Tall building sighting distance

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Discussion Overview

The discussion centers around the visibility distance of tall buildings, specifically the Burj Khalifa, from the ground. Participants explore the mathematical relationships involved in determining how tall a building must be to be seen from a distance of 250 kilometers, involving concepts of geometry and the curvature of the Earth.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the claim that a building needs to be 4.9 km tall to be visible from 250 km away and requests an explanation of the equation used to arrive at this figure.
  • Another participant asserts that their calculations indicate the height required is slightly more than 4.9 km, using the polar Earth diameter and the distance as line of sight.
  • A participant provides their own equation for calculating the height needed for visibility, suggesting that the initial approach may have been inaccurate.
  • Further calculations are presented, with one participant refining their earlier approximation to 4.90058148 km based on a more precise Earth radius at a specific latitude.
  • Another participant discusses using the same formula to determine the distance from which the Burj Khalifa can be seen, based on its actual height and the Earth’s radius.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions regarding the height required for visibility. There is no consensus on the correct height, and multiple competing calculations are presented.

Contextual Notes

Participants rely on various approximations and definitions of the Earth's radius, which may affect their calculations. Some assumptions about the visibility conditions and the geometry involved are also not fully resolved.

Echoing Stars
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From what distance can a person on the ground see the Burj Khalifa?

Or: How tall does the building have to be to be seen from 250 kilometers away?
 
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##4.9 \,\text{km}##
 
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What you said: The building needs to be 4.9 km high to be visible from 250 km away?
I would like an explanation of the equation that produced this result.
 
Echoing Stars said:
What you said: The building needs to be 4.9 km high to be visible from 250 km away?
I would like an explanation of the equation that produced this result.
See. And I would have liked to have seen some effort from your side to solve the problem.
 
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russ_watters said:
I scribbled it on a tiny kid's toy so I may have made a mistake, but I don't think so.

I used the polar Earth diameter from Wikipedia, ##12713\,\text{km}## and the distance ##250\,\text{km}## as line of sight. I reviewed it, and it is a little more than ##4.9\,\text{km}.##
Solve: ##0=(r+h)^2-r^2-(250)^2##
 
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You are a champion.
This is my equation:
h=63712+2502−6371h = \sqrt{6371^2 + 250^2} - 6371
 
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Echoing Stars said:
You are a champion.
This is my equation:
h=63712+2502−6371h = \sqrt{6371^2 + 250^2} - 6371
I think you have been a bit sloppy in the first equation. The same formula (Pythagoras) also allows you to solve for the distance ##x## if you use the radius 6374.344 km and height (peak) 0.8298 km of the actual building. Then ##x=102.857\,\text{km}.##
 
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  • #10
fresh_42 said:
אני חושב שהיית קצת מרושל במשוואה הראשונה. אותה נוסחה (Pythagoras) מאפשרת לך גם לפתור את המרחק ##x## אם אתה משתמש ברדיוס 6374.344 ק"מ ובגובה (שיא) 0.8298 ק"מ של הבניין בפועל. ואז ##x=102.857\,\text{km}.##
And it's just like that..
Look here
 
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  • #11
fresh_42 said:
I scribbled it on a tiny kid's toy so I may have made a mistake, but I don't think so.

I used the polar Earth diameter from Wikipedia, ##12713\,\text{km}## and the distance ##250\,\text{km}## as line of sight. I reviewed it, and it is a little more than ##4.9\,\text{km}.##
Solve: ##0=(r+h)^2-r^2-(250)^2##
Sorry, there were two questions asked: I thought you were answering the first and I skimmed past the second.
 
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