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[TowerTalk] Ginpole for 20 foot sections

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Subject: [TowerTalk] Ginpole for 20 foot sections
From: (David Robbins)
Date: Wed, 13 Jun 2001 19:20:03 +0100
"Brown, James E LRDOR" wrote:
> And what if there are no workers applying "downforce", but instead the end
> of the line is tied off to a tree, and the load is suspended static?
> Jim   W4LC

its the same as if someone was holding it, you still have that downward force
exerted to keep the load where it was.  one basic misconception is that a force
being applied doesn't mean that anything is moving.  when all the forces exactly
balance everything stands still, which is what i have been calculating in all
these cases.  to get a load to start moving upwards you have to exert MORE than
the force required to hold it still... note this means that in the basic gin
pole arrangement with just the single pulley at the top the MINIMUM downward
force it has to be able to support is double the weight of the load.  it will
always take more force to get the load started depending on how good the bearing
is and how quickly you pull to get it moving (overcoming the inertia)... add in
extra forces caused by the normal jerking on the end of the stretchy ropes we
mostly use and it can be lots more than the weight of the load.

one interesting case is what happens if you have a ratchet in the system
attached to the pole itself or at the top pulley.  i have a small block and
tackle system that has this type of arrangement.  in this case when you let go
of the rope the ratchet at the top pulley grabs it and prevents the load from
falling.  now if you figure this in, when you let go of the lifting rope you
were pulling down on and the ratchet grabs it the load on the pole actually
drops to be just the weight being lifted.  this makes sense because the extra
force downward you were applying is now gone so all the top attachment point is
doing is holding the load up in the air by itself... but what replaced your
downward force???? friction.  if you were to measure the forces inside the
ratchet you would see the same force you were exerting is now being applied by
the ratchet mechanism, but since this is not pulling down there is only the dead
weight of the load seen on the top attachment point.

> -----Original Message-----
> From: Paul Finch []
> Sent: Wednesday, June 13, 2001 2:24 PM
> To: Tower Talk;
> Subject: RE: [TowerTalk] Ginpole for 20 foot sections
> Hello,
> So the weight exhibited to the Gin pole at the top pulley mount point is the
> down force of the load line plus down force it takes to hold the load static
> at the workers end of the load line?  No matter how many times the
> advantage.  Right?

right.  and the down force required by the workers is the weight/mechanical
advatage.  so the general formula would be    
weight*(1+1/mechanical_advantage)     for the downward force on the pole.

David Robbins K1TTT
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