The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Debate => Topic started by: hoppy on January 17, 2014, 04:57:02 PM
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I just took this image today, how can this house be visible if the earth is a sphere?
This image was taken from 4.4 miles as I found on google earth. There is no other proof for you roundies, you just have to take my word for now about this. I will work on better documentation later. But what I have stated here is true.
(http://imgur.com/iUBPPWM.jpeg)
(http://imgur.com/Otx5xG1.jpeg)
Use the google lat long co ordinates to locate this house. It is the only octagonal house on this shore. Location is Cecil County, Maryland.
(http://imgur.com/AEuNITM.jpeg)
I was next to the pier sitting on rocks,at Perry Point VA hospital, the camera was 2' above the water. I was at one end of the yellow line, the octagon house was at the other end.
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Since the earth has an equatorial bulge there is no perfect way to determine the drop but we can come up with some sort of average.
Suppose that the earth is a sphere of radius 3963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile to point Q then you can form a right triangle. Using Pythagorean theorem
a^2 = 3963^2 + 1^2 = 15705370 and thus a = 3963.000126 miles.
3963.00126 - 3963 (earth's radius) = 0.000126 miles above surface at point Q
Convert to inches:
12*5280*0.00126 = 7.98 inches.
Hence the earth's surface curves at approximately 8 inches per mile.
So for one, a 12 ft drop is erroneous. For 4 miles we're looking at under 3 ft and this is if you are looking from ground level. At 2 ft above the water it's even less.
It's 32 inches - 24 inches = 8 inch drop if 2 ft above water.
And that's if you were actually 2 ft above the water which it doesn't appear that way from your pic.
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An fe'r who is finally posting pictures of observations. Well done.
First my question; do you have a picture from the same distance taken from a higher elevation? Or a picture from much closer?
Where are the bushes and steps between the house and the water? There's another small structure lower on a beach just off the side of your picture (I can make out another house further back under the trees) was that lower structure also visible, but just not framed in the shot? GE shows the parking lot (and I assume the first floor) to be 9 feet.
Also, you claim the drop is 12.9 feet, I see no problem with this, and according to the ENaG chart that's correct. However, that's based off a starting point elevation of zero. If you were 2 feet above the water, that puts the drop starting point out further (1.5miles I guess?) to where your line of sight intersects the water, and from there the drop starts. I'll just go with a 1.4 mile advantage from the 2 foot camera elevation just to make an even 3ft. That means there was a drop over 3 miles from where your line of sight would intersect the water. A drop of 6 feet. 6 feet seems to correspond with what is shown in the aerial picture.
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Hey Hoppy if you get the chance to go back there sometimes when the Wind is at least 20-30 MPH you will see the Water magically rise up at least 10-15 feet.
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Since the earth has an equatorial bulge there is no perfect way to determine the drop but we can come up with some sort of average.
Suppose that the earth is a sphere of radius 3963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile to point Q then you can form a right triangle. Using Pythagorean theorem
a^2 = 3963^2 + 1^2 = 15705370 and thus a = 3963.000126 miles.
3963.00126 - 3963 (earth's radius) = 0.000126 miles above surface at point Q
Convert to inches:
12*5280*0.00126 = 7.98 inches.
Hence the earth's surface curves at approximately 8 inches per mile.
So for one, a 12 ft drop is erroneous. For 4 miles we're looking at under 3 ft and this is if you are looking from ground level. At 2 ft above the water it's even less.
It's 32 inches - 24 inches = 8 inch drop if 2 ft above water.
And that's if you were actually 2 ft above the water which it doesn't appear that way from your pic.
12 foot drop is not erroneous, 8" drop the first mile is correct. The drop is progressively more in each following mile.
Refer to this chart, it is correct.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
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Hey Hoppy if you get the chance to go back there sometimes when the Wind is at least 20-30 MPH you will see the Water magically rise up at least 10-15 feet.
Have you ever been here? The water will only rise 10-15 feet during a strong hurricane. The tides in this upper Chesapeake Bay region rise and fall only about 3 feet daily. 30 MPH winds don't affect the tides as much as you say.
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An fe'r who is finally posting pictures of observations. Well done.
First my question; do you have a picture from the same distance taken from a higher elevation? Or a picture from much closer?
Where are the bushes and steps between the house and the water? There's another small structure lower on a beach just off the side of your picture (I can make out another house further back under the trees) was that lower structure also visible, but just not framed in the shot? GE shows the parking lot (and I assume the first floor) to be 9 feet.
Also, you claim the drop is 12.9 feet, I see no problem with this, and according to the ENaG chart that's correct. However, that's based off a starting point elevation of zero. If you were 2 feet above the water, that puts the drop starting point out further (1.5miles I guess?) to where your line of sight intersects the water, and from there the drop starts. I'll just go with a 1.4 mile advantage from the 2 foot camera elevation just to make an even 3ft. That means there was a drop over 3 miles from where your line of sight would intersect the water. A drop of 6 feet. 6 feet seems to correspond with what is shown in the aerial picture.
When factoring in elevation, it is correct to subtract the elevation of the camera from the total drop expected. The 12.9' drop would be 10.9' after factoring in my elevation. I didn't shoot any pictures from a higher elevation.
I also plan on getting more of these types of shots with better documentation. My friend has a boat that we take out in these waters, I will work up some good proof then as well(during boating season).
29 also about the other structure in my picture, it appears after looking on GE live that it maybe the house in back of the octagonal house. I don't know if you were able to find this place on GE live.
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Hey Hoppy if you get the chance to go back there sometimes when the Wind is at least 20-30 MPH you will see the Water magically rise up at least 10-15 feet.
Have you ever been here? The water will only rise 10-15 feet during a strong hurricane. The tides in this upper Chesapeake Bay region rise and fall only about 3 feet daily. 30 MPH winds don't affect the tides as much as you say.
I think he means the illusion of the water rising.
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Hey Hoppy if you get the chance to go back there sometimes when the Wind is at least 20-30 MPH you will see the Water magically rise up at least 10-15 feet.
Have you ever been here? The water will only rise 10-15 feet during a strong hurricane. The tides in this upper Chesapeake Bay region rise and fall only about 3 feet daily. 30 MPH winds don't affect the tides as much as you say.
I think he means the illusion of the water rising.
Yeah, even if the water only rises 3 feet per day high winds will create a higher optical illusion. I don't live near anyplace with a lot of water to test this out though so my knowledge of tides and such are mostly limited to what I've read or seen in videos. It's hard to perform water experiments without any large bodies of water available.
I've done a few light fringe experiments and it seems to be like the darker it gets the higher the ground seems to rise quicker in the distance this effect seems to be amplified by water as it diffracts light so theoretically you should see the water level appear higher in low light conditions.
Wind experiments I'm yet to participate much in but I think if wind is blowing in your direction you should be seeing the water appear to rise higher optically while if it is blowing away from you it should appear to be lower on your horizon.
Wind blowing around light is a strange phenomenon but I have seen it happen before.
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Refer to this chart, it is correct.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
Only up to a certain point, beyond which the chart is useless. I posted the appropriate calculations in another thread recently, and will put them here when I remember/find them.
When factoring in elevation, it is correct to subtract the elevation of the camera from the total drop expected.
No, it is not. The method 29silhouette gave is correct. Doing it your way will lead to confusion and incorrect answers.
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Refer to this chart, it is correct.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
Only up to a certain point, beyond which the chart is useless. I posted the appropriate calculations in another thread recently, and will put them here when I remember/find them.
When factoring in elevation, it is correct to subtract the elevation of the camera from the total drop expected.
No, it is not. The method 29silhouette gave is correct. Doing it your way will lead to confusion and incorrect answers.
The truth is, thinking the earth is round leads to confusion.
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Refer to this chart, it is correct.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
Only up to a certain point, beyond which the chart is useless. I posted the appropriate calculations in another thread recently, and will put them here when I remember/find them.
When factoring in elevation, it is correct to subtract the elevation of the camera from the total drop expected.
No, it is not. The method 29silhouette gave is correct. Doing it your way will lead to confusion and incorrect answers.
The truth is, thinking the earth is FLAT leads to confusion.
Damn that freudian slip.
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Do any FE'ers care to chime in with some thoughts?
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I think the chart you posted made sense. I can agree with that but what I can't agree with is the stuff that silhouette brought up regarding the obvious differences between the picture you shot and the topographical features of the area shown on google earth. For this to be truly convincing we'd need a couple shots closer to the house so we can compare what's there with the shot you took from across the bay. We also need some verification that you were indeed 2 feet above the water and for whatever height above the water you are we need to determine what the correct drop would be corresponding to that. The way I see it, for every bit you raise the height the drop changes significantly because the angle at which your line of sight meets the horizon changes. So I think Scintific is right in that we don't simply subtract the height from the drop. This presents a problem for the math because we no longer have a right triangle. Your best bet would be to put the camera as close to the water as possible.
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Which means you subtract the distance 1.73 miles from your 4.4 miles to get 2.67 miles.
Using rowbothams method for deriving the drop:
To find the curvature in any number of miles not given in the table, simply square the number, multiply that by 8, and divide by 12. The quotient is the curvation required.
2.67^2 * 8 / 12
This makes the new drop 4.75 feet.
And without a good image to indicate what that beach really looks like it is difficult to determine what is going on but I would guess that a typical beach like the one we see in the picture is at least over 5 ft.
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Another variable is the fact that we don't know how high the tide was at the time of the picture.
In the last week it appears that tides vary between 1 and 2 feet for the water in this bay http://www.myforecast.com/bin/tide.m?city=19127&zip_code=21902&metric=false&tideLocationID=T8148 (http://www.myforecast.com/bin/tide.m?city=19127&zip_code=21902&metric=false&tideLocationID=T8148)
What this means is that if you do wish to verify to us that a picture from across the bay matches the water of a picture you take of a close up then these pictures need to be taken at the same time. With a drop as low as 4.75 and a tidal variation of up to 2 feet, we are looking at some major differences in the drop if you show us pictures at times with different tides.
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When factoring in elevation, it is correct to subtract the elevation of the camera from the total drop expected. The 12.9' drop would be 10.9' after factoring in my elevation. I didn't shoot any pictures from a higher elevation.
I also plan on getting more of these types of shots with better documentation. My friend has a boat that we take out in these waters, I will work up some good proof then as well(during boating season).
Let's go with a hypothetical elevation of 3 feet for a couple questions.
The drop starts from an elevation of 'zero elevation' (eyeball or camera is on the ground), correct? Y/N
At a height of 72" (6'), one's straight line of sight angles slightly downward and intersects the horizon 3 miles away where it is now at 'zero' elevation, correct? Y/N
What would be the drop from the line of sight at the 4 mile mark?
29 also about the other structure in my picture, it appears after looking on GE live that it maybe the house in back of the octagonal house. I don't know if you were able to find this place on GE live.
Yeah, the house in the background shows up through the trees, but I was curious about a small structure right on the beach just south of the house that I saw on GE when I looked myself. Probably just wasn't framed in the shot. It's even lower than the main house.
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Thanks for that site. I could only estimate about 1.5ish but didn't know how to get an exact number because I'm horrible with math. (been working on that though)
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Which means you subtract the distance 1.73 miles from your 4.4 miles to get 2.67 miles.
Using rowbothams method for deriving the drop:
To find the curvature in any number of miles not given in the table, simply square the number, multiply that by 8, and divide by 12. The quotient is the curvation required.
2.67^2 * 8 / 12
This makes the new drop 4.75 feet.
And without a good image to indicate what that beach really looks like it is difficult to determine what is going on but I would guess that a typical beach like the one we see in the picture is at least over 5 ft.
You beat me to it. Once you raise your camera above the circle, you are no longer at the peak of the circle, you are then moved away from that peak.
You can even work the equation backwards to figure out how far the peak is.
Take your elevation, multiply it by 12 then divide it by 8. Take the number you get and take the square root of it.
2*12/8=3
sqrt(3)=1.73
So that makes the horizon 1.73 miles away.
Which corresponds to the answer that the calculator gave.
On a side note, it seems that Rowbotham's table actually supports round earth...
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Which means you subtract the distance 1.73 miles from your 4.4 miles to get 2.67 miles.
Using rowbothams method for deriving the drop:
To find the curvature in any number of miles not given in the table, simply square the number, multiply that by 8, and divide by 12. The quotient is the curvation required.
2.67^2 * 8 / 12
This makes the new drop 4.75 feet.
And without a good image to indicate what that beach really looks like it is difficult to determine what is going on but I would guess that a typical beach like the one we see in the picture is at least over 5 ft.
You beat me to it. Once you raise your camera above the circle, you are no longer at the peak of the circle, you are then moved away from that peak.
You can even work the equation backwards to figure out how far the peak is.
Take your elevation, multiply it by 12 then divide it by 8. Take the number you get and take the square root of it.
2*12/8=3
sqrt(3)=1.73
So that makes the horizon 1.73 miles away.
Which corresponds to the answer that the calculator gave.
On a side note, it seems that Rowbotham's table actually supports round earth...
ah, very clever on doing Rowbotham's method backward.
To be fair to FE'rs, Rowbotham only uses the round earth table because his attempts at proving the earth is flat is by looking at round earth data and attempting to show that the RE data is false. He probably actually did believe the earth is flat.
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Which means you subtract the distance 1.73 miles from your 4.4 miles to get 2.67 miles.
Using rowbothams method for deriving the drop:
To find the curvature in any number of miles not given in the table, simply square the number, multiply that by 8, and divide by 12. The quotient is the curvation required.
2.67^2 * 8 / 12
This makes the new drop 4.75 feet.
And without a good image to indicate what that beach really looks like it is difficult to determine what is going on but I would guess that a typical beach like the one we see in the picture is at least over 5 ft.
Using that calculator I put in 12.9= the height, result is 4.4 miles. The same result as Rowbathoms chart. So if I dropped down 10' feet from the 12.9 that would put the camera at 2.9'. That would indicate 10' of water should be blocking the rocks and part of that house.
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Which means you subtract the distance 1.73 miles from your 4.4 miles to get 2.67 miles.
Using rowbothams method for deriving the drop:
To find the curvature in any number of miles not given in the table, simply square the number, multiply that by 8, and divide by 12. The quotient is the curvation required.
2.67^2 * 8 / 12
This makes the new drop 4.75 feet.
And without a good image to indicate what that beach really looks like it is difficult to determine what is going on but I would guess that a typical beach like the one we see in the picture is at least over 5 ft.
Using that calculator I put in 12.9= the height, result is 4.4 miles. The same result as Rowbathoms chart. So if I dropped down 10' feet from the 12.9 that would put the camera at 2.9'. That would indicate 10' of water should be blocking the rocks and part of that house.
No it, doesn't work that way. You don't just add or subtract numbers from the height to get the amount of drop. These equations for drop are based on the angle toward the horizon at the height in question. That angle changes depending on the height and that significantly alters the drop.
The drop for where you were was 4.75 ft. You can't use/take advantage the calculator to confirm a height of 12.9 ft and then turn around and disregard the calculator for the height of 2 ft in the very next sentence.
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Thanks for that site. I could only estimate about 1.5ish but didn't know how to get an exact number because I'm horrible with math. (been working on that though)
That site gives the same calculations as Rowbothams chart.
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Thanks for that site. I could only estimate about 1.5ish but didn't know how to get an exact number because I'm horrible with math. (been working on that though)
That site gives the same calculations as Rowbothams chart.
We don't disagree with Rowbotham's chart. The problem is that his chart is using numbers based on the observer being at 0 ft high.
You said you were 2 ft high and that changes the spot at which the starting point for the drop is.
So... we still used rowbotham's chart to get 4.75 ft. It's just that the distance (4.4 miles) from you to the house is irrelevant when you are 2 ft high. Since you are 2 feet high you now have to come up with a new distance from a (the horizon relative to you location) to b (the house).
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Which means you subtract the distance 1.73 miles from your 4.4 miles to get 2.67 miles.
Using rowbothams method for deriving the drop:
To find the curvature in any number of miles not given in the table, simply square the number, multiply that by 8, and divide by 12. The quotient is the curvation required.
2.67^2 * 8 / 12
This makes the new drop 4.75 feet.
And without a good image to indicate what that beach really looks like it is difficult to determine what is going on but I would guess that a typical beach like the one we see in the picture is at least over 5 ft.
Using that calculator I put in 12.9= the height, result is 4.4 miles. The same result as Rowbathoms chart. So if I dropped down 10' feet from the 12.9 that would put the camera at 2.9'. That would indicate 10' of water should be blocking the rocks and part of that house.
No it, doesn't work that way. You don't just add or subtract numbers from the height to get the amount of drop. These equations for drop are based on the angle toward the horizon at the height in question. That angle changes depending on the height and that significantly alters the drop.
The drop for where you were was 4.75 ft. You can't use/take advantage the calculator to confirm a height of 12.9 ft and then turn around and disregard the calculator for the height of 2 ft in the very next sentence.
You are wrong on the point you are making. You can subtract the elevation. At 4.4 miles the curve of the earth's sphere is 12.9' no matter what. The beach and at least half of the first floor should be invisible due to water blocking it.
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Here we go: Using the calculator on this page (http://www.cohp.org/local_curvature.html) we can determine how far the horizon is for a given elevation. So for 2 ft the distance to the horizon is 1.73 miles.
Which means you subtract the distance 1.73 miles from your 4.4 miles to get 2.67 miles.
Using rowbothams method for deriving the drop:
To find the curvature in any number of miles not given in the table, simply square the number, multiply that by 8, and divide by 12. The quotient is the curvation required.
2.67^2 * 8 / 12
This makes the new drop 4.75 feet.
And without a good image to indicate what that beach really looks like it is difficult to determine what is going on but I would guess that a typical beach like the one we see in the picture is at least over 5 ft.
Using that calculator I put in 12.9= the height, result is 4.4 miles. The same result as Rowbathoms chart. So if I dropped down 10' feet from the 12.9 that would put the camera at 2.9'. That would indicate 10' of water should be blocking the rocks and part of that house.
No it, doesn't work that way. You don't just add or subtract numbers from the height to get the amount of drop. These equations for drop are based on the angle toward the horizon at the height in question. That angle changes depending on the height and that significantly alters the drop.
The drop for where you were was 4.75 ft. You can't use/take advantage the calculator to confirm a height of 12.9 ft and then turn around and disregard the calculator for the height of 2 ft in the very next sentence.
You are wrong on the point you are making. You can subtract the elevation. At 4.4 miles the curve of the earth's sphere is 12.9' no matter what. The beach and at least half of the first floor should be invisible due to water blocking it.
No matter what? Are you saying that if I increase my elevation then the horizon doesn't get further away? This is a common argument RE'rs make against FE'rs and an established fact. When I'm in an airplane I can see farther away. The same goes if you are 2 ft high. The higher you go, the farther you can see.
Try going back to the same spot and put your camera right next to the water. Put then lens as close to the water as possible. If you do that THEN you'll get a drop of 12.9 ft.
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Dude, the the drop is 12.9 feet because it is 4.4 miles. Subtract my 2' camera elevation 10.9 feet should be blocked.
You are making this more difficult than it needs to. The way I am showing you is correct.
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Dude, the the drop is 12.9 feet because it is 4.4 miles. Subtract my 2' camera elevation 10.9 feet should be blocked.
We have showed you step by step why this is incorrect. If you wish to ignore the facts then good for you.
I'll try one more time:
Observe this image that Rowbotham uses to illustrate this:
(http://www.sacred-texts.com/earth/za/img/fig01.jpg)
The line of sight toward the horizon in this image is resting right on top of the earth. This means that Rowbotham is using a height of ZERO FEET to calculate the distance to the horizon.
From the height of ZERO FEET and from that height only is it acceptable to point the camera exactly parallel to the surface of the earth for these calculations. When you increase the height then the angle toward the horizon is shifted downward and shifts downward more so the higher you go.
Because your camera was 2 feet high, we no longer consider the distance of 4.4 miles because 2 ft high from your location 4.4 miles away from the house would yield the same results for drop as putting the camera at water level 1.73 miles from your location at Perry Point.
You cannot simply subtract the height in the manner you are suggesting. Changing the height changes the angle and changing the angle changes the distance to the horizon. You follow?
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Hoppy, it isn't a matter of making it difficult. It's about doing things correctly.
You can insist that you are correct all you want but I have laid out why it is that you are not. I hope you can comprehend this because it's basic geometry.
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Hoppy, it isn't a matter of making it difficult. It's about doing things correctly.
You can insist that you are correct all you want but I have laid out why it is that you are not. I hope you can comprehend this because it's basic geometry.
I suspect the confusion comes because Hoppy is looking at it as seeing an object over the top of a ball. if you were on one side of a ball and an object on the other and you moved the ball down 2 inches you'd see 2 inches more of the object.
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Hoppy, please observe this image I just threw together:
https://drive.google.com/file/d/0BwWwYmdlMY6KZ1BCLWRzZ0JuMlk/edit?usp=sharing
The line at the top of the circle is a 12 foot observer trying to view the house in the distance.
The dashed red line is Rowbotham's line of sight from ZERO FEET.
The dashed green line is the actual line of sight from the height of 12 feet.
Compare that red line to the green one and you'll notice a couple important factors:
1) The higher the observer the more the angle changes by shifting downward.
2) The more the angle changes (by shifting downward) the farther the observer will be able to see.
So when you make the elevation 2 feet that is not the same as ZERO FEET.
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Hoppy, it isn't a matter of making it difficult. It's about doing things correctly.
You can insist that you are correct all you want but I have laid out why it is that you are not. I hope you can comprehend this because it's basic geometry.
I suspect the confusion comes because Hoppy is looking at it as seeing an object over the top of a ball. if you were on one side of a ball and an object on the other and you moved the ball down 2 inches you'd see 2 inches more of the object.
Which would be fine if the lines of sight toward the horizon were parallel to each other but as height changes the lines are no longer parallel.
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Just trying to help :)
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Like I said earlier with an example using 6 feet for the observation height. At 4 miles, one doesn't simply subtract that height from the drop (10-6), as that would indicate a 4 foot drop over the distance of 1 mile. It needs to be recalculated using the distance advantage of added height from the starting point.
Using Rowbotham's diagram:
(http://imagizer.imageshack.us/v2/800x600q90/853/eieo.png)
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I'm assuming this is because if you imagine your sight line as a big lever going from you to the house, as you move it higher at your end you are not only moving the far end of the lever but also shifting the pivot point towards the far side?
Meaning each small movement at your end would create a progressively greater shift at the other?
EDIT: wait, do I mean lesser shift? but initially far greater on the far end than the near?
It's not something I'd considered before, never had cause.
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I certainly don't doubt the veracity of Hoppy's images.
Having said that, we may be fine-tuning the maths for the earth's fall due to the earth's curvature a little too over-enthusiastically. We're only talking about 4 miles or so, and there's a few variables that could distort our calculations unrealistically, from refraction cause by temperature inversions, to a swell, to a large zoom factor, to a low-resolution sensor in the camera. The image is a little low-res for my liking.
Could you please let us know what camera you were using? And its lens focal length and optical zoom factor?
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You are wrong on the point you are making. You can subtract the elevation.
No, you really can't. Hopefully this image I just spent 10 minutes on will make it a little clearer for you:
(http://imageshack.com/a/img513/2302/f843.png)
The blue triangle represents drop given by Rowbotham's chart, whereas the red triangles represent line of sight and height differences using your camera (NOTE: the image is absolutely not to scale! Everything has been exaggerated for clarity). Note the huge difference between d1 and d2 created by the small difference c. This is what is happening with your camera 2ft above the water level. See how c cannot be subtracted from d1 in order to get d2. I hope this helps to clear up your confusion on the matter.
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I certainly don't doubt the veracity of Hoppy's images.
Having said that, we may be fine-tuning the maths for the earth's fall due to the earth's curvature a little too over-enthusiastically. We're only talking about 4 miles or so, and there's a few variables that could distort our calculations unrealistically, from refraction cause by temperature inversions, to a swell, to a large zoom factor, to a low-resolution sensor in the camera. The image is a little low-res for my liking.
Could you please let us know what camera you were using? And its lens focal length and optical zoom factor?
The camera is Samsung HMX-F80. 52X optical zoom.
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Rot,
The problem is you are using a SE model. You need to use the real thing...you know, the FE model.
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Rot,
The problem is you are using a SE model. You need to use the real thing...you know, the FE model.
Because he is asking how it is possible in an RE?
We are even using Rowbotham's own numbers from his own chart out of his own book to show what is going on.
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Rot,
The problem is you are using a SE model. You need to use the real thing...you know, the FE model.
Because he is asking how it is possible in an RE?
We are even using Rowbotham's own numbers from his own chart out of his own book to show what is going on.
If the world is flat then why can't I see the empire state building from California?
That'd be the test if we used a flat model.
Game over.
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You think you are so smart. I just laugh. You will never see that far from your angle. The Empire State building would have to be extremely high......I'm sure you mathematicians can work out the details.
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You think you are so smart. I just laugh. You will never see that far from your angle. The Empire State building would have to be extremely high......I'm sure you mathematicians can work out the details.
only on a round earth.
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Rot,
The problem is you are using a SE model. You need to use the real thing...you know, the FE model.
Because he is asking how it is possible in an RE?
We are even using Rowbotham's own numbers from his own chart out of his own book to show what is going on.
If the world is flat then why can't I see the empire state building from California?
That'd be the test if we used a flat model.
Game over.
Well because the Rocky Mountains are in the way ::) ;D
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Rot,
The problem is you are using a SE model. You need to use the real thing...you know, the FE model.
Because he is asking how it is possible in an RE?
We are even using Rowbotham's own numbers from his own chart out of his own book to show what is going on.
If the world is flat then why can't I see the empire state building from California?
That'd be the test if we used a flat model.
Game over.
Your eyes are only going to see so far, the house in this thread was not even visible to the naked eye. I could only see it through the zoom lens maxed out.
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Rot,
The problem is you are using a SE model. You need to use the real thing...you know, the FE model.
Because he is asking how it is possible in an RE?
We are even using Rowbotham's own numbers from his own chart out of his own book to show what is going on.
If the world is flat then why can't I see the empire state building from California?
That'd be the test if we used a flat model.
Game over.
Your eyes are only going to see so far, the house in this thread was not even visible to the naked eye. I could only see it through the zoom lens maxed out.
Nothing a good telescope can't solve.
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Having said that, we may be fine-tuning the maths for the earth's fall due to the earth's curvature a little too over-enthusiastically. We're only talking about 4 miles or so
I tried photographing curvature drop over 3/4 of a mile once across calm water from a couple different elevations. It was hard to tell with those pictures. 4 miles isn't too bad. Shots from multiple elevations would have been much more telling.
You think you are so smart. I just laugh. You will never see that far from your angle. The Empire State building would have to be extremely high......I'm sure you mathematicians can work out the details.
Have you been making progress yet on a drawing on how a tennis ball is obscured across a living room at least?
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Rot,
The problem is you are using a SE model. You need to use the real thing...you know, the FE model.
Because he is asking how it is possible in an RE?
We are even using Rowbotham's own numbers from his own chart out of his own book to show what is going on.
If the world is flat then why can't I see the empire state building from California?
That'd be the test if we used a flat model.
Game over.
Your eyes are only going to see so far, the house in this thread was not even visible to the naked eye. I could only see it through the zoom lens maxed out.
Nothing a good telescope can't solve.
Previously I have borrowed my brother inlaw's $400.00 used telescope. It comes in a big metal box and weighs about 25-30 lbs. A big huge telescope that you connect a camera to. My little camera seems to zoom in as good as a big expensive telescope.
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The camera is Samsung HMX-F80. 52X optical zoom.
Thanks. I was just trying to equate the image quality with the camera. It has a very small 5MP sensor, which at 52x zoom ain't real great. Is it possible to borrow a mate's still camera with (say) an APS-C sensor of around 12MP or so? The zoom's not as critical as the camera's image resolution capabilities for this experiment.
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The camera is Samsung HMX-F80. 52X optical zoom.
Thanks. I was just trying to equate the image quality with the camera. It has a very small 5MP sensor, which at 52x zoom ain't real great. Is it possible to borrow a mate's still camera with (say) an APS-C sensor of around 12MP or so? The zoom's not as critical as the camera's image resolution capabilities for this experiment.
May I borrow yours?
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Previously I have borrowed my brother inlaw's $400.00 used telescope. It comes in a big metal box and weighs about 25-30 lbs. A big huge telescope that you connect a camera to. My little camera seems to zoom in as good as a big expensive telescope.
Yep! It's funny how many people on Youtube have used digital cameras and gotten excellent footage of the moon AND the sun by zooming in. These Spherical Earthers need to get real.
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Previously I have borrowed my brother inlaw's $400.00 used telescope. It comes in a big metal box and weighs about 25-30 lbs. A big huge telescope that you connect a camera to. My little camera seems to zoom in as good as a big expensive telescope.
Yep! It's funny how many people on Youtube have used digital cameras and gotten excellent footage of the moon AND the sun by zooming in. These Spherical Earthers need to get real.
Yeah, because those people are amazing photographers. Are we going to seriously give props to hoppy's camera? The image is awful.
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Yeah, because those people are amazing photographers. Are we going to seriously give props to hoppy's camera? The image is awful.
The point is, from a low angle (human's perspective on Earth), you simply cannot see more than a few miles, with or without a telescope/excellent camera. The atmosphere is too dense (just like it is when seeing the sun at sun rise and set). The object has to be higher than the atmosphere or the observer has to be higher than the atmosphere.
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Yeah, because those people are amazing photographers. Are we going to seriously give props to hoppy's camera? The image is awful.
The point is, from a low angle (human's perspective on Earth), you simply cannot see more than a few miles, with or without a telescope/excellent camera. The atmosphere is too dense (just like it is when seeing the sun at sun rise and set). The object has to be higher than the atmosphere or the observer has to be higher than the atmosphere.
I can see trees on hills well over 5 miles away on a clear day.
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Yeah, because those people are amazing photographers. Are we going to seriously give props to hoppy's camera? The image is awful.
The point is, from a low angle (human's perspective on Earth), you simply cannot see more than a few miles, with or without a telescope/excellent camera. The atmosphere is too dense (just like it is when seeing the sun at sun rise and set). The object has to be higher than the atmosphere or the observer has to be higher than the atmosphere.
This entirely depends on the weather features of the day. On some days, you can see for hundreds of miles which you can back up with any experience looking out of cabin windows on airplanes.
I can also verify this with my experience as a weather observer aboard naval ships. Every hour we would go outside and do a weather observation and one of the items to be filled out on our observation sheet was visibility. We literally had to determine the distance in statute miles that one could see that day. Over time, my skills at guessing the visibility got better and better but there were times that we could actually verify the visibility and test accuracy. We have a radar display in our office connected to the Combat Information Center (the dark office on naval ships that you normally see in movies) that would show the distance to objects in the vicinity using a multitude of different surface search radars at our disposal. If there was a ship nearby then it was a perfect opportunity to test our ability at guessing the visibility and I can tell you for a fact that I have been able to see ships as much 15 statute miles away from vulture's row of our amphib.
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So, before this whole thread wanders, I'd just like to say this.
Despite being taken with 52 optical zoom through a small lens (anybody who knows scopes, binos, cameras, and such should understand the problem with this) and with a smaller MP low-end sensor, There's plenty of detail to see windows, gutters, railing, porch supports, even another house back in the trees. Which means, there should be enough detail to see the embankment of bushes with the set of steps leading down to a path to the beach. Looks like a thin layer of mirage between the water and the first floor however. Another picture from a higher elevation would have helped a lot too.
Some people make clearer pictures from video by stacking hundreds of frames and tweaking it. I just got the software to do this a few days ago and I'm experimenting with it. I more or less got my webcam adapted to my new 20-60x80 spotting scope. Also finished modifying my point&shoot camera mount for the scope. I'll be able to use my video camera with it too. (last night I should say. Haven't tested it yet.)
Hoppy, did you use manual focus/exposure? If not, these might help.
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I just took this image today, how can this house be visible if the earth is a sphere?
This image was taken from 4.4 miles as I found on google earth. There is no other proof for you roundies, you just have to take my word for now about this. I will work on better documentation later. But what I have stated here is true.
(http://imgur.com/iUBPPWM.jpeg)
(http://imgur.com/9U9cxIE.jpeg)
I went over to the house in the picture and snapped a shot of the house and beach, there is a 10' max from the top of the water to the 1st floor of the building. I risked life and limb being on private property, but nobody said anything to me.
(http://imgur.com/Otx5xG1.jpeg)
Use the google lat long co ordinates to locate this house. It is the only octagonal house on this shore. Location is Cecil County, Maryland.
(http://imgur.com/AEuNITM.jpeg)
I was next to the pier sitting on rocks,at Perry Point VA hospital, the camera was 2' above the water. I was at one end of the yellow line, the octagon house was at the other end.
Bumped to add a close picture of the house.
I'm sorry FE'ers 10.9 feet is cut off from the water line to the bottom of the house.
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I'm sorry to say hoppy but it appears that the entire first floor of the picture you took from across the bay is missing.
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So, before this whole thread wanders, I'd just like to say this.
Despite being taken with 52 optical zoom through a small lens (anybody who knows scopes, binos, cameras, and such should understand the problem with this) and with a smaller MP low-end sensor, There's plenty of detail to see windows, gutters, railing, porch supports, even another house back in the trees. Which means, there should be enough detail to see the embankment of bushes with the set of steps leading down to a path to the beach. Looks like a thin layer of mirage between the water and the first floor however. Another picture from a higher elevation would have helped a lot too.
Some people make clearer pictures from video by stacking hundreds of frames and tweaking it. I just got the software to do this a few days ago and I'm experimenting with it. I more or less got my webcam adapted to my new 20-60x80 spotting scope. Also finished modifying my point&shoot camera mount for the scope. I'll be able to use my video camera with it too. (last night I should say. Haven't tested it yet.)
Hoppy, did you use manual focus/exposure? If not, these might help.
I was lucky to get this shot I think, 4.4 miles is long way away. I used auto focus and exposure. Maybe I'll figure out how to use the manual settings for better pictures. I am impressed by what this has done on auto, even though some complain about it. I could barely see the house on my small monitor screen when I took the picture, due to glare on the screen. As someone who dragged around a huge telescope with a camera mount, i am impressed with this little samsung. I think the pictures are just as good.
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I'm sorry to say hoppy but it appears that the entire first floor of the picture you took from across the bay is missing.
It looks like you're right, at least half of the 1st floor is not there. That is due to looking at an angle up from the water level and the bushes blocking part of the first floor.
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You can always see farther the higher you are especially as air density thins out. You can also see farther when you are out at sea because light refracts and creates a large circle of visibility especially when wind is blowing in your direction. When wind is blowing away from you your visibility goes down when wind is blowing towards you visibility goes up. 11 MPH is average Wind Pressure and about 1.1-2.2 Miles is about normal viewing range from static elevation. Viewing range goes up as 1 Mile of Elevation will net you about 4 miles of optical distance from eyesight alone. This can be increased substantially a good Camera can do at least 42x Optical Zoom which will net you about 42/4 = 10.5 Times the normal viewing distance from your angel. So if I had a good 42x Camera and I can see 1 Mile from my location I should be able to see as much as 10 miles if elevation remains fairly static. However remember that after 45 Degrees Optical Zoom your returns will increase substantially up to about 90x Zoom which will put you about at 6 Times the viewing range so with a Stories High Earth telescope you might be able to increase 1 mile of viewing distance to about 60 miles of viewing distance in low Earth atmosphere with no elevation advantage.
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Also wanted to show you that apparently according to Google Earth this House should be significantly above the Water Line so the water cutting off part of the house can't be due to curvature but rather optics, however in this case this round Earth geometry is clearly very distorted. In fact a lot of Globe mapping is very wrong to look at.
(https://scontent-b-sea.xx.fbcdn.net/hphotos-ash3/t1/1012370_568493113234738_1989674956_n.jpg)
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I'm sorry to say hoppy but it appears that the entire first floor of the picture you took from across the bay is missing.
It looks like you're right, at least half of the 1st floor is not there. That is due to looking at an angle up from the water level and the bushes blocking part of the first floor.
Supposing that 1st floor is about 20 ft tall and using this trig calculator (http://www.carbidedepot.com/formulas-trigright.asp) I found the angle to only be 0.0443923822686798°.
Make side a 20 ft
Make side b 4.4 miles (23232 ft)
This is not enough an angle to do what you are suggesting.
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I'm sorry to say hoppy but it appears that the entire first floor of the picture you took from across the bay is missing.
That's what I was mistaking for a mirage. It's actually the upper few feet of the first floor below what I previously thought was the first floor.
It looks like you're right, at least half of the 1st floor is not there. That is due to looking at an angle up from the water level and the bushes blocking part of the first floor.
Actually you're 2 feet above the water, so if you had been aimed at a point two feet above the waterline, assuming flatness (RE you would be looking at a slight downward angle), you would have been looking level, not up at an angle. The rest of the house would have been at a slight upward angle, true.
If the first floor is being obscured by bushes (They don't look thick enough to obscure that much), then why can't we see all the bushes, steps, rocks, and trail down to the water. You could try adding a line where you feel the waterline is, but it looks a lot like it's well above the bottom of the first floor.
I risked life and limb being on private property, but nobody said anything to me.
Well we appreciate it. The picture clears up some things.
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Also wanted to show you that apparently according to Google Earth this House should be significantly above the Water Line so the water cutting off part of the house can't be due to curvature but rather optics, however in this case this round Earth geometry is clearly very distorted. In fact a lot of Globe mapping is very wrong to look at.
(https://scontent-b-sea.xx.fbcdn.net/hphotos-ash3/t1/1012370_568493113234738_1989674956_n.jpg)
The elevation is not distorted. The textured images placed on top of the elevation appear distorted from that viewing angle because only buildings in major metropolitan areas are vectored.
Without the house being vectored and with the scale in Google Earth not showing which part of the beach is the same eye level as Google Earth's ground view (which is 7 ft) it is impossible to tell if Google Earth is accurate here.
I'd agree that optics might be playing a role via a mirage as silhouette29 suggested. It appears that there is a tan line above the water across the whole beach despite the fact that neither Google Earth or the close-up can confirm that the beach extends across the entire waterfront. I'm more convinced that the tan line is a mirage too because it is located where part of the first floor should be but the bottom half of the first floor would appear to belong below the waterline if you ask me.
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The Conspiracy know's you're a FE believer so they raised all the houses around you.
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In light of the close up picture of the house, i might have to reconsider my FE stance. In the far picture it seems that about 10' of water is blocking most of the first floor,
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In light of the close up picture of the house, i might have to reconsider my FE stance. In the far picture it seems that about 10' of water is blocking most of the first floor,
Good for you hoppy. :)
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In light of the close up picture of the house, i might have to reconsider my FE stance. In the far picture it seems that about 10' of water is blocking most of the first floor,
Its only early days yet Hoppy.
In time you will see the different variables.
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Hoppy, are you absolutely sure the photo you took from across the bay is the same one on Google? I ask because in the Google image, there are no houses visible on either side. Only trees and driveway. I know that Google doesn't always have current photos either.
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Spaceship, I thought the same thing when he first posted. It does look like an octogon house but I just wanted to check for myself so I searched the entire shoreline on Google earth the other day and there are no other houses that look anything like that.
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I just did that as well only found one other octagon shaped home but it was taken directly above the house and the rooftop is gray/black. I also noticed in my search on Google, that the buildings next to the house are not on the shoreline, they are behind the house. The only thing I can suggest is that this effect (10 ft. disappearing) is caused by the ocean water and moisture/dense air. Would we see this same thing over land? The next experiment should be to take a photo of an object at the same distance on land. Anyone near open pasture? Or better yet, a long road or abandoned airport runway on a cool day so there are no mirages.
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Or the world is round. I keep trying to tell you that. Maybe you'll have to prove it to yourself like hoppy. Nothing like doing it yourself.
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This says the unaided eye can see objects 3 miles away on a flat surface:
http://answers.yahoo.com/question/index?qid=20080429143254AAQAJjv (http://answers.yahoo.com/question/index?qid=20080429143254AAQAJjv)
I want to test it myself. I'm going to meet a friend somewhere and park one of our cars in the open and drive the other's car 4 miles away and take a picture of it with my dig cam.
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I just did that as well only found one other octagon shaped home but it was taken directly above the house and the rooftop is gray/black. I also noticed in my search on Google, that the buildings next to the house are not on the shoreline, they are behind the house. The only thing I can suggest is that this effect (10 ft. disappearing) is caused by the ocean water and moisture/dense air. Would we see this same thing over land? The next experiment should be to take a photo of an object at the same distance on land. Anyone near open pasture? Or better yet, a long road or abandoned airport runway on a cool day so there are no mirages.
Search out my "measure a mountain" experiment, it's not exactly what you're after I guess, but it's close, and it avoids the mirage problem. Also, it's hard to find a really flat piece of land. Salt lakes are about the best you can do, and even they aren't completely flat.
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This says the unaided eye can see objects 3 miles away on a flat surface:
http://answers.yahoo.com/question/index?qid=20080429143254AAQAJjv (http://answers.yahoo.com/question/index?qid=20080429143254AAQAJjv)
I want to test it myself. I'm going to meet a friend somewhere and park one of our cars in the open and drive the other's car 4 miles away and take a picture of it with my dig cam.
I'm kind of confused about the purpose of this. You'd essentially just be testing the human eye. What does that have to do with the shape of the earth?
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How about a Fe'er chiming in, what is with the 10' wall of water?..... Bishop? .....Thork?.... Anyone?
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Refraction of light over a body of water is a well known phenomena. The light from the bottom portion of the house simply hits the water before it reaches your eyes. The light reflected off of the second story travels farther because it is higher above the water.
I hope this makes you feel better, hoppy.
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Refraction of light over a body of water is a well known phenomena. The light from the bottom portion of the house simply hits the water before it reaches your eyes. The light reflected off of the second story travels farther because it is higher above the water.
I hope this makes you feel better, hoppy.
It doesn't really make me feel better. Rowbothams experiments always saw through the refraction. My experiment did not, just by chance the refraction is equal to RE measurements of curvature. So far this is not looking good for FET
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hoppy, take the same picture several times, with different atmospheric and tidal conditions. Let's see if that picture does not change.
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I'm kind of confused about the purpose of this. You'd essentially just be testing the human eye. What does that have to do with the shape of the earth?
No, I'll be testing whether the bottom of an object disappears over land like it does over water. I suspect it will not.
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hoppy, take the same picture several times, with different atmospheric and tidal conditions. Let's see if that picture does not change.
Take the picture at night during a full moon and with the lights on in the house.
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I'm kind of confused about the purpose of this. You'd essentially just be testing the human eye. What does that have to do with the shape of the earth?
No, I'll be testing whether the bottom of an object disappears over land like it does over water. I suspect it will not.
How would you determine if the ground is flat and level over any great distance?
hoppy, take the same picture several times, with different atmospheric and tidal conditions. Let's see if that picture does not change.
This is actually a very good idea whichever belief you hold.
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You guys are in luck if you want to know what to expect with refraction because that is my primary job. Just tell me the date and time that you want to take the picture and I can run an atmospheric profile for the project. I will have a graph which will display whether there are any trapping, super refractive or sub refractive layers at the time asked. Furthermore I can run a multitude of radars and or light scenarios against that atmospheric profile and tell you whether the refractive indeces will or will not produce refractive effects, by what angle and for however many statute miles.
Also, I have already posted the expected tidal data based on climatology but I can also forecast the tidal data for that data using top of the line Naval meteorological tools.
For reference my refraction program is called AREPS (Advanced Refractive Effects Predictions System) and my Tidal/Lunar program is called GFMPL (Geophysics Fleet Mission Program Library).
I won't be back to work until Wednesday so if hoppy is interested I'd prefer to have Wednesday to set that up for this thread. I just need the time (no more than 3 days after Wedenesday due to forecast accuracy) and the exact lat/long of the point at which hoppy will be taking the picture from.
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Don't do it Hoppy, He's part of the conspiracy, He'll call in a drone strike!
;D :D
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How would you determine if the ground is flat and level over any great distance?
Yeah, same goes with water.
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How would you determine if the ground is flat and level over any great distance?
Yeah, same goes with water.
No, do we really need to explain why water isn't like the ground?
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This says the unaided eye can see objects 3 miles away on a flat surface:
http://answers.yahoo.com/question/index?qid=20080429143254AAQAJjv (http://answers.yahoo.com/question/index?qid=20080429143254AAQAJjv)
I want to test it myself. I'm going to meet a friend somewhere and park one of our cars in the open and drive the other's car 4 miles away and take a picture of it with my dig cam.
Are you going to take the pictures from eye-level? If so, most of the car should still be visible 4 miles away (assuming you're using some fairly high-powered zoom lens)
Refraction of light over a body of water is a well known phenomena.
Wouldn't that require the light coming from the house to bend downward and then back up to to the camera with no major distortion in order to produce what is seen in the picture?
The light from the bottom portion of the house simply hits the water before it reaches your eyes. The light reflected off of the second story travels farther because it is higher above the water.
This part sounds like round earth properties.
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No, do we really need to explain why water isn't like the ground?
That's my point! How can you determine the water in that photo by the bay is level? In all actuality, it wouldn't be on a spherical Earth. Explain that one.
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No, do we really need to explain why water isn't like the ground?
That's my point! How can you determine the water in that photo by the bay is level? In all actuality, it wouldn't be on a spherical Earth. Explain that one.
Are you talking about a slight curvature of the surface that would block viewing of the bottom part of distant objects.
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No, do we really need to explain why water isn't like the ground?
That's my point! How can you determine the water in that photo by the bay is level? In all actuality, it wouldn't be on a spherical Earth. Explain that one.
This is the entire point of these experiments. Th FE view is that the world is flat and the RE view is that it is not. The experiment is done to prove one way or the other. When we talk about the "flatness" of the ground or the water in these experiments there is a few things both sides can agree about:
1) The term "flatness" simply means an elevation of zero, preferably throughout the entire area in which the experiment is conducted.
2) Water is better for this than the ground because we all know that ground can have topographical features which prohibit said flatness.
3) When such an experiment is conducted over water it is preferable to use water that is as still as possible.
The main point of my comment to you about the differences between water and the ground is that the ground will not be at an even elevation for a 4 mile stretch in a majority of cases. This is why everyone conducts these experiments over water, including Rowbotham.
When water is not a river and connected to the ocean then the elevation is ZERO.
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But you must allow for the effects of water....evaporation/vapor/density in the air near the water. This is extremely important, and in my opinion, what causes the bottom of objects to disappear from a distance. Water is not a solid (obviously) and is taking of space by volume and in the air above it.
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But you must allow for the effects of water....evaporation/vapor/density in the air near the water. This is extremely important, and in my opinion, what causes the bottom of objects to disappear from a distance. Water is not a solid (obviously) and is taking of space by volume and in the air above it.
Yes you do and guess what? Refractive effects such as trapping layers over water are always present but never less than 25 ft. I know this because it is my job. What else?
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Spaceship, on the other forum Tintagel was posting almost the exact same excuses and here was my response:
How about a Fe'er chiming in, what is with the 10' wall of water?..... Bishop? .....Thork?.... Anyone?
A couple of things at play - I don't see a ten foot wall of water, I see the lower section of the house sort of "compressed" - this is partially an optical trick which occurs close to the horizon (sunsets do it too, the sun sort of collapses into itself when its image is on the horizon, and partially due to the way light travels on a Flat Earth. See the Electromagnetic Accelerator thread. The lower areas of the house are hidden behind water because that light "dips" before bending upward again.
Oh boy, an actual experiment grinds against your world view and instead of taking this data and reconsidering your views, you treat your views as infallible and come up with an excuse.
Let's be clear, there is no such thing as bendy light. The closest real world example of such a thing is refraction, which, while nearly always present over water, it does not work in light of this picture.
First off, I'd like to mention that my job is refraction. I create atmospheric profiles for various locations around the planet and determine how those profiles will effect the propagation of radar signals and light. Over water there is ALWAYS an effective trapping layer that causes propagation for as much as 256 miles but the problem with your analysis is that this trapping layer in my experience doing literally 1000's of profiles is never less than 25 ft and 9 times out of 10 it is 30 ft. It is never 10 ft which is exactly what it would have to be for your suggestion to even have potential. Furthermore, if what we are seeing is compression then there would be a couple of things wrong with that. The index of refraction between a medium of air above the trapping layer and a medium of air in it would differ by small amounts causing changes that could not be as dramatic as what you are seeing. Secondly, compression suggests that the refraction would cause the refracted light to change direction at both the top and the bottom of the trapping layer but that is not how this works because the refracted light would be unaffected within the trapping layer medium. Refracted light is the effect that causes light to change direction from one medium to the next but not within the medium itself.
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Let's be clear, there is no such thing as bendy light. The closest real world example of such a thing is refraction, which, while nearly always present over water, it does not work in light of this picture.
Isn't "bendy light" called refraction?
And no, I didn't say it was caused by light. I said water vapor....air density.
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Let's be clear, there is no such thing as bendy light. The closest real world example of such a thing is refraction, which, while nearly always present over water, it does not work in light of this picture.
Isn't "bendy light" called refraction?
No. Bendy light is called Electromagnetic Acceleration. Essentially, bendy light causes light to bend the opposite way that you would expect refraction to bend light.
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No, I'll be testing whether the bottom of an object disappears over land like it does over water. I suspect it will not.
Could you clarify why you believe a car will not disappear from the bottom up when viewed from ground level (you're going to be using a height of 5-6 ft correct?) 4 miles away, but believe a tennis ball will disappear from the bottom up when viewed from close to the floor (ground level) across a living room?
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29,
With enough distance, everything disappears. So it remains to be seen, I suppose.
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Let's be clear, there is no such thing as bendy light. The closest real world example of such a thing is refraction, which, while nearly always present over water, it does not work in light of this picture.
Isn't "bendy light" called refraction?
No. Bendy light is called Electromagnetic Acceleration. Essentially, bendy light causes light to bend the opposite way that you would expect refraction to bend light.
Not to mention that bendy light is theorized to have a continuous bend and is mentioned to be caused by aether by many FE believers. In refraction, light doesn't continue to bend and the cause is simple... it goes in a straights line and then refracts when it enters a new medium and when it does enter that medium its bending is abrupt. Entirely different concepts, one of which is verified to occur.
So 2 major differences. The way it works and what causes it. Both of which are completely understood by modern science.
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Let's be clear, there is no such thing as bendy light. The closest real world example of such a thing is refraction, which, while nearly always present over water, it does not work in light of this picture.
Isn't "bendy light" called refraction?
And no, I didn't say it was caused by light. I said water vapor....air density.
No, it is not and refraction is caused by incident light encountering a new medium which can be caused by water vapor and density, but like I said, Refractive layers with these capabilities over water are never EVER less than 25 ft.
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(http://i.imgur.com/n2HZiUs.jpg)
This image is from the same rock as the original image, camera 32" above the water.
(http://i.imgur.com/SnDtIuy.jpg)
This image is from a pier , camera is 8' above the water.
Here are two pictures from today, Today the pictures look entirely different. It looks flat today. ??? ??? ??? WTF
I also want to add, the water level was within 6" of the first image. Low wind conditions, but the haze in the area seemed thicker today. There was heavy chemtrail spraying today.
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Wow, that is very interesting! Thank you for posting those Hoppy! More of the house is visible than before and yet, the water looks different in all three pictures. Could it be that the camera picks up waves on the water that the unaided eye cannot?
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(http://imgur.com/iUBPPWM.jpeg)
I reposted the original image for ease of comparison.
(http://i.imgur.com/n2HZiUs.jpg)
This image is from the same rock as the original image, camera 32" above the water.
(http://i.imgur.com/SnDtIuy.jpg)
This image is from a pier , camera is 8' above the water.
Here are two pictures from today, Today the pictures look entirely different. It looks flat today. ??? ??? ??? WTF
I also want to add, the water level was within 6" of the first image. Low wind conditions, but the haze in the area seemed thicker today. There was heavy chemtrail spraying today.
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What time of day were they taken? Looks like the sun was more to the right side of the house in the newer pics.
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What time of day were they taken? Looks like the sun was more to the right side of the house in the newer pics.
The newer pics were in the morning, original was late afternoon. This side of the house faces west.
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Things optically will move up until they can't move more up then will move down. Today I did an experiment where I looked at a house at 11AM with strong Southern wind and noticed I could see a house on the Horizon about a quarter mile away (At eye level) and I could also see below a window (about 5 feet off the ground) that was facing my direction. I checked the House again five hours later and noticed that the wind direction has shifted from South to North and also noticed that I looked upon the house again and could only barley see below the roof-line (5 feet above the window starting point on a 12 feet tall house) that means the light from the hill was reaching me 5 feet above at 4 PM then it was at 11AM.
Now I'm not sure what the cause of this was cause I was standing in the same spot for both observations but it seems clear to me that I can see farther in the direction that the wind is blowing (And I see things as more flat) then when the wind is blowing away from me.
Read this quote from me and go back and pay very close attention to wind some day if you have time and see if wind blowing upon or away from you matches with my own experiment. (Also if you can try to take a picture at Noon and just before it gets dark to see if that changes anything). Either that or just believe that wind can make at least 5 foot of difference in optics my experiment was only from a quarter mile away with my unaided eye but if your using a camera the camera will offset the difference itself.
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You didn't feel the need to go back to the original height and take another picture?
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How would you determine if the ground is flat and level over any great distance?
Yeah, same goes with water.
water finds it's own level.
Waves etc. excepting.
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Also, I have already posted the expected tidal data based on climatology but I can also forecast the tidal data for that data using top of the line Naval meteorological tools.
Gee rottingroom, you're taking a risk on this forum!
You're actually gonna post some irrefutable scientific evidence to support your claims? Are you fully prepared for the shit-storm of hocus-pocus that the FEs are gonna bring down on you?
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Thanks for the new images hoppy, but you desperately need to use a camera with a higher resolution sensor. 5MP just won't do the job I'm afraid.
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Thanks for the new images hoppy, but you desperately need to use a camera with a higher resolution sensor. 5MP just won't do the job I'm afraid.
Dude, I'm going to go back there and get more images. I'll try manual focus, different times of day, whatever. But that house is so far away that you can't see it with a naked eye. When I push the button to take the image, the camera shakes and takes the house all the way out of view. Let's see your pics of something from 4.4 miles.
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Thanks for the new images hoppy, but you desperately need to use a camera with a higher resolution sensor. 5MP just won't do the job I'm afraid.
Dude, I'm going to go back there and get more images. I'll try manual focus, different times of day, whatever. But that house is so far away that you can't see it with a naked eye. When I push the button to take the image, the camera shakes and takes the house all the way out of view. Let's see your pics of something from 4.4 miles.
Yeah, so there Geoff! Maybe you need to delete that sarcastic post above to Rot about showing evidence.
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Hoppy is still subtracting the elevation the picture is taken at from the drop. This is incorrect.
The new drop at 32 inches in elevation is 3.47 ft while the 8 ft elevation shot is .589 ft.
You figure this out by using this calculator (http://www.cohp.org/local_curvature.html).
You use the first calculator by putting in your elevation. It gives you distance which you subtract from your 4.4 miles.
Then you take that distance and put it in the second calculator which gives you your drop.
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Hoppy is still subtracting the elevation the picture is taken at from the drop. This is incorrect.
The new drop at 32 inches in elevation is 3.47 ft while the 8 ft elevation shot is .589 ft.
You figure this out by using this calculator (http://www.cohp.org/local_curvature.html).
You use the first calculator by putting in your elevation. It gives you distance which you subtract from your 4.4 miles.
Then you take that distance and put it in the second calculator which gives you your drop.
Of course you keep getting the wrong answer, you keep using the wrong formula. See the example I gave Markjo above.
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Hoppy is still subtracting the elevation the picture is taken at from the drop. This is incorrect.
The new drop at 32 inches in elevation is 3.47 ft while the 8 ft elevation shot is .589 ft.
You figure this out by using this calculator (http://www.cohp.org/local_curvature.html).
You use the first calculator by putting in your elevation. It gives you distance which you subtract from your 4.4 miles.
Then you take that distance and put it in the second calculator which gives you your drop.
Of course you keep getting the wrong answer, you keep using the wrong formula. See the example I gave Markjo above.
First of all... Wrong forum. Markjo responded on the other forum.
Secondly... NO!!!
Elevation makes the distance to the horizon exponentially further away.
Here is the conversation between you, Markjo and I to help people follow along:
Even your calculator gave the same answer as Rowbotham's chart.
Yes, because Rowbotham's chart is completely correct for an elevation of 0 FT.
Just because I move the camera up 2' feet doesn't change the drop over 4.4 miles. So the end of the drop is going to 10.9' instead.
It absolutely does change the drop. It isn't a matter of simple subtraction.
Think of it like this, if the drop over a distance was 10', and blocked the entire view of a 10' bldg sitting right at the water. If you climbed a 10' ladder you would be able to see the entire building.
Yes because climbing to 10 ft in elevation would account for the entire height of the building but the problem is that between the elevation of 0 and 10 ft there is an exponential curve.
I have made 2 charts representing the relationship in Rowbotham's chart between height and distance. \
HERE THEY ARE (https://docs.google.com/spreadsheet/ccc?key=0AgWwYmdlMY6KdDdnNDdIV3F1RjZTVGVmMlU2dVVMc2c&usp=sharing)
One of them shows the relationship for elevations 0-10 and the other shows relationship for elevations 0-20. In both examples distance grows exponentially. For what you are saying to hold true... that we simply subtract a number then the growth in distance should not be exponential but linear. This should tell you that there is more involved that just subtracting the height.
I hope this clears things up.
You have to think of the total distance, don't divide it up to 2 different distances. If you do divide it, you will get the wrong answer.
Incorrect. You must divide the total distance into two different distances to get the correct answer. First there is the distance from the observer to the horizon, then there is the distance from the horizon to the house.
Marrrrkjo. That how to get the wrong answer.
That is why the drop over 3 miles is 6' = 72". Correct
Not a drop of 1 mile(8") +(8") + (8") = 24" Incorrect
That isn't what Markjo is saying. When you raise the elevation of the observer. You get a new distance to the horizon. Then you measure the drop from that horizon to the target. You don't just divide things to your liking in the same way as your crude example. You have to try and think about this.
Have a good look at 29silhouette's picture again:
(http://imagizer.imageshack.us/v2/800x600q90/853/eieo.png)
For a distance of 4 miles there should be a drop of 10.7 ft
The top of the image shows the observers horizon (when viewing from an elevation of 6 ft) to be 3 miles.
From this point we can use Rowbotham again because NOW we are at 0 elevation for the remaining mile giving us the 8 inch drop.
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Either that or just believe that wind can make at least 5 foot of difference in optics my experiment was only from a quarter mile away with my unaided eye but if your using a camera the camera will offset the difference itself.
You're saying that If i look at a house a quarter of a mile away with the wind in different directions it will seem to have a 5ft difference in location but if I take a photo it won't but will instead appear to be in the same place both times?
Presumably that place not being the same as it looks to be with eyesight?
Believing that would be a very big ask.
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Hoppy is still subtracting the elevation the picture is taken at from the drop. This is incorrect.
The new drop at 32 inches in elevation is 3.47 ft while the 8 ft elevation shot is .589 ft.
You figure this out by using this calculator (http://www.cohp.org/local_curvature.html).
You use the first calculator by putting in your elevation. It gives you distance which you subtract from your 4.4 miles.
Then you take that distance and put it in the second calculator which gives you your drop.
Of course you keep getting the wrong answer, you keep using the wrong formula. See the example I gave Markjo above.
Break the formula down into what it's actually saying.
1. On a round earth the distance to the horizon will (optically) show 0 drop (you will be able to see the very bottom of any object at any point up to the horizon).
2. the higher you are the further to the horizon
3. once an object is over the horizon then drop will begin and gradually increase.
Unless you disagree with one of those three points then Rot's method of calculation is correct. I can't speak for the numbers merely the method.
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Thanks for the new images hoppy, but you desperately need to use a camera with a higher resolution sensor. 5MP just won't do the job I'm afraid.
Dude, I'm going to go back there and get more images. I'll try manual focus, different times of day, whatever. But that house is so far away that you can't see it with a naked eye. When I push the button to take the image, the camera shakes and takes the house all the way out of view. Let's see your pics of something from 4.4 miles.
From what I read, your camera does video at 720p (I'd have to double check). If you have issues with movement when taking a picture, try simply recording video for a few seconds. You can let go and it will be completely still while recording (I'd turn off stability, because on my camera, with stability on, the image continues to move around a tad settling. It's steadier without it in this case.... unless it's too windy). If you have the editing software (or whatever you view the video with) you can save individual frames.
Whether frame captures will be higher res than photo mode, I don't know without comparing side by side. I've never really done a good comparison with mine even to be sure (1080 video, 3.3mp stills)
I probably don't need to mention late afternoon pictures will also be better as the sun will be more 'behind your back', thus illuminating the house and shoreline better.
Yeah, so there Geoff! Maybe you need to delete that sarcastic post above to Rot about showing evidence.
And where's your evidence, the pictures and diagrams you've claimed to be working on?
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Thanks for the new images hoppy, but you desperately need to use a camera with a higher resolution sensor. 5MP just won't do the job I'm afraid.
Dude, I'm going to go back there and get more images. I'll try manual focus, different times of day, whatever. But that house is so far away that you can't see it with a naked eye. When I push the button to take the image, the camera shakes and takes the house all the way out of view. Let's see your pics of something from 4.4 miles.
Hoppy, the roundies will try to have you believe a photo from sea level then another from a higher level is what is needed.
Much more interesting are photos with variations of all weather conditions, that's when odd things occur.
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Hoppy, I posted this back on page 2 (full post (http://theflatearthsociety.org/forum/index.php?topic=60895.msg1576269#msg1576269)), but just as a reminder:
(http://imageshack.com/a/img513/2302/f843.png)
d1 is the drop you get from Rowbotham's chart, ie with zero elevation of the eye (or camera). d2 is the drop you get with elevation c. As you can see, d1 - c =/= d2.
Much more interesting are photos with variations of all weather conditions, that's when odd things occur.
This would be good for comparison, but it might be more work than hoppy is willing to put in. To be properly informative, hoppy would need to take notes of the exact weather conditions each time he/she went to take photos. If you are willing to put in the work though hoppy, it would be very informative!
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Hoppy, the roundies will try to have you believe a photo from sea level then another from a higher level is what is needed.
Much more interesting are photos with variations of all weather conditions, that's when odd things occur.
Feel free to take some pictures yourself of your observations Tappet, and share them here.
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Sci,
Your diagram is not what is happening. Draw a picture of a FLAT circle and show the angles as the altitude above the flat plane increases.
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Please add that diagram to the list of diagrams you need to draw Earth.
BTW, how is that diagram of the tennis ball being obscured by the floor from across the living room going?
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Sci,
Your diagram is not what is happening. Draw a picture of a FLAT circle and show the angles as the altitude above the flat plane increases.
Why would we do a flat circle. The point of the experiment is to test whether or not we are on a sphere. So we test against data about sphere's.
There really isn't much to do if we did the experiment in reverse as you are saying. I should be able to see downtown San Diego from my house, but it doesn't work that way.
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Sci,
Your diagram is not what is happening. Draw a picture of a FLAT circle and show the angles as the altitude above the flat plane increases.
Show us how it's done then.
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Sci,
Your diagram is not what is happening. Draw a picture of a FLAT circle and show the angles as the altitude above the flat plane increases.
What would be the point of that? Visible distance over a flat circle from any altitude: infinite. Portion of object obscured when viewed over a flat circle: zero.
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Please add that diagram to the list of diagrams you need to draw Earth.
BTW, how is that diagram of the tennis ball being obscured by the floor from across the living room going?
It's going great. ;D I'll post it tomorrow.
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Why would we do a flat circle. The point of the experiment is to test whether or not we are on a sphere. So we test against data about sphere's.
There really isn't much to do if we did the experiment in reverse as you are saying. I should be able to see downtown San Diego from my house, but it doesn't work that way.
I give up trying to explain it to you Rot. You would have to be able to see THROUGH all the stuff in your line of vision that is higher than you are.
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Why would we do a flat circle. The point of the experiment is to test whether or not we are on a sphere. So we test against data about sphere's.
There really isn't much to do if we did the experiment in reverse as you are saying. I should be able to see downtown San Diego from my house, but it doesn't work that way.
I give up trying to explain it to you Rot. You would have to be able to see THROUGH all the stuff in your line of vision that is higher than you are.
I don't disagree with that. Don't pretend that you don't like talking to me for any reason other than that I pwn you.
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Pwn? I had to look that one up. Sorry, I don't play video games enough to know terminology of that sort. And the closest I've come to computer hacking is the movie Hackers.
Oh and Wargames.
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Pwn? I had to look that one up. Sorry, I don't play video games enough to know terminology of that sort. And the closest I've come to computer hacking is the movie Hackers.
Oh and Wargames.
I'm just joking anyway. Have a good night spaceship.
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Draw a picture of a FLAT circle and show the angles as the altitude above the flat plane increases.
Uh... we're not talking about a flat, circular plane here. The diagram illustrates the phenomenon as it occurs on a sphere (or the earth).
It's counter-intuitive to shift the goalposts midway through the argument.
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But the Earth is NOT a sphere. So it IS counterintuitive.
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But the Earth is NOT a sphere. So it IS counterintuitive.
Famous FE last words. No need to investigate and test things because we can just look out the window eh? Why are you here responding to any arguments then?
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Famous FE last words. No need to investigate and test things because we can just look out the window eh? Why are you here responding to any arguments then?
Poor old EarthIsASpaceship obviously failed to notice—or has chosen to ignore—that I said "the diagram illustrates the phenomenon as it occurs on a sphere".
Flat earthers are so desperate to defend their claims that they often miss the salient points of a round-earth argument. As in this case. FE's also have a common tendency to avoid considering hypotheticals; everything that they believe in is supported by nothing more than their own—often misguided—personal opinions, and more often than not misconceptions of what they're actually seeing with their own eyes (as in "seeing" that the earth immediately surrounding them is "flat").
And I also note that despite me asking a few days ago for the FEs to cite a couple of accredited, contemporary scientists who subscribe to the flat earth theory, no names are yet forthcoming. Why is this?
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Since the earth has an equatorial bulge there is no perfect way to determine the drop but we can come up with some sort of average.
Suppose that the earth is a sphere of radius 3963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile to point Q then you can form a right triangle. Using Pythagorean theorem
a^2 = 3963^2 + 1^2 = 15705370 and thus a = 3963.000126 miles.
3963.00126 - 3963 (earth's radius) = 0.000126 miles above surface at point Q
Convert to inches:
12*5280*0.00126 = 7.98 inches.
Hence the earth's surface curves at approximately 8 inches per mile.
So for one, a 12 ft drop is erroneous. For 4 miles we're looking at under 3 ft and this is if you are looking from ground level. At 2 ft above the water it's even less.
It's 32 inches - 24 inches = 8 inch drop if 2 ft above water.
And that's if you were actually 2 ft above the water which it doesn't appear that way from your pic.
12 foot drop is not erroneous, 8" drop the first mile is correct. The drop is progressively more in each following mile.
Refer to this chart, it is correct.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
No, that chart and equation are NOT correct, it is an equation for a parabola, not a circle!
You might want to check who created it, turns out it was that moron Samuel Robotham, the same one who did the totally debunked "Bedford level experiment". Might interest you to know that he dropped out of school at age 9. In short you are depending for your information on a grade school dropout. He also wrote an insane book on 100 reasons why the earth is not a sphere. In short, you listen to absolute morons!
You might want to consider how absolutely insane this flat earth fairy tale is. You know that the FACT that the earth is spherical was recognized by the Greeks around 2,500 years ago. That means that for this fact to be concealed would require that virtually every physical scientist, astronomer, and geographer for the last 2,500 years has been part of this massive conspiracy. That does not sound even remotely sane, which means that anyone actually believing this crap is delusional and probably in need of psychiatric care.
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And that's if you were actually 2 ft above the water which it doesn't appear that way from your pic.
12 foot drop is not erroneous, 8" drop the first mile is correct. The drop is progressively more in each following mile.
Refer to this chart, it is correct.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
No, that chart and equation are NOT correct, it is an equation for a parabola, not a circle!
You might want to check who created it, turns out it was that moron Samuel Robotham, the same one who did the totally debunked "Bedford level experiment". Might interest you to know that he dropped out of school at age 9. In short you are depending for your information on a grade school dropout. He also wrote an insane book on 100 reasons why the earth is not a sphere.
Without entering into the debate between hoppy and rottingroom I'm afraid that you are incorrect in a few factual points:
- You say that the "chart and equation are NOT correct, it is an equation for a parabola, not a circle!"
Now that is technically correct.
The "chart and equation", are however, an excellent approximation to the drop of the surface of the
below the local horizon.
Have a look at Walter Bislin's Eight Inches per Miles squared Formula Derivation (http://walter.bislins.ch/bloge/index.asp?page=Eight+Inches+per+Miles+squared+Formula+Derivation) which shows that the formula is within 0.1% up to almost 500 miles!
This formula is an approximation: until 550 km or 342 mi the error stays within ∓0.032%. Until 800 km or 497 mi the error is less than 0.1%.
(https://www.dropbox.com/s/luepwn5h9q7bobt/8%20ins%20per%20mile%20squared%20error%2C%20Walter%20Bislins.png?dl=1)
So, it is only an approximation to the geometric drop but it is quite accurate enough over reasonable distances.
- Nobody should call Samuel Robotham a "moron"!
I would claim that he was a very "smart cookie" at what he did. For one thing, I suspect that he knew about "atmospheric refraction" and was smart enough to perform his "Bedford Level Experiment" very close to the water to give the result he wanted.
And he paved the way for the modern flat-Earth movement which, like it or not, has had a significant impact.
- You say "You might want to check who created it" but it would seem that Hoppy already knew that.
If you bothered to check the reference Hoppy referred to you might have noted that it was written by Samuel Birley Rowbotham!
Have a look at: Zetetic Astronomy, by 'Parallax' (pseud. Samuel Birley Rowbotham), CHAPTER II. (http://www.sacred-texts.com/earth/za/za05.htm).
- Then you claim "He also wrote an insane book on 100 reasons why the earth is not a sphere".
Would you please show evidence for that because the closest that I know of is this book:
One Hundred Proofs that the Earth is Not a Globe by William Carpenter (https://www.gutenberg.org/files/55387/55387-h/55387-h.htm).
If you want to debate with flat-Earthers it might be better to get your facts right or you might be the one thought a moron.
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No, that chart and equation are NOT correct, it is an equation for a parabola, not a circle!
No, the chart and equation are correct. They are approximations.
If you notice a parabola is approximately the same as a circle for small displacements.
If you need an example, look here:
https://www.desmos.com/calculator/s7as1l6oaf
For a more math based approach, consider this image:
(https://i.imgur.com/KG51eWw.png)
There are a few different options for distance and height.
Lets take the simplest one as distance along the surface (d1) and height perpendicular to the surface (h1).
Well, for a given distance we get the angle as a=d1/R.
Cos(a)=R/(R+h1).
so sec(d1/R)=(R+h1)/R
R+h1=R*sec(d1/R)
h1=R*(sec(d1/R)-1)
You could computer this exactly, or alternatively just use an approximation.
sec(x)~=1+x^2/2.
Using that instead:
h1=R*(d1^2/(2*R^2)+1-1)
h1=1^2/(2*R)
Notice that this is a parabola.
If instead you want to use h3, that has:
cos(a)=(R-h3)/R
R-h3=R*cos(a)
h3=R*(1-cos(a))
Then using the small angle approximation for cos of cos(x)=1-x^2/2 we end up with:
h3=R*(1-(1-d1^2/R^2)/2)
h3=d1^2/(2*R)
Just like the first.
And then if you want to use d2 or d3, the small angle approximation is that x~=sin(x)~=tan(x), so they are effectively the same, and with d3 and d2 effectively the same, h3 and h2 are as well.
Based upon the diagram, Row Boat appears to be using d2 and h1.
This gives us yet another way to calculate it.
R^2+d2^2=(R+h1)^2=R^2+2*R*h1+h1^2.
Noting that h1 is tiny, this simplifies to:
d2^2=2*R*h1
h1=d^2/(2*R).
So no, that parabola is just fine.
As for why it is 8 inches per mile squared, we want to output a height in inches, using a distance in miles.
If we use a distance in miles and radius of Earth in miles, we will get a height in miles which needs to be converted to inches by 63360.
So the number we need is 63360/(2*R). The radius of Earth is 3,958.8 miles, which gives us 8.002424977 inches per mile squared. So the 8 actually underestimates it until you get to quite some distance.
As a comparison of it all:
| d2 | a (milli radians) | a (degrees) | d1 | d3 | H Row Pg | h Row | h par | h1 | h2 | h3 |
| 1 | 0.25 | 0.01 | 1.00 | 1.00 | 8 | 8.00 | 8.00 | 8.00 | 8.00 | 8.00 |
| 2 | 0.51 | 0.03 | 2.00 | 2.00 | 32 | 32.00 | 32.01 | 32.01 | 32.01 | 32.01 |
| | | | | | | | | | | |
| 3 | 0.76 | 0.04 | 3.00 | 3.00 | 6 | 6.00 | 6.00 | 6.00 | 6.00 | 6.00 |
| 4 | 1.01 | 0.06 | 4.00 | 4.00 | 10 | 10.67 | 10.67 | 10.67 | 10.67 | 10.67 |
| 5 | 1.26 | 0.07 | 5.00 | 5.00 | 16 | 16.67 | 16.67 | 16.67 | 16.67 | 16.67 |
| 6 | 1.52 | 0.09 | 6.00 | 6.00 | 24 | 24.00 | 24.01 | 24.01 | 24.01 | 24.01 |
| 7 | 1.77 | 0.10 | 7.00 | 7.00 | 32 | 32.67 | 32.68 | 32.68 | 32.68 | 32.68 |
| 8 | 2.02 | 0.12 | 8.00 | 8.00 | 42 | 42.67 | 42.68 | 42.68 | 42.68 | 42.68 |
| 9 | 2.27 | 0.13 | 9.00 | 9.00 | 54 | 54.00 | 54.02 | 54.02 | 54.02 | 54.02 |
| 10 | 2.53 | 0.14 | 10.00 | 10.00 | 66 | 66.67 | 66.69 | 66.69 | 66.69 | 66.69 |
| 20 | 5.05 | 0.29 | 20.00 | 20.00 | 266 | 266.67 | 266.75 | 266.75 | 266.75 | 266.74 |
| 30 | 7.58 | 0.43 | 30.00 | 30.00 | 600 | 600.00 | 600.18 | 600.17 | 600.19 | 600.16 |
| 40 | 10.10 | 0.58 | 40.00 | 40.00 | 1066 | 1066.67 | 1066.99 | 1066.96 | 1067.02 | 1066.91 |
| 50 | 12.63 | 0.72 | 50.00 | 50.00 | 1666 | 1666.67 | 1667.17 | 1667.11 | 1667.24 | 1666.97 |
| 60 | 15.15 | 0.87 | 60.00 | 59.99 | 2400 | 2400.00 | 2400.73 | 2400.59 | 2400.87 | 2400.31 |
| 70 | 17.68 | 1.01 | 69.99 | 69.99 | 3266 | 3266.67 | 3267.66 | 3267.40 | 3267.91 | 3266.89 |
| 80 | 20.21 | 1.16 | 79.99 | 79.98 | 4266 | 4266.67 | 4267.96 | 4267.52 | 4268.40 | 4266.65 |
| 90 | 22.73 | 1.30 | 89.98 | 89.98 | 5400 | 5400.00 | 5401.64 | 5400.94 | 5402.33 | 5399.54 |
| 100 | 25.25 | 1.45 | 99.98 | 99.97 | 6666 | 6666.67 | 6668.69 | 6667.62 | 6669.75 | 6665.50 |
| 120 | 30.30 | 1.74 | 119.96 | 119.94 | 9600 | 9600.00 | 9602.91 | 9600.71 | 9605.12 | 9596.30 |
Notice how it holds quite well? Even at the 120 miles, the way he is expressing it still has it underestimating.
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just took this image today, how can this house be visible if the earth is a sphere?
This image was taken from 4.4 miles as I found on google earth. There is no other proof for you roundies, you just have to take my word for now about this. I will work on better documentation later. But what I have stated here is true.
Owing to there being an atmosphere surrounding Earth it is quite possible to see a lot further than 4.4 miles, 10 miles or even 50 miles under the right conditions owing to the refractive properties of air.
It would be a different story if there was no atmosphere since then the rate of curvature would be the only one of two things (your elevation and the rate of curvature) that determined how far you can see. As it is there are three. Your photo is showing you something that seemingly supports what you choose to believe. Therefore you overlook or simply ignore the other possible explanations for what might make it possible but which do not support your belief.
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Wow. This is a random 6 year old thread. But at least the OP is still actually here.
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I must admit I don't often take any notice of the posting dates. Just titles. I was going to post a reply to a discussion in the FE repository that goes back to 2009!
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Wow. This is a random 6 year old thread. But at least the OP is still actually here.
Yes, hoppy's still here but seems suffer dementia now because few of his posts are more than two words.
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Wow. This is a random 6 year old thread. But at least the OP is still actually here.
And look at this photo showing oil "Platform Habitat" 9.41 miles away, yet visible from (supposedly) one foot above sea-level:
(https://i.postimg.cc/Pqb5yrM4/Ocean-Horizon-Rising-atmosphere-Blocking-Visibility-at-2-41.jpg) (https://postimages.org/)
Ocean Horizon Rising (atmosphere) Blocking Visibility at 2.41
Then from the same location and (supposedly) the same height above sea-level but on a different day and much of oil "Platform Habitat" hidden by "something":
(https://i.postimg.cc/sXwDF6qG/Ocean-Horizon-Rising-atmosphere-Blocking-Visibility-at-1-00.jpg) (https://postimages.org/)
Ocean Horizon Rising (atmosphere) Blocking Visibility at 1.00
See what severe refraction, looming and towering can do.
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That is some crazy distortion. Compare the close up:
(https://i.imgur.com/1DZLJi6.png)
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That is some crazy distortion. Compare the close up:
(https://i.imgur.com/1DZLJi6.png)
It's "some crazy distortion" alright yet a number the "top YouTube FEers" are claiming that as the ultimate "Globe Killer".
Then the one who presented the video explained away that second photo as due to the ocean Horizon rising due to the atmosphere blocking visibility.
Here is the video that those screenshots came from and another be the same maker.
Ocean Horizon Rising (atmosphere) Blocking Visibility..Not Curvature by bmlsb69
Just compare 1:00 and 2:45 in that video.
This video by the same person might also be relevant:
Atmospheric MAGNIFICATION: Why objects are behind horizon, even after zoomed in with P900 by bmlsb69
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Wouldn't this same sort of distortion account for the strange shapes you see in the Sun when it is sitting in the horizon? Sometimes the top of the Sun even becomes detached as it disappears over the horizon and leads to the 'green flash' effect.
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Wouldn't this same sort of distortion account for the strange shapes you see in the Sun when it is sitting in the horizon? Sometimes the top of the Sun even becomes detached as it disappears over the horizon and leads to the 'green flash' effect.
Yes, weird effects like that over water are not that rare especially what the water temperature is much colder than the air above and there are other effects when the surface is warner than the air.
Try this one from Argentina:Astronomy Picture of the Day (https://apod.nasa.gov/apod/ap111010.html)
Discover the cosmos! (https://apod.nasa.gov/apod/archivepix.html) Each day a different image or photograph of our fascinating universe is featured, along with a brief explanation written by a professional astronomer. 2011 October 10
(http://apod.nasa.gov/apod/image/1110/strangesunrise_argerich_900.jpg)
A Strange Sunrise Over Argentina Image Credit & Copyright: Luis Argerich (http://www.luisargerich.com/about.html)
Explanation: Why would a rising Sun look so strange? (http://asterisk.apod.com/discuss_apod.php?date=111010)
Weird!
Here is the map of a site with some discussions, examples and explanations of these and other atmosphere effects:
The Weather Doctor: Site Map (https://www.islandnet.com/~see/weather/doctor.htm)
And the section Optical Phenomena: Sunsets, Halos, Mirages, etc. has links to many of these effects with these being the most relevant:
Mirages: A Primer (https://www.islandnet.com/~see/weather/elements/mirage1.htm)
The Inferior Mirage: Not Just For Deserts Anymore (https://www.islandnet.com/~see/weather/elements/infmrge.htm)
The Superior Mirage: Seeing Beyond (https://www.islandnet.com/~see/weather/elements/supmrge.htm)
"Seeing too far" does not necessarily prove the Earth to be flat and "things hidden" is not of itself proof of curvature.
One needs to check how changing atmospheric conditions (even just at different times if the day) affects the distance.
Seeing things become hidden as either they move away (over the curve) or becoming hidden as the observer goes lower is far more significant.
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I just took this image today, how can this house be visible if the earth is a sphere?
This image was taken from 4.4 miles as I found on google earth. There is no other proof for you roundies, you just have to take my word for now about this. I will work on better documentation later. But what I have stated here is true.
(http://imgur.com/iUBPPWM.jpeg)
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Your "expected values" are not the expected values:
Earth's Curve Horizon, Bulge, Drop, and Hidden Calculator
Distance = 4.4 miles (23232 feet), View Height = 2 feet (24 inches) Actual Radius = 3959 miles (20903520 feet)
With the refraction approximation* giving an effective radius of 4618.83 miles
Refracted Horizon = 1.87 miles (9876.73 feet)
Refracted Drop= 11.07 feet (132.79 inches)
Refracted Hidden= 3.66 feet (43.88 inches)
Refracted Horizon Dip = 0.023 Degrees, (0.0004 Radians)
Note: Not accurate for observations over water very close to the horizon (unless the temperature and vertical temperature gradient are accurate)
Geometric results (no refraction)
Geometric Horizon = 1.73 miles (9144.07 feet)
Geometric Drop = 12.91 feet (154.92 inches)
Geometric Hidden= 4.75 feet (56.97 inches)
Geometric Horizon Dip = 0.025 Degrees, (0.0004 Radians)
Pure Angular Size results (Explanation Here)
Angle between eye level and the horizon = 0.02506 Degrees, (0.000437 Radians)
Angle between eye level and the bottom of the target= 0.03677 Degrees, (0.000642 Radians)
Angular size of hidden amount = 0.01171 Degrees, (0.000204 Radians)
And in that picture the beach and the bottom of the house are clearly missing.
.
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I just took this image today, how can this house be visible if the earth is a sphere?
This image was taken from 4.4 miles as I found on google earth. There is no other proof for you roundies, you just have to take my word for now about this. I will work on better documentation later. But what I have stated here is true.
(http://imgur.com/iUBPPWM.jpeg)
...
Your "expected values" are not the expected values:
Earth's Curve Horizon, Bulge, Drop, and Hidden Calculator
Distance = 4.4 miles (23232 feet), View Height = 2 feet (24 inches) Actual Radius = 3959 miles (20903520 feet)
With the refraction approximation* giving an effective radius of 4618.83 miles
Refracted Horizon = 1.87 miles (9876.73 feet)
Refracted Drop= 11.07 feet (132.79 inches)
Refracted Hidden= 3.66 feet (43.88 inches)
Refracted Horizon Dip = 0.023 Degrees, (0.0004 Radians)
Note: Not accurate for observations over water very close to the horizon (unless the temperature and vertical temperature gradient are accurate)
Geometric results (no refraction)
Geometric Horizon = 1.73 miles (9144.07 feet)
Geometric Drop = 12.91 feet (154.92 inches)
Geometric Hidden= 4.75 feet (56.97 inches)
Geometric Horizon Dip = 0.025 Degrees, (0.0004 Radians)
Pure Angular Size results (Explanation Here)
Angle between eye level and the horizon = 0.02506 Degrees, (0.000437 Radians)
Angle between eye level and the bottom of the target= 0.03677 Degrees, (0.000642 Radians)
Angular size of hidden amount = 0.01171 Degrees, (0.000204 Radians)
And in that picture the beach and the bottom of the house are clearly missing.
It's a very fuzzy photo for 4.4 miles and I've wondered but haven't been sure enough to say it but I suspect a "haze" line along the apparent beach and stretching (towering) above that.
There's not enough information to know for sure - not like the YouTube "Black Swan".