I want anyone here to give me a logical reason as to why the earth is flat.

Quote from: Lolflatdisc on May 25, 2013, 06:40:54 AM

Quote from: sceptimatic on May 25, 2013, 05:52:32 AM

How much of a 1000 ft sky scraper would I be able to see if I was viewing it across the water from say, 50 miles away, assuming the earth was a globe?

You can do the math yourself. Use the Pythagorean theorem.

I'll just wait to see if I can get an answer. It doesn't matter if you don't want to do it.

At 50 miles you wouldn't be able to see it at all. Check this chart.
http://www.sacred-texts.com/earth/za/za05.htm

Do you round earth indocs accept the figures what Hoppy has put up?

Those figures are pythagorean theorem.

Depending whether you base the 'drop' off of a right angle from given distances along your line of sight from the starting point, or base it off being level at given distances along the surface of the sphere, it works for me.

Usually when you ask things like this scepti, you're about to attempt to get a point across.  We're all dying to know what it is as usual.

Quote from: hoppy on May 25, 2013, 09:10:49 AM

Quote from: sceptimatic on May 25, 2013, 08:58:15 AM

Do you round earth indocs accept the figures what Hoppy has put up?

Those figures are pythagorean theorem.

Yes Hoppy. I just want to make sure that these round earth indocs accept these answers as being correct by their view and if not, I'd like them to say why not.

Quote from: sceptimatic on May 25, 2013, 06:43:02 AM

Quote from: Lolflatdisc on May 25, 2013, 06:40:54 AM

Quote from: sceptimatic on May 25, 2013, 05:52:32 AM

How much of a 1000 ft sky scraper would I be able to see if I was viewing it across the water from say, 50 miles away, assuming the earth was a globe?

You can do the math yourself. Use the Pythagorean theorem.

I'll just wait to see if I can get an answer. It doesn't matter if you don't want to do it.

At 50 miles you wouldn't be able to see it at all. Check this chart.
http://www.sacred-texts.com/earth/za/za05.htm

Hoppy  says the chart used the Pythagorean theorem. Here is my explanation in full detail.

 I will use the theorem of Pythagorean as well to calculate how much you will be able to see at a certain distance and what the maximum distance is at which you can see something.

Before we go on and calculate, we first need to determine some things.

We know the earth's radius is 6371 km or 3959 miles. I prefer km, so I will go ahead and calculate using the metric system.
In the picture I can see the Willis Tower (Sears Tower), which has a height 442 meters (total height of 552 meters with antennas).

Now for the distance across Lake Michigan we need to know the location. Looking into the source tells us that the picture was taken from either Indiana Dunes National Lakeshore or Mount Baldy. This is a distance of approximately 60 km. If we would be right across Chicago on the other side of Lake Michigan, this would be a distance of approximately 80 km. The latter is about 50 miles, the first one is about 37 miles.


The formula =  c^2 = a^2+ b^2

The following diagram shows you what we're doing.



With Pythogerean we can calculate one side of a right angle triangle if we know only two. We know the radius of the earth. This is A. Side C is the radius + the height of the building. Side B is the distance between the viewer and the building(s).
The green line and the green A is the same as the red A and we will need to substract this later on.

We can calculate how much of the lower section of the building is falling behind the horizon. This is C^2 = 6371^2+60^2
C^2 = 40,593,241  sqrt of 40,593,241 = 6371.282. All we need to do is substract the radius and we have know how much of the tower is not visible.

6371.282 - 6371 =  282 meters. Since the Willis Tower is 442, we now know that we only see the top 162 meters. 


We can also calculate the maximum distance at which the top would just barely be below the horizon. This is 6371 + height of the building = 6371,442 ^2 = 6371 ^ 2 + B^2

40,595,273.16 = 40,589,641 + B^2 ?

We just have to solve this
-40,589,641  at both sides and we are left with

5632.16 = B ^2
Therefore B = sqrt 5632.16 = 75 km that is 46 miles. So at a distance of 46 miles the top of the Willis tower is not visible. The antenna would still be, but that is so small that it is hard to see with the naked eye. With antennas the height is 552 meters and doing the same calculations except with 6371,552 tells us that you will need to be 81,5 km away or 50.64 miles. Right across Lake Michigan you would thus be able to see the antennas of the Willis tower using binoculars, and if the weather is clear.

But, from the place where the picture was taken 162 m of the top is still visible.

Quote from: 29silhouette on May 25, 2013, 10:08:54 AM

Depending whether you base the 'drop' off of a right angle from given distances along your line of sight from the starting point, or base it off being level at given distances along the surface of the sphere, it works for me.

Usually when you ask things like this scepti, you're about to attempt to get a point across.  We're all dying to know what it is as usual.

Correct I am, so here it is.

If the earth is a globe, could this be seen from 48.5 miles away?
I cannot verify the distance but I'm sure somebody on here will have seen this and know the distance.
The point being. Would these building be seen like this from this distance?

Do you actually care to read the entire description or only the parts of your interest? It says "I was standing on dunes ". I do not know the heights of the dunes in that area, but I know dunes in my country can reach about 50 meters in height.
This changes the calculations and you would be able to see Chicago just like that from that distance. Please provide me the link to the source so I can see exactly where he stood.

horizon distance 50 miles altitude=1,600 feet.

Tallest building is about that height, so even given a couple hundred foot dune height (which is ridiculous) only the highest parts of the buildings should be visible.

Flat earth confirmed.

horizon distance 50 miles altitude=1,600 feet.

Tallest building is about that height, so even given a couple hundred foot dune height (which is ridiculous) only the highest parts of the buildings should be visible.

Flat earth confirmed.

Clearly you haven't read my calculations.

The 80 meter attenta on top of the Willis tower is visible 4 miles further than the top of the willis tower building at 442 meters. Such a small increase in height can have huge effects on the calculations. I can only do the calculations if I know the place from where the photo was taken. He provides the link in the description. All sceptimatic has to do if provide the link to the source of the image.

Clearly you haven't read my calculations.

The 80 meter attenta on top of the Willis tower is visible 4 miles further than the top of the willis tower building at 442 meters. Such a small increase in height can have huge effects on the calculations. I can only do the calculations if I know the place from where the photo was taken. He provides the link in the description. All sceptimatic has to do if provide the link to the source of the image.

You're right, I just jumped in. My apologies.

Quote from: Lolflatdisc on May 25, 2013, 11:46:58 AM

Clearly you haven't read my calculations.

The 80 meter attenta on top of the Willis tower is visible 4 miles further than the top of the willis tower building at 442 meters. Such a small increase in height can have huge effects on the calculations. I can only do the calculations if I know the place from where the photo was taken. He provides the link in the description. All sceptimatic has to do if provide the link to the source of the image.

You're right, I just jumped in. My apologies.

I found the source of the photo http://www.flickr.com/photos/metroblossom/171641577/# and I could click on the link to location of which he said he was. I assume he was at the beach, since the houses would block his view. The height of that piece of land is 186 meters AMSL. Chicago on the other hand lies at about 180 meters AMSL. This means the photographer is 6 meters higher.

The distance from his appointed location to skyline is 73,74 km after measuring in google earth. This is actually 45,8 miles and NOT 48,5 miles as he claimed.

The radius at the photographers location is set at 6371.186 km.

C^2 = 6371.186^2+ 45.8^2 = 40,594,108.69
sqrt 40,594,108.69 =  6371.350617

Substract 6371.18 and we are left with 0.170 km or 170 m. That is 557.74 feet. So 170 meter or about 557 feet is not visible. The Willis tower is 442 meters or 1,354 ft (top floor). That leaves us with a visible 272 meters or 797 feet willis tower.
 

Oh, and it would defenitely help if you could do these kind of math yourself. That way you can verify things yourself without relying on others to explain it to you.

Did a quick look at that photo compared to another of the skyline.  the buildings look slightly compressed, which would be from refraction, which would allow them to be further exposed over the curvature, but I'll look closer at some comparisons later.

I have seen what you are talking about from a high rise ocean front condo. The limits of your vision cause this distortion, say you're standing on the balcony at ocean front. Just say your eyes have the ability to see 250 miles. 250 miles from you stretched out forward and to the left and right of you, forms an arc. It is not the horizon ending, just the limits of your vision.

You're saying our vision is limited to 250 miles? ... so all the stars are within 250 miles of us then?

As I expected no FE is stepping forward admitting these calculations are accurate and that they were wrong, as clearly, with the curvature of the earth in mind, the buildings can still be visible at that distance.

Still. Would this not render the smaller buildings invisible?

No, they would appear somewhat flattened, but visible as the light is curved slightly over the horizon.

Quote from: hoppy on May 22, 2013, 09:25:51 PM

I have seen what you are talking about from a high rise ocean front condo. The limits of your vision cause this distortion, say you're standing on the balcony at ocean front. Just say your eyes have the ability to see 250 miles. 250 miles from you stretched out forward and to the left and right of you, forms an arc. It is not the horizon ending, just the limits of your vision.

You're saying our vision is limited to 250 miles? ... so all the stars are within 250 miles of us then?

  No that is not what I'm saying. Your eyesight obviously has limits, I was just using 250 as an example.

Quote from: 29silhouette on May 25, 2013, 12:59:10 PM

Quote from: sceptimatic on May 25, 2013, 12:32:11 PM

Still. Would this not render the smaller buildings invisible?

No, they would appear somewhat flattened, but visible as the light is curved slightly over the horizon.

Would they appear as big as they do in that picture?

They'll appear however big the camera's magnification allows them to appear.  The flattening will occur vertically.

Oh, and it would defenitely help if you could do these kind of math yourself. That way you can verify things yourself without relying on others to explain it to you.

Math? I'm a flat earther, I dropped out of middle school.

Quote from: sceptimatic on May 25, 2013, 11:06:17 AM

Quote from: 29silhouette on May 25, 2013, 10:08:54 AM

Depending whether you base the 'drop' off of a right angle from given distances along your line of sight from the starting point, or base it off being level at given distances along the surface of the sphere, it works for me.

Usually when you ask things like this scepti, you're about to attempt to get a point across.  We're all dying to know what it is as usual.

Correct I am, so here it is.

If the earth is a globe, could this be seen from 48.5 miles away?
I cannot verify the distance but I'm sure somebody on here will have seen this and know the distance.
The point being. Would these building be seen like this from this distance?

Do you actually care to read the entire description or only the parts of your interest? It says "I was standing on dunes ". I do not know the heights of the dunes in that area, but I know dunes in my country can reach about 50 meters in height.
This changes the calculations and you would be able to see Chicago just like that from that distance. Please provide me the link to the source so I can see exactly where he stood.

This picture is proof for FE.
 45.8 miles on RE the drop is 1398'
The roof of sears tower is 2046' above sea level.
 If the photo was taken near sea level 648' of Willis Tower could be visible.
Compare the picture of wiki picture of the tower   http://en.wikipedia.org/wiki/Skydeck_Ledge
Comparatively you can nearly see the whole tower.

Quote from: hoppy on May 22, 2013, 09:25:51 PM

I have seen what you are talking about from a high rise ocean front condo. The limits of your vision cause this distortion, say you're standing on the balcony at ocean front. Just say your eyes have the ability to see 250 miles. 250 miles from you stretched out forward and to the left and right of you, forms an arc. It is not the horizon ending, just the limits of your vision.

You're saying our vision is limited to 250 miles? ... so all the stars are within 250 miles of us then?

INDEED! I have good reason to think the stars and planets are MUCH closer and MUCH smaller than we are led to be LIE ve.

This picture is proof for FE.
 45.8 miles on RE the drop is 1398'
The roof of sears tower is 2046' above sea level.
 If the photo was taken near sea level 648' of Willis Tower could be visible.
Compare the picture of wiki picture of the tower   http://en.wikipedia.org/wiki/Skydeck_Ledge
Comparatively you can nearly see the whole tower.

Looks can be quite deceiving when you have a look at other pictures. I found another picture taken much closer to the skyline. I tried to copy paste it, but obviously it did not work quite as intended. The long distance shot has to deal with the perspective, which the closer shot does not. Anyway, my calculations tells you, from the location of the photo you only are not able to see the bottom 170m of the skyline. So it is not impossible.

It's a mirage.  You can see the "haze" between the bottom of the buildings and the ocean.  But the camera is too far away to see the the "reflection" in the haze.

http://mintaka.sdsu.edu/GF/explain/simulations/inf-mir/Kaplan_photos.html

It's a mirage.  You can see the "haze" between the bottom of the buildings and the ocean.  But the camera is too far away to see the the "reflection" in the haze.

http://mintaka.sdsu.edu/GF/explain/simulations/inf-mir/Kaplan_photos.html

Interesting link, but if you take a closer look at the chicago buildings. They stick out quite a distance above the horizon. The sun in the last picture, just above the horizon is reflecting just a tiny bit, but you can still see that it does not touch the horizon completely. A tail-tale sign of a mirage, but that tail-tale sign is completely missing in the chicago skyline picture.

There is obvious, visible curvature in this image, even though there shouldn't be. It's almost certainly the camera, and not the horizon.

Quote from: Lolflatdisc on May 25, 2013, 11:11:52 AM

Quote from: sceptimatic on May 25, 2013, 11:06:17 AM

Quote from: 29silhouette on May 25, 2013, 10:08:54 AM

Depending whether you base the 'drop' off of a right angle from given distances along your line of sight from the starting point, or base it off being level at given distances along the surface of the sphere, it works for me.

Usually when you ask things like this scepti, you're about to attempt to get a point across.  We're all dying to know what it is as usual.

Correct I am, so here it is.

If the earth is a globe, could this be seen from 48.5 miles away?
I cannot verify the distance but I'm sure somebody on here will have seen this and know the distance.
The point being. Would these building be seen like this from this distance?

Do you actually care to read the entire description or only the parts of your interest? It says "I was standing on dunes ". I do not know the heights of the dunes in that area, but I know dunes in my country can reach about 50 meters in height.
This changes the calculations and you would be able to see Chicago just like that from that distance. Please provide me the link to the source so I can see exactly where he stood.

This picture is proof for FE.
 45.8 miles on RE the drop is 1398'
The roof of sears tower is 2046' above sea level.
 If the photo was taken near sea level 648' of Willis Tower could be visible.
Compare the picture of wiki picture of the tower   http://en.wikipedia.org/wiki/Skydeck_Ledge
Comparatively you can nearly see the whole tower.

@ lol, more than 31% of the building would be blocked by the earth"s curve if it was round.

Actually 31% would be visible and 69% would be blocked, which is clearly not the case here.

Also, moved to Q&A.

Quote from: Lolflatdisc on May 23, 2013, 09:37:11 AM

There is obvious, visible curvature in this image, even though there shouldn't be. It's almost certainly the camera, and not the horizon.

No, it is hazy, so you can't see the real horizon. The Google Earth image I provided gives conclusive evidence though. It is taken from almost the the same altitude, the same location, same perspective and it too shows a curvature like we see in the high altitude picture.

@ lol, more than 31% of the building would be blocked by the earth"s curve if it was round.

True. The building with the light is the 311 S Wacker Dr building. 293 m high. You only see about 120 m. The building to the left of that CNA Plaza which is 183 meters heigh. In the picture you can only see the very top. Here is a zoomed in picture with the two buildings I mentioned.

Wait a minute. Chicago average height above sea level is 579', the waterfront is listed as 577.
http://en.wikipedia.org/wiki/Aon_Center_(Chicago)

The overall grade of the city's central, built-up areas, is relatively consistent with the natural flatness of its overall natural geography, generally exhibiting only slight differentiation otherwise. The average land elevation is 579 ft (176 m) above sea level. The lowest points are along the lake shore at 577 ft (176 m), while the highest point, at 735 ft (224 m), is a landfill located in the Hegewisch community area on the city's far south side.

This throws off the previous calculation, of the picture across the lake.

The Willis tower is 1451' high, RE drop is 1398. You not see the building at all if earth is round. You definitely would not see any of the other buildings either.

Wait a minute. Chicago average height above sea level is 579', the waterfront is listed as 577.
http://en.wikipedia.org/wiki/Aon_Center_(Chicago)

The overall grade of the city's central, built-up areas, is relatively consistent with the natural flatness of its overall natural geography, generally exhibiting only slight differentiation otherwise. The average land elevation is 579 ft (176 m) above sea level. The lowest points are along the lake shore at 577 ft (176 m), while the highest point, at 735 ft (224 m), is a landfill located in the Hegewisch community area on the city's far south side.

This throws off the previous calculation, of the picture across the lake.

The Willis tower is 1451' high, RE drop is 1398. You not see the building at all if earth is round. You definitely would not see any of the other buildings either.

I already took care of the heights in my calculations. I use meters instead of feet. I told you Chicago is at an height of 180 meters (590 feet) however, the photographer was at 186 meters (610 feet). Look at my calculations. I took care of it already.

Quote from: hoppy on May 26, 2013, 10:27:56 AM

Wait a minute. Chicago average height above sea level is 579', the waterfront is listed as 577.
http://en.wikipedia.org/wiki/Aon_Center_(Chicago)

The overall grade of the city's central, built-up areas, is relatively consistent with the natural flatness of its overall natural geography, generally exhibiting only slight differentiation otherwise. The average land elevation is 579 ft (176 m) above sea level. The lowest points are along the lake shore at 577 ft (176 m), while the highest point, at 735 ft (224 m), is a landfill located in the Hegewisch community area on the city's far south side.

This throws off the previous calculation, of the picture across the lake.

The Willis tower is 1451' high, RE drop is 1398. You not see the building at all if earth is round. You definitely would not see any of the other buildings either.

I already took care of the heights in my calculations. I use meters instead of feet. I told you Chicago is at an height of 180 meters (590 feet) however, the photographer was at 186 meters (610 feet). Look at my calculations. I took care of it already.

Right a difference of 20'. None of smaller buildings would be visible at all.