How far can a 100 watt linear broadcast?
Anonymous
wattage? There must be some kind of math formula out there. Just
curious, Paul
Bill Nelson
: Any idea how to figure out how far you can broadcast with a particular
: wattage? There must be some kind of math formula out there. Just
: curious, Paul
Nope, no formula. In general, you just get the same distance as with a
legal rig. Your signal is just louder.
Of course, if the person at the other end does not have comparable
equipment, your linear is useless - as you would not be able to hear
the other person.
Bill
vinny
>Anonymous (nob...@REPLAY.COM) wrote:
>: Any idea how to figure out how far you can broadcast with a particular
>: wattage? There must be some kind of math formula out there. Just
>: curious, Paul
>
>Nope, no formula. In general, you just get the same distance as with a
>legal rig. Your signal is just louder.
Louder.. that means that is can be received where the legal signal
becomes too weak.
>
>Of course, if the person at the other end does not have comparable
>equipment, your linear is useless - as you would not be able to hear
>the other person.
Very important point: They can hear you, but the you hear them?
And remember, you have to double your output-power, to improve your
signal with 1 s-point.
>
>Bill
-------------------------------
The early worm has a deathwish.
Einstein is dood, Newton is dood,
en ik voel me ook al niet zo lekker.
E-mail :vi...@pi.net
Bill Nelson
: On 7 Jan 1996 07:46:46 GMT, bi...@PEAK.ORG (Bill Nelson) wrote:
: >Anonymous (nob...@REPLAY.COM) wrote:
: >: Any idea how to figure out how far you can broadcast with a particular
: >: wattage? There must be some kind of math formula out there. Just
: >: curious, Paul
: >
: >Nope, no formula. In general, you just get the same distance as with a
: >legal rig. Your signal is just louder.
: Louder.. that means that is can be received where the legal signal
: becomes too weak.
Not necessarily - or not enough to make much of a difference. 11 Meters
is basically line of sight. Once you can talk that distance, extra power
is not going increase that range.
Bill
Bert Hyman
> 11 Meters
>is basically line of sight.
I beg your pardon? Or doesn't the planet you live on have an ionosphere?
--
Bert Hyman W0RSB St. Paul, MN be...@winternet.com
David Toste
> In article <4cpfgq$c...@odo.PEAK.ORG>, Bill Nelson <bi...@PEAK.ORG> wrote:
>> 11 Meters
>>is basically line of sight.
>
> I beg your pardon? Or doesn't the planet you live on have an ionosphere?
>
And how often is 11m open?
It doesn't matter if you have 1 watt or 1kw, if the conditions are open
your going to get out.
--
David Toste [VE3TOS] Internet - aa...@freenet.toronto.on.ca
Don Mills, Ontario. SWLOGit - The Ultimate Shortwave Listeners
ftp.virginia.edu /pub/swlogit/ Software. (Fidonet: 1:250/930)
http://www.io.org/~saturn/SWLOGit.html (SWLOGit Web Page)
Drew Durigan
>your going to get out.
It matters a great deal! The more wattage you have, the better your
chances of being heard over the hundreds of other stations that will be
competing with you on the same frequency and are also attempting to be
heard!
-Drew in Charlotte-
Bill Nelson
: In article <4cpfgq$c...@odo.PEAK.ORG>, Bill Nelson <bi...@PEAK.ORG> wrote:
: > 11 Meters
: >is basically line of sight.
: I beg your pardon? Or doesn't the planet you live on have an ionosphere?
Certainly, but we were talking about regular propogation, and not skip.
The same thing applies. More power will not necessarily produce any more
distance. It WILL just make the already received signal louder.
There are cases where more power will produce multiple hops, where lower
power may not.
Bill
Roger
{:->>>On 7 Jan 1996 07:46:46 GMT, bi...@PEAK.ORG (Bill Nelson) wrote:
{:->>>>Anonymous (nob...@REPLAY.COM) wrote:
{:->>>>: Any idea how to figure out how far you can broadcast with a particular
{:->>>>: wattage? There must be some kind of math formula out there. Just
{:->>>>: curious, Paul
{:->>>>
{:->>>>Nope, no formula. In general, you just get the same distance as with a
{:->>>>legal rig. Your signal is just louder.
No formula is correct however you DO get more distance with MORE POWER.
Linears increase signal strength not modulation strength
It may not be the distance you may think when comparing a 100 watt to the
legal 4 or so but you can broadcast farther. Linears will help you talk skip
easier.
{:->>>Louder.. that means that is can be received where the legal signal
{:->>>becomes too weak.
{:->>>>
{:->>>>Of course, if the person at the other end does not have comparable
{:->>>>equipment, your linear is useless - as you would not be able to hear
{:->>>>the other person.
Not true, The linear won't be "Useless" Your equipment may be able to receive
better in an area with less noise than what's on the other end. Your linear
would be VERY useful in this situation allowing the other person to hear
you over the noise that would normally be washing you out.
As a point..I use a linear in the city where I live when I deliver papers.
my wife has a radio with a poor receive section. I sometimes have to
turn on my 100 watt for her to hear me. I can hear her just fine in almost
any area of the town.
{:->>>Very important point: They can hear you, but the you hear them?
{:->>>And remember, you have to double your output-power, to improve your
{:->>>signal with 1 s-point.
True.. Therefore you get more distance with your signal with more power.
example: 5 watts - S-3
10 Watts - S-4
20 Watts - S-5
40 Watts - S-6
80 Watts - S-7
A 4 S-unit increase in signal on a receivers end is a TREMENDOUS amount.
{:->>>>
{:->>>>Bill
{:->>>-------------------------------
{:->>>The early worm has a deathwish.
{:->>>Einstein is dood, Newton is dood,
{:->>>en ik voel me ook al niet zo lekker.
{:->>>E-mail :vi...@pi.net
Roger Randolph - Cheyenne Wyoming
rand...@tcd.net
Web Page -
http://www.tcd.net/~randolph/
Country Artist Fan Club listings and
Links to HitMakers WWW pages
Soon to have Scanner Freqs and Links
Jason Goldring
> In article <4cpfgq$c...@odo.PEAK.ORG>, Bill Nelson <bi...@PEAK.ORG> wrote:
> > 11 Meters
> >is basically line of sight.
>
> I beg your pardon? Or doesn't the planet you live on have an ionosphere?
>
> Bert Hyman W0RSB St. Paul, MN be...@winternet.com
Actually, the line of sight would apply in a weird type of way with respect to
the surroundings. Tests performed and documented by Ralph Engels, a professor
from the University of Toronto, has shown that smog, pollution, and other
man made anomalies can reflect a large portion of radio signals back to the ground
before it has a chance to bounce off the ionosphere, thus limiting the
communication to just past the line of sight. In actual applications, a 100 watt
amplifier broadcasting on 10 meters, compared to the same unit with an RF output
of 5 watts showed that their was only an increase of about 2 "s" units.
Regards...
N9QPI
(Drew Durigan) writes:
>Subject: Re: How far can a 100 watt linear broadcast?
>From: VUB...@prodigy.com (Drew Durigan)
>Date: 8 Jan 1996 21:50:11 GMT
One thing you all miss is that Clarity is the big difference from being
heard and not being heard. You can run all the power you want but if the
person can't understand you then you are basically not being heard. Alot
of guys are running Talkback equipment, Powermics that are cranked wide
open, Echo mics, etc. and they think this will make them be heard better.
A friend and I have been on 10 meters living only 5 miles apart me running
100 watts and he was running QRP 200mw and we were both heard in
Pennsylvania and the fella in Penn. said there was no difference in our
signals.
Your antenna is what makes the difference sometimes. You can run all the
power you want but if you antenna is not setup correctly then you just
will not get out like the other stations.
Jim, N9QPI
ROBERT WEBB
modulation is not there,It won't make any difference.
Search for mod,not wattage.
Chad Payne
There is not some math formula out there becuase conditions are always
different. There is so much more to how far your signal travels than
just wattage. In fact, simply putting a 100 watt linear on a radio is
not going to do a whole lot for you. The single biggest factor in how
far you signal goes is your antenna. Other factors are, the weather, the
terrain between you and the receiver, their antenna system, your antennas
height, and the modulation of your signal.
I know a guy who had a cheap radio and a cheap antenna. He could only
get about 2 or 3 miles. He went out and bought a 100 watt linear instead
of spending less money on a good anteanna, and because of his poor radio
and antenna, the linear just distorted his signal and he sounded like
crap. His signal still only made it 2 or 3 miles, but now no one would
talk to him because he sounded so bad.
Take my advice. Get a good antenna like an Antron 99, it will do far
more for you than a linear. If you are running mobile, a linear will do
even less for you.
Chad
5MT 127
Dan Alfonsi
>
> Any idea how to figure out how far you can broadcast with a particular
> wattage? There must be some kind of math formula out there. Just
> curious, Paul
The basic rule of thumb for power is: 4 times power increase = 1 S-UNIT
increase on the receive end (also = 6dB Gain). If you add a 100W amp, this
is about a 25 times power increase which is roughly 3 1/2 S-Units increase
on the receiver. This essentially means that if Joe was receiving you with
an S3 barefoot, he'd receive you at a little over S6 running 100W. If you
didn't move his needle barefoot, you'd move it to S3 with the amp. If you
wanted to add another S-Unit, you'd need to do another 4 times power
increase to 400W (it's logarithmic), so it won't buy you much to go above
100 (unless you go to a few thousand). Just for info, a good, 3 element
Yagi beam has about 9dB of gain (1 1/2 S Units increase) so you could put
up a good 5 element beam and get a few S-UNITS leagally without causing
havoc on your neighbors with RF interferrence.
As an amateur radio operator, we have power limits up to 2000 watts, but
very few actually use more than 100. I spend a great deal of time on VHF,
and usually run 10 watts. Many VHF and UHF operators brag on using less
than 1 watt on Walkie Talkies.
Also, you are looking at the 11 meter radio band. This is a fun frequency
range. 11 meters is basically line of site at night and during low
sun-spot activity cycles (such as now). Although 11 meters is not too
reliable for long-distance skip communications now, give it a few years,
and it will open up.
As far as range is concerned, it depends greatly on terrain (water, dirt,
wet dirt, desert, mountains, buildings, etc.). I've been able to talk 100
miles on a few watts from high on top of a mountain, and at the same time
only 20 miles with 100 watts in the city. There are Amatuer operators that
communicate several hundred miles on a few milliwatts using highly
directive antennas.
I know I didn't answer youre range question, but this may offer some
insight.
Dan
Dave O'Neal
>
>vinny (vi...@pi.net) wrote:
>: On 7 Jan 1996 07:46:46 GMT, bi...@PEAK.ORG (Bill Nelson) wrote:
>
>: >Anonymous (nob...@REPLAY.COM) wrote:
>: >: Any idea how to figure out how far you can broadcast with a
particular
>: >: wattage? There must be some kind of math formula out there.
Just
>: >: curious, Paul
>: >Nope, no formula. In general, you just get the same distance as
with a
>: >legal rig. Your signal is just louder.
>: Louder.. that means that is can be received where the legal signal
>: becomes too weak.
>
Meters
>is basically line of sight. Once you can talk that distance, extra
power
>is not going increase that range.
>
>Bill
sight. CB being around 27MHZ and lowest freq. thought of as
line-of-sight being 150 or 200 MHZ! CB is capable of talking around
the world. My formula indicates that 100 watts at 27MHZ should talk
about 50-60 miles with a 60 foot antenna. A mobile would be much less.
CB has a great deal of limitations to it's range because of all the
interference, but on a good day you will talk much further than line
of sight!
Bruce Herder
>
>vi...@pi.net (vinny) wrote:
>
>{:->>>On 7 Jan 1996 07:46:46 GMT, bi...@PEAK.ORG (Bill Nelson) wrote:
>
>{:->>>>Anonymous (nob...@REPLAY.COM) wrote:
particular
>{:->>>>: wattage? There must be some kind of math formula out there. Just
>{:->>>>: curious, Paul
>{:->>>>
a
>{:->>>>legal rig. Your signal is just louder.
the receiving end, you must QUADRUPLE your power... not just double it.
So... if 4 watts = 5 "S" units, then
16 watts = 6
64 watts = 7
256 watts = 8
etc.etc.etc.....
>
Robert V. Grizzard
: Nope, no formula. In general, you just get the same distance as with a
: legal rig. Your signal is just louder.
Actually, there is a formula. It incorporates transmission line loss,
antenna gains, transmitter power out, coupling loss and path loss, and the
result is Received Signal Level. It is pretty much useless at HF because
of the vagaries of propagation and the noise level.
: Of course, if the person at the other end does not have comparable
: equipment, your linear is useless - as you would not be able to hear
: the other person.
Yep. "Ya' Can't work 'em if ya' can't hear 'em."
: Bill
KBKJ5838
P2-11-46502
PG-GB-02791
KG7YY
ADOCS
Codys PC
question. It is only the answer to that question, that is difficult...
ESSAY=On.
THESIS=Enabled.
CREDIT_HOUR=Disabled.
FINAL_EXAM=Ask_Teacher_for_reconsideration.
Anonymous wrote:
>
> Any idea how to figure out how far you can broadcast with a particular
> wattage? There must be some kind of math formula out there. Just
> curious, Paul
Well, I'm no scholar, but I may be able to help answer part of the
question.
You must understand that there are many variables that affect the outcome.
There is no true equation, only "Fuzzy Logic". A basic answer is yes, how
far? Then read on.
A radiated signal contains both Magnetic and Electrial fields that
cris-cross
themselves perpendicularly as they propagate. The waves are a combination
of
these fields, that when they strike on, pass through, or conduct to a
metalic
surface, small currents and voltages are produced in the metal - we will
refer to this as the receiving antenna wires' signal. The transmitter,
produces the waves. Distance is the waves' free air space - which includes
ambient noises both man made and natural as well as obstacles that can
affect
the waves propagation through free space. The receivers antenna, and the
receiver itself, must amplify the incoming signal and the noise present at
that frequency.
Since we've established that, lets go onto some simple observations...
Light, -even though it is not being discussed here-, can be used as a
reference to the part I wish to address, that is, distance from the source
to
the objective [receiver].
Lumen developed a standard of Luminous Flux, called a Lumen/Candelas/Lux -
It
measured the amount of light striking a surface at a given distance from a
light source of known intensity.
What does this have to do with how far a radio can broadcast? Absolutely
nothing ;-), er at least with the properties of light versus radio waves.
However, I am referring that radio waves follow a similar process -
Strength
of Signal as a given point of distance from the source, as measured in dB,
as
a ratio of strength to the square of the distance. The dB of a signal is
not
measure from a 0 volt/amp point, but more of the ratio of the received
signal
from the noise level.
Signal [in dB] =20 Log[10] ([Power received from TX station in volts or
amps]
/ [Power received from ambient noise in volts or amps])
NOTE: This is assuming that both TX/RX impedances of the antennas are the
same, 50 Ohms. This measurement is in microvolts/microamps measured at the
RX
antenna at the frequency you want received.
We don't know how effective an antenna can radiate a signal, except for
physical measurement of signal strength, commonly referred to as field
strength from transmitter power, or actual Wattage the transmitter
produces
to force power from the radio itself into the coax then to antenna and out
to
the air. Also, Free air space resistance is affected by antenna height
from
both the transmitter and receiver standpoints. The attenuation of the
radiated signal of a given frequency in free air space and gain of TX/RX
radiated/received signal as referenced to an isotropic source also come
into
play.
So, we have several problems here. How much energy is being radiated into
free space? Plus, the distance traveled [How far?] - ambient noise [How
strong?]. And, how much gain is available from both the transmitter and
receiver antennas? But we do know how much signal we do receive from a
point
of reference as the distance from the transmitters antenna.
Codys PC
Many radios today fail at proper measurement of S-units in reference to
noise
level - they all measure RF from a 0 signal standpoint - they do not
factor
in noise level per se, but they are commonly used to refer to signal
strength
and the operator does not need to factor in these other elements. The
meters
use a Logarithmic scale measured in dB, but doesn't necessarily mean
actual
power radiated being received, it is only a reference.
If you're still reading this, ;-) there is a way to compute the amount of
received signal, then using that level, find the difference in reading a
signal strength of a stronger signal at the same transmitter station. And
then compute the measured dB increase that the amp makes using the same
distance. Since you didn't specify frequency, I can only give you an
example
for reference. If you need more info on formulas computing the actual
attenuation of a signal at a given frequency and distance, there are some
older trade journals still around, one is "Electronic Databook" by Rudolf
F.
Graf, from TAB Books, INC. ISBN 0-8306-1538-5.
For example. If your friend is receiving you at 1 S-unit on a 4 watt power
signal [referenced to the TX stations abiltiy to produce 4 watts], and you
wish to know the increase in dB as heard on your friends receiver, you can
read the following formula.
dB=10 Log[10][Pout/Pin] NOTE: Log[10] is used as common logarithm.
When you increase the transmitters wattage output [from 4 watts to 100
watts]
your increasing the output reference 96 watts, or 100w-4w=96w. Gain is
already computed as the receiving signal is at 1 S-unit, as well as
distance
and propagation ability for your friends receiver.
The ratio then is 100w:4w
100/4=25 then 10xLog[25]=13.9794 dB <- is increase of signal strength as
the
multiplier.
But you don't see that at the receiving antenna. Why?
Because of the INCREASE of 96w increase in output as referred to original
4
watts, so we really should recompute for 96 watts. However the formula
refers
to the TX station, not the RX station.
Because, he only increased the Actual Signal 13.9 dB, at his station, as
referenced to 0dB at 4 watts. Not at your location, he only made this
increase at the TX antenna, your receiver will pick up accordingly to a
log
of the increase at the same distance traveled to the receiver station.
Or, if you figure for log, RX S-units = Log[13.9 dB] = 1.143 <- this is
your
multiplier for the S-units received. Huh? You don't see that? Wait!
Please!
Read on.
If he hears you at 6 S-units [with amp off], 6x1.1143 [with amp on] should
be
6+ S-units. But you don't see that either. Remember Lumen?, the
brightness
of a light source decreases proportionately to the square of the distance.
So
you have to factor the actual dB is the difference in signal received from
the TX station at 4 watts versus 100 watts at a given distance or,
Received S-Units increase in dB =
10 Log[10][Watts[on]-Watts[off]]/Distance^2
Distance is measured any way you want, Feet, Miles, Kilometers.
For instance, in 2 miles of distance,
dB=10 Log[10](100-4)/2^2 or compute parentheses to (96w)/4.
96/4 = 24, taking 10xLog[24]=13.8 as the dB factor. You should see an
increase of this amount from the original measurement taken with the amp
off.
Remember, an S-meters accuracy fails at higher levels due to is
logarithmic
reading action, it is not linear.
For 4 Miles,
(96)/16 = 6 take 10xLog[6] = 7.78 as the dB factor.
The dB factor also shows a decrease in watts/volt/amp readings in field
strength as you increase the distance. You may not see this due to
variations
in the RX antenna field reception because of limiting factors, however,
this
will help give you a guide to figuring out what you need. [I hope].
As you can see, there is no real way to determine the actual amount of
power.
Only reference to it. The increase of signal from the amp is marginal at
lower receptio
Bill Nelson
: In article <4cpfgq$c...@odo.PEAK.ORG>, bi...@PEAK.ORG says...
: >
: >Not necessarily - or not enough to make much of a difference. 11
: >is basically line of sight. Once you can talk that distance, extra
: power
: >is not going increase that range.
: >
: >Bill
: Boy is Bill wrong!!!! He must be a ham! CB is far from being line of
: sight. CB being around 27MHZ and lowest freq. thought of as
: line-of-sight being 150 or 200 MHZ! CB is capable of talking around
: the world. My formula indicates that 100 watts at 27MHZ should talk
: about 50-60 miles with a 60 foot antenna. A mobile would be much less.
Before you spout off any more - I suggest that you do some research.
Certainly, it is possible to talk around the world on 11 meters. So
what? That is skip - not normal propogation - which is slightly over
line of sight.
Your "formula" is about right for normal line of sight propogation,
including a slight amount of "over the horizon" distance. The power
has nothing to do with it - except to make the signal louder.
If you don't believe me - dig out a reference and calculate the distance
to the horizon for an "eye height" of 60 feet. If you cannot find the
information anywhere else - look in a Nautical Almanac - which includes
tables.
: CB has a great deal of limitations to it's range because of all the
: interference, but on a good day you will talk much further than line
: of sight!
See above. I am well aware of the propogation characteristics of all
the HF bands. I have communicated with both Australia and Germany using
12 watts SSB and a good antenna.
Bill
Codys PC
Here is the last of the topic, there is some overlap so you can pick up
from
the last one. I didn't know of the truncation of this file until I u/l
the echo from the newsgroup. Without further ado...
The dB factor also shows a decrease in watts/volt/amp readings in field
strength as you increase the distance. You may not see this due to
variations
in the RX antenna field reception because of limiting factors, however,
this
will help give you a guide to figuring out what you need. [I hope].
As you can see, there is no real way to determine the actual amount of
power.
Only reference to it. The increase of signal from the amp is marginal at
lower reception levels - due to noise and distance, hence the logarithmic
scale. As you approach the TX station, you'll see the level
increase[naturally].
The dB level is in reference to distance from the station at given points,
with the amp on. With the amp off, the level with drop by that factor
divisible by the difference in received signal with amp on versus with the
amp off. Logarithmicaly speaking. ;-)
S-units, once again, are used as a reference, not a true dB meter to what
you
want to see. Your results will vary because of the ambient noise level at
the
receiver. S-units themselves are difficult to discuss as well for they too
have problems in presenting the operator the correct level received. As in
linearity, presence of adjacent channel intereference, noise level, gain
of
RF-amp and antenna, ground conductivity, propagation, etc, add to the
problem - hence the reference to Fuzzy Logic. ;-).
Another problem stems from propagation and free air space attenuation of a
signal at a given frequency. Lower frequencies tend to refract better,
and
have less attenuation. Higher frequencies tend to become line of sight
type
of transmissions and attenuation becomes much sharper. If you were to
figure
out how far a certain wattage goes, well this might help.
3 MHz - 60 dB of attenuation at a distance of 1 mile
30 MHz - 68 dB of attenuated signal at a distance of 1 mile.
300 MHz - 89 dB of attenuation at that frequency for 1 mile distance.
This information doesn't help unless you know how many dB of transmitter
power that is being transmitted and gain factors of TX/RX antennas, along
with Frequency and distance. There is much more information needed from
you to really answer your question.
If anyone sees something here that they wish to discuss further, please
let
me know. No flames please!
I hope I helped to answer a question.
Thank You.
:+> Andy <+:
Codys-You-Know,-I-Learned-Something-Today, PC
... Disclaimer:
This message does not promote or discourage the use of amplifers - Please
use them with common sense and with proper installation and operating
procedures.
vi...@pi.net
>It's a long dissertation...Please bear with me...
>
>Received S-Units increase in dB =
>10 Log[10][Watts[on]-Watts[off]]/Distance^2
>
>Distance is measured any way you want, Feet, Miles, Kilometers.
huh?
>
>For instance, in 2 miles of distance,
>dB=10 Log[10](100-4)/2^2 or compute parentheses to (96w)/4.
So if I use 3.2 kilometers instead of 2 miles, I get the same result? (none of
the other variables change)
>
>96/4 = 24, taking 10xLog[24]=13.8 as the dB factor. You should see an
>increase of this amount from the original measurement taken with the amp
>off.
>Remember, an S-meters accuracy fails at higher levels due to is
>logarithmic
>reading action, it is not linear.
>
>As you can see, there is no real way to determine the actual amount of
>power.
>Only reference to it. The increase of signal from the amp is marginal at
>lower receptio
I knew that.
jeffrey...@gmail.com
crin...@gmail.com
On 11m ssb I used to, in my cb days talk around the world on 6W.
Clownshoes
Exactly 12 feet, nine and three-quarter inches.
You're welcome.
- Clownshoes the CB Dude