Fall Force Calculator?

Woodwork

Active Member
Looking for an online calculator to tell if mass x falls distance y, it creates z lbf of impact force.

Tried
but it is so virused and/or popuped out, I can't use it, and I don't think it even tells me what I want to know, anyway.

Trying to figure out how strong rigging needs to be to drop weight A distance B and brake over distance C. In other words, how many kN of force is generated when it stops/brakes.

I thought I'd found the answer here

but when I use all the same figures but increase the fall from 10m to 20m to 30m (not real world drops, was just inputting numbers to see if the calculator behaved as I expected), the fall energy in kN went up proportionally. (The calculator shows kN for 30 m fall being triple that for 10 m fall.) but I think it goes up exponentially – not proportionally – due to the acceleration of gravity.

If I'm not mistaken, tripling the fall distance should make the fall force 9X higher (3²), and quadrupling the fall distance should make the fall force 16X greater (4²) ... right? (All assuming a sudden stop, again, not real-world.)

Does anyone know where I could find a calculator for this, or maybe a simple equation that I could plug numbers into?

(This all started when I wondered how much force on my roof anchor I'd apply if I fell x feet before I unslacked and my prusik grabbed, the rope stretched and I stopped in y feet of deceleration.)

Thanks for any help.

Jeff
 
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Tuebor

Well-Known Member
...but when I use all the same figures but increase the fall from 10m to 20m to 30m (not real world drops, was just inputting numbers to see if the calculator behaved as I expected), the fall energy in kN went up proportionally. (The calculator shows kN for 30 m fall being triple that for 10 m fall.) but I think it goes up exponentially – not proportionally – due to the acceleration of gravity.

If I'm not mistaken, tripling the fall distance should make the fall force 9X higher (3²), and quadrupling the fall distance should make the fall force 16X greater (4²) ... right?
The increase in velocity over time will be linear, but since it's falling faster as it falls, it takes less time to cover the next distance interval - so the increase in velocity (and kinetic energy) over distance will be inversely exponential with distance. I think that's making the impact force linear.
 

RBJtree

Well-Known Member
Thus, a 100 lb log falls 1 foot = 100lb force, a 100 lb log drops 3 ft= 300, a 100lb log falls 5 ft = 500... I once saw a live test with a impact scale that confirmed that a 100 lb log dropped 5 foot made 500 lbs of force. But dropping it 1 foot only making a force equal to static weight seems wrong to me.
 

Woodwork

Active Member
Thanks guys. I don't know why this is giving me such a brain fart.

I guess what's throwing me is that I know (or seem to remember?) that the energy of a moving object is

E = mv²
kinetic energy = mass x velocity squared

But I guess that means that the distance fallen per unit time would increase the longer you fall
At the end of one second you'd be falling at 10 m/sec
At the end of two seconds you'd be falling at 20 m/sec
At the end of three seconds you'd be falling at 30 m/sec

But if you break it down into actual distance fallen, I guess it's more linear.

Thank you for the replies. I think I understand better now. Guess I'm making it more complicated than I need to, again. Appreciate the help.

Jeff

But dropping it 1 foot only making a force equal to static weight seems wrong to me.
Me too!

I also have trouble grasping that a 100# object hanging from a pulley applies 200# of force on the pulley (because both the load and the effort apply 100# of force to the pulley). It seems like the load on the upper anchor should be 100# whether it's tied directly or on a pulley, but the diagram makes sense...

 
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RBJtree

Well-Known Member
The info from yale also includes calculations on how to determine how much impact a rope of a certain amount of elasticity can absorb. Unfortunately, the only company that makes the energy absorption data easily accesable is yale. Theoreticly, with yales data and calculations I think I could determine how much a rope would stretch when loaded when over a target. That way when we are close we can determine if the rope would or would not stretch enough to allow the object to hit the target. But I haven't had the opportunity to figure that out and test it yet. And I currently have samson ropes for rigging.
 

Tuebor

Well-Known Member
RBJtree said:
But dropping it 1 foot only making a force equal to static weight seems wrong to me.

Me too!

I also have trouble grasping that a 100# object hanging from a pulley applies 200# of force on the pulley (because both the load and the effort apply 100# of force to the pulley). It seems like the load on the upper anchor should be 100# whether it's tied directly or on a pulley, but the diagram makes sense...
Not 100 lbs, but 100 ft-lbs.

And in the diagram, if you weighed exactly 100 lbs, lift your feet off the ground - the pulley would be supporting 200 lbs.
 

RBJtree

Well-Known Member
Not 100 lbs, but 100 ft-lbs.

And in the diagram, if you weighed exactly 100 lbs, lift your feet off the ground - the pulley would be supporting 200 lbs.
I don't fully understand lbs vs ftlbs. in this application. 1 pound is one pound. One ft lb is the amount of torque produced by 1 lb applied to a 1ft lever. So, I don't understand how that applies to the force of a falling log. I used the ftlb measurement because that is how yale stated it.
 

evo

Well-Known Member
Thus, a 100 lb log falls 1 foot = 100lb force, a 100 lb log drops 3 ft= 300, a 100lb log falls 5 ft = 500... I once saw a live test with a impact scale that confirmed that a 100 lb log dropped 5 foot made 500 lbs of force. But dropping it 1 foot only making a force equal to static weight seems wrong to me.
Close.. you need to then add the weight. 100# falling 1 foot is 200# of force
 

evo

Well-Known Member
Interesting. That makes more sense.

So if a 100# log fell 3 feet, would the force be (3x100) + 100 = 400 ft-lbs?
I suck at math, but that is the basics of what I was taught. Add swing, subtract rope stretch, rope lenght, and figure pre load and it all gets complicated quick.
 

RBJtree

Well-Known Member
Close.. you need to then add the weight. 100# falling 1 foot is 200# of force
This is directly from the yale article:
"First, we will calculate the ft lbs of energy needed to arrest the 500 lb trunk section falling 5 ft. The simple equation of the weight multiplied by the fall will get the result within 1%, so 500 lb x 5 ft = 2500 ft lbs."
I am guessing there is more math to do at less than 2 feet of drop to come out with the right number. So at one foot of drop it is more than 1x static weight, but less than 2x... or yale is wrong, but I did see the drop test on a digital scale of the ~100 lb log making ~500 lbs of force, not 600. But it was about 15 years ago, my memory could also be faulty.
 

evo

Well-Known Member
This is directly from the yale article:
"First, we will calculate the ft lbs of energy needed to arrest the 500 lb trunk section falling 5 ft. The simple equation of the weight multiplied by the fall will get the result within 1%, so 500 lb x 5 ft = 2500 ft lbs."
I am guessing there is more math to do at less than 2 feet of drop to come out with the right number. So at one foot of drop it is more than 1x static weight, but less than 2x... or yale is wrong, but I did see the drop test on a digital scale of the ~100 lb log making ~500 lbs of force, not 600. But it was about 15 years ago, my memory could also be faulty.
Trying to come up with hard numbers is beyond the point. What is 500#'s plus or minus when you're shock loading your system at 2500# . The take home is we are working with best guesses, over estimate than under, and don't shock load your shit...
Not calling Yale wrong, but I cant understand if your dropping 1000lbs one foot, you only get 1000 of force. Perhaps there is something in the wording where you dont add the weight to the foot lbs of force, but its implied.
 

RBJtree

Well-Known Member
220 lbs making 1421lbs force at 6.5 ft. (220x 6.5 = 1420) your right of course, estimate the weight high and the system strength low, its just fun to think about. And I think its good to know where you are at in the scale of high and low. I never broke a rope while rigging. I've broken trees, but not ropes slings, ect. And I have only gained knowledge and become more cautious over the years. But I also strive for efficiency and sometimes that means take a big peice, IF ITS SAFE. So we have to know how to figure out appoximately what is safe. 3,000 lbs of force vs 2,500 lbs doesn't sound like much, but it could be the difference. 500 lbs is a lot to some people.
 

Jehinten

Well-Known Member
Looking for an online calculator to tell if mass x falls distance y, it creates z lbf of impac
I've been using a free app on my phone, the name of the app is in the screenshot, I paid a one time fee of $1 to unlock the force multiplier at the bottom of the screen.
Screenshot_20190426-055639.jpg
 
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Bart_

Member
Here's to bending one's brain a little from the norm.

Hanging the 356 lb log from the block, the groundie holds just hard enough so the up rope coming out of the porty has 355 lbs tension. The log comes down real slow and steady. Pretend the pulley is perfect to ease the level of brain hurt.

Or the groundie slacks his grip so the up rope coming out of the porty has 178 lbs tension. Now the log accelerates at 1/2G (because it has 356 lbs worth of mass but is not getting 356 lbs of acceleration but has half applied as "brakes". It picks up speed at 16 ft/sec squared. At some point the groundie reigns it in, grabs harder and gets the up line to tension at 712 lbs. Now the log sees 712 lbs on its rope and feels 2 Gs. The net acceleration is 1 G gravity downward minus 2 Gs from the rope, leaving 1 G of "slowdown". That cuts the dropping speed not bad without killing the rope, porty or the gloves.

But, our groundie is actually the bionic man and has metal legs and unfortunately is situational awareness challenged. So his foot gets wrapped in the rope, sucked into the porty and the rope instantly locks to a stop. Enter the realm of Samson rope energy absorption for shock loading.

Our conservative groundie used 1" rope with practically zero stretch and the log stopped dead in 1". Quiz question: how big was the tension spike in the rope? I know I didn't say how far it fell and how long he put the brakes on for, but I just squeaked out of teaching college , ok? Multiple choice answers: a) holy sh_t b) Holy sh_t or c) Holy Sh_t. or d) the block sling broke.

A little humour, hope you enjoyed.

There is as real answer to these calculations. Take the log mass and get 1/2 mVsquared as its energy at the lockup instant. Then do that same work as force x distance (stopping distance) assuming constant force over the 1". Stick to metric to avoid mass vs force errors. Then realize factors like samson rope spike aborption alter the result a bit, basically by elongating the 1" dimension.

The other way is ratios of acceleration. 1 G freefall for ten feet and you want to stop it in ten feet you need 1G net backwards which is 2 G worth in the rope. If you want to stop it twice as fast (5 feet), you want 2 G's net which means 3 G from the rope. Stopping 4 times as fast (2 1/2 feet) would need 4 G's net from the rope or 5 G's total in the rope because it first has to cancel th 1 G of gravity. So 5x the log weight worth of rope tension starts to add up. So don't let it free fall very far and have lots of room to catch it.

Best wishes to all.
 
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