The earth has a circumference of approximately 25000 miles and you're going to put up a structure of some particular height, perhaps an antenna. How far will it be from the top of that structure along the earth's surface to the horizon?

The calculation, or perhaps more properly, the estimation is as follows:

While it's true that the algebra would be simpler using the arc cosine instead of the arc tangent, I wrote the code for this in GWBASIC and I only had the ATN function available to me, at least so far as I know.

Hi John,

The "radio horizon" distance tends to be approximately 1.3 times greater than the optical horizon because of RF refracting or bending as it follows the curvature of the earth. This is an established fact.

Your 1000 feet AGL transmitting antenna example should present a fringe "radio horizon" of 50 to 51 miles -- and that is not taking into account the height of the receiving antenna, which could extend the range much further.

Best Regards,

Steven Gehring

Posted by: Steven Gehring | May 31, 2011 at 02:04 PM

As Steven mentions, the radio horizon is further than the optical horizon. However if we are looking at radio signal propagation, then the partial fresnel zone obstruction by the earth is also an issue.

Posted by: Richard Urmonas | June 01, 2011 at 04:43 PM

Hi John,

For all practical radio links you only need to be able to calculate power (square) and Squareroot. The atn function is so close to 1 that it does not matter. Look at it this way: the tower is leaning slightly backwards from the perspective of the receiver and the atn is compensating for that to calculate the difference between the distance from receiver to the foot and the top of the broadcast tower. However, at a few dozen miles distance, who cares about a few feet less distance?

I tried posting the simple formula here, but apparently the

comments filter thinks I am swearing, so it just replies that

I cannot post that data here...

The essence is Pythagoras formula so the antenna height is

the difference between the earth radius (the long side opposite to the 90 deg angle) and the next longest side being the squareroot of the squared earth radius minus the squared distance of the link to the horizon (or halfway a point to point link)

Since typically for long distance backhaul both antennas are

elevated, you can find the minimum elevation due to earth bulge

underneath the link by entering half the total distance

in the formula to find minimum elevation for a horizontal link

(touching earth in the middle). This is just the optical model and you may find that the signal can reach further depending on its frequency, but you also need to consider the fact that a radio signal does not travel along a pencil-thin line, it uses a certain area to travel, creating what is called Fresnel zone around the optical Line of Sight.

To give full LoS for a radio link you need to add the first Fresnel zone clearance to the minimum height, which makes the height freq dependent because the first Fresnel zone depends on the freq.

Posted by: Cor van de Water | June 09, 2011 at 08:51 PM