>>198
AA would last 0.7416 hrs ( (2.255 Wh)/ 3W, you do not combine each cells' in the overall Wh = C, since they running at the same (variable) current).
What?
No!. The cells' energies ("Wh") add up. If one cell has 3 Wh and I give you another cell that has 3 Wh, then you have 6 Wh of energy that happens to be in two cells. They add up.
Also you mistyped "2.255" as "2.225" in your calculator, the correctly-calculated but theoretically incorrect answer would be "0.7516".
Another way of seeing this:
The number of coulombs (or amount of current) that will exit the battery of cells (in series) is the same as the one exiting a single cell (because they are mounted in series). However, the voltage of the battery of cells is the sum of the voltages of each cell (once again, because they are mounted in series).
Therefore the energy in a battery of (identical) cells is the number of coulombs provided by each cell (Ah) multiplied by the sum of the voltages, and thus is equal to the number of coulombs per cell times the voltage per cell times the number of cells, which is then equal to the energy per cell times the number of cells.
Mathematically: let
Ucell be the nominal voltage of a single cell and let
Qcell be the charge ("Ah") the cell can deliver. Then:
Ecell =
Ucell ⋅
Qcell
Now suppose we have a pack of
n cells mounted in series:
Upack =
Ucell ⋅
n
Qpack =
Qcell
Epack =
Upack ⋅
Qpack
= (
Ucell ⋅
n) ⋅
Qcell
= (
Ucell ⋅
Qcell) ⋅
n
=
Ecell ⋅
n
Therefore the energy of the pack is the energy of a single cell times the number of cells. QED.
It still means you will consume 13.53 Wh on your house bill when you recharge AA, than PP3 at 1.395 Wh.
Yes, and the AA 6-cell battery pack it will last 9.7 times more than the single PP3. In other words what you're saying is that you have to use more energy to charge larger capacity batteries that last longer. I think that's rather obvious. Also you have to recharge the 6-cell battery pack 9.7 times less often than the single PP3.
If you are too set to use AA, it shall be.
I am really set on it because I am absolutely convinced that PP3 is a really terrible choice as justified in my previous posts. PP3 generally used for applications that need higher voltage but not that much energy. It is a tradeoff between higher voltage and diminshed capacity. In fact, a PP3 cell is larger and bulkier than a single AA cell yet it has only 60% of its energy (as I stated previously).
So no ethernet‽
Wait, didn't you say "a single gigabit ethernet"?
Only 4 USB hosts‽
So 2 internal and 2 external? What if someone wants to have wifi *and* GPS *and* CDMA/GSM/mobile? Though I guess they could always just stick one of them in the external one or use a hub... Also make sure you leave enough room inside for particularly bulky dongles.
What any more IO ports?
If you want this to be a hacker's computer you might as well expose the onboard serial port for debugging (not as a full DB9 connector, just something tiny). That would also help with early development.