# Earth:Earth bulge

Test Earth bulge is a term used in telecommunications, meaning the curvature of the Earth's surface, which limits the range of communication methods that require a line-of-sight path, like high frequency radio waves, microwaves, or lasers.

## Distance to horizon

R is the radius of the Earth, h is the height of the transmitter (exaggerated), d is the line of sight distance

Assuming the Earth is a perfect sphere with no terrain irregularity, the line-of-sight distance to the horizon from a transmitter located at a given altitude above the surface can readily be calculated. Let R be the radius of Earth and h be the altitude of the transmitter's antenna. Line of sight distance d of this station is given by the Pythagorean theorem;

$\displaystyle{ d^2=(R+h)^{2}-R^2= 2\cdot R \cdot h +h^2 }$

Since the altitude of the station is much less than the radius of the Earth,

$\displaystyle{ d \approx \sqrt{ 2\cdot R \cdot h} }$

The mean radius of the earth is about 6,378 kilometres (3,963 mi). (See Earth radius) Using the same units for both the altitude of the station and the radius of the earth,

$\displaystyle{ d \approx \sqrt{ 2\cdot 6378 \cdot h} \approx 112.9 \cdot \sqrt{h} }$

But the radius of Earth is given in kilometers. Converting the equation for distance in kilometers and height in meter,

$\displaystyle{ d_{km} \approx \sqrt{ 2\cdot 6378_{km} \cdot h_{km}} = \sqrt{ 2\cdot 6378_{km} \cdot h_m \cdot \frac{1_{km}}{1000_m}}=\sqrt{ \frac{2\cdot 6378}{1000}}(\sqrt{h_m}) km }$

where $\displaystyle{ d_{km} }$is the distance in kilometers and $\displaystyle{ h_m }$is in meters. Thus, if the height is given in m. and distance in km,

$\displaystyle{ d \approx 3.57 \cdot \sqrt{h} }$

If the height is given in ft. and the distance in miles,

$\displaystyle{ d \approx 1.23 \cdot \sqrt{h} }$

Of course the presence of hills or mountains between the transmitter and receiver may reduce this distance.

## Effect of atmospheric refraction

The above analysis doesn’t take the effect of atmosphere on the propagation path of the RF signals into consideration. In fact, the RF signals don’t propagate in straight lines. Because of the canalizing effects of atmospheric layers, the propagation paths are somewhat curved. Thus, the maximum service range of the station, is not equal to the line of sight distance. Usually a factor k is used in the equation above

$\displaystyle{ d \approx \sqrt{2 \cdot k \cdot R \cdot h} }$

k > 1 means geometrically reduced bulge and a longer service range. On the other hand, k < 1 means a shorter service range.

Experience has shown that, under normal weather conditions k is 4/3.[1] That means that, the maximum service range increases by 15%. (square root of 4/3=1.15)

$\displaystyle{ d \approx \sqrt{17h}\approx 4.12 \cdot \sqrt{h} }$

for h in m. and d in km.

$\displaystyle{ d \approx \sqrt{2h}\approx 1.41 \cdot\sqrt{h} }$

for h in ft. and d in miles ;

But in stormy weather, k may decrease to cause fading in transmission. (In extreme cases k can be less than 1.) That is equivalent to a hypothetical decrease in Earth radius and an increase of Earth bulge.[2]

For example, in normal weather conditions, the service range of a transmitting station at an altitude of 1500 m. with respect to receivers at sea level can be found as,

$\displaystyle{ d \approx 4.12 \cdot \sqrt{1500} = 160 \mbox { km.} }$

## Altitude dependence

Let d be the distance in km from the observer to the maximum earth bulge, h be the height in m of earth bulge and k the bulging factor. [3] The height of the earth bulge is

$\displaystyle{ h=\frac{d_1d_2}{12.75k} \,\! }$

which shows a dependency on the observer distance. This gives zero bulge at the location of the observers and maximum at its midpoint. To prove the equation, taking the equation for d with k budge factor,

$\displaystyle{ d \approx \sqrt{2 \cdot k \cdot R \cdot h_m}=\sqrt{2\cdot 6378 \cdot k\cdot h_m}=\sqrt{12.756 \cdot k\cdot h_{km}} }$

Having $\displaystyle{ d_1=d_2=d }$and $\displaystyle{ h_1=h_2=h }$,

$\displaystyle{ h= \frac{\sqrt{12.756 \cdot k\cdot h_{km}}\sqrt{12.756 \cdot k\cdot h_{km}}}{12.75k}=\frac{12.756 \cdot k\cdot h_{km}}{12.75k}\therefore=h }$