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## [TowerTalk] Wind torque balancing

 To: "'TowerTalk'" [TowerTalk] Wind torque balancing "Tod - Minnesota" Tue, 11 Jan 2005 21:54:28 -0600
 ```A few days ago I said that I would load some pictures showing the wind effect on a beam which had a single element at one end of the boom and where the boom was mounted so that the mast (pivot) went through the center of the boom. I have completed a short note with pictures that shows an experiment run in April of 2002 to demonstrate what happens. The note may be accessed at the following URL. http://www.k0to.us/2Wind%20Torque%20Experiment%20-April%2020 02.mht For those who don't wish to bother with looking at the pictures and write up here is the summary equation. ********************* If one is dealing with a boom that is mounted so that the boom length on one side of the mast is greater than the boom length on the other it is possible to quickly calculate the dimensions of a ?wind sail? that will compensate for the wind loading on the long side of the boom. This presumes that the elements pierce the boom rather than being mounted on the top or the bottom of the boom. In those cases the result will be close but there will be a slight error. Probably the rotator gears can handle the error. Let the longest side of the boom be X1 and the shortest side be X2. Let the area of the sail be A and the distance from the mast to the center of the sail be L. All lengths should be in the same units (feet, centimeters, etc.) The case below is for the situation where the sail is mounted on an arm extending out from the mast at the same angle as the short side of the boom. The sail is mounted so that its plane is the same as the plane formed by the mast and the boom. In essence it will be parallel to the boom and the sail could even be mounted on the boom if that is convenient. The product L x A is equal to [ ½ x (X1)^2 - ½ x (X2)^2 ] <<. Trust me or I will inflict a series of algebraic equations on you.>> You may select any combination of L and A that fits the equation which allows one to tailor the sail and the arm to suit your circumstances. Note that the diameter of the boom does NOT need to be known since it does not appear in the equation. Also, the wind velocity does NOT appear in the equation. We are balancing things and can conveniently drop the wind force terms since they are the same on all elements. Purists will know that my equation is not exactly correct, but I hope they will agree that the error is trivial in the context of the problem being solved -- how to keep the stress on the rotator gears below the breaking point. Tod Olson, KØTO January 11, 2005 _______________________________________________ See: http://www.mscomputer.com for "Self Supporting Towers", "Wireless Weather Stations", and lot's more. Call Toll Free, 1-800-333-9041 with any questions and ask for Sherman, W2FLA. _______________________________________________ TowerTalk mailing list TowerTalk@contesting.com http://lists.contesting.com/mailman/listinfo/towertalk ```
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