Mechanical avantage pulling trees

treejames64

I was wondering if anyone new a formula or an answer to this? It has always bugged me. If I'm pulling a say 100 ft. tree from a hill that's 100ft high or a direct level pull with the top with say 200 lbs force. What is the loss of pull if the height of the hill is halved to say 50 ft? Is it directly proportional? Would the force be halved to 100lbs. I was trying to figure out on tall trees how if you could do a quick calculation as to the angle between pull rope and tree for ex. 30 degrees. Does that mean 70 percent of your pull is lost? How does back leans affect it?

Eric Frei

The amount of force on the rope is constant, how that force applies to the tree when the rope is not horizontal is called vectors, broken into to axis of force (vertical and horizontal) using triangulation (sin, cosine).

In this website is a java application that allows you to drag a given force through various angles.

Vector Calculator

Note that a magnitude "10" force at 45degree angle applies approximately "7" force to both the vertical and horizontal plane (X and Y axis).

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treejames64

Thanks a lot. Great calculator though I'm still haveing trouble gettion what I want out of it ( I took trig over 20 years ago). Though what I hear you saying at is with a 100 lb pull at a 45 degree angle I would have 70 lbs. pull vertical. I'm still trying to figure a tree with a back lean. I know being off center even a little has to change this pull because now the entire weight of the tree would be putting force against you. Does weight of the center vector change these equations?

Eric Frei

No. The entire weight of the tree will not be putting force against you.

But what you will have is a small horizontal vector force pulling in the opposite direction (back leaning tree) as most of the wieght of the tree will be vertical (down the trunk).

Using trig you can calc out that force, as long as the pull line horizontal force exceeds it the tree comes toward the pull rope.

Using the same calculator if the tree weighed 10tons and was on a 10 degree lean (that's a lot) there would be around 9.9ton of force down the trunk (vertical axis) and 1.7ton of force horizontally away from the pull rope.

To pull the tree opposite with the rope at a 45degree angle and anchored above the centre of mass of the tree you'd need 2.5ton+ of force ... that would mean 1.7ton+ opposing the horizontal component.

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treejames64

Thanks, I use trig with a compass to calculate distance for crane work when there's a house in the way so I guess I'm going to have to figure this out by getting the books out. My wife has a doctrate in business and doesn't know how to do this...Thanks a bunch for the example. I'm always trying to guess this stuff in a tree and want to know approx horizontal pulls.

newguy18

I was taught the for every degree of angle that is not inline with the load your pulling you lose one percent of the pulling power.

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old schooler

moray

Change the last paragraph to "at the center of mass..." and it is exactly correct. Keep the 45 degree angle but move the anchor point further up the tree and the required pull force goes down. The horizontal pull force, correctly calculated by Ekka, multiplied by the moment arm (anchor height above felling cut) must exceed horizontal force of leaning tree multiplied by its moment arm (height of center of mass above felling cut). It sounds complicated to say it, but most people understand it more or less instinctively. But to calculate it you need a bit of trig.