[TowerTalk] Weight of tower on the main bearing of a rotating tower?

[Roger (K8RI) on TT]

If this is the Joe Subich who wrote the electronic series, he  was an 
expert on this stuff, but I do not like rule of thumb approaches to 
towers unless the dimensions are followed exactly.

Don't forget the weight of the guy wires which with EHS can be 
substantial. Pick two, 100 foot coils of 1/4" or 3/8ths inch EHS
At the top most of the guy weight is supported by the tower, while the 
bottom tier has a roughly half the weight of the guys supported by the 
tower.
To get more specific on the download pressure Vs tension.
(I'm going by memory so I really need to look up the function) You are 
working with a right triangle, so, take the angle the guy forms with the 
tower.
Don't know that angle. Right triangle, angles form a total of 180 
degrees.  The one at the base is 90 deg. Measure the guy angle at the 
guy anchor, add to 90 degrees and subtract that total from 180 and you 
have the angle the guy forms at the tower.

Take the Cos of that angle and multiply it by the tension and you will 
have the downward force X 3 for that level.
I don't like rule of thumb, because the force is about ?# and a guy 
radius of  ?Feet.  That radius can change the down force by a lot for 
the top guys , while a big change in the guy radius at low levels makes 
very little difference.

At least with a rotating tower you are forced to use guy rings, or 
brackets which are often ignored in non rotating towers with the guys 
hooked to tower legs.  This puts a lot of side force (strain) on the 
tower, while a guy bracket, or ring takes the horizontal force instead 
of the tower.

73

Roger (K8RI)

On 5/31/2015 9:25 AM, Joe Subich, W4TV wrote:
>
> > Any idea how much weight the bearing will actually have to take?
>
> As a starting point, about 78% of the pretension of the top guy wires
> (assuming 80% guying - or 80 foot radius for a 100 foot tower) will
> be converted to downward force on the tower.  For your 6700# guys
> that's 6700 * 0.10 * 0.78 * 3 ... about 1560# for the top set of guys.
> Add a second set of guys at 2/3 height and around 65% of the pretension
> becomes downward force ... 1300#.  For a third set at 1/3 height, about
> 40% of the pretension is downward force ... 800#.
>
> That means 3700# give or take for the static load of the guy wires.
> *HOWEVER* that's the minimum is calm winds and no ice load.
>
> How much would 1/2" or 3/4" of radial ice covering the entire surface
> weight?  If you had winds that loaded the guy wires to 50% of their
> breaking strength, the downward force imparted by the guy wires would
> be five times as great as the no wind condition.
>
> It's those dynamic forces that you need to design for <G>.
>
>
> 73,
>
>    ... Joe, W4TV
>
>
> On 2015-05-31 8:48 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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-- 

73

Roger (K8RI)


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