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

[Boll, Martin Dr.]

Steve and Drew

First thanks for the link. It's the first time I saw there such a windlass.
Fascinating simple!

 

Turn the calculation and ask if you could lever a weight of 540 kg a 2meter
high distance with an electrical motor, having a mechanical output power of
3 Watt.

I did not calculate that, but I have the feeling that it does not work.

 

On the other hand: Two months ago in Burgundy I saw a medieval fire-place
with a mechanical (by weight) driven spit to roast meat of pig-size.

If they had to lever the weight every 5 minutes to make the mechanic to work
by itself, I think they had not built this, but had driven it constant by
hand. 

I could not get nearer information about that spit, though I found it
interesting to know. 

 

Imagine: 

-The drive of an old feet-driven sewing-machine would drive a small flywheel
connected -in the right gear- with a small radial-fan (instead of a
sewing-machine). That would certainly work, and if not, 

-you could use a sort exercise machine (bicycle-type) to generate even more
energy than needed.

By each of these two things, I think, you could manage the 540Kg being
lifted 2m within 2 hours.

-Was it so bad "cycling" during the cooking procedure? -A lot of our
dietary-girls would like to do so-. But the exercise-machine would be too
expansive.

 

By the way "high tech":

Try once really to make fire by friction in the old fashioned manner: You
will feel yourself *that* old low tech is "high" tech in the 21st century.

Possibly some wires of different material, putted into some apples
-connected in series- could generate the electricity needed for our
mini-fan, as the to-days low tech solution. :-)

 

Best wishes

 

Martin

 

 

drew wrote:
>     So a simple electric fan alternative that might be explored is a fan 
> based on a differential windlass.    Differential windlass systems were 
> often the motive force behind grandfather clocks (taking the high torque 
> low rpm energy from a slowly dropping weight and turning it into a high 
> rpm low torque) . 
Lets say we need 3W for a blower (underrated, in my estimation), and 
that we are going to dissipate the energy over a burn of say 2 hours, we 
need to hold
3x3600 joules =10.800 joules. Lets raise a weight say 2 metres (just 
under 7 feet), what weight do we need ?
10,800=MxgxHor M=10800/2/10 =540 kg, half a ton. Is that practical ? 
Double the height (14feet), 270 kg, double it again (28 feet), 123kg.
Is that practical ?
Electricity in the 21st century is not high tech.
 
Steve