fall vs energy

i don't think this Load X (Drop + 1) trick will hold at 10'; as speed is increasing exponentially with the increase in height. Speed is the operating distance multiplier here i believe.
 
i think rope type/ tensile and length should be noted; as well as pre-tension of line device; to be compareable to other numbers given. Also, setup measures double hit at block or just single line.

Assume no rope run or relief from support. Angle of line for load and control legs of pull on block would make minor differances especially in spreads from 0-30 degress, where the change in direction of pull on support would be of more relevance and measureability than the change in line tension measured by Dyno above block.

i would use larger sheaves on axle to factor out some friction degradation of force, but not bearing$.
 
I haven't seen a formula for free falling weights. I can guarantee you that it will be much higher than 2,220 lbs. Our dyno only goes to 5,000 lbs. We can drop a specific weight from different distances to see if we can come up with a formula.
 
Something I see missing and will likely always be missing in these calculations is the mass of the trees resistance to movement. It would useful to know the duration of the peak loads that the dyno is measuring.

It is likely that continual loads that would cause tree failure would not even come close to causing failure if applied for only a few hundred milliseconds. Not to be over looked is the fact that rope stretch has absorbed short duration peaks before the dyno even gets a chance to read the peak load.

Mass dampening has been discussed quite a bit as it relates to calming down the climbers ride when rigging the top of a tree back onto itself.

No doubt it is a factor in these discussion too.

Dan
 
The tree absorbing enery is huge. I used my dyno today. I thought the peice would come in around 3000#. It was boxelder and had quite a lean. The tree took a ton of the force the dyno said 1350#. Every time I use the dyno I am always amazed.
 
This most certainly is an interesting topic of discussion. Too bad there's not an easy answer.

One can go to http://en.wikipedia.org/wiki/Kinetic_energy and find
e02161cd0e8a6a7655e816c5e23a15ee.png
. In order to determine what the "v" is in that, you could go to http://en.wikipedia.org/wiki/Equations_for_a_falling_body and choose
1a6e5238d2ab226044c4f776d6ee7e97.png
"Instantaneous velocity v of a falling object that has travelled distance d".

Going through all the steps (thank God for programmable calculators!), I find that our (200# in this case, but immaterial for anything without great aerodynamic drag, which is yet another factor) item dropping 10' has achieved a velocity of 17.3 mph. Plugging that velocity in, I am able to discover that, surprise, the available energy has become 2000 ft-lbf! For this exercise it's apparent that all the intermediate steps can be omitted and we can know for a certainty that the 200# chunk of wood falling 10' onto the driveway will hit with 200# × 10' = 2000 ft-lbf of energy.

(Compare that with some of the tabled figures in http://en.wikipedia.org/wiki/Muzzle_energy.)

The problem is in determining the length of time it takes to come to a come to a complete stop, thus dissipating all of the energy. That's the whole enchilada in terms of how violent the energy exchange is.

http://en.wikipedia.org/wiki/Impact_force describes it all pretty well, I guess, and this article at "roofingcontractor.com" goes into great detail figuring terminal loads generated when arresting a falling worker. I even tried plugging "our" values into their equations, but how much distance do you want to conclude "our" fall is "arrested" in? They picked 6" for their example. I'd like to say 0" for the chunk hitting the driveway, but that creates a "divide by zero" situation which doesn't go anywhere.

Determining how much energy is in the system is the easy part. The length of time or distance over which the energy is spent is the entire key. Zero time is infinite (rather, undefined: divide by zero) energy and infinite time is zero energy.

But then we've come full circle because we already knew that :)
 
[ QUOTE ]
The tree absorbing enery is huge. I used my dyno today. I thought the peice would come in around 3000#. ... The tree took a ton of the force the dyno said 1350#.

[/ QUOTE ]
So the tree flexed and spread the load over time/distance.

This means that whenever you do your series of tests to plot a graph you'll need to mount the dyno to something totally rigid or it will be relatively meaningless.

Once you get the data gathered, you'll be able to both determine what the loading will be worst-case and the effect flexibility of the rigging point has on the system.

We're still going to be back 'round to having only a "rule of thumb" and the requirement for a nice touch by the groundie.
 
This really is an interesting question that may not have an easy answer to it.

It is the kind of question that our late colleague Dr. Peter Donzelli, who was a tree worker and mechanical engineer would have found fascinating, and probably would have had some good answers to it.

I will still probably use the NAA formula for small drops that are not in critical locations as well as cross checking it with the ArborMaster software in more dicey situations. Though to be honest with you most of the time all this calculating of loads is not necessary, as experience will usually be my best guide.

Chris
 
i think that speed increases up until the terminal velocity of the shape's air friction as a multiplier of the speed overcomes the load's weight, so it can't increase in speed anymore. The speed factor as a multiplier of the weight gives potential force, so can't be a static/ across the board measurement per foot of drop; cuz that would be too easy and wouldn't fry your brain. A lever is calculable like that, because it assures that the load moved does so in the same amount of time that the effort is input, given the lever doesn't flex (then there is loss of the potential). So there the load speed is relative to the input speed, so speed is cancelled out. But, the speed in free fall increases on the graph at these lower falls, then flatlines, so is variable and can't be put into such a neat formulae as drop X weight. (Drop + 1) X weight being a neat formulae for low drops; but the speed overtakes that as you go higher.

Force Calculator
 
that force calculator was exactly what i was looking for!!! Thank you!


I have spent hours(because i was not paying too much attention in physics) figuring what the force potential is on some of our rigging systems, and this helps....

it is crazy how much energy our ropes dissapate(and good ground men). one thing that i am trying to teach my guys is that if you shock load your ropes they lose their ability to dissapate all of this energy(or force).... i need a dyno, anyone want to sell one cheap?

Rob
 
wikipedia has a great article on "terminal velocity".

For our purposes, terminal velocity might come into effect for something small in diameter with many leafed branchlets. But most folks don't start getting into load concerns with the periphery of the tree, right?

The force calculator link you provided precisely matches the values I'd given in my previous post. A 200# item dropping 10' produces the same potential energy as a 400# item falling 5'; 2000# and 1'; etc.

How far does the driveway flex (for the value of "d")? If you put 0 there, nothing happens, as I'd said. If you put .01 there (~ 3/8") you get a pretty large value in return. Try .001 (~ 1/32") and the value is 10 times greater. The former is likely too great a distance just as the latter is likely too short. Even a little bit of speculation here gives a very wide range of difference in results.

200# = 90.8kg
10' = 3.05 m

try it and see...
 
Rob, see also the "external link" in the "Impact force" wikipedia link I provided above. It may serve even better to illustrate to your crew what you're trying to say to them.
 
[ QUOTE ]
... what is the formula to find out how much energy gravity and weight can produce...

[/ QUOTE ]
The formula you need to use is the one for potential energy which is m*g*h (mass * gravity * height). Since weight is m*g, a force, it simplifies to w*h, so the simple answer is multiply the weight by the height to get the energy.

Cary
 
Glen, you need to remember both bodies will likely deform, but you are correct that as the time or distance an energy is dissipated over approaches zero the resulting force of stopping/impact becomes infinite. This is why you should never use a static rope as fall protection or to stop a falling object. Though static ropes do make getting into a tree easier since they don't bounce as much (wasted energy that is dissipated as heat). The ideal rope for lowering object that may be dropped a small distance would be very stretchy, but would have minimal bounce.

I can elaborate more on the rest of this discussion latter if needed, but the key thing to remember is keep track of the units to make things work out right.

Cary
 
[ QUOTE ]
The tree absorbing enery is huge. I used my dyno today. I thought the peice would come in around 3000#. It was boxelder and had quite a lean. The tree took a ton of the force the dyno said 1350#. Every time I use the dyno I am always amazed.

[/ QUOTE ]

I agree the tree's flex absorbs a lot of energy but I was talking about the 'mass' of the tree resisting sudden movement.

As an analogy picture a weight of 200lbs. sitting still on a swing at bottom dead center. This 200lb. weight will not even move when 'shock loads' of 1,000lbs are applied for periods of a few milliseconds. It is this resistance to movement I was talking about.

This is not the same resistance to force as the tree flexing to absorb energy but instead is an at rest body's restistance to movement.
 
[ QUOTE ]
... potential force ...

[/ QUOTE ]
That would be potential energy. Force and energy are two different beasts. For rope loading it is the force that dissipates the potential energy that needs to be calculated. Which likely has both a static and dynamic component.

Cary
 
[ QUOTE ]
Glen, you ... are correct that as the time or distance an energy is dissipated over approaches zero the resulting force of stopping/impact becomes infinite.

[/ QUOTE ]
Ha! Not quite. I'd said (along with "come to a come to a") the resulting energy would be either infinite or zero. I suppose you understood what I'd meant and are just being kind in not pointing out the error in terminology :).

I actually have a really good "intrinsic" understanding of this kind of stuff but have spent several minutes trying to find particulars to share. I'd rather dig up the equations and work them than find a web page with input boxes and which (might? I'm not blindly trusting) gives the results. Who cares what the answer is? The fun is in the process of determination!
 
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