theoretical radio range....

"COLIN LAMB" wrote in message
link.net...
"I know a ham who used to routinely work moon-bounce on VHF, with a rig
powered by a single 9v transistor radio battery. I think he had something
like 60 _milliwatts_ on transmit."

Although the statement does not give all of the facts, I am sure there is
an error. The moon is 250,000 miles away - which means a total distance
travelled of 500,000 miles. The signals must refect off a less than
perfect reflecting surface (moon dust).

I am an amateur operator and have heard signals off the moon. With modern
digital modes, there is an improvement in single signal performance - and
if the station on the other end has a giant antenna (such as the giant
radio telescope in Puerto Rico, used on occasion by amateurs having fun),
it is possible to work moonbounce with a 100 watt rig and a long single
yagi.

That is a far cry from 60 milliwatts. There is about 32 db difference
between 60 mw and 100 watts. That would mean the antenna, instead of 15
db gain for a long yagi would need to have 47 db gain. An antenna that
size might raise some neighbor's objections (blocking the sun). And,
operating it would not be routine, as an antenna with such high gain needs
to track the sun. Might need something about the size of a locomotive to
move it.

Colin


Colin,

I did some moon-bounce work back in the '70s. At the time I was using a
full 1,000 watts on 2 meters with a stack of eight 15 element circular
polarized yagi antennas. Of course the receivers are far better today and I
am sure that it can currently be done with far less power. However, like
you, it is hard for me to believe that it can be done with 60 milliwatts.

On the other hand, as you know, 60 milliwatts is adequate for a dedicated CW
QRP operator to communicate world wide using code on the 20 or 40 meters
frequency bands.

Wayne
.....................
H Wayne Paul
W7ADK
.......................
HP-14 "6F" N990
http://www.soaridaho.com/

Hi Wayne:

!,000 watts and 8 antennas is about right for state of the art 1970s to hear
your own signal (but only sometimes (due to Faraday rotation) and even then
very weakly. The secret to using less was to "talk" to another station with
a better antenna. Today, if the other station has a gigantic antenna,
stations with 100 watts and a good single boom yagi can talk to others.

On occasion, hams have "borrowed" the radio telescope at Arecibo, PR. It is
a 1,000 foot diameter dish with 50,000 square feet of capture area. When it
was used on vhf, the other stations can use mediocre equipment and bounce
signals off the moon. I have not been able to find the actual gain of
Arecibo on vhf - but contacting them would not be a routine event and I
question whether 60 mw would do it into a moderate antenna.

Maybe on 1296 MHz or 10 GHz - but that would not meet the claimed vhf
guideline.

Back to the original post, there is a distinct advantage to using a 5 watt
transmitter over a 10 watt transmitter when you do not have a big fan out in
front of you. As a fellow pilot who routinely flies with the fan off, I use
a 5 watt transmitter. If people cannot hear me, they are often too far away
to be meaningful at the present moment.

Colin

In article .net,
COLIN LAMB wrote:
"I know a ham who used to routinely work moon-bounce on VHF, with a rig
powered by a single 9v transistor radio battery. I think he had something
like 60 _milliwatts_ on transmit."

Although the statement does not give all of the facts, I am sure there is an
error. The moon is 250,000 miles away - which means a total distance
travelled of 500,000 miles. The signals must refect off a less than perfect
reflecting surface (moon dust).

I guarantee the accuracy of the power source. I don't guarantee my
recollection of the power level, but one battery was 'more than sufficient'
for an entire evening's operation.

The rig was home-built transistor stuff, about the size of a pack of cigarettes.

I'm drawing a blank on the guy's call-sign, he lived outside Ogden Dunes, In.

I am an amateur operator and have heard signals off the moon. With modern
digital modes, there is an improvement in single signal performance - and if
the station on the other end has a giant antenna (such as the giant radio
telescope in Puerto Rico, used on occasion by amateurs having fun), it is
possible to work moonbounce with a 100 watt rig and a long single yagi.

That is a far cry from 60 milliwatts. There is about 32 db difference
between 60 mw and 100 watts. That would mean the antenna, instead of 15 db
gain for a long yagi would need to have 47 db gain. An antenna that size
might raise some neighbor's objections (blocking the sun). And, operating
it would not be routine, as an antenna with such high gain needs to track
the sun. Might need something about the size of a locomotive to move it.

Antena was a 22' solid (not mesh) parabolic dish , on a heavy-duty equatorial
mount.

He had two other, smaller, dishes, as well.

In article .net,
COLIN LAMB wrote:
Hi Wayne:

!,000 watts and 8 antennas is about right for state of the art 1970s to hear
your own signal (but only sometimes (due to Faraday rotation) and even then
very weakly. The secret to using less was to "talk" to another station with
a better antenna. Today, if the other station has a gigantic antenna,
stations with 100 watts and a good single boom yagi can talk to others.

On occasion, hams have "borrowed" the radio telescope at Arecibo, PR. It is
a 1,000 foot diameter dish with 50,000 square feet of capture area. When it
was used on vhf, the other stations can use mediocre equipment and bounce
signals off the moon. I have not been able to find the actual gain of
Arecibo on vhf - but contacting them would not be a routine event and I
question whether 60 mw would do it into a moderate antenna.

Maybe on 1296 MHz or 10 GHz - but that would not meet the claimed vhf
guideline.

Back to the original post, there is a distinct advantage to using a 5 watt
transmitter over a 10 watt transmitter when you do not have a big fan out in
front of you. As a fellow pilot who routinely flies with the fan off, I use
a 5 watt transmitter. If people cannot hear me, they are often too far away
to be meaningful at the present moment.

Colin

Well, that is true of an FM signal where the "capture effect" will let
you hear the strongest signal only...on AM, you'll still hear the squeal
if another signal comes on frequency. Aviation radios are AM, so....

My personal feeling is that the 5W will do marginally better than the
10W radio that if there is a significant cost difference, go with 5W.

Scott

Kyle Boatright wrote:


One thing to consider is that a 10w radio will have an easier time
overpowering a distant signal, so your transmissions get "stepped on" less.

KB

Except for one factual problem, I can accept the statement. The problem is
the VHF part. A 22 foot dish is a poor antenna for VHF. But, as you go up
into the microwave region the gain gets higher. The gain of the antenna on
the other end are higher, too.

So, substitute uhf or microwaves for the VHF and I will not argue with you.

My directory of EME stations for 1974 shows the only station from the 9 call
district was W9WCD, who was reported to have a 16 foot dish. In 1982, the
only statin listed in IN was K9CA.

I have not got my gain chart out for a 22 foot dish, but it still seems
marginal for 60 mw at any frequency - especially using old technology.

Colin Lamb

Oops, switch that to read the 10W will do marginally better...

Scott


Scott wrote:
Well, that is true of an FM signal where the "capture effect" will let
you hear the strongest signal only...on AM, you'll still hear the squeal
if another signal comes on frequency. Aviation radios are AM, so....

My personal feeling is that the 5W will do marginally better than the
10W radio that if there is a significant cost difference, go with 5W.

Scott

Kyle Boatright wrote:


One thing to consider is that a 10w radio will have an easier time
overpowering a distant signal, so your transmissions get "stepped on"
less.

KB

RST Engineering wrote:
Assuming a 1 microvolt (pretty numb these days) receiver at the other end
and quarter wave vertical whips at both ends, a 5 watt transmitter has a
THEORETICAL range of about 3000 miles. Doubling the power increases the
range by (sqrt(2)) or a THEORETICAL range of about 4300 miles for the 10
watter.

Now since most of us will operate somewhere below the oxygen limited 12000
MSL altitude, and presuming you are over the ocean, your range will be
horizon ("line of sight") limited by the old familiar equation that horizon
(in miles) is equal to 1.4 times (sqrt (altitude in feet)) or something on
the order of 150 miles. You may get a BIT of refraction, but not enough to
make a difference in the basic equation.

The real answer is that 5 or 10 watts really doesn't make a difference in
quiet spectrum range. It only helps "punch through" when there is a lot of
interfering garbage on the frequency.

Jim


1.4 Jim??

Thought it was 1.17.



http://www.boatsafe.com/kids/distance.htm

In article .net,
cavelamb himself wrote:

RST Engineering wrote:
Assuming a 1 microvolt (pretty numb these days) receiver at the other end
and quarter wave vertical whips at both ends, a 5 watt transmitter has a
THEORETICAL range of about 3000 miles. Doubling the power increases the
range by (sqrt(2)) or a THEORETICAL range of about 4300 miles for the 10
watter.

Now since most of us will operate somewhere below the oxygen limited 12000
MSL altitude, and presuming you are over the ocean, your range will be
horizon ("line of sight") limited by the old familiar equation that horizon
(in miles) is equal to 1.4 times (sqrt (altitude in feet)) or something on
the order of 150 miles. You may get a BIT of refraction, but not enough to
make a difference in the basic equation.

The real answer is that 5 or 10 watts really doesn't make a difference in
quiet spectrum range. It only helps "punch through" when there is a lot of
interfering garbage on the frequency.

Jim


1.4 Jim??

Thought it was 1.17.

Square root of 2 = 1.414

The equation may be written in either of two common formats:

h = horizon (in statute miles)
a = altitude (in feet)

h = sqrt (2*a)

or

h = 1.414 * (sqrt (a))

If we are ballparking instead of surveying, we generally drop it to 1.4.


Jim




"john smith" wrote in message
...
In article .net,
cavelamb himself wrote:


horizon
(in miles) is equal to 1.4 times (sqrt (altitude in feet))


1.4 Jim??

Thought it was 1.17.

Square root of 2 = 1.414