## 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.

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'.

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.