CONVERTING PFS TO KILOWATTS
Since I have no idea what FASA used as the basis for the "Power
Factor" this all a guess. Conversion is made even more difficult
by the fact that baseline items in Rigger 2 (such as electronics)
consume no power on their own. This is highly odd in my mind,
but was likely done for simplicity.
I based my conversion on reviewing the power output of Rigger 2 solar cells (p. 121, Rigger2) as compared to Vehicles 2 solar cells (p. 96, Vehicles2). R2 solar cells produce .007 PF/second x Body of vehicle compared to Vehicles2 (TL9) .08 kW x square feet.. Since body apparently also represents surface area AND mass it makes things difficult.
Based on TL9 solar cells I can assume a Body represents SOMEWHERE in the area of 10 square feet as far as surface area. Since a TL9 solar cell gives you .08 kW per square feet then we can figure out what a PF is (.08 x10 (1 Body point)=.8; .8 x3600 (number of seconds in an hour)=2880. 2880/15 (number of PFs 1 Body point gives you)=192. Voila! You just figured out how many kilowatts a PF is! Rounded down a bit you get 190 kilowatts per PF!
This is also borne out by Jon Szeto himself (the following quoted material is unedited from Mr. Szeto's original post):
To estimate the performance of a PF, I would benchmark it against a "core" electric vehicle that appeared in the BBB: the Mitsubishi Runabout (which appeared in the BBB1 and BBB2). Looking up the stats of the Runabout on p. 162 of Rigger 2, we find out the following stats for the Runabout:
Speed: 75 mpt
Body: 3
Fuel: 150 PF
Economy: 0.5 km/PF
Based on the Fuel and Economy, we can figure the operational range of the Runabout to be 75 km. To keep the math simple, let's assume that the economy is "highway" and that the Runabout is operating under "highway" conditions. (If you want to be specific, say we're running it on a test track.)
One of the descriptions of energy is the capability to do work. As basic high school (pre-uni for the non-Americans) physics teaches us (or did in my day) is that work is defined as the change in the total energy (potential + kinetic) of a system. In this case, the amount of work performed by the Runabout's engine is what's needed to bring the subcompact from full stop to full speed.
The basic formula for the kinetic energy of a body is 1/2*m*v^2. We know that v is 75 meters per combat turn, which translates into 25 meters per second (1 combat turn = 3 seconds, so to convert to m/s, you divide the Speed Rating by 3). Looking on the Body Ratings Table on p. 23 of Rigger 2, we find that Body 3 covers a range of weight from 200 to 750 kg. We estimate the gross vehicle weight of the Runabout to be somewhere in the middle, say 500 kg. So the kinetic energy of a Runabout running at its maximum sustained speed is (0.5)*(500)*(25)^2, or 156,250 joules.
But wait, there's more.
Not only is the electrical engine getting the Runabout from 0 to 75, but it's also fighting off the natural deceleration caused by various different factors (friction, air resistance, and so on). In my Honda Accord, I have noted that when I let the car coast, it decelerates at a rate of approximately 5 mph in 3 seconds (both in city and highway driving). That translates to a deceleration rate of:
-5 mph/3 s * 1610 m/mi * 1/3600 s/hr = 0.74537 m/s^2 (rounded to 5 places)
Of course, to counteract this natural deceleration, the engine has to provide enough force to provide equal forward acceleration. Again, harking back to basic physics, Force = mass * acceleration, so in this case, the force needed is:
(500 kg) * (0.74537 m/s^2) = 372.69 newtons
Also, basic physics tells us that another definition of work is force exerted across a distance. Since we've already figured the operational range to be 75 km, a simple way to figure work (if we ignore the effect of the Runabout accelerating to full speed) is
(372.69 newtons) * (75,000 meters) = 27,951,750 joules
So the total work done by that 150 PF engine is 156,250 plus 27,951,750, or 28,108,000 joules (rounded down to 28 megajoules).
Dividing this by the fuel quantity (150 PF), we come to a rate of 1 PF = 187 kilojoules (rounded off).
OK, granted, this is a very simplistic calculation and ignores a lot of extenuating factors. (Not to mention the mechanical efficiency of the drive system.) But it's probably close enough to get you in the ballpark.
Also, if you tried this same calculation on other sorts of Shadowrun vehicles, you will probably not get a close result. I suppose that you could blame it on the fact that, before Rigger 2, there really wasn't a consistent system for designing vehicles (or if there was, nobody told the current staff about it.) When I went about developing the construction system, I ended up benchmarking the various chassis and power plants against "core" vehicles that appeared in the BBB. Of course, YMMV.
So, 1 PF = approximately 187 kJ. So what? What does this mean in layman's terms?
Well, 1 watt-hour of electrical energy is equal to 3600 joules (power = work/time, so work = power * time, and 1 joule = 1 watt * 1 second). So 187 kJ = about 52 watt-hours. Which is slightly less energy consumed than a 60-watt light bulb shining for an hour. (Incidentally, the power units on a Pentium-type PC, so I'm told, are rated at 250-300 watts, so 1 PF would run a PC for about 10-12 minutes.)
In automotive terms, one horsepower is equal to about 745.7 watts, so 1 PF is the same amount of energy consumed by a 4-hp motor running for about a minute.
WALL CURRENT
This data is based on the recharge rate for the Redline laser
pistol (2 shots per hour). Since we have already figured out its
beam power and power requirements in kilowatts we can also use
that data to find out how many PFs/kilowatts you can pull from
a wall outlet (since we can easily assume the unnamed recharge
source is an power converter hooked to a wall socket).
Figuring PFs from Wall Current
Assuming the Redline is 400 kJ and requires 800 kilowatts of power
per shot we can figure out how much power wall current provides.
It recharges 2 shots per hour. So obviously the wall power must provide 1600 kilowatts of power over that hour. That's 8 PF if you use my 1 PF=190 kilowatts conversion.
But how much per second? Divide the amount by 3600 and there you go! In this case you get .4 kilowatts per second (400 joules). Or approxomantely .002 PFs/sec
Sources of Data
Guns Guns Guns gives some sample figures both for conventional
batteries and wall power that I found helpful. In this case the
amounts given above are on the REALLY low end for what they have
for household current.
G3 suggests (assuming current-tech power grids) 1800 joules/sec from wall current. That's 6480 kilojoules per hour, or in Shadowrun terms: 34PF per hour and .009PF/second.
Possible Reasons
SR notes that they use peak-discharge batteries for lasers. Since
obviously the batteries are not directly dumping into the beam
we have to assume that they power an HPG or a set of capacitors.
I assume the recharging limitations are a function of the batteries
themselves. But it's somewhat odd that you can't top off the HPG
or capacitors (which vehicles would probably do) on the Redline
or other manpack lasers.
Final
For "realism" I'd suggest
using the G3 suggested values from wall current. So 34 PF per
hour can be gathered from an average wall outlet before you trip
the circuit breakers. Areas with worse power grids (such as the
Barrens) will give you proportionally less PF and factories or
other locations with specialized power outlets can give you more.