*Technical musings*

This page is a bit rambling I admit, but I hope you find something educational in here.

I recently discovered that things are so much easier if you stick to one way of describing
power. Use decibels relative to 1mW, known as dBm, and the difference between transmitted
power and recieved power neatly sums up all the losses along the path.

**Transmitted Power** :
500mW is a power output of +27dBm, 27dB higher than 1mW (a milliwatt, a 1000th of a Watt; 1000mW = 1W).
This is 3dB less than (half of) a whole Watt,
i.e. 500mW is 0.5W... 1W would be +30dBm because 30dB is a power ratio of 1000.
Amateurs are used to filling in their logbooks with powers expressed in dBW,
just add 30 to convert to dBm ... 2W = 3dBW = 33dBm.

**Received Power** :
Typical levels will vary from -10dBm nearby another 446 radio, down to levels of -120dBm
where the power is so low compared to the general background noise around you (radio noise at that frequency)
and the noise levels in the radio's circuitry itself, that reception is marginal and would be lost
at further low levels. This difference between these extremes covers a range of 110dB, or in
other words one level is 100,000,000,000 times as big as the smallest. The way that radios cope with
this large range of signal levels is what makes radio possible in the first place, along with the way
that radio signals don't lose power travelling through space - if signal power died
away more rapidly with distance then radio would not be anywhere near as useful as it is. The only reason why
signals get weaker away from the transmitter (in freespace) is because the greater the distance the total amount
of power is spread over a larger area.
*Our ears cope with
a similar dynamic range of power levels of air pressure vibrations, 0dB being the threshold of hearing,
110dB being very loud and damaging to your hearing, but it's a good thing that energy is lost as sound
travels through the air (air must be compressed along the wave path) otherwise we'd be able to hear for
miles just like radio and the world would be a very noisy place!*
Radio waves propagate through free space
(even the vacuum of space, which makes you wonder what exactly is passing on the wavefront!) without loss,
so it's not taking any energy to propagate the wave. If 1W ERP is radiated in all directions, and a cube shaped
box surrounds the transmitter (in free space), then the total power 'illuminating' all 6 faces of the cube will be 1W
regardless of how big the box is (in fact even if the antenna is directional, the sum total is still 1W because
you don't gain any power overall). Radio waves never die, they simply spread out. Instead of using a cube, a
theoretical sphere is actually used to work out how much power 'density' there is at any given distance over
a square meter area, for theoretical field strength calculations.

**Path losses** :
The best radio path you can hope for is line of sight,
more or less "freespace" (but complicated by reflections). Freespace path loss represents
the decay of signal strength over distance, a curve on a linear graph, following an
inverse square rule whereby twice the distance gives four times less signal etc. This
means that there isn't a fixed number of dB loss for any given number of EXTRA distance of
freespace due to the logarithmic way that dB decibels work - the loss over the first 1km of a freespace
path is constant, but adding an EXTRA 1km to any given paths of various lengths will have
differing effects depending on the ratio of the lengths before and after the extra km.
The following formula gives the freespace loss from a transmitter TO AN ISOTROPIC antenna
(2.15dB less gain than a dipole, even radiation in all directions, impossible to achieve!) :

*Freespace Loss (dB) = 32.45 + (20 x log(FreqInMHz)) + (20 x log(kmDistance))*

**Formula Explanation**

The 32.45dB is what the loss is for 1MHz at 1km distance ( log(1) = 0 ), and the rest of it simply
compensates for different frequencies and distances with a squared relationship. If the
distance doubles then the loss quadruples, and a similar thing happens at higher frequencies
due to the smaller aperture of a simple antenna at higher frequencies. Let's see what happens
with 1W at 1MHz at 1km (in freespace). You're spreading 1W over a sphere of 4 x pi x 1000 x 1000 square metres,
12566370.61 of them. Each square metre gets 1/12566370.61 of the Watt, -71dB down from 1W (and indeed
the Power Density would be 0.08uW/m).
However, our 1MHz frequency has a wavelength of
300 metres, and even a dipole would be 150m long!
The aperture (wavelength squared, divided by 4 pi) of an isotropic will be 7161.97, giving us an advantage
of 38.55dB. -71dB + 38.55 means we have -32.45dB compared to 1W; the loss is 32.45dB

The formula is sometimes given for miles instead of km, using 36.6 instead of 32.45 - in this
case because miles are 1.609 bigger than km, it makes 2.59 times difference (1.609 squared),
which is 4.13dB ... 32.45 + 4.13 = 36.58

**Aperture?**

Aperture (or capture area) is a measure of how effective an antenna is at picking
up all of the power available in the space around it. The higher the frequency, the smaller
the wavelength and the size of a dipole, so a dipole will pick up less of the power 'in the air'
at higher frequencies.

It's interesting to note that frequency is irrelevant in terms of ERP and field strength transmitted.
Whatever the frequency, a dipole in freespace will put out the same ERP and the field at any fixed distance
(say 1km) will be the same regardless. 1W ERP on 80m will give the same field strength as 1W ERP on 70cm at
1km. However, a receiving dipole at that 1km point will pick up less signal at 70cm then it will at
80m because the antenna is smaller. This is why the freespace path loss formula (from one isotropic to
another) only includes the frequency term once - if the TX antenna size mattered it would be 40 x log(F), or
20 x log(F) + 20 x log(F), or 20 x log(F squared) ... all three are equal.

Although field strength is one dimensional (measured
in Volts per metre), power flux density 'in the air' is over a square area (as in Watts per
square metre). A dipole is one dimensional too (works along a line) and so there will be a square
relationship.. as wavelength doubles, a dipole for the relevant frequency has 4 x the aperture etc.

The formula is:
aperture = wavelength squared, divided by 4 pi. In a quest to understand this, I wondered
what would be the wavelength where the aperture is 1, and why. At 2m (150MHz) the aperture
is 0.318 (about -5dB). Halving to 1m (300MHz), the aperture is 0.079
(-11dB; halving the wavelength makes a 6dB loss). At 4m (75MHz) the aperture is 1.273 (+1dB),
so I was getting closer. Looking at the formula it became obvious that if the wavelength was
the root of 4 x pi, then squaring it gives 4 x pi, and any fraction with the same number above
and below the line equals 1. So the aperture is 1 when the wavelength is the root(4 x pi),
which is 3.545m (about 84.6MHz). Why?

The power at distance d is spread over a sphere of
surface area 4 x pi x (d squared), so obviously 4 x pi is important somehow. No books or webpages
really explained just what was so special about 3.545m, why the aperture was then 1! It took a day or two
of puzzling to figure it out, helped by a webpage that said the capture area of a 1/2 wave dipole
is a square with 1/2 wave sides (dipole along the middle, sides extending a quarterwave out).
I compared two formulas at 3.545m, where wavelength is the root(4 x pi), which can also be
expressed as 2 x root(pi). At this wavelength, the surface area of the smallest sphere into which
a dipole could be fitted, works at a pi squared (quarter wave is root(pi)/2, square it and
you get pi/4, times 4 x pi, equals pi squared). A dipole would be root(pi) long, and the square
around it would be pi square metres; if this is squared again, the two match - does this explain
the aperture of 1?

I tried the two formulas for 4m. Sphere area = 4 x pi. Dipole square = 4, squared again = 16.
How nice it was to find that 16/(4 x pi) = 1.273 ... 1dB more than an aperture of 1! It's hard
to intuitively understand what's going on, but at least the numbers fit :o)

Along the way I discovered that aperture = (sphere surface area around the dipole) / (pi squared)
... which added to my confusion for a while! Also the formula for the sphere surface area can be
boiled down to (pi x wavelengthsquared)/4, and by coincidence the surface area of a cylinder at 1/4 wavelength
around a dipole (minus the end circles) is the same. Who said maths was dull, LOL!

Aperture is therefore the 'absolute gain' of an antenna, however the gain of an antenna is usually compared to
a dipole or isotropic at any given frequency. Apertures and capture areas are seldom mentioned
anywhere (the freespace path loss formula takes care of most concerns) so there's no need
to lose any sleep over it.

Freespace path loss at 446MHz is 85.4dB at 1km. Comparing to 1km, the loss is
20dB less at a 10th of the distance (65.4dB at 100m), and another 20dB less when another 10 times
closer (45.4 at 10m). Further away from 1km, the freespace loss is 6dB more at 2km, 12dB more at 4km, 20dB
more at 10km, 40dB more at 100km, 54dB more (139.4dB) at 500km. With +27dBm to play with, and
a sensitivity of -119dBm, we could stand to lose 146dB at a maximum in space.. which would take 1069km!

So PMR446 should work perfectly well to and from the space shuttle in Low Earth Orbit (500km approx?),
assuming you had the channel clear - no-one else using it at all - because the shuttle would probably
just receive a load of noise on PMR446 from all the concurrent transmissions. It may work to the shuttle
in theory, but not very well to the geostationary satellites (like Sky TV's Astra) at 35,786 km. Here's where the
square law actually applies for once : 35 times the range (35,000 versus our max of approx 1,000km)
would require 1225 times the power - 600W on a handheld would be a bit daft!
Or an improvement in antenna gain of 31dB would achieve the same, although that would take quite a dish..
a typical satellite TV dish had a gain like that (but only for its 11,000MHz band), so now you can see how
*that* works, I hope (the dish would have to be 25 times the size of a TV dish to have that gain at 446MHz!).

At any given distance from a transmitting isotropic antenna, say 10km (IN FREESPACE), the
total amount of output power is distributed around an imaginary globe of the example 10km radius, regardless
of frequency. The overall loss only increases with frequency because the receive antenna is shorter the
higher the frequency, so less of the available power is captured.
I don't like the way the above formula is called 'path loss' formula, think of it more
as 'path loss to a simple antenna'.

For comparison, freespace-to-simple-antenna loss at 145MHz over the same 1km freespace
path would be only 75.6dB,
some 10dB better (almost 2 S-points). 50MHz is another 9dB better, and 27MHz yet another
5dB better. 934MHz is 6dB worse than 446MHz. In fact from one location to another
visible location a genuine 1W ERP on 934MHz should in theory be received a staggering 30dB lower
than a genuine 1W ERP on 27MHz on similar gain antennas - 5 S-points - simply because a 934 quarterwave
is 34 times smaller than a 27 one, the effectiveness different by 34 squared, = 1154, = 30dB.
This is part of the reason why lower frequencies are more effective for mobile comms in large rural areas.
Despite the band in use the antenna is probably going to be about 1m in size where possible, but even though
a UHF antenna this long is a collinear with 'gain', and an HF antenna this long is very lossy, the low
bands win because of the squared frequency aspect. Also, lower bands get about hilly areas
and through vegetation more easily too. The amount of 'flutter' on mobile 934 was horrendous compared to
the more 'solid' 27MHz (things are different in urban areas where there may be more noise on low bands,
and higher bands gain an advantage from being reflected more easily from buildings, but even that didn't
help 934 much). Mobile phones on 950 MHz only succeed because of well sited base stations, and often large
numbers of fill-in cells in troublesome areas. *If you can access the hidden 'netmonitor' screen on your
GSM phone which shows you dBm signal strength (and channel numbers, what fun) of the current cell and the
next few strongest, you will find this is more educational than years of using CB and ham radio. It ranges
from a maximum end-of-scale reading of -20dBm standing right by a hidden cell, to -40 to -50 driving past
a cell tower, rapidly diminishes away from line of sight, all the way down to -110dBm which is the cut-off
point where a cell is deemed too weak to use.
It really
is an eye-opener and I wish more FM ham radios would give dBm readings rather than S meters. For more information
try a Google search for "netmonitor" and your make/model of phone, some need software to enable it whereas others
are just a few secret button pushes away... i.e. http://www.gsm-support.net/shop/page.php/phone_codes.inc/_61/en
... it's nice to see how much signal you're bathing in at any location, and how you get more 'cancerous radiation'
(LOL!) walking past a picocell hidden in an urban street than you get when 100m away from macrocells on a big
conspicuous tower! Anti-mast campaigners should try it, they'd probably find that the mast half a mile from their
kid's school is giving much less signal then they think, thousands of times weaker than that transmitter hidden
in the petrol station forecourt sign.
It's also nice to see radiation patterns at work - more signal 100m from a collinear pole than you get when
standing underneath it. Walk away from picocells and watch the dBm falling while you count paces... it's a better
illustration of UHF propagation and the inverse square rule than most of what amateur radio
has to offer without a lot a mucking about!
*

Fans of the old 934MHz CB system should note that 0.5W on 446 should perform as well as 2W on 934
(and only half an S-point lower than 4W on 934) - but 446 gets through terrain far better than 934.
*(you can compare frequencies
using 20 x log(FreqInMHz) ; i.e. about 53dB for 446MHz, 43.2dB for 145MHz)*

Note that radio waves never die in freespace, they are
simply distributed over increasingly large areas such that a given antenna picks up
less of the available signal at further distances. In fact, 'path loss' is called
'spreading loss' in books dealing with RF link budgets.

If you calculate the surface
area of the sphere for any given radius (4 x pi x radius squared), then divide the total
power output by the number of square metres, that gives you an idea of the probable
maximum field density in watts per square metre (discounting additive reflections in the real world).
You may even see square centimetres used
instead of square metres, just to give us 40dB more conversion fun with 10,000 sq.cm to the sq.m :o)

At 10 metres from an antenna the signal would be covering a sphere of area 1256 sq. m.
Any one square metre section of this sphere (slightly curved but don't worry about that)
will be getting 1/1256 worth of the signal; 0.079 percent of it - a loss of 31dB already
(calculate 1/1256, [log], x 10). So a 446 radio putting out 500mW equally in
all directions would generate at most 0.4mW/sq.m at 10m.
The aperture of even an istropic, 2.15dB less than a dipole, (aperture = wavelength squared
divided by 4 pi) at 446MHz is about
-14dB, so from the starting point
of +27dBm the signal level (in the receiver) 10m away is already 45dB lower at -18dBm.
(*plug 446 and 0.01km into the freespace formula and you'll get a result of some 45dB too*)
This is
still 75dB more than S9 (-93dBm) level! At 1km away (100 times as far) the signal will
be 10,000 times less (-40dB), still S9+35dB. At 100km away, 10,000 times less again (40db less),
and we've still got -98dBm : about S8 (-99dBm). No wonder DXing from mountain tops seems to work!

**Power Density v Field Strength**

We saw above that any 1W ERP in freespace gives 0.08uW/m power density at 1km distance,
due to spreading over a sphere of 4 x pi x 1000 x 1000 square metres.
Instead of working in
square meters we can lose a dimension and work with the Field Strength in a line between two
points a metre apart, a voltage per metre. O level physics may/would have taught you that

Power equals Current times Voltage, and that :

Voltage equals Current times Resistance (or impedance of course).

The latter can be rearranged such that Current equals Voltage over Resistance. This can then be substituted into
the first formula - replace Current with V/R and you get : Power equals V/R x V ... or in other words :

Power equals Voltage squared divided by Resistance.

A rearrangement gives you Voltage squared equals Power times Resistance, and so finally :

Voltage equals the square root of (power times resistance).

So we can convert from
power flux density (Watts per square metre) to field strength (Volts per metre) because
we know that the impedance of free space is 120 x pi, 377 Ohms. With 8x10-8 W/m then, multiply by 377
and get the square root...
this gives the Electrical Field Strength (E) of about 5.5mV (0.005477 V) per metre.

There is a direct formula E = ( SquareRoot(30 x WattsPower x NumericalGain) ) / MetresDistance, the 30 can be
explained because there would originally have been a '120 x pi' on the top of the formula (impedance of
freespace) and a '4 x pi' on the bottom (a circle thing!). Naturally, the root of 30 is 5.477 and
divided by 1000 this is 0.005477, so
you can see it agrees with the indirect method of starting from a power density and converting.

**Antenna Factor**

Antenna Factor is the ratio of the field strength (volts per metre) to the load voltage (usually 50 Ohms).
If the level in the coax is half the 'voltage in the air' then the AF is 3dB.
So this is similar to aperture (or capture area) with smaller antennas at higher frequencies picking up less
voltage from the field, but now working with volts/m instead of power levels in dB directly.
A formula I've found says the antenna factor and the gain can be related to each other :

G = 20 Log (F) - AF - 29.79

Where G = Isotropic gain in dBi, F = Frequency in MHz, AF = Antenna Factor. A rearrangement gives

AF = 20 Log (F) - 29.79 - G

As you can imagine, I'm keen to find out why there's 29.79 in there! A 145MHz dipole (2.15dBi) would
apparently have an AF of 11.29dB, and an isotropic at 446MHz would have AF = 23.2dB (it's a loss so it's
really -23.2dB).

Any given power will give a different voltage depending on the impedance.
The conversion from 377 Ohms in freespace to the voltage in a 50 Ohm load gives
an apparent drop of voltage of around 8.75dB in power terms - our 5.5mV/m field strength is
0.08uW which would be 2mV in 50 Ohms
(try 0.08uW times 50, then take the square root), from 5.477 down to 2 mV is a 'drop' of 8.75dB (the power
hasn't dropped, only the voltage has because of the impedance change, and we've converted a voltage change
to a power change in dB - remember to multiply the log(2/5.477) by 20, not 10!)

So, combine the apperture dB figure with this 8.75dB and you get the AF.
Aperture at 446 = -14.44dB, -8.75 = -23.19 etc. With
slight rounding errors, aperture at 145MHz = -4.67dB, -8.75 + 2.15 = -11.27 ... so that explains it well enough.
The puzzle over the 29.79dB figure is solved now : the aperture at 1MHz is 38.5443, take away 8.7506 .. result!
It's closer to 29.790 than to 29.800 only if you calculate wavelength using the real speed of light (i.e. 299.79)
rather than 300 when dividing by MHz (which gives 38.5503 for aperture instead). If you truly appreciate dB, you
will laugh at 0.01dB accuracy, but it was all part of the pleasure of unravelling the formula, hehe.

However, one mystery remains. I found an AF formula that says that :

AF = 9.73 / (metresWavelength x root(numericalGain))

and this gives 14.4 for 446MHz and 4.7 for 145MHz isotropics - which look remarkably similar
to the aperture figures!!

After all that, lets look at the case at 2 metres wavelength, using a dipole. In a field of X mV over
a distance of 1 metre, with an antenna 1 metre long... even taken into account the gain of the antenna
and the impedance conversion, it's still a few dB from what you might first expect. Ah, the joys of radio!

Freespace losses under line-of-sight conditions are all very well, but
actual losses at ground level in typical terrain are much higher, leading to the kind
of results presented in the above table. Just as you get more Miles Per Gallon on the open
road than you do around town, so does 446 perform worse in built up areas - although obviously
for far different reasons!.
Getting a signal from inside a car to another
car just one mile away can involve the kind of losses associated with hundreds of miles
over a visible path such as mountain top to mountain top or to/from a satellite.
It's only when there's nothing in the way that really impressive distances are achieved.

*More power won't help as much as some say!*

You might get S4 level (-123dBm; S-points are 6dB) from someone 1km away in real life,
so compared to what the signal would be in freespace (-58dBm) - somehow that 1km of
real world has cost you 65dB! At this point, the myth is that four times the power gives
you twice the range... not really! Another load of similar terrain will cost you another
65dB, so 6dB more power isn't really going to help (a forest will attenuate
by about 100dB/km at 70cm :
*"According to a CCIR report (1145), the attenuation is of the order of 0.05 dB/m at 200 MHz,
0.1 dB/m at 500 MHz, 0.2 dB/m at 1 GHz, 0.3 dB/m at 2 GHz and 0.4 dB/m at 3 GHz.
At lower frequencies, the attenuation is somewhat lower for horizontal polarization than
for vertical, but the difference disappears above about 1 GHz."*
- http://www.tapr.org/tapr/html/ve3jf.dcc97/ve3jf.dcc97.html
). You also have to bear in mind the
'radio horizon' from one height at one end, to the other height at the other end, due
to the curvature of the Earth. If the two ends wouldn't be able to see each other on a
smooth featureless Earth, you've no chance; no amount of power will help much. It's no
wonder they put broadcast sites up on high towers, and successful VHF/UHF amateurs use
beams on towers too!

Range in nautical miles = 1.23 * (sqrt(height1) + sqrt(height2) )

This gives 5.5 n.miles for two 5 foot heights (handhelds).

- 10.1 km, 6.3 miles.

*(1n.mile=1.852km=1.1508miles, 1mile=1.609km)*

5W may seem wonderful with 10 times more power than 446, but you might not get
the extra 3.16 (square root of 10) times the range! That inverse square rule
ONLY APPLIES TO FREESPACE spreading. As soon as you have losses on this planet, the losses
mount up per obstacle.

Think of it this way... you have a light and a piece of dark glass. However strong
the light is, the other side of the dark glass gets a 1000th of the intensity.
So if the light is a metre away from the glass, at the other side of the glass you've
got 30dB less light for starters. Quite apart from spreading loss, if you put another piece of
dark glass another metre away, you'll have at least -60dB behind THAT. Quite clearly,
(or dimly, rather!) the signal 2m away is certainly not 4 times less than the
signal 1m away from the light. The extra losses all add up in the same way as loss
per metre in coax cable. If you double the length of a piece of cable, the overall
loss is not 6dB less.. it depends on the loss per metre, and the lengths. So, boosting
your power to 5W is going to drain your batteries a lot quicker, but the effect on
range is going to still depend upon the terrain. Yes, at the point where things are
getting tricky with half a Watt, the extra 10dB will give you 10dB more signal (an S
point and a bit) but that's soon likely to be eaten up by a bit more distance over
similar conditions. However, if that extra distance is unobstructed open land, it
may well help, but then again there wouldn't have been much to block the borderline
500mW signal either!

*'Tropo' helps..*

Long distance contacts on UHF happen by either (a) being very high up, with more line-of-sight
range, or (b) having some help from unusual weather conditions down near ground level.
Atmospheric conditions can sometimes bend and focus
signals within the first km or two of air above us, back down to the ground - when they
were otherwise heading up into space. Such 'lifts' are quite common in the summer (being associated
with high
pressure) although some of the best have occured at all different times of the year. We may
get a couple a really good lifts per year on average, with a few more average ones (but still
worth having).

This happens at VHF and UHF (30 to 3000MHz) via layers of air in the lower atmosphere
(troposphere). The air (with its moisture content) has different densities (directly related to temperature) at
different altitudes.
The large volume of air between you and
a distant radio station already slightly bends radio signals to help them reach just over the
horizon, no matter what the weather's doing. The rare conditions simply have
a better arrangement of air densities that can provide a signal
path that isn't normally possible - often refered to as good 'tropo', or a 'lift' in
conditions. This is due to the refraction (change of direction) that occurs when a ray of light or radio wave
passes through a change of density - as with mirages, or as with light exiting from water into the air whereby
trying to spear a fish
becomes tricky because the fish isn't exactly where it appears to be.
In a high pressure weather system there is a downward
movement of air which leads to the more rapid change of density with height, and more refraction occurs.
Similarly at sea the warm evaporating air can be held near the surface leading to 'ducting'
which keeps a signal at a
certain level when it would otherwise head upwards as the Earth/sea curves away. Larger
than normal distances have been achieved over sea paths in this way.

For an idea of whether a 'lift' is likely, see the excellent forecasts at
home.cogeco.ca/~dxinfo/tropo_nwe.html

UHF signals are not known to be affected by 'skip' from the D, E or F layers (over 80km up) that are
so useful at lower frequencies, due to free electron densities varying markedly at certain heights.
Also, you're unlikely to enjoy Meteor Scatter or Auroral propagation
on PMR446 radios - and certainly no moonbounce!

**Signal Levels**

This table shows what the actual power levels are, using the common measurement
of dBm (decibels relative to one milliwatt). S-meter points are spaced 6dB apart.

dBm
+27 max 500mW ERP 446 output
+17 typical 446 with short antenna
0 1mW (reference for dBm)
-18 max 446 from 10m (freespace)
-58 max 446 from 1km (freespace)
-93 S9
-99 S8
-105 S7
-110 max 446 from 400km (freespace)
-111 S6
-117 S5
-119 typical 446 sensitivity for 12dB SINAD
-121.4 thermal noise floor (FM Wide 180kHz)
-123 S4
-129 S3
-135 S2
-133.6 thermal noise floor (FM 11kHz bandwidth)
-139.5 thermal noise floor (SSB 3kHz bandwidth)
-141 S1
-147 thermal noise floor (CW 500Hz bandwidth)
(Geocities upload mangles a column around here..
- it looks fine here!)

Reception is limited by background noise. Even a simple resistor is always
generating tiny amounts of noise, due to all those electrons naturally whizzing around within it.
Boltzmann's Constant says that the lowest noise level we can hope to achieve
is -198.6 dBm per Hz at 0 degrees Kelvin. If we use 17 degrees C for our normal world purposes,
that's 290 deg.K (24.6dB more noise power), so the noise is now higher at -174dBm per Hz.
If we assume that PMR 446
allows a maximum modulation freq of 3kHz, with 2.5kHz deviation, the bandwidth is 11kHz (just fits
in the 12.5kHz channel raster). So, 10 x log(11000) is 40.4, so the best noise floor we'd ever
get for 446 would be 40.4dBm more than -174dBm, which works out at -133.6dBm.

Connecting an antenna will cause the noise level to increase even more.
The additional noise can come from the sky, the ground, local man made noise and a combination
of mush from all other transmissions around.
It is all added to the receiver input noise floor - and then we have to consider the
noise factor of the front end stage of the radio too!

FM Reception is possible when the incoming carrier signal is 2 to 6dB or so higher
(as in Carrier-to-Noise) than the receiver's noise floor (*if all the webpages I've read could
agree, life would be simpler!*).
If a receiver has a noise figure of 10dB, then the incoming carrier has to be 12dB above our -133.6dBm
mininum noise level. This required level of signal higher than the noise refers to the radio carrier signals,
not the S/N Signal-to-Noise
ratio of the recovered (demodulated) audio which is much higher due to the way audio is recovered
from the FM carrier.

Receiver specifications usually mention a '12dB Sinad',
SINAD meaning Signal over Interference, Noise And Distortion - 12dB of difference between DEMODULATED
signal and other noises, gives comfortable readability.
Other specs may use a 20dB SINAD, or specify
a number of microvolts intead of -dBm, just to confuse!
UHF receivers generally resolve signals as weak as -120dBm or less.

Sensitivity equivilances
(probably in this range for a 12dB SINAD
or several dB higher for a 20dB SINAD)
dBm ---(50 Ohms)--- (75Ohm) dBW
-115 0.398uV -8dBuV 0.487uV 5dBf -145
-116 0.354uV -9dBuV 0.434uV 4dBf -146
-117 0.316uV -10dBuV 0.387uV 3dBf -147
-118 0.282uV -11dBuV 0.345uV 2dBf -148
-119 0.25uV -12dBuV 0.306uV 1dBf -149
-120 0.223uV -13dBuV 0.273uV 0dBf -150 (femtowatt)
-121 0.2uV -14dBuV 0.245uV -1dBf -151
-122 0.178uV -15dBuV 0.218uV -2dBf -152
-123 0.158uV -16dBuV 0.194uV -3dBf -153
-124 0.141uV -17dBuV 0.173uV -4dBf -154
-125 0.126uV -18dBuV 0.154uV -5dBf -155
-126 0.112uV -19dBuV 0.137uV -6dBf -156
(last column goes from -145 to -156dBW ... don't
blame me for the Geocities upload which somehow corrupts the file
and repeats columns in PRE-formatted sections!)

(note that hifi FM tuners usually have 75 Ohm inputs, and Power=VoltageSquared/Impedance, so microvolt levels
will be different; 0.2uV would be -122.7dBm. Instead of 12dB SINADs, the IHF (Institute of High Fidelity) rate
for weak mono signals, 'useable sensitivity'
with a -30dB level of total harmonic distortion plus noise (THD+N), that is a mono signal with 3% distortion.
)

(note that TV reception requires between 60 and 80 dBuV (75 Ohms) (or 0 to 20 dBmV ... a mV being 1000 times
higher than a uV means a 60dB power difference) for analogue, and about 20dB less for digital. Ideally all channels
should be within 6dB of each other, and 80dBuV is the maximum before overloading. From 60dBuV downwards picture
quality suffers. Analogue needs a 44dB C/N ratio. Thermal noise floor in a 6MHz channel is about 0dBuV.
)

So, given the +27dBm output of PMR446, losses of up to a hundred or more dB still allow enough
signal to permit communication.

(Here's a link to more about receiver specifications
- missing, so feed that URL into http://www.archive.org !)

Real long distance stuff will all be down amongst the noise, which is why radio amateurs
use CW and SSB on VHF and UHF - better than FM because you don't have to have all those extra dB
of carrier strength before you can hear the DX. An typical FM amateur with
a lowish gain vertical antenna will hear 'anything that's up there' so long as it's a
certain strength. The more enthusiastic amateur with better modes and a beam will be able
to focus in on weak signals that may not even be detectable on FM, and they will work a LOT more distant DX.

If you're really serious about radio, you will want to get an amateur licence. However,
it's still a valid and enjoyable exercise to try and squeeze the most out of PMR446!
Good paths from high locations greatly extend the range, and tropo lifts can bring in
signals from hundreds of miles away with signal strengths strong enough for FM, so
DX is quite possible.

On 446, the 0.5W ERP specification means that larger antennas would allow a
lower power requirement for the same output, with less battery drain, and improved
reception as well. The better the antenna, the better.
Some reviews suggest that the really expensive professional 446 radios will make the distance
to up to 1 mile when these are making it to but half a mile. Other reviews claim that the kind
of typical 'rubber duck' antenna on a cheap 446 radio performs 10dB worse than a full 16cm
quarter wave as used on the more professional radios.
"This means for our PMR446-radio set (5 cm aerial) that its range with the original spiral aerials
is only 30 % of what it would be if it were fitted with lambda/4-antennas." -
http://home.foni.net/~michaelbosch/umbau_gb.doc

-10dB compared to a quarterwave would indeed reduce one radio's range by the square root
of 10 in FREESPACE, but here on the ground it just means you've got 10dB less to play with.
(The lousy antenna also reduces the reception abilities, so a poor pair in freespace would
suffer x 100 power wise, having just a tenth of the range of good pair of quarterwave-endowed radios!)

It all gets a bit subjective and not especially scientific, because the terrain is still
the most important and variable factor. But if this really is the case it would mean that the Effective
Radiated Power of the worse radios would be just 50mW - a 10dB reduction that equals a drop of one
and a half S-points on a radio signal meter. *(Recent discoveries of FCC testing of FRS radios shows
this is indeed the case, with many radios (models used on 446) having quite pitiful ERPs of less than 50mW!)*
Bear in mind that at the
half a mile point where one radio is breaking up, the other radio only really has 1.5 S-points
advantage, which is soon lost as range increases. Moving a radio around slightly for
better reception (and better transmit back too) can have far more effect.
Bear in mind that air-air users have reported 40km range with just 10mW radios (433MHz LPD),
which are 17dB less powerful than the maximum allowed 500mW on 446!

If radio A has four times the effective output than radio B, then at any distance from them
both you'd get four times more signal from radio A (one S-point better). The only time this really helps is
when the signal from radio B is JUST too noisy, at which time the signal from radio A can still JUST be heard.

It isn't ideal to listen out for other 446 activity, it has to be said you'd pretty
soon tire of all that noisy, choppy reception anyway. In most areas you have to be quite
close to 446 users to get consistent signals without it breaking up all the time. 446 is
not ideally suited to mobile use unless you're keeping in touch with another vehicle heading
the same way - it could work well to base stations having rooftop antennas if that was allowed,
but handheld-to-handheld is always much more limited on any frequencies. Having said that,
446 beats the living **** out of 27MHz CB handhelds.

Some people worry about ill effects from cellphones. PMR446 uses frequencies
half those of 890-960MHz GSM - the wavelengths being twice as large. But the power
output is constant while you transmit, and of the same order of field strength.
Best not to put the antenna too close to your head, just in case.

At 446MHz, the ICNIRP limit for RF exposure is 2.23Watts per square metre.
We calculated earlier that a 446 radio putting out 500mW equally in all directions
would generate at most 0.4mW/sq.m at 10m. At 1m there would 100 times as much,
40mW/sq.m. Unfortunately we can't zoom in any more, because the 'path loss'
formula only applies to 'far fields' and not the 'near field' right by the
antenna. At this point I'm stumped for now, but if I find out more I'll add to
this page for sure! I doubt that sensible use of a 446 at least 10cm from your
grey matter is going to cause any problems.

An interesting PDF document all about 460MHz links is quite educational :
http://www.woodanddouglas.co.uk/products/download/70000000.pdf