[TowerTalk] Tension on tramline

[Howard Klein]

David,
I am plannng  the raising of an 90# antenna on a 32 foot boom to 82 feet on 
a self supporting tower. The hitch is I have a similar antenna at 40 ft. 
necessitating a tramline to pass the lower antenna. What magnitude of force 
would I expect on the mast? Could the tramline be untensioned (my word) once 
past the obstacle to release lateral force on the mast? I will give the math 
a try if I can set up the necessary parameters.

Howard...K2HK


From: David Robbins K1TTT <k1ttt at arrl.net>

I have an old article about this from the yccc newsletter for oct 1994.
unfortunately this one hasn't been converted and put on our web site
yet.  I will try to get this scanned and put on my web site as I can't
find the original for it here either.

The formulas for the static case are not too bad.  Imagine the angle 'a'
is the angle from the tower to the tram wire, where 0 is straight down
and 90 is horizontal from the tower.  And angle 'b' is the angle from
the ground to the wire, where 0 would be laying on the ground and 90
would be straight up. W = weight of antenna hanging from the wire.

]\
]a\
]  \
]   \
]    \
]     \
]      \
]       W-_
]           -_
]              -_
]                 -_
]                  b -_
=========================================
(don't you just hate ascii graphics!)

P=tension on ground end=W*sin(a)/cos(a+b)
L=tension on tower end=(W+P*sin(b))/cos(a)

Note that when the antenna is stationary the L tension is shared between
the lifting rope and the tram wire, assuming the lifting rope is
attached near the same point the tram wire is attached.

The important things are to note that as a+b approaches 90 degrees the
tension becomes infinite... so the more sag in the wire the better.
trying to design it generically is tough since so much depends on the
tower arrangement and how much room is available on the ground.  I found
it easiest to make a scale drawing of the side view, figure out how much
clearance is needed between the antenna being lifted and the tower at
various points(usually worst case is near the top of course), this will
give you a minimum value for the 'a' angle and the closest point you can
anchor the tram to the ground.  then determine how far out I had to put
the ground anchor to give a reasonable 'b' angle.

other notes:
The 'tiller' method of tilting up the antenna elements as it nears the
tower is a great way to reduce the needed clearance over the top guy
wires.

A back stay on the tram wire is almost always a good idea, the side
forces on the top of the tower for most reasonable angles get huge
quickly.  An extra clamp above the rotor to make sure the extra down
force doesn't end up crushing the rotor doesn't hurt either.

A quick example of the forces from the calculations I did for lifting a
170# telrex up a 150' tower with the ground anchor about 150' from the
tower base at about mid point of the lift:
Angles a=40, b=40
P= 629#
L= 750# (this is shared between wire and lifting rope)
Horizontal force on mast = 482#
Vertical force on mast = 1409# (tram+backstay+down force on lifting line
pulley)

More from old discussions on this topic are at:
http://www.k1ttt.net/technote/tram.html

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