theoretical radio range....

[Andy[_3_]]

i have a radio rated at 10W and a radio rated for 5W output. mine 5
watter isn't a handheld but this is typical output for that genre.

assuming they are using the same antenna what is the theoretical range
difference between the two and what is the practical range
difference? it seems the price difference is 2X to 3X. is the price
difference justified?

i guess i'm asking "should i ebay the 10W unit and find a better use
for the remainder?"

[Philippe Vessaire]

Andy wrote:

> i have a radio rated at 10W and a radio rated for 5W output. mine 5
> watter isn't a handheld but this is typical output for that genre.
>
> assuming they are using the same antenna what is the theoretical range
> difference between the two and what is the practical range
> difference? it seems the price difference is 2X to 3X.

> is the price difference justified?
No.... you will get similar range... If you want the best range you can,
just install a "perfect" antenna.

I may acheive a 40Nm range with 2500' above grouns with an old Icom A20.
The "secret" is a good and large ground pannel inside the wood and
fabric fuselage. It's made from metalic mosquito mesh.

by
--
Volem rien foutre al païs! minicab F-PRAZ
Philippe Vessaire Ò¿Ó¬

[Kyle Boatright]

"Andy" > wrote in message
oups.com...
>i have a radio rated at 10W and a radio rated for 5W output. mine 5
> watter isn't a handheld but this is typical output for that genre.
>
> assuming they are using the same antenna what is the theoretical range
> difference between the two and what is the practical range
> difference? it seems the price difference is 2X to 3X. is the price
> difference justified?
>
> i guess i'm asking "should i ebay the 10W unit and find a better use
> for the remainder?"

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.

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

[Morgans[_2_]]

"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.

But it will receive no better than the 5w radio.
--
Jim in NC

[Vaughn Simon]

"Kyle Boatright" > wrote in message
...
>
> 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.

>
> 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.

That is a small (but valid) consideration. Another valid consideration is
that the higher power radio will cause more interference. Particularly when
transmitting on Unicom channels that are reused by nearby airports.

[Robert Bonomi]

In article . com>,
Andy > wrote:
>i have a radio rated at 10W and a radio rated for 5W output. mine 5
>watter isn't a handheld but this is typical output for that genre.
>
>assuming they are using the same antenna what is the theoretical range
>difference between the two and what is the practical range
>difference?

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.

Good antenna's (and proper installations) make a bigger difference than RF
power. :) Years ago, I had a base-station installation that outperformed
virtually every other installation in the territory -- who were almost all
running 2-4.5x the power I was.

Now, "all else being equal", and for the same recieved RF signal level,
range will chage proportionally to the square-root of the change in
power level.

Caveat: 'all else' is *rarely* equal. <wry grin>

That said, the 10-watt rig would be expected to have an approximately 40%
greater working range than the 5-watter. *Assuming*, of course, that the
transmitter on the _far_end_ has sufficient power to reach _you_ at that
distance.

> it seems the price difference is 2X to 3X. is the price
>difference justified?

Depends on 'how badly' you need the extra range, doesn't it? *grin*

Only -you- can evaluate your needs/requirements.

>i guess i'm asking "should i ebay the 10W unit and find a better use
>for the remainder?"

[Gerry Caron]

"Vaughn Simon" > wrote in message
...
>
> "Kyle Boatright" > wrote in message
> ...
>>
>> 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

[COLIN LAMB]

"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

[RST Engineering]

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



"Andy" > wrote in message
oups.com...
>i have a radio rated at 10W and a radio rated for 5W output. mine 5
> watter isn't a handheld but this is typical output for that genre.
>
> assuming they are using the same antenna what is the theoretical range
> difference between the two and what is the practical range
> difference? it seems the price difference is 2X to 3X. is the price
> difference justified?
>
> i guess i'm asking "should i ebay the 10W unit and find a better use
> for the remainder?"
>

[RST Engineering]

Square root.

jw


"Kyle Boatright" > wrote in message
...

>
> My recollections from physics 20+ years ago is that radio wave strength is
> determined by the cube root of the transmitter strength.

[Wayne Paul]

"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/

[COLIN LAMB]

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

[Robert Bonomi]

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.

[Robert Bonomi]

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
>
>

[Scott[_6_]]

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
>
>

[COLIN LAMB]

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

[Scott[_6_]]

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
>>

[cavelamb himself]

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

[john smith[_2_]]

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

[RST Engineering]

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

[Wayne Paul]

Here are a few more references to add to the discussion.
http://en.wikipedia.org/wiki/Horizon
http://www.firestik.com/Tech_Docs/dist2horizon.htm
http://www.rudhar.com/vizhoriz/vizho-en.htm

Wayne
HP-14 "6F"
http://www.soaridaho.com/


"RST Engineering" > wrote in message
...
> 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
>

[cavelamb himself]

Wayne Paul wrote:
> Here are a few more references to add to the discussion.
> http://en.wikipedia.org/wiki/Horizon
> http://www.firestik.com/Tech_Docs/dist2horizon.htm
> http://www.rudhar.com/vizhoriz/vizho-en.htm
>
> Wayne
> HP-14 "6F"
> http://www.soaridaho.com/
>
>
> "RST Engineering" > wrote in message
> ...
>
>>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
>>
>
>
>

Ok, I think I got it figured out...

Nautical miles vs Statute!


Thanks guys.

Richard