0
\$\begingroup\$

Very curious to know the range of 50 W transmitter of UHF range in nautical miles? The antenna used is 1/4 wave antenna of Jaybeam company. Can someone tell the exact formulation to find out the answer? Just to add further, height of Tx and Rx antenna is around 60 meters.

\$\endgroup\$
  • \$\begingroup\$ Is thin on aircraft? Or urban area? Open land? What's the exact gain in dBi of the antenna? How about the length and losses of the coaxial cable powering the antenna? What modulation type are you using? What's the receive sensitivity of the receivers? We need a lot more info to make this question answerable and a lot more context, if we need to assume. \$\endgroup\$ – AndrejaKo Aug 27 '18 at 12:39
  • 5
    \$\begingroup\$ The transmitter of the Mariner IV spacecraft, that relayed the first shots of Mars taken from an interplanetary probe, had a power of 10 watts... \$\endgroup\$ – xxavier Aug 27 '18 at 12:45
  • 1
    \$\begingroup\$ Range is meaningless without knowledge of the receiver - an EM wave will carrry on until infinity gradually blending in with the background noise of space. Question is under-constrained. \$\endgroup\$ – Andy aka Aug 27 '18 at 12:53
  • 1
    \$\begingroup\$ Hey D Duck how you calculated the horizon? \$\endgroup\$ – Rumi Aug 27 '18 at 15:49
  • 1
    \$\begingroup\$ Wikipedia says it all en.wikipedia.org/wiki/Horizon but there are online calculators for this sort of stuff. You need also to assume that the Earth is not flat. \$\endgroup\$ – D Duck Aug 27 '18 at 16:05
1
\$\begingroup\$

The horizon is ~ 30 km away if you're 60 m tall. So you expect a line of sight of 60 km. And then with refraction (for UHF) you get abut 1/3 more so about 80 km.

80 km = 43 nautical miles.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ How do certain frequencies refract more than others? Interaction with moisture? \$\endgroup\$ – PJazz Aug 27 '18 at 12:44
  • \$\begingroup\$ Link to website about VHF/UHF/Microwave Radio Propagation tapr.org/ve3jf.dcc97.html \$\endgroup\$ – D Duck Aug 27 '18 at 12:46
  • \$\begingroup\$ How did you conclude that you would get a 1/3 more at UHF? Most articles I've read say that beyond 30 MHz the game is up for ionospheric reflection. \$\endgroup\$ – Andy aka Aug 27 '18 at 13:09
  • 1
    \$\begingroup\$ 4/3 factor is the "earth bulge" or "effective radius" of the earth and a refraction effect. en.wikipedia.org/wiki/Line-of-sight_propagation glossary.ametsoc.org/wiki/Effective_earth_radius \$\endgroup\$ – D Duck Aug 27 '18 at 13:15
  • \$\begingroup\$ Horizon is an important issue. How much beyond that you can go depends on things like temperature-driven ducting along the curve of the earth, or flukes like meteor ionization trails, as normally UHF would not be reflected by the ionosphere. And of course the earth isn't barren and round, especially in interesting areas. Power comes in to play when there's a marginal path, or interference, but by 50 watts most of what is achievable is probably achieved. \$\endgroup\$ – Chris Stratton Aug 27 '18 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.