[Stoves] Using a differential windlass as motive power for stove

[Steve Taylor]

drew wrote:
> Hi Steve, 
>
> Thanks for the comments,   lets reframe your evaluation a bit. 
>
> If your fan is 3w what is it's efficiency?  50% electricity to torque 
> let's be generous and call it 70% and what's your charger to battery 
> ef
I am ignoring efficiency in all these calculations.  The fan output is 
mechanical power. 3 W of blower is probably understating it.
>    What's the windlass eff, I think likely greater than 90%.    
>   
I doubt it. There ain't no such thing as a free lunch. The stores are 
always out of light, inelastic string, friction less pulleys etc
;-)

The Efficiency in the required pulleys  to turn the high torque, low 
speed into high speed low torque and drive a fan is defined for a pulley 
ratio of n:1 by Efficiency = mechanical advantage/ velocity ratio, if 
you want to get really technical. Just at the minute I really can't be 
bothered about doing all the maths.

> What is the average burn time of these types of stoves?   somewhere less 
> than 1/2 hour, not the two hours you have used.   What if someone had to 
> reset the weight once or twice or even 3 times in that 1/2 hour,  then 
> really the time it needs to run is 10 minutes not your stated 2 
> hours.    Looking at it that way might be more helpful. 
The calculations work like this.

Watts = Joules per second.

Total burn time = 10 mins = 600 seconds.

Total energy required = Watt X seconds. = 3W x 600 = 1800 Watt seconds = 
1800 Joules.
     
Now, if we use Drew's weight system, all THAT energy is stored in a 
lifted weight. The energy required to lift a weight of mass M, through 
height h = Mgh, where g is the gravitational acceleration constant, g = 
9.81 m/s/s
Lets assume we lift it

Energy = Mgh or  1800J = M x 9.81 x 2

Rearranging for M, 1800/2/9.81 ~ 90kg,

200 lbs , raised 7 feet, dropped over 10 minutes generates 3 Watts.


Steve