# Pulling Strategy



## TheTreeSpyder (Jun 12, 2002)

i was wondering what you guys thought of the choices here to optimize security and pull!

The picture won't load to answer the question here, so it is below.


----------



## TheTreeSpyder (Jun 12, 2002)

Picture for Poll


----------



## treeclimber165 (Jun 12, 2002)

Going by the diagrams you provided, any of the 4 will work perfectly. The amount of leverage available is 20X more than needed to get the log heading in the right direction. But I prefer using either the 'C' or 'D' setup in order to make it easier to untie the rope and keep the knot from getting buried under the log.


----------



## geofore (Jun 12, 2002)

*pull poll*

Before I can tell anything, how high off the ground is that open faced cut? Are you topping out a tree or dropping a tree from ground level? It makes a difference. Do you want a slow fall or fast? What kind of tree is this? Is your landing zone clear?

If you are on the ground and you want to aim this tree I would warn you to use a different cut, not an open face. The tree may come back at you. An upside down directional cut will be better to aim and with a and with a downward felling cut instead of a level felling cut the tree can not come back at you. As the tree falls it will jump off the stump in the right direction being unable to climb up agianst the downward cut and come back at you.

If you are topping a tree use the top half a directional cut on all four sides and leave a 4"x4" or 6"x6" square in the center and then cut the bottom half of the directional cut in the side you want the tree to fall. DO NOT MAKE A FELLING CUT,at this point you can climb out of the tree and pull on the rope, then pulling the top out. As for your rope I would use two not one because if your truck stalls that tree can land anywhere in the 180 degree arc you set with one rope. The second rope minimizes the arc the tree can swing.

If some one else where there to pull out the top I would use a different cut.

I will recommend you use two ropes or a block and tackle to pull over a tree just to limit where the tree can fall.

You do it your way and I do it mine.You simply do not give enough imformation in your pictures and discription to make a fair decision.


----------



## rbtree (Jun 13, 2002)

It seems to me, geofore, that thou dost complicate matters a bit too much.

The question was well presented, at least for me. I agree with Brian, all pulling scenarios are fine, and overkill compared to wedges. But C and D are better, and D is great, something I never did til a couple years ago.

I've been doing big, and I do mean big, stuff for many years. Both falling logs, straight and leaning, and some whole trees, as well as log lengths from the air. And that is PNW export log lengths, not dinky 12-20 footers.

I use whatever technique is convenient and best for the situation, including the appropriate felling cuts, whcih might be open face, level backcut and face, Humboldt, conventional, or very shallow wedge but in about 50% when I want the log to pop off the stump and rotate less.

But, make an hinge cut, partially set up the backcut, then descend to pull it over----never!!!!

And, as you can tell from this picture, I like rat's nests!! LOL. If you ask nicely, I'll explain what the heck each line is for........uh, well, p'haps....


----------



## geofore (Jun 13, 2002)

*poll pic*

There is no back cut, all the cuts are down at 30 -45 degrees and only the directional is cut out in the front. I don't think OSHA would like the drop and pull it over yourself either even if you tie it off. Never had one fail but then I never tried one in a wind storm. It takes a 35 - 45 mph wind to knock it out.

That is a lot of lines, Do you use colour coded lines so you can tell them apart or is your ground crew really good in the memory dept? Did one like that, took nearly four hours to do the set up and the whole thing was down in less than an hour. A crane would have been nice and faster.

Do have to agree D is good.


----------



## TheTreeSpyder (Jun 13, 2002)

We can go any way with this. But, i was looking at this as a scenario for pre-setting for felling previous to a backcut, with maximum power and security; irregardless of input pulling force. 

i see a few things here that i was wondering if others did too.



P.S. Irregardlessly- Well, the last topic i started that led into this, (that i was tiptoeing around) was "re-guarded" by some as 'absurd', so i chose that word rather specifically..............; for it is deemed as unacceptable by the book, but exists in dictionaries and is used by many and powerfully! Though painting in the color of sarcasm, i'm trying to poke fun at myself mostly, as my way of keeping it light around hear................... Glad someone was awake, makes it more fun!


----------



## treeclimber165 (Jun 13, 2002)

"irregardless" is not a word. But it IS one of my pet peeves. If yer gonna use big words, at least use big words that exist.


(On my first cup of coffee.....)


----------



## Treeman14 (Jun 13, 2002)

INDUBITABLY!


----------



## Dave (Jun 13, 2002)

You're all just a bunch of antidisestablishmentarianismists!


----------



## John Paul Sanborn (Jun 13, 2002)

I'm just a jingoist myself.

irregardless
adv : in spite of everything; without regard to drawbacks; "hecarried on regardless of the difficulties" [syn: regardless,irrespective, disregardless, no matter, disregarding]


----------



## Rob Murphy (Jun 14, 2002)

Geofore I dont mean to be polemic but you said:

"An upside down directional cut will be better to aim and with a and with a downward felling cut instead of a level felling cut the tree can not come back at you."

Are you refering to a sloping backcut? I was wondering what the reasoning was behind this techinque. The problem I see with it is the lack of leverage/lift that you would get if you had to use wedges.It is considered an illegal techinque here in Tassie, inspectors would actually take your Fallers ticket off you for using it.
As for the one man topping out technique.....I am stunned........where did you lean that one?


----------



## Kevin (Jun 14, 2002)

Rob, I read between the lines and I think it`s an upside down Conventional face cut which translates to a Humbolt but I wouldn`t hang my rope on it!

Downward felling cuts are Taboo.


----------



## geofore (Jun 14, 2002)

*cut*

The post is to get a reaction and see how many different ways it could be done to avoid what happens next. I have seen fly by night crews doing things and I know some read this B board and will see themselves. I don't feel my post was anymore out of line than pics A or B for pulling a tree down. We need a do not do this Smilie for the bad ideas but also the need for why you don't do this needs to be explained to the new guys before they get into trouble. What happens next is not something that is top of the list on new guys in a hurry. 

Yes the safety folks would shut you down if they saw this done and it would save your life. The important thing is to explain why it is wrong, so from time to time you will see me post the absurd. If I say it is absurd when I post it no one finds out why it is wrong. I do expect to get blasted for the post, the reason why you don't do this is responded to. 

Who in their right mind works alone??? Who would cut a top half way out and then climb out of the tree to pull the top out ??? Why would you under cut??? I've seen it done by guys that know better. I tell them they should think about what can happen next before they do it. If I post it, it gets a good responce. 

I use a block and tackle to pull trees down for more control. To top trees I use the groundies to pull it over and usually tie it off so it can be controled on the way down. I did not live to be this old on luck.


----------



## treeclimber165 (Jun 14, 2002)

*geofore*

Thanks for clarifying. I was too stunned when I read your first post to even respond. I figured if you were actually doing all the things you mentioned in your post, nothing I could say would convince you to change. 

Removed a 60' laurel oak today, 3' from a house. The trunk was hollow from top to bottom with about 1"-2" of good wood except for where the big holes were. I was able to stick my entire leg inside one hole, even with my size 10 1/2 boots and gaffs on. DBH was about 28".
We gingerly roped 3 limbs from over the house, me in the tree, cut man on the roof, rope man on the ground. Smooth as silk, no jerking on the trunk. Then we ran a running bolen up the handline to the top of the trunk, I came down and pulled out the Flop-O-Matic (Stihl 046 w 28" bar) and dropped the tree between the houses. 
$850 before lunch (12:15) with 4 guys, not too bad.


----------



## Rob Murphy (Jun 14, 2002)

*Point taken*

Kevin 
Yes I thought he meant a Humbolt for the face cut , with a sloping backcut (which geofore called a felling cut).
OK Geofore
I gave one reason why a sloping backcut was potentially dangerous . 
Are there any other reasons? Any comments anyone?

On the other subject: I sometimes come down inverted, using my feet to control the friction hitch, this I do slowly and only when Iam playing.
Also I have been thinking about getting 2 petzel swivels and set up 2 lines so I can do spins !!!
Just for fun!!!


----------



## mikey (Jun 14, 2002)

*pulling stragety*

With the info provided,I agree with treeclimber


----------



## geofore (Jun 14, 2002)

*cut*

What do you do if it slides off and spikes into the ground then tips back??? 

Yesterday I warned the guy 8 houses up the street from me to pay up his homeowners ins. . The cherry tree in the lot next to his house is leaning towards his house and the black ants have it hollowed out to about 22' and the croutch is split to the ground. A good wind will set it in motion and with the lean it has it ought to hit the peak of his roof. The tree is tall enough it will land the top on the lot on the other side of his house. Should take out the entire house.The tree was planted in 1910 and the lot owner passed away in 1929 and no one has claimed the lot since. Last time I told someone on my street their tree was about to fall it was 5 days later it fell on the house and it cost the homeowner $4,200 to have it removed and the roof fix was extra. 

If all goes right tomorrow I have five trees overhanging a parking lot that were struck by lightning. Not a leaf on them. I should be able to set ropes and blocks without climbing them.


----------



## TheTreeSpyder (Jun 28, 2002)

i have been trying to present this idea in diffrent ways for a year or so, so tried this picture deal. i think i see something here, but have not all the proper names to call it out by to be recognized.

i am looking to maximize securty and pull. The reason for maximizing the pull is to bend the spar over slower and more controlled, maximizing the load on the hinge. So, i don't think that it will come straight off and dig ito the ground, the whole idea is to lean it as far as possible (safely) before coming off the stump.

i like lacings that minimize the load on the knot with a wrap or half hitches previous to the knot. Therefore, i drop out choice A).

The leverage is greater the higher the line, and lower the cut, so B) kinda drops out.

But what if we trenched across the top, so that the height of the pull of C),D) was equal to B)? Well i think that there is something in the positioning of C),D) that might make them greater pulls slightly. i'm looking at the cocked back postion that the pull comes from, perhaps putting a more effective pull somehow? And then....... Does that bend want to come out with more pressure? Does it want to open up and 'dump its bucket'? Does this add something extra? If i take a tight line and pull it (sideways like a bowstring)i can get fantastic pressure and sweat it tighter like this on an anchor, can that same effect be werking in a diffrent form here?


But, then a step further i think i see more...........

In defining between C) and D), the reach is diffrent down the spine, could this length make a diffrence?

If i had on a helmet that also braced my neck, and a line came over my head and under bowlined under my arm pits, would there be less action than if it came down and tied at my ankles (ok some of you'se guys are liking this example a lil'too much...........). It seems to me in this example, i can feel the line pushing my head down as it grabbed what it was reaching for (ankles) and tried to pull there. i think it would try to flip me around more, and deliver more 'action' per same pull compared to C). It also seems the angle of the bend (more acute/more pronounced effect) ie. pull closer to tree, bend wants to be straightened more, makes a diffrence.

i witness this i think in that simple example, in dropping spars, and in this self-tourquing rig. pic i made a while back. With this i can make a horizontal spar tighten its own line, and sweep to the side with very little drop if done write. If you make a hinge that faces down at 4'oclock, and leave a long line of fibre ripped at hinge finish, the torque will just spin the limb around. But if you eliminate all but a 'round' spot of holding wood at the top of the hinge face, the limb will flip over in right conditions (as there is no long hinge to leverage against such rotation action). Now, one good flip over, and the head is usuall off the roof; but you might have to trim branches pointing up, so they don't bang anything during the flip. Once the head is off the roof, you should be ok with the green (more soft and forgiving) end heavier, as now the machine action will lift the butt clear off the roof. i more clearly see where the lenght of the line between the bend and knot makes more diffrence.

So it seems to me that the bend wants to come out of the line the more tension that is placed on the line; that it matters in your choice for hitch positioning. i play with it a lot; but it isn't always needed so i must go as i can, trying to gather these things and test them against themselves as the situations arrive. i seek to find these things, tweak them; and compound them altogether. i think the more items i can gather on my side the more i can usher in more discriminating circumstances.

So........... anyone else ever wonder or refute such things?
Am i asking for it or what?
Is Joe out thar? Or does JP have to shake his head alone?


----------



## geofore (Jul 4, 2002)

*tree pull*

I used the (D) on two of the cherry trees, on a hill, after topping them. It worked but control of direction of fall is touchy on a slope.One up the hill and one down the hill. The trees landed well in the 5' of space between the new trees already planted. I still like two ropes for more control and I told my son if he ever tries this to do it on a tree with more of a landing zone to practice first. We used block and tackle on the others. With a bigger landing zone this is less work than block and tackle but in a tight spot I would use two lines or the block and tackle.
We had folks watching, I know they were there to see a screw up but there were none. We got four requests to do other trees from the onlookers so I'll be out looking at other jobs. Yes, I got that question," How much to cut my tree?". Sight unseen I don't have a clue, I have to see the tree first. The only place I've seen a flat rate offered is in the Penny Saver and I doubt the guy's able to do any tree for one price.


----------



## Stumper (Jul 4, 2002)

Assuming that all ropes are rigged at the same height, notched the same,and identical tension on the rope is applied to identical trees there isn't any appreciable difference-the leverage pulling the tree into the notch will be the same. The rigging illustrated in fig D. does reduce the danger of fracture of the spar (due to neutral plane shifting ala cable-backed bows in archery), however if the rope is simply passed over a limb or through a crotch there is a possibility(albeit slight) of fracturing the limb due to the downward pressure generated. As far as a flipping action is concerned ,it sounds reasonable BUT remember that every ounce of "lifting" pressure is couteracted exactly by the downward pressure generated at your upper 'redirect'. In actuality it is a wash. 
If there seems to be a possibility that the tree will break up when subjected to the pressure you deem necessary the D cofiguration offers a little security if rigged properly (tied off above then strung down the back). Practically speaking I usually use A. Since I frequently pull non-single leader trees I sometimes run another line from stem to stem so that a fracture won't eliminate control.

Geofore, I am in my right mind and I choose to work alone-but there are jobs which I will not attempt alone.


----------



## TheTreeSpyder (Jul 4, 2002)

Hmmmmmmmmm, okay, next view..........

In D) The final tie off could help bind/fortify a fracture whereby a running bowline, rather than clove could be constricting right over the fracture, binding it per force applied through this machine of rope.

Also, If D) was a metal pole, with a pulley at the top, the final pull would be lifting up, but............................. the redirect point's downward pressure would be ?............2x pull (with vertical lines) when, there is no movement in base? 

So the pressure on top would be of tree might = 2xPull - friction -whatever for indirect pull? And the final termi-nation/hithchpoint (clove, runningB.), would be pulling up, but the hinge itself, would be tring to convert that somewhat to forward, as the line went up the tree's spine, pushing it forward at all points with the tense line, from zero leverage on up? Still not sure of all werking elusive proper-tease, as they continuosly dance before me, and i try to name them. i seek this so i can sight them out and see them in other things; assembling and compounding their forces to usher gravity powered motion through our machines manipulating these giants of the earth! 

As i lace a self-torquing rig deal i mentioned, applying this in the air it uses the same placements and lacing, seeming to have and use the same properties. While, watching it's motion for clues; i seem to witness that the length from the redirect 'curve' to the hitchpoint on the limb itself matters, for i seem to get more 'action' if it is longer, so suspect that holds here also.

That is about it, there is a lil'more; it all has kept me awake some, trying to unlock it's secrets.


----------



## Stumper (Jul 4, 2002)

Treespyder, You are right about the multipying factor downward a the redirect point. My statement about the pressures couteracting exactly was technically inaccurate -- A.due to friction and B. in this scenario because we are rigging on the subject tree. In affect we are compressing the spar longitudally. This may account for perceptions of flipping etc. since we have stored energy in the wood that is later released causing reactions which may or may not be significant/perceptible. 
You have an interesting post going. I suspect that most of us have a decent knowledge of "applied" physics from our work experience but a limited knowledge of the science and associated terminology.


----------



## TheTreeSpyder (Jul 14, 2002)

Well, a while back ol Joe, gave me the words of forced vectors and triangles to research, but; all the greek is like another language............ And i feel fairly comfy with math! i thought!

But, in the end, i think it was about what i thought, only sorting here and some of others reflections bring me back to that tracing under the belly and back, or likewise to the side and under, will put 2/1(minus)friction(minus)indirect angle of pull error. 

Whereby it is a 2/1 maximum, if the control line pulls and fights friction going around load to hitchpoint some of that 2/1 is lost per that friction, do tho this innefficiency.

Then, only a direct inline pull can be maximum, so the wider the angle of the bend aover top, the less direct the pull, so even more eficiency is subtracted from the 2/1. So lacing up the back spine over the top, and down a sloped face in front (for more acute angle, so more direct/inline power possible) might be maximum.

Now the thing about the length from the hitchpoint to the bend in the riggingline over the load, having determining characteristics- i'm still crunching that, trying to visualize angles having effect in there somehow too. 

But all this can apply to the 2 self torquing setups, and maximizing pull for bringing tree into face with thicker hinge, or in rigging too. It would have that anyway, if it was the same size and weight, only at more of a cocked angle, that gravity would be pulling on it more directly, instead of this rig. With that thicker hinge i can bring more down safer, within the realms of space measured by the 'stik-trik'. Sometimes i'll set a rig in tree and have guys throwline the end of the load, for most leveraged pull, so it folds the hinge with more fiber to usher it threw prescribed arc of hinge. Like forcing the hinge to be stronger by doing this, then using that strength.


----------



## Lumberjack (Sep 25, 2003)

According to my physics teacher both C and D would give you the most leverage. Appon further thought I relized that if the rope slips out of the notch, on D you would lose all leverage whereas on C the rope would come down to the knot and stop, keeping most of your leverage. My teach also pointed out that A would be the simplist way, and if the pic is to scale, you would only lose around 9% of your leverage, not your pulling power. To answer the poll C is the most secure and powerfull, but if you moved A up to the top (without it comming off) it would save time and permit almost all the power of C, losing only around 2.5-5% of your max leverage, while saving time not having to cut your notch and run the rope through it.


----------



## TheTreeSpyder (Sep 25, 2003)

i think that coming over the top does 2 things definitively:
A) Grabs most leveraged position of lever at farthest possible end.
B) Breaks some of the direct pressure over the top, taking load off of weakened area leading to knot, making it more secure.

On slipping, i dog in pretty good for everything, leading half hitch stop wrap comes above pulling line and repeats (clove) then 2 half hitches. This gets dogged into natural or made (mini humboldt) imperfection resisting the upward pull. All half hitches are oversteered beyond wher they go, then pounced on in 2/1 formation to set into place. Rear, leg is sweated in hard as line is pulled. Sometimes top of log might even accidentally get a mini groove in it to trace line over the top of (stuff with moss first if pine!) Or might be braced with krab. Or huge, positive natural fork.

Where are the pulls and pushes in the change of position C and D? At that change,where is everything in consideration to C.o.B.?
i oculd use a physiccs teacher to sort and bounce osme of this off.....

Thanx.


----------



## TheTreeSpyder (Sep 25, 2003)

i always 2/1 lock down my half hitches (bend backwards over them selves and impact with whatever massive body weight/distance ya can habitually) and set up against branch, swell, imperfection or cut that locks agianst the direction of pull forming a positive mechanical stop. In these over the hill configurations, the main stress/ shocking on the hitching is releived anyway going over the top methinks. Also, i'll clove (alternatively 1/2 hitch running bowline, or choked sling; always looking for 2 grab system); always with the first turn coming above the line, to lock it against the upward pull, then again, then 2 1/2 hitches, bury the tail under main line pull. So most secured point of line is further trapped under mainline tension as a knotting strategy all over the place, and meticulous impacting, leverageng, sweating in.....

i think there is something going on here, that i have witnessed time and again with the "Over the Hill" type lacing. 

Dropping the pull point below the C.o.B., while spreading out the 2 compounding forces created by line under tension (push at top, pull at knot) to draw the C.o.B. forward. i think these could be relevant aspects. Anway it works for me!


----------



## DDM (Sep 26, 2003)

Spidy, i just want to know what software your drawing diagrams with.


----------



## Kneejerk Bombas (Sep 27, 2003)

Microsoft Paint, it's on your computer.

Here's an improvment on Spyder's Lacing:


----------



## TheTreeSpyder (Sep 27, 2003)

MS Paint(price is right.....), windows freebie for all the pix and drawings. Prolly need to go with something with text box editing though.

i think that one different thing here, on this 'Over the Hill' lacing for felling is in the way that it is being looked at. i am not looking at that lacing to achieve pulling the tree down, that would be laced to just the front, perhaps just slightly over the top.

Right or wrong, to me i see that farther down lacing as trying to flip the log (even if it can't), so this is where i see like this tourque. Just as a line laced through pulleys to make a 5x1 pull on a load, when pulled with 100#, can load successive places with that line tension wherever that power flows, till it is dead ended, i see the unrestricted force in the line flowing. Even a loaded line just rebounding off of successive pulley redirects (as traced through a winding tunnel to bring pulling power from end to the other) can place load on whatever it touches before restriction of first choking half hitch, i see the extra leg of line stretching that leveraged length as delivering power to more points than to just a straight pull forward. 

i see the line coming over the top (compressing into hinge), pushing forward at the top (in the direction you want to go, from highest leverage point), pushing forward (and bracing/steadying)all the way down the back spine of the tree, to finally pull at the first choking half hitch. i think even in a half hitch running bowline configuration, you can plot most of the pull ending at the half hitch, still some trace bracing etc. extending to the running bowline. Like electric force running down line, lighting anything it touches.

So, what if the request of this rope machine, is as any other, to pull that first choking half hitch to you? What if arching it around something takes a longer route to do that? And that asks something stiff to arc? 

The amount of line force lost in the curve over the top is due to friction, not a subtraction of power used up, but still it is reflected at that point at 2 different angles and carries down the line (left over from friction) to place pull at first choking half hitch (in addition to any other non cancelling push pulls it gave enroute). So i kinda see more working towards target. Some of the power loss tracing down the line can be handled very well with good'ole sweatin'in hard (a'la Brion Toss). IN fact any leveraged line pull gotten from sweatin in across the friction buffer over the top, would now use that friction to stay between the top and first choking half hitch. Especially as steady tension was kept on pull line. This would render more pull to first choking half hitch. Placing that pull further down would take more leverage to invoke spin?

Complimenting directional pulls and pushes are asking the tree to arc? So that the first choking half hitch is pulled to you? Then i think spreading the compounding pull up bottom/push down top positions, would give more spin power? So take further leveraged command of the end (bottom or top) they (forces) are trying to move? Even if we stop as tree fells, and just look at are line to pull it down, What is the line thinking it has to do? Relieve tension by pulling first choking half hitch to source of pull? i don't think that soft fibers recognize push as an output of themselves, if they happen to do it, there is no loss, only relief as they reach for the end of their job (first choking half hitch)?


----------



## Stumper (Sep 27, 2003)

Mike, You forgot to draw the line coming up through the floorboard where it is held with one hand while backing away from the stump. 

Ken, We have had many good exchanges about felling techniques and how hinges work. I'm convinced that your understanding in that field is pretty advanced beyond the norm. On this "over the hill" lacing I just can't see it. One of the things you've mentioned -the rope as an energy storage device has merit. A longer length of rope lets you store more energy. Going to the tip top of a spar gets you all the leverage you can grab so that is a plus compared to a lower anchor. Since the spar can flex under load, some energy storage in the spar takes place but I have difficulty seeing how that would be greater in a fishpole lacing compared to a top anchor. As far as multiplying factors due to bends in the rope- not much can happen due to friction- then add in the fact that all the rigging is taking place on the spar itself and it comes out to a wash. I know that you and Brian like this technique and I have no doubt that it works well but I suspect that its greater efficacy is really just a matter of your expectations. There is a real tendency even among honest observers to see what we are expecting to see. You mention spinning possibilities when a back branch is traced . You are soooo right about that. Occassionally that can be useful but it introduces a lot of extra forces on the hinge so I usually try to avoid such rigs. I'm not trying to rain hard on your parade but I don't think that science supports you on this one.:angel:


----------



## TheTreeSpyder (Sep 27, 2003)

> _Originally posted by Stumper _
> *Since the spar can flex under load, some energy storage in the spar takes place but I have difficulty seeing how that would be greater in a fishpole lacing compared to a top anchor. *



If friction free top at bend, then as much power transfered down to first choking half hitch as started. If, we fly over head and pull up at 1000#, (that could be the same pull up at base with forward 1000#pull laced over friction free top) and log is standing on end vertically, freely; it will flip, top will come down, bottom up. The farther from the C.o.B. the pull comes from, the more leveraged the spin; @ C.o.B. hanging balance; above C.o.B. no flip as Goldilocks's 3 choices here.

Pulling @ 1000#forward/down only (running bowline to front of tree no over the hill positioning) on free standing vertical telephone pole will pull top down/ bottom up too/ more dramatically with more leveraged positiong from C.o.B.

Both of these actions invoke bottom up/ top down direction; if truck pull and 'copter lift where slammed on to compound (geometric rather than arithmetic gain?) each other at the same time would logically be more force than either seperately to flip pole; especially if they leveraged off each other to the same directional end. The motion is not linear, but spinning IMLHO by the mechancial command of the lacing that you have given. 

In fact the inneficient 2/1 (open angle of lines of pull reduction in 2/1 potential even before introducing friction back in to puzzle) can give more force at top than original pull, while still continuing down to other end and pulling again to twist/flip in the same direction. So, more than original force at top, and help in leveraged spin?

Jacking freewheeling tire up on truck: pull down on top lug nut, pull down on top lug nut and up on bottom lug nut, down on top of tire and up on lower lug nut, down on top of tire and up on bottom of tire? Each spins tire, each with different affect by leveraging and angling of pulls and their realtionship to each other? Even rooted in the ground and can't move, same forces, even if impossible task to spin tire, same forces are exerted? In fact,if you had a chance to break the friction of the weight of truck on tire off jack, you would want to grab and power from the outside of the tires, not lugs for best chance (without wrench on lugs).

Hinge pressure at first movement sets hinge strength, attatching our free standing telephone pole to the ground and trying to flip with 'copter and truck (or laced line pressures) still tries to flip/move C.o.B. forward (from gravity pull at flip) placing more tourqued pressure on hinge at movement making stronger hinge? Even though connected to hinge and impossible task force is built to pick up at first choking half hitch, while simultaneously pushing down on top? The closer to the end that pick up at first choking half hitch, the more power to exert picking up at that end can be delivered?

All the rope knows is it pulls, if it pushes something out of the way without friction, it kinda doesn't erode the transmision of power, and pushes whatever out of it's way to do it's job of pulling with least strain automatically? Sometimes building forces to work for or against us?

Orrrrrr something like that......

i think!

:alien:


----------



## Stumper (Sep 27, 2003)

I'm not convinced-I'd like to agree with you but I still think that the fishpole lacing is a static system. every bit of "UP" is counteracted by down pressure. You may induce bend in the spar and store energy there that will be released upon felling the tree but you aren't getting any lift. Grab a real fish pole and try it with the line tied to the tiptop and compare to when you have the line threaded through the guides. There is no difference in force "flipping" it out of your hand. The guides do help distribute pressure evenly along the rod and thus help prevent breakage-if you only run the line through the tiptop and then down to the butt you can overload the tip if the departure angle of the line gets too acute. IF you hung a pulley on your helicopter, or on another tree you could generate some butt wrenching lift but with the rope bending on the spar or on a pulley tied to the spar you either get no change (the tag line would need to run straight out from the tree 90* to the trace section-I've done that working on mountain slopes but you probably don't do it often in Florida) or you generate more downward pressure than lift. It can't help make it fall but if the spar is slinky enough to bend a good bit you can get more violent tearoff and some funky gyrations as the tree topples. Come on somebody prove me wrong so I can apologize to Spydy.:angel:


----------



## murphy4trees (Sep 28, 2003)

I AM with Stumper on this one.... Both in that I recognize Spidy's thinking ahead of the pack in hinge mechanics and other areas. And in that I don't see any mechanical advantage or added pulling power in the lacing technique.... I do however think its a great technique due to its easy application. Good to have in your bag of tricks both in felling and rigging limbs....


----------



## TheTreeSpyder (Sep 28, 2003)

As this is on the far reaches, every mind that has chimed in here has made me look again. Especially as ye 2 that seemed to see how to leverage hinge with me. Then to use that leverage to carry the off balance lean to side of fall, so that all/more of the force of lean forward and line pull farward can be used to force stronger hinge (as work of direction to face was handled my hinge, and these 2 forces where more freed up to help force stronger hinge forward).

Just as learning over time from everyone here, and then have someone expand the view you try to lend is really what it is all about; along with the helping, comradre.

All along, i've been assuming that the idea of applying horizontal (to correct lean) tourque/spin pressure was more seen than the same bend vertical. i also hope that many see this as possibly a strategy to take the highest leveraged point easily, offer it more bracing than a straight pull line too, offering even more realistic choices of pull ing points. i don't think that is disputed, non-slipping hitch, strength and security at bend, and i beleive this technique has enough merit to give ya more in many occassions, totally apart from the rest of what happens with the pressures on the final leg as it goes over the top, and if any appreciable help is gained from that part of it. In short there is enough good reasons to use this technique i think, even if ya don't buy the last 'leg' of the 'arguement'.

A telephone pole laying flat on the ground to be dragged by a truck will react different to differnt lacings? Wherever the line of force goes till first choking half hitch, force is transmitted, and contact points of that force have to be accounted for? The line will pull to the first choking half hitch, in considerately pushing something out of it's way if it has to? Even the load itself, if so laced, like pulling telephone pole (dragging it as it lays flat on the ground) in over the hill type lacing will be different than straight pull from top? 'Long legged over the hill ' lacing would spin telephone pole more than short legged? Short legged might not really give appreacible spin, just best leverage for turn?

Something is making my spyder sense tingle, that something is different with the different lacing somehow, even if just smoother, better braced from tigh line down spine pushing forward and pull coming from different point, somehow seems to make a difference...... 

So, even i have questions, where are wee at here fellers?:Eye:


----------



## TheTreeSpyder (May 19, 2004)

Though this is from a few years ago, i still think this is another multiplier for some situations. It simply takes the present line force, and applys it to the target differently;running the objjective of maximizing any helpful multipliers that i can chain together to help to task.

My latest explanations of how this more arching pull on the spar or anything can help leverage are :

8th post in Stacking Wedges 

Pulling Directions on a Fall 

i think the application easily gives more securing and problem solving leveraging. i have then turned the effect sideways. very easy to place on limbs, throw out to a better leveragd crotch, boomerang or drops back to you (steel carabiner in eye); mount in front of cut. Need loaded tourque? Bend it around from over far side around and under tie to close to support side. After tear off it will "unroll" dropping some, unless there is a branch or carabiner stop, that let the pressure flow before, bot doesn't let it unroll after tearoff. The breadth of the hinge across is the leverage that doesn't allow unroll/twist before tearoff; if you want to flip off roof etc. in one move, take hinge down to a small circle, then no leveraged length to fight twist. In using a longer hinge the twist is trapped and the force goes to turn, but is all automatic in the rig, once set.

Set, msut be tight, as to capture all movement, and use it for immediate, intense leveraging, wasting nothing.

Because this rig is a multiplier, tightly set, to start, rather than trying to press available gravity force towards target inititially; i wish to run that force throuhg the multiplier of this leveraging. So i bluff like i am aiming down, cutting down, loading the line in this leveragaed position purposefull, knowing it ain't putting up with it, there is no way this branch is going down! The leveraged response is to multply the force that i was going to send towards target, then pull with that multiplied force to target!

i really think that if a simple twsit or easier placement of a line that comes out better, ya would want to know about it!

Orrrrrrrrrrr something like that!
:alien: :alien:


----------



## TheTreeSpyder (May 20, 2004)

In a leveraging machine, without play , that would lose the force, send it unfocused etc.; the long way, the arc condneces force to a smaller area of more power. There is only so much power/movement available, don't waist a drop, multiply what ya can and set it back against the target that caused/powered it. Then, the control forces are self adjusting,as the load sets them itself.

Same arguemeants for rigging torqued limbs, pressing down with load on a leveraged line that has no play; knowing that the force will be multiplied, then the multiplied force will pull to real target.

The arching, takes more distance to arrive at the same point; so it takes whatever leveraged/multiplied force handed to it and adds it's multiplier i think.

Or something like that...
:alien: :alien:


----------



## The Best GM (May 21, 2004)

OOOOHHHH yAAAA treeclimber165 . 

Thats a pageright out of my book. A great rope man letting them fly , running bowline, thats THE 
STUFF BRO. MY father is the same way. Not much wood, climb her anyway. Unless some one else feels like paying the bills. We all no the answer to that one. A and d look fine. Running bowline low, running bowline high up.


----------



## rahtreelimbs (May 21, 2004)

> _Originally posted by The Best GM _
> *OOOOHHHH yAAAA treeclimber165 .
> 
> Thats a pageright out of my book. A great rope man letting them fly , running bowline, thats THE
> STUFF BRO. MY father is the same way. Not much wood, climb her anyway. Unless some one else feels like paying the bills. We all no the answer to that one. A and d look fine. Running bowline low, running bowline high up. *



What???


----------



## The Best GM (May 21, 2004)

Cant anybody see those letters all side by side that form a name and identify the person that i was talking to.


----------



## rahtreelimbs (May 21, 2004)

> _Originally posted by The Best GM _
> *Cant anybody see those letters all side by side that form a name and identify the person that i was talking to. *




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.


----------



## TheTreeSpyder (May 22, 2004)

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:


----------



## TheTreeSpyder (May 26, 2004)

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:


----------



## glens (May 26, 2004)

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.&nbsp; 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.&nbsp; If you pull 100_lb. on the rope that's all the force there is.&nbsp; It's the same 100_lb. both sides of the crotch (assuming zero friction); your 200_lb. figure exists only in your mind.&nbsp; 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).&nbsp; 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


----------



## TheTreeSpyder (May 26, 2004)

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 (May 26, 2004)

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 (May 26, 2004)

B and D seem incorrect to me.


----------



## glens (May 26, 2004)

"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.&nbsp; In "A" and "C" you have the fulcrum independent of both the load and motive force.&nbsp; 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.&nbsp; 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.&nbsp; If split equally sideways and downward (45&deg, the 100 will net 70.7 each downward and sideways <i>at the point of contact</i>.&nbsp; 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


----------



## TheTreeSpyder (May 26, 2004)

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:


----------



## glens (May 26, 2004)

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.&nbsp; 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.&nbsp; You obviously put considerable thought into your work and there's no doubt you get the results you're trying for.&nbsp; I perceive you're plagued with generally having a clear mental image without an equivalent capability to convey it (the image) to others.&nbsp; I also gather you achieve certain results for reasons other than you'd envisioned.

Glen


----------



## Stumper (May 26, 2004)

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.


----------



## TheTreeSpyder (May 26, 2004)

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?


----------



## Tim Gardner (May 26, 2004)

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.


----------



## TheTreeSpyder (May 26, 2004)

> _Originally posted by Stumper _
> *It loads the supension line with its full weight but no more on that leg. *



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:


----------



## glens (May 26, 2004)

Kenny,

Set 180_lb on a small tray with handles and pick it up.&nbsp; How much force do your muscles need to exert?&nbsp; 180_lb total.&nbsp; 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.&nbsp; Then pick up as hard as you can on the handles.&nbsp; How much force will your muscles exert on the handles before the tray elevates?&nbsp; It's the same type of loading from your shoulders to the handles both ways so how can your muscles tell the difference?&nbsp; 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").&nbsp; 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.&nbsp; What happens when you pull on the right-hand rope with 200_lb of force?&nbsp; Obviously the load will rise, but how much weight is present at the redirect?&nbsp; 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.&nbsp; Pull down with 50_lb and that's what the redirect sees.&nbsp; Make it 500_lb and <i>that's</i> what the redirect sees.&nbsp; 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?&nbsp; You're assuming counter-action where none exists.

Glen


----------



## Tim Gardner (May 26, 2004)

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.


----------



## TheTreeSpyder (May 26, 2004)

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.


----------



## Tim Gardner (May 26, 2004)

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.


----------



## TheTreeSpyder (May 26, 2004)

> _Originally posted by Tim Gardner _
> *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. *



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?




> _Originally posted by Tim Gardner _
> *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.*
> 
> With no opposite pull the limb will fall? With no friction, the pull must come from hitch?


----------



## TheTreeSpyder (May 27, 2004)

> _Originally posted by glens _
> *Kenny,
> 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.&nbsp; What happens when you pull on the right-hand rope with 200_lb of force?&nbsp; Obviously the load will rise, but how much weight is present at the redirect?&nbsp; It's still (basically) only 200_lb.
> Glen *



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.


----------



## Tim Gardner (May 27, 2004)

> _Originally posted by TheTreeSpyder _
> *
> 
> With no opposite pull the limb will fall? With no friction, the pull must come from hitch? *



There is no way you can exert 200# anywhere on a rope with a 100# limb tied and hanging like in your illustration. Where is the extra 100# coming from? The limb only weighs 100#. If it is still attached I can see you being able to PULL the rope and get 200# but not with the limb just hanging from an anchor point.


----------



## Tim Gardner (May 27, 2004)

> _Originally posted by TheTreeSpyder _
> *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?
> 
> 
> *



I can not see how applying tension to the stick along its axis ABOVE the hinge will in any way apply any force at the hinge. 

I can see however, by applying tension along the axis of the stick above the hinge would use some of the force exerted on the rope that would otherwise wind up at the hinge. That means that you would get less pull at the hinge.


----------



## Nickrosis (May 27, 2004)

I spent waaay too much time making this, but it demonstrates the principle I believe Spidey is using.


----------



## Tim Gardner (May 27, 2004)

> _Originally posted by Nickrosis _
> *I spent waaay too much time making this, but it demonstrates the principle I believe Spidey is using. *




What I got out of Ken’s illustration is that there is 200 # at the pulley when the pulley is attached to the 100 # weight. In your illustration Nick there would be 200 # at the pulley.


----------



## glens (May 27, 2004)

> _Originally posted by TheTreeSpyder _
> i think that in non friction, non leveraged circumstance, you cannot exert 200# against 100#.


Sure you can.&nbsp; The result will be that the 100# load will accelerate, allowing marginally greater weight to be present at the pulley than that of the static load.

Any time you use a pulley you have a leveraged circumstance of some type.

Here's the answer to your previous question.

Glen


----------



## TheTreeSpyder (May 27, 2004)

Well, i guess where even Glens; i don't know where ya git the 17#, 77# and 200# values with zero friction, no angle factor in line. Just looking at the 3 positions of Load, Pulley , Anchor; i think the diagrams are equivalent.

To place the 200# pull on a line to a friction free pulley to lift 100# i think you would have to move pretty fast with a hard snatch; for in general with only 100# resistance, you can't pick up 200# on it, any more than you can walk over to a canary feather fluttering across the ground lightly, and exert 5# of effort picking it up un tethered. Now, perhaps you can exert 200 foot pounds, to lift 100# 2'.

Tim, i think that there is the same pull at the hinge before the spar starts to move. At the precise moment of movement, i think an upset happens; it is the only way i can de-scribe with these standard numbers, what i think i have observed for years, and tried to explain here in the last few years here.

Nice drawing Nick; i say Spyder with a "Y" for the quiet watcher, the rope thang we do, and the (inter)net. Made it up for email to 'civilians'; started using it on ISA board outta frustration when it wouldn't let me be just "KC"; in a time of Tom's fireside chats there.


----------



## Nickrosis (May 27, 2004)

> _Originally posted by Tim Gardner _
> *What I got out of Ken’s illustration is that there is 200 # at the pulley when the pulley is attached to the 100 # weight. In your illustration Nick there would be 200 # at the pulley. *


Which is what's illustrated.


----------



## TheTreeSpyder (May 27, 2004)

i thought my foam example would answer the query here.

Did you visualize it like this?


----------



## TheTreeSpyder (May 27, 2004)

Applied directly to puzzle


----------



## Tim Gardner (May 27, 2004)

Ken are you trying to illustrate a reversed single whip where the pull is coming from the weight itself?


----------



## TheTreeSpyder (May 27, 2004)

i guess that is what puzzle would be, if i said it better!


----------



## Tim Gardner (May 27, 2004)

Then I believe the power would be 1/2x at the foam or crotch. If you look at a reversed single whip the leg that is pulled on only requires 1/2x the weight of the load to lift. If the load were applying 2x the weight on the pulley (foam or crotch) then it would require 2x the power to lift the load.


----------



## TheTreeSpyder (May 27, 2004)

is not the line tensioned by the weight just as if you pulled it?

If the upside down slingshot was nailed to the ground, and the line laced in the suggested "J" pattern, and the line tension set to 100# on long end, no friction, would not the foam have 2x pull on it as line crushed it down?

Would not the pull leg and the anchor leg both have to shorten 1', for a total of 2' of line slack taken out, to compress foam 1'?

Take out nails, float the 100# slingshot, so that the line tension ont he long leg of line was once again 100#. Does not anchor leg and pull leg have to shorten 1', for a total of 2' of line reeled in,inorder to compress foam 1'?

In putting up 2' of distance in effort, to be funneled into 1' of work in compressing foam would always increase the power beyond the line tension especially in friction free etc.

edit: just caught pic; Tim i think that anchor marked 1/2couldn't be to the load itself, it would have to be to a seperate anchor, then each seperate anchor would hold each 1/2 load; to match each other through pulley. Here, on a single leg, the line tension is 100/full load, for the load is borne on a single support. The weight of the load is the power for the system, like in rigging etc. i think.


----------



## glens (May 27, 2004)

> _TheTreeSpyder: _
> Well, i guess where even Glens; i don't know where ya git the 17#, 77# and 200# values with zero friction, no angle factor in line. Just looking at the 3 positions of Load, Pulley , Anchor; i think the diagrams are equivalent.


By "no angle factor in line" I take it to mean the "pulls" are purely vertical, even though shown differently.&nbsp; The sideways components arise from the offset from center of stem to the point the pulley is anchored.&nbsp; If you didn't intend such, why the distinction between the first two as a group and the third?



> To place the 200# pull on a line to a friction free pulley to lift 100# i think you would have to move pretty fast with a hard snatch; for in general with only 100# resistance, you can't pick up 200# on it...


If you place 100# of tension you will not move the load, you will merely counterbalance it.&nbsp; You must exert <i>more</i> or <i>less</i> force to move it.&nbsp; You could make it be only an ounce different and move it slowly or, for a brief moment, double or triple the force (or halve or third it) and get the thing moving rapidly.&nbsp; Certainly you will not be able to indefinitely keep more than the counterbalance force applied; that's not what I suggested.

Why do you think there will be any load on the stationary line when it's anchored on the same immovable object as the redirect?&nbsp; Don't you think that the "load" would rather be either zero or infinite?&nbsp; When it's a 100_lb weight, there is no load on the line until slack is removed and then it will ever be 100_lb (though inertia will effectively add or subtract load momentarily when direction of travel is changed).&nbsp; When the line is affixed to the same member as is supporting the redirect, the assumed "load" is only present whenever the free end is being pulled and immediately goes to zero when pull ceases.&nbsp; Think of the bald kid in the matrix having said, instead of "there is no spoon", "there is no load".

In my answer to your "riddle", the weight of the entire assembly is given as 200_lb, 100_lb, and 100_lb.&nbsp; That is correct as stipulated.

You may crush your "foam" the same amount between the first two, where one has an actual load hanging, and the other has no load, but the first will weigh 100_lb more at the base of the support than will the second because the first is supporting a load at 2:1 and the second has no load to support.

Really, Ken, have a look at my recent post in off-topic, <a href="http://www.arboristsite.com/showthread.php?postid=164333#post164333">29=30</a>.

You need some scales or load cells to see that although you might crush the foam the same as if there were 200_lb of total force, there is in fact only 100_lb of total downward force in your examples which have the one end tied to the redirect support itself.&nbsp; There is no equal and opposite weight present.

Glen


----------



## glens (May 27, 2004)

> _TheTreeSpyder: _
> Would not the pull leg and the anchor leg both have to shorten 1', for a total of 2' of line slack taken out, to compress foam 1'?


Sure, you've got 2:1 against the foam, but you do not have 2:1 against the slingshot.&nbsp; The pull leg will shorten 2' to compress the foam 1', with the result being the anchor leg shortening 1'.&nbsp; The compressing foam causes the anchor leg to shorten, not the other way around as you suggest.&nbsp; If the opposite leg from the pull leg had an actual load on it of 100#, then both legs would shorten by 1' causing the foam to compress (as the result) 1'.&nbsp; I guess I have to ask, just what the hell is your point?



> Take out nails, float the 100# slingshot, so that the line tension ont he long leg of line was once again 100#. Does not anchor leg and pull leg have to shorten 1', for a total of 2' of line reeled in,inorder to compress foam 1'?


Again, the anchor end shortening is strictly a <i>result</i> of the foam compressing, not a cause.



> In putting up 2' of distance in effort, to be funneled into 1' of work in compressing foam would always increase the power beyond the line tension especially in friction free etc.


NO!&nbsp; The 2' of rope pulled will compress the foam, resulting, <i>finally</i>, in a total load of 100# on the rope.



> edit: just caught pic; Tim i think that anchor marked 1/2couldn't be to the load itself, it would have to be to a seperate anchor, then each seperate anchor would hold each 1/2 load; to match each other through pulley. Here, on a single leg, the line tension is 100/full load, for the load is borne on a single support. The weight of the load is the power for the system, like in rigging etc. i think.


I'm still waiting for the image to arrive...

Okay, the image is correct, but does not apply to the upside-down slingshot example, where the anchored end is to the load itself and the figure in the image as "x/2" would be simply "x".

Glen

Instead of 634,014 bytes as BMP format, try 672 bytes as PNG...


----------



## Stumper (May 28, 2004)

Tim, Technically a single whip is an anchored pulley to redirect force. One pulley floating in the system is called a runner-and it does provide a 2/1.


----------



## Tim Gardner (May 28, 2004)

> _Originally posted by Stumper _
> *Tim, Technically a single whip is an anchored pulley to redirect force. One pulley floating in the system is called a runner-and it does provide a 2/1. *



Stumper, yes I know a reversed single whip is called a runner. And yes I know that it gives a 2:1 MA. In my illustration it shows that it is 2:1 just in a different way. I just assumed everyone knew that. If you look at #3209 in The Ashley Book of Knots you will see my source: “A runner is the same as a single whip, reversed. The figure 2 means that the lift is double the effort spent.”

My reason for labeling the illustration they way I did was to show that the pull (the leg that you actually pull with your hands) is 1/2x that of the load being lifted.


----------



## Tim Gardner (May 28, 2004)

> _Originally posted by glens _
> *
> 
> Instead of 634,014 bytes as BMP format, try 672 bytes as PNG... *



Thanks Glen. I realized that I had posted a bmp and converted it to jpg for my post in Ken’s other thread relating to this subject.


----------



## Tim Gardner (May 28, 2004)

> _Originally posted by glens _
> *
> 
> Okay, the image is correct, but does not apply to the upside-down slingshot example, where the anchored end is to the load itself and the figure in the image as "x/2" would be simply "x".
> ...



Glen, I think my image does apply to Ken’s. If you look at the image I posted of the reversed single whip, the working end (the part of the rope coming out of the pulley that is not anchored but pulled on) only requires a pull 1/2 of the weight of the load to lift the load. If you tied the working end of that reversed single whip to the pulley above the shive you would have the same set up as Ken’s “slingshot”. Ken’s “slingshot being the load and pulley in one. I believe that even though the working end is tied off to the pulley it still would only bear 1/2 of the load. The opposite leg leading to the anchor would carry 100% of the load. I could be wrong believing this but until I see proof otherwise I am sticking with it. Maybe Ken will work the details out on how to measure the loads and settle this. 

I have included a revised image of the reversed single whip with the parts labeled to hopefully illustrate how it applies.


----------



## glens (May 28, 2004)

Hey Tim.

How is it that you expect the pulley arrangement to work when one end of the rope is attached to the pulley itself?&nbsp; That's like taking the fulcrum away from a lever system and expecting it to still work, isn't it?

Rather like grabbing your ankles and pulling up as hard as you can, I think.&nbsp; Try as you might, it just ain't gonna work.

See my upcoming reply to the "Condensed Mayhem" thread.&nbsp; I've put together an image you might be able to follow.

Glen


----------



## glens (May 28, 2004)

Tim, 

I just more carefully read your last post and scrutinized the image.&nbsp; You have 1&frac12; &times; the load accounted for!&nbsp; With the "working end" supporting half the load, the "anchor" end would carry the other <i>half</i> of the load.&nbsp; Fastening either end to the load itself would cause that end to bear <i>none</i> of the burden and the half it had previously carried would be shifted to the other end, which would then carry the <i>full</i> load.

Please look at my "29=30" thread in off-topic (linked earlier in this thread, I think) to see how junk math can cause confusion.

Glen


----------



## Tim Gardner (May 28, 2004)

> _Originally posted by glens _
> *Tim,
> 
> I just more carefully read your last post and scrutinized the image.&nbsp; You have 1&frac12; &times; the load accounted for!&nbsp; With the "working end" supporting half the load, the "anchor" end would carry the other <i>half</i> of the load.&nbsp; Fastening either end to the load itself would cause that end to bear <i>none</i> of the burden and the half it had previously carried would be shifted to the other end, which would then carry the <i>full</i> load.
> ...



Glen, somehow I knew you were going to say something like that after I made the post. I apologize. I was not implying that those numbers applied to the reversed single whip but to Ken’s “slingshot”. Thank you for pointing it out though.

As far as the “working end” leg supporting weight I should have written load applied to. It is obvious that that portion of the rope would not support any of the load.


----------



## Tim Gardner (May 28, 2004)

> _Originally posted by glens _
> *Hey Tim.
> 
> How is it that you expect the pulley arrangement to work when one end of the rope is attached to the pulley itself?&nbsp; That's like taking the fulcrum away from a lever system and expecting it to still work, isn't it?
> ...




I never wrote anything to lead you to believe that I expect the pulley arrangement to work when one end of the rope is attached to the pulley itself. I wrote: “If you tied the working end of that reversed single whip to the pulley above the shive you would have the same set up as Ken’s “slingshot”. Ken’s “slingshot being the load and pulley in one.” Once the working end is tied off it is no longer a reversed single whip. I did not say it would function as a reversed single whip. I might not be a very good writer or know all the technical terms to label my drawing but that part was pretty clear. Read my post again.

Ken’s argument has been that there is 2 times the weight of the load applied to the crotch of his “slingshot”. My argument has been that there is 1 times or less the weight of the load at the crotch and between the crotch and tie off point. Where you got the idea that I thought I could grab my ankles and lift myself I don’t know but it was not from my writings.


----------



## TheTreeSpyder (May 28, 2004)

Mayhem in my mind, test setup.


----------



## Tim Gardner (May 28, 2004)

> _Originally posted by glens _
> *
> Please look at my "29=30" thread in off-topic (linked earlier in this thread, I think) to see how junk math can cause confusion.
> 
> Glen *



Glen, correct me if I am wrong, because it will help me be a better person, but I think that if X = 1 then 1X = 1 and 1/2X = 0.5. Is this correct?


----------



## glens (May 28, 2004)

<b>Tim:</b>&nbsp; I apologize.&nbsp; For some reason I thought you were agreeing with what Ken was saying; not that you were simply trying to restate it in a (more?) clear manner.&nbsp; That is what you're saying, right (not that <i>you</i> believe X = 1.5X)?

<b>Ken:</b>&nbsp; I was wondering if you'd done that with the setup.&nbsp; It was not exactly clear by the photos.&nbsp; That's some sort of hybrid between a (what do you call it?) "single whip" and what you have shown with your inverted slingshot.&nbsp; Try it again, only this time anchoring the pulley so it is immobile like the actual equivalent would be.

Glen


----------



## a_lopa (May 28, 2004)

have you ever done the test 3 pulleys 3 ropes you can pull a forklift with hand brake on with one hand,that will put alot of minds at rest.


----------



## TheTreeSpyder (May 28, 2004)

The pulley is anchored to the broomstick, or system would just hang by hitch, and not route lower through pulley first. It pulls 2x load from this position.

The silver hook below it was to mount the pulley, then alternateively the pulley with scale in between as pictured. i have tested this in my mind for years, observed in my rigs, now bought scales to prove even further. i believe i can setup the system as fairly as possible, trading only the pulley and free travel on the line to assimulate the fairly friction free requirements.


----------

