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Old 04-05-2018, 05:09 AM
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rickoff rickoff is offline
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Join Date: May 2008
Location: Maine, USA
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Response to BroMikey

Quote:
Originally Posted by BroMikey View Post
So lets say 5 pounds per tube. It looks like to me you will have 15 lbs of available force. If we used 5 gallons per tube that might be 150 lbs available and 55 gallon drum size tubes would 1600 lbs torque.

There are only 3 tubes pulling down all of the time. Not a bad analysis
I thought it would be harder than this.
Sorry, Mikey, but your reasoning is flawed. The actual force is dependent not only upon the applied weight in each tube, but also the distance the center of that weight lies from the wheel's vertical centerline. Multiplying each distance by the weight provides us with the foot-pounds of available force, or torque, for each tube. I'd suggest that you rethink your rapid analysis and take the time to properly calculate the torque in the manner which I described in post #2.

If you go back and look at the drawing, you will see that, at the stage of rotation depicted, there are 4 tubes (A, B, C, and D) providing rotational force on the left side of the center line. Tube D will continue aiding rotation until its marked center of water fill weight falls directly behind the vertical centerline. At that point it will have zero torque, and then there will be 3 tubes at the left side providing rotational force. Tubes L and F will be level and at the mid-point of transitioning at that same moment, which means that they will have a neutral effect until the next one degrees of rotation, when their water weights will be shifted further left, meaning that L will then become overbalanced to F and will begin aiding the rotational force of tubes A, B, and C. In other words, the wheel will always be in an overbalance condition at the left side of the centerline.
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Last edited by rickoff; 04-05-2018 at 11:13 PM.
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