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Let us in on your secret code..........If you have something to say then say it.........I am sure us little boys can take it.
Pulling Strategy
- Thread starter TheTreeSpyder
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Let us in on your secret code..........If you have something to say then say it.........I am sure us little boys can take it.
i think the mechanichs have not changed, and wondered if anyone else agfter several examples in different applications and explanations now sees the direction as another force to maximize, not jsut steer.
By using whatever force you muster for most power; then taking that a step further; and now maximize how that force enacts upon the target. By applying it specifically higher for more power, but also look at direction as power beyond steering fall, but also the arc on the hinge imposed. That a straighter across pull is less efficient use of your force than arching on the hinge.
It is an element i try to point out in a few things for easily making things more easy and powerful. An arching force pulls to felling target less directly, but using a leveraging machine; therefore the force is not wasted but increased, by rebounding the force off the leveraging, rather than pulling directly to target.
Orrrrrrrr something like that..
:alien:
By using whatever force you muster for most power; then taking that a step further; and now maximize how that force enacts upon the target. By applying it specifically higher for more power, but also look at direction as power beyond steering fall, but also the arc on the hinge imposed. That a straighter across pull is less efficient use of your force than arching on the hinge.
It is an element i try to point out in a few things for easily making things more easy and powerful. An arching force pulls to felling target less directly, but using a leveraging machine; therefore the force is not wasted but increased, by rebounding the force off the leveraging, rather than pulling directly to target.
Orrrrrrrr something like that..
:alien:
Part of this principal intersects with the "Mayhem Puzzle" in Talking at H.S. about Arboriculture Thread
i think this technique for rigging and felling introduces the compressive force that Stumper spoke of a while ago. But at the first hair of movement the inertia of the forces is still pulling at the original angles; but the spar has shifted slightly, now it is not compressive, but rotational force on the load. With hitch pulling up at rotational angle as the top is pused down at a rotational angle, compounding each other's actions. The first folding i beleive sets maximum strength of hinge, then should leave it alone unless it stalls for slowest, dribbling, in granny gear power.
The rotational force appearing suddenly at first upset can pulse through a higher loading pull; just at that first folding(??)
i am just seeking the explanation for what i believe i've witnessed, a helpful momentary force, at the right instant.
Orrrrr something like that
:alien: :alien:
i think this technique for rigging and felling introduces the compressive force that Stumper spoke of a while ago. But at the first hair of movement the inertia of the forces is still pulling at the original angles; but the spar has shifted slightly, now it is not compressive, but rotational force on the load. With hitch pulling up at rotational angle as the top is pused down at a rotational angle, compounding each other's actions. The first folding i beleive sets maximum strength of hinge, then should leave it alone unless it stalls for slowest, dribbling, in granny gear power.
The rotational force appearing suddenly at first upset can pulse through a higher loading pull; just at that first folding(??)
i am just seeking the explanation for what i believe i've witnessed, a helpful momentary force, at the right instant.
Orrrrr something like that
:alien: :alien:
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In both the first attachment to this thread, and the last one, once you get to the initial point of contact between the tree and rope, anything else you do is totally meaningless. Except, that is, for the situations where the rope is passed back down and fastened to a lower point, when you will be creating the related detrimental situations of having to pull more rope as you take up all the slack, and (thus) creating vibrations in the spar as the rope slides over/through the contact point.
In fact, in this last image, the bottom section is plain wrong. If you pull 100_lb. on the rope that's all the force there is. It's the same 100_lb. both sides of the crotch (assuming zero friction); your 200_lb. figure exists only in your mind. In the left side nothing happens except trying to compress the tree and in the right nothing happens any differently (any at all) than if you moved the lower hitch to the top of tree, just below the crotch (except that then you'd be taking up much less slack). Compression of the tree is the same in all of the situations and arises solely from the pull on the free end of the rope.
You may certainly induce a rotational component to the pull by tying (or passing the rope) off-center, but once that initial contact point is established, the rest of what you're trying to do is all smoke and mirrors.
Saying the tree will be lifted by the lower attachment point is like saying you can lift yourself off the ground by pulling up on your boot straps.
Here's a link that looks both easy and comprehensive enough: http://id.mind.net/~zona/mstm/physics/mechanics/vectors/findingComponents/findingComponents.htm
Glen
In fact, in this last image, the bottom section is plain wrong. If you pull 100_lb. on the rope that's all the force there is. It's the same 100_lb. both sides of the crotch (assuming zero friction); your 200_lb. figure exists only in your mind. In the left side nothing happens except trying to compress the tree and in the right nothing happens any differently (any at all) than if you moved the lower hitch to the top of tree, just below the crotch (except that then you'd be taking up much less slack). Compression of the tree is the same in all of the situations and arises solely from the pull on the free end of the rope.
You may certainly induce a rotational component to the pull by tying (or passing the rope) off-center, but once that initial contact point is established, the rest of what you're trying to do is all smoke and mirrors.
Saying the tree will be lifted by the lower attachment point is like saying you can lift yourself off the ground by pulling up on your boot straps.
Here's a link that looks both easy and comprehensive enough: http://id.mind.net/~zona/mstm/physics/mechanics/vectors/findingComponents/findingComponents.htm
Glen
i have funned about one laying next to me, sleepless over Demi Moore's makeup. But honestly many sleepless knight tumbling with this query.
Tim Gardner
ArboristSite Guru
Ken it seems like you could set up an experiment on a smaller scale to test your theories. In an environment where you could control all forces and repeat them would lead to definitive answers. I think your test results would be very interesting.
Tim Gardner
ArboristSite Guru
B and D seem incorrect to me.
"B" and "D" <i>are</i> incorrect.
I think I see where you're going in "A" and "C", but in "B" and "D", you're essentially trying to pick yourself up by standing on your hands and lifting your feet. In "A" and "C" you have the fulcrum independent of both the load and motive force. In "B" and "D" you have the equivalents of "A" and "C" but with the rope between the load and "sheave" passing through a full-length steel pipe. No work is accomplished unless the load moves relative to the pulley in these cases.
In "B", if the item weighs 200, the load on the free rope will be 200 to suspend the item.
In "D", if you're pulling straight down at 100, the downward force will be 100. If split equally sideways and downward (45°
, the 100 will net 70.7 each downward and sideways <i>at the point of contact</i>. The downward force is always the same in any given situation "irregardless" [a non-word which is a humorous combination of irrespective and regardless] of whether the stationary end of the rope is anchored at the top or anywhere else along the stem.
Glen
I think I see where you're going in "A" and "C", but in "B" and "D", you're essentially trying to pick yourself up by standing on your hands and lifting your feet. In "A" and "C" you have the fulcrum independent of both the load and motive force. In "B" and "D" you have the equivalents of "A" and "C" but with the rope between the load and "sheave" passing through a full-length steel pipe. No work is accomplished unless the load moves relative to the pulley in these cases.
In "B", if the item weighs 200, the load on the free rope will be 200 to suspend the item.
In "D", if you're pulling straight down at 100, the downward force will be 100. If split equally sideways and downward (45°
, the 100 will net 70.7 each downward and sideways <i>at the point of contact</i>. The downward force is always the same in any given situation "irregardless" [a non-word which is a humorous combination of irrespective and regardless] of whether the stationary end of the rope is anchored at the top or anywhere else along the stem.Glen
Here is a stripped to basic components view. i believe the same pattern exists; i beleive we have all tested these things out in the more familiar forms.
i think some of the power that i see in self torquing limbs, hinge pocket, over the hill, leg on load, hinge strength etc. ideas hinge on this concept.
i am assuming zero friction, zero deflective angles right now, for catching sight of this at peak to trace as it goes to lesser power (but still helpfull) as friction and angle are added.
In the Self Tightening Torque Thread; you have the friction at the turn of the branch that lessens the pull; but any tension you can get on B side still helps, so i sweat that point in to leverage a higher purchase of power from the line. Then sweat the rest tight from climber's post; this also makes sure ground control commands the control leg of the right line. Now, in leveraging right; if the line is tight enough, i will begin by cutting down, and finish by sliding across. i do this so i empower the leveraging system of the line by loading the line by cutting down, not the hinge by cutting across. Then try to catch at just when the line gets tired of the 'abuse' and responds/leverages back, pulling across more powerfully, forcing the strength in the hinge as it does.
Orrrrrrrr something like that
:alien:
i think some of the power that i see in self torquing limbs, hinge pocket, over the hill, leg on load, hinge strength etc. ideas hinge on this concept.
i am assuming zero friction, zero deflective angles right now, for catching sight of this at peak to trace as it goes to lesser power (but still helpfull) as friction and angle are added.
In the Self Tightening Torque Thread; you have the friction at the turn of the branch that lessens the pull; but any tension you can get on B side still helps, so i sweat that point in to leverage a higher purchase of power from the line. Then sweat the rest tight from climber's post; this also makes sure ground control commands the control leg of the right line. Now, in leveraging right; if the line is tight enough, i will begin by cutting down, and finish by sliding across. i do this so i empower the leveraging system of the line by loading the line by cutting down, not the hinge by cutting across. Then try to catch at just when the line gets tired of the 'abuse' and responds/leverages back, pulling across more powerfully, forcing the strength in the hinge as it does.
Orrrrrrrr something like that
:alien:
If the anchors are all physically independent of the redirects (that is, connected <i>only</i> by the rope and by no common solid material), then all of your last examples are the same. The problem you're getting yourself into is when you connect anchor to redirect with solid material; you no longer have the simple (or any?) machine in such cases.
I'd love to watch you work a few problems in the field. You obviously put considerable thought into your work and there's no doubt you get the results you're trying for. I perceive you're plagued with generally having a clear mental image without an equivalent capability to convey it (the image) to others. I also gather you achieve certain results for reasons other than you'd envisioned.
Glen
I'd love to watch you work a few problems in the field. You obviously put considerable thought into your work and there's no doubt you get the results you're trying for. I perceive you're plagued with generally having a clear mental image without an equivalent capability to convey it (the image) to others. I also gather you achieve certain results for reasons other than you'd envisioned.
Glen
Stumper
One Man Band
KC, In the tree trunk images B is incorrect if it is showing the tree suspended upside down with the rope passed through the crotch then tied on the main stem. Regardless of the rope routing a suspended load only weighs what it weighs. It loads the supension line with its full weight but no more on that leg.
Glens-i guess i fail to see where if you have the same pattern, that how the rope knows how it is mounted and to what beyond the assumed pattern ?
Stumper-i think the pattern is the same, in each case the line is tensioned to the 100# (If B 'load' weighs 100#); that sets the line tension in A to 100#; to run the same pattern of the Jay hook; assuming the coveneant of the arc power, if ye will.
i think, if in C(line looped over top of slingshot/tree tied off to own base and pulled on other leg) had a hardened foam material that you needed to crush, and only had own body wieght as power source, there would be a point where the line would crush it as a 2/1 where your body weight on top wouldn't? Assuming ZeroFriction, ZeroLineAngle, Equal footprint of pressure on foam between the 2 methods.
i don't think anchoring to self matters; i think the foam would be 2/1 crushed by the line whether the end was anchored to itself or another anchor. i don't think the line cares! Then, just becuase the line isn't crushing the tree with out the foam, does not mean that the line isn't loaded, just that the tree can well ressit the force, but it still exists.
So, kind sirs; i offer another view, an explantaion to the free hanging piece perhaps having 2x at the Jay hook as any other system it resembles. Where does that extra 100# come form, how can it be there without lifting?
?Stumper-i think the pattern is the same, in each case the line is tensioned to the 100# (If B 'load' weighs 100#); that sets the line tension in A to 100#; to run the same pattern of the Jay hook; assuming the coveneant of the arc power, if ye will.
i think, if in C(line looped over top of slingshot/tree tied off to own base and pulled on other leg) had a hardened foam material that you needed to crush, and only had own body wieght as power source, there would be a point where the line would crush it as a 2/1 where your body weight on top wouldn't? Assuming ZeroFriction, ZeroLineAngle, Equal footprint of pressure on foam between the 2 methods.
i don't think anchoring to self matters; i think the foam would be 2/1 crushed by the line whether the end was anchored to itself or another anchor. i don't think the line cares! Then, just becuase the line isn't crushing the tree with out the foam, does not mean that the line isn't loaded, just that the tree can well ressit the force, but it still exists.
So, kind sirs; i offer another view, an explantaion to the free hanging piece perhaps having 2x at the Jay hook as any other system it resembles. Where does that extra 100# come form, how can it be there without lifting?
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Tim Gardner
ArboristSite Guru
I can not see how you could get 2 times the load on a rope when the anchor point and load are the same. It seems to me the load on the rope at the point where it runs thru the crotch would be less than the total load up to the point where it exits the bend in the rope. The way a round turn on a ring spreads the load out and strengthens the hitch.
Yes, Leg A, that is source of line pull here, instead of arm as source. Then, with no friction; must B, not likewise load?
i think that the lesson is; that C must be the sum of the pulls of A+B; even in this case. (in land of no friction, zero degrees bend)
If the foam had to be crushed down 5'; how much rope would have to be pulled?
The rope doesn't know how that it is anchored or pulled, just that it is, and the coveneant of the arc of power is imposed in between.
Or something like that
:alien:
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Kenny,
Set 180_lb on a small tray with handles and pick it up. How much force do your muscles need to exert? 180_lb total. You can pull with 1000_lb but the tray will (basically) see only 180_lb of that force.
Now, place the tray on the floor and stand on it. Then pick up as hard as you can on the handles. How much force will your muscles exert on the handles before the tray elevates? It's the same type of loading from your shoulders to the handles both ways so how can your muscles tell the difference? That type of argument is like my example of the three hotel rooms the other day in off-topic (I forget which thread, maybe the "official joke thread"). It's that kind of "physics" you're working with here.
If you have an independent load of 100_lb on the left-hand rope coming from your frictionless redirect point and you place a load of 100_lb on the right-hand rope you have a total of 200_lb on the redirect and the load can be held aloft. What happens when you pull on the right-hand rope with 200_lb of force? Obviously the load will rise, but how much weight is present at the redirect? It's still (basically) only 200_lb.
If the left-hand leg is affixed to the support for the redirect, there is no load hanging from that side, so any force applied to the right-hand leg is the total present at the redirect. Pull down with 50_lb and that's what the redirect sees. Make it 500_lb and <i>that's</i> what the redirect sees. It's basically the same as if you'd tied a large knot which jams in the redirect point, or if you'd tied directly <i>to</i> the redirect point (except with the lower tie-in you have more slack to take up; there's still <i>no load</i> on that side).
Get it? You're assuming counter-action where none exists.
Glen
Set 180_lb on a small tray with handles and pick it up. How much force do your muscles need to exert? 180_lb total. You can pull with 1000_lb but the tray will (basically) see only 180_lb of that force.
Now, place the tray on the floor and stand on it. Then pick up as hard as you can on the handles. How much force will your muscles exert on the handles before the tray elevates? It's the same type of loading from your shoulders to the handles both ways so how can your muscles tell the difference? That type of argument is like my example of the three hotel rooms the other day in off-topic (I forget which thread, maybe the "official joke thread"). It's that kind of "physics" you're working with here.
If you have an independent load of 100_lb on the left-hand rope coming from your frictionless redirect point and you place a load of 100_lb on the right-hand rope you have a total of 200_lb on the redirect and the load can be held aloft. What happens when you pull on the right-hand rope with 200_lb of force? Obviously the load will rise, but how much weight is present at the redirect? It's still (basically) only 200_lb.
If the left-hand leg is affixed to the support for the redirect, there is no load hanging from that side, so any force applied to the right-hand leg is the total present at the redirect. Pull down with 50_lb and that's what the redirect sees. Make it 500_lb and <i>that's</i> what the redirect sees. It's basically the same as if you'd tied a large knot which jams in the redirect point, or if you'd tied directly <i>to</i> the redirect point (except with the lower tie-in you have more slack to take up; there's still <i>no load</i> on that side).
Get it? You're assuming counter-action where none exists.
Glen
Tim Gardner
ArboristSite Guru
Ken, in your last illustration if that forked limb is still attached there would be 2x the load of pull at C if there were no friction. If the limb is hanging free I can not see how there would be more than the weight of the limb between B and C.
Again, I think you should test your theories. A set of fishing scales would be useful. I have used them to calculate the loss of pull in an 8 fold (16 pulleys) block and tackle due to friction.
Again, I think you should test your theories. A set of fishing scales would be useful. I have used them to calculate the loss of pull in an 8 fold (16 pulleys) block and tackle due to friction.
B would be whatever friction didn't reduce from A, or load falls. In any other use of this same pattern of loading C is A+B. i think it is just that the mount is also the power source, that it doesn't look so.
i have thought of trying the scales, they would not catch the real point beyond the power positions.
The pattern of the lacing is the same, we have all tested it for years. i have used this in the torquing limbs etc. and tried to explain them before.
i think there is enuff power here to hang around and take a chance and see.
The climber lifting self is a closed system, not as much as Glens, whereby lifting yourself up by bootstraps does not react on an outside position at all, therefore is not productive....
But still the DdRT fool the eye at first for the same reason i think, the system is closed, and the load position is not just serving one purpose but 2, so is hard to follow, but power to be had, none the less.
i have thought of trying the scales, they would not catch the real point beyond the power positions.
The pattern of the lacing is the same, we have all tested it for years. i have used this in the torquing limbs etc. and tried to explain them before.
i think there is enuff power here to hang around and take a chance and see.
The climber lifting self is a closed system, not as much as Glens, whereby lifting yourself up by bootstraps does not react on an outside position at all, therefore is not productive....
But still the DdRT fool the eye at first for the same reason i think, the system is closed, and the load position is not just serving one purpose but 2, so is hard to follow, but power to be had, none the less.
Tim Gardner
ArboristSite Guru
Any advantage that would be gained at the pulley would be lost at the tie off point, which is above the hinge while pulling a stick over. Thus it would not help pull the stick over anymore than if the rope was tied off at the top.
If that forked limb is hanging free there is not an opposite pull as there would be if it were attached. This is what I am suggesting you test.
If that forked limb is hanging free there is not an opposite pull as there would be if it were attached. This is what I am suggesting you test.
Very good on both counts; the extra leg of pull down on pulley on same spar is neutralized by the pull up along the major axis of the spar. But the pressure at the redirect position, still follows the pattern of loading topping out at 2/1 here, as outside constraint. Now, what happens the second that spar starts to move; and seeing as that is the same time frame as Forcing Hinge Strength is there usable force operating un-neutralized?
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i think that in non friction, non leveraged circumstance, you cannot exert 200# against 100#.
i think pulling on a line against any anchor, creates a passive, reflective force, mirroring your pull; matching in power, reversing in direction of pull. Friction free bend that line to parallel 0 degrees (2/1); you pull 100#(A), anchor responds back with a 100# pull(B),and the union of those lines with no deflected angle/direct/zero degrees /parallel to each other sustains both of the pulls :200#(C). In any form that you put that bent, frictionless line too loaded. In this way, an anchor point on 1 leg and 100# on the other is equivalent to 100# on either side, both loading the pulley, hitching and support point @2/1(200# in example).
i think Tim's observation that in a vertical spar the forces are cancelling each other on leg B, also is key to why puzzle does not rise with 200# on the redirect point of a 100#floating load; the hitch is pulling down with 100# to match the extra 100#pull up at pulley on puzzle. With that negated, your down to 100# of support on line for 100#load.
i think that is a lil'more rational explanation, than the loaded pattern yields these forces in any form except upsidedown.
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