DUBAI | Burj Khalifah (Burj Dubai) | 828 M / 2,716.5 FT - Pinnacle | 162 FLOORS

[graham]

Originally Posted by towerguy3

Is it only in an equilateral triangle that the angles add up to 180 degrees? So the unknown angle is 90 degrees minus 0.65 degrees? Two sides (representing the Earth's radius) have the same length, 6391 km...

The central cross section of the earth is a circle which is 360 degrees.
the circumferance is 25,000 miles.
25,000/360 = 69.4 miles per degree.
90 mi. ( 150kms [the distance in question]) is 1.3 degrees.
The 0.65 degrees is just to show that the measurement for x is taken at the center of the 1.3 degree section.

So what is the value of x? That's the part I can't get.
Where are all the math geniuses?

[MikeM]

Originally Posted by graham

Good point for the 1st steel floor. ok.
But the conversion is done! It's just steel on steel now. And it still seems to be taking forever.
The method and material has been perfected over many decades.
I'm sooo confused - lol.

The method may have been perfected but rarely, if ever, done at that altitude. Do you know what the winds are like up there? Even on an absolutely still day there will be significant air movement at the top. Just hoisting the members up there is a challenge. Now imagine a large steel member weighing a few tonnes swinging in the wind and you have to hand-guide it to a precise location. All this while the building itself is probably moving a significant amount. One small miss-calculation and ...

I'd be very, very careful if I was up there - I'm rather attached to my fingers.

Not to mention hands, arms and neck!

[MikeM]

Originally Posted by towerguy3

Is it only in an equilateral triangle that the angles add up to 180 degrees? So the unknown angle is 90 degrees minus 0.65 degrees? Two sides (representing the Earth's radius) have the same length, 6391 km...

Any triangle on a planar surface has angles that add up to 180°

[graham]

Originally Posted by MikeM

The method may have been perfected but rarely, if ever, done at that altitude. Do you know what the winds are like up there? Even on an absolutely still day there will be significant air movement at the top. Just hoisting the members up there is a challenge. Now imagine a large steel member weighing a few tonnes swinging in the wind and you have to hand-guide it to a precise location. All this while the building itself is probably moving a significant amount. One small miss-calculation and ...

I'd be very, very careful if I was up there - I'm rather attached to my fingers.

Not to mention hands, arms and neck!

Yeah - sounds like your speaking from experience.
So - wind related safety issues due to extreme altitude - makes sense.
Thanks for your input.

[towerguy3]

We can ignore the h^2 term for small h and get only a tiny error in the process. To a very good approximation, h = (d^2)/2R.

For d = 150 km, I get h = 1.65km, which is about 2.5 times the height of the building currently (2000 feet). A mountain of that height would have its summit just reaching the horizon.

The longest distance at which you could just see a light on the top of Burj Dubai as it currently stands (2000 feet high) would be about 93 km. That is assuming no atmosphere. Actually the atmospheric refraction would extend the range a bit.

[towerguy3]

The 69.4 mile figure is equivalent to 60 Nautical Miles, each Nautical Mile being 6080 feet as opposed to 5280 feet for a Statute Mile. There are 60 Nautical miles in one degree of latitude.

The term Nautical Mile is defined as the arc length of 1/60th of a degree of Latitude, or one arc minute.

The length of a degree of Longitude varies as the Cosine of the Latitude. At the equator one degree of longitude is 69.4 miles whereas at 60 degrees North it's one half of that (Cosine of 60 deg is 0.5)

[graham]

Originally Posted by towerguy3

We can ignore the h^2 term for small h and get only a tiny error in the process. To a very good approximation, h = (d^2)/2R.

For d = 150 km, I get h = 1.65km, which is about 2.5 times the height of the building currently (2000 feet). A mountain of that height would have its summit just reaching the horizon.

The longest distance at which you could just see a light on the top of Burj Dubai as it currently stands (2000 feet high) would be about 93 km. That is assuming no atmosphere. Actually the atmospheric refraction would extend the range a bit.

wow - so your saying from only 100kms the entire building ( at 2000' now) would be cutoff, atmospheric lensing notwithstanding. But is that calculated with the eyeball at 5 or 6 feet above the ground?

Can you calculate for the value of "x" in my diagram?

[graham]

Originally Posted by towerguy3

The 69.4 mile figure is equivalent to 60 Nautical Miles, each Nautical Mile being 6080 feet as opposed to 5280 feet for a Statute Mile. There are 60 Nautical miles in one degree of latitude.

The term Nautical Mile is defined as the arc length of 1/60th of a degree of Latitude, or one arc minute.

The length of a degree of Longitude varies as the Cosine of the Latitude. At the equator one degree of longitude is 69.4 miles whereas at 60 degrees North it's one half of that (Cosine of 60 deg is 0.5)

Ok - but what do you get for x in my diagram?

[towerguy3]

yeah we're back to square one; all I've shown is from a height of 600 meters you can see a light on the horizon 93 km away... not the other way around... I'm stuck, help

[MeTP]

I hope nobody takes too much offense at this, but would it be asking too much to take the math conversation to a new thread? This really is only peripherally related to the topic of this thread and it's getting a bit old for those of us who aren't really interested in it.

[towerguy3]

using Quadratic formula:

h = R (+/-) sqrt (R^2+1) that's for d=1

here it is in general: h = R (+/-) sqrt {(R^2)-(d^2)}

[graham]

Originally Posted by towerguy3

using Quadratic formula:

h = R (+/-) sqrt (R^2+1) that's for d=1

here it is in general: h = R (+/-) sqrt {(R^2)-(d^2)}

That's greek to me.

But I just did my best calcs I could do and now think that x = 1425 feet.
With this number in hand I think you could see the top 600 feet of the finished 2650' Burj Dubai tower. So - about 2050 feet cut off wehn viewed from 150 kms.
That seems to fit closer to yours and some other forecasts for 150kms.

Sorry if it's getting old MeTP - just that progress is almost at a standstill and at least this is something related to chat about.

You want to change the subject - go ahead.
If you do, I'll stop posting about this stuff right away.

[MeTP]

Originally Posted by graham

You want to change the subject - go ahead.

Again, no offense intended, but this thread already has a subject and you're way off of it. It's just a big let down to keep coming to this thread thinking there's some news about Burj Dubai construction only to find a continuation of page-after-page-after-page of this snooze fest. I'm sure it's very interesting to you but it has nothing to do with the topic of this thread. Anyway, I'm not going to say any more about it. If the moderators don't think it's worth saying something it's certainly not up to me to do anything about it. I'll just go read something else.

[towerguy3]

h = R (+/-) sqrt {(R^2)-(d^2)} Here I'm letting h represent the height above the ground of Burj Dubai visible above the horizon from a distance d.

It works both ways: from 93 km out you could see the top of a 600 meter tower (where BurjDubai is right now) or from the top of a 600 meter tower you could see to the horizon. h = 600 m or 0.6 km

so from 10 km out, the bottom 7.8 meters of BurjDubai should be cut off by the Earth's curvature

20 km out = bottom 31 meters cut off

30 km out = bottom 70 meters cut off

40 km out = bottom 125 meters cut off

50 km out = bottom 195 meters cut off

60 km out = bottom 281 meters cut off

70 km out = bottom 383 meters cut off

80 km out = bottom 500 meters cut off

as you can see it's not a linear variation

[malec]

graham, I'm more than halfway through a physics degree and that involves a shit load of math. The formula that I used to calculate those numbers is the correct one based on geometry and to deny it's right means you either deny the size of the Earth, or Pythagoras's theorem. In fact it's the same formula on wikipedia: http://en.wikipedia.org/wiki/Horizon

The question is what external factors contribute to the viewing distance (not including haze). One is if you're on a hill. If you are standing 50km away from the tower but on a hill of 10m only. Then you'd see as much of the tower as if you were standing about 40km away at ground level. This makes a HUGE difference. Also refraction of light will play a role. I have no idea how much this affects but if you want a detailed analysis it's impossible because it involves understanding the turbulence in air and turbulence is an unsolved problem.

[Rise To The Top]

Originally Posted by MeTP

Again, no offense intended, but this thread already has a subject and you're way off of it. It's just a big let down to keep coming to this thread thinking there's some news about Burj Dubai construction only to find a continuation of page-after-page-after-page of this snooze fest. I'm sure it's very interesting to you but it has nothing to do with the topic of this thread. Anyway, I'm not going to say any more about it. If the moderators don't think it's worth saying something it's certainly not up to me to do anything about it. I'll just go read something else.

x2. Lets see some updates please.

[johnandahalf]

Originally Posted by Rise To The Top

x2. Lets see some updates please.

x3

[towerguy3]

sorry actually h represents the amount of the tower cut off by the Earth

[towerguy3]

So in the image of Toronto taken from 50 km out in St. Catherines, what we're seeing is the effect of the atmosphere bending the light (refraction). According to our calculations the bottom 600 feet or so of the CN Tower should be cutoff by Lake Ontario's water and that's clearly not the case.

So rather than saying the photo is an excellent example of the curvature of the Earth, rather it's an awesome example of the power of the atmosphere to act as an optical lens!

[graham]

Originally Posted by malec

graham, I'm more than halfway through a physics degree and that involves a shit load of math. The formula that I used to calculate those numbers is the correct one based on geometry and to deny it's right means you either deny the size of the Earth, or Pythagoras's theorem. In fact it's the same formula on wikipedia: http://en.wikipedia.org/wiki/Horizon

The question is what external factors contribute to the viewing distance (not including haze). One is if you're on a hill. If you are standing 50km away from the tower but on a hill of 10m only. Then you'd see as much of the tower as if you were standing about 40km away at ground level. This makes a HUGE difference. Also refraction of light will play a role. I have no idea how much this affects but if you want a detailed analysis it's impossible because it involves understanding the turbulence in air and turbulence is an unsolved problem.

yeah - it's incredibly complicated and frought with undeterminable variables.
Thanks for the Wikpedia link. Some good info there.
I figure the only way to really understand it is to do real world tests with
gps equipment and very good telephoto lenses and photoshop and using good weather reports and a whole lot of patience.
I wonder what the results would be.
I guess we'll know soon since the Burj Dubai is almost finished and there will be peeps taking shots from all over the place.
Looking forward to it.

Is it not funny that some guys seem to blame us curvature chatters because there are no construction updates recently.
Like we told the updaters to shutup while we're posting our stuff? lmao
There are no updates because there are none. It has nothing to do with our "view from distance chat". What a laugh eh?