How to calculate the autonomy of a laptop ?

[WarMachine]

Hiya all !

I hope you will be able to help me.

Here's the question :

Knowing the power consumption of a lappy (x Watts), and the capacity of a battery (y mAh), is that possible to determine the autonomy (in hours), of this machine ?

Example : My lappy consumes 80W, the battery is a 3800 mAh one, how long will I be able to use the lappy before using the power cord ?

Is there a relationship between these two data ?

Thx for your answers.

W.

[rkawakami]

I would say assuming a LINEAR consumption rate (which is not what the battery or the system will do in real-life), the calculation would be:

Watts/10.8V = amps of current/battery AH rating = hours of run time

[WarMachine]

Yeahhh, so it's possible ?

Well !

Forgive me, but I don't totally understand the formula.

(I have had a lot of difficulties with numbers all my life, and it's not better nowadays).

How do you find "10.8V" ?

Could you explain me with the data of my example ? It will be more easy to me.

Forgive me again. You help is greatly appreciated.

Thx again.

W.

PS : On the other side, I totally understand that the result is approximative, as the autonomy will change depending on the use of the machine.

[rkawakami]

Sorry for not making the 10.8 clear... It is the voltage of the battery. The formula for calculating power in watts is:

Power (Watts) = Voltage x Current

Since you have given an example of 80 watts and the voltage of the battery is known to be 10.8 volts (and is assumed to be constant), then the current can be calculated by this formula:

Current (amps) = Power (watts) / Voltage (volts) or Current = 80(w) / 10.8 (v) or 7.4 amps

Since you have also given the battery's capacity as 3800mAH (milliamps-hours) or 3.8AH, the time that the battery can supply the rated power (or current) of 80 watts is:

Time (hours) = Current (amps) / Capacity (AH) or Time = 7.4 (A) / 3.8 (AH) or 1.95 hours

Of course, all of these figures are assuming a constant, linear power draw and a battery that is TOTALLY exhausted. As the power circuits in the laptop will automatically shutdown at some point prior to total drain of the battery, the calculated time is only approximate.

reference:
http://science.howstuffworks.com/question501.htm

[WarMachine]

Ab-so-lu-te-ly wonderful !!!

Big thanks for your explanations and your patience !

For sure, I will keep your answer in my bookmarks for a while.



W.

[WarMachine]

Sorry for not making the 10.8 clear... It is the voltage of the battery. The formula for calculating power in watts is:

Power (Watts) = Voltage x Current

I'm trying to understand, but I still don't.

Where does the 10.8 come from ?

Value "Power" = 80W, that is OK, but you don't know the value "Current", so, how do you calculate the value "Voltage" ?

I'm not dumb, I'm not dumber, but I certainly am dumberer !

Thanks again.

W.

[anthean]

Just using the parameters you used, shouldn't the answer be half an hour ?

I figure:

P * t = V * (At)

where:
t = time (what we will solve for)
P = power = 80 Watts
V = voltage = 10.8 V
At = capacity = 3800 mAh

Solving for t

t = V * (At) /P = 10.8 * 3.8/40 = .51 hours

If this sounds rather low, I think the main reason is that 80 Watts is far too large, and that 40 Watts peak is more reasonable, with an average of half that or even less.

Thus, if your average draw (including periods of inactivity) was only 15 Watts, you would get almost three hours.

To see some actual measurements (admittedly, with a Dell ultraportable), see:

http://www.codinghorror.com/blog/archives/000562.html

[tomh009]

If this sounds rather low, I think the main reason is that 80 Watts is far too large, and that 40 Watts peak is more reasonable, with an average of half that or even less.

Thus, if your average draw (including periods of inactivity) was only 15 Watts, you would get almost three hours.

The battery properties applet (you can run it through the IBM Access utility) will show your actual power consumption (when on batteries), enabling you to calculate actual numbers for your ThinkPad.

As a point of reference, my slightly undervolted X31, with wireless and full display brightness, will consume about 13W with the disk running but CPU idle.

[rkawakami]

Just using the parameters you used, shouldn't the answer be half an hour ?

DUH!

I had inverted the Current and Capacity figures in my earlier formula and didn't catch the obvious error. If the battery can only supply 3.8AH it stands to reason that a 7A drain can only be sustained for 1/2 hour!

And yes, you are correct in saying that 80W is WAY too much power. I was going with the example that the OP wanted explained.

[WarMachine]

Hi,

All that is not easy.

If I take my TP600E for example, i have these data :

The battery :
FRU P/N : 10L2158
10.8V
3.2AH

I've googled a little to find other specs.

I've found that the battery offers 43.2W.

I will suppose that's the maximum power used by the Thinkpad (but I don't know if I'm right).

If I use the formula given by anthean :

P * t = V * (At)

I will have :

P = 43.2W
t = we search it...
V = 10.8V
(At) = 3.2AH

t = V * (At) / P
t = 10.8 * 3.2 / 43.2
t = 0.8



I should find something like 3 hours.

What is wrong ?

Or the result is good, and gives us the autonomy of the lappy when used intensively ?

W.

[tomh009]

If I use the formula given by anthean :

P * t = V * (At)

I will have :
P = 43.2W
t = we search it...
V = 10.8V
(At) = 3.2AH
t = V * (At) / P
t = 10.8 * 3.2 / 43.2
t = 0.8
I should find something like 3 hours. What is wrong ?

The problem is your power consumption figure of 43.2W -- where did you find this? I could believe that as the maximum power consumption from the power supply, but then that would be reached only when charging the battery.

I would expect the system power consumption to be in the range of 10 to 15W, but I don't have a 600E to test it. Is the battery properties applet compatible with the 600E? If it is, you should run it to find out the real power consumption figures.