Why can you RE'ers see this house.

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hoppy

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Why can you RE'ers see this house.
« on: January 17, 2014, 04:57:02 PM »
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.




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.


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.
« Last Edit: January 17, 2014, 06:17:49 PM by hoppy »
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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #1 on: January 17, 2014, 07:31:09 PM »
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.
« Last Edit: January 17, 2014, 07:47:04 PM by rottingroom »

Re: Why can you RE'ers see this house.
« Reply #2 on: January 17, 2014, 07:51:06 PM »
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.

Re: Why can you RE'ers see this house.
« Reply #3 on: January 17, 2014, 08:03:19 PM »
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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hoppy

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Re: Why can you RE'ers see this house.
« Reply #4 on: January 17, 2014, 08:29:09 PM »
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
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hoppy

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Re: Why can you RE'ers see this house.
« Reply #5 on: January 17, 2014, 08:35:48 PM »
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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hoppy

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Re: Why can you RE'ers see this house.
« Reply #6 on: January 17, 2014, 08:41:49 PM »
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.
« Last Edit: January 17, 2014, 09:00:27 PM by hoppy »
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tappet

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Re: Why can you RE'ers see this house.
« Reply #7 on: January 17, 2014, 10:06:56 PM »
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.

Re: Why can you RE'ers see this house.
« Reply #8 on: January 17, 2014, 10:28:45 PM »
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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Scintific Method

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Re: Why can you RE'ers see this house.
« Reply #9 on: January 18, 2014, 04:17:21 AM »
Refer to this chart, it is correct.
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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...the FE'ers still found a way to deny it. Not with counter arguments. Not with proof of any kind. By simply denying it.

"Better to keep your mouth shut and be thought a fool, than to open it and remove all doubt."

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hoppy

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Re: Why can you RE'ers see this house.
« Reply #10 on: January 18, 2014, 06:09:38 AM »
Refer to this chart, it is correct.
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.
« Last Edit: January 18, 2014, 07:25:03 AM by hoppy »
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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #11 on: January 18, 2014, 07:31:28 AM »
Refer to this chart, it is correct.
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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hoppy

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Re: Why can you RE'ers see this house.
« Reply #12 on: January 18, 2014, 08:24:45 AM »
Do any FE'ers care to chime in with some thoughts?
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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #13 on: January 18, 2014, 08:48:01 AM »
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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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #14 on: January 18, 2014, 09:16:10 AM »
Here we go: Using the calculator on this page 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:

Quote
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.

« Last Edit: January 18, 2014, 09:18:41 AM by rottingroom »

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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #15 on: January 18, 2014, 09:26:11 AM »
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

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.

Re: Why can you RE'ers see this house.
« Reply #16 on: January 18, 2014, 09:30:44 AM »
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?

Quote
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.

Re: Why can you RE'ers see this house.
« Reply #17 on: January 18, 2014, 09:36:53 AM »
Here we go: Using the calculator on this page 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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BJ1234

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Re: Why can you RE'ers see this house.
« Reply #18 on: January 18, 2014, 09:47:21 AM »
Here we go: Using the calculator on this page 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:

Quote
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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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #19 on: January 18, 2014, 10:05:15 AM »
Here we go: Using the calculator on this page 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:

Quote
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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hoppy

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Re: Why can you RE'ers see this house.
« Reply #20 on: January 18, 2014, 10:16:58 AM »
Here we go: Using the calculator on this page 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:

Quote
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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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #21 on: January 18, 2014, 10:22:06 AM »
Here we go: Using the calculator on this page 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:

Quote
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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hoppy

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Re: Why can you RE'ers see this house.
« Reply #22 on: January 18, 2014, 10:22:15 AM »
Here we go: Using the calculator on this page 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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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #23 on: January 18, 2014, 10:28:09 AM »
Here we go: Using the calculator on this page 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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hoppy

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Re: Why can you RE'ers see this house.
« Reply #24 on: January 18, 2014, 10:28:19 AM »
Here we go: Using the calculator on this page 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:

Quote
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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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #25 on: January 18, 2014, 10:31:58 AM »
Here we go: Using the calculator on this page 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:

Quote
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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hoppy

  • Flat Earth Believer
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Re: Why can you RE'ers see this house.
« Reply #26 on: January 18, 2014, 10:40:22 AM »
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.
« Last Edit: January 18, 2014, 10:46:16 AM by hoppy »
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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #27 on: January 18, 2014, 10:53:25 AM »
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:



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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rottingroom

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Re: Why can you RE'ers see this house.
« Reply #28 on: January 18, 2014, 10:56:16 AM »
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.

Re: Why can you RE'ers see this house.
« Reply #29 on: January 18, 2014, 11:10:07 AM »
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.