# oh for pete's sake



## ramanujan (Jan 16, 2003)

So my boss bought me a new ms200t two days ago. yesterday i was pruning a large fir, working near the top just with a hand saw, 200t on my belt. rappel down a few feet and the saw sits on a limb beside me and i gently pull it off, (being very nice with this new saw you know) and drop it down to my side, expecting it to fall the usual 2". nope. 60' down, never hits a limb even, right onto a large large rock. arghh. ????ed screwgate biners i'll never use another.


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## tshanefreeman (Jan 16, 2003)

About a year ago, I was just about to start my way up a tree to do a deadwooding job. Just before I was to leave the ground I realized that the delta link (pictured below for those that are confused) which attaches my Grillion of my lanyard setup to my belt had come open. Made me wonder how long it had been like this......... but for the sake of my nerves, I figure it must have rattled open in the truck on the way to the job. Ya right! Needless to say, my lanyard is now attached with a Petzl Ball Lock 'biener.............hopefully that sucker won't come open.


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## Kevin (Jan 16, 2003)

My buckingham lanyard has two rings.


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## Ryan Willock (Jan 16, 2003)

Kevin, thats the same type of laynard that i use and so far i have been very happy with it


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## ramanujan (Jan 16, 2003)

don't those lanyards have a warning on them "do not use with a saw that weighs more than 15lbs" or something like that? i would be concerned that it would break in a situation where you would not want it to.


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## Kevin (Jan 16, 2003)

Yes, that's true. 
15lbs. max.
That`s plenty for any top handled saw, it's rated for 200lb. breaking strength.


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## TREETX (Jan 16, 2003)

My question is can you cut and then chunck the saw without worrying about it breaking??


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## treeclimber165 (Jan 16, 2003)

Tex,
If the saw lanyard breaking strength is 200 lbs and the saw weighs 8 lbs, that is better than 20-1 margin. 

My current Echo has survived 2-3 falls in the last 3.5 years. It's still running strong. I do not like the long saw lanyards, they always end up inbetween the bar and the log. I use a short lanyard with ring girth hitched on the handle. Unhook to use, then hook it back up. If it gets jerked out of my hand it won't pull me off a limb with it. Saws are cheap, my body is expensive. 
If my saw gets caught, I can secure it to a limb with a loop runner or lanyard so I can use both hands to free it without worrying about it dropping.


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## Kevin (Jan 16, 2003)

I haven't experienced any problem with it and I`ve never heard of anyone else having a problem.

What force does a 7lb. saw being dropped 50" develop?


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## treeclimber165 (Jan 16, 2003)

> _Originally posted by Kevin _
> *
> What force does a 7lb. saw being dropped 50" develop? *


I do not know the exact formula, but for a general rule of thumb I've assumed a load's force could double for every foot of freefall. 

7-8 lb dropping 4 feet could be as much as 120 lbs impact force. But you would subtract from that since the saw would not freefall the entire distance. The lanyard is stretchy and would slow the saw for most of the fall. Depending on the elasticity of your lanyard, it could be as little as 30-40 lbs force. Still more than I want jerking on my saddle.


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## monkeypuzzle (Jan 16, 2003)

I took my first chainsaw cut because of a lanyard, I think. Tension build-up in three 12'' to 16'' dbh wind blown sweetgums. I was tied in to a tall pine a few steps away,and working from the tip down,no big deal.


When I found the magic spot they exploded and it was every man for himself.If I could have abandoned the saw she would have hit the ground and not the top of my thigh.

I only use the saw "accessory'' strap now. 


The swing from a chainsaw connected by a lanyard keeps me a little off balance, banging around below.

If I need a lanyard,i'll throw a loop runner on too.


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## TREETX (Jan 17, 2003)

My lanyard never gets in the way of my cutting. It is just a section of rope with a snap and a twist-lock at the other end.

Holds my 020 great. 

I like having it so I can cut and chunk pieces of spar, etc. If not chuncking, pushing.

I can cut, flip the saw out of my hand, grab the leash(lanyard) to cushion the fall and push. I know that took a while to say but it only takes a half second to do one handed.

I have had some terrible saw snatchers with both the 020 and the echo. Both on pieces of spar. Jumping things that I had no business jumping. For me, the ring popped off the saws, but not before making me feel drawn and quartered.

Now, I have noticed that the snap comes of the saw if I am cutting big wood. That way, it is the best of both, free saw and attached saw.

I have noticed there are leash people and variations of clip people. I have seen the leashless drop saws on many occasions.

165 - many of the guys here use a "U" bolt on the echo handle and slide that over a hook on their side. A good one handed setup.

No right or wrong, just my .02

Nathan


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## kf_tree (Jan 17, 2003)

any improved saftey stats about saw leashes? i've never used one with the feeling that, i'd rather let the go than have it tied to me if some thing happened. why do you guys choose to use them?


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## John Paul Sanborn (Jan 18, 2003)

The lanyard must allow the saw to hang below the climbers feet per ANSI (FWIW).

I have it so i can easily discard it when working, iethe the panic throw or just a quick release whwn chunking or such.

I find it much mor convieniant when making multiple cuts to let it idle on the strap when hanging. Especialy when freefalling a lot of brush. make a few cuts, reposition, make a few cuts, turn it off hook it and climb.

Have you looked at Greame's vids where the saw is flopping around on his lanyard when he jumps off the spar onto the line set in an adjacent tree? Now that is too much!


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## TREETX (Jan 18, 2003)

Simply put, after making a cut and letting off the throttle, I can do other things like chunck wood or just move to the next limb. I don't have to re-hook my saw.

double for every foot of free fall?? No way. 10lb saw

Foot 1 = 2*10lb = 20lb

Foot 2= 2*20lb=40lb

Foot 3= 2*40lb= 80lb

Foot 4= 2*80lb = 160lb

Foot 5= 2*160lb = 320lb

Foot 6= 2*320lb = 640lb

I think that is what you meant.

I dropped my 020 when pulling up my climbing line - it fell about 25 feet before hitting the end of my line - it didn't tear me in half and tear the tree down.

I understand what you are getting at but can't answer it myself.

Terminal velocity should play a role.

Please excuse any excess sarcasm

Nathan


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## Jumper (Jan 18, 2003)

*Screw Links*

Looking at Shane's attachment of the Delta Link brought back a VERY unpleasant memory of 1989- we use similar screw gate links on parachutes to connect the chute's numerous suspension lines to the canopy risers. I had just finished a jump and went to flake out the canopy when all the lines (similar to 550 cord) flew off one of the 4 screw gated links, which had opened sometime after I touched down. I had a long heart to heart with the Team Rigger who had had the whole thing apart a few days earlier for some routine maint and then never secured the gate. I shudder to think what would have happened if all those lines had parted company with the link (and riser) close to the ground, little was holding them on except tension. Since then I do not trust the d*mn things, even though this was a situation due to human error. Lock biners seem the way to go here... but we still use the screw gated connector links on chutes, albeit covered with a clear plastic protector.


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## Dan Flinn (Jan 18, 2003)

I used to use the hook, climb, unhook, cut, hook method until one day, I was is a good rhythm. Then my ground guy asked me a question in between the cut and hook and $210 later I am using my breakaway lanyard again.....

Dan


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## The Climber (Jan 19, 2003)

I have one of those lanyards and I forget, How much pull is it suposed to take for the break-away feature to go


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## Dan Flinn (Jan 19, 2003)

Mine is supposed to break away at 200 lbs.

Dan


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## TheTreeSpyder (Jan 19, 2003)

i believe at short drops the rule of thumb is :
[({Drop in feet}+1) x {weight in pounds} = 'Z' Foot/Pounds of force]. Going beyond 8' or so things getting more complicated, speed becoming more of a component to increase the force more dramatically, and with higher math. So 10', 15' you can know it is gaining quickly towards failure for anything of real mass on a 6k line. 

So that at zero drop, 10# has the force of it's own weight [({0' Drop}+1) x {10#}=10'# of force.). Whereby, you could slide it over a 9# sensitivity switch and it would trigger it. And the further you drop it, the more force it will have; but it will always weigh the same, strictly speaking "pounds" only refers to weight, not force; kilonewtons and foot pounds refer to force, taking motion into consideration.

If you drop 10#, 1 foot, it will have ~20foot pounds of force [({1'drop} + 1) x {10#} = 20'# of force]. 4' drop with 10# would be ~50'#s <read : approximately 50 foot pounds (of force)>. 

It is similar to: if you drop that force in a 1 square foot area = 50 '#s of force per square foot there. Drop that into 2 square feet landing flatly, then 25'#s of force/square foot etc. In both examples there is a finite calculable amount, accounting for distance and weight, nothing coming about, or disapearing magically. So that a 500# log that drops 5' has ~3000'#s [(5+1) x 500] on a line (probably dropping even further as the line stretches, adding to things!!!!). Now, if that is in a simple pulley redirect, that ~3000'#s of force; powers a 2/1 machine on the redirect/support system (pulley redirects braking force to above load), so the pulley, support/anchor and attatchments take a 6000'#s hit, as the line and friction device take a 3000'# hit! Now that is only realized if you offer that much resistance in opposition (friction) so that running it, would not invoke that much force, only as much as the total 'drag' on the line called out from it.

The Law of Energy Conservation commands that these numbers accumulate like this, for the same reasons as in Mechanical Advantage levers, pulleys, ramps etc. That all that energy x time, distance must be accounted for, arcing a prybar 30"s to move the other end 1" lift gives ~30/1 leverage, a pulley system that you pull 5x as much line (with 5 pulls on load), yields ~5/1 power, a 4'drop increase force adding about 4xload before speed (+ its own force of just being a certain mass sitting still) must be realized strongly. All exactly the same but diffrent, ruled by the same law. That is so pervasive that it demands that all of the archived calories, pressure etc. of fossil fuels, create this dynamic explosive force, for that energy of calories didn't just disapear, energy changes over to diffrent forms, it doesn't appear and disapear. Even 'fertilizer' is an encapsulate of energy, and 'passes it ' to new growth, it all must be accountable, and in balance, for even E=MC squared, for everything is in this balance. 

Breakaway lanyard can have 2 systems i think; 1 to deacellerate by having so much give through stretch to buffer the force, also stitches that shear open disapating whatever force that takes. System 2 would be to be a mechanical fuse that would shear/ seperate at so much force; so that you would only incur the weight of a saw drop on your bod, not one caught in a top (then it would shear or 'blow the fuse"). So of the total force of pull, you only oppose 200'#s; so thereby only incur that, and it is trying to deaccelerate it as it goes into that.


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## Newfie (Jan 19, 2003)

". I shudder to think what would have happened if all those lines had parted company with the link (and riser) close to the ground, little was holding them on except tension."

Sounds like a close call Mitch. Guess we would be calling you Bouncer instead of Jumper?


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## treeclimber165 (Jan 19, 2003)

Nah, we would probably be calling him 'Spot'.


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## John Paul Sanborn (Jan 19, 2003)

I think velosity is distance is multiplied by 3 with each incremtn of time something like V=(d3*t)t, but that does not look right.

One of Galileo's laws, he used an incline plane, marbles and his pulse to time it all.

So the saw fall 1 ft in 1 second 3 feet in the second, in the third second it falls 9 in the third...

What all this has to do with impact force id beyond me though


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## Joe (Jan 19, 2003)

I'm sorry you guys can't handle a simple energy equation. I really thought I put enough understandable information out here(in cyber space) in the past where you guys could at least calculate the energy in a lanyard from a falling saw. I'm speaking of you guys who've posted with me over the years.

You know, I needed to be corrected about the terminology used in this circumstance. The great thing about it was I could easily look it up to confirm it was correct. Hell, Ken posted the simplest link I've come across at this site some time ago. I guess nobody cares to read something as important as calculating energy from a freely falling object. The rope manufacturers had to do this to determine the strength of the ropes used in this business. Just think what ropes would be like if these people didn't know this stuff. They just made ropes and somebody else gave them a machine which broke ropes they made and said to themselves," that seems to be strong enough". It doesn't work this way guys. 

Here you go guys, this will be the energy from a saw which falls freely. 

A saw which weighs 7 lbs held in your hand is 7 lbs. Since there is no distance for which the saw is falling, multiply the weight of the saw times 0. The saw still weighs 7 lbs but because there is no distance of fall, there is no kinetic energy in the system just potential energy or energy that can be spent; like a full tank of gasoline. So 7 lbs times 0 distance is 0 energy. Understand? Your gasoline tank is full. 

O.K., now the saw is dropped and falls 1 foot. Multiply the distance of fall by the weight. This gives 7 ft›lbs of energy. It is kinetic energy, energy spent.

7 lbs × 2 ft = 14 ft›lb

7 lbs × 3 ft = 21 ft›lb

7 lbs × 4 ft = 28 ft›lb

etc., etc, etc. 

This is like the gasoline used-kinetic energy. The gas tank is empty when it can no longer fall. Notice this is considered energy and not force. 

Joe :crying:

????, you guys are so bad at this you made me cry. 

Edited many times because I have that option open to me at any time.


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## treeclimber165 (Jan 19, 2003)

Well, Thank God we have Your Royal Highness to straighten us poor saps out on this matter. Too bad your attitude is so condescending, you actually have some knowledge that more might actually want to hear if you weren't so full of yourself.


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## Kevin (Jan 19, 2003)

_the saw is dropped and falls 1 foot. Multiply the distance of fall by the weight. This gives 7 ft›lbs of energy. It is kinetic energy, energy spent._

7 lbs × 2 ft = 14 ft›lb

7 lbs × 3 ft = 21 ft›lb

7 lbs × 4 ft = 28 ft›lb

*This is all we really needed, thanks.*


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## Joe (Jan 19, 2003)

> _Originally posted by treeclimber165 _
> *Well, Thank God we have Your Royal Highness to straighten us poor saps out on this matter.*



Those who consider me royalty and themselves saps are welcome for the information too.



> *Too bad your attitude is so condescending, you actually have some knowledge that more might actually want to hear if you weren't so full of yourself.  *



We all hear things from people we don't want to hear, and hear from people we don't want to hear from. There are 2 people who posted to this thread I really thought knew what I posted. There is at least 1, maybe 2 other poster besides the other 2 I know who knows it, not the plane jumper, who are keeping their mouths shut. 

I'm betting after reading the post it is clear without question how to calculate energy in a system as simple as being described. That was 1 of my goals.

Joe


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## The Climber (Jan 19, 2003)

I thought the condecention was good. There are a few know it all M.F.'s here who need to take some now and then.:blob5:


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## Stumper (Jan 19, 2003)

As Spydy pointed out in his post, ft.pounds as weight times fall distance in feet is only close to accurate on short drops. Joe (or should I say "Your Royal Snottiness") is close in his figures but only the 7lbs x 1 ft = 7ft.lbs is actually correct. Accelleration due to gravity (32feet per second per second) is adding up the further the fall. On the short fall the time is so short that significant speed hasn't developed and it doesn't skew the figures too far to use the shorthand.


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## Joe (Jan 19, 2003)

> _Originally posted by Stumper _
> * is close in his figures but only the 7lbs x 1 ft = 7ft.lbs is actually correct. Accelleration due to gravity (32feet per second per second) is adding up the further the fall. On the short fall the time is so short that significant speed hasn't developed and it doesn't skew the figures too far to use the shorthand. *



O.K. Still don't believe me?

http://theory.uwinnipeg.ca/mod_tech/node32.html

Joe (or should I say "Your Royal Snottiness")


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## John Paul Sanborn (Jan 19, 2003)

The reason Joe is crying is because he is pulling out what little hair he has left, and the smoke form his ashtray is getting in his eyes.

 

I freely admit to being mathmaticaly challanged. It takes me a while, with regular use, to to absorb and remember formulae.

But then I'm an arborist so the symantics of force and energy is just a parsing argument to me. They are interchangabel to me.

But I still cannot get over velocity not being a factor in the impact force, but I'll take Joe's word on it. most of the falls I would need to estimate are short fall catches blocking out wood.


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## Gypo Logger (Jan 19, 2003)

Thanks JPS: the provided link says it all. 

BTW: velocity and acceleration are there, they just aren't needed for this specific calculation.

Joe


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## TheTreeSpyder (Jan 19, 2003)

In "Arborist Equipment" ( bottom of page 97 here) Don Blair uses the example about shock loading that reads:

"A rough rule of thumb that probably does more good than harm is: For every foot a falling object falls, it gains a unit of weight pllus one.

Example: 500 pounds falling four feet will hit the rigging at about 2,500 pounds."

Then he goes on about how that is a rule of thumb for a few feet of drop, due to exponential increases in force due to the speed component increasing so.

i think that means the 500# load has it's own force of 500# sitting on your chest with no drop, dropping it would increase that. So (y1-y2) at a drop of 1 foot the gravitational potential energy is 500# as stated in Joe's example, but that doesn't take in consideration the force of that 500# sitting still. So, there fore with no drop 0 x 500#=0# on your chest invalid.

Therefore, in my example above i tried to show that a 10#weight, with no drop, could trigger a 9#sensitivity switch positively. It also would take 10#of force to lift it, for it has the potential to resist just that much force. It also has enough force to lift/maintain 10# in a seesaw on the opposing seat. etc.

In rigging, i take it a step further and say well if we were doing that in pounds per inch.............. and minimize all drops and pretigheten everything, using the hinge to hand off gradually as possible rather than shock load.


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## Joe (Jan 19, 2003)

I posted a web link which simply repeats what was written. It's all accounted for, acceleration, mass, distance. Don Blair was trying to simplify it in a way he felt others would understand it. As another poster pointed out all that was needed was an example. It's easy stuff if one doesn't try to think beyond it. Reread the beginning of the example I provided, it does account for the weight of the object sitting on one's chest. Read the link too. The guys who write the information in these links are the guys we all learn this stuff from including Don Blair. Don's example isn't reaching his audience in a way he intended, the web link I posted should do this.

Joe


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## Stumper (Jan 19, 2003)

I stand corrected. A little knowledge is a dangerous thing! Knowing that velocity is squared in kinetic energy equations caused me to wrongly conclude that as distance/time increased that energy gains would become greater. I forgot that the accelleration of gravity is a constant. I had to go put myself through the mathematical gymnastics to get things straight in my own mind.. 
Technically it is true that impact force will be less than the potential energy figure-due to atmospheric resistance using up some- but those figures ar TINY in our short/slow falls. 

Sorry!


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## Joe (Jan 20, 2003)

> _Originally posted by Stumper _
> * Knowing that velocity is squared in kinetic energy equations caused me to wrongly conclude that as distance/time increased that energy gains would become greater.*



Well, 1/2 × (m v^2) is an equation to compute kinetic energy. See physics link in a later post. Also, see the part about burning other posters with it. It appears I burned myself. Therefore, this correction.



> * Technically it is true that impact force will be less than the potential energy figure-due to atmospheric resistance using up some- but those figures ar TINY in our short/slow falls.*



The 1st is the concept of impulse, the 2nd is drag. There's no relationship between the 2 in the way you're stating it as I understand it.

Joe


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## TheTreeSpyder (Jan 20, 2003)

Mr. Joe,

You have cleared many queries before, you have previously cleared this for me; for Don Blair is not the only quotable source simplifying approximations with that rule of thumb for drops of short distances. Indeed, i see in all the physics pages we've shared that : Force = Mass x Speed; and that speed accelerates in falling.

So i had to wonder about the justification and misleading expression of the rule of thumb of: [({Drop in feet}+1) x {weight in pounds} = 'Z' Foot/Pounds of force]. i thought of it as in safety factors etc. of the extra unit of weight to cover the building force from speed accelerating etc. At some point i think i wanted to cast more fault to that formulae than due.

But kinda in a barefoot, uneducated, working physics i came to see more gift and sense in it. For i understand, that as an arguement insamuch that i see force pressing on my chest with 500#, you say that there can't be force till it moves for Mass x Velocity, so if either componenent (velocity here) is zero the force is zero. That is true, it won't crush me till it moves; but............. it will move,therefore it will crush me! If you shoot me with a 5# ball from the side at the same speed as it would have if dropped 4'; it will have by definition have the same force as if you dropped it on to me from 4'; but after taking that punch i would rather it not be on top of me! If a saw drops 4' and stops it has {~4 x weight} of force, if removed from the air at that point, like the classroom example. But in the tree, i can't isolate that force, and take that saw out of the air; it's inertia would carry to the next step and force. So i felt less fault with the rule of thumb as Blair stated.




> _Originally posted by Stumper _
> *On the short fall the time is so short that significant speed hasn't developed and it doesn't skew the figures too far to use the shorthand. *



With a mixture of this fact as expressed, extra 'slop' going to more forgiveness (SWL), and inertia carrying a load beyond the theoretical measurement of its isolated force if it doesn't move; i find plenty of room for exercise of this understanding, through application of "Blair's" rule of thumb in the real world.

i wonder if Gypo will ever give out his password again.


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## Stumper (Jan 20, 2003)

Spydy, My "fact" while sounding right is, in fact, in error. The simple reason is the conservation of energy principle that Joe led us to with his link. That seemed too simple so I had to work some equations to satisfy myself. The key is that gravitational acceleration is a constant. The speed is constantly building as an object falls but the time it has to fall is constantly growing shorter. --An object falls for 1 second it now has a velocity of 32 feet per second it did not however fall 32 feet rather approximately half that since it started from rest (0) and built to 32fps.:as it drops the second second it will travel 32 feet plus the gain due to acceleration(a bit less than 16feet in distance) and end the second interval traveling at 64 fps :third second means an additional fall of almost 80 feet (64 plus the 15 point whatever) and ending velocity of 96 fps etc etc. The velocity is gaining at a constant rate but the total distance fallen increases at a rate that keeps the energy figures in foot pounds matching.

Joe, My point, poorly stated no doubt since I am talking over my head, was that technically the energy figures we quote are only absolutely correct if operating in a vacuum. Due to drag, the acceleration due to gravity does not actually bring a falling object to a velocity of 32 fps per second when we are operating in atmosphere. Drop a 5 pound leafy branch 60feet and a 5lb chunk of wood the same distance. Which will impact harder? Drag increases more and more rapidly as velocity increases so if you drop any object from a great enough height it can reach a point where it no longer accellerates while falling. The conservation of energy still applies but the energy (gravitational) that is being used in overcoming drag is not present upon impact having been expended previously. This is pretty irrellevant in the case of dropping a saw on a lanyard or dropping chunks of wood but technically, calculating absolutely correct figures is more complicated. Thanks for the education.


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## Joe (Jan 23, 2003)

http://www.sciencejoywagon.com/physicszone/default.htm

Here is that link Ken posted here some time ago. It's very easy to understand, great for continued review, and a sure fire way to burn the next guy that shows up and spews material he doesn't understand. One needs to keep the link though.

Joe


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## Tom Dunlap (Jan 23, 2003)

Shane,

Changing from a delta to a ball lock isn't any better prevention. Both connections can come open. Proper protocol is to do continuous gear checks. I've been using screw links for many years. The most that I have ever found one unscrewed is three "flats" or about a half turn from snugged down. That's because I check all of my gear when I saddle up and then continue doing checks as I climb. My routine is to do a check everytime that I use a piece.

If your delta was open far enough so that the sleeve was clear of the threads you must not have checked it for a long time. How often do you open and close the biner? Should be at least at the beginning of each day.

The formula that Don Blair has promoted for free fall loads is to take the weight of the piece, multily by the fall distance and add the weight of the piece. There IS NOT a doubling. For a twelve pound saw dropping three feet...you do the math. The bungee in the B'ham will act as a partial decelerator too before the tear away starts to take effect.

Fresco has an interesting saw lanyard system too. They have a double buckle setup on the saw so that the lanyard can be detached or changed to a different saw.

http://www.frescoarborist.com/


Tom


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## tshanefreeman (Jan 23, 2003)

Tom,

While I'm not trying to defend myself, I think that my description sounded a little more devistating than it actually was. The gate had come open about 3/4 to 1 revolution. The gate itself was still on 4+ treads. But it still sent a chill down my spine and created a lump in my throat. The one thing that I can say that I learned from this was that my few years of experience had conditioned me to visually inspection not a physically inspect, you now actually turning, snapping, and clicking things. 

Point taken!


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## John Paul Sanborn (Jan 23, 2003)

I'd have to use it for a while before saying anything, though the concept does give a V8 moment. I think the connector is a tad over built right now. They do say that this is the first incarnation and better renditions are in the works.


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## Joe (Jan 23, 2003)

> _Originally posted by Tom Dunlap _
> *The formula that Don Blair has promoted for free fall loads is to take the weight of the piece, multily by the fall distance and add the weight of the piece. There IS NOT a doubling. For a twelve pound saw dropping three feet...you do the math.
> 
> Tom *



Hi, Tom;

Ken correcly used the Don Blair rule of thumb earlier in the thread. Ken also quoted the rule of thumb along with Don's example earlier in the thread, but later than in the 1st example.

I tried rereading some of the replies in this thread and don't see where some1 went so far off track they doubled something. I'm scratching my head here.

Joe


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## hillbilly (Jan 23, 2003)

Made I'm mistaking, but wasn't the original question really:
What force is generated on the saw lanyard if the saw was
dropped l meters ?

Joe wrote:
"O.K., now the saw is dropped and falls 1 foot. Multiply the distance of fall by the weight. This gives 7 ft›lbs of energy. It is kinetic energy, energy spent."

Joe, you can't say that kinetic energy is measured in ft*lbs.
Check out the links you posted yourself.
Ft vs. meters and lbs vs kg might be a difference between countries. However kinetic energy is not.
Kinetic energy is measured in kg * m^2/s^2.
kg means kilgrams, m means meters, s means seconds.

The formula for kinetic energy is 0.5 * m * v^2.
In words that is: half the mass multiplied with the square of
the objects velocity.

Kinetic energy in a free fall could also be interpreted as the
differnece in potential energy.
KE = m * g * l
m is the mass of the object
g is the gravitational constant of the earth, approx. 9.82
l is the length of the fall.
Note that this yields the same unit as above: kg * m^2/s^2

Now an example:
A chainsaw weighing 3.5 kg (7 lbs) is being dropped 1 meter
will have the kinetic energy = 3.5 * 9,82 * 1 = 34.37
This is NOT the force that will work on the lanyard, it is infact NOT
a force at all, it is the kinetic energy that the chainsaw now posesses.

This kinetic energy must now be converted into some other form
of energy to stay at 1 meter below the original position.
It must be converted to heat energy.
How this is done depends heavily on the material of the lanyard.
How this is done also dictates the actual force that will work on
the lanyard.

This is where my knowledge in mechanics start to run out.
So I'll leave it to Joe to calculate the actual force...good luck!


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## Joe (Jan 23, 2003)

> _Originally posted by hillbilly _
> *Made I'm mistaking, but wasn't the original question really:
> What force is generated on the saw lanyard if the saw was
> dropped l meters ?
> ...



We're still cool on this. You aren't wrong and neither am I. It's just the difference of using the metric system by you and the english system by me. The link uses the metric system.



> *The formula for kinetic energy is 0.5 * m * v^2.
> In words that is: half the mass multiplied with the square of
> the objects velocity.*



...which can be measured using the metric system like in the links or the english system in the U.S.A.



> *Kinetic energy in a free fall could also be interpreted as the
> differnece in potential energy.
> KE = m * g * l
> m is the mass of the object
> ...



In the english system we use the slug or I think also pound mass as the unit of mass. Since converting from lbs(force) to kilograms(mass) is easier tonight, 1 kg = 2.205 lbs (approximately) so a 7 lb saw has a mass of 3.175 kg.

1 lb = 4.45 N = .454 kg.

1 ft = .3048 m

Joe


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## Tom Dunlap (Jan 23, 2003)

Ok, Joe and Hillbilly,

Now that the physics is working, let us know what it's going to feel like to a climber who were to drop that seven pound saw, three feet. Let's say that the lanyard is made out of a material that has no stretch. One of the high mod fibers available and there is no break-away. Would I be heading off to my chiropractor after dropping the saw? Would I be passing blood from internal organ damage? Or would I be wishing for a little bit of bungie in the lanyard?

Thanks!

Tom


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## hillbilly (Jan 24, 2003)

Joe, sorry if I came on too hard on you, but you were pretty
bald a couple of posts back, and you still didn't come up with
an answer.

My mistake about the lbs, I could have calculated the
lbs to kg a bit more accurate than by roughly dividing
by 2.

The english system uses lbs to measure force as well as mass !?
It sure is a strange system!
I still can't see how you can ignore the gravitational constant
of the earth when doing the kinetic energy calculations.

No material is truly static, even steel stretches, in fact
steel has pretty good stretching capabilities, but not
compared to rope. Of course you know this.
This really has to do with spring forces and the characteristics
of the lanyard material. I would love to see a rope technichinan
come on the board and clear this out.
I rest my case as I find my mechanics knowledge hinders
me. I also feel that I'm really just messing this thread up
with my metric system.

How I wish your countries could convert to
the metric system.
1 foot = 12 inches !?
1 yard = 3 feet = 36 inches !?
How about 1 meter = 100 centimeters = 1000 millimeters
and 1 cubic meter = 1000 liters


Joe, I would love to see the resulting force and I will try
to keep my metric mouth shut


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## Stumper (Jan 24, 2003)

No doubt about it Hillbilly, the metric system is much more logical and manageable overall. Unfortunately, whatever is familiar seems easier so we have resisted the conversion to the better system. The gravitational constant is there in the english system, it just isn't always made obvious. The falling object and foot pounds calculation is one of the rare 'easy' ones on the english system-a fact I wasn't aware of prior to this thread. A typical projectile energy calculation in the ES would be Projectile weight in grains (7000 grains per pound) x velocity in feet per second squared. . divide result by 450240 = energy in ft lbs. The 450240 number is the place where the gravitatonal constant is hidden-it was used in calculating that number so that several steps can be skipped. The constant is there in Joe's simple calculation because the english unit of force ( the foot pound) is actually based upon the force of a pound FALLING one foot. The gravitational constant was included from the beginning.


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## John Paul Sanborn (Jan 24, 2003)

Soooo, since the saw falls 3 feet, the work is 21 ft/lbs. So this is waht the climber feels jerk on his lanyard?

So I drop a 15# saw on a 5 foot lanyard I feel 75# of force?

If I drop a 100 # chunk 60 ft there is 600# of force in the impact?


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## Stumper (Jan 24, 2003)

100# x 60'= 6000 OUCH!


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## The Climber (Jan 24, 2003)

Thank God my lanyard isn't 60' long!


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## Joe (Jan 24, 2003)

> _Originally posted by hillbilly _
> *Joe, sorry if I came on too hard on you, but you were pretty
> bald a couple of posts back, and you still didn't come up with
> an answer.*



I'm not sure how to interpret this statement. I'm pretty bald on all my posts, a little smokey too. 



> *My mistake about the lbs, I could have calculated the
> lbs to kg a bit more accurate than by roughly dividing
> by 2.*



I figured it was another poster who wasn't accustomed to the english system of measure. I offered the unit conversions as a courtesy so the thread could move forward and we could see our similarities. 



> *The english system uses lbs to measure force as well as mass !?
> It sure is a strange system!*



If the post is reread, it will be shown the unit of mass is the slug. I think the term slug is being phased out in favor of the term "pound mass".
I'm not sure of this.



> *I still can't see how you can ignore the gravitational constant
> of the earth when doing the kinetic energy calculations.*



The pound is a unit of force. The gravitational constant is there.



> *No material is truly static, even steel stretches, in fact
> steel has pretty good stretching capabilities, but not
> compared to rope. Of course you know this.*



I wouldn't be so sure of this. Some high modulus fibers are pretty stiff.



> *This really has to do with spring forces and the characteristics
> of the lanyard material. I would love to see a rope technichinan
> come on the board and clear this out.*



It really has to do with a lot of things. What we are discussing in this thread is the tip of the iceberg. I'm thankful I know even a little. I doubt if a rope technician would ever show up and dump on a public site like this one. It would be nice if 1 did show up and dump.



> *I rest my case as I find my mechanics knowledge hinders
> me. I also feel that I'm really just messing this thread up
> with my metric system.*



My mechanics knowledge hinders me also. There are people here who can handle the metric system and people who cannot. Perhaps the exposure of the metric system will at least let those who aren't familiar with it know it when they see it.



> *How I wish your countries could convert to
> the metric system.
> 1 foot = 12 inches !?
> 1 yard = 3 feet = 36 inches !?
> ...



I guess you "are" familiar with the english system of measure. You weren't trying to pull my leg into thinking you weren't were you? It worked.



> *Joe, I would love to see the resulting force and I will try
> to keep my metric mouth shut  *



What force are you asking me to calculate?



> *This is NOT the force that will work on the lanyard, it is infact NOT a force at all, it is the kinetic energy that the chainsaw now posesses.*



Do you want me to calculate the kinetic energy force?



> *This kinetic energy must now be converted into some other form
> of energy to stay at 1 meter below the original position. It must be converted to heat energy. How this is done depends heavily on the material of the lanyard. How this is done also dictates the actual force that will work on the lanyard.*



Really? How does this work?

*



This is where my knowledge in mechanics start to run out.

Click to expand...

*
Me too!



> *So I'll leave it to Joe to calculate the actual force...good luck! *



What is it I need to calculate that you can't? I still don't understand what I'm supposed to calculate?

Joe


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## ramanujan (Jan 25, 2003)

the actual force that a climber will feel when he drops his saw will be much less than any simple calculation gives.

Final Velocity=sqrt(2 x 9.8m/s² x distance)


sqrt(2 x 9.8m/s² x 1)=4.427m/s


so the saw would be moving 4.427m/s (assuming, unreasonably, that the saw fell straight down, didn't bump anything during it's fall, nor did it tumble.)

4.427m/s=9.9mph

now for the force felt by the climber.

Force=Mass x Acceleration

In this case the acceleration is in fact deacceleration as the saw is slowing down, and we'll assume it's at a constant rate. One more thing we'll assume is that it takes the saw about 8 centimeters (3.15 inches) to stop once it starts slowing down (due to stretch of the lanyard, swing of the saw, movement of the climbers belt, etc.)


(Final Velocity)²=(Initial Velocity)² + (2)(Acceleration)(Distance)

Acceleration=(4.427²)/((2)(0.8))=11.81m/s²

Force=Mass x Acceleration

Force= (3.18kg)(11.81m/s²)=37.56kgm/s²=0.369 kN=82.8 lbs


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## Joe (Jan 25, 2003)

> _Originally posted by ramanujan _*
> (Final Velocity)²=(Initial Velocity)² + (2)(Acceleration)(Distance)
> 
> Acceleration=(4.427²)/((2)(0.8))=11.81m/s²*



My hat comes off for you sir, "*but*"

4.427² = 19.598 m²/s²

1 meter = 100 cm then 8 cm = 0.08 m

2 × 0.08 m = .16 m

(19.598 m²/s²)/ .16 m = 122.49 m/s²




> *Force=Mass x Acceleration
> 
> Force= (3.18kg)(11.81m/s²)=37.56kgm/s²=0.369 kN=82.8 lbs *



(3.18 kg)(122.49 m/s²) = 389.5 kgm/s²

Isn't kgm/s² the equivalence of a N(newton)? 

So 389.5 kgm/s² = 389.5 N = .3895 kN 

then (389.5 N / 1 N) × 0.2248 lbs = 87.6 lbs ?

It's a minor thing but only in the sense of computational correctness.

It was great to read this post!

Joe


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## TheTreeSpyder (Jan 25, 2003)

Wow, am i understanding that a 7# saw dropping 4' hits with 82.8# of force?

So a 4' drop on a calculated table would almost be @4' drop force=~12 x weight of object? For the speed @ 4' is the same regardless of weight, only resistant forces (air friction to shape of weight, landing resistance)? Then even a 10:1 SWL sounds risky to me, if it can be overrun in 4'; for, certainly as higher drops would yield a higher force multiplier.

So in topping and catching on the same spar; where the center of balance moves from rest (precut) to catch; at a change of 4', if i determine that i want to add the SWL factor in after calculation of force (83# force for 7# saw,after 4' drop x SWL of 10 = 830#+ test line). If i seek to calculate that guardedly, i have to deal with a 120SWL @ 4' drop. So the safe capacity of a 6000 test line would be 50# a 4' drop(50# x 120SWL=6000)??? @ 4' drop factor of 12x, arbor plex 5400 will break @450# load, how many cycles to failure?

i get lost in these higher calculations; but i know that the same math must work in both examples of falling saw, man or log. They all teach the same lesson.

i have googled trying to find a simple chart that says @ 1' drop force will be Y x Weight, @2' etc. If Force = Mass x Velocity; and everything falls at the same speed from the same height irregardless of weight; then the Factor of Velocity (F=MxV) should be a recognized accepted constant given per foot of drop from a given elevation (all this changes with altitude i would think)!!! i don't know if that is to simple to do or what!! Seems like putting it on a level like that would make it 3rd grade math with a fixed table?? i have faith in Don Blair's rep enough that i don't think such a formulae would be written, to be so far off in 4'!??!

i'm lost , willing to learn!

edit-Now Joe has upped the ante to ~87# ??


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## Stumper (Jan 25, 2003)

Hmm... Ramanujan and Joe, There is a mistake or 2 there somewhere. The right answer in foot pounds can't change when worked in the metric system and converted to english. Among other things I noticed that R's formula uses initial velocity squared- Fine. But there is a serious boo-boo- in your working of the problem INITIAL velocity was 0.


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## Stumper (Jan 25, 2003)

Incidentallly, Rananujan's point about the sensation of impact being lessened by spreading it over time/distance is correct but the total force or energy expended doesn't change. Decelleration is not a force multiplier.


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## treeclimber165 (Jan 25, 2003)

As the title says, "Oh, for Pete's sake!" Go find a fish scale or something, clip your 4' saw lanyard to it and drop your saw off of it. Read the highest number indicated. 
I was always good at math in school, but I haven't used calculations like shown above in 20 years. Gives me a headache trying to wade through it all.


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## TheTreeSpyder (Jan 25, 2003)

Perhaps someone has a dynomike-meter...........for calculculations of such moving maas? Using non-dynamic cable?

i think that out of the tree is the time for such checks on the strategies we use for decision models in the tree/field; to make sure wee are within realistic bounds. If we are loading lines and gear, with more force than we thought, possibly bringing it outside the safety perimiter i think that reflection is needed. Including deeper respect for the said devices, and work carried!

As the numbers of this force climb with length of drop. Several things become more well defined peering deeper into this world, that is before me all day. Mostly that the dynamic elasticity to absorb all this energy in a line might be more important than a dynamic/elastic component rather than the static weight numbers expressed? Making Nylon more valued than polyester for these extremes? How much more pretightening, solid supports, stronger hinges, load control etc. is important as contorllling forces become more precious and appreciated, as the numbers of dynamic forces express themselves increasing so quickly.

If this here, on this particular website for examination to our particular applications are not the place for this; then where? Where else could one of ours come to seriously look at these things, save outside of bugging a Rope lab? 

i stand by that this is thee place on thee web, and these things and there implications to all we do are relevant; further from any kind of impatience of censorship than other topics previously addressed. i find these things worth our time and minds. A responsibility to understand them, and this place the best vehicle on the planet to share them. As being and increasingly looked to as this premiere place; perhaps this space has the responsibility for offering these topics in the mix; the headache is quantafiably less than overhead support failure!


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## The Climber (Jan 25, 2003)

I agree, althow I haven't taken part in the discussion( because my math skills fall short of those of the scholars above ) I have been reading it a couple times a day because I am interested in the outcome. the resulting knowladge is something I can put to use in my every day life and I'll be safer because of it.


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## John Paul Sanborn (Jan 25, 2003)

Mathmaticaly challenged.


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## ramanujan (Jan 26, 2003)

thanks for making those corrections Joe.. it was too late and I was in too much of a hurry 

Stumper I didn't use the same values that others did in their calculations if that explains why my values differ from others.

I agree with you that the potential energy of the saw before it is dropped must equal the energy expended when it has stopped but I think that the original question was what regarding what a breakaway lanyard will take before it breaks. ie, not total energy but peak force.

also someone mentioned that the energy disappates in the form of heat. this is incorrect, only a very small fraction of the energy will be transformed into heat, most will be used to move the climber and the tree and maybe tear the lanyard etc.


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## Stumper (Jan 26, 2003)

ramanujan, What values did you use? You had a bunch of fomulae and figures but where did you start? The discussion had been centering around a 7 pound saw dropped 4 feet. The figures you show would match up to a fall of over 12 feet. I wonder if there is an engineer lurking out there laughing at all of us (myself included) for going on and on about things we obviously don't fully understand.
Despite our dismal math performance I think that most of us know what we can do safely in the trees.:angel:


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## Joe (Jan 27, 2003)

> _Originally posted by Stumper _
> *I wonder if there is an engineer lurking out there laughing at all of us (myself included) for going on and on about things we obviously don't fully understand. Despite our dismal math performance I think that most of us know what we can do safely in the trees.:angel: *



So what if there are engineers, mathematicians, and physicists laughing at us. We still need to learn the stuff, right? Are there any engineers, mathematicians, or physicists speaking up? No. So what is going to be done to fill in the blanks? I wouldn't waite for an engineer, a mathemtician, or a physicist to show up and explain things. Remember, these guys have spent a lot of time, money, and effort
to learn what they know. Don't expect any of them to give it away for free.
Forget about them and focus on what needs to be done.

Joe


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## Joe (Jan 27, 2003)

http://www.motionmountain.org

I think the guy who put this link together is an amateur and enthusiast.

Joe


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## hillbilly (Jan 27, 2003)

Ramanujan, thanks for your post, it makes sence in many ways!
However:
"In this case the acceleration is in fact deacceleration as the saw is slowing down, and we'll assume it's at a constant rate"

Maybe the assumation that the deacceleration (and in turn
the force) is linear is a too big assumtion. Would expect
the force to be smaller in the beginning of the stretch and larger
at the end of the stretch.

"also someone mentioned that the energy disappates in the form of heat. this is incorrect"

What I meant was that energy is never destroyed, if the
potential energy is reduced and the kinetic energy is to
be 0 at the end of the fall, that "lost" energy must go somewhere.
If it be the tree, the climber, the saddle and the
lanyard that gets heatend, it still must happen.
This energy transfer might very well be negliable when
calculating the force on the lanyard, but it still must
take place.

I studied mathematics 1,5 years at the university, sadly
it is very theoretical and if not combined with other fields,
pretty useless. However if anyone were to have any mathematical
questions, I will be more than happy to help you. Just send
me a PM.


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## CJ-7 (Jan 27, 2003)

Ok, I am really asking for it here, I am an engineer, but 34 years since my last physics class, and I don't use dynamics (that is what you are talking about here) in my work. I am a civil engineer not a mechanical. Here's my interpretation out of a book provided by a colleague, who refuses to get involved in this discussion (many of these formulas have already been given in prior posts):

F=force (pounds, force (lbf) as opposed to pounds, mass)
m=mass (pounds, mass (lbm))=7
a=acceleration (feet per second per second)
g=gravitational constant (feet per second per second)=32.2
v=velocity (feet per second (fps))
Ek=kenetic energy (feet-pounds force(ft-lbf)
s=distance (feet)=4

Saw weight = 7 pounds
Drop = 4 feet
assume no friction and initial velocity = zero
assume lanyard evenly stops saw over the distance of 6"

v=sqrt(2as) =sqrt(2x32.2x4) = 16 fps (here g=a)

after falling 4 feet the 7 pound saw has gained kenetic energy

Ek=((m/g)(v squared))/2 =((7/32.2)(16x16))/2 = 27.82 ft-lbf

now how much force, evenly distributed over time, is required to decelerate the saw in 6" (0.5 feet)

first calculate the acceleration (actually deceleration) needed to stop the saw from 16 fps over a distance of 6".

a=(v squared)/(2s) = (16x16)/(2x0.5) = 256 ft per second per second

F=(m/g)a = (7/32.2)x256 = 55.6 lbf not that far from the force noted in other posts. 

I have made some simplifications here to get some of the basics across.


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## TheTreeSpyder (Jan 27, 2003)

i've been searching for some kind of table that says something falling from 1' has a 'fall factor' of........, from 2' of....... to 10. Seems that we could use such a table for a lot of things and understandings. Seeing that 2 things fall to earth at the same speed from the same height; and that speed x mass = force would be pretty simple for many examinations. 

i have found these 4 calculators that may help, but are even a bit beyond my confidence as they present more functions, many leaning towards the energy absorbtion properties of the lines etc.

Fall Calculator 

The Splat Calculator 

Terminal Velocity Calculator & Graph 

Climbing Fall Calculator

Again and again, minimizing force by minimizing acceleration seems to bear out, bringing me back to tight lines and meaty hinges, minimizing fall etc.


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## Joe (Jan 27, 2003)

Here's my personal favorite.

http://www.glenbrook.k12.il.us/gbssci/phys/class/BBoard.html

Joe


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## Jock (Jan 29, 2003)

Just dont fall


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