[j.p.s]

"He who refuses to do arithmetic is doomed to talk nonsense." - John McCarthy
"... as simple as possible, but not simpler" - Einstein Principle
"garbage in, garbage out" - Computer culture proverb

The topic of electronic book reader battery life comes up often on the mobileread.com forums, but all the discussions I can find read like the blind men arguing about the nature of an elephant. Everyone is looking for a sound bite, but there is no nice single round number that will give a good idea of how many days a fully charged battery will last for a particular reading habit. Part of the problem is knowing when various causes of battery drain are insignificant and when they matter a lot.

I think that three terms are enough to allow anyone to easily estimate how long a reader battery charge can last and whether one reader can last significantly longer than another for a particular reading schedule. Trying to present this in a way simple to understand and include supporting detail has made this a long post. I have tried to put as much of the critical information close to the front as I can.

x. Battery charge per page turn, in per cent, over and above power on usage
a. Battery charge per power on hour, in per cent, over and above sleep usage
b. Battery charge per hour in suspend, or sleep, in per cent

battery_per_day (pages_per_hour*x + a)*hours_reading_per_day + 24*b
days_per_charge = 100/battery_per_day

Let's calculate these from the numbers measured or provided for the Nook STR in various messages. This will take only a little bit of the very simplest algebra. When manufacturers can be persuaded to provide x, a, and b, only a little bit of simple arithmetic is needed to calculate battery life for anyones average daily reading schedule, taking into account pages per hour as well as hours per day.These numbers may not be exact, but they are good enough for the purposes of this message. Let's say these are official, accurate numbers for some Brand X reader.

1. 25000 pages at 1 second per page.
2. 150 hours at 1 page per minute.
3. 60 days at one half hour reading per day.

Please note that this does not mean that you can read for anywhere near 250 days at 100 pages per day or 30 days at 5 hours per day. Similarly, if an Amazon kindle battery is drained after 30 days at 1 hour per day, it will not last anywhere near 60 days at half an hour per day of reading. This is because although turning a page uses over 30 times more power than when not turning a page, simply being on adds up over time, as do days spent in suspend or sleep.

25000 page turns at 1 per second is about 6.9444 hours and 150 hours of 1 page per minute is 9000 page turns. If we have y = a + b, then

25000x + 6.9444y = 100%
9000x + 150y = 100%

16000x = 143.0556y or y = 111.8446x, so

1 page turn is 0.003879% battery charge
1 hour "on" is 0.4338% battery charge

Now let's solve for a and b

30 hours is 1800 minutes or page turns, so 2 months 30 page turns per day uses 6.9822% of the battery and there are 1440 hours in 60 days, so

30a + 1440b = 93.0178%
a + b = 0.4338%

30a + 30b = 13.014%, 1410b = 80.0038

a = 0.3771%
b = 0.0567%

1 hour "sleep" (or suspend) time is 0.0567% battery charge

So a fully charged Brand X put to sleep and disconnected from the charger will run out of battery sometime during the 74th day without ever being unsuspended or any pages turned, much less 15 pages per day for 120 days.

As a check on work, lets calculate the days per charge for half an hour per day, one minute per page:

battery_per_day = (60*0.003879 + 0.3771)*0.5 + 24*0.0567 = 1.6657%
days_per_charge = 100.0/1.6657 = 60.03 days

Within 0.03 days is pretty close, 43 minutes of suspend, 5 minutes of "on", or 10 pages at 1 page per second.

So let's go to one hour per day.

battery_per_day = (60*0.003879 + 0.3771)*1.0 + 24*0.0567 = 1.97064%
days_per_charge = 100.0/1.97064 = 50.7 days, a lot more than 30.

So how many hours per day to get 30 days battery life?
30 = 100/battery_per_day, battery_per_day = 3.3333%
3.3333 = (60*0.003879 + 0.3771)*hr_per_day + 24*0.0567
3.3333 = 0.60984*hr_per_day + 1.3608
hr_per_day = 3.23, a lot more than 1.

Let's cut that in half and see how many days we get.

battery_per_day = (60*0.003879 + 0.3771)*1.615 + 24*0.0567 = 2.3456%
days_per_charge = 100.0/2.3456 = 42.63, a lot less than 60 (we already
know that we have to cut back to 30 minutes per day to get 60 days).

As far as page turns, 1 battery charge gives us:
25000 in 7 hours, or
9000 in 150 hours, or
5800 in 30 days, or
4100 in 42 days
3000 in 50 days, or
1800 in 60 days

What I have tried to show is that one number is not enough information to calculate electronic ink reader battery lifetime for a different reading speed or hours per day of reading from the exact case used to specify that single number. But three numbers work just fine, provided there is not a lot of wifi, indexing, etc thrown in. If you want to take those into account, you need more numbers. But good luck, because battery use under those conditions varies a lot.

Please take away from this that the more spread out your reading time is, the fewer total page turns per charge you get, that cutting your reading time per day in half does not nearly double your battery charge lifetime, nor does doubling your reading time per day anywhere near cut the number of days per charge in half.