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

[drew]

Steve Taylor wrote:
> 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.
Ignoring electrical loss seems like misrepresentation to me, as did your 
2 hour figure.    Do you know the word obfustication Steve, consider 
looking it up if you don't.   If a fan is rated at 3w, my understanding 
is that's it's draw, not it's output?   The efficiency of these little 
motors is in my understanding not great, so your 3w of current draw is 
probably less than 2w of shaft hp?  And if you were running that off a 
battery what would the storage loss be, almost as much as the three 
watt's if it's a cheap battery and charger so your 3 watt blower is 
costing you probably 5 or 6 watts, and giving you hopefully when 
everything is good 2 watts shaft hp......
>>    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.
You seemed in such a rush to do math, sorry if my request that you do 
more is a burden.  If you looked at the pic in the link you would see 
that there is one pulley required not pulleys, and that it is going to 
move relatively slowly, therefore very little loss around 1 pully and 
less as the cord wraps around the drum, hemp or sisle twine is very 
cheap and at the loads I am talking about almost stretch free.
>
>> 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.
So lets say the cook is using a common t-lud, it simmers well, but needs 
to bring water to a boil early on in cooking or the fuel is a little 
damp, so the cook needs to up the output of the stove for part of it's 
burn, maybe at the begining, or end of the cooking cycle?    How many 
cooks using stoves like this put food on, then walk away?   Not many I 
would imagine, in places where these stoves are likely used food is 
relatively very expensive.   So the cook is likely going to be close by 
anyway, therefore if the cook had to raise a weight every 3 minutes, 
while the desired higher output, would that really be a problem?  Might 
this even help them to save fuel, one of the goals of these stoves?   In 
fact if you had a fan assist on a tlud wouldn't it be wise to build into 
the circuit a timer to slowly reduce output, to save fuel and keep cooks 
from forgetting that they are burning fuel faster than needed?     In 
blacksmith shops the electric blowers were often removed after the boss 
found out how much his coal consumption went up.     Many forges were 
ruined by the electric blowers that were a "flash in the pan" as the 
heat didn't automatically turn down when the smith turned away so if 
they left the blower on full, and a large pile of coal on the fire, it 
could melt out the bottom of the tyre, and would certainly reduce 
anything left in the fire to oxide.   I am not saying that the 
differential winch is necessarily the best system, but it certainly is a 
good candidate, and for many users (perhaps most) in the third world it 
might be much better than the system you seem to be advocating.   
    I suspect that there are a couple of problems here, one of the 
problems here is the typical reductionist model engineers use, by 
focusing closely on one part of the picture they produce "solutions" to 
the minutia.   This is a very useful pattern in that it allows for the 
optimization of working systems.   The problem often becomes when you 
ask an engineer to solve a multifaceted problem, in my opinion their 
solutions are sometimes more destructive than the problem (consider 
nuclear power).    Just like technology, engineers are very useful, but 
when left undirected, often problematic.
    The other problem is the adversarial model that they are taught in 
school.   Attack the weakness of the competitive model, and hide the 
weaknesses of your own model.   Not really that useful compared to 
cooperative problem solving, and one of the reasons that to spur 
development all kinds of races, and competitions are needed.  
> 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
>
>
>

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