0
$\begingroup$

Random question that popped into my mind after a 4-hours power outage. Let us assume that I am eating an extra dessert (250 kcal) and that I am using a bike and a generator to power my laptop (it consumes 10W while idle). How much time can I use the laptop?

I computed this as follows: $$ t = \frac{E}{P} = \frac{250 kcal}{10W}=\frac{250 * 4,18KJ}{10J/s} = 1.045 * 10^5s=30h $$

The above calculation assumes 100% efficiency. But even if we assume 10% efficiency, I would still be able to power my laptop for $3h$ and all this by eating just one extra dessert.

Questions: Are my calculations correct? Are my assumptions reasonable? Why don't we see a proliferation of bike-laptop devices?

$\endgroup$
1
  • $\begingroup$ Turns out that there is something like this already available for the OLPC: wiki.laptop.org/go/Peripherals/Hand_Crank. The efficiency of a human muscle is ca. 15-25% so a total efficiency of 10% is within reason $\endgroup$
    – Alexander
    Commented Sep 12, 2012 at 13:49

2 Answers 2

4
$\begingroup$

There's some discussion on this subject by AlanSE and me at How fast would someone have to run to run over water?.

A typical human i.e. not a trained athlete can manage a mechanical output of about 200W, and at this speed you burn about a 1,000 Calories per hour. Assuming it scales linearly (it probably doesn't!), and assuming 100% efficient conversion of mechanical power to electricity, powering your 10W laptop would consume about 50 Calories per hour. Your estimate of 3 three hours is in line with this given all the approximations involved.

As for why we don't see a proliferation of bike powered laptops, it's kind of hard to take the bike generator along with me on the train :-)

$\endgroup$
0
1
$\begingroup$

A litre of gasoline costs around \$1 and contains around 35MJ of energy. A Mars bar costs \$1 and contains 250Kcal = 1MJ Cycling at 15km/h uses about 30W so 1MJ = 9 hours of gentle cycling = 140Km

So the question is why don't we see more cycle powered cyclists!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.