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Re: [TowerTalk] Tram rope

To: Jim Lux <>
Subject: Re: [TowerTalk] Tram rope
From: Eric Scace K3NA <>
Date: Tue, 19 Dec 2006 15:38:12 -0500
List-post: <>
Hard to draw a diagram on a text-oriented reflector.  If you read this 
message with a fixed width font it should work reasonably well.

First, the tram line in this diagram is like the gondola lift you sited, 
Jim.  Tram wire sits overhead, and the sheave of a pulley rides on it.  
The antenna dangles below, and attached to, the frame of the pulley.

Sketch at the point where the pulley rests on the tram:

<-- tower is towards the left.

  |  '-._
(ant)    '-._ tram line continues off to right toward ground anchor.

There are three force vectors meeting at the pulley:
   -- antenna weight, points straight down.  100 lb in this example.
   -- up toward the tower to the left.  I used 45 degrees in my quickie 
statics example.
   -- to the right down to the ground.  I used 30 degrees below horizontal.
If you resolve the last two vectors into vertical and horizontal 
components, you can create two equations with two unknowns, and solve.

Of course, I did the algebra without the benefit of a morning cup of 
coffee... and got wrong answers.  The answers for a 100 lb antenna 
should be:
   333 lb up toward the tower
   273 lb down toward the ground anchor

The 237 lb force up and to the left at 45 degrees gives two parts:
   sin(45) x 333 = 236 lb component pointing straight up.
   cos(45) x 333 = 236 lb component pointing to the left.
The 273 lb force down and to the right resolves into:
   sin(30) x 273 = 137 lb down
   cos(30) x 273 = 236 lb right
The two down forces (ant of 100 lb + 137) = the up force (within 
round-off error).
The left and right forces also balance (within round-off error).

As others have mentioned, additional forces at play in the real world 
include weight of the tram line, changing angles as the antenna moves up 
the tram, and transient forces used to overcome friction, start the 
antenna moving, swaying, etc.  So 333 lb is just a starting point...

-- Eric (the algebraically challenged) K 3 Numerical Analysis

on 06 Dec 19 Tue 10:10 Jim Lux said the following:
> At 09:32 PM 12/18/2006, Eric Scace K3NA wrote:
>> Yes, I saw a tram line part and the results were not pretty.  The person
>> operating the winch narrowly escaped injury by the antenna and by the
>> parted line.  This occurred early in my ham antenna career as was a real
>> wake-up call to the forethought required for safe installation.
>> A tram line's forces act against the dead weight and wind load of the
>> antenna being installed.  Because the tram lines come off at an angle,
>> those forces are much larger.  (Draw out a force diagram using high
>> school physics.)  Assume a tram line leaves the pulley holding up the
>> antenna at a 45 degree angle up and toward the tower, and on the other
>> side continues on at a 30 degree angle below horizontal in the opposite
>> direction toward the ground.  For a 100 lb antenna, the force on the
>> line to the tower is 738 lbs and on the line to the ground is 602 lb.
> I'm having a bit of trouble visualizing the forces here.  Is the 
> antenna supported by pulley hanging from the tram line (a'la a gondola 
> lift at a ski area), so you're talking about the forces on the support 
> line.  Or are you skidding the antenna up a guy wire, and the tram 
> line is just the hauling line pulling it up at the angle of the guy?
> Either way, I have a hard time getting 45 degrees to turn into 7.38 or 
> 1/7.38.
> Can you point me to a diagram?
>> That is the static case.  Loads will vary when one attempts to start the
>> antenna moving from that point.
>> -- Eric K3NA

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