# Trees and Physics



## ddhlakebound (Aug 9, 2007)

What formula or computations are necessary to go from knowing how much an object weighs and how far it falls, to knowing how many ft/lbs of force are generated in the fall?

What about variables like falling in an arc?

When dealing with rigging trees, natural crotch rigging does not multiply forces, but rigging off a fixed block creates a x2 multiplier on the TIP of the block, and rigging off a block set as an adjustable false crotch results in a 4x multiplier. Is this correct?

It seems that much of the time that basic rigging jobs are almost entirely based on judgement, with little or no math involved, and I've been wanting to grasp a higher understanding of the forces created and applied to all links in a rigging system.


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## clearance (Aug 9, 2007)

Good you are thinking about this. I read this stuff every once in a while here-"My rope is rated at 20 000K!!!!!!!!" Right on retard, what is the scummy tree you are lowering big wood from rated at? There have been many talks here about this, about lowering wood and letting it run, about death caused because "the groundsman didn't let it run", about rigging that you can pull up the Titanic with and so on. What slays me is guys who condem me for one handing risk thier lives to save some ones landscaping by using a completly unrated tree for a spar, one they are tied into, and have guessed its strength. I never lower logs, ever. The forces are huge, the multiplications of these forces are massive. There are methods to estimate shock loads, methods of estimating wieghts of wood, ways to let it run, all dependent on variables you cannot totally control, ever. I like pushing off big chunks or logs, they make dents, sure, whatever.


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## woodchux (Aug 9, 2007)

I've heard that a 100# log can generate 1000# of force if it falls 10 feet and is jerked to a halt. Not sure of how accurate that is though.


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## safeT1st (Aug 9, 2007)

*Dynamometer drop test*

A book of mine "The Art and Science of Practical Rigging" goes into this subject in some detail. It describes two scenarios where a dynamometer was used to measure lbs of force generated . The first case was a 650 lb chunk rigged to block below cut and allowed to run before stopping . Peak load at anchor point (porta-wrap) was 2,189 lbs -force and peak load at rigging point (arborist block) was 4,554 lbs-force . A 631 lb chunk was snubbed off and it generated 3,517 lbs-force at porta-wrap and7,326 lbs-force at rigging block ! As Clearance described the forces are immense and multiply rapidly ! At rest the chunk has "potential energy'. As it falls , potential energy becomes kinetic energy due to gravity . As the rope stretchs the kinetic energy is converted to elastic energy ( some of it ) . If the chunk is allowed to run the kinetic energy will be converted to heat through friction . There are 5 variables in this equation: type of rope , length of rope allowed to stretch , weight of chunk , distance of fall and angle of rope at rigging point . Forces generated can be reduced by smaller chunks , reducing length of fall by tightening rig line prior to drop or raising rig point towards cut , allowing piece to run , increase angle at rigging block and use more line by placing friction device on seperate and distant spar . I know this is not news to most of you out there but interesting none the less .


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## BobEMoto (Aug 11, 2007)

What lengths of rope were used in the examples?


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## safeT1st (Aug 11, 2007)

*Details of load test*

Believe it or not the book doesn't specify but I have a set of VHS tapes that mirror the text , I will plug the tape in tonite and get that answer .As I recall the snubbed drop was only about 10 ft and the other was allowed to run about 20 ft . I will confirm this tonite as I have to run out the door right now .


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## safeT1st (Aug 11, 2007)

*Drop distances for load tests*

Reviewed video and they do not specify drop distances as they are only trying to emphasize the fact that rigging point will feel about twice the load as anchor point and the effectiveness of letting a piece run to lessen loads . Again , I would estimate the snubbed chunk to be about a 12 foot drop and the piece that ran about 25 feet.


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## Tom Dunlap (Aug 12, 2007)

Everyone should own...and read, this seems obvious but it isn't...The Tree Climber's Companion and The Art and Science of Practical Rigging before they go off the ground. 

These two books are the foundation of knowledge for what we do and cover the topics very well.


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## clearance (Aug 12, 2007)

I refer everyone to a thread titled "Lighter/thinner or heavier/thicker?"
In this thread, Tom, on 08-30-2005, at 12:50 pm made some comments about the knowledge of the author of this book. These comments are well worth looking at, for they illustrate my concerns well.


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## ddhlakebound (Aug 12, 2007)

Clearance your concerns are legitimate, and Tom, your advice on reading material is well worth heeding too. 

But I'd really like it if this thread was about learning how to accurately determine the forces involved in a rigging system. There will be lots of people who tie a rope to a limb and start cutting with nothing but a guesswork estimation that "it'll be ok". 

People have made mistakes in the past, and died for it. People will mistakes in the future, and die for them too.

Instead of pointing fingers, or saying a certain book will answer your questions, lets discuss the math and physics of rigging so that the people reading and learning here now can benefit from that information. 

Is a ft/lb as simple as "You drop a one pound weight one foot, and it has 1 ft/lb of energy"? 

Drop a 500 lb weight 3 feet, and develop 1500 ft/lbs of force?

Are the terms energy/force even interchangeable here?

How does ft/lbs of energy or force relate to tensile strength?


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## Tom Dunlap (Aug 12, 2007)

Your desire for knowledge is commendable, that will keep you alive longer.

This topic has been discussed VERY thoroughly in the past on the old I S A forum, Treeb*zz and also here. Taking the time to go back and do a literature review or Search will find lots of great commentary.

Somewhere in an obscure corner of a storage medium there is a record of a thread called 'Forces on a speedline' from the original I S A forum. If anyone ever finds a copy of that thread you will have struck gold. But, I fear, it is like the Lost Dutchman Mine in the Supersition Mountains...found once but now lost in obscurity.

Pete Donzelli told me that there is no good engineering formula that can be applied to the way that we rig. There are too many variables to account for. But...there are some good general principles and rules of thumb available...you'll find these in the discussion forums as well as in books on rigging for rescue, caving and rock climbing.


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## TheTreeSpyder (Aug 12, 2007)

As an approximation; in short drops; we say that the force increases a unit of it's own weight per foot. This, actually has more to do with speed, than distance dropped. So, as the speed increases at higher drops it is my understanding that the seams start to show on that rule of thumb. Or, on the flip side too;-as the leaves/sail increases; the speed is slowed, so thereby the force.

We do tend to get macho and brag on tensile strength only. Such as Clearance's 20k boast example. But, there is another side to this coin. As we speak of dynamic force, we must recognize it's countering dynamic absorbtion forces too. So, id moving/dynamic force; as we go from a 10k line to a 20k; to be more safe; we are actually increasing the forces!!! For, as the strength of the line goes up and the force stays the same; we are losing elasticity in trade. So, the rope, knots, supports etc. are all taking a bigger dynamic hit.

The 2:1 on the redirect pulley; is only a true 2:1; if the load and control legs of line to the pulley are pairallell. Any spread lessens this effect. A pulley, within a pulley system can give the 4:1 - spread of the legs angle effect. Also, - any friction; whereby the total of load leg + control leg that makes the 2:1; will only have full load on the load leg, then a reduction of that force to the control leg; for a lil'less than 2:1. Thus, a frictional support/ no pulley; will have 1+ multiplier; a total of the full load leg + whatever force is on the control leg after friction.

But, we do have some further complications. A shorter load line length + control line length combination; gives less elastic absorbtion too. A 2:1 rig (2 legs of support on load) statically (no movemeant) will give 1.5 x load; loading on support; whereby a simple 1:1 as noted will give 2:1. And a 3:1; will give 1.33x loading statically; or 1 + 1/legs on load. But; this inverses on dynamics; especially on short lengths(notice the total amount of rope length between support and load doesn't matter, just the length from support to load). Whereby, each leg on a 3:1 will carry less load; so has less elasticity value. So statically and at low impacting the 2:1, 3:1 etc. will give less loading on support; but under a good impacting; you can give support much more force. Especially, while very little line in system.

Now, the downside; of using lower strength lines to absorb more shock; is less cycles to failure....

ArborMaster / Sherrill have made a type of calculator for some of this i posted: Peak at program along with a spreadsheet of some of the outputs i copied.

Also, as we load all of this onto a support, with slings, knots etc.; each component becomes part of said system. So, the dynamic absorption that a rope doesn't give, can still be given by flex of the support etc.

Orrrrrrrrrrrrrrrrrrrrrrrr something like dat...


edit: swooped by Tom a-gain; i'll see if i can find such a legendary th-read in the musty archives'ere


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## BobEMoto (Aug 12, 2007)

Here are some force calculations using realistic rope stretch factors. Values are for end of the rope, at the pulley would be more (up to double). This are simple drop calculations without any swinging or bashing into the trunk.


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## safeT1st (Aug 12, 2007)

Is a ft/lb as simple as "You drop a one pound weight one foot, and it has 1 ft/lb of energy"? 

Drop a 500 lb weight 3 feet, and develop 1500 ft/lbs of force?

Are the terms energy/force even interchangeable here?

How does ft/lbs of energy or force relate to tensile strength?[/QUOTE]

ft lbs does not apply to these scenarios . Ft lbs is used to measure work . Work =force X distance . Work = 400 lbs x 10 ft . Work = 4,000 ft lbs .

Energy has 2 main forms : potential and kinetic ( for our purposes ) 

Potential energy = weight x height , releasing potential energy causes motion which is kinetic energy . Energy is often expressed in the unit of horsepower :33,000 ft lbs of work done in one minute . The key being that time is now a factor in the equation . I can spend all day moving one log and I have only accomplished that much work . If I move the same log in 3 minutes I have now entered a specific time into the equation and can determine how much energy ( expressed as horsepower ) I have expended . Again , work = force x distance . Energy (or power ) = force X distance over time (divided by ) .

Newtons second law states that acceleration produced in a mass by a given force is directly proprtional to the force and inversely proportional to the mass . Force = mass x acceleration .


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## clearance (Aug 12, 2007)

For a better understanding of these issues there is a thread "Drop zone safety", from 2006. Tom posted a great link and I dug up a couple as well. There is also an excellent post by Sunrise Guy on page 2. Lowering heavy peices of wood cause me great concern, I never do it. It enrages me that people are advised to risk thier lives to save landscaping. I have had many arguements here about it, its not a popularity contest. As it always takes longer to lower than to hammer it down, people will lower bigger and bigger pieces to make it go faster, exposing themselves to greater and greater risk.


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## safeT1st (Aug 12, 2007)

*Continued*

If an object is allowed to fall freely with the acceleration of gravity it will accelerate uniformly at 32.17 ft per second every second it falls. Thus you can see that the forces generated by a falling object will increase over time spent falling . Again : Force = mass x acceleration . It is important to know the difference between weight and mass . Weight is the force with which gravity attracts a mass . Thus the further an object is from the centre of the earth the less it weighs . Mass is the amount of material in an object and never changes . An astronaut has the same mass on earth and in space but weighs less in space . The mass of an object is determined by the formula

Mass = Weight
__________

Acceleration due to gravity ( 32.2 ft per second per second ) 

I think you now have all the formulas required to calculate the force of any particular object (known weight and distance of fall ) .


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## ddhlakebound (Aug 12, 2007)

safeT1st said:


> ft lbs does not apply to these scenarios . Ft lbs is used to measure work . Work =force X distance . Work = 400 lbs x 10 ft . Work = 4,000 ft lbs .
> 
> Energy has 2 main forms : potential and kinetic ( for our purposes )
> 
> ...




I disagree about the first part. Foot pounds do apply to these scenarios, just as they apply to the energy a bullet transmits upon impact. The block or log we cut is the bullet, and gravity is the gunpowder. 

As for the second part, I may be incorrect, but I think that force = mass x velocity. The wall does not care how long it took the cannonball to reach it's impact velocity, only what that velocity is at the time of impact.


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## safeT1st (Aug 12, 2007)

*I'll say it again*

Ft lbs is an expression of WORK . No time involved period. Every object dropped or moving only falls or moves for so long (acted upon by outside forces ) Because TIME is now part of the equation this is a whole different ballgame . You move that bullet to the target by hand and you use exert less energy but take longer . Fire it from your rifle and you accomlish same amount of work but expend more energy .Work never changed , only amount of energy to accomplish it . This is a hard concept to grasp but thats how it is . Ask Sir Isaac Newton .


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## ddhlakebound (Aug 12, 2007)

safeT1st said:


> Ft lbs is an expression of WORK . No time involved period. Every object dropped or moving only falls or moves for so long (acted upon by outside forces ) Because TIME is now part of the equation this is a whole different ballgame . You move that bullet to the target by hand and you use exert less energy but take longer . Fire it from your rifle and you accomlish same amount of work but expend more energy .Work never changed , only amount of energy to accomplish it . This is a hard concept to grasp but thats how it is . Ask Sir Isaac Newton .



Is the dissipation of kinetic energy before impact not work?


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## safeT1st (Aug 12, 2007)

*Work?*

Work = force x distance . Period . No other factors in equation . If you introduce time then it becomes energy to accomplish that work . If I ask you to move the wood pile and say you have a yr to do it no problem right ? I f I say it has to be moved before coffee you think '" holy crap , I going to need alot of energy to get that done in that time . Energy cannot be created or destroyed, only changed . If I carry a rock up the mountain I have created potential energy in that rock but I sweat my B-lls of doing it and burnt up my breakfast . Rob from Peter to pay Paul sort of idea . Losses are not work. Losses are energy transfers . I have to go out but would enjoy talking about this with you later tonite .


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## TheTreeSpyder (Aug 12, 2007)

Nice chart Bob; but i'd think we'd have to say certain rope constructions are more static or elastic?

i think just because we speak in a rule of thumb for every 1 foot of drop adding 1 unit of load force to the load force; we should not confuse that with "foot pounds" in a free fall. Even when we speak of all that force landing on end of a 1x1 cant vs. the same force/fall on the 1x8 side of the cant; in calculating a "ground psi" of the concussion on hard ground - at least formally.

But, we we lift the big rock with hoist, or carry it up so high; then i think we can talk in foot pounds; without time factor involved as pointed out.

Terminal velocity is were the air friction increase from the speed increase creates enough drag to not allow further acceleration; only constant falling rate. Though we won't see this; it is the bigger picture of the scenario; defining operatives more.

The total force in the rig; would be load x acceleration/deacceleration by the responding deformations (elastic and none). This is all really just leverage etc. Force x distance. Distance of lift, deformation, spread out of impact etc. Even the fall that is load x speed is the same argueably; in that the speed in itself is how much distance over how much time. So really, time is another distance factor itself(only working inversely here; whereby less is a higher multiplier); only set to a multiplier of 1; when we aren't measuring it IMLHO.

As Tom pointed out in the words of the late Pete Donzelli; there are so many factors we can't put hard numbers on it. But, we can sit here and theorize the patterns and variables to get a feel for it, and/or make sure we are tuning in to the correct feel as we work. Because mostly it is that feel and experience that will get you through the day. 2 men can appear to be doing the exact same thing; but one will rig better time, direction, prestretch(line tension and angle/direction being the ballast of load force to achieve float etc.) and impact etc. into the line subtlely; that will make all the differance. One will shockload the line, support and crew more; testing everything more; inviting Murphy's Law to step closer... On the flipside, one will know when to use the amplified force to advantage also! 

Most just try to l-earn from their miss-takes; but if you study the successes just as hard, sifting the info from them; like that is the real paycheck for the day; that will 'fund' all your other days!


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## clearance (Aug 12, 2007)

So, after all the semantics, theory and assorted b.s., you are back to trusting your lives to variables you can't control, am I wrong?


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## TheTreeSpyder (Aug 12, 2007)

i think it is fairer to say that we try to minimize the load multipliers; until we hit the inverse of where we make them work for us; not against. As we at the same time maximize the force dividers; until we hit the curve where they work against and not for us. Using time, distance, angle and direction of force etc. to control forces to our success; like we were dancing wit'em of like driving a car.

And as i told Tom once; i don't understand why people can't control these things with hydraulics in a car; when we control these forces with nothing but the direction/angle and tension in a half inch of line! But, i guess you can't control the variable of the other driver totally; just do your best when it's your turn at bat. i've repounded in daily lessons of force, direction, friction reduction, shock absorption etc. many times driving home; by dropping the wall between these 2 volumes of information we take in. i guess it is all exactly the same; but different!

Even conversation can be viewed as the same dynamic in this framework i think; the real thing is to l-earn what you can as an investmeant for next time; and to be able to healthfully shake it all off as the dust from ye sandals if it brings you down/ hold on to it if it lifts ya higher; and go home at the end of the day and turn off the key(ki).


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## deercatcher (Aug 12, 2007)

Most of us here offer our services for hire. We develop our own niches in the marketplace; and often a reputation can be built by the jobs we refuse. I have studied dynamics; and see a place for a guy like clearance that refuses to do any heavy rigging. But no one has mentioned the degradation of rope strength in tying a knot, which adds in another 50% reduction in rope strength. I admit to being surprized in this work, and surprises are bad. I eagerly awaited the third part of a technical paper written on tree rigging dynamics by Dr. Peter Donzelli, mentioned in this thread by Tom Dunlap. In the magazine, was Pete's obituary. Scared the h-ll out of me.

If you want to calculate wood rigging, remember first, the knot reduces about 50% with a bowline. If you rig a marl (half twist) between the bowline and the anchor, you reduce the rope bend to 90 degrees and keep more fiber bearing the load. Figure your block weight. Then multiply by drop distance for up to only 5 feet. Ad your block weight back in, and that is a safe impact calculation. Acceleration increases on an exponential curve too quickly to use this linear calculation past a drop of 5 feet. If you can layer your systems you can spread the load around some, but not as much as you might think. Swinging a block between two neighboring trees will apply total force to both trees! The two ropes will not(!) half the weight between them. I have butt hitched and used a large speed line at the same time to arrest the fall and stop the movement quickly.

Use everyone's mind that is on the crew...
Expirament in a safe place sometime to familiarize yourself with technique.

Watch out for surprizes!


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## BobEMoto (Aug 12, 2007)

TheTreeSpyder said:


> Nice chart Bob; but i'd think we'd have to say certain rope constructions are more static or elastic?


Thanks. The type of rope is definitely variable. I used specs from a static kermantle rope. I think this has stretch factors that are typical, but its hard to find publishing specs on all types of ropes.



ddhlakebound said:


> As for the second part, I may be incorrect, but I think that force = mass x velocity.


Force = mass x acceleration
but note that it is the acceleration of the stop.
The impact velocity sets the starting conditions (of the stop).


I think the biggest use for the table is not individual values, but to get an idea of when you're in the dieing range. I included 1000 pounds, but I wouldn't want to be anywhere close to dropping something that big.


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## moray (Aug 13, 2007)

*Not Quite Right*



safeT1st said:


> Ft lbs is an expression of WORK . No time involved period. Every object dropped or moving only falls or moves for so long (acted upon by outside forces ) Because TIME is now part of the equation this is a whole different ballgame . You move that bullet to the target by hand and you use exert less energy but take longer . Fire it from your rifle and you accomplish same amount of work but expend more energy .Work never changed , only amount of energy to accomplish it . This is a hard concept to grasp but thats how it is . Ask Sir Isaac Newton .



I've been away or I would have jumped into this thread sooner--I love this stuff! My first reaction is that no one is going to correctly calculate anything if they are confused about the quantities involved. If you are clear about the definitions, then the calculations for a particular problem are usually pretty straightforward, though I doubt anyone is going to actually do them at the job site except in very special circumstances. 

We need to understand mass, force, energy, work, weight, time, power, velocity, and acceleration. With all due respect, the statement--"Work never changed , only amount of energy to accomplish it ."--is simply wrong. Work and energy are equivalent. What I think you are describing is the rate at which work is performed, (or energy expended), which is the definition of power. 

I hope I am not insulting anyone by suggesting that a good way to get a solid grasp of this stuff is to study a good high school physics text. And even if one never actually calculates a rigging problem, it is very satisfying to understand this stuff.

Probably no one is going to follow my suggestion to brush up on their high school physics, and even those with an iron grip on the physics will rarely run through the calculations. Calculations or not, the strength of the wood at the rigging point can't be measured, and everything depends on that. So we fall back on seat-of-the-pants judgment, just like Clearance does every time he decides a tree is strong enough to support his weight.

We are probably naturally pretty talented at that. Our cousins, the great apes, do all sorts of fabulous dynamic moves in the trees, making great leaps across large gaps and safely landing in some surprisingly small wood. Even with all our math and physics, we are hopelessly inept compared to a gibbon.


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## clearance (Aug 13, 2007)

moray said:


> Calculations or not, the strength of the wood at the rigging point can't be measured, and everything depends on that. So we fall back on seat-of-the-pants judgment, just like Clearance does every time he decides a tree is strong enough to support his weight.



Thank you.


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## TheTreeSpyder (Aug 13, 2007)

Clearance's worries amuse me; not out of disrespect, but rather the flipside cuz i won't play around electricity. So, i guess we each have these great forces we think we can dance with intelligently; and just do the next most right thing to keep things on our side.

Here is the saved page from ol'ISA Tom spoke of. i did not have the foresight to also copy the pix before the forum took it's final crashing nosedive.

But, that is where i got my start on this: http://www.mytreelessons.com/Pages/Rope%20Angle%20Leverage%20Calculator.htm just reapplying the formula in the spreadsheet Dave pro-vided from the thread.

Which he then took an made a formulae for a DWT for lifting or lowering on spread supports like DeerCatcher speaks of. Notice it is the same, but adds another angle; that adjusts larger as the other does smaller! 

i prefer to think of each leg of the speedline support as bearing half the load; but just at a leveraged angle; of how many degrees off inline with the flow of force(gravity)/vertical. So, directly overhead support is inline, no leveraged multiplier, so gives us a 2:1. As we spread the supports apart, there is still 2 supports bearing each half the load x the leveraged multiplier for degrees from vertical. Pulling from the end, we still have decreasing leverage over the bend until 120(where each leg is 30 degrees off of vertical) or 1/.5 is the multiplier for half the load or each leg of support line has a tension of the load... Flatter than that and you'd rather reverse your strategy and pull from the bend to have leverage over the end. This is the force for raising tension of line in sweating/swigging; that you then capture as a "purchase". And do it again.


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## moray (Aug 13, 2007)

TheTreeSpyder said:


> i prefer to think of each leg of the speedline support as bearing half the load; but just at a leveraged angle; of how many degrees off inline with the flow of force(gravity)/vertical. So, directly overhead support is inline, no leveraged multiplier, so gives us a 2:1. As we spread the supports apart, there is still 2 supports bearing each half the load x the leveraged multiplier for degrees from vertical.



Correct, as usual. 

For those who know their math, the technical language of vectors and trigonometry is a bit easier to understand than SpyderSpeak. 

But there is a shortcut to getting the numbers right that doesn't require any talk of leverage or degrees, and doesn't require any math beyond the ability to use a ruler. You simply draw the problem to scale (see one of Spyder's links, above, for a diagram). From the load, draw a horizontal line to the right-hand vertical support, and a similar line to the left-hand support. You now have two triangles. The lengths of the lines are proportional to the loads (ignore the imaginary horizontal lines). The longest lines are the two legs of the rope and represent tension. The shorter vertical lines at the supports represent the downward force from the load in the center. This simple method works when the load is halfway between the two supports. With minor adjustment, it will work for any configuration of load and supports.


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## safeT1st (Aug 13, 2007)

[
We need to understand mass, force, energy, work, weight, time, power, velocity, and acceleration. With all due respect, the statement--"Work never changed , only amount of energy to accomplish it ."--is simply wrong. Work and energy are equivalent. What I think you are describing is the rate at which work is performed, (or energy expended), which is the definition of power. 

Well said Moray . Indeed I was trying to convey that without considering time,only work can be determined .I mistakenly infered that energy mirrors power which you have corrected me on . Reason being is that the most common unit for both energy and power is the horsepower which is calculated using time . I too enjoy these topics despite what others might think of them . I refer to my old texts constantly and find it re-sets the information in my head . Remember learning this stuff for the first time and thinking "when will I ever use this stuff " ? Every step we take each day is actually a calculated risk and we are always playing the odds . If nothing else, I think having a better handle on the forces at work around us gives us a better chance to beat the dealer in life ,or hold him at bay .


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## ddhlakebound (Aug 14, 2007)

So we've arrived at:

It's nearly impossible to accurately calculate the forces involved in a dynamic rigging scenario. 

Even if we could calculate the forces accurately, we still don't have any idea how much stress the rig point will support before failure. 

If the terminology isn't used properly, it affects the acceleration of gravity.







Sorry for the sarcasm, I'm somewhat joking, I do understand that the proper use of the terminology in critical to understanding and applying the math. And I'm working on understanding and using it correctly. 

I understand (somewhat) the difficulty in accessing the effect of all the variables involved, (part of the reason we use a 10:1 saftey factor), but I'm unwilling to accept that we can't regularly arrive at a (close to) accurate figure for our loads. 

It may take a long time to reach the level of understanding I'd like to be at, but......watcha gonna do?

Will it be worth it? To be able to put a numerical figure on the load I'll apply to my rigging set? Maybe not, but more information has got to be better than less. But I guess even with the math, it all boils down to judgement. 

Could someone please explain F=MxA? I'm having a hard time grasping how Acceleration matters instead of Velocity. When we drop that block or log into the rigging, the acceleration in freefall will be constant (32 ft/sec2), and from this we should be able to calculate the velocity at impact. How is it that acceleration matters instead of velocity?


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## safeT1st (Aug 14, 2007)

*Study this*

I will quote from a text of mine to minimize confusion :Newton's second law states that the acceleration produced in a mass by the addition of a given force is directly proportional to the force and inversely proportional to the mass . When all forces acting on the body are in balance the object remains at a constant velocity. However, if one force exceeds the other, the velocity of the object changes. Newetons second law is expressed by the formula:
Force=Mass x Acceleration . Speed and velocity : speed and velocity are often used interchangeably but are actually quite different . Speed is simply a rate of motion or the distance an object travels in a given time : mph, feet per second , knots ,kilometers per hr etc. Speed DOES NOT take into consideration any direction. For instance I can walk around the block ( 4 different directions ) and still apply a final figure of speed to my walk:say 3 mph. Velocity on the other hand is the rate of motion in a given direction and is expressed in terms like 500 ft per minute downward or 300 knots eastward.An increase in the rate of motion is called acceleration and a decrease is deceleration,both measured in terms like feet per second per second or metres per second per second. Now my velocity's on that walk around the block will be expressed in 4 seperate figures: 2.5 mph east,1 mph north 4mph west and 4.5 mph south . Do you see the difference ? Now you simply have to accept that to calculate force you use acceleration rather than velocity Going back to Newtons law , the only time the forces are in balance and velocity is constant is terminal velocity. I don't recall but it's about 120 mph ? We will not see this in every day rigging so as a result we are always dealing with acceleration . Again , it's hard to explain and sometimes you just have to say "alright,I dont really get it but I will accept it "


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## dc59222 (Aug 14, 2007)

We must remember two things when we talk about objects falling. One is that terminal velocity is determined by three different factors. The first is the density of air the object is falling through, the second is the amount of drag imparted upon the object by the air through which it is falling (i.e. the size and shape of the object that is falling dictates how fast it can fall through a given density of air), and the third is the weight of the object. For example, the terminal velocity for the average human is about 120 mph at sea level. Terminal velocity for a penny is about 60 mph at sea level. the further away from sea level you rise the higher the terminal velocity becomes for either object due to the decrease in the density of air. Second, we must remember that Newton's second law applies only to objects in a vacuum (i.e. a space with no air and subsoquently no drag). According to Newton two objects of different weights and sizes dropped from the same height will fall at the same height, but that only works in a vacuum because the differences in size are negated by the lack of drag. For an extreme example, if you drop a baseball and a sheet of paper off of the same building undoubtedly the baseball will hit the ground first. Conversely, do the same experiment in a vacuum and both will hit the ground at the same time. Sorry for the high school physics lesson, but there were some over generalizations being made. 
Back to the topic at hand, I subscribe to the philosophy of "use strong rope, place your block in the strongest place available, and cut small pieces." This way may take longer, but there is absolutely no reason to toy with the envelope in this business when lives are on the line.... especially when it is your own. The bottom line is that there are so many variables in tree rigging that no matter what mathmatical formula you use, you never get any closer than an "educated guess" as to how much weight is safe to rig out of any given tree. My motto is make an educated guess and cut smaller just to be safe. Maybe that is stupid, but I am not aware or any climber who has been killed because a tree failed from a load that was too light.


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## TheTreeSpyder (Aug 14, 2007)

i maid this lil'widget a while back. You can adjust the force at the pulley by typing. Also, you can adjust the ends or bend (pulley) position by dragging either end or pulley; to ultimately adjust the angle (multiplier of the minimal/nominal/ inline force) with mouse drags. Then read out the nominal force per leg (half the load); angle of each leg from inline, sine of that angle and the tension force in the system from this leveraging as outputs.

i'm kinda rough cutting it; but i'm taking half the load X leveraged multiplier; leveraged multiplier being 1 divided by the sine of each leg from inline. Stay on the pulley or center of the end; so you don't slip off; and move slower than the 12/second frame rate and it works best on the drags.

http://www.mytreelessons.com/Flash/knots/lineBend.swf

i purposefully did this pulling across with produced rather than gravitational force; to show it is the direction of the load force in comparison to the opposing direction of the support force that gives the multiplier for the nominal (load\supports as minimal loading). we just usually see that as vertical force on a speedline; but that is just because gravity makes the load force up and down against the nearer horizontal supports (giving leveraged angle multiplier to the equal and opposites). Direction is very important to the relationship/marriage of the equal and opposites; another multiplier like time and distance; sometimes just set as a multiplier to 1 when inline or not treating/looking at it; but all ways there!

Notice if this was a rig catching force; the catching of the load would be while ropes where in the most leveraged angle multiplier of that force; then less of an angle as it lowered. The catch is also the speed/impact moment; and the point when there is less elastic rope length in the system to take all this sudden dynamics; especially if redirect to a control leg is frictional/not a pulley. This is especially where running load a bit can ease forces; as long as you were snubbing force out, not building it with the added distance of fall.

Great Discussion!


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## moray (Aug 14, 2007)

safeT1st said:


> [
> 
> Well said Moray . I too enjoy these topics despite what others might think of them . I refer to my old texts constantly and find it re-sets the information in my head . Remember learning this stuff for the first time and thinking "when will I ever use this stuff " ? Every step we take each day is actually a calculated risk and we are always playing the odds . If nothing else, I think having a *better handle on the forces at work around us *gives us a better chance to beat the dealer in life ,or hold him at bay .



Very well said, yourself! You have nailed what I consider the heart of the matter: that the real benefit of a working competence at math and science is that it deepens your judgment and appreciation of things, not that it allows you to calculate complicated formulas. If only it were taught that way in school! A few nerdy types like myself like to do the calculations as well, but as you suggest, it is not really about numbers and calculations. It adds a whole dimension to things, like adding color to a black-and-white world. This may sound a little over the top, but having known a few people really well who had no numerical sense at all, I don't think so.

I remember that Harrison Ford movie where he is making a mid-air transfer from one plane to another, and ends up at the end of a rope trailing from the cargo plane. The rope is nearly horizontal with a slope of maybe 1:20. The non-mathematical among us would probably see nothing wrong with this picture, but I instantly knew it was absurd because he would be supporting something like 20 times his weight. Even if his wrists had been shackled to the rope, the force would have pulled his shoulders from their sockets.

It doesn't just happen in movies. I saw a video a few years back of a dramatic and tragic stunt, again involving a rope and some people with no mathematical sense. Six or seven people were planning to do a long swinging jump from a high bridge over water. The rope was maybe 150 feet long, and the jumpers were standing on the bridge deck, clipped in together at one end of the rope. The other end was anchored to the bridge deck 150 feet away. The jumpers climb over the rail and start a countdown. 

Since the rope is horizontal, the jump is going to be a huge swing, not a bungie jump. One, two, three...they all jump together. Everything is going according to plan until they reach the bottom, at which point the rope breaks. They hit the water going about 90 mph, which results in at least one death and several grievous injuries, if I remember right. Apparently they didn't reckon on centrifugal force dramatically increasing the load on the rope.


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## moray (Aug 14, 2007)

dc59222 said:


> The bottom line is that there are so many variables in tree rigging that no matter what mathmatical formula you use, you never get any closer than an "educated guess" as to how much weight is safe to rig out of any given tree. My motto is make an educated guess and cut smaller just to be safe.



Right on. But the "educated" guess is way better than an uneducated one, so having a good grasp of the math and physics, even without outright calculations, is all to the good.



ddhlakebound said:


> It may take a long time to reach the level of understanding I'd like to be at, but......watcha gonna do?
> 
> *Could someone please explain F=MxA? I'm having a hard time grasping how Acceleration matters instead of Velocity*. When we drop that block or log into the rigging, the acceleration in freefall will be constant (32 ft/sec2), and from this we should be able to calculate the velocity at impact. How is it that acceleration matters instead of velocity?



I admire your desire to better learn this stuff. Bravo!

To answer your questions, I would first note you are on the right track to think velocity matters (it does), but by itself it tells you nothing about the force your rope is going to experience. The complete picture of what happens when you drop a load on a rope involves the velocity when the rope goes taut, the length of the rope supporting the load, the stretchiness of the rope, the mass of the falling piece. In a complicated scenario, other factors like pulley friction, swing angle, rope runout through a friction device like a Porty, etc., might be involved. These don't make the problem hopelessly complicated at all; they just mean your final estimate is going to be a bit rougher than in a simpler case.

Let me just deal with one thing here: F = M x A. (Incidentally, this is not very useful in thinking about this problem. What you really want to know is how much is my rope going to stretch, because that directly tells you the force it is experiencing.) If you were an astronaut in orbit, where things are weightless, and you were asked to take an ordinary baseball and throw it at a target, you might think, because the ball is weightless, this would require no effort. But it requires the same effort (force) that it does on earth. Changing the state of motion of an object is acceleration. Catching or throwing a baseball are examples. To do this requires force, as your own experience shows. Catching or throwing a basketball takes more force, because the mass of the basketball is greater.

When your rope catches a falling load, it continues to stretch until the load is stopped. At that point, because the rope is like a spring, it is applying the maximum force to the dropped load and it is experiencing the maximum tension. The tension in the rope varies from zero when the rope first goes taut to this maximum value when the load has stopped. In theory, the tension in the rope should be proportional to the stretch at each point in this process. This linear relationship makes it easy to calculate how much the rope will have to stretch to absorb all the energy of the falling load. But as you will have noticed, we aren't really using F = M x A at all.


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## TheTreeSpyder (Aug 14, 2007)

i think a lot of good info and feel has/is being given for the subject.

Any sitting back a lil'lost or whatever and wondering why worry about all this etc.; should hang in there along with everyone else.

Because, as we've said these are just guesstimations; in other words proper assessmeant and feel, building on experience is the way one progresses hear. And this information is how you digest your experiences to understand and compare what forces went right or wrong, not quite as expected etc. to get the most out of the experiences; wringing them dry! These things also suggest how and why to minimize what forces present that stand against ye and maximize what forces are working for ya. They also show in time; when just a little correction makes a lot of difference-or not worth the time/fight to correct, and working from the weakest link to upgrade the whole system immediately etc.

We build mostly machines that work by friction and tension; McGuyvering along with the simplest of tools in remote places; no 2 exactly alike. The more you know about the distilled out bare operating properties of each item and their linked/paired relationship to each other; the more things you can pull off safely IMLHO.

But, mostly i think; Sir Francis Bacon said it best in my sig....


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## BobL (Aug 29, 2007)

Greetings all, ddhlakebound suggested I take a look at this thread because I am a physicist to see if I could contribute.

I should perhaps preface any remarks with the fact that I am a home miller (076AV with 52" mill) and know very little about the practicalities of bringing down heavy lumps of wood from a tree. My only direct experience in this regard was some 20 years ago when I was in a hurry and tied up a 20 ft long x 8" branch with a piece of old rope before cutting the branch with a bow saw. What I didn't take into account was the levering action of the length of the branch so the rope snapped and the branch fell - only about 3 ft, flat onto the neighbors wooden shed. The shed made a couple snapping sounds and wobbled a bit but stayed upright. I managed to get the branch off the shed without any further damage. A year or so later we had a big storm and the shed collapsed!

OK - that aside, I can still help with concepts and formulas. To really understand these problems with, diagrams would be useful - you would all get marks docked if you were in my Physics 101 class.  Moray's post on scale drawings is a good one but would really benefit from a diagram or two.

Bobemoto, I'm not quite agreeing with the results on your table .
Lets start by seeing if my diagram represents what you are calculating.
For a 100 lb lump (or for that matter any lump) falling freely from rest over any distance S will reach a velocity of SQRT(2 x g x S).
g = 32.2 ft/s/s/ for most placed on the earth.







When S = 12 ft , v = SQRT(2 x 32.2 x S) = 27.8 ft/second

To decelerate this to zero velocity over distance d =1.27 ft requires a deceleration of v^2/(2d) = 28^2/(2x1.27) = 309 ft/s/s

Now here is where you use F = m x a. 

The force F, generated on the rope during the deceleration = mass x deceleration = 100 x 309 = 30900 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 959 lbs PLUS the actual 100lb mass of the lump = a total of 1059 lbs (Bobemoto gets 1043 lbs) Bobemoto , are you taking rope/pulley friction - ie non free falling, into account?

I hope this helps. BTW there is some very good physics software out there that simulates these situations including angles and rope elasticity very well, see something like http://www.knowplay.com/science/interactive-physics.html

Like others have said, none of this will predict super accurately what happens in reality but you will gain some idea of what to expect. I believe a good experiment under controlled conditions will also give you other ideas and help build your experience.

Happy to try to answer other questions

Cheers


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## BobEMoto (Aug 30, 2007)

Nice picture.



BobL said:


> Bobemoto, I'm not quite agreeing with the results on your table .


I think the difference can be explained by the precision used in some of the fundamental constants. I used:
3.28083989501312 feet/meter
0.45359237 kg/pound
9.80665 m/s^2, but gravity really varies from 9.789 to 9.832.


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## BobL (Aug 30, 2007)

BobEMoto said:


> Nice picture.
> 
> 
> I think the difference can be explained by the precision used in some of the fundamental constants. I used:
> ...



Wow - you did it in SI units and the converted! I went to all the trouble of doing it in ft/lb/s from scratch. I haven't done that since high school in teh 60's.

Anyway - glad you liked the diagram.
Cheers


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## ddhlakebound (Aug 30, 2007)

BobL said:


> Greetings all, ddhlakebound suggested I take a look at this thread because I am a physicist to see if I could contribute.
> 
> I should perhaps preface any remarks with the fact that I am a home miller (076AV with 52" mill) and know very little about the practicalities of bringing down heavy lumps of wood from a tree. My only direct experience in this regard was some 20 years ago when I was in a hurry and tied up a 20 ft long x 8" branch with a piece of old rope before cutting the branch with a bow saw. What I didn't take into account was the levering action of the length of the branch so the rope snapped and the branch fell - only about 3 ft, flat onto the neighbors wooden shed. The shed made a couple snapping sounds and wobbled a bit but stayed upright. I managed to get the branch off the shed without any further damage. A year or so later we had a big storm and the shed collapsed!
> 
> ...



Sweet!

Thanks for joining us, and I think you've done a great job with your whole post, you've cleared alot of things up, and communicated VERY clearly in doing it. 

Now let me give it a try. I'm sure I'll be missing some of the variables, but this first run is basic. 






When S = 3 ft , v = SQRT(2 x 32.2 x S) = 13.86 ft/second

To decelerate this to zero velocity over distance d =1 ft requires a deceleration of v^2/(2d) = 13.86^2/(2x1) = 96.05 ft/s/s

To decelerate this to zero velocity over distance d = 2 ft requires a deceleration of v^2/(2d) = 13.86^2/(2x2) = 48.02 ft/s/s

In the 1ft deceleration, the force F, generated on the rope during the deceleration = mass x deceleration = 425 x 96.05 = 40821.25 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 1267.74 lbs PLUS the actual 425lb mass of the lump = a total of 1692.74 lbs 

In the 2ft deceleration, the force F, generated on the rope during the deceleration = mass x deceleration = 425 x 48.02 = 20408.5 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 633.8 lbs PLUS the actual 425lb mass of the lump = a total of 1058.8 lbs 

Not taking rope/pulley friction - ie non free falling, into account. I wouldn't even know how to begin to calculate/apply those variables....??

Is this correct?


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## BobL (Aug 30, 2007)

ddhlakebound said:


> Not taking rope/pulley friction - ie non free falling, into account. I wouldn't even know how to begin to calculate/apply those variables....??



You got it!  

The example you chose is a good one because it demonstrates that a 100 lb log falling 12 ft and slowing to zero in 1.27 ft (my example) applies close to the same effective force on a rope as a 425 lb log falling 3 ft and stopping to zero in 2 ft (your example).

Rope and pulley friction will vary. In all cases it will reduce the falling velocity which in turn will reduce the effective final force on the rope in the deceleration phase. I've seen arborists wrap rope around other branches to uses as pulleys (I don't know if this is considered OHS appropriate or not) in this case the friction can be very significant.

It sure would have been nice if the guy who cut down my 60 ft gum tree in my back yard had even stopped to think about how much damage he was likely to do by bouncing 50 lb lumps of tree off my brick paving.  

Now here's a physics type question for ya. What is the force on a rope *while it is being used* to lower a 100lb piece at 2 ft/s and at 10ft/s?

Cheers


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## Pilsnaman (Aug 31, 2007)

*May be too late but*

Sorry I didn't get into this earlier and all this may have already been taken care of, I didn't go through each post word for word probably because I forgot my Adderall today (problem with ADD is you get distracted and can't remember to take your ADD medication).
Anyway, I figured as a mechanical engineer and someone that did tree work for a year this may be a subject I can help with. Plenty of others have already shown how the basic calcs are done so here is a little more info. As has already been stated the base equation used is Force = Mass x acceleration. It should be noted that in dynamics you rarely hear the word deceleration, instead it is called a negative acceleration. So what we care about is the rate at which velocity is changed (in our case reduced) from the time force is applied to the lowering line to the time our log reaches a velocity of 0 ft/sec. What we need is the maximum acceleration during this time to determine the maximum force applied to the rope, tree, and block. Here is where we just don't have enough information to perform the calculations required for this problem. A rope is acting as a spring when it begins to deform (stretch) but like most springs in the real world the rope does not act as a linear spring, the force it applies increases as it elongates. In the end we only have the ability to do basic approximations due to this lack of information. Something I learned a long time ago is books and calculations will only bring you so far, experience is required to bring you the rest of the way. When I did tree work my first thought was to try and calculate this but quickly realized the experienced climbers knew what to use when and how much wood to take safely. This leaves room for human error but thats part of the job, we can't calculate how strong the dead tree we are removing will be.
One other thing, if anyone is confused about something or has any other engineering related question feel free to ask. Not saying I will always know the answer but the sun shines on a dog's a$$ every now and then. 
While this post probably doesn't give the answers you guys want, the numerical kind, hopefully it is somewhat insightful.


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## moray (Aug 31, 2007)

Good thread, it is nice to see people doing equations!

The calculations in the last few posts show how to figure the velocity of the falling load at the moment the rope goes taut, and from that and the stopping distance how to calculate the acceleration required to stop the load. Knowing the mass of the dropping load, it is then straightforward to derive the force. 

But if you then go climb your tree and drop the 425 lb chunk, expecting, as calculated, that the rope will experience a force of 1692.74 lbs, you are in for a rude surprise. The actual maximum force, which is the thing we really want to know, is twice that, or 3385 lbs. Why? Because the unspoken assumption that the falling load will be stopped by a constant acceleration (96.05 ft/s/s) is false. We are talking about catching the load with a rope, are we not? We are also talking about the uncluttered simple case of the rope alone doing this work--there is no ground person with a PortaWrap letting the load run. 

The rope acts like a spring. Over a usefully large range, from zero up to about 1/3 of the breaking strength of the rope, it is a pretty good spring in that there is a very good correspondence between applied force and amount of stretch (a linear relationship).






In the diagram the upper weight is positioned where the rope first goes taut. The rope begins to stretch as the load continues downward, but the further the load travels, the more the rope stretches, and the more upward force the rope applies to the load. When this force exceeds the weight of the load, the load begins to slow down. The force, and therefore the acceleration of the load that is slowing it down, continues to increase in linear fashion until the load has stopped, represented by the lower weight in the diagram. At this point the rope experiences the maximum force and the load experiences maximum acceleration upward. The load will bounce, but because of large heat losses in the rope, the bounce height will be very low compared to drop height.

The triangle diagram to the right of the two weight positions shows how force on the rope (horizontal scale) varies with degree of rope stretch (vertical scale). The maximum force, at the bottom, is twice the average force.



Pilsnaman said:


> ...
> What we need is the maximum acceleration during this time to determine the maximum force applied to the rope, tree, and block. Here is where we just *don't have enough information *to perform the calculations required for this problem. A rope is acting as a spring when it begins to deform (stretch) but like most springs in the real world *the rope does not act as a linear spring*, the force it applies increases as it elongates. In the end we only have the ability to do basic approximations due to this lack of information.



I have to partly disagree with the above statement. The rope manufacturer's web sites (Samson Rope, for example) do provide enough information to show that the rope is a roughly linear spring, and give the numerical relationship between force and stretch. I think these numbers are plenty good enough to use in our calculations.


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## ddhlakebound (Aug 31, 2007)

BobL said:


> You got it!
> 
> It sure would have been nice if the guy who cut down my 60 ft gum tree in my back yard had even stopped to think about how much damage he was likely to do by bouncing 50 lb lumps of tree off my brick paving.
> 
> ...



Thx BobL...

Examples like your 60' gum tree are why were here trying to understand this stuff better. The customer wants "no damage", but not damaging the LZ means we've either got to build some sort of a cushion to absorb the damage, or rope it to the ground. In big wood, weight compounds quickly. A 34" diameter, 18" long piece of red oak weighs in at roughly 576 lbs. A 16" 6 foot log weighs in at 510#. 

So anytime that there is no overhead rig point, we're faced with dynamic rigging to lower it. And as has been noted several times in the thread, even if we can calculate all the forces in the system, there's no way to "rate" the tree. 

On lowering a 100# piece @ 2fps, or at 10 fps, since the velocity is constant, only the actual weight of the piece would factor into the force on the rope, right?

So on both 2 & 10 fps.....100# on the working leg, and 100#-friction in the rig point on the running leg. 

Hope I'm still passing.....:biggrinbounce2:


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## lxt (Aug 31, 2007)

All these charts are wonderful, math seems to be on, however noone has considered how old is the rope? & the cut force placed upon the rope by certain tree`s; example clove hitch while butting down a sycamore log as opposed to butting down a shag bark hickory! would this matter?

I always try to give a best guess & if worried take a smaller peice or use a 5/8 instead of a 1/2 rope. just wondering!!

LXT...............


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## Joe (Aug 31, 2007)

http://www.arboristsite.com/showthread.php?t=7385

The information in this thread may also help. I had fun in it.

Joe


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## moray (Aug 31, 2007)

Joe said:


> http://www.arboristsite.com/showthread.php?t=7385
> 
> The information in this thread may also help. I had fun in it.
> 
> Joe



Good link, this is a fun read!

I have to confess I couldn't keep from smiling while reading many of the posts, but I was openly admiring the posters for trying heroically to come to grips with a tricky topic.

Besides a persistent confusion about the meaning of things (mass, energy, velocity, force, acceleration) the biggest problem seemed to be that very few people understood that the impact force from dropping a load cannot be calculated without knowing the distance used up in stopping it. I especially didn't like the several references to some rule of thumb connecting impact force with weight of object and distance of drop. Such a rule could only work if everyone always used the same type of rope.

But here is a rule of thumb for calculating dynamic loads when dropping chunks of wood on a pulley where the groundie can let it run. For a load, W, that drops a distance, D, and then runs a distance, R, the load on the rope will be W + (D/R) x W. 

Two examples: Drop a load 5 feet and let it run another 5 feet. The maximum load on the rope will be twice the weight of the load.
Drop a load 5 feet and let it run 2.5 feet. The maximum load on the rope will be 3 times the weight of the load.

The basic idea here is that if gravity has 10 feet in which to accelerate the load, and you have 5 feet in which to stop it, you have to supply twice the acceleration of gravity, or 2g, to stop it because you have half the distance in which to do it. If you must stop it in 2 feet, then you will need 5g of acceleration to do it. Since you must still support the load when it has stopped moving, you must add W at the end to get the correct value.

This simple method has an optimistic assumption behind it: the groundie is able to supply completely constant acceleration to stop the load. This certainly won't be the case, but it might be pretty close in the case of an experienced operator.


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## ddhlakebound (Sep 3, 2007)

moray said:


> But if you then go climb your tree and drop the 425 lb chunk, expecting, as calculated, that the rope will experience a force of 1692.74 lbs, you are in for a rude surprise. The actual maximum force, which is the thing we really want to know, is twice that, or 3385 lbs. Why? Because the unspoken assumption that the falling load will be stopped by a constant acceleration (96.05 ft/s/s) is false. We are talking about catching the load with a rope, are we not? We are also talking about the uncluttered simple case of the rope alone doing this work--there is no ground person with a PortaWrap letting the load run.
> 
> The rope acts like a spring. Over a usefully large range, from zero up to about 1/3 of the breaking strength of the rope, it is a pretty good spring in that there is a very good correspondence between applied force and amount of stretch (a linear relationship).




Moray, I'm glad you brought this up. I hadn't considered that the acceleration wasn't constant. I do think you've made a couple minor errors after some reflection. I may be wrong. 

I think that the first error is that you can't just double the final load number, you must remove the static 425 lbs, then double, then add the 425 lbs back in. So we get 1692 - 425 x 2 + 425 = 2959 lbs. 

We can double check this by changing the acceleration distance to zero in the equation. 

When S = 3 ft , v = SQRT(2 x 32.2 x S) = 13.86 ft/second

To decelerate this to zero velocity over distance d = 0 ft requires a deceleration of v^2/(2d) = 13.86^2/(2x0) = 192.1 ft/s/s

In the 0ft/instant deceleration, the force F, generated on the rope during the deceleration = mass x deceleration = 425 x 192.1 = 81642.5 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 2535.48 lbs PLUS the actual 425lb mass of the lump = a total of 2960.48 lbs

So the absolute peak load possible in dropping a 425 lb block 3 feet, then stopping it instantly is 2960.48 lbs. 

And since the rope we are using does have some stretch, we won't be stopping the load instantaneously. So now I guess we need to apply some other equation or algorithm to determine the actual peak load, which will be somewhere between 1692 and 2960 lbs. 

Any way you look at it, that's an awful lot of weight generated dropping 425# a mere 3 feet.


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## moray (Sep 3, 2007)

ddhlakebound said:


> We can double check this by changing the acceleration distance to zero in the equation.
> ...
> So the absolute peak load possible in dropping a 425 lb block 3 feet, then stopping it instantly is 2960.48 lbs.



ddhlakebound, good reply--I'm glad you're still trying to wrap your mind around this. It is also good you are trying to attack this from various angles, like a true mathematician!

Before I show how I would prefer to understand and solve this problem, let me first comment on your second sentence quoted above. To stop something instantly as you suggest means instantly changing its velocity from some positive number to zero. To change the velocity of something is to accelerate it. To do it in zero time requires an infinite acceleration. Since F=M x A, your proposed situation implies infinite force! I am not sure what you intended here, but this cannot be it!

How would I solve this? We know that when the 425# block has finally stopped, it has dropped 4 feet. During the last foot, the rope was stretching and catching it. We know the rope is like a spring, and that the distance it stretches is proportional to the force applied to it. This force reaches its maximum value when the block is at the 4-foot position, and it is zero at the 3-foot position. At intermediate positions the force is intermediate also. The graph of this relation is a right triangle (see my diagram in post #44). 

When the block has descended the full 4 feet, it has given up 4ft x 425# of energy, or 1700 ft-lbs. Where does this energy go? Into stretching the rope. The equation connecting energy and force is E = F x D, or energy equals force times distance. The rope absorbs 1700 ft-lbs of energy while stretching one foot. The _average _force on the rope during its one-foot stretch will be 1700 lbs. That gives the right answer of 1700 ft-lbs. But we know force starts at zero and increases linearly to some maximum value. This means the maximum must be 3400 lbs.

Another way to say this to say the area of my triangle represents the energy stored in the rope. The only way for this area to be 1700 ft-lbs is for the bottom leg of the triangle, the maximum force, to be 3400 lbs.

And just to make sure you really have a headache, try this experiment: Take the very same rope and the same 425# block of your hypthetical problem. Only this time use 50 feet of rope. Drop the block 50 feet. The rope goes taut and catches the block. Maximum force on the rope? 3400 lbs! (Only assumption here, not quite true in your diagram, is that the length of freefall and length of rope are the same...)


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## Joe (Sep 3, 2007)

Actually, the correct and simplified procedure for figuring the force or tension in a line from a falling anything on a rope is outlined in post #53 of the A.S. link a few posts back. The metric units for distance and acceleration are used there. There are also a few physics links in that thread which one can use to reference introductory concepts applied in that post to calculate the final force. It's there but one needs to spend the time with it. 

Sherrill, a sponsor of this site, sells rigging software which can show how loads affect different rope constructions when rigging. It's on page 47 of the 2007 master catalog. It will help.

Here's another link which shows how to figure fall factors into the anything falling while tied to a rope problem.

http://student.kuleuven.be/~m9916724/physics/physics.htm

Joe


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## moray (Sep 4, 2007)

Joe said:


> Actually, *the correct* and simplified *procedure *for figuring the force or tension in a line [...] It's there but one needs to *spend the time with it*.



Sorry, I can't let this go by. There are as many correct ways of solving a problem as you can dream up, and the more the better! In fact, WAY better. I like your other comment about spending time with it. To me this means running through the problem various ways, changing parameters, checking the relationship of various quantities, etc. In other words, solve it and related problems many times in many different ways and pretty soon you will truly understand it.

While there are usually many different ways to solve a problem, that doesn't mean they are all equally good. In regard to the problem at hand, even though it can be solved using F = M x A and a little knowledge about springs, it is much cleaner and easier to use the energy relations, as I did above. But do it both ways! The more the better. The results should be the same. Knowing how to attack the problem by more than one route just deepens one's understanding.

Now I'm sure you didn't mean to attach any particular significance to the word "procedure" in your post, but it brought up bad memories of high school math classes where a poor student would ask a good student "What's the formula for such and so?" The useful but unwelcome answer would be "You need to actually understand this stuff so you don't have to memorize some formula or procedure. How will you know which formula to use and when?" Who would you trust to calculate some rigging problem for you--some one who had memorized some formulas or procedures but didn't really understand the physics, or someone who really understood the subject but had memorized nothing?


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## Pilsnaman (Sep 4, 2007)

*Just a little note*

Moray, I am sorry to say that the rope does not act as an ideal spring just many springs don't. Where on the rope companies' websites do you see such information given (like a spring constant perhaps)? Going to the Sampson rope website and looking at the Arbor-Plex they only give information on the elongation percentage with respect to breaking strength percentage. Now take those three data sets and plot them, do you see a straight line? If you answered yes then you did something wrong because that is the only way this is linear. In reality the relationship between elongation and force is not linear but parabolic or exponential. What you are thinking of is Hooke’s Law of elasticity, which you can look up on-line as the idea that the force applied by a spring is linear to the spring’s deformation. The problem is that this approximation is only applicable to certain kinds of material and synthetics (like rope made of polyester and ployolefin) are not applicable. This is reaffirmed by the following calculations I made:

Lets say we have a 200 lbm (pound mass) piece of wood that falls 15 ft, at which time the rope is held tight. The block is located at the same point as the wood’s starting point and 50 ft from the ground. Assume the wood is dropped perfectly vertical and the rope is tied off to the base of the tree and is perfectly vertical. Also, lets assume there is no friction in the block, that the rope going through the block will have no effect on its elongation, and the rope’s weight is negligible compared with the wood.
The point when all slack has been removed from the rope will be 0 ft. According to the Sampson webpage, ½” Pro Master has an average strength of 6300 lbf (pounds force) and shows a table of percent elongation with respect to percent breaking strength. Using this table you can find the following:
x0 = 0 ft; x1 = -1.3 ft; x2 = -2.08 ft; x3 = -2.54 ft
F0 = 0 lbf; F1 = 630 lbf; F2 = 1260 lbf; F3 = 1890 lbf

Where ‘x’ is the distance that the rope has stretched and ‘F’ is the force applied by the rope.
Now we know all know the equation F = m * a, and solving for acceleration gives us a = F/m. we can now solve for ‘a’ at the four points above to determine the acceleration of the mass. Here ‘F’ is the force of the rope in addition to the force of gravity. Note that the gravitational constant must be multiplied to convert units and a positive acceleration will be in the upward direction.
a = F/m
a1 = (630 lbf*200lbf / 200 lbm) * (32.17 lbm*ft/lbf*sec^2) = 69.17 ft/sec^2
a0 = -32.17 ft/sec^2 ; a1 = 69.17 ft/sec^2 ; a2 = 170.5 ft/sec^2 ; a3 = 271.84 ft/sec^2

Using Excel we can plot these accelerations with respect to ‘x’ and have the software extrapolate an equation for this. Acceleration of the wood piece as a result of the rope elongation can be represented as,
a(x) = 35.71x^2 + 26.86x + 31.23

While an exponential equation may also work it does not allow for negative values which is a problem so I went with the parabolic one. Either way this is not a linear spring and the calculations are much more difficult then most are making them out to be.


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## ddhlakebound (Sep 4, 2007)

moray said:


> ddhlakebound, good reply--I'm glad you're still trying to wrap your mind around this. It is also good you are trying to attack this from various angles, like a true mathematician!



Thanks.



moray said:


> Before I show how I would prefer to understand and solve this problem, let me first comment on your second sentence quoted above. To stop something instantly as you suggest means instantly changing its velocity from some positive number to zero. To change the velocity of something is to accelerate it. To do it in zero time requires an infinite acceleration. Since F=M x A, your proposed situation implies infinite force! I am not sure what you intended here, but this cannot be it!



You are correct, I didn't mean to say that we could stop the load "instantaneously". I did mean that *if* we could set up the system to be totally static, so that when the load reached 3 ft, it stops there, with *nearly* zero stretch, that the absolute peak load (425#, 3ft drop) would be 2960 lbs. 



moray said:


> How would I solve this? We know that when the 425# block has finally stopped, it has dropped 4 feet. During the last foot, the rope was stretching and catching it. We know the rope is like a spring, and that the distance it stretches is proportional to the force applied to it. This force reaches its maximum value when the block is at the 4-foot position, and it is zero at the 3-foot position. At intermediate positions the force is intermediate also. The graph of this relation is a right triangle (see my diagram in post #44).



This is the part I'm not very clear on. It seems to me, and again, I could easily be wrong, that the more you stretch the rope, the more difficult it becomes to stretch it more, so this doesn't seem to be a linear relationship, but rather an upward curve, from zero stretch, to the breaking point, where no more stretch is possible. 



moray said:


> When the block has descended the full 4 feet, it has given up 4ft x 425# of energy, or 1700 ft-lbs. Where does this energy go? Into stretching the rope. The equation connecting energy and force is E = F x D, or energy equals force times distance. The rope absorbs 1700 ft-lbs of energy while stretching one foot. The _average _force on the rope during its one-foot stretch will be 1700 lbs. That gives the right answer of 1700 ft-lbs. But we know force starts at zero and increases linearly to some maximum value. This means the maximum must be 3400 lbs.



This doesn't seem logical, even though the avg works out perfectly. When the rope catches the load at 3 ft, and begins to stretch, it's consuming energy to stretch. That energy is no longer available for "weight" when we finally bring the load to a stop, so everything which has been consumed in stretching the rope would not be felt when the load comes to rest. 



moray said:


> Another way to say this to say the area of my triangle represents the energy stored in the rope. The only way for this area to be 1700 ft-lbs is for the bottom leg of the triangle, the maximum force, to be 3400 lbs.



True if the stretching relationship is linear. How can we determine that it is?



moray said:


> And just to make sure you really have a headache, try this experiment: Take the very same rope and the same 425# block of your hypthetical problem. Only this time use 50 feet of rope. Drop the block 50 feet. The rope goes taut and catches the block. Maximum force on the rope? 3400 lbs! (Only assumption here, not quite true in your diagram, is that the length of freefall and length of rope are the same...)



This does not jive at all. Lets drop the variables into the formulas. 

When S = 50 ft , v = SQRT(2 x 32.2 x S) = 56.75 ft/second

To decelerate this to zero velocity over distance d = 0 ft requires a deceleration of v^2/(2d) = 56.75^2/(2x0) = 3220 ft/s/s

In the 0ft/instant deceleration, the force F, generated on the rope during the deceleration = mass x deceleration = 425 x 3220 = 1368500 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 42500 lbs PLUS the actual 425lb mass of the lump = a total of 42925 lbs

You would have to get up around 15+ feet of stretch from a 50' rope to get the peak load down to anything reasonably manageable. Basically, dropping 425 lbs 50 feet onto any arborist rope will assuredly result in a broken rope, failed rigging, or failed rig point, unless it is allowed to run a whole whole lot. 

Maybe we need the physics professor to clear some more stuff up...:help:


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## moray (Sep 5, 2007)

Pilsnaman said:


> Moray, I am sorry to say that the rope does not act as an ideal spring just many springs don't. [...]
> Either way this is not a linear spring and the calculations are much more difficult then most are making them out to be.



This is an excellent post, Pilsnaman! Thanks for checking on me.

First off, I completely agree that my assertion that the rope acts like a spring is just an approximation. As you point out, even a real spring only behaves approximately like an ideal spring, but the numerical error in treating it as ideal is usually small. In the case of rope, it certainly shows spring-like behavior in the sense that it takes an increase in force to produce an increase in stretch. But how linear is this relationship?

I put together the chart below from the Samson Web site. I have looked at these numbers many times because I own 5 of the ropes and have been interested in the others. For each rope, the three numeric columns represent, in order, the % rope stretch at 10%, 20%, and 30% of ultimate break strength. They are, effectively, 3 data points on the curve of stretch vs. force. None of them will plot as a straight line (though Amsteel-Blue and Tree-Master are pretty close). An ideal spring must, of course, plot as a straight line.






One thing jumps out at me. In every case, the first number is larger, often much larger, than it should be. If the spring behavior is linear, the first number should be 1/3 of the 3rd number, but it is always more than that, and in some cases more than 1/2. (Arbor-Plex, by the way, your choice of rope, is by far the worst rope of the lot in terms of linearity.) I think the explanation for this is that up to about 10% maximum load or so the rope fibers are stretching (they are loaded, after all), but the rope is stretching as well. By this I mean the weave is tightening and aligning more with the tension, spaces between the yarns are closing up, and all the slack and slop in the construction is being eliminated. This causes a lengthening of the rope quite separate from the stretching of individual fibers. Once this initial construction slack is largely gone, the rope behaves from then on (higher loads) in a more linear fashion, as the table shows, because from then on almost all of the stretch comes from stretching individual fibers.

If a particular rope actually behaved like a spring, the linear relation between force and stretch means the difference between columns 1 and 2 must equal the difference between columns 2 and 3. Each increment of force must produce the same increment of stretch.

Look at the first 4 ropes. In all of them the second increment is smaller than the first. The rope is growing stiffer under more tension. In the next group of 5, the second increment is larger than the first. The rope is growing less stiff under more tension! The two groups deviate from linearity in opposite directions! This is why I chose to split the difference and treat them all as roughly linear. 

The two figures below show the stretch characteristic of first an ideal rope that behaves like a perfect spring, and second a close approximation of a real rope. The horizontal axis in each case represents force as a percent of ultimate break strength, and the vertical axis represents amount of stretch in arbitrary units. In the second figure I took pains to make the stretch at the 10% load equal to half the stretch at the 30% load--somewhat worse than the behavior of most of the real ropes.






What can we conclude? When I look at the two figures I don't see a huge amount of difference, so I am inclined to continue thinking of the rope as a spring. Because, in every case, the rope is much softer at the beginning of its stretch, we need to allow for that in any calculations we make. The area in the upper triangle and the area in the lower "triangle" (with curved hypotenuse) represent the work (energy) done to stretch the rope. Since the area of the lower figure is clearly less than that of upper figure--the ideal case--less energy has been stored or consumed by the real rope. We must stretch it still more by applying more force in order to store as much energy as the ideal rope. The peak force when the real rope catches a load will thus be somewhat higher than what we calculate for an ideal rope. It looks to me if we allowed for the peak force to be 25% greater than calculated, we would be in the ballpark.

Once again, excellent discussion!


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## moray (Sep 5, 2007)

*Whew!*



ddhlakebound said:


> I did mean that *if* we could set up the system to be totally static, so that when the load reached 3 ft, it stops there, with *nearly* zero stretch, that the absolute peak load (425#, 3ft drop) would be 2960 lbs.



How fast are you stopping it with "nearly" zero stretch? The governing equation is _F=m*a_. If _a_ is huge, the _F_ will be correspondingly huge. Inescapable. There is no such thing as an absolute peak load, to put it bluntly.



ddhlakebound said:


> This is the part I'm not very clear on. It seems to me, and again, I could easily be wrong, that the more you stretch the rope, the more difficult it becomes to stretch it more, so this doesn't seem to be a linear relationship, but rather an upward curve, from zero stretch, to the breaking point, where no more stretch is possible.



You are absolutely right about the rope. I think you are confusing the word _constant _and the word _linear_. The relationship is linear if each equal increment of force causes equal increments of stretch. If you keep adding weight to the rope, 10 lbs at a time, and each time the rope stretches one more inch, you have a linear relationship.



ddhlakebound said:


> This doesn't seem logical, even though the avg works out perfectly. When the rope catches the load at 3 ft, and begins to stretch, it's consuming energy to stretch. That energy is no longer available for "weight" when we finally bring the load to a stop, so everything which has been consumed in stretching the rope would not be felt when the load comes to rest.



Don't mix up weight and energy. They may be related in a particular situation, and it is enormously useful to understand what the relationship may be, but they are different things. The phrase "energy no longer available for weight" seems to imply you expect one to be converted into the other. The _motion _of the load goes to zero when the rope is fully stretched, all the energy has been transferred to the rope, the force on the rope reaches maximum, and the accleration of the load also reaches maximum.



ddhlakebound said:


> True if the stretching relationship is linear. How can we determine that it is?



See my long post above as regards linearity. Others may disagree, but I think it is a useful and easy approximation, and we can add a fudge factor to allow for the softness of the rope during early stages of stretch.



ddhlakebound said:


> This does not jive at all. [...]
> a total of 42925 lbs
> 
> You would have to get up around 15+ feet of stretch from a 50' rope to get the peak load down to anything reasonably manageable. Basically, dropping 425 lbs 50 feet onto any arborist rope will assuredly result in a broken rope, failed rigging, or failed rig point, unless it is allowed to run a whole whole lot.


 
I was sure this would get your attention! This idea, known as Fall Factor, is well-known to rock climbers. It basically says, for a given falling mass, it is the _ratio _of fall distance to length of rope catching the load that matters. Distance of fall does not matter. So, in our case, dropping 425# 3 feet and catching it with 3 feet of rope is exactly the same, as far as the rope is concerned, as dropping it 50 feet and catching it with 50 feet of rope.

And you are right about the 15+ feet of stretch! Your rope from your example stretched 1 foot from the 3-foot fall; it will stretch 16.67 feet from the 50-foot fall.

This is terrific stuff, but you have to live with it for awhile before it clicks.


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## Pilsnaman (Sep 5, 2007)

The fact is we only have three data points for each rope type, not nearly enough to make any conclusion. That being said, we can not assume this to be linear because the change is negligible in comparison. Take a look at the graph below of the acceleration with respect to distance, this for Pro-Master, and note that I have included the trend line. You are correct in saying the acceleration will have a more drastic change at the beginning of the stretch but what happens near the rope's breaking strength? This data is not given by the manufacturer as they only go up to 30%, and we both know people hear bring their ropes past this. We also know that Hooke's Law is not normally applicable to synthetic springs, like the rope we are evaluating. There is a reason for this, synthetics tend to have spring characteristics that are too far from being linear. The simple comparisons you did are not enough to assume linearity, we are not talking high school physics here.

All this aside here is my engineering evaluation of this system... The only way to correctly identify the forces seen at different parts of a lowering system is to use measuring tools and do tests. This is because of the large number of unknown variables seen in the real world. Just to name a few we have the friction at the block, the angle of the rope going from the block to where it is held/tied off, the angle from the block to where the wood first takes up rope slack (no way the wood falls exactly vertical and/or in a vertical line with the block), the amount the tree gives with this load, how much the groundy lets the rope run...
I just don't see how we can assume all of these variables are negligible, especially ones like angle and tree movement.
On a final note, most of the guys on here don't like when John Doe homeowner or landscaper tries to tell them they don't do something right. That's because a professional has knowledge beyond that of a non-pro. As someone with a Mechanical Engineering degree from Virginia Tech and that does mechanical engineering as a job I have spent years working on things like dynamics, statics, mechanics of materials, vibrations, and deformation. When it comes to something involving mechanical engineering I have more knowledge and expertise then say a software developer. I am not trying to brag or say I am smarter then anyone here, there are A LOT of things I am ignorant on but that is why we have professionals. This just happens to be the subject I am a professional in.
P.S. Sorry for the large image but things were not coming through clear when scaled down.


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## ddhlakebound (Sep 6, 2007)

moray said:


> How fast are you stopping it with "nearly" zero stretch? The governing equation is _F=m*a_. If _a_ is huge, the _F_ will be correspondingly huge. Inescapable. There is no such thing as an absolute peak load, to put it bluntly.
> 
> You are absolutely right about the rope. I think you are confusing the word _constant _and the word _linear_. The relationship is linear if each equal increment of force causes equal increments of stretch. If you keep adding weight to the rope, 10 lbs at a time, and each time the rope stretches one more inch, you have a linear relationship.
> 
> ...



OK, I finally understand that there's no absolute peak load. Had to run a couple fractional equations to get it. 

I didn't mix up weight and energy. Thats why weight was in quotes. When we bring the load to a stop, the kinetic energy is felt as lbf....basically simulated weight. 

I too disagree that the rope works in any linear fashion as a spring. I would like the link to where you got those stretch/load figures, I did a quick search and couldn't find them. But based on the limited info from 3 points, it shows to me that the relationship is not linear. 



moray said:


> I was sure this would get your attention! This idea, known as Fall Factor, is well-known to rock climbers. It basically says, for a given falling mass, it is the _ratio _of fall distance to length of rope catching the load that matters. Distance of fall does not matter. So, in our case, dropping 425# 3 feet and catching it with 3 feet of rope is exactly the same, as far as the rope is concerned, as dropping it 50 feet and catching it with 50 feet of rope.
> 
> And you are right about the 15+ feet of stretch! Your rope from your example stretched 1 foot from the 3-foot fall; it will stretch 16.67 feet from the 50-foot fall.
> 
> This is terrific stuff, but you have to live with it for awhile before it clicks.



Fall factor has alot more to consider than the distance of the fall. It also includes the TOTAL amount of rope in the system. So to figure the fall factor, we also need to know how high up the rig point is. 

Rockers use fall factors because they climb on dynamic rope. In the trees most of the rope we use is much more static, so trying to compare the two is definately apple and oranges. 

Now, we are not going to get 1 foot of stretch from a 3 foot rope, and we are not going to get 15 feet of stretch from a 50 foot rope. So a large percentage of the distance needed to stop the load from either fall comes from letting the rope run. Is it easier to let it run one foot or fifteen? So how is it the same on the rope?

The mass and the distance of the fall determine how much kinetic energy we must dissipate to stop a load. The distance we stop it in determines how much force we apply to the rope. The distance of the fall matters very very much, no matter how you look at it. More distance = more energy. For it to matter less, you'd need some very tall trees, and a whole lotta rope in the system, to make your ratio's work out. 

I'd be very interested to see you successfully drop 425 lbs 50', then catch it on a 1/2 inch arborist rope. Just make sure if you try it that nobody is near the dropzone, cause I bet the rope, the rigging, or the rig point breaks. 

Even discussing it muddies the waters, clouding what we're trying to learn here.


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## moray (Sep 6, 2007)

Pilsnaman said:


> The fact is we only have three data points for each rope type, not nearly enough to make any conclusion. That being said, we can not assume this to be linear because the change is negligible in comparison. Take a look at the graph below of the acceleration with respect to distance, this for Pro-Master, and note that I have included the trend line.



I, too, am saddened by the fact we have only 3 data points per rope. I guess we have to make the best of it. Sorry, I don't understand your second sentence. I like your chart with the best-fit second-degree polynomial. I show it again, below, this time with two straight lines fitted by hand. This shows how we can do a pretty good job fitting the high-load part of the curve with one straight line, and the low-rope-stiffness, low-load part of the curve with another. But for someone just trying to get a general sense of how all this works, I would suggest a single straight line from the y-intercept to the furthermost data point.








Pilsnaman said:


> You are correct in saying the acceleration will have a more drastic change at the beginning of the stretch but what happens near the rope's breaking strength? This data is not given by the manufacturer as they only go up to 30%, and we both know people hear bring their ropes past this. We also know that Hooke's Law is not normally applicable to synthetic springs, like the rope we are evaluating. There is a reason for this, synthetics tend to have spring characteristics that are too far from being linear. The simple comparisons you did are not enough to assume linearity, we are not talking high school physics here.



The Samson data actually show the opposite: there is relatively little change in acceleration (equivalently, force) at the beginning of the stretch. Who cares what happens near the rope's breaking strength? No one in their right mind operates their rope at the edge of failure, and I have no idea how to calculate what happens in that region. You keep raising the notion that synthetics are not true Hookeian materials, but nowhere do I claim such a thing nor rely on it. My claim is that if you look at Samson's data sheets, and assume a smooth curve connects the (admittedly sparse) 3 data points, you can clearly fit a straight line to them without too much error. For an even better fit, you can model the low-load and high-load sections with two separate lines. It makes no difference whether the underlying rope material is springlike or not--we have to live with real rope behavior, and the Samson data, sparse as they are, tell us something about that. (But see the citation at the bottom of this post for some really good rope data.)



Pilsnaman said:


> All this aside here is my engineering evaluation of this system... The only way to correctly identify the forces seen at different parts of a lowering system is to use measuring tools and do tests. This is because of the large number of unknown variables seen in the real world. Just to name a few we have the friction at the block, the angle of the rope going from the block to where it is held/tied off, the angle from the block to where the wood first takes up rope slack (no way the wood falls exactly vertical and/or in a vertical line with the block), the amount the tree gives with this load, how much the groundy lets the rope run...
> I just don't see how we can assume all of these variables are negligible, especially ones like angle and tree movement.



To take your last sentence first, who is assuming anything is negligible? This whole thread has been almost exclusively devoted to the simple case of a rope catching a vertically dropped load. No angles, no run-out, no wood, no friction. It might be useful to get a good handle on this simple case before we take on a blizzard of additional factors. As to the utility of measuring tools and tests, I heartily agree (see citation below). But if ddhlakebound wants to develop a mathematical sense for the behavior of a rope catching a load, he, like the rest of us, is going have to live by his wits, tying together the limited data out there with solid math, plausible assumptions, and good approximations.

And then we come to the following amusing and regrettable passage:


Pilsnaman said:


> On a final note, most of the guys on here don't like when John Doe homeowner or landscaper tries to tell them they don't do something right. That's because a professional has knowledge beyond that of a non-pro. As someone with a Mechanical Engineering degree from Virginia Tech and that does mechanical engineering as a job I have spent years working on things like dynamics, statics, mechanics of materials, vibrations, and deformation. When it comes to something involving mechanical engineering I have more knowledge and expertise then say a software developer. I am not trying to brag or say I am smarter then anyone here, there are A LOT of things I am ignorant on but that is why we have professionals. This just happens to be the subject I am a professional in.



Since, no matter how hard I tried, I just could not find anything constructive or useful in this paragraph, I was at first inclined to ignore it. Maybe I should have. But then I decided I would at least wonder out loud why anyone would interrupt a perfectly good and lively discussion of complex ideas with a sexist and self-inflating paragraph from a poorly written job application?! And why spoil even that by including a completely gratuitous and invidious comparison with a total stranger? Employer to Applicant: "Don't call us, we'll call you. And while you're waiting to hear from us, learn some manners!" 


But back to the subject and back to the world of facts and reason. Here is a link to some excellent real-world testing of various synthetic ropes: http://www.xmission.com/~tmoyer/testing/Qualifying_a_Rescue_Rope.pdf
I have run into other papers by the author, who works in search and rescue for the Salt Lake City Sheriff's Office, and does some very nice experiments in his spare time. Every time I read one of his pieces I come away impressed by the quality of both his experiments and his analyses. In this paper there is an excellent chart on page 3 that plots force vs stretch for a number of ropes with something like 15 data points each. All of the plots, save one, look just about as linear as real-world data ever get. The one exception is only mildly curved. Again, this bears out my contention that over reasonable load ranges a rope will behave approximately like a spring.


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## moray (Sep 6, 2007)

*good questions*



ddhlakebound said:


> I didn't mix up weight and energy. Thats why weight was in quotes.


I saw the quotes. Point taken. By the way, please don't take anything I say personally. I sincerely admire your desire to develop a mathematical understanding of this stuff. If I sound abrupt or severe at times, that is just my crappy style, nothing personal at all. I want to be helpful, and I love a good technical discussion...



ddhlakebound said:


> When we bring the load to a stop, the kinetic energy is felt as lbf....basically simulated weight.


This is the kind of expression that makes me wince a little, even though I know what you mean. The kinetic energy is spent by stretching the rope. The rope resists this stretching because it takes force to stretch the rope. When the load has been brought to a full stop, the kinetic energy is all gone and the rope is exerting the maximum force against the load. I think this force is the "simulated weight" you speak of. Right?



ddhlakebound said:


> I too disagree that the rope works in any linear fashion as a spring. I would like the link to where you got those stretch/load figures, I did a quick search and couldn't find them. But based on the limited info from 3 points, it shows to me that the relationship is not linear.


 
Follow the link in my previous post. There are some beautiful data sets there. I wish the manufacturers of rigging rope would publish nice big tables of rope-stretch tests so we could make more accurate calculations of various rigging scenarios.



ddhlakebound said:


> Fall factor has alot more to consider than the distance of the fall. It also includes the TOTAL amount of rope in the system. So to figure the fall factor, we also need to know how high up the rig point is.


I thought I defined it correctly. It is the ratio of the distance the object falls to the length of rope that will arrest the fall. If the rope goes over a biner at a protection point and down to a belayer, that is part of the rope length that must be counted. But this complicates things a bit, because the load on that leg of the rope will be significantly less than the leg attached to the falling load (climber). In my comment about fall factor, I thought it was clear I was comparing a 3-ft fall arrested by a 3-foot rope to a 50-ft fall arrested by a 50-ft rope. In other words, no block, no biners, no belayer, just a knot attaching the rope to some theoretical anchor. In both cases, obviously, the fall factor is 1.



ddhlakebound said:


> Rockers use fall factors because they climb on dynamic rope. In the trees most of the rope we use is much more static, so trying to compare the two is definately apple and oranges.



Easy to say, but is either statement true? I think climbers use the fall factor concept because it beautifully captures the physics of falling on a rope in a very simple and useful form. As to the second statement, why should it matter how stretchy the rope is? If you look at the math behind the fall factor concept, the stretchiness of the rope doesn't enter into it.



ddhlakebound said:


> Now, we are not going to get 1 foot of stretch from a 3 foot rope, and we are not going to get 15 feet of stretch from a 50 foot rope. So a large percentage of the distance needed to stop the load from either fall comes from letting the rope run. Is it easier to let it run one foot or fifteen? So how is it the same on the rope?



Wait a minute. I never claimed this was a realistic rope. I merely took your original example, for which I had calculated the maximum rope tension at 3400 lbs and extrapolated it to a 50-ft fall on the same rope. If you prefer a more realistic example, then drop 20 lbs first 10 feet on 10 feet of arborist rope, and then 50 feet on 50 feet of the same rope. The maximum tension on the rope, whatever that turns out to be, will be the same in both cases because the fall factor is the same. 



ddhlakebound said:


> The mass and the distance of the fall determine how much kinetic energy we must dissipate to stop a load. The distance we stop it in determines how much force we apply to the rope. The distance of the fall matters very very much, no matter how you look at it. More distance = more energy.



Perfect! Can't argue with this.



ddhlakebound said:


> I'd be very interested to see you successfully drop 425 lbs 50', then catch it on a 1/2 inch arborist rope. Just make sure if you try it that nobody is near the dropzone, cause I bet the rope, the rigging, or the rig point breaks.
> 
> Even discussing it muddies the waters, clouding what we're trying to learn here.



I can see why discussing a highly unrealistic example may not appeal to you, but it doesn't affect the math. There have been plenty of times in the past when I have struggled with some mathematical or physical problem that suddenly became clear when I picked highly extreme or unrealistic examples to run the math on. But if you prefer to use nothing but highly realistic examples from here on out, I promise I won't stop talking to you.


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## Pilsnaman (Sep 6, 2007)

TreeCo said:


> Here it is reduced in file size and turned into a jpg.



Thank you TreeCo. The only reason I didn't do that was so the equation and numbers could be easily read, found that parts of letters got lost when reduced.


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## Pilsnaman (Sep 7, 2007)

The article on rescue rope is very interesting, I plan on reading it in more depth later on. As for Hooke's Law, you are saying the rope abides by this in saying the amount it deforms is linearly proportional to the force required to cause deformation (approximately). Look up Hooke's Law on wikipedia, its always good to do some research before making a comment right? Also, saying that something is "spring like" is not synonymous with saying deformation and force have a linear relationship, there are plenty of springs in the world made specifically so this relationship is non-linear.
The ropes we use for rigging may very well exhibit a linear relationship but the only way to know this is through additional information. All I know about the Samson rope is the difference in elongation between 20% and 30% breaking strength is almost half of that between 10% and 20%. Using a straight line on this will result in more error then you imply, which is increased if the value at zero is included. I am not saying there could not be a linear relationship after some point, actually if the parabolic trend line I made is accurate there will be a point at which linearity can be assumed (until plastic deformation occurs but that would require loading close to breaking strength which is not good). I don’t think the two lines you drew on the graph would work for this as it will be further up on the parabola that linearity begins.
Why is it unreasonable for an arborist to load a rigging rope beyond 30% of its breaking strength? I am not saying they should bring it to 80 or 90%. A piece of oak 3’ long and 20” in diameter will can weigh around 420 lbs, think of how much force that translates to when coming to a stop in 3 ft.
You said we are only looking at the simplistic case of the piece falling vertical and no other variables but looking at the original post of this thread I disagree. The question included how falling in an arc will change loading. Pointing out all the variables is important as they can dramatically change the forces seen on these ropes. The simple calcs you are doing will not be true representations of what is seen on the job and I wanted to make sure it was stated, kind of like a little disclaimer. What I think you are trying to say is that we can still work on the ideal scenario and with that I fully agree.
As for the explanation of my background, again I am not trying to toot my own horn or anything. That was also not meant to be a job application, don’t really know where that came from. All I was trying to do is show a bit of my background to give my information some credibility. Figured my information would mean a bit more if you it was coming from a mechanical engineer, but I guess that’s not the case with you (moray). Why should anyone believe something without ensuring credibility of its source? The only thing I know about you (moray) is that you are a software designer and to me that does not inspire confidence in doing engineering analysis.
I am out of this pi$$ing contest now, further discussions on loads are more then welcome though.


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## moray (Sep 7, 2007)

Pilsnaman said:


> As for the explanation of my background, again I am not trying to toot my own horn or anything. That was also not meant to be a job application, don’t really know where that came from. All I was trying to do is show a bit of my background *to give my information some credibility*. Figured my information would mean a bit more if you it was coming from a mechanical engineer, but I guess that’s not the case with you (moray). [/b]Why should anyone believe something[/b] without ensuring credibility of its source? *The only thing I know* about you (moray) is that you are a *software designer* and to me that *does not inspire confidence *in doing engineering analysis.
> I am out of this pi$$ing contest now, further discussions on loads are more then welcome though.



Pilsnaman, thank you for your measured response. I don't like this kind of exchange at all, but I feel it is a price I have to pay to preserve good high-level discussions about ideas, including the one we were having until the offensive paragraph came along. I want to actually explain myself this time rather than simply express displeasure with the paragraph in question. I also want to leave you out of the equation entirely, and make a general philosophical case that is addressed to everyone, but first I need to identify the two particular items that I think are both illegitimate and harmful in any field of discourse. The first is the argument _ad hominem_ in which you conclude, from two words in my Bio ("software developer") that anything I say in this thread must be discounted. The second is the related argument from Authority in which you explicitly claim anything you say, by contrast, has to be taken much more seriously. End of personal references.

Facts stand on their own feet. They are true or false or something in between regardless of who presents them. Arguments are the same--either spot on, a little off the mark, or irrelevant and worthless, no matter who the author may be. A mathematical relation is true or false or misapplied or whatever, again, irrespective of who writes it down. Part of the joy of a site like AS is the challenge of navigating a thicket of facts and pseudo-facts and good and bad arguments and good advice and not-so-good advice to reach your own understanding of something you didn't understand before. This is an experience that cannot simply be delivered by an "authority." I have learned a lot of good stuff here, and had hours of enjoyment in the bargain, and never once did I need to consider whether someone was an "authority" or not.

To shine a really bright light on the argument from Authority, I refer everyone to the "Monty Hall problem," which is well-described on Wikipedia. This was briefly quite famous when the _Parade_ magazine column ran in 1990, and several hundred mathematicians, and thousands of normal folks, wrote the magazine to claim the published answer was wrong. What the Wikipedia article doesn't say, but I remember reading in some follow-up a few years later, was that some 100 or so of the mathematicians decided to indulge themselves by trotting out the argument from Authority: "I've been teaching college-level probability courses for 40 years and I am appalled that your columnist would actually get this elementary problem wrong." When it finally became clear that the mathematcians were wrong, and the columnist right, I am sorry to report that only 2 or 3 of the hundred who had enjoyed publicly waving their credentials actually had enough character to write back and apologize for their unworthy behavior.

I'm done with this, and ready to get back to rope calculations.


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## moray (Sep 9, 2007)

*A Real Experiment: Background*

Throughout this thread there has been a vigorous debate about the true nature of the stress-strain relationship for arborist rope, and whether that relationship is linear, or close enough to linear, that the rope can be treated like a spring for calculation purposes. Everyone seems to agree that the data published by the rope manufacturers, Samson in particular, are woefully abbreviated, and that much more extensive data sets would solve the problem. A parallel issue has been the nature of the underlying material from which the rope is made. Because the rope is a complex structure, the stress-strain relationship for the rope may not precisely mirror the stress-strain relationship of a single fiber of the rope. But it would be reasonable, if actual data for a rope are unavailable, to assume that the rope behavior won't be wildly different from that of a single fiber. The more we know about the actual behavior of the rope, the less we have to fill in the data gaps with assumptions and reasoning and calculations.



Pilsnaman said:


> In reality the relationship between elongation and force is not linear but parabolic or exponential. What you are thinking of is Hooke’s Law of elasticity, which you can look up on-line as the idea that the force applied by a spring is linear to the spring’s deformation. The problem is that this approximation is only applicable to certain kinds of material and synthetics (like rope made of polyester and ployolefin) are not applicable.



I did a bit of online searching to find some evidence for or against the categorical statements above. As to the contention that Hooke's law doesn't apply to synthetics, my Web search pretty much came up empty. Hard data on rope stress-strain relations weren't jumping out at me at every turn, either, but the following link is excellent: http://www.xmission.com/~tmoyer/testing/Qualifying_a_Rescue_Rope.pdf.
And here is the Samson link:
http://www.samsonrope.com/index.cfm

The first reference shows that for a number of ropes the stress-strain relationship is quite linear up to 1000 lbs, which is as far as the testing went. Data points from Samson, which I tabulated in an earlier post, are clearly not linear over the load range from zero to 30% of ultimate strength; it would be an approximation to treat them as if they were. 

What I would _really_ like to do is some experiments like those performed in the first reference above, but on arborist ropes. Naturally I don't have the big expensive machinery I would need for such a project, much as I would love to play with such stuff. But suddenly it occurred to me that I could perform a mini-experiment with simple materials ready at hand. I could measure the stress-strain relationship for a single yarn of an arborist rope!

The rope I chose was a length of 1/2 inch Samson Tree-Master. This is a 3-strand polyester rope, sometimes used for rigging, with a published break strength of 7000 lbs. There are 117 yarns in the rope, each of which is a mildly twisted bundle of dozens of individual parallel fibers. The yarns are bundled into mini 3-strand ropelets, twisted in the reverse direction. These ropelets are then bundled into groups of 13 and given a left-hand twist. These bundles of 13 comprise the major strands of the rope, which are given a right-hand twist. I mention all this to indicate how complicated even a "simple" 3-strand rope can be, and why we should hesitate to assume that the stretching behavior of a complex structure like this would necessarily be as simple as that of an individual fiber.

Since there are 117 yarns in the rope, each one should have a break strength of about 60 lbs. Since a single fiber of the rope, if you could see it all by itself, would take the form of a complicated set of spirals within spirals, it is clearly not lined up directly with tension in the rope, and it won't be as strong in that orientation as it would be if were simply a straight fiber under tension. This means one yarn from the rope will have to be a bit stronger than 60 lbs for the whole rope to have 7000 lbs of tensile strength. In any event, properly testing a yarn of about 60 lbs tensile strength was going to be well within my experimental means.


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## moray (Sep 9, 2007)

*A Real Experiment: Methods and Materials*

I retrieved 3 yarns from an unused piece of rope, each about 5 feet long. The idea was to anchor the top of the yarn to a low tree limb and then add succesive units of weight to the bottom. Measuring the stretch of the yarn would be as simple as noting the position of a marked point on the rope against a background ruler. With each addition of weight, the rope would stretch a certain amount, the mark would descend a short distance, and the new location of the mark could be recorded.

Since I didn't have a whole box of identical weights, I decided use identical volumes of water for my weights. I suspended a 5-gallon plastic bucket from the bottom of the strand being tested. Each time I added water to the bucket, I would record the new position of the mark against the ruler. Since I wanted to be able to do this quickly and uniformly, I chose to fill a large tub with about 10 gallons of water, and scoop water from the tub with a pitcher. The pitcher, full to the brim, would then be slowly poured into the bucket to avoid applying a significant dynamic load to the tested strand. A full pitcher of water represented slightly more than one liter.

I decided to record the readings with a camera. This provided several advantanges. One, measurement-taking was very uniform. This meant no significant parallax effects, which would not have been the case if I had taken the readings by eye. Two, I could carefully study the photos later to locate the position of the index mark as accurately as I could. This could take several seconds, which I didn't want to waste while the experiment was in progress. Three, it very fast and easy.

I did a full experimental run on two separate strands. The first run was a practice run that allowed me to prove the method was viable and to eliminate a couple of small bugs in the method. The second run was the one from which I recorded the data.

The final setup was this: The empty bucket, which weighs about 2 lbs, was attached to the bottom of the yarn. The distance from the upper anchor to the bail of the bucket was a bit over 3 feet so there would be no interference between the ruler and the system under test. The ruler was also suspended from the same tree limb just behind the yarn anchor. Any sag or movement of the limb, or the whole tree for that matter, would be experienced equally by the yarn and the ruler, and should have no effect on the measurements. A marker pen was used to create an index mark about 30 inches below the upper anchor. Thus it was the stretching of this 30-inch yarn segment that was being measured by the experiment.

Once the empty bucket was suspended from the yarn, it was allowed to hang there for about 10 minutes before I started loading on weight. This allowed the yarn to slowly untwist and equilibrate, and gave me some time to finish preparations. The camera was set up on a tripod about 5 feet away. The tripod was adjusted so the camera lens would be about an inch below the index mark on the yarn. I knew from the first run that the index mark was going to move about 2 inches by the time the bucket was full, and I wanted the camera to be positioned about in the middle of that range. Then the camera zoom was set to give me good readability for the real measurements. With the camera locked in place, I was ready to begin. The photo below is the first photo of the run, taken when the bucket was empty and just before I started adding water. The index mark is visible at the 705-mm position.







It took about 20 seconds to go through one complete cycle, and I tried to keep this as uniform as possible: Scoop a full pitcher of water. Slowly pour it into the bucket over 3 or 4 seconds. Stabilize the bucket with one finger to dampen any swing. Put the pitcher on the ground. Take two steps over to the camera. Half-depress the shutter button and wait for the ready "beep." Take the shot. Return to the pitcher and water tub ready to do it all again.

The last data point was recorded when the bucket was full and on the verge of overflowing. I later measured the weight of an overflowing bucket at about 42 lbs. I guess it really does hold 5 gallons. Just to satisfy my own curiosity I went ahead and loaded the yarn to failure by pressing down on the full bucket with two hands. My subjective impression was that I pressed down with about 15 to 25 pounds. I am very sure I didn't use 40 lbs. I just wanted to get some sense of how far the experiment had pushed the yarn towards failure--probably something like 60 or 70 percent. If there were a compelling reason to know this more accurately, anyone could easily repeat the experiment using a 10-gallon bucket.


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## moray (Sep 9, 2007)

*A Real Experiment: Results*

The photos were studied carefully to extract the readings, which were then entered into Excel. For readings that appeared to fall pretty squarely on a millimeter tick mark, I entered the reading as a whole number, as in 109. If the reading appeared to be about halfway between 109 and 110, I entered it as 109.5. So the resolution of the measurements was .5mm. What does this really mean? It means that if I were shown two photos, and in one of them the index mark was actually precisely .5mm higher than in the other, I could always detect the difference.

The only adjustment to the data was to subtract 705 from every reading. Since the empty bucket was the starting point for the experiment, and I wasn't interested in absolute numbers but in actual stretch, starting from zero at the starting point, after the subtraction step, the numbers along the vertical axis represent mm of stretch. For example, if one of the readings had been 725, it would now show up on the chart as 20, and that would indeed be the amount of stretch from the starting point.

The units along the horizontal axis are units of force--pitchers of water!! You won't find this in the physics books, I know. But we would have exactly the same curve if I had measured the weight of one pitcher to be, say, 2.18281 pounds, and then noted in the chart legend that 1 unit on the horizontal axis equals 2.18281 pounds. Same chart, same data, different legend.

Finally I let Excel draw in the best-fit straight line. This is hardly necessary; as anyone can see, the raw data show the stress-strain curve is almost perfectly linear.


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## Pilsnaman (Sep 10, 2007)

Moray, interesting test and the methods are well thought out. Other then my uncertainty in the use of a single strand vs the original rope configuration I would agree that the relationship is linear. I do not agree that braided line is lower in strength then straight line. It has always been my understanding that braiding rope increases strength over similar diameter mono-filament rope, I will see if this can be verified. Also, it is strange that your rope section didn't have the initial curve but was linear from the start, unlike the data on Samson's website. That does bring up a red flag but what do I know, according to you the people that study a topic are going to get it wrong and don't know what they are talking about. I guess we should ask a journalist to work this problem out for us. In the end a linear relationship is looking likely, I am the kind of person that doesn't believe things until they are truly tested. While I did contact Samson to try and obtain additional information they have not yet replied, not that I am expecting them to. If anyone has a scale that can read forces up to around 1500 lbs let me know so rope in its normal configuration can be tested, figure using a 3 ton floor jack would do the trick for adding weight.
I have enjoyed this bit of conversation and learned some stuff, always interesting, but don't feel much was gained from an arborist standpoint. What do we learn from examining an over simplified problem? Any force we calculate will be totally different with a groundy that lets it run and pieces that swing around instead of moving exactly vertical. In all this did the original question asked in post #1 ever get answered?


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## moray (Sep 10, 2007)

*Another Experiment: Nylon*



Pilsnaman said:


> Moray, interesting test and the methods are well thought out



Thanks.

I decided to do another test, this time on the nylon core from some Arbor-Master rope, left over from a splice I had done. The six yarns that make up the core are much larger than the polyester yarn in the experiment described above, but there are a number of smaller yarns that comprise the big one, and I was able to unwind and peel them out, one at a time.

I ran 3 tests. The small, slippery nylon yarn was much more difficult to handle than the polyester. I had a problem with slippage where I had wrapped it around the anchor limb. It also slipped on the bail of the bucket. By the final run I had solved both problems by creating and testing a loop of yarn. The loop ran through the same screw link from which the metal ruler was suspended, and at the lower end it ran through a keychain carabiner that was clipped to the bucket bail. The loop was formed by placing the ends of the original yarn together and tying a surgeon's knot in both at once. When it was pulled tight, it appeared to be rock solid. 

Water was added in 1/2 liter increments using a glass measuring cup. The last data point was not of my choosing. The yarn held for several seconds, then broke while I was filling the cup for another go. In all three runs I noted the same behavior (and I got soaked every time)--the stress-strain behavior seemed smooth and uniform till the yarn broke without warning.

The results are plotted in the two charts below. As in the previous experiment, the starting point represents the empty bucket, or about 2 lbs.






Without question, the curve is not linear. In the upper figure, Excel has fitted a second degree polynomial (a parabola), and in the lower figure Excel has fitted two straight-line segments.

I show both of these to make the following point: In both figures the mathematical curves are approximations. We don't know what the "true" curve is, and cannot know. We can only take lots and lots of measurements and then try to connect the dots with a smooth curve. This is eminently reasonable and practical, but it is an approximation. For my money, both approximations I show look about equally good in terms of closeness of fit to the data points. That does not mean that both are equally close to describing the underlying physics. There is no way that the true relationship between force and stretch is linear in this case. It is certainly curvilinear and quite well matched with the parabolic curve I show. 

If I want to use these data for the purpose of finding the answer to a stress-strain problem, I am not interested at that point in the underlying physics. I may just want my calculation to be easy to do and to give me an answer that is not far off the mark. For this purpose, using an approximation that does not mirror the underlying physics may work just fine.


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## Pilsnaman (Sep 10, 2007)

Moray, the first graph looks like a better representation may be a parabolic trend line up to about 8 jugs and linear after that. Just an observation.


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## moray (Sep 10, 2007)

Pilsnaman said:


> Moray, the first graph looks like a better representation may be a parabolic trend line up to about 8 jugs and linear after that. Just an observation.



Agreed, that would give an excellent fit.

After I plotted this, it occurred to me that the curvilinear characteristic of the Arbor-Master core might help explain something that I had always wondered about--why make the core out of nylon? One of the reasons I bought a bunch of Arbor-Master for my climbing line was that it seemed to be the only one out there where you didn't have to milk the slack out of the cover a few times before the rope would stabilize. 

The curve in the plot looks a whole lot like the general curve for most of the Samson ropes--relatively soft in the beginning, but much stiffer at higher loads. The core in Arbor-Master is not braided or twisted--only the yarns have a mild twist. The curve I measured should be pretty similar to the curve for the core as a whole, because there is little in the way of constructional slack to complicate the equation. So the cover, which has constructional slack but is woven from a linear fiber, more or less matches the stretch curve for the core, which lacks constructional slack, but is built from non-linear fiber! Probably totally wrong, but maybe Samson will tell us...


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## moray (Sep 11, 2007)

Pilsnaman said:


> I guess we should ask a journalist to work this problem out for us.



Boy oh boy did you ever give me a scare, Pilsnaman! I was just settling down with a nice cup of coffee to enjoy some of your peevish remarks, when this one brought me up short. Holy Sh1t, I thought, I know one! A journalist, that is. 

I was sitting in the local coffee shop a couple of years ago, practicing tying knots with a bit of cord, when she approached me with a huge smile on her face. She said she couldn't help noticing all the knot tying, and just had to ask what it was for. When I explained, she allowed as how she knew a bit about knots, herself. Turns out she was part of the local search and rescue team, specializing in high angle rescue. 

Over the next few weeks we became friends, and she taught me quite a lot of cool stuff about ropes and knots and rigging. The trunk of her car was full of carabiners and slings and quickdraws and belay plates and other stuff I had never seen before or even heard of. Even though I never got to watch her practice rescue rigging scenarios, she came over and climbed a big tree with me one time so she could see first hand what that was all about. 

But what a close call, eh?!! If only your wise observation about the competence of journalists had been available to me a couple of years ago, I could have avoided this dangerous encounter! Whew! I just pray to God I haven't suffered any permanent damage.

But this has set me to thinking. I realize my upbringing has been sorely deficient in preparing me to deal with all the different kinds of people out there, and I need to grow up and learn a real system. The swiftness and certitude with which you can divine that a journalist can't do math just takes my breath away, and I want to see if I can learn your system. But as I am just a total beginner at this, and probably have no natural aptitude at all, I need your help. Could you maybe send me a list (you can PM me if you want) that shows all the categories you use (like homeowner, journalist, etc.) and what each one means? I think it would take me forever to figure this stuff out on my own, so I would much prefer to learn it from a pro like yourself. 

Many thanks,

moray


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## Pilsnaman (Sep 11, 2007)

Don't think you quite got the point there Moray. I was making a sarcastic remark on the following:


moray said:


> To shine a really bright light on the argument from Authority, I refer everyone to the "Monty Hall problem," which is well-described on Wikipedia. This was briefly quite famous when the _Parade_ magazine column ran in 1990, and several hundred mathematicians, and thousands of normal folks, wrote the magazine to claim the published answer was wrong. What the Wikipedia article doesn't say, but I remember reading in some follow-up a few years later, was that some 100 or so of the mathematicians decided to indulge themselves by trotting out the argument from Authority: "I've been teaching college-level probability courses for 40 years and I am appalled that your columnist would actually get this elementary problem wrong." When it finally became clear that the mathematcians were wrong, and the columnist right, I am sorry to report that only 2 or 3 of the hundred who had enjoyed publicly waving their credentials actually had enough character to write back and apologize for their unworthy behavior.


I understand that you are trying to point out that because someone says they are a mathematician, engineer, etc doesn't mean they are always going to be right or knowledgeable on a topic within their field. That being said, would you rather have someone with an english or math degree work a math problem for you? While it would depend on the people involved, generally the mathematician will be a better choice.

Also, the more I think about it the more I don't think testing single fibers or groups of fibers will result in data that represents a complete rope. Between the braiding of a rope and combination of a core and sheath there is no way we can assume the single strands will be good representations. Now I am not saying the data won't be a representation of the rope but there is no data or information to support this assumption.


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## moray (Sep 11, 2007)

Pilsnaman said:


> Also, the more I think about it the more I don't think testing single fibers or groups of fibers will result in data that represents a complete rope. Between the braiding of a rope and combination of a core and sheath there is no way we can assume the single strands will be good representations. Now I am not saying the data won't be a representation of the rope but there is no data or information to support this assumption.



Well put; I totally agree. I wish the manufacturers would give us more data, or that I had some way of doing my own tests.


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## BobEMoto (Sep 11, 2007)

Just when I thought the movie opcorn: was over you guys get all mussy again.

There is more information out there. It's just hard to find. I've haven't kept track of all the links, but from what I recall: rope stretchiness has a little nonlinearity at the beginning indicated by some of the curve matching not hitting zero. This is probably caused by interfiber friction. Then it enters a near-linear region for the safe working load area and a little beyond. Then it starts to stretch less and less up to the breaking point. I would guess this has to do with the molecular characteristics of each fiber and that when it quits stretching, it breaks.

Treating the rope as a spring doesn't really tell us anything new. The force equations give the maximum and the smooth stretchiness is the spring constant in action.

What is more interesting is how to deal with the dynamic loads. I tried approximating the effects in the earlier chart I published. I now think my approximation wasn't very close. The issue I had was how to deal with the kinetic energy gained by the fall. This extra energy equates to an increased weight. The key seems to be that the potential energy initially gained is linear with distance. This is shown in Yale's somewhat cryptic chart:
http://www.yalecordage.com/html/pdf/industrial_marine/low/Pg8.pdf

So the forces previously published show more force then will really be achieved because the extra energy causes the rope to stretch more which causes the stopping force to be lower.


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## Pilsnaman (Sep 12, 2007)

BobEMoto said:


> What is more interesting is how to deal with the dynamic loads. I tried approximating the effects in the earlier chart I published. I now think my approximation wasn't very close. The issue I had was how to deal with the kinetic energy gained by the fall. This extra energy equates to an increased weight. The key seems to be that the potential energy initially gained is linear with distance. This is shown in Yale's somewhat cryptic chart:
> http://www.yalecordage.com/html/pdf/industrial_marine/low/Pg8.pdf



If you need help with this just let me know what the weight is and length of fall. I can calculate the resulting velocity and post how the work was done. Also the energies can be calculated if you would like that too.


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## moray (Sep 12, 2007)

BobEMoto said:


> Just when I thought the movie opcorn: was over you guys get all mussy again.



Yeah, sorry about that... Maybe we can get the Oscar in the Public Brawling category.


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## Pilsnaman (Sep 12, 2007)

moray said:


> Yeah, sorry about that... Maybe we can get the Oscar in the Public Brawling category.



I would like to thank my wife and parents for all the practice throughout my life, I couldn't have done it without them.


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## moray (Sep 14, 2007)

BobEMoto said:


> This is shown in Yale's somewhat cryptic chart:
> http://www.yalecordage.com/html/pdf/industrial_marine/low/Pg8.pdf



What a great link--the Yale site is a gold mine! You can easily spend an hour or two with the tables and charts, the engineering descriptions, the discussion of working loads, and so on. I especially liked the charts in the Industrial section (why can't Samson do this?).

Comparing the charts for a number of different ropes several things become apparent. For one, the ropes are quite different. The stiffer the rope, in general, the less energy storage capacity it has. The range of values in energy storage per lb. of rope was more than 5:1. The static ropes arborists like to use for rigging are near the bad end of the scale in this regard. (Even though there is only one chart for each rope type, and energy storage per lb. is indicated on the chart, Yale notes that this quantity varies by size of rope for certain specific ropes.)

The working load for the rope, which Yale discusses at some length, also varies from rope to rope, but the energy storage at the working load is always just a small fraction of the rope's maximum.

The shape of the stress-strain curve varies a bit from rope to rope, from a nearly perfect straight line to somewhat curvy. Since the working load region of the curves is so short, in nearly every case a straight-line approximation for that section of the curve should work fine for rough calculation purposes.

In the second Engineering Index page on the site Yale goes through the dynamic load calculations for a scenario just like the one we have been discussing in this thread. It is based on the straightforward energy absorption method, but things are simplified a bit by the use of a quantity derived from the charts--energy absorption per lb. It occurred to me that for someone who uses the same rigging rope over and over again, this could be further simplified by converting this to energy absorption per foot of rope. I decided to do this for the rigging rope I have been using, 1/2 inch Samson Stable Braid.


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## moray (Sep 14, 2007)

*Stable Braid*

The following calculation applies to 1/2 inch Samson Stable Braid, a low-stretch double-braided polyester rope meant for rigging operations. The specs and construction of the rope seem quite similar to Yale's Double Esterlon, though the latter rope seems quite a bit stretchier.

The chart below shows the data points published by Samson; the trend line is Excel's best-fit straight line. If you ask Excel to fit a second-degree polynomial (a parabolic curve) instead, it looks almost exactly like the straight line--for much of the chart you can hardly tell there is more than one line. Why make life complicated? Choose the straight line.






The equation of the line is shown. You can actually figure a lot of things out using the chart directly, but it is easier to do the calculations numerically. Without the chart, though, it would be really really hard to understand this stuff.

The number I want to find is energy absorbed per foot of rope at the working load limit (WLL). Working load is a more or less arbitrary fraction of breaking load; For illustration purposes I chose 25%, which is the number used by Yale for Double Esterlon. If you choose a different number, obviously you will get a different value for absorbed energy.

The calculation is actually quite easy. The published break strength of Stable Braid is 10,400 lbs. Then the WLL is 1/4 of that, or 2600 lbs. The arrow on the chart points to the fake data point I installed at 2600 lbs. From the equation, or just by reading the chart, 2600 lbs. corresponds to a stretch of 2.25%. In a 100-ft piece of rope, that would be 2.25 feet. Just imagine the units along the X axis are feet instead of per cent; in this case the chart shows the stretch characteristic of 100 feet of 1/2 inch rope. The energy needed to stretch the rope 2.25 feet when the force increases linearly from 0 to 2600 lbs is 2927 ft. lbs. This is just the shaded area of the chart. The shaded area is the area of the triangle, which is given by the height times the base divided by two. If 100 feet of rope absorbs 2927 ft. lbs. of energy at the WLL, then 1 foot absorbs 1/100 as much, or 29 ft. lbs. This is the magic number I want to remember. (If you use Samson's recommendation of 20% break strength for WLL, the magic number is about 18.5 ft. lbs. per foot of rope.)

What is so cool about this number is that it allows you to solve fairly tough dynamic loading problems, the ones where you drop a load and the rope alone stops it, without any need for F = M x A, or velocity, or mass, or the gravitational constant. The three things needed for the calculation can be estimated by eye: how heavy is the load, how far will it drop before the rope catches it, and how much rope will do the catching.

The energy the rope must absorb is just the product of the load weight times the distance it drops, i.e., E = F x D. The length of rope doing the catching is the length of the rope from the load to the anchor. (If the rope passes over a pulley, we still measure all the way to the anchor. If there is a friction device in the line, or the load is allowed to run, then this calculation will not apply.) Now divide the energy by the rope length to get energy per foot. If it is less than 29, you have not exceeded the WLL; if it is more than 29 you have.

It is shockingly easy to exceed the WLL. Tie some Stable Braid to a low limb. A couple of feet below the limb, tie on a 30-lb barbell. Lift the weight up to the limb and drop it. You have just applied more than 2600 lbs of force to the rope and stressed it beyond the WLL.

The math is really easy. You are about to drop 100 lbs 5 feet. But the rope from the anchor through the block and down to the load will be 25 feet long. The 500 ft lbs of energy will be divided by 25 ft of rope, or 20 ft lbs per foot. This is well below the WLL at 29 ft lbs per foot of rope.


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## moray (Sep 14, 2007)

*Stable Braid--a real event*

This is the embarrassing report of the first piece I ever dropped on a rope, and it shows the folly of not knowing what you are doing. It was an experiment inasmuch as it wasn't necessary at all, but I wanted to see what it would be like, and it seemed like a good piece to practice on.

The piece was a 6-foot long vertical piece of a recently dead elm, and I guess it weighed about 60 lbs, more or less. The rope (1/2 in. Stable Braid) was hitched to the wood about 1 foot above the cut. A block was fastened about 6 in. below the cut. I figure the center of mass of the spar was about 3 feet above the cut as it was a bit bottom heavy. By the time the rope caught the falling wood, the center of mass had fallen about 7 feet. This amounts to 420 ft. lbs of energy the rope must absorb.

The length of rope catching the load was the 1.5 feet from wood to block, and roughly another 4 feet to where the rope took half a wrap around a limb, or 6.5 feet in all. Using the Samson recommended 20% maximum load for the WLL, we have, from the previous post, a figure of 18.5 ft. lbs. of energy absorbed per foot of rope, or 120.25. So the load actually delivered 420 ft. lbs. of energy, but only 120 was permitted. Oops. Note this does not mean the maximum force experienced by the rope was nearly 4 times the WLL. It would actually have been the square root of 420/120 times WLL, or a bit less than twice the WLL.

This was bad and I won't do it again. What about the end of the rope, you ask? I had a ground person holding the end who had never done this before, but had been instructed to let it run a bit. Assuming it did run at least a few inches, and accounting for a little slack in the sling holding the block, and some slack and slip where the rope was hitched to the falling wood, it seems likely that the rope got a significant amount of help from all these other elements and actually experienced somewhat less maximum force than the worst case I calculated. Nevertheless, a very simple calculation would have told me this was going to overload the rope.

Since the cut was about 25 feet above the ground, the same calculation would have shown me that the exact same setup described above, but with the rope anchored to the base of the tree, would have given me about 27 feet of rope to stretch. At 18.5 ft. lbs. per foot, this could have absorbed about 500 ft. lbs. of energy, safely within the WLL.


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## oldugly (Sep 14, 2007)

Calculating the physics of rigging is essential in some ways, very dangerous in others. 
Noting the shock load and figuring out the physics to reduce it is, slow or stop the decent of falling wood, brush etc. controlling the landing zone, to eliminate damage, while safely securing the load away from the climber..etc.

These are essential in rigging, but coming up with a formula to push limits is the essential recipe for disaster. Your variables in rigging have very little to do with the rope strength...(any embicle can figure out when a load is too heavy for a rope)..your variables are the tree itself, and the working load and strength of your rigging point/s.

Ways of reducing stress on your rigging points, ways of reducing shock load on your spar, and different methods of strengthening your rigging points.ways of directing your load etc. are very helpful.

Formulas that allow you a false sense of security, and a major potential for injury...very damming, and you are playing with fire. I would much rather trust what I know is overkill, than to calculate what I do not know, and try to base the limits of my work on variables of which I am unclear.

I have very seldom seen men that consider themselves unlearned, or underinformed prior to a big decision, make dumb choices. Its almost always the one that thinks he has it all figured out. (Myself included).


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## TheTreeSpyder (Sep 17, 2007)

Wow, quite a th-read!

While O.Ugly's post might carry some wait hear on the tests and theories; let's not miss the point.

Maximizing safety can sometimes only be done by examinations of pushing the limits to find where they lie, curve and are minimized too. So, even when we study where something breaks; the real study is how not to reach that point after defining it. Also, what variables have different effects, in what directions; and at what times altering said variables has maximum or conversely marginal effects for the expense of your time and efforts; also; when they are actually needed or not.

In the field; we can't stop and calculate everything; but in playing with these things just for observations, we can find and define the patterns of loading and changes; that then we can take the patterns and not the precise math into actual field work.

i see in Moray's test he used fairly inelastic StableBraid; at a very high tensile for the loading; over a very short piece of line before frictions. All 3 factors construction, loading to tensile and short length of line offer very little deformation response. The inversion doubles the fall impact. In this case, a less strong rope; even of the same construction would have given less loading because of more elastic absorption/ dampening of force. Most folks try to place the first catch/ Half Hitch just over the cut; though it might seem counter intuitive; placing this higher/farther from cut is a way to sneak more elastic line length into the system(but all catches/ Half Hitches and Running Bowls should be before the CG). i also like a separate tag line tied to the top of the load, to flex the load farther over on the hinge; to start it's drop later; or perhaps sometimes get total inversion before tearoff; from being 'muscled over'; but this is hard to due with anything of any size. And unless the rigging line was pre-stretched or is slack is being taken up along the way; can still give some impact force. If the line is pretightend well; and the first hitch higher; this can also give a stronger hinge. The more vertical a CG is from the compressed part/pivot of the hinge; the less it forces it's own hinge strong! A stronger hinge with no dutching, can give a much softer catch impact by the line. 

Just as fall length gives distance to build force; deformations (elastic and non) give the inverse; distance to dissipate said dynamic forces. Sometimes on short lengths, we have placed a pulley block at base instead of a Porty, then stretched the redirected line across the yard to the Porty. Thus, more elastic line length in the formulae to absorb forces. This can also help to facillitate more tightening, by providing a very good sweating/ swigging in point; and even creatively a http://www.mytreelessons.com/farmer%20snob.jpg type strategy; once again using the less thought of science of perpendicular leverage force on a line. The tighter the line is; the more it resists bending, the more multiplier it presents to the application of perpendicular force to a straight line.

Another factor of deformations; is that of the wooden supports itself; once again whether permanent or elastic deformation; it all absorbs forces/ releases peak steam from the built up pressures. So, what works with one setup; may not work in the next tree, even if the setup is exactly the same!

Another counter-intuitive point; is that though a pulley redirect will give 2:1 on the support; and making it a redirect + 2:1 support on the load will give only 1.5:1 on support (and same as a 3:1 on load + redirect will give 1.3333:1 on support etc.); that is just statically. But, dynamically; especially under short lengths (which at the initial catch/top is shorter length and of peak force); the forces of the more legs to load are increased on the support, not decreased. This is due to less loading per line strength/per leg, once again receiving less elastic response/ dampening. So; in dynamic loading; less tensile can be better; in that it can be deformed more; so absorbs more impacting from all connected parts (knots, supports etc.); but at a cost of cycles to failure. So, the 'macho' view; of the more strength the better can fall short; in neglecting the 'softer' side of rope dynamics; it's elastic response, the wide band between pure static hold and pure failure.

Thanks for a good read all!


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## cityboy (Sep 20, 2007)

ddhlakebound said:


> What formula or computations are necessary to go from knowing how much an object weighs and how far it falls, to knowing how many ft/lbs of force are generated in the fall?



Hi there, found this thread doing a search on an unrelated topic, thought I'd try to help out a bit. Don't know anything about tree trimming, but can help a bit with the physics part.

Weight is simply a force. A force in physics is equal to mass*acceleration. Gravity causes acceleration, at a rate of about 32 ft/sec^2. This means anything falling for one second (from rest) with have a velocity of 32 ft/sec, after two seconds 64 ft/sec, and so on. This of course ignores air resistance and any other forces that would reduce the acceleration, so in real world applications its not always completely accurate unless you understand the exact nature of the opposing forces.

The part that isn't all that intuitive is (in a vacuum) no additional force is required to maintain a given velocity once something has been accelerated to that velocity. No force is required unless something is accelerated (or decelerated).

So the force you're trying to calculate has to do with how quickly the falling object is decelerated to a stop. Let's say the falling log gets to 64 ft/sec velocity before you try to stop it with the rope. If you stop it in two seconds, you would have applied the same force as gravity (but in the opposite direction) to stop it, or 1G. During that time the force on the rope was twice the normal weight of the object, because you will always have to overcome the 1G force of gravity. If the log is being held stationary in the air by a rope, with no acceleration, the log will always have 1G of gravitational force on it, which is counteracted by 1G of force on the rope. So whatever the deceleration force is, add 1G to that to get the force on the rope. So, if you stop it in 1 second, that would require 2Gs of deceleration force, so the rope sees 3Gs, or 3 times the normal weight.

In practice its going to be difficult to plan exactly how far the object falls before you halt it with the rope, and exactly how much force you apply on the rope, and for how long, and how the force on the rope might vary over time, rope stretch, etc. I think the best advice that was given is to not try to calculate the force precisely, because you will likely be wrong given the lack of control over the situation to reproduce the exact conditions used in the calculation.

If you're interested in a bit more math than I've given here, PM me and I can go into more detail if you like.


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## TheTreeSpyder (Sep 21, 2007)

For those wondering about the Rigging Software by Sherrill and ArborMaster; i have dug up some pics.

Screen shots of Rigging Program

Some Outputs of the Rigging Software with manipulations of key variables

Also, i found these in my stash; that someone posted earlier:


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## moray (Sep 21, 2007)

TheTreeSpyder said:


> For those wondering about the Rigging Software by Sherrill and ArborMaster; i have dug up some pics.



Spydie, I really like your old tire shock absorber--very clever. 

I assume you own a copy of the rigging software. You could run an interesting experiment. By choosing your input variables carefully, you could run several simulations for a single type of rope, say 1/2 in Stable Braid, and determine whether the software assumes the stress-strain relationship is linear or not. I would lay odds that it does assume linearity.


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## oldugly (Sep 21, 2007)

While not as scientific as the previous posts I have learned and presently use some methods to lessen the load on the rigging point.
By taking one natural friction wrap of the lowering rope around the trunk before the rigging point, I can reduce the force on the block, crotch, etc. by spreading the dynamic of the opposing force and lessening the 1:2 ratio to approx. 1:1.5. 
By my groundsperson using a soft-stop I can reduce the impact energy to near nil. Would I trust my life in a situation that would require my predictions to be perfect no. 
Using a redirect before my rigging point can also reduce the load on the point of contact...but again, maximizing the load you can lower on any point requires the control of all variables. The variables of the internal dynamics of the tree itself are sometimes completely unknown to you, and therefore beyond any calculation or control.
Regardless of how you calculate, redirect, or reduce load, rigging limbs, and false crotching spars are some of the most dangerous aspects of our game...and should never be pushed to its limits, or approached in a manner of production over safety, rather it should be approached in a reverse order of safety over production.


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