Experiment conducted April
2002 by KØTO and KØLR to demonstrate that the boom of a
horizontally oriented beam is the major contributor to wind torque. If that
conjecture is correct a beam which has one element located at an end and is
fastened to the mast at the center of the boom will orient itself in a wind so
that the boom is perpendicular to the wind direction.

A model was made of a single
element beam where the single element is located at one end of the boom. The
element pierces the boom so that there will be no effect from clamps or offset
elements. The pivot point is in the exact center of the boom. The end which
does not have the element has a weight fastened to it to provide a
counterbalance to the element weight.

Lyle, KØLR, holds the ‘beam’ in the wind.
The boom has moved to a neutral position so that the wind is normal
(perpendicular) to the boom.

The picture below graphically shows that the wind is indeed
blowing and that the boom is normal to the wind.

After running the previous
experiment we decided to check out what happens if the beam element is oriented
vertically. This time the beam behaves like a weather vane.

Below is an equation that will let you calculate how to
balance a boom that is not mounted with the center of the boom at the vertical
mast.

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 (X2)] 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.

*********************************************************

[(X1 x D) x
(½  x
(X1)) x  F)] - [(X2 x D) x (½
x (X2)) 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.

*********************************************************

[(X1 x D) x
(½  x
(X1)) ] - [(X2 x D) x (½ x (X2)) -  ( 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. 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 you use 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.

Tod Olson,  K0TO       January
11, 2005