In telecommunication, radio horizon is the locus of points at which direct rays from an antenna (electronics) are tangential to the surface of the Earth. If the Earth were a perfect sphere and there were no atmospheric anomalies, the radio horizon would be a circle. To compute the radius of the circle drawn on the earth in such a case use the formula: :Radio Horizon(statute miles) = 1.23 * sqrt(Antenna Height(feet)) This is the geometric, straight line of sight horizon. For an equivalent formula for an antenna height in metres and a radio horizon in kilometres would be :Radio Horizion (kilometres) = 3.56 * sqrt(Antenna Height(metres)). The radio horizon of the transmitting and receiving antennas can be added together to increase the effective communication range. Antenna Heights above 1 million feet (1966 miles - 3157 kilometres) will cover the entire hemisphere and not increase the radio horizon. Very high frequency and Ultra high frequency radio signals will bend slightly toward the earth's surface. This bending effectively increases the radio horizon and therefore slightly increases the formula constant. The ARRL Antenna Book gives a constant of 1.415 for the feet/miles formulat for weak signals during normal tropospheric conditions. This can usefully be approximated as: :Radio Horizon(statute miles) = sqrt(2* Antenna Height(feet)) In practice, radio wave propagation is affected by atmospheric conditions, and the presence of obstructions, for example mountains or trees. The simple formula above gives a best-case approximation of the maximum propogation distance but is not sufficiently adequate for determining the quality of service at any location. References: * The ARRL Antenna Book, 19th Edition, R. Dean Shaw, pp 22-25 * Federal Standard 1037C, in support of MIL-STD-188 Radio frequency propagation Wireless communications
16 Jun 2005, Boon Phing write: I think the conversion here is more for optical horizon rather than radio horizon. aren't we suppose to use r = 4/3 r0 for earth radius? then D [NM] = 1.23 * sqrt(H [ft]) <= radio horizon instead of D [mi] = 1.23 * sqrt(H [ft]) <= optical horizon ======================================================================= someone wrote: "GF: I believe that the conversion here from ft to meters is wrong. Should have taken the square root of the conversion factor?" : I agree. : Horizon (mi) = 1.23 * sqrt(Height (ft)); : take an antenna 100 ft high. sqrt(100)=10; Horizon = 12.3 miles. : Horizon (km) = 1.6 km/mi * 1.23 * sqrt(3.28 ft/m) * sqrt(Height (m)); : 1.6 * 1.23 * 1.8113 = 3.5646384... chop precision, to get : therefore Horizon (km) = 3.56 * sqrt(Height (m)); also, "The ARRL Antenna Book gives a constant of 1.415 for weak signals during normal tropospheric conditions." : strangely, the ARRL Handbook condradicts this: 1.15 is the factor quoted there (page 21.20 of the 1999 edition.) so which is correct? User:Waveguy
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Radio horizon
In telecommunication, radio horizon is the locus of points at which direct rays from an antenna (electronics) are tangential to the surface of the Earth. If the Earth were a perfect sphere and there were no atmospheric anomalies, the radio horizon would be a circle. To compute the radius of the circle drawn on the earth in such a case use the formula: :Radio Horizon(statute miles) = 1.23 * sqrt(Antenna Height(feet)) This is the geometric, straight line of sight horizon. For an equivalent formula for an antenna height in metres and a radio horizon in kilometres would be :Radio Horizion (kilometres) = 3.56 * sqrt(Antenna Height(metres)). The radio horizon of the transmitting and receiving antennas can be added together to increase the effective communication range. Antenna Heights above 1 million feet (1966 miles - 3157 kilometres) will cover the entire hemisphere and not increase the radio horizon. Very high frequency and Ultra high frequency radio signals will bend slightly toward the earth's surface. This bending effectively increases the radio horizon and therefore slightly increases the formula constant. The ARRL Antenna Book gives a constant of 1.415 for the feet/miles formulat for weak signals during normal tropospheric conditions. This can usefully be approximated as: :Radio Horizon(statute miles) = sqrt(2* Antenna Height(feet)) In practice, radio wave propagation is affected by atmospheric conditions, and the presence of obstructions, for example mountains or trees. The simple formula above gives a best-case approximation of the maximum propogation distance but is not sufficiently adequate for determining the quality of service at any location. References: * The ARRL Antenna Book, 19th Edition, R. Dean Shaw, pp 22-25 * Federal Standard 1037C, in support of MIL-STD-188 Radio frequency propagation Wireless communications
Radio horizon
16 Jun 2005, Boon Phing write: I think the conversion here is more for optical horizon rather than radio horizon. aren't we suppose to use r = 4/3 r0 for earth radius? then D [NM] = 1.23 * sqrt(H [ft]) <= radio horizon instead of D [mi] = 1.23 * sqrt(H [ft]) <= optical horizon ======================================================================= someone wrote: "GF: I believe that the conversion here from ft to meters is wrong. Should have taken the square root of the conversion factor?" : I agree. : Horizon (mi) = 1.23 * sqrt(Height (ft)); : take an antenna 100 ft high. sqrt(100)=10; Horizon = 12.3 miles. : Horizon (km) = 1.6 km/mi * 1.23 * sqrt(3.28 ft/m) * sqrt(Height (m)); : 1.6 * 1.23 * 1.8113 = 3.5646384... chop precision, to get : therefore Horizon (km) = 3.56 * sqrt(Height (m)); also, "The ARRL Antenna Book gives a constant of 1.415 for weak signals during normal tropospheric conditions." : strangely, the ARRL Handbook condradicts this: 1.15 is the factor quoted there (page 21.20 of the 1999 edition.) so which is correct? User:Waveguy
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