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  #5601  
Old Posted Feb 9, 2008, 9:25 PM
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Nicko999 Nicko999 is offline
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Quote:
Originally Posted by towerguy3 View Post
How far can you see from the top of Burj Dubai assuming the top is 850 meters above sea level?

Assuming flat earth or ocean, the maximum line of sight (distance to the horizon) is calculated by:

d = SQRT((2*R*h)+h*EXP(2))

d = farthest distance you can see to the horizon

R = Earth's radius in km

h = your height above sea level in km

d = SQRT((2*6356.75km*0.850km)+0.850km*EXP(2)) = 104 kilometers

So even with the curvature of the Earth if your boat is out on the ocean and you're 104 km out you will still see the top of Burj Dubai !
Nice, you are probably very good in math! 104 km is alot. I have been at the top of the CN Tower but I can't imagine a tower that would allow you to see more than 100km.
     
     
  #5602  
Old Posted Feb 10, 2008, 1:05 AM
aaron38 aaron38 is offline
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Speaking from experience with Sears Tower in Chicago, you need exceptionally clear weather to see Sears from 50km out. Most of the time it's invisible.
Atmospheric haze won't let you see the tower from 100km.
     
     
  #5603  
Old Posted Feb 10, 2008, 2:43 PM
gsgeorge gsgeorge is offline
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Quote:
Originally Posted by aaron38 View Post
Speaking from experience with Sears Tower in Chicago, you need exceptionally clear weather to see Sears from 50km out. Most of the time it's invisible.
Atmospheric haze won't let you see the tower from 100km.
I distinctly remember being able to see the CN Tower and most of the Toronto skyline from 55km across Lake Ontario from the docks at Saint Catharines. In fact, here's a photo to prove it.



It should be no problem to see the Burj from several dozen more kilometers away on a clear day.

Last edited by gsgeorge; Feb 10, 2008 at 2:43 PM. Reason: spelling
     
     
  #5604  
Old Posted Feb 10, 2008, 5:06 PM
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cn tower

Quote:
Originally Posted by gsgeorge View Post
I distinctly remember being able to see the CN Tower and most of the Toronto skyline from 55km across Lake Ontario from the docks at Saint Catharines. In fact, here's a photo to prove it.



It should be no problem to see the Burj from several dozen more kilometers away on a clear day.
This looks to be a shot from the island with a slightly wide lens.
distance - 3 kilometers maybe.

See next post for real world examples, not pie in sky math thats not relevant for earth curvature at these short distances.
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  #5605  
Old Posted Feb 10, 2008, 5:44 PM
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CN tower from 50 kms

Burj dubai at 2600 feet will be seen from great distance given clear atmospheric conditions and of course binoculars or telescope will help.
Since above 2000 feet the tip will be of a small diameter, one may need binoculars to see anything at all, but there will be line of sight for almost the entire building available from well over 100kms. Earth curvature won't apply here.

image 1 - this is a crop ( magnified section ) of CN tower



made from this image below which shows a natural/realistic view as if you were standing there (how your eyes would see it).



shot from exactly 50.8 kilometers


and from an altitude of 725ft above sea level.


The CN tower base is at 250 feet above sea level.
So the camera was positioned at 50 kms from and at 475 feet above the base.

This image shows the distance from St. Catherines docks as 48.8 Kms


So gsgeorge's shot does not represent realistically what the tower looks like from St. Catherines Docks.
I'm pretty sure that shot is with a wide angle lens from the island only 3 kms away.

Sine the Burj will have an extra 800 feet on the CN tower, The Burj could be seen from 100+ kms given very clear conditions. Binoculars may be needed to magnify the thin upper spire sections. The earths curvature will barely cut into the base of the building. Given telescopes and unrealisticly clear atmosphere one could see the tower from much further than 100 kms, maybe 150kms, especially if slightly elevated ( for example a 50 story rooftop ). In the end it is haze that will limit the viewing distance, not earth curvature.

What I'm trying to say is that empirical evidence will squash math theory most of the time and then some.
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Last edited by graham; Feb 10, 2008 at 6:05 PM. Reason: spelling errors
     
     
  #5606  
Old Posted Feb 10, 2008, 7:49 PM
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^^ I don't know what you mean empirical evidence will squash math theory. It'll only squash it when the theory is not correct.

Assuming the Earth is spherical (in reality it's not exactly but doesn't matter here), its radius 6373km and the burj being 808m high then if your eye is at ground level you'll see the tip appearing over the horizon 101.486km away from it. If you assume your eye is 1.8m above the ground then you'll see it 106.276km away (see how much difference 1.8m can make?). Then if you stand on a 200m tall hill you'll see it at 151.976km.

So the conclusion is you're right, and the math proves you right also! When standing on a 50 storey rooftop you'll see it when you're 150km away (if there's no haze).


Now here's a picture by Tom Green on SSC taken from 130km away in a plane (as he claims)

     
     
  #5607  
Old Posted Feb 10, 2008, 9:47 PM
Nowhereman1280 Nowhereman1280 is offline
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^^^ Sometimes atmospheric conditions actually extend the distance at which you can see a skyline. There is an effect called lensing that happens on Lake Michigan sometimes that allows you to see the Milwaukee Skyline (tallest building is like 200m) from clear across the lake which is like 80 miles or 150km or something like that.

What happens is that cold, dense air, gets on top of the relatively warm non-dense air generated by the relatively warm lake water and acts as a lens bending the light coming from Milwaukee back down by a few degrees allowing you to see it from great distances.

PS: When the Chicago Spire is complete there will be an easy line of sight between the top of CS and the top of the tallest building in Milwaukee. I have a feeling that you will often be able to see Milwaukee from the top of CS whenever there is a slight lensing effect from the lake. That would be sweet, seeing another city from nearly 100 miles away.
     
     
  #5608  
Old Posted Feb 10, 2008, 10:58 PM
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[QUOTE=malec;3345226]^^ I don't know what you mean empirical evidence will squash math theory. It'll only squash it when the theory is not correct.

My point exactly! It is almost always not correct for one reason or another.
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  #5609  
Old Posted Feb 10, 2008, 11:15 PM
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Quote:
Originally Posted by malec View Post

Now here's a picture by Tom Green on SSC taken from 130km away in a plane (as he claims)

This picture is taken from 100km maximum! I really don't think that the Burj Dubai is 130km away in this picture.
     
     
  #5610  
Old Posted Feb 11, 2008, 4:31 AM
towerguy3 towerguy3 is offline
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When h << the Radius of the Earth (as it is with buildings) a rough approximation is d = SQRT (13h) where h is in meters. If you prefer using feet it's d = SQRT (1.5r) where h is in feet and d is in miles.

My understanding is when you factor in the lensing effect of the atmosphere you can add about 20 % to the derived d value. This lensing effect is also the reason why when you see a sunrise or sunset the Sun is actually below the horizon at the moment you "see" the Sun on the horizon. The atmosphere bends the light upwards to your eyes.

White mountaintops covered in snow and glistening in sunshine are easier to see from distance. Here in Vancouver you can stand atop Grouse Mountain 4000 feet and easily glimpse snow covered Mt. Baker 60 miles away which is 10,800 feet high.

When you're 4000 feet up looking at another mountain 60 miles away that's almost 11,000 feet high, you're above most of the haze anyways. I would think the air around Dubai is quite clear being a dry arid climate...
     
     
  #5611  
Old Posted Feb 11, 2008, 4:34 AM
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sorry, it's d = SQRT (1.5h) where h is in feet and d in miles

from the top of Mt. Everest the line of sight distance to the horizon is an unbelievable 335 kilometers
     
     
  #5612  
Old Posted Feb 11, 2008, 5:28 AM
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In the first image below taken from 55 km (St. Catherines) you can clearly see the effect of the curvature of the Earth as the bottom halves of the other towers such as the TD towers are cut off by the water. The Earth's curvature is very apparent in this photo. If the Earth were truly flat you'd see those towers right down to their bottoms!

This is a great visual example of the curvature of the Earth! :



In the second one below the observer is elevated so the curvature is not noticable and you can see the Toronto towers down to their bottoms:



Keep in mind that a typical person 5.6 feet high can only see a horizon distance of 2.9 miles. That distance increases rapidly as your elevation increases. At an elevation of 100 feet you can see 12.2 miles.

Mathematically d varies as the square root of 13 times the height h in meters or the square root of 1.5 times the height h in feet.
     
     
  #5613  
Old Posted Feb 11, 2008, 7:00 PM
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Quote:
Originally Posted by towerguy3 View Post
In the first image below taken from 55 km (St. Catherines) you can clearly see the effect of the curvature of the Earth as the bottom halves of the other towers such as the TD towers are cut off by the water. The Earth's curvature is very apparent in this photo. If the Earth were truly flat you'd see those towers right down to their bottoms!

This is a great visual example of the curvature of the Earth! :



In the second one below the observer is elevated so the curvature is not noticable and you can see the Toronto towers down to their bottoms:



Keep in mind that a typical person 5.6 feet high can only see a horizon distance of 2.9 miles. That distance increases rapidly as your elevation increases. At an elevation of 100 feet you can see 12.2 miles.

Mathematically d varies as the square root of 13 times the height h in meters or the square root of 1.5 times the height h in feet.
All this reliance on math to make assumptions that are actually quite complicated when pure empirical evidence is staring us right in the face to refute it, is amazing to me.

Let's take the two photos above.

1) shot from 50km at lake level elevation ( about same level as base of tower) in St. Catherines supposedly. I don't think so, but never the less the above poster assumes this. He says you can see the curvature cutting stuff off.

2) Here the view is from an elevation of 475 above tower base.
According to his math there should be a big difference.
As you can clearly see in both photos - there is the same amount of section below the restuarant level can be seen as referenced from the relative size of the section above the restaurant. Restaurant at 1100 feet - top observation at 1500 - tip at 1800. Its all there in both photos.
But as I said, I think the St. Catherines view is really from only 3 kms away with a short lens.

Anybody have images showing curvature blocking an object at 100km??
Why not just use the empirical evidence to make a deduction?
Using math just introduces all sorts of innacuracies, assumptions, half truths etc and the end result it bogus. When you have empirical evidence all theory is usually bogus or half bogus.

The initial assertion is - no matter 50 , 100 or 150 km, earth curvature is not the limiting factor by a longshot. Haze is.

Also - it's interesting - I've never seen the same equation quoted twice.
The stories about tall ships dissapearing below the curvature of the earth and leading to theory about a round earth i think is bogus. It was huge 50 to 60 foot swells, the ship and observer in swell troughs.

According to one equation with some apparent credibility, the Burj would have the bottom 140 feet cutoff when viewed from 150Kms. Since the Burj Dubai is 2650' tall, this cutoff would be barley noticable. And you would only need to view from the 15th story of a building to see it all.

Anybody watching car racing when Al Unser was an announcer might remember this........
The other announcer was going to great lengths to communicate the gap between the cars in seconds and fractions of seconds and calculating gain per lap etc all very complicating and distracting, when Al pointed out that the gap was small enuff that you can now see the gap between the cars, for crying out loud. So stop calculating. hehe LMAO
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Last edited by graham; Feb 11, 2008 at 10:11 PM. Reason: made an error
     
     
  #5614  
Old Posted Feb 11, 2008, 11:13 PM
towerguy3 towerguy3 is offline
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I've been to Toronto and that view in photo 1. is definitely not from 3 km away.

Are you suggesting the Earth is not round?
     
     
  #5615  
Old Posted Feb 11, 2008, 11:35 PM
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Quote:
Originally Posted by graham View Post
This looks to be a shot from the island with a slightly wide lens.
distance - 3 kilometers maybe.

See next post for real world examples, not pie in sky math thats not relevant for earth curvature at these short distances.
Nope, that looks about right for 50 klicks away.

The tallest buildings in my city are about 120-130 meters high. You can see them from fifty kilometers away while driving into the city during the day. Keep in mind that the city in question is Winnipeg and it's located on a flood plain. It is flat as a pancake here and visitors are always amazed at just how flat the terrain is when they see it from the air or drive out of the city.

The suggestion that the shot in question was taken from 'maybe three kilometers away' is downright laughable. It suggests that you're either not very familiar with the metric system (unlikely) or that you aren't used to being able to see buildings from very away.
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  #5616  
Old Posted Feb 12, 2008, 1:06 AM
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I know from personal experience that you can see Torontos skyline from Niagra Falls. Not sure of the distance but clearly visable across the lake.
     
     
  #5617  
Old Posted Feb 12, 2008, 1:28 AM
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I live across Lake Ontario from Toronto, in Niagara, and you can definitely see the Toronto skyline and it looks exactly like that picture. I can't see how that picture could possibly be from the islands.
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Last edited by dougtheengineer; Feb 12, 2008 at 1:29 AM. Reason: spelling
     
     
  #5618  
Old Posted Feb 12, 2008, 2:59 AM
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Why are you both asking and answering the same question?
Do you just like showing people what you know and being smug about it?
For your information those results have been posted many times for interest sake over the last 3 years, it has also been deduced that due to its proximity to the sea, the sand particles in the air and the extreme ambient heat haze pretty much year round, the maximum visible distance achievable in Dubai on the clearest of days will be no more than 60kms.

Also, as the coastline goes East to West, any vista would continue in a path with the equatorial radius of 6378.135 kms not the polar radius of 6356.75 kms as you have provided.

Also, in maths do not use the letter Q as it omitted for the more popular and correct usage of the Greek θ (theta)....and capital letters are never used in algebraic formulae.
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  #5619  
Old Posted Feb 12, 2008, 5:15 AM
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Here are views of downtown from Toronto Islands, a distance of 3 km. You judge for yourself



     
     
  #5620  
Old Posted Feb 12, 2008, 5:21 AM
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This is off topic, get back to the Burj Dubai please.
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