Tall building sighting distance

  • Thread starter Thread starter Echoing Stars
  • Start date Start date
AI Thread Summary
The discussion centers around the visibility of the Burj Khalifa from a distance of 250 kilometers and the necessary height for a building to be seen at that distance. Calculations indicate that a building would need to be approximately 4.9 kilometers tall to be visible from 250 kilometers away, based on the Earth's curvature and using the Pythagorean theorem. The Earth’s radius at Dubai's latitude is about 6374.344 kilometers, which factors into the calculations. The conversation also touches on the accuracy of the initial equation used to derive this height, with confirmations that the approximation was correct. Additionally, it is noted that the actual height of the Burj Khalifa is 0.8298 kilometers, allowing for further calculations regarding visibility distances based on its height.
Echoing Stars
Messages
4
Reaction score
2
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?
 
Last edited:
Earth sciences news on Phys.org
##4.9 \,\text{km}##
 
  • Informative
Likes Echoing Stars
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.
 
  • Like
Likes Echoing Stars
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##
 
Last edited:
  • Like
Likes Echoing Stars
You are a champion.
This is my equation:
h=63712+2502−6371h = \sqrt{6371^2 + 250^2} - 6371
 
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}.##
 
  • Like
Likes Echoing Stars
  • #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
 
  • #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.
 
Last edited:
Back
Top