[TowerTalk] HF winch

[Jim Lux]

On 2/3/14 6:26 AM, Mike Reublin NF4L wrote:
> Thanks David,
>
> The plan is to have a 10' tall mast, and attach the cable 10' from
> the bottom of the tower, giving a circa 45 degree pull when the tower
> is horizontal.
>

How tall is the tower?
The "down" force at your 10 foot pick point is tower weight* length/10

As an example, say you've got 50 feet of Rohn25, which weighs about 200 
pounds.

So, if you had a scale and were lifting straight up on the horizontal 
tower at the 10 foot mark, it would be about 1000 pounds (= 200 lbs * 50 
ft/10ft)

But you're not picking straight up, you're pulling at a 45 degree angle, 
so the actual pulling force is more like 1400 lbs.  (= 1000lbs/sin(45))

When you're half way up (tower at 45 degrees) the angle for the pull is 
better (22.5 degrees) and the load is less (because some of the weight 
is being carried by the base of the tower).  The pull is now

1000lb*cos(45)/sin(67.5) = 1000 * 0.707/0.924 = about 770 lbs

And then it rapidly gets better.


For your specific case of a pick point at 10 ft and a 10ft high pole 
with the winch, the formula is:

load in lbs = weight of tower * length of tower/10 ft * cos(angle of 
tower with respect to ground) / sin(angle of cable with respect to tower)

At the beginning, with tower flat on ground:
angle of tower = 0
angle of cable = 45

At the end, with tower vertical
angle of tower = 90
angle of cable = 90





> Do you know a formula a non-engineer could use to calculate the pull?
> The ones I find are all based on Newtons, And not the one I know,
> Fig.
>


4.5 Newtons = 1 pound  (approximately)