karlmarx
join:2006-09-18
iraq
 ·Fairpoint Communic..
| reply to Kompressor
Re: hmmm.....
well, considering in general terms, the radio distance is defined as O(ptical) = sqrt(2h(r))+sqrt(2h(t)), where O = distance to optical horizon, and h=(r)receiver height, and h(t) is transmitter height. So assume the antenna is 20' at both sides. The optical distance would be approximately sqrt(2*20) + sqrt(2*20) = Appx 12.5 miles.
Now we calculate the loss.
PL = 117 + 20log10fMhz-20log10hthr+40Log10D
Where 117 = Constant
2.4Ghz
20' and 20'
Distance of 12 miles
So it's about 175db loss over that distance. Thus, the antenna is the key, because with a 175db loss, the antenna is going to have to be pretty tight.
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RadioDoc
58ef2c0
Premium,ExMod 2000-03
join:2000-05-11 | reply to Kompressor
Well, them frequencies are not used for plain vanilla 802.11 WiFi...
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Toolmaster of La Grange. |
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Kompressor
Premium
join:2002-02-12
Huntington Beach, CA | reply to RadioDoc
Depending on what frequency you are transmitting on, it is possible for a simplex signal to go around the world. |
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RadioDoc
58ef2c0
Premium,ExMod 2000-03
join:2000-05-11
·AT&T Midwest
| reply to listenup
Except that unless you have an extremely high transmit and/or receive location you are not going to get anywhere near 60 miles LOS. The curvature of the earth makes sure of that. The 'world record' referenced by Karl Bode  used two mountaintops. How many regular users have a couple of 7500 foot mountain peaks laying around?
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Toolmaster of La Grange. |
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