Where does the 3.57 come from in the Line-of-Sight formula???
Figure 1. The geometry of the visible line-of-sight distance
From the figure 1, we have:
re = radius of the earth = 6370 km
hkm = antenna height in km
dkm = visible line-of-sight distance in km
The triangle in figure 1 is a right triangle. Recall that Pythagorean’s theorem for a right
triangle is c 2 = a 2 + b 2 where c is the hypotenuse and a and b are the two other sides of
the triangle.
Notice in figure 1 that one side is the radius of the earth, another side is the visible LOS
distance and the hypotenuse consists of the radius of the earth plus the height of the
antenna. Substituting:
Side 1:
Side b:
Hypotenuse:
into c 2 = a 2 + b 2 yields:
a = re
b = dkm
c = re + hkm
(re + hkm )2 = re 2 + d km 2
2
Re-arrange to solve for d km :
d km = (re + hkm ) − re
2
2
2
Perform square root to solve for d km :
(re + hkm )2 − re 2
d km =
Expand the (re + hkm ) term to get:
2
d km = re + 2re hkm + hkm − re
2
2
2
The terms re and − re cancel to get:
d km = 2re hkm + hkm
2
(1)
Consider typical antenna heights. In general the antenna height is much less than the
radius of the earth:
hkm << re
Multiplying both sides by the antenna height gives us:
hkm << re hkm
2
Since re hkm < 2re hkm , we can write:
hkm << re hkm < 2re hkm
and simplify to:
hkm << 2re hkm
2
2
2
Since hkm is much less than 2re hkm , we can ignore this term in equation (1) to give us:
d km = 2re hkm + hkm ≈ 2re hkm
(2)
d km ≈ 2re hkm
(3)
2
Or simply:
Substituting radius of the earth = re = 6370 km into (3), we get:
d km ≈ (2 )(6370)hkm
(4)
Substituting radius of the earth = hkm = hmeters / 1000 into (4), gives us:
d km ≈
(2)(6370)⎛⎜ hmeters ⎞⎟
⎝ 1000 ⎠
(5)
This computes to:
d km ≈ 12.74hmeters
≈ 12.74 hmeters
≈ 3.5693 hmeters
Rounding off two places, we get:
d km ≈ 3.57 hmeters