oh for pete's sake

[Newfie]

". 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?

 

[treeclimber165]

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

 

[John Paul Sanborn]

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

 

[Joe]

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.

 

[treeclimber165]

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.

 

[Kevin]

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.

 

[Joe]

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

 

[The Climber]

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:

 

[Stumper]

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.

 

[Joe]

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")

 

[John Paul Sanborn]

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.

 

[Gypo Logger]

Thanks JPS: the provided link says it all.

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

Joe

 

[TheTreeSpyder]

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.

 

[Joe]

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

 

[Stumper]

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!

 

[Joe]

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

 

[TheTreeSpyder]

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.

 

[Stumper]

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.

 

[Joe]

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

 

[Tom Dunlap]

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