# Curvature of the Earth

First off, don't worry this is not a home work assignment . I am just hoping someone can satisfy my curiosity.

At 6'5", how far would I be able to look over a large body of calm water or a perfectly flat land mass until an item that was moving a way from me would drop below the curvature of the Earth?

 CaptainRich Posts: 6,080, Reputation: 2749 Cars & Trucks Expert #2 Jul 6, 2007, 07:29 PM
Hold on, buddy! This question is giong to draw a crowd! "Perfectly flat AND curvature of the Earth...?"

Let's start with: what is the heigth of the object departing? What latitude are you veiwing from? Altitude where you're standing...?

We want to know it all...

Effects of the Earth's Curvature and Atmospheric Refraction: on estimating a target's position
 letmetellu Posts: 3,153, Reputation: 1708 Ultra Member #3 Jul 6, 2007, 07:35 PM
I was taught that the line of sight on the ocean is 12 miles, I have no idea if this is a proven fact but it is a starting place.
 Wangdoodle Posts: 220, Reputation: 257 Full Member #4 Jul 6, 2007, 07:41 PM
Well, lets say a truck at 39 degrees, altitude 2000 ft.
 Wangdoodle Posts: 220, Reputation: 257 Full Member #5 Jul 6, 2007, 07:54 PM
Quote:
 Originally Posted by letmetellu I was taught that the line of sight on the ocean is 12 miles, I have no idea if this is a proven fact but it is a starting place.

That's the term I should have used."Line of sight". That's what I was curious about.

Thanks
 ebaines Posts: 10,054, Reputation: 5539 Expert #6 Jul 7, 2007, 06:12 AM
You can use the Pythagorean Theorem to determine the line of sight for a person whose eyes are $h_e$ meters above the horizon. Assuming a perfectly round earth, the distance to the horizon is:

$
D= sqrt {2R_e h_e + h_e^2}
$

To just be able to barely see the top of the truck of height $h_t$ slip below the horizon, you add
$
D_2 = sqrt {2R_e h_t + h_t^2}
$

If you use 2 meters for both $h_e$ and $h_t$, the total distance is about 17.5 Km, or 10.8 miles.

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