[TowerTalk] RE: Product Improvement - Wind torque balancing
Tod - Minnesota
tod at k0to.us
Wed Jan 12 12:15:46 EST 2005
Let me start by telling a true story.
A very long time ago I was taking a college math class which
had an instructor who was more focused on proving that those
of us in the class were intellectually incompetent than on
proving the equations he wrote on the board. (I suspect
others reading this list had the same class but at a
different University).
One day he put an equation on the board. Then, below it, he
put another equation. He turned to the class and stated, "
The transition from equation 1 to equation 2 is obvious".
After saying that he turned and looked at the board for at
least a minute. Then without saying a word he left the
classroom. Those of us in the class tried to guess what was
going on.
After twelve minutes he returned and once again addressed
the class saying, "Yes, it is obvious!".
------------------------------------------------------------
------------------------------------------
Last night I posted an equation saying that was "obvious"
only it is not. In fact, there is an error in it.
So, in an effort to escape with minimum ridicule from the
audience I am trying again.
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.
Let the longest boom side be X1 and the shortest boom 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.
If the diameter of the boom is D, then the area of the boom
on each side is X1 x D = A1 and X2 x D = A2. The units for D
should be the same as those you use for X1, X2 and L and
should be squared (e.g. square feet) for A.
The torque in foot-pounds (or m-kg if you live outside the
US) is the area, A1 or A2, multiplied by the wind force, F,
and by the midpoint of the distance from the mast to the end
of the boom. (½ x X1 or ½ x X2). This gives us an equation
that shows the imbalance -
A1 x [½ x(X1)] x F – A2 x [½ x(X1)] x F = imbalance in foot
pounds ( or m-kg)
[(X1 x D)x (½ x (X1)) x F] - [(X2 x D) x (½ x (X2)) x F) =
imbalance
Obviously, when X1 =X2 the imbalance is zero.
Imbalance correction = L x A x F where L is the
sail arm
and A is the sail
area for the counter torque balance.
************************************************************
************
[½ x (X1) x (X1 x D x F)] - [½ x (X2) x (X2 x D x F) - (L x
A x F) = 0
************************************************************
************
For all cases we can divide both sides of the equation by F
and remove it from consideration.
************************************************************
************
[½ x (X1) x (X1 x D)] - [½ x (X2) x (X2 x D) - (L x A) = 0
************************************************************
************
The diameter,D, however, still is in the equation and must
be if we want to get a correct value for the imbalance and
the product L x A which is the correction for the imbalance.
The fact that we have a minus sign in front of the quantity
(L x A) tells us that the correction must be on the same
side of the mast as the X2 section of the boom.
The units for D should be the same as those for X1, X2 and
should be squared (e.g. square feet) for A.
The sail may be mounted on an arm extending out from the
mast at the same angle as the short side of the boom. In
essence it will be parallel to the boom. It could also be
mounted on the boom if that is convenient.
This time I think it may actually be obvious -- if not
please send your correction to me quickly to minimize my
exposure to embarrassment.
The web site has also been corrected to the above. The URL
again is
http://www.k0to.us/2Wind%20Torque%20Experiment%20-April%2020
02.mht
The 202002.mht ending is correct.
Tod, KØTO
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