom line on this website is to not take any "flat earth fantasies" seriously.
So the bottom line on this website is to not take any "flat earth fantasies" seriously.
I wasn't. I don't take the FE supporters seriously either. I think the dumb is an act.
No RE-er would expect to see curvature on such a small FOV, sorry to bust your bubble.
lol, so that was a joke then? That was a good one. You are one of the funnier RE'ers here. Most of you are militant and not funny at all. I suppose the shill pay grade lets you say things like this?
The horizon is always a distinct line where earth - or sea - meet the sky. It is never blurred except in rare occasions when there might be haze, mist, fog, darkness or some other condition to cause the horizon to not be distinct. The horizon is most distinctly defined on open seas in the middle of the ocean on a clear day.
DECEIVE MUCH?
Stating that the horizon is crisp and clear is not proof of RE, and you know it. You people use half-truths to make your lies appear reasonable. "The Devil is in that lack of details", in the things that aren't said, the things that are purposely withheld. Half truths are half lies.
Photo #1
The water horizon is crisp; but what's that in the background? Why it's land. "LAND AHOY!" Isn't that interesting that the waterline is crisp and seems to end before the land - and the land is hazy. If the earth was a ball, the curve of the water would block our view of the land.
http://www.mlewallpapers.com/image/16x9-Widescreen-1/view/St-Lucia-Horizon-I-321.jpg
Photo #2
Waterline is crips; land is hazy. Again this proves that the waterline can be crisp while being able to see beyond it.
http://images.forwallpaper.com/files/images/b/b984/b984965a/106188/sea-ocean-water-sky-horizon.jpg
Photo #3 & 4
As I STATED above, when you raise the perspective, it allows you to SEE FARTHER, because when you are close to sea-level the waves themselves rise into your view - not permitting you to see past the waves. As a kayaker I know that even on lakes the waves can get several feet high on windy days—it's really fun I must add!
Below we can see that when we take a photo from high up, we can see farther, and the crisp horizon becomes blurry because we can see farther.
http://static.panoramio.com/photos/large/36665053.jpg
https://c2.staticflickr.com/2/1288/1308812562_5a099e6d7d_z.jpg
Mountain of Water
Well I think we've debunked your debunkery, and debunked RE, and proven you're deceiving people, because on a BALL you would not be able to see the land past the mountain of water in the middle.
See the following for the MOUNTAIN of water evidence:
http://forum.tfes.org/index.php?topic=3211.0
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After you do that, go the the ocean with a telescope and watch a ship "go over the horizon." Once the ship is totally gone "over the horizon" zoom in with your telescope and try to explain why the ship is still there, and not over the horizon.Perhaps you could show us proof of this. I have yet to see anything showing this.
After you do that, go the the ocean with a telescope and watch a ship "go over the horizon." Once the ship is totally gone "over the horizon" zoom in with your telescope and try to explain why the ship is still there, and not over the horizon.Perhaps you could show us proof of this. I have yet to see anything showing this.
A video that clearly shows a ship or object in the process of 'sinking' beyond the horizon (disappearing because the camera resolution isn't enough to make it out doesn't count), and then rising back up to full unobstructed height as magnification is further increased, would suffice.
After you do that, go the the ocean with a telescope and watch a ship "go over the horizon." Once the ship is totally gone "over the horizon" zoom in with your telescope and try to explain why the ship is still there, and not over the horizon.Perhaps you could show us proof of this. I have yet to see anything showing this.
A video that clearly shows a ship or object in the process of 'sinking' beyond the horizon (disappearing because the camera resolution isn't enough to make it out doesn't count), and then rising back up to full unobstructed height as magnification is further increased, would suffice.
Nice demonstration of the camera's resolution limitations and the video creator's spelling ability limitations, but I don't see a distinctly sunken ship returning to full height with increased magnification.After you do that, go the the ocean with a telescope and watch a ship "go over the horizon." Once the ship is totally gone "over the horizon" zoom in with your telescope and try to explain why the ship is still there, and not over the horizon.Perhaps you could show us proof of this. I have yet to see anything showing this.
A video that clearly shows a ship or object in the process of 'sinking' beyond the horizon (disappearing because the camera resolution isn't enough to make it out doesn't count), and then rising back up to full unobstructed height as magnification is further increased, would suffice.
Nice selection of music. "You just don't get it, just don't get it, just don't get it. . . You're just so pathetic!"
https://www.youtube.com/watch?v=VFhhCYYkILw
Watch to the end where it zooms all the way out and the ship is COMPLETELY gone from the naked eye POV.
See Scene 2:52No, I didn't notice the entire city skyline. It's the same video again. ::)
Notice the entire city skyline is gone when looking with the naked eye; but it miraculously comes back into view when you zoom in. Well it's not really magic, it's REALITY!
https://youtu.be/VFhhCYYkILw?t=1m47s
But if the surface is indefinite and goes on forever (???), then there should be NO HORIZON as one goes further up. One will only see wider view of the indefinitely wide surface without any horizon. The furthest point from view may blur up due to infinite distance, but there should not a horizon.
Exactly, that's the point of this thread. So far, all attempts to defend the flat Earth on this matter has been a pitiful fail. Special mention for Pongo, with the wave.
Another big factor in how far one can see is the amount of contrast.
For example in the daytime there is a lot of contrast between earth and sky. Particularly at sun set or rise on the sun side you will see high contrast and on the opposite side you will see low contrast. In that case you can see land that is otherwise obscured by atmosphere.
The primary reason that the horizon is a sharp line is that it tends to be fairly close 4-10 miles whereas the horizon on the flat earth model tends to be thousands of miles away so that it would be obscured by the atmosphere even on the clearest day.
In the flat earth model a ship which is viewed in front and below the FE horizon will remain in front and below that horizon no matter what distance it is viewed from it will never appear to sit on the horizon and never appear behind the horizon
This SketchUp model shows a ship sized rectangle -100 ft long and 25 ft high spaced one mile apart
The conclusions are:
• drivel drivel, more drivel drenched in self-importance
Thanks Giants Orbiting, for that remarkably long winded journey to a non conclusion.
And to call me a flat earther is you're attempt at discrediting me, even though nowhere in my post did I even make an inclination of my stance.I know you weren't talking to me, but you seem to "not a flat earther" and "not a globe supporter" (since all you arguments seem to be against the very idea of a rotating globe.)
There is an excerpt from a Navy manual for lookouts showing how far you can see to the horizon dedepending on your height. It's on another thread.
There is an excerpt from a Navy manual for lookouts showing how far you can see to the horizon dedepending on your height. It's on another thread.
Please explain to me how that same effect wouldn't be observed on a flat earth? I'll wait.
Yes, most of your first bit is essentially correct, but then you say " The horizon would technically be traveling exponentially away from you.". . . . . . . . . . . . . . . . . . . . . . . . . . . .You can see further the higher up you go, until eventually your vision fails or the atmospheric conditions inhibit your viewing distance. This would be true on a flat or a round earth. Except on a round earth, if the horizon is the curvature itself, going up you shouldn't be able to see as far as you would on a flat earth. The horizon would technically be traveling exponentially away from you.
In a plane the horizon appears faded, and not the sharp line you guys are saying it is. Why is that the case?
Yes, most of your first bit is essentially correct, but then you say " The horizon would technically be traveling exponentially away from you.". . . . . . . . . . . . . . . . . . . . . . . . . . . .You can see further the higher up you go, until eventually your vision fails or the atmospheric conditions inhibit your viewing distance. This would be true on a flat or a round earth. Except on a round earth, if the horizon is the curvature itself, going up you shouldn't be able to see as far as you would on a flat earth. The horizon would technically be traveling exponentially away from you.
In a plane the horizon appears faded, and not the sharp line you guys are saying it is. Why is that the case?
The distance to the horizon does increase with altitude, but certainly not "exponentially" as you claim. It increases as the square root of distance, much, much different.An approximate expression is: (https://upload.wikimedia.org/math/f/c/9/fc99cd32db01b85175e2a304c6441940.png)Where h is in metres and d is in km.
You do take everything to extremes! Who said that the horizon is sharp from a plane at high altitude? Surely if someone says that the horizon is sharp looking out over the ocean on a clear day, it is a bit rich criticising them if it not sharp from a plane.
Now, as you know very well the horizon is blurred from a plane because we looking through so much atmosphere. The expression above would put the horizon at 357 km for a plane at 10,000 m altitude. Even in the clearest air this is about the "Rayleigh limit" and the horizon would always look a bluish haze.
For a "not Flat Earther" you certainly do a lot of arguing against the idea of a Globe!
Curvature of the horizonSeems even Wikipedia takes issue with facts being presented without sources, even one something as mundane as the curvature of the earth in relation to the horizon.
This section has multiple issues. Please help improve it or discuss these issues on the talk page.
This section does not cite any sources. (June 2013)
This article's factual accuracy is disputed. (June 2013)
This section requires expansion with: examples and additional citations. (June 2013)
The Earth does technically "move away" from the observer exponentially... if it is round. The "curve" isn't linear, as in, a flat slope.Seems even Wikipedia takes issue with facts being presented without sources, even one something as mundane as the curvature of the earth in relation to the horizon.
If you square the whole equation you provided you end up with something very similar to the 8" of curvature per distance squared formula you've undoubtedly seen used numerous times here.
On a flat earth, viewing distance is a lot simpler, using only a triangle and Pythagoras theorem you can calculate how far per height rather easily. This would be a good way to test both theories if some interested parties would like to partake in an experiment to test these variables.
And don't forget, the magical wild card we call refraction.
https://en.wikipedia.org/wiki/Horizon#Effect_of_atmospheric_refraction
What's more interesting to me is the section immediately following the one above.
Curvature of the horizon
This section has multiple issues. Please help improve it or discuss these issues on the talk page.
This section does not cite any sources. (June 2013)
This article's factual accuracy is disputed. (June 2013)
This section requires expansion with: examples and additional citations. (June 2013)
Elevation | Horz dist d = 3.57xh0.5 | Horz (Exp) |
0 m | 0.0 km | 0.7 km |
2 m | 5.0 km | 5.0 km |
5 m | 8.0 km | 101.4 km |
7 m | 9.4 km | 749.3 km |
10 m | 11.3 km | 15050.1 km |
| Of course the equation I gave essentially agreed with the 8"/(mile squared) - I never claimed any differently! I'm not getting into "curvature of the horizon" apart from:
On "vanishing point", it is only a drawing aid! The horizon is not any "magic" vanishing point. The ultimate vanishing point is an indefinite distance away. Of course in drawing we (not me - can't draw for nuts!) draw the horizon as FLAT, it looks FLAT! Imagine sitting it a boat on a perfectly smooth lake. On a Globe of Flat Earth the horizon would be exactly the same all around us - flat. I would contend that (on a clear day) the horizon: on the Globe would be sharp (and close), while on the Flat Earth it would be quite indistinct and an indefinite distance away. Look at the picture at the right (from a Flat Earth Video I might add), taken at quite high zoom. Quite apart from any other issues, clearly the horizon is not the vanishing point! The buildings are certainly further away than the horizon, yet show quite clearly. | (http://i1075.photobucket.com/albums/w433/RabDownunder/Horizon%20Zoom%20Boom%20Earth%20Flat_zpsgrsg64nz.jpg) From Horizon Zoom Boom Earth Flat (https://www.youtube.com/watch?v=8athT6tfRIg&feature=youtu.be) |
As it is written:
When you've gotta go, you've gotta go!
All I'm saying is that you can express yourself with a lot less verbiage and still be concise.
So far I see a lot of filler designed to look like a coherent, original thought.
Explain why perspective is flawed.
Explain why you think atmospheric conditions aren't a reasonable explanation for viewing distance.
Also tell me why you think cognitive dissonance only affects flat earthers and not round earthers.
Not trying to just be rude, but I think it's rude to assume someone has 30 minutes to carefully read a post that should be at least half as long and twice as easy to follow.
And to call me a flat earther is you're attempt at discrediting me, even though nowhere in my post did I even make an inclination of my stance.
Firstly, I am afraid you are completely mistaken with "The Earth does technically "move away" from the observer exponentially... if it is round." NO it DOES not! Do you know what "exponentially means"? Obviously not. It means a variation with some constant, commonly e. So an exponential variation would have:
distance = (some constant) x e(height/constant).
Study up on your math a bit!An exponential variation and a square root variation are completely different animals (or whatever - they are as different as chalk and cheese!)
You really do have a sense of humour with your "Now let me help you out rabinoz", or more like a blown up sense of you own self-importance.Firstly, I am afraid you are completely mistaken with "The Earth does technically "move away" from the observer exponentially... if it is round." NO it DOES not! Do you know what "exponentially means"? Obviously not. It means a variation with some constant, commonly e. So an exponential variation would have:I'm so glad that Orbisect brought this thread back up. I almost missed how you tried to school me on math and exponential equations and completely blew it.
distance = (some constant) x e(height/constant).
Study up on your math a bit!An exponential variation and a square root variation are completely different animals (or whatever - they are as different as chalk and cheese!)
Now let me help you out rabinoz, I know you haven't been to school in about 40 or 50 years, and maybe things have changed since the 50's.
Get on with it! Of course I know the elementary stuff.
First the difference between something that is linear and exponential.
A linear equation represents a line that travels along a slope in a straight flat line.
Example of a straight line formula
(https://upload.wikimedia.org/math/d/2/4/d24ebc87176b242c935535a363c5fc10.png)
Example of a straight line plotted on a Cartesian coordinate system:
(http://mathsfirst.massey.ac.nz/Algebra/StraightLinesin2D/images/intercept2.gif)
Yes, no problem with that! But the critical thing is that to be an "exponential variation" the exponent must contain the independent variable!
An exponential function describes a line that travels in an increasingly curved line:
(https://upload.wikimedia.org/math/8/a/e/8aeddf13a124333c1160595b2eaf8660.png)
Graphed:
(https://upload.wikimedia.org/wikipedia/commons/thumb/c/c6/Exp.svg/800px-Exp.svg.png)
Sure, I guess any equation can be written in parametric form, and most can be written in non-parametric form, but often these have multi-valued solutions as the case for a circle where:
A parametric equation, is one that can describe a curved line, it can be parabola, a circle, or countless other odd lines.
Example of the equation for the circle:
(https://upload.wikimedia.org/math/9/c/d/9cd7a600b187da351f13b7ec10a93fb8.png)
A circle on the same type of graph:
(http://www.mathportal.org/analytic-geometry/conic-sections/circle_files/circle.gif)
Now you are starting to get screwed up. Just because an equation has exponents does not make the variation exponential!
As you can see the equation for a straight line and something else is obviously very different. One has exponents, and one doesn't. In case you don't remember what an exponent is: y2<-- this is a exponent. Different types of lines will have different exponents, it's not always squared. But in our case, a curved line equation deals with squared variables. This is why you can "square" that cute little equation you gave and it will still technically resolve into the "8 inches per mile squared for a 6' tall observer" little ditty you see floating around.
Technically, the horizon on a curved Earth would move away from you exponentially, in the sense that it doesn't move away from you in a linear fashion.
Maybe you should study up on your math a bit. Break out that trust abacus you used to use in Senior High.
Even Wikipedia would set you straight: Quote from: Wikipedia Exponential growthFrom Wikipedia, the free encyclopedia, Exponential_growth (https://en.wikipedia.org/wiki/Exponential_growth) | Spacer | (https://upload.wikimedia.org/wikipedia/commons/thumb/6/64/Exponential.svg/718px-Exponential.svg.png) The graph illustrates how exponential growth (green) surpasses both linear (red) and cubic (blue) growth. (red) linear growth (blue) cubic growth (green) exponential growth |
blah blah blah... 8th grade math... blah...more 8th grade math... blah...
Do you know what "exponentially means"? Obviously not. It means a variation with some constant, commonly e. So an exponential variation would have:
distance = (some constant) x e(height/constant).
Study up on your math a bit!
A parametric equation, is one that can describe a curved line, it can be parabola, a circle, or countless other odd lines.
Example of the equation for the circle:
(https://upload.wikimedia.org/math/9/c/d/9cd7a600b187da351f13b7ec10a93fb8.png)
blah blah blah... 8th grade math... blah...more 8th grade math... blah...
Yes yes, both of you are very smart.Do you know what "exponentially means"? Obviously not. It means a variation with some constant, commonly e. So an exponential variation would have:
distance = (some constant) x e(height/constant).
Study up on your math a bit!
*ahem* TheTruthIsOnHere was correct in stating your original description of an exponential equation was wrong.A parametric equation, is one that can describe a curved line, it can be parabola, a circle, or countless other odd lines.
Example of the equation for the circle:
(https://upload.wikimedia.org/math/9/c/d/9cd7a600b187da351f13b7ec10a93fb8.png)
*ahem* This description of parametric equations is also wrong. This isn't a parametric equation. Not that it is relevant anyway...
Point being, the horizon curves away from you on a round earth. On a flat earth I feel you would technically be able to see further with increased altitude, and then as shown in the photos posted in the thread is when the atmosphere would have the most pronounced effect on your visibility.
The fact still remains that there are countless photographs on the internet that clearly exhibit being able to see further than you *mathematically* should be able to.
And yes, I accept photographs as evidence. Unless it's a photograph mathematically impossible to take, like one of the Earth from a million miles away.
. . . . . . . . . . . . . . . . . . . . . .Yes, I was "trying to be smart", simply because TheTruthIsOnHere raved on for pages, "showing off how smart he was" claiming I knew nothing of maths. He was completely false in claiming that the variation in distance with elevation was exponential, and it most certainly is not, not even in the "colloquial sense" of "very rapidly".
Yes yes, both of you are very smart.
Do you know what "exponentially means"? Obviously not. It means a variation with some constant, commonly e. So an exponential variation would have:*ahem* TheTruthIsOnHere was correct in stating your original description of an exponential equation was wrong.
distance = (some constant) x e(height/constant).
Study up on your math a bit!
Where claims " The horizon would technically be traveling exponentially away from you." he is completely incorrect!Except on a round earth, if the horizon is the curvature itself, going up you shouldn't be able to see as far as you would on a flat earth. The horizon would technically be traveling exponentially away from you.(and I replied)[/i]
Yes, most of your first bit is essentially correct, but then you say "The horizon would technically be traveling exponentially away from you."
The distance to the horizon does increase with altitude, but certainly not "exponentially" as you claim. It increases as the square root of distance, much, much different.An approximate expression is: (https://upload.wikimedia.org/math/f/c/9/fc99cd32db01b85175e2a304c6441940.png)Where h is in metres and d is in km.
The funny thing is I really maxed out around this level of math. I actually failed Calculus in 12th grade. Rab obviously went on to learn more insanely complicated mathematic1, well beyond my definition of practicality. I don't even know why he decided to talk about "exponential variation," whatever that means. I certainly didn't bring that up. I don't even remember how this conversation devolved into a history lesson on the graphing calculator, and really what any of this math even has to do with the OP.Well, I brought "exponential" up because this statement in an earlier post of yours:
I did obviously interpret "traveling exponentially away from you. as being an exponential variation, so I made what I thought was a reasonable post saying that it was not an exponential variation. You then come back with your pages stuff saying I was wrong and needed to study some maths. Well, I guess I reacted a bit, but all I did was answer each point you made. If you don't have much maths knowledge, that's OK, not everyone does - but in that case be a little careful how you criticise.The horizon would technically be traveling exponentially away from you.
You can see further the higher up you go, until eventually your vision fails or the atmospheric conditions inhibit your viewing distance. This would be true on a flat or a round earth.
. . . . . . . . . . . . . . . . . . . . .
In a plane the horizon appears faded, and not the sharp line you guys are saying it is. Why is that the case?
On a flat earth it is a matter of using the Pythagorean theorem to calculate viewing distance. The relationship between the observed, the observer, and the angle of viewing is represented very neatly by a triangle.I will try to tackle the rest later, after you show just how you use "Pythagorean theorem to calculate viewing distance" on the flat earth.
On a flat earth it is a matter of using the Pythagorean theorem to calculate viewing distance. The relationship between the observed, the observer, and the angle of viewing is represented very neatly by a triangle.I will try to tackle the rest later, after you show just how you use "Pythagorean theorem to calculate viewing distance" on the flat earth.
Honestly I have never seen that and cannot see how it could ever be done. So, I am very interested.
Sorry, I misapplied Pythagoras, It's actually a trigonometric application, thanks for jostling my brain a little. If you know your altitude, and you know your angle of viewing, then you can solve the bottom leg of the triangle. Assuming the earth is flat, let's represent it as the sea being flat to the horizon, and if we are looking down at an object before the horizon, say tilted POV of 45degrees, and you know your height, you can solve how far away that object is. But as I said, interestingly enough, if you are looking at the horizon, from any altitude your viewing angle will be 90 degrees. So it really is a tricky problem to solve.
Tell me exactly how one calculates the distance to the horizon on a curved earth, without navy manuals. Is it some obscure geodetic equation of curved triangles or what?
No one can answer me as to why, if the horizon is the tangent of your position and the Earth's curvature, why would the horizon rise to meet your eye.
Or, if the horizon is literally the earths curvature, why is the horizon not curved to a similar degree all around you. Is the Earth really a cylinder?
Your first picture shows no curve either. Wouldn't you expect to find that on a round-earth?
As you say if, on a flat earth you are looking down a point on the ground at a know angle you calculate your distance to it. Actually, since the "curvature" of the globe is actually quite small, you can do the same thing for objects on the Globe with little error for distances up to a few kilometres.On a flat earth it is a matter of using the Pythagorean theorem to calculate viewing distance. The relationship between the observed, the observer, and the angle of viewing is represented very neatly by a triangle.I will try to tackle the rest later, after you show just how you use "Pythagorean theorem to calculate viewing distance" on the flat earth.
Honestly I have never seen that and cannot see how it could ever be done. So, I am very interested.
Sorry, I misapplied Pythagoras, It's actually a trigonometric application, thanks for jostling my brain a little. If you know your altitude, and you know your angle of viewing, then you can solve the bottom leg of the triangle. Assuming the earth is flat, let's represent it as the sea being flat to the horizon, and if we are looking down at an object before the horizon, say tilted POV of 45degrees, and you know your height, you can solve how far away that object is. But as I said, interestingly enough, if you are looking at the horizon, from any altitude your viewing angle will be 90 degrees. So it really is a tricky problem to solve.
Tell me exactly how one calculates the distance to the horizon on a curved earth, without navy manuals. Is it some obscure geodetic equation of curved triangles or what?
(http://www-rohan.sdsu.edu/~aty/explain/atmos_refr/figs/dip1.gif) | Spacer | Distance to the Horizon It shows a vertical plane through the center of the Earth (at C) and the observer (at O). The radius of the Earth is R, and the observer's eye is a height h above the point S on the surface. (Of course, the height of the eye, and consequently the distance to the horizon, is greatly exaggerated in this diagram.) The observer's astronomical horizon is the dashed line through O, perpendicular to the Earth's radius OC. But the observer's apparent horizon is the dashed line OG, tangent to the surface of the Earth. The point G is the geometric horizon. Elementary geometry tells us that, because the angle between the dashed lines at G is a right angle, the distance OG from the observer (O) to the horizon (G) is related to the radius R and the observer's height h by the Pythagorean Theorem: (R + h)2= R2 + OG2 orOG2 = (R + h)2 − R2 .But if we expand the term (R + h)2 = R2 + 2 R h + h2, the R2 terms cancel, and we find OG = sqrt ( 2 R h + h2 ) . It's customary to use the fact that h << R at this point, so that we can neglect the second term. ThenOG ≈ sqrt ( 2 R h ) is the distance to the horizon, neglecting refraction.Numerically, the radius of the Earth varies a little with latitude and direction; but a typical value is 6378 km (about 3963 miles). If h is in meters, that makes the distance to the geometric horizon 3.57 km times the square root of the height of the eye in meters or about 1.23 miles times the square root of the eye height in feet. From: ROHAN Academic Computing at San Diego State University: Distance to the Horizon (http://www-rohan.sdsu.edu/~aty/explain/atmos_refr/horizon.html) |
Height | Spc | Dip angle |
0 m | Spc | 0.00° |
10 m | Spc | 0.10° |
100 m | Spc | 0.32° |
1,000 m | Spc | 1.02° |
10,000 m | Spc | 3.20° |
d = R cos-1(R/(R+h))