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  1. #1
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    Default A little maths help

    Evening gentlemen,

    I have a sizable(for me) gum tree that has blown 1/3 over, taking a fair root ball with it. Its into the top of another leaner and a vertical dead tree.
    Of course the vertical dead tree is down hill in the direction of the lean of the leaner.
    Felling the leaner sideways maybe an option, but for now I'm thinking about the possibility of winching the tree back up, then winching into another lay.
    My question is, When winching from a point say 10m up the trunk to a block 10m up a neighboring tree, what % of the trunks weight is on the winch and what % is still carried by the ground?
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  2. #2
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    It depends on where the centre of mass of the tree is. If it is where you have tied your rope, the horizontal component of the weight is all on the rope. Above or below, it's in proportion (distance centre of mass to pivot*/ distance attachment point to pivot)

    *rootball

    Michael

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    ok Thanks

    Lets assume the center of mass is the center of the tree so 12m. (ok that's a WILD guess, but you have to start somewhere.)
    You can assume the tree weighs 5t(if that helps, another wild guess)

    With a 30m rope I make the angle about 145deg. That gives me a 0.6 factor on the block right?

  4. #4
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    If the centre of mass is at 12m and your rope at 10m, then you would be lifting 12/10 of the mass (that is, the horizontal component of the 5t mass). If your tree was at 135 degrees, my basic maths says you would have to be able to pull about 6t (I don't work in tree removal either). As the tree was pulled upright, that force would lessen, but the adjacent tree with the block on it would need to be up for that initially. If it is further over (145 degrees), the force would be greater

    Michael

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    Hi Michael,

    I wonder if my drawing has lead to a misunderstanding of what I am trying to do, looking at it again I couldn't blame you.

    Pictures may do a better job.

    Red arrow is the tree the pulley will be in.
    Blue arrow is the tree I want to pull upright.
    Green arrow is the leaning tree that the blue tree is hanging up in.
    Purple arrow is the dead standing tree.(its behind the green tree in the first pic and I'm not sure which one it is in the third pic)
    Off to the left of my drawing is a track going down hill at 6deg.

    Can't cut the green tree down ATM as it will just hang up in something else(wouldn't be that thrilled at the idea anyway).


    Does that change anything?
    Or did I just not get the answer I wanted? lol (I was guessing the pull would be around 1/3 the weight of the tree)
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  6. #6
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    Quote Originally Posted by Stustoys View Post
    Does that change anything?
    Or did I just not get the answer I wanted? lol (I was guessing the pull would be around 1/3 the weight of the tree)
    I think it is more the latter and not much of the former.

    In all these cases I'm thinking of the root ball as a fucrum and the trees as lever arms. You could do a little practical experiment with a broom and a piece of string I guess but the mass of the leaning tree (due to gravity) is a force downwards. The pulling force is roughly horizontal and if the tree is at 45 degrees, to resolve the vectors they are going to be around equal.
    Looking at the photos, apart from you being in a very attractive spot there, I would suggest that the tree you want to pull up and the pulley tree are roughly equal in size, so a force on the pulley tree able to pull up the leaning tree is likely to damage the pulley tree.
    Seeing the photos I think I'd leave it if it did not have to be cleared. Short of throwing chainsaws at it and hoping for a lucky hit, it's going to be an awkward thing to try to clear I think.

    Michael

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    Quote Originally Posted by Michael G View Post
    I think it is more the latter and not much of the former.
    Thought you might say that lol

    Quote Originally Posted by Michael G View Post
    You could do a little practical experiment with a broom and a piece of string I guess but the mass of the leaning tree (due to gravity) is a force downwards. The pulling force is roughly horizontal and if the tree is at 45 degrees, to resolve the vectors they are going to be around equal.
    I have done a little of the broom handle thing. I might scale it up a little(wasn't 100% this would scale, sure my guess is it does.. but that's not worth much)

    Quote Originally Posted by Michael G View Post
    Looking at the photos, apart from you being in a very attractive spot there, I would suggest that the tree you want to pull up and the pulley tree are roughly equal in size, so a force on the pulley tree able to pull up the leaning tree is likely to damage the pulley tree.
    Thank you, we like it.

    Sort of operating on the idea that an force I can put on the tree is peanuts to what 100kph wind can do.(yet another guess, but that's also when I wasn't thinking in the 6t range and I thought the rope angle gave me the (load x 0.6) on the tree)

    Quote Originally Posted by Michael G View Post
    Seeing the photos I think I'd leave it if it did not have to be cleared. Short of throwing chainsaws at it and hoping for a lucky hit, it's going to be an awkward thing to try to clear I think.
    There is no hurry on this(in fact I'd like it to stay up for a few weeks yet), but it does need to go. Its just a bit to close to the house(close in the sense of people "going for a wander", not the "falling on the house" sense). Though granted ATM we don't have to worry about visitors going for a walk.

    The problem with real winching gear is once I start putting gear up the tree, getting it back is tricky at best until the tree is on the ground.

    I might have option using telstra rope, but I'd need to get the load down to about 600kgs. I might see just how far I can get a throw line up the tree.

  8. #8
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    Exclamation Rusty Maths Help

    Quote Originally Posted by Stustoys View Post
    Hi Michael,

    I wonder if my drawing has lead to a misunderstanding of what I am trying to do, looking at it again I couldn't blame you.

    Pictures may do a better job.

    Red arrow is the tree the pulley will be in.
    Blue arrow is the tree I want to pull upright.
    Green arrow is the leaning tree that the blue tree is hanging up in.
    Purple arrow is the dead standing tree.(its behind the green tree in the first pic and I'm not sure which one it is in the third pic)
    Off to the left of my drawing is a track going down hill at 6deg.

    Can't cut the green tree down ATM as it will just hang up in something else(wouldn't be that thrilled at the idea anyway).


    Does that change anything?
    Or did I just not get the answer I wanted? lol (I was guessing the pull would be around 1/3 the weight of the tree)

    Hi Stu

    My maths is a little rusty, but it seems to give you an answer closer to your guess.

    If the tree is leaning at 20 deg, and assuming the root ball acts as a fulcrum (a big assumption) and your rope is horizontal AND the centre of gravity is at your rope tie point, the force (F) needed to rotate the tree back to vertical should be M* tan 20 =0.36 M. Not far from your guess of 1/3.

    If the centre of gravity is higher than your rope tie point, the force required would increase by the lever ratio (1.2 in the example you gave).

    If your winch is a fair way from the red tree so that rope is at about 6 deg from horizontal, the force on the red tree will be downwards, and roughly F * sin 6 = 0.1 * F.

    So your 5 tonne tree with 12 m. centre of gravity and 10 m. tie point would need a winch force of 2.2 tonnes, and the red tree and its block would need to withstand about 220 kg downwards.

    Best to apply plenty of safety factor to the numbers, and stand well away from that rope when pulling, as you don't really know any of those weights or how much the tangle in the other tree will require, or if the root ball will bend back easily.

    Hope this helps save that tree. Your area does look pretty good with them all standing.

    Cheers

    Al

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    Some rough results.
    Not exactly consistent but I will have a think and repeat the tests with a better setup.

    2400mm x 75 x 50 hardwood, weight 7.80kgs at 70 degs

    Lifting point 1400mm from floor.
    Pulley height 1000 1200 1400
    Load on rope 3.4kg 2.6kg 2.07kg

    Lifting point 1200mm from floor.
    Pulley height 1000 1200 1400
    Load on rope 3.0kg 2.6kg 2.5kg

    Lifting point 1000mm from floor.
    Pulley height 1000 1200 1400
    Load on rope 3.1kg 3.35kg 3.7kg


    The root ball should help(but then I have already used that to say the mass is balanced). We have a couple of tree stumps 8 to 10 ft high and I had wondered "why would someone bother to climb up there to cut it?", it finally dawned on me (I'm guessing) they are blow downs that have "stood back up" when cut up. I hope the guy on the chainsaw was expecting it.

  10. #10
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    Quote Originally Posted by Alh01 View Post
    Hi Stu
    Ok thats strange.
    I got the email for this 10:08am 29 Aug.
    Did you really post at 26th Aug 2021, 11:26 PM?
    Sorry if you did.
    Got to duck out I will read later.

  11. #11
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    Quote Originally Posted by Alh01 View Post
    If your winch is a fair way from the red tree so that rope is at about 6 deg from horizontal, the force on the red tree will be downwards, and roughly F * sin 6 = 0.1 * F.
    Can't do that. Even on flat ground I make that 95m. The ground goes down at 6 degs for awhile and then gets worse. Don't have that much winch cable either.

    Quote Originally Posted by Alh01 View Post
    Best to apply plenty of safety factor to the numbers, and stand well away from that rope when pulling, as you don't really know any of those weights or how much the tangle in the other tree will require, or if the root ball will bend back easily.
    Both my winches have remote control so I can be well out of the way. Most of my gear is 9ton WLL or more so the rope will fail first. Though when I want some spring I've been known to run out a couple of hundred meters of telstra rope double it up, then use a scale to put 300kgs on each leg, can't see that helping this time.



    Quote Originally Posted by Alh01 View Post
    Hope this helps save that tree. Your area does look pretty good with them all standing.
    Well it wont save this tree but it might just save the 4 trees "to the right" that I would have to cut down to clear this mess if I cant lift it back up.

    I'll start by getting some telstra rope up as high as I can and pulling it down the hill, never know your luck.(hill gets pretty steep so its going to be pulling down as much as sideways in all likelihood).

  12. #12
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    Hi Stuart,

    You could anchor a pulley as high as you can get on an adjacent tree in order to give you a better pulling angle !
    Best Regards:
    Baron J.

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    Quote Originally Posted by BaronJ View Post
    Hi Stuart,

    You could anchor a pulley as high as you can get on an adjacent tree in order to give you a better pulling angle !
    Hi Baron,
    As I understand it that doesn't actually help, unless you can also get the anchor point on the leaning tree up to the same height.
    What I think is going on you can see in my figures.
    The lowest force* is when the pulley and anchor are at the same height. If you move the pulley higher than the anchor some of the force goes into trying to left the tree vertically. If you move the pulley lower than the anchor some of the force is pulling the tree into the ground.

    *ok the 1200 and 1400 test at 1200 high pulley dont show that but my second set of tests did.

    I didn't put them up because I'd used three G clamps to mount anchors and I realised that my weigh was no longer uniform over the length of the "tree". Anyway here they are.
    2400mm x 75 x 50 hardwood, weight 9.77kgs at 70 degs

    Lifting point 1400mm from floor.
    Pulley height 1000 1200 1400
    Load on rope 4.2kg 3.3kg 2.84kg

    Lifting point 1200mm from floor.
    Pulley height 1000 1200 1400
    Load on rope 3.9kg 2.2kg 3.3kg

    Lifting point 1000mm from floor.
    Pulley height 1000 1200 1400
    Load on rope 3.9kg 3.9kg 4.3kg

  14. #14
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    Golly all these calcs are making my brain hurt, but one thing I can say that is when lifting power poles (made from tree wood), if you choke hitched the pole at a point 1/3 along the length from the butt end, it would almost hang vertically, certainly vertical enough to place in the hole prepared for "the planting." At a casual glance the poles looked parallel, with no noticeable taper. Of course the taper was there alright if you looked carefully, so I would imagine that Stuarts tree will not have anything like a linear c of g relationship either. Another thing to factor in would be wind load. If the canopy is subject to any significant wind, the sail area could make a big impact on any attempts to manipulate that tree with winches.
    We have just cleared quite a few trees from around our home which were getting dangerous both from a falling and impacting aspect and also a fire risk which would have made the house undefendable should they have caught alight. We got a contractor with a fair sized excavator, with a three tyned grab with a thumb function, and limited rotation, which could reach 8-10 metres up and control the fall of the tree when the tree feller did his work. The cost to us was $250/hr for the excavator and tree faller. Something more to ponder Stu, but good luck with your project.

  15. #15
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    Quote Originally Posted by Ropetangler View Post
    Golly all these calcs are making my brain hurt,
    A man needs a hobby
    Really just doing the cals to be sure there is at least a chance of success. As I said the c of g is an a wild guess, the tree still has its top so that moves the cg up, but it also has a root ball that moves it down, the root ball has some weight lifting the tree, very likely there are some roots trying to help out as well. But I hope you are right, the lower the cg the better, but then the weight is a wild guess as well.
    As far as wind load goes there isn't a lot to the top of this tree, but what there is does have a hell of a lever. I certainly wont be trying to lift it if there is any wind. As I said I am in no hurry so I can afford to wait for the right day. What I might do is run the rope as per my last post, load it up, then leave it and see if the rope + the wind pulls it out.

    Hey its pretty windy today maybe it will sort itself out

    Calling in the big guns is an option, we shell see.

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