[TowerTalk] Wind torque balancing

[Tod - Minnesota]

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