|
"Vaughn Simon" <vaughnsimonHATESSPAM@att.FAKE.net> wrote in message
news:v1%Mh.179946$5j1.176834@bgtnsc04-news.ops.worldnet.att.net...
>
> "Kyle Boatright" <kboatright1@comcast.net> wrote in message
> news:-_GdneniFpJ585nbnZ2dnUVZ_oernZ2d@comcast.com...
>>
>> My recollections from physics 20+ years ago is that radio wave strength
>> is determined by the cube root of the transmitter strength. So, a 10w
>> radio has twice the power of a 5w radio. The cube root of 2 (twice the
>> power) is 1.26, meaning that the higher powered radio should have 26%
>> more range than the low powered radio.
>
> Another way of saying the same thing is that to double the range, you
> must quadruple the power.
>
> I think Kyle has stated the theoretical difference. The actual
> difference is that you likely won't notice any difference caused by the
> difference in output power. The antenna and the modulation quality
> probably matter more than the output power.
>
The real limiting factor is that VHF is line-of-sight. Over relatively flat
terrain either radio is capable of reaching the horizon until you get above
8500-10000 feet. In mountainous terrain, the horizon is typically much
less.
At 10000 feet the horizon is a little over 100 nm which the 5W radio is
capable of reaching. If you get above that, you'll get more range out of
the 10W, but practically, it will be about 10-20% more. And even it will
max out at about 130-140 nm, regardless of altitude. The 5W will max out
around 105-115 nm.
Most of the time, you'll be talking to someone less than 50 nm away.
Gerry
|
|
Re: theoretical radio range....
