It's easy or complicated, depending on your point of view. Here's a quickee
estimate.
You can sum up the point loads of the antenna, rotor, mast, and multiply that
by 35/30 to get the
vertical lifting force needed for that stuff at the attach point. (note it's
more than their weight
because of the relative position of them, attach and pivot point.
The weight of the tower is small (100 lbs I'm guessing?) but by lifting at the
end, you can say
that's a point load at the halfway point of the tower (think levers...it's all
about making moments
match at the pivot point at the base). If you didn't lift at the end, you'd
break it into two point
loads.
So your total vertical lifting force is maybe 200 lbs required? But your cable
is at a shallow
angle. think of a triangle and pythagoras for resolving the force vectors. with
the house 30 more
feet away, and 15 feet high, the cable will need see about 4x that 200 lbs in
it..i.e 800 lbs.
But the pulley is seeing 2x that, since the cable is going back to the tower:
the pulley sees both
the tower and the winch forces. I'm assuming the extra pulley at the tower
isn't providing much of
an extra compound pulley effect since it goes to the winch at the base, so I'm
ignoring that
benefit.
So the house pulley and attach will see 1600 lbs pullout force, which can be
decomposed into
horizontal and vertical components. (which might be interesting to understand,
depending on how you
attach to the house)
You can reduce this a lot by lifting the tower some other way from perfectly
horizontal for a
couple of feet. You won't be fully horizontal anyhow because the antenna is on
there.
If the forces are too high, you need to raise the height of the cable where it
crosses the pivot
point. You can do this by attaching at the house higher, or some other way.
But compared to a steel crankup, your forces aren't that high. However: it
calls into question what
you consider an adequate attach at the 15 ft point at the house.
Note also the tower attach sees more than just the vertical force, it sees the
force the cable
sees.
So there's no one equation or number. You have to analyze all the components
separately. Changing
distributed loads (tower) to point loads (before and after the attach
separately if necessary, then
using equal moments to figure out resulting forces. Then working out how the
angles affect the
force vectors
And don't forget some safety factor. Also, it's easy to overestimate how strong
a wall attachment
to a house is. (plus you have to worry about leaks/wood damage)
Separate analysis would be questions like: are you going to bend the tower or
base attach? You can
imagine that at some very low angle the answer is yes (probably not for you).
You can get that by
figuring out the horizontal component of the cable forces on the tower.
-kevin
ad6z
------- Original Message -------
>From : Alfred Frugoli[mailto:ke1fo@arrl.net]
Sent : 5/13/2010 12:10:18 PM
To : towertalk@contesting.com
Cc :
Subject : RE: [TowerTalk] Calculating load when raising tower
Hello Everyone,
Is there a simple equation for figuring out the forces that would be
experienced in tilting up a tower? I'm buliding a raising fixture, and want
to make sure that pulleys and the winch can handle the associated loads.
The tower is a 30 foot tower, and will be raised by a cable attached near
the top of the tower, that will run through a pulley 15 feet above ground
about 30 feet away from the tower (on an appropriately reinforced point on
the house), then back to a pulley near the top of the tower then to a winch
mounted at an appropriate height at the tower base. The tower is a 30 foot
aluminum tower with a rotor, 5 foot aluminum mast, and a small tribander
just above the top of the tower.
A 2nd question is would it be better to have the cable routed in a different
way, or attach at a lower point on the tower (level with the raising point
at ~15 feet)?
73 de Al, KE1FO
K3 #3055
K3 #4094
-----
Check out my Amateur Radio Contesting blog at ke1fo.wordpress.com.
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