Trees and Physics

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.

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.

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.

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 .

What lengths of rope were used in the examples?

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 .

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.

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.

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.

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?

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.

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

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.

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 .

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.

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 ) .

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 )


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 .

Click to expand...

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.

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 .

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 .

Click to expand...

Is the dissipation of kinetic energy before impact not work?

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 .