48 Watt Solar Panel Arrives

My 48 watt solar panel arrived today. Hooray! I need to make a power harness terminated with an Anderson Power Pole but other than that, it is ready for action.
The unit is well-built and seems rugged, much like my 15 watt panel.
I had walked through the mathematics of these panels in a post back in October of last year, but it might be fun to quickly crank through the calculations again with the new panel. Here goes.
The panel has 12 areas with photo voltaic (PV) cells. Each area is approximately 31.5 cm wide by 18.5 cm high. Multiplying height by width we see that each PV cell area is 582.75 cm2. There are twelve such areas so 12 x 582.75 = 6993. We'll round that off to 7000 cm2 of active area for this panel.
There are 100x100=10000 cm2 in a square meter. Converting our area to square meters is done by dividing the active area (7000 cm2) by 10000 giving us 0.7 m2.
Assume that 1000 watts fall on each square meter of Earth on a nice day. The amount of power falling on the active area of the panel is 1000 w/m2 times 0.7 m2 = 700 watts.
It would be great if we could convert all that sunshine into electricity, but the panel is not 100% efficient. In fact, panels of this sort are only about 7% efficient. If we multiply the total power falling on the active elements of the panel (700 watts) by our efficiency (7%) we get 49 watts. Close enough!
What will I do with all this power? Operate portably a very long time. Certainly there will be some Georges Island operations. I also like to operate portably from hilltops and picnic tables. I managed a few QSOs from atop Mount Washington last July (though it seemed more like January up there!). This power means I can operate QRP all day with my K2, or operate most of a day at 25 watts with the IC7000.