OK, so here's a crash course "back-of-the-envelope" solar estimation. "Back-of-the-envelope" means in engineer-speak a very rough approximation to quickly estimate whether something is even close to feasible. It uses arbitrarily chosen reference points, and even some outright guesses. But that's the point: it's fast and easy. If the numbers land close to being feasible, then a proper design can be done in which components are chosen more carefully and the numbers are more reliable.

First, the fridge. It's the biggest consumer. I couldn't find any power consumption numbers for mini fridges on Home Depot's web site; the only data point I have is my own fridge sitting in the shop waiting for its day to be installed in my bus. It has been plugged into a Kill-A-Watt meter for a "long time" which now reads 99.5 kWh and 1342 hours -- its average is 1.8 kWh per day. Maybe a mini fridge will be 70% of that; 1.8*0.7=1.26 kWh.

Lights. Having no other data, I'll guess the bus needs interior light from 6-10 pm and 6-8 am. Maybe three of the 8-watt LED lamps will do. 24 watts times 6 hours is 0.144 kWh.

Phone. Negligible.

Computer. Again, very little data.. suppose 60 watts for 3 hours a day. 0.18 kWh.

Total daily load: 1.584 kWh. It's all 120 volt load, so it runs through an inverter. Suppose 85% efficiency, then 1.86 kWh are needed at the input (ie, from the battery).

Battery sizing. Make it big enough to run through two days, in case of cloudy weather and no solar production. Battery needs 3.7 kWh output. Suppose it's a 12v bank, then 3.7 kWh/12 v=310 Ah.

Battery type. Cheap lead acid? Don't discharge them fully. Normally the rule of thumb is to size it to double the anticipated load. In this case I already doubled the load to account for a cloudy day, so let's let the battery go to 33% state of charge on these infrequent days (you're not in Seattle, right?). Multiply up by 1.5; bank should be around 450 Ah. Check the instant power output. Trojan seems to recommend not more than 0.2C output, 450*0.2=90 amps. Times 12v is 1080 watts. The regular stuff, is well below this, but the occasional toaster oven or hair dryer, could be pushing it a little.. Charge efficiency. Lead acid battery charge efficiency varies inversely with state of charge (

http://windandsunpower.com/Download/Lead Acid Battery Efficiency.pdf). Suppose we go with 40% efficiency; then the 3.7 kWh output from this bank is going to require 3.7/0.4=9.25 kWh input. But it's only that bad if we need to catch up a full day's consumption with no solar input; let's allow two days for the catch-up so we need 1.584 kWh*1.5/0.4=6 kWh charging input.

Solar panels. Suppose we use 315 watt monocrystalline panels and we have 5 hours of good sun each day. The 6 kWh/315/5=4 panels. These panels are about 78x40 inches, so they'll cover 87 square feet and cost about 350*4=US$1400. Plus racks, wiring, charge controller, batteries, inverter......

Booyah and others -- how'd I do? Any corrections?