# Radio Horizon Distance

Hi I wonder if anyboday could help me. I have got 2 questions

1. Given that the radius of the earth is 3963 miles, How  i can show/illustrate that the radio horizon can be approximated by sqrt(2*h).
Where h is height of antenna. what is acutally radio horizon

2.In text books, a formula is given to calculate  line of sight  d= 3.57* sqrt(h)
where d is distance between antenna and horizon.
my question is what 3.57 vaule represents in this formula
thanks
Rafi-muqaddar
Asked:
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Commented:
The basic diagram for this problem is as follows.  First, draw a quarter-circle representing the earth's surface, including a point for the center of the earth.  Draw an antenna at the center of the arc, with a height about 1/10 the radius of the earth.  Draw a line from the tip of the antenna to a tangent point on the surface.  Then draw another line from the tangent point down to the center of the earth.

You should have a right triangle here, with the hypoteneuse between the tip of the antenna and the center of the earth, and the other vertex at the tangent point.

The hypoteneuse has length (3963 + h_miles), the leg from the center of the earth to the tangent point is a 3963 mile radius, and the other leg is approximately the radio horizon Z in miles (perhaps the length of the arc between the antenna and tangent point is the true horizon distance, I'm not sure).
Using the Pythagorean theorem:
(3963 + h_miles)^2 = 3963^2 + Z_miles^2
2*3963*h_miles + h_miles^2 = Z_miles^2
Z_miles = sqrt(2*3963*h_miles + h_miles^2)
Since the height of the antenna will be much smaller than 3963 miles, we can ignore the h^2 term in the sqrt.
Z_miles = sqrt(2*3963*h_miles)
Antenna heights are usually measured in feet.
Z_miles = sqrt(2*3963*h_feet / 5280)
Z_miles = sqrt(1.501*h_feet)
This is approximate to sqrt(2h).

If we go metric (earth radius=6378km), the same derivation goes like this
Z_kilometers = sqrt(2*6378*h_kilometers)
Z_kilometers = sqrt(12.576*h_meters)
Z_kilometers = 3.571*sqrt(h_meters)

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Author Commented:
Thanks NovaDenizen excellent answer I give you A.
Thanks
Rafi-Muqaddar
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