Fall Force Calculator?

evo

Well-Known 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.
What if said log had a 30 lbs eagle attached, this bird held on for the initial loading, spread its full wing span and then bailed when the groundies leg temporarily locked the porty before all the blood acted as a lubricant?
 

TreeFan

Member
A lot of good physics insight in these responses, but some approaches aren't 100% right. The original hyper-physics link is correct, but for some goofy reason the distance of fall must include the entire distance fallen even after the rope starts pulling on it. Here is the correct formula, if you don't mind doing just a bit of math.

Average Stopping Force = weight *(distance of free fall/stopping distance) + weight

(An earlier post was right to add in the weight, but don't do that before multiplying by the distance ratios.)

So, for example, suppose that a 1000 lb log freely falls 5 ft before the rope starts acting on it and then the log continues to fall 15 feet before it stops. The average stopping force would be:

Average Stopping Force = 1000 * (5/15) + 1000 = 1333 lbs

Also, note that lbs is a measure of force while ft-lbs is a measure of energy (or work, or torque). The Android app gives the correct energy right before the rope tension is applied, but that is not the force needed.

I'll try linking to a spreadsheet I made that does the calculation. Let me know if it doesn't work.

 
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RBJtree

Well-Known Member
I like to know the force applied at a dead stop in order to account for if there is a mistake made somewhere. Idealy, a 1000 lbs log will never drop 5 feet and then be stopped dead with no slowdown except for rope elasticity, but mistakes happen. Ground guys take too many wraps or the rope gets jammed up somewhere. I have seen a rope get crossed over itself on the brake and not run. If we rig for the worst case senerio, it is safer.
 

TreeFan

Member
I like to know the force applied at a dead stop in order to account for if there is a mistake made somewhere. Idealy, a 1000 lbs log will never drop 5 feet and then be stopped dead with no slowdown except for rope elasticity, but mistakes happen. Ground guys take too many wraps or the rope gets jammed up somewhere. I have seen a rope get crossed over itself on the brake and not run. If we rig for the worst case senerio, it is safer.
If you use the stretch of the rope as the stopping distance, then the formula above works fine. If the elasticity of the rope is known, then that distance could be calculated as well, but as an estimate put something like 1 foot in for the stopping distance and watch the force get huge.
 
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TreeFan

Member
I like to know the force applied at a dead stop in order to account for if there is a mistake made somewhere. Idealy, a 1000 lbs log will never drop 5 feet and then be stopped dead with no slowdown except for rope elasticity, but mistakes happen. Ground guys take too many wraps or the rope gets jammed up somewhere. I have seen a rope get crossed over itself on the brake and not run. If we rig for the worst case senerio, it is safer.
So you really ignited my inner physics geek and I just had to calculate some more realistic numbers for the case of a dead stop. This problem is very similar to a homework problem I used to give my physics students in which they had to calculate the maximum force on a rock climber who fell a distance before a rope would start decelerating her. I gave them some idealized data, but you can get actual specs about elongation at different loads from Yale or Samson sites. I used Samson 5/8 Arbor-plex as a test case, but you can use others. I revised that spreadsheet above to include separate cases where the end of the rope is fixed.

If you don't want to wade through the spreadsheet use, (where you can alter the inputs), here is an example case. Suppose that a 1000 lb log falls 5 feet before it tensions a 50 ft long, 5/8 Arbor-plex rope. I estimate that it will stretch the rope 3.7 feet and create a maximum tension of 4700 lbs. (The break strength of the rope is 9000 lbs, so better not to do his, duh.)
 
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RBJtree

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So you really ignited my inner physics geek and I just had to calculate some more realistic numbers for the case of a dead stop. This problem is very similar to a homework problem I used to give my physics students in which they had to calculate the maximum force on a rock climber who fell a distance before a rope would start decelerating her. I gave them some idealized data, but you can get actual specs about elongation at different loads from Yale or Samson sites. I used Samson 5/8 Arbor-plex as a test case, but you can use others. I revised that spreadsheet above to include separate cases where the end of the rope is fixed.

If you don't want to wade through the spreadsheet use, (where you can alter the inputs), here is an example case. Suppose that a 1000 lb log falls 5 feet before it tensions a 50 ft long, 5/8 Arbor-plex rope. I estimate that it will stretch the rope 3.7 feet and create a maximum tension of 4700 lbs. (The break strength of the rope is 9000 lbs, so better not to do his, duh.)
So the elasticity of that particular rope softened the blow by 300 lbs?
 

TreeFan

Member
So the elasticity of that particular rope softened the blow by 300 lbs?
Well, it is just a coincidence that in my example the value of the maximum stopping force in pounds is close to the value of the energy of free fall in foot-pounds. When I use 3 feet of free fall, the maximum stopping force is almost 4000 lbs, which is actually greater than the energy of 3000 foot-lbs. Again, energy in foot-lbs is a completely different beast compared to force in lbs. Sort of like the difference between distance and speed. (You can easily insert different log weights, rigging line lengths, and free fall lengths in the spreadsheet to see other values of stopping force.)
 
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RBJtree

Well-Known Member
Well, it is just a coincidence that in my example the value of the maximum stopping force in pounds is close to the value of the energy of free fall in foot-pounds. When I use 3 feet of free fall, the maximum stopping force is almost 4000 lbs. while the energy would be 3000 foot-lbs. Again, energy in foot-lbs is a completely different beast compared to force in lbs. Sort of like the difference between distance and speed. (You can easily insert different log weights, rigging line lengths, and free fall lengths in the spreadsheet to see other values of stopping force.)
It's nice to have a spread sheet, but I want to understand why.
1 lb applied to a 1 foot lever = 1 foot lb of force.
1 lb falling 1 foot =1 foot lb of force. The falling distance is = to the lever?
And that is why you can multiply the lbs by the fall in feet to get the foot lbs of energy? And the difference in lbs vs ft lbs can be seen in the difference between the breaking strenght numbers yale publishes in lbs vs the ones they publish in foot lbs.
 

TreeFan

Member
In some cases, calculations are so complex that I just end up relying on calculators and trying to gain insight from various results. Finding the maximum stopping force in a given situation is such a case. But in response to your questions:

1 lb applied to a 1 foot lever = 1 foot-lb of moment (engineering language) or torque (physics language), not force.

1 lb falling 1 foot = 1 foot-lb of energy, not force. And it is weird that energy and torque have the same units, even though they are used in different situations.

I'm surprised that Yale publishes breaking strength numbers in foot-lbs. I'd have to see the context to figure out why they are doing that. It is certainly not a force.
 

RBJtree

Well-Known Member
In some cases, calculations are so complex that I just end up relying on calculators and trying to gain insight from various results. Finding the maximum stopping force in a given situation is such a case. But in response to your questions:

1 lb applied to a 1 foot lever = 1 foot-lb of moment (engineering language) or torque (physics language), not force.

1 lb falling 1 foot = 1 foot-lb of energy, not force. And it is weird that energy and torque have the same units, even though they are used in different situations.

I'm surprised that Yale publishes breaking strength numbers in foot-lbs. I'd have to see the context to figure out why they are doing that. It is certainly not a force.
Here is the graph they provide for double esterlon
Screenshot_20190428-083625.png
Note the working load and ultimate strength listed is ftlbs per pound of rope.
Here is a link to the product page for double esterlon. Go to the tab for "data" to find the breaking strength, elongation under load, and weight per 100ft of rope , ect.
 

TreeFan

Member
Thanks. Twas very interesting (for a geek) reading. I am reluctant to quibble with Yale, but I really think their approach is a bit goofy. It is certainly possible to state the breaking strength of any rope in terms of energy; you just multiply the breaking force times the distance the rope stretched. But that stretch distance, of course, depends on the rope length, or, in Yale's usage, the weight of the rope. Thus they must end up dividing by the weight.

So (really geeky alert) here is what I think Yale is doing. They have that apparatus that allows them to drop a known weight from a given height. They do that for several thicknesses and lengths of rope and come up with an energy of free fall (in ft-lbs) that breaks each rope and then they divide by the weight of the rope. That's an easy test, and the number can be used in their (misguided imo) calculations so they just state that number.

My big quibble with them is that their calculations (found at http://www.yalecordage.com/featured-industries/arborculture/dynamic-energy-arborist-rope ) completely ignore the gravitational energy released as the rope stretches. It is MUCH more than the 1% effect that they claim. To see this, take an extreme case. Suppose you tie off a 3000 lb log (which may be below the working strength) such that it did not fall any distance before starting to tension the rope. The energy of free fall is 3000 * 0 = 0, which Yale claims is the same as the rope's energy absorption and which the rope could certainly handle. In reality when you released the log, it would greatly stretch the rope until a restoring force of 6000 lbs was reached, which could be above the working strength. 1% is just silly.

(Finally, Yale's sample calculations are pretty sloppy. They either have serious typos that leave out a decimal point, or they are multiplying by 100 instead of dividing. Their units are wrong (using ft-lb instead of ft-lb/lb) and the data they use for the polydyne rope energy absorption (1040 ft-lb/lb) is inconsistent with the value in the rope specs (576 ft-lb/lb). Not their best engineering foot forward here.)
 
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RBJtree

Well-Known Member
Thanks. Twas very interesting (for a geek) reading. I am reluctant to quibble with Yale, but I really think their approach is a bit goofy. It is certainly possible to state the breaking strength of any rope in terms of energy; you just multiply the breaking force times the distance the rope stretched. But that stretch distance, of course, depends on the rope length, or, in Yale's usage, the weight of the rope. Thus they must end up dividing by the weight.

So (really geeky alert) here is what I think Yale is doing. They have that apparatus that allows them to drop a known weight from a given height. They do that for several thicknesses and lengths of rope and come up with an energy of free fall (in ft-lbs) that breaks each rope and then they divide by the weight of the rope. That's an easy test, and the number can be used in their (misguided imo) calculations so they just state that number.

My big quibble with them is that their calculations (found at http://www.yalecordage.com/featured-industries/arborculture/dynamic-energy-arborist-rope ) completely ignore the gravitational energy released as the rope stretches. It is MUCH more than the 1% effect that they claim. To see this, take an extreme case. Suppose you tie off a 3000 lb log (which may be below the working strength) such that it did not fall any distance before starting to tension the rope. The energy of free fall is 3000 * 0 = 0, which Yale claims is the same as the rope's energy absorption and which the rope could certainly handle. In reality when you released the log, it would greatly stretch the rope until a restoring force of 6000 lbs was reached, which could be above the working strength. 1% is just silly.

(Finally, Yale's sample calculations are pretty sloppy. They either have serious typos that leave out a decimal point, or they are multiplying by 100 instead of dividing. Their units are wrong (using ft-lb instead of ft-lb/lb) and the data they use for the polydyne rope energy absorption (1040 ft-lb/lb) is inconsistent with the value in the rope specs (576 ft-lb/lb). Not their best engineering foot forward here.)
The energy absorption numbers for both examples they provide are way off of what is in the individual rope data. That sucks because I really want to understand this stuff. The manufacturer should be a trust worthy source of info. Maybe you are a trust worthy source and you know exactly what you are talking about, but I have no proof of that and I wouldn't bet my livelihood and the saftey of my crew and my customers property on it. I want verifiable sources of info that can be trusted.
 

TreeFan

Member
I'm right; Yale's wrong. What else do you need to know? (Just kidding.) But seriously, Yale is the only company I've found that provides rope specs in such an unusual fashion.

My calculations and Yale's calculations won't disagree by much, and in the real world other factors will be more important than the differences between our equations. You have to accurately estimate the weight of the log, the length of the rope, and the length of free fall. The rope's properties change with temperature, humidity, age, and use. Then you need to consider also the strength of the spar, the anchor sling holding the pulley, and the knots in the ropes. The original post wanted a calculator, which my spreadsheet sort of provides, but I think any calculator should be used with caution. (If there really is demand for such, I could make an app with a pull-down box to choose a rope, but I doubt it is worth my time.)

Teufelberger produced a really useful document with great references dealing with this stuff. (I think it is required by the European Union). https://www.teufelberger.com/pub/media/contentmanager/content/downloads/6800504_02-2016_Herstellerinformation-und-Gebrauchsanleitung-Rigging-Seile.pdf
They helped sponsor a guy doing his master's thesis on this topic and here is the simplest conclusion (there are more technical ones) drawn from that work:

--- "Accordingly, to ensure that cordage components do not fail when subjected to impact loads, the breaking load of the anchor sling in the chosen configuration must be more than 9-20 times the mass of the log and the breaking load of the rigging rope in the chosen configuration must be more than half the breaking load of the anchor sling. Choose a sufficient safety factor on top! " ---

(They acknowledge in a footnote that by mass they really mean gravitational force, which is what pounds measures anyway, so we don't need to worry about that.) They suggest a safety factor of seven or so if using the breaking load limit. Or just use the working load limit with no safety factor. I think this is the procedure I would follow if I didn't really just rely on the experience of others on the team.

Sorry about these long, dry posts. Rain day today and research is fun.
 
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Keeth

Active Member
For the record, TreeFan is correct. With the Conservation of Energy the calculation really is that simple.

There is a factor we have neglected: the size (distribution of mass). So far the log has been treated like a point mass that is dropped straight down. However, we are felling cylinders, often times with an initial effort from a tag line pulling the piece over. In essence this is a physical pendulum problem. For example: a 6 ft poplar log with a diameter of 14 in and a 3 ft log with a 22 in diameter both weigh around 300 lbs. If the block is set 2 ft below your cut, one acts like a pendulum of 5 ft while the other has a length of 3.5 ft with respect to the Center of Mass.

In general, if you multiply the distance from the center of mass of the log to the lowering point by two (the fall distance) and the weight of the log, you’ll have the amount of work you’ll needed to bring the log back to rest. For our 6ft, 300 lb log above, we will have 5 ft x 2 x 300 lb or 3000 lb-ft of energy. With 500 lbs of tension in the lowering line we will have 200 lbs available to slow the log to a stop needing only 3000 ft-lb/ 200 lb or 15 ft to do so.

By comparison, the shorter log will have a fall distance of 3.5 ft x 2 = 7 ft. The work needed to stop it will be 2100 lb-ft. With the same 500 lb of tension, the log will only need 2100 lb-ft/200 lb = 10.5 ft to stop.

The force exerted on the rope is pretty much determined by the way you limit the length of the lowering line. If you lock it off, the stretch of the rope is your distance and the force is huge. If you “let ‘er run” and smooth it to a stop, the force won’t need to be much larger than twice the log’s weight to stop it in a reasonable distance.
 

Bart_

Member
So just to be clear, does anyone on this thread wear a shirt pocket nerd-pack with pens, rulers, mini slide rule etc? Seems like a can of smart has been opened. Kind of antithetical to sweat, noise, engine hp and the rending of wood fibers. But then again we all know what others see is not what's going on mentally when we do tree work.

To me the voodoo of it is the variability of the rope stretch and energy absorption and the chancery of rigging tip, spar(s) and hidden rope faults. There is artistry in being a good ground man who fiddles these scenarios on the fly, including tuning the fall to the transients of the tree and load swing. A good safety factor has basis in real stuff. Neat to see bonafide research on it and that the research reinforced the rules of thumb.

Best wishes.

By the way, the bionic man thing is because you know what would happen to a limb sucked into a portawrap stopping a log - no dead stop, but a lot of damage to the guy.
 

Bart_

Member
Wonder if one could work backwards from the rigging rope. Most recommend roughly 5:1 breaking to working load and 1/2" lines are about 10,000 lbs giving 2,000 lbs working. Using the simple ratio method, a 1000 lb log could drop any arbitrary distance free fall as long as you had that much distance again to catch it it applying 2,000 lbs to the portawrap up line. If you want to stop more aggressively, you have to derate the log size. Just use the ratio with the +1G onto the stop force.

For the yale 1/2" 11,000 lb Polydyne their 2,200 lb working load has about 5% elongation at that point. With 100 ft in play (portawrap to block to log) 5% is 5 feet!! of elastic stretch in addition to letting the rope run. At 7.6 lbs/100 ft it is 7.6 lbs of rope and at 576 ft lbs per lb of rope gives 4378 ft lbs max allowed energy absorption as the work of force x distance in stretching the rope. Looking almost dead linear, the area under the 2,200 lb point at 5%/5 ft is 1/2 x 2,200lbs x 5 ft = 5500 ft lbs which is in the ball park vs the by weight number. These numbers are the rope imploding, number of cycles to failure issue. This elongation occurs in a normal gentle rig (if sized right) or a shock loading catch (if sized right e.g. much derated log size with matching more abrupt stop). This is the rope manufacturer guiding you to not ride the rope into a heavy load cycles premature death of the rope.

The other issue to respect is heavy point loading with contact motion causing local instant heating weakening and breakage. You may be within max break strength and have the rope break anyway. That was an initial concern with rigging rings because they were a sharper bend and small thermal mass that might quickly heat at the contact surface.

Short answer: find the weight number for each size rope like 1000 lb log 1/2" rope equal fall and arrest distances and derate log sizes from there and all is good.
 
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