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  • From: "WILLIAM JONES" <billj AT harborside.com>
  • To: <homestead AT lists.ibiblio.org>
  • Cc:
  • Subject: [Homestead] Wirtz Bike
  • Date: Mon, 25 Jun 2012 14:30:15 GMT

It's important to realize that an ordinary Archimedean screw gets water and
air alternately simply by the fact that the "snout" (the end of the hose)
dips into the water and then back into the air. If you want to use one as a
motor, you have to somehow feed it both air and water.
However, there's a simple alternative to the fairly complex mechanism which
I've described. Why not just inject a foamy mixture of air and water, and let
gravity sort it out after they're in the hose? I'll draw up a diagram for the
mechanism that this alternative permits.
Now let's turn to the question of how much torque a Wirtz engine can produce.
Basically each semicircular loop of hose produces a little bit of torque, and
when you add them all up it amounts to a rather large torque. To find the
torque generated by a single hose loop, first find the center of gravity of a
semicircle (which of course does not lie on the curve itself, but inside).
This is equivalent to finding the average value of one hump of the cosine
function. This curve is common enough that its c. g. appears right inside the
front cover of my engineering text. The c. g. is located 2r/Pi units to the
right of the center of the (semi)circle. We just need to multiply this by the
weight of the water in this half a hose loop, which is easily seen by the
theorem of Pappus to be w = rdgAPi. So after simplifying, we obtain Torque =
2Adgr^2, after cancelling Pi, where A the cross-sectional area of the hose, r
is the radius of a hose loop, d is the density of water, and g is the
acceleration due to gravity.
So in order to have a successful bike trip, the sum of all these little
torques has to be greater than the maximum torque required to move the bike
forward for every combination of hills and gears that the cyclist opts to use.



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