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Re: [TowerTalk] HF winch

To: towertalk reflector <towertalk@contesting.com>
Subject: Re: [TowerTalk] HF winch
From: Mike Reublin NF4L <nf4l@comcast.net>
Date: Mon, 3 Feb 2014 10:14:35 -0500
List-post: <towertalk@contesting.com">mailto:towertalk@contesting.com>
Great Jim, thanks!

Lets say 1200 lbs., length of 32' including mast.

So 1200 x 32 / 10 = 3840.

Cos(0) = 1, sin(45) = .707

3840 x 1/.707 = 5430 lbs. initial pull, correct?

Halving that with a 2:1 pulley system = 2715 lbs., a bit over the winch rating. 
A 3:1 pulley arrangement drops it to 1810. Do-able but close to the edge.

Mike NF4L

On Feb 3, 2014, at 9:48 AM, Jim Lux <jimlux@earthlink.net> wrote:

> 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)
> 
> 
> 
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