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== Radio horizon == {{see also|Radar horizon}} The ''radio horizon'' is the [[locus (mathematics)|locus]] of points at which direct rays from an [[antenna (electronics)|antenna]] are tangential to the surface of the Earth. If the Earth were a perfect sphere without an atmosphere, the [[radio]] horizon would be a circle. The radio horizon of the transmitting and receiving antennas can be added together to increase the effective communication range. <!-- Antenna heights above {{convert|1,000,000|ft|mi km|0|abbr=off}} will cover the entire hemisphere and not increase the radio horizon.{{Citation needed|reason=This not only contradicts the formulae presented elsewhere, it seems axiomatically wrong - no matter how high an antenna/observer is, it is impossible to (fully) 'see' an entire hemisphere, so unless I'm missing something, this is nonsense. |date=April 2016}}{{Citation needed|reason=If I rearrange the equation, the satellite would have to be over 3179km above the Earth's surface. |date February 2017}} --> [[Radio propagation|Radio wave propagation]] is affected by atmospheric conditions, [[ionospheric absorption]], and the presence of obstructions, for example mountains or trees. Simple formulas that include the effect of the atmosphere give the range as: :<math>\mathrm{horizon}_\mathrm{mi} \approx 1.23 \cdot \sqrt{\mathrm{height}_\mathrm{feet}}</math><!--As per geometric distance shown below--> :<math>\mathrm{horizon}_\mathrm{km} \approx 3.57 \cdot \sqrt{\mathrm{height}_\mathrm{metres}}</math> The simple formulas give a best-case approximation of the maximum propagation distance, but are not sufficient to estimate the quality of service at any location. === Earth bulge === In [[telecommunications]], '''Earth bulge''' refers to the effect of [[earth's curvature]] on radio propagation. It is a consequence of a circular segment of earth profile that blocks off long-distance communications. Since the vacuum line of sight passes at varying heights over the Earth, the propagating radio wave encounters slightly different propagation conditions over the path.{{citation needed|date=October 2022}} === Vacuum distance to horizon === {{main|Horizon distance}} [[File:RadioHorizont_h_d.jpg|thumb|318x318px|''R'' is the radius of the Earth, ''h'' is the height of the transmitter (exaggerated), ''d'' is the line of sight distance]] Assuming a perfect sphere with no terrain irregularity, the distance to the horizon from a high altitude [[Transmitter station|transmitter]] (i.e., line of sight) can readily be calculated. Let ''R'' be the radius of the Earth and ''h'' be the altitude of a telecommunication station. The line of sight distance ''d'' of this station is given by the [[Pythagorean theorem]]; : <math>d^2=(R+h)^{2}-R^2= 2\cdot R \cdot h +h^2</math> The altitude of the station ''h'' is much smaller than the radius of the Earth ''R.'' Therefore, <math>h^2</math> can be neglected compared with <math> 2\cdot R \cdot h</math>. Thus: : <math>d \approx \sqrt{ 2\cdot R \cdot h}</math> If the height ''h'' is given in metres, and distance ''d'' in kilometres,<ref>Mean radius of the Earth is β 6.37Γ10<sup>6</sup> metres = 6370 km. See [[Earth radius]]</ref> : <math>d \approx 3.57 \cdot \sqrt{h}</math> If the height ''h'' is given in feet, and the distance ''d'' in statute miles, : <math>d \approx 1.23 \cdot \sqrt{h}</math> [[File:RadioHorizont h H.jpg|thumb|318x318px|''R'' is the radius of the Earth, ''h'' is the height of the ground station, ''H'' is the height of the air station ''d'' is the line of sight distance]] In the case, when there are two stations involve, e.g. a transmit station on ground with a station height ''h'' and a receive station in the air with a station height ''H'', the line of sight distance can be calculated as follows: <math>d \thickapprox \sqrt{2 R} \, \left( \sqrt{h} + \sqrt{H}\right) </math> === Atmospheric refraction === {{main|Atmospheric refraction}} The usual effect of the declining pressure of the atmosphere with height ([[vertical pressure variation]]) is to bend ([[refraction|refract]]) radio waves down towards the surface of the Earth. This results in an '''effective Earth radius''',<ref name="ITU 2021">{{cite web | title=P.834 : Effects of tropospheric refraction on radiowave propagation | website=ITU | date=2021-03-05 | url=https://www.itu.int/rec/R-REC-P.834/en | access-date=2021-11-17}}</ref> increased by a factor around {{frac|4|3}}.<ref>Christopher Haslett. (2008). ''Essentials of radio wave propagation'', pp 119–120. Cambridge University Press. {{ISBN|052187565X}}.</ref> This ''k''-factor can change from its average value depending on weather. ====Refracted distance to horizon==== The previous vacuum distance analysis does not consider the effect of atmosphere on the propagation path of RF signals. In fact, RF signals do not propagate in straight lines: Because of the refractive 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 vacuum distance. Usually, a factor ''k'' is used in the equation above, modified to be : <math>d \approx \sqrt{2 \cdot k \cdot R \cdot h}</math> ''k'' > 1 means geometrically reduced bulge and a longer service range. On the other hand, ''k'' < 1 means a shorter service range. Under normal weather conditions, ''k'' is usually chosen<ref>Busi, R. (1967). ''High Altitude VHF and UHF Broadcasting Stations''. Technical Monograph 3108-1967. Brussels: European Broadcasting Union.</ref> to be {{frac|4|3}}. That means that the maximum service range increases by 15%. : <math>d \approx 4.12 \cdot \sqrt{h} </math> for ''h'' in metres and ''d'' in kilometres; or : <math>d \approx 1.41 \cdot\sqrt{h} </math> for ''h'' in feet and ''d'' in miles. But in stormy weather, ''k'' may decrease to cause [[rain fade|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.<ref>This analysis is for high altitude to sea level reception. In microwave radio link chains, both stations are at high altitudes.</ref> For example, in normal weather conditions, the service range of a station at an altitude of 1500 m with respect to receivers at sea level can be found as, : <math>d \approx 4.12 \cdot \sqrt{1500} = 160 \mbox { km.}</math>
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