Wherein I try some Math (Flamestower edition)


Thank you to everyone who has in one way or another found themselves at the slowlorisblog during its two-week launch extravaganza. I hope you continue to find it interesting enough to visit on a weekly basis as we settle into a rhythm here. To start our regular programming, I’d like to do a little math. Not really my wheelhouse, but occasionally something piques my interest and provides a fun diversion until I realize I’m out of my depth.

To business: A few months ago I learned about the kickstarter for Flamestower, an off-the-grid way to charge your cell phone using a plain vanilla fire:


This is certainly not the first MacGyver-style charger on the market, nor is it the first or best to use thermal energy to generate electricity. But, in passing it along to the group of friends with which I game regularly each week, one happened to ask the question: How many of these would we need to run 3 desktop computers playing Sid Meier’s Civilization V?” Excellent question, I thought. I figured with my high-school level math skills, I could put away this question in twenty minutes. Two hours later, far more humbled, and finally finished, I concluded I needed more math in my life. In any case, below is my proof of the problem, because maybe someone out there has once thought something like How many solar backpack chargers would I have to daisy chain together to keep PrimeGrid running after the apocalypse?


The Flamestower charges via USB 3.0 (Thank god it’s not 2.0). At 2 amps and 5 volts, the max USB 3.0 is rated at is 10 watts, so the short answer is you’re looking at 65 of them per computer if that computer is running a 650 watt PSU. So my two friends and I would need 185 between us. The next logical, and far more interesting question, is how much wood would you need to power it? This is where we go down the rabbit hole.

The kickstarter doesn’t give exact specs, but here’s what we know.

1 watt = 4.1868 calories/sec

1000g (1 kg ((standardized amount of wood one might collect)) mass of oak)  * .00048 (specific heat of oak in cal/gram Celcius) * 482 (combustion temp, in C, of oak):

= 231.6 calories contained in the wood

231.6/4.1868cal/sec = 55.32 seconds of power at 1w. But this is also assuming 100% efficiency.

Assuming a thermal efficiency of 8% (this is the top end of efficiencies for a thermoelectric generator):

= 4.43 seconds of power at 1 watt, or 1.48 seconds at 3 watts (since we want to minimize the number of units we’d have to buy).

So, for every 1 kg of wood we collect, we can power this thing for 1.48 seconds. Weak. But it gets worse.


We need 1650 watts for a 10-hour game of Civ 5.

10 hours = 36000 seconds.

We need 1650 watts continuously.

550 (minimum number) units pumping 3 watts continuously, each burning 1 kg of wood, lasts for 1.48 second. So we’d consume:

550kg wood/sec * 24324.3 (36000 seconds * 1.48 seconds of 3 watt-rate contained in each):

13,378,378 kg of wood. Off the grid, indeed.


The only problem I ran into that I can’t resolve is that, while the thermoelectric generator runs at 8% efficiency, this doesn’t include the heat lost to the surrounding air, which I assume is a lot. So this experiment is assuming that if you can find a way to burn a certain mass of wood at an exactly controlled rate, you could also devise a way to minimize heat loss to the convecting air. Otherwise, you’d probably have to multiply that number by 100 or something on the assumption that only 1% of a fire’s thermal energy gets trapped by the Flamestower.

And that’s that. I’m exhausted.

Feel free to point out any mistakes you see, but be warned I reserve the right to incorporate your corrections, delete your comment, and pretend I knew what I was doing all along.


5 thoughts on “Wherein I try some Math (Flamestower edition)

  1. A couple of questions and comments:

    Regarding the “short answer”: You state that USB 3.0 is rated for 10 W and hence if we have a 650 W PSU we need 65 units to run a computer. Per the flamestower website the flamestower is only rated for 3 W maximum, with 2.5 W of average power. So if we assume the unit can continuously supply 2.5 W of power we are actually looking at 260 units per computer, no? It is possible you address this further below in the long answer?

    It also really isn’t that simple (which I am sure you know). If you were actually going to connect a bunch of flamestowers together in an effect to run a desktop computer you’d need to look at the current and voltage demands of the computer and construct the array accordingly. For example, if your gpu requires 25 A at 12 V (quick googling suggests this is in the ballpark) and each flamestower is outputting .5 A at 6V (it is really only rated for 5 V but lets call it 6) you would need 100 units just to run the video card (each mini-array with two units in series giving you 6*2=12V at .5 A and 50 mini-arrays giving you .5 A *50 = 25 A which all together gives you the needed 25 A at 12 V).

    • absolutely. the psu requirements of the computer isn’t going to be simple wattage, but if you have the know-how to come up with a daisy-chain like this, I’m assuming you’re bringing a voltage regulator/inverter along for the ride 🙂 -rmm

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