It is not that tough to calculate. The tension in your guys can be
treated a a vector and resolved into two orthogonal components. An
illustrative example:
Suppose the height of the connection of a guy to the tower is 4 units
and the distance from the tower to the guy anchor in the ground is 3
units. Then as you may recall from school, the guy length is the
hypotenuse of a right triangle and is 5 units long. Forces are
distributed in proportion to this triangle. For simplicity lets say the
guy tension is 5 units. Then the down force on the tower due to this
guy is 4 units.
Another approach is to work with sines and cosines. For an example
lets assume you have your guys at a 45 degree angle to the ground, i.e.
the ground anchor is as far from the tower as the guy to tower
connection point is above grade. As you may recall at 45 degrees the
sine and cosine are both 0.707, equal to each other but any angle could
be used and te force be calculated. Anyway, in this example the added
down force on the tower due to this one guy is equal to 0.707 times the
tension in the guy. In practical installations the side forces generated
by the guys cancel each other and the net result is a down force of all
the guy's down forces added to the weights of tower and other equipment.
Patrick NJ5G
On 5/31/2015 7:48 AM, Richard Thorne wrote:
I'm finishing up the rebuild/refurbish of the parts for a 55g rotating
tower.
I was pondering and I'm curious. How much weight will the main
bearing have to take? It's one thing to add up all of the components
on the tower but then there's the downward force of the guy wires.
I did some quick math and I'll probably have less than 2500 pounds of
tower, rotating rings, antennas, coax assembles etc. I'll be using
1/4" guy material which is rated at 6700 lbs and requires 10%
tension. The rotating base will be at ground level.
Any idea how much weight the bearing will actually have to take?
Thanks
Rich - N5ZC
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