[TowerTalk] Calculating Forces for Tilting tower

[terry burge]

Dave,
I loved the way you described that. What about if instead of 10' up you were say 20' up? Does that lower the weight on the rope or increase it and by how much? Guess I was incorrect in my explanation to the original questioner.

Terry
KI7M

> On May 29, 2019 at 1:34 PM Dave Hachadorian <k6ll.dave at gmail.com> wrote:
> 
> 
> Maximum force in the rope is 743 pounds.
> 
> This is a "sum of the moments" problem.
> 
> The tower won't start to move until the rotating force in foot-pounds about 
> the hinge in the upward direction exceeds the foot-pounds in the downward 
> direction
> 
> Foot-pounds in the downward direction = (50#)(40') + (160#)(20') = 5200 
> ft-lbs.
> where:
> 50# is the weight of the load 40' from the hinge
> 160# is the weight of 4 sections of Rohn 25 with CG at a distance of 20' 
> from the hinge
> 
> If you attach a line to the tower at 10' out and to the house at 10' up, the 
> closest approach of that rope to the hinge is 7'.  That is the distance you 
> use to figure the leverage of the rope.
> 
> Force in the rope = 5200 ft-lbs / 7' = 743 pounds
> where
> 5200 is the downward moment
> 7' is the leverage distance of the rope.
> 
> Dave Hachadorian, K6LL
> Yuma, AZ
> 
> 
> -----Original Message----- 
> From: Dave Tipton
> Sent: Wednesday, May 29, 2019 6:37 AM
> To: towertalk at contesting.com
> Subject: [TowerTalk] Calculating Forces for Tilting tower
> 
> I have a 40 foot Rohn 25, with a tilt base section.   It will be sitting 
> next to a detached framed (not block) garage with siding.
> 
> I can attach a winch to an extra guy bracket I have at the ten foot level.
> 
> There will be roughly 50 lbs at the top of the tower.  (Rotator, Hex Beam, 
> GP3)
> 
> How much is the maximum force required to raise this from a cable at the 10 
> foot level?
> 
> Thanks,
> Dave, W5DMT
> 
> Sent from my iPhone
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