Hey, that simple threat becomes very interesting battery knowledge source!

Most of ndi post is very useful for me, but I think that shadowjk got a point here. If you let the phone charge overnight, in fact it is just charged to full in few hours, then discharged to 95%, then charged again to 100, etc... It's true that RGB diode is green all the time, also battery-eye show it as "full" part of cycle. Unfortunately, N900 doesn't use power directly from charger when plugged - it just use the battery, charging in the same time. When battery reaches "full", power is cut off until battery drop to 95% or so. You can easily monitor it with bq27200 kernel-power module, by:

cat /sys/class/power_supply/bq27200-0/current_now

...when plugged in, you got value under 0 (for example, "-2475") which is exactly (charge current)-(current used by phone to "live"). When battery reach "full", current_now report just positive (i.e used actually) current - exactly the same as if plug is disconnected = non charging.

So, that would mean You got 10-20 cycles every night...

Also, after short logic check i totally disagree with Your calculations about powercycle drain. Precisely, your calculations are correct, but upside down - it is not 180W in couple of seconds, but 0.180W = 180mW!

Look ndi. N900 battery rounded up got 5Wh - You're right at that. It's capable to safe deliver for our needs 1315 mA constant current @ 3.8V (rounded from 3.3V-4.2V, doesn't matter for our calculations) for time of 1 hour. Also, it can deliver 657 mA constant current for 2 hours, etc. Every fluctuation of current change total time, but after C (current)*V (voltage)=5W or higher, battery is already at 3.3V and is discharged, aka "dead".

Then, if we want to calculate how much power could be drained if powercycle would use 3% of battery capacity, we need to keep in mind that Wh = Ch*V, so (Wh = Ch*V) / h -> W=C*V - we need to divide by h, not multiply by h as you did.

Using known values, we want to get 3% power during, lets say, 15 seconds of bootup - it is W*3/100=C*V, V is constant four our calculation (3.8V), W after /h is just 5. 5*3/100=0.15.

We want C (current), so C=W/V (or, if you prefer that, Ch=Wh/V, doesn't matter at all for us now), we get 0.039=0.15/3.8. Our 0.039 is in A, so it is 39mA. When we don't use /h, we get 3% of battery power = 39mAh. If You want to check for mistakes, 39mAh *33.33 = 1299.87 mAh - thats our 1315 mAh capacity rounded down.
@ 3.8V it is 5Wh.

So, we already know and proofed (if anyone understand my not so self-explanatory calculations ), that 3% of battery power is 150 mW (yea, it's logical, but we wanted hard proofs, doesn't we?). 10%, then, is 500 mW, 10% battery capacity use is just 500mW. No need for blue smoke and other things.

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Also, if we treat serious Your calculations about 10% beign 180W, that would mean we got 1800W at 100% - real beast. That would allow us for powering up high-power driller for an hour, just using our little battery, no need for wall plug I got nothing against that idea, but for now we must be happy with 5W thing...

I tried to explain calculations as easy as it can be, but I'm again after sleepless night, so sorry for any hard to understand shortcuts. We can pull here dr_frost from "3000mAh battery mod" thread, he use high W batteries in his RC machines, so maybe he will be able to explain it more user-friendly