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Old 06-12-2011, 06:32 PM   #1
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Battery life, simple and easy specification, calculation and measurement

"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.
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Old 06-12-2011, 11:24 PM   #2
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Umm, I'm glad you did the math and all.
but by about 1/4 of the way down the post, my eyes glazed over and I was wanting to go to sleep.

Keep it up
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Old 06-12-2011, 11:37 PM   #3
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great, now do the equations for LCD screens...

also, rarely I read a book without some wall of soundtrack to isolate me from external interference. A perfect silent library with my own ambient music in plain urban jungle... so, you should account for that too.
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Old 06-13-2011, 12:15 AM   #4
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Hmmm yasss... but does that take into consideration the electro-chemical impedance differential between a fully charged and a partially charged battery?
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Old 06-13-2011, 02:50 AM   #5
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Quote:
Originally Posted by j.p.s View Post
"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.
Uh oh. I see a major typo. Can you spot it?
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Old 06-13-2011, 12:27 PM   #6
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Thank you for this post, it helped me rationalize the maddening debate about battery life of the various book readers.

Your math is impressive but hard for me to follow. My take-away is that when I see an absolute value given for the battery life of a book reader I'm seeing Jabberwocky penned by someone's marketing department, not by their engineers.

Thank you again for clarifying this issue.

Last edited by 6charlong; 06-13-2011 at 12:30 PM.
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Old 06-13-2011, 01:33 PM   #7
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The essence of the post is true for almost every battery operated device. As consumers we never get that sort of analysis given to us by the manufacturers.

Take digital cameras for example. You'll see some manufacturers claim, "takes 400 images on one charge." But that assumes a fixed 'test' amount of idle on-time, only so much focusing for each photo, some average flash use (a HUGE variable on battery life), only so many on/off cycles, and only so much chimping (staring at a stored image on the display). The only thing that is normally consistent is they don't leave the camera off for long periods (letting the battery self-discharge).
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Old 06-13-2011, 03:02 PM   #8
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Quote:
Originally Posted by Ravensknight View Post
Umm, I'm glad you did the math and all.
but by about 1/4 of the way down the post, my eyes glazed over and I was wanting to go to sleep.
Quote:
Originally Posted by 6charlong:
Your math is impressive but hard for me to follow. My take-away is that when I see an absolute value given for the battery life of a book reader I'm seeing Jabberwocky penned by someone's marketing department, not by their engineers.
OK. Less words, more intermediate arithmetic.

Definitions:
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

Data:
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.

Arithmetic:

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

25000x + 6.9444y = 9000x + 150y

16000x = 143.0556y

y = 111.8446x

9000x + 150*111.8446x = 100%
9000x + 16776.69x = 100%
25776.69x = 100%
x = 0.003879% battery per page turn

y = 111.8446x = 0.4338% battery per hour on time

y = a + b = 0.4338%

60days * 0.5hours/day = 30 hours power on time while reading
30hours * 60minutes/hour * 1page/minute = 1800pages
60days * 24 hours/day = 1440hours in 60days

1800pages * 0.003879% = 6.9822% battery used on page turning in 60 days

100% - 6.9822% = 93.0178% used for idle on time + suspend (sleep) time

a + b = 0.4338%
30a + 30b = 13.014% = 30*0.4338%
30a +1440b = 93.0178%

1410b = 80.0038% = 93.0178% - 13.014%

b = 0.0567%
a = 0.3771% = 0.4338% - 0.0567%

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 for 30 minutes reading per day

From previous post, you can read over 3 hours per day and still get 30 days between charges.

That is the main point. If an Amazon Kindle gives 30days on a charge at 1 hour per day reading, it can not give 60 days at 30 minutes per day. You can not get twice as many days by reading half as long per day on any of these readers unless you completely power off after reading. Even then, the power used by booting up would still keep you from getting twice the number of days. In addition, it would either cut into your reading time or you would need to deal with all the hassel of making sure to turn it on early, but not too early.

Last edited by j.p.s; 06-13-2011 at 03:04 PM. Reason: change incorrect 6000x = 143.0556y to correct 16000x = 143.0556y
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Old 06-13-2011, 03:07 PM   #9
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Hmmm yasss... but does that take into consideration the electro-chemical impedance differential between a fully charged and a partially charged battery?
No.

If you think that would materially affect the results, please show some evidence.
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Old 06-13-2011, 03:19 PM   #10
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Quote:
Originally Posted by Penforhire View Post
The essence of the post is true for almost every battery operated device. As consumers we never get that sort of analysis given to us by the manufacturers.

Take digital cameras for example. You'll see some manufacturers claim, "takes 400 images on one charge." But that assumes a fixed 'test' amount of idle on-time, only so much focusing for each photo, some average flash use (a HUGE variable on battery life), only so many on/off cycles, and only so much chimping (staring at a stored image on the display). The only thing that is normally consistent is they don't leave the camera off for long periods (letting the battery self-discharge).
Taking a picture on a digital camera is much more demanding on a battery than turning a page on an e-ink device, and has more complicated effects on battery life. See the crack by RichL above.

I really like the long battery life on these things and hope the one day (tops) tablets don't wipe out the 1 week (minimum) readers. I just thought it worth while showing a more accurate picture of how battery life varies with reading time and the very real but very worthwhile cost of using suspend mode. It is very nice to be able to just put the thing down when finished and pick it up the next day, or whenever, and be on the same page in a couple of seconds, and still have days of battery life left.
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Old 06-13-2011, 03:24 PM   #11
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Uh oh. I see a major typo. Can you spot it?
No. Would you please point it out so that it can be fixed?
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Old 06-13-2011, 05:39 PM   #12
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Originally Posted by j.p.s View Post
No.

If you think that would materially affect the results, please show some evidence.
I'm retired and you want me to work?

On a serious note the following will have an affect on pages turned etc.

electro-chemical impedance... , ambient temperature, humidity and age of battery (that is was it sitting on a shelf for a couple of years before being installed in the reader)

My guess though is that we would be looking at less than 1% of total pages turned and maybe a millisecond difference in page turn time. Pretty minor points unless you are designing an ereader or making a flippant reply.
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Old 06-13-2011, 06:11 PM   #13
Rainmaker
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Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.Rainmaker can program the VCR without an owner's manual.
 
Posts: 103
Karma: 185586
Join Date: Apr 2011
Device: Kindle 3G
Its well known that for cameras the 2 most important power draining operations are:
1. Using flash (even old cameras had to charge a capacitor)
2. LCD screen

And i 've heard (i can't find it on google right now), that there are some camera models, which include an "off" mode, in which you can take pictures the old way.
Tha is, LCD screen off, but you can still take pictures.
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