falling log weight formula

Discussion in 'Rigging and Roping' started by texasmossball, Jun 22, 2013.

  1. texasmossball

    texasmossball Member

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    After all these years, I have to say I still do not understand the commonly quoted rule of thumb, often attributed to Don Blair, that says "For every foot an object falls, it gains a unit of weight plus one". Sounds very straightforward. But then the example often given with it is what confuses the hell out of me. That example is: 500 lbs falling four feet = 2,500 lbs.

    Now, either I don't understand math or I don't understand English, but if you say that for every foot an object falls it gains a unit of weight, plus a unit of weight, to me that seems to mean that a 500 lb log falling four feet would hit rigging with an impact of 4,000 lbs (500 plus 500 times 4).

    I realize that is not right, but isn't that what that verbage is saying?

    I also realize that is a crude rule of thumb.

    Any input from those who do a lot of rigging?
     
  2. jomoco

    jomoco Active Member

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    Re: falling log weight formula *DELETED*

    Post deleted by jomoco
     
  3. jomoco

    jomoco Active Member

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    Can't be right or I'd pound a hole in the ground jumping from ten feet!

    There's a progressive element to the calculation though.

    Spidey knows!

    Reg, Tom, somebody!

    jomoco
     
  4. Tom Dunlap

    Tom Dunlap Administrator

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    A long time ago Dave Spencer and Ken Casey/Spidey, did a very accurate calculation of impact loads on rigging points. The accuracy of the math was confirmed using their dynamometers.

    Don's published rule of thumb is plenty accurate to figure out impact loads and size the rigging gear, with proper safety factors.

    Weight of piece/500#
    Drop of 4'

    [500x4]+500=2,500 impact on the rigging point. The actual in the field numbers were very close to that point.

    I've done some demos using small blocks of wood and a 150# game scale. The numbers reflect the same results.

    I wish that ISA had backed up the original discussion forums. Their server crashed and destroyed all of that amazing information.

    Over the years other people have replicated this research. I'd bet that a call to ISA would get you a lead to any articles or research that was supported with ISA funds.
     
  5. joe

    joe Active Member

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    distance unit of weight+1

    0 ft. 500#

    1 ft. 500 +500=1000#

    2 ft. 1000 +500=1500#

    3 ft. 1500 +500=2000#

    4 ft. 2000 +500=2500#
     
  6. moray

    moray Member

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    [ QUOTE ]
    A long time ago Dave Spencer and Ken Casey/Spidey, did a very accurate calculation of impact loads on rigging points. The accuracy of the math was confirmed using their dynamometers.

    Don's published rule of thumb is plenty accurate to figure out impact loads and size the rigging gear, with proper safety factors.

    Weight of piece/500#
    Drop of 4'

    [500x4]+500=2,500 impact on the rigging point. The actual in the field numbers were very close to that point.

    [/ QUOTE ]

    This "rule of thumb" business seems to come back to life every so often no matter how many times I try to shoot it dead.

    What is the rule of thumb for how fast a motor vehicle can accelerate? How far it can go on a tank of gas? How much runway an airplane needs to get airborne? Everyone can see these are ridiculous questions because there are lots of variables that must be specified before any of the questions could possibly be answered.

    For the rigging question, for a given distance of drop, by far the most important variables are the stretchiness of the rope and the total amount of rope available to absorb the energy of the fall and stop the load, yet no one has mentioned either of these!

    Let's reverse analyze the example given by assuming the load starts falling right at the anchor point and falls 4 feet. At that point the rope goes taut and begins to stretch. The rope keeps stretching and the load keeps slowing down until finally it stops. At that point we measure 2500 lbs. force on our dynamometer. Since we know nothing about the rope we have to make some reasonable assumptions. Most ropes behave more or less like springs, that is, the amount of stretch is directly related to the force applied (look on the Yale Web site for some excellent graphs). What this means is the rope's stretch at 2500 lbs. is twice what it is at 1250 lbs. We can now say the average force the rope felt over the full distance, x, that it stretched, is 1250 lbs. Applying a little algebra we can solve for x and find that it is 2.67 feet.

    The "rule of thumb" is looking really bad right about now, because none of us owns a rope that can stretch 66% of its length without breaking. That would take a bungee cord. This result also implies that the actual force on a real rigging rope that could withstand the 4-foot drop would be much much greater than 2500 lbs.

    There is no rule of thumb. You simply must have at least a ball-park idea of the stretchiness of the rope, usually clearly indicated in the online catalogs, the expected distance of fall, the weight of the load, and the total length of rope available to stretch. There are just too many factors here to lead to a useful rule of thumb.
     
  7. Tom Dunlap

    Tom Dunlap Administrator

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    Moray,

    If I had saved more info in files...and then be able and willing to retrieve the info I'm sure that I could make a case for the rule of thumb in question is good enough for our discussion.

    For more exacting conclusions you're right, more variables would need to be accounted for. Tree flexing, friction at the rigging point...myriads of small details. In whole number terms this rule of thumb holds up...for me anyway.

    I've never broken a rigging rope or rigging anchor point without using it to lift a load using a machine.
     
  8. Norm_Hall

    Norm_Hall Active Member

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    Rope construction, diameter and fiber all come into play, along with the amount of rope 'in the system'.
    During demonstrations, Todd Kramer and I have gotten readings up to 7X dropping a 15 pound log 1 foot on 1/2" stable braid with about 15 feet of rope 'in the system, measuring with a Dillon ED2000 digital Dynamometer.
     
  9. tuttle

    tuttle Active Member

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  10. texasmossball

    texasmossball Member

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    Thanks everyone. I understand there are many variables here, including the ones mentioned such as rope stretch, but really, I am just trying to understand why either the math or the wording just doesn't seem right in this commonly quoted chestnut.

    Thanks, Joe, for "doing the math". That is how one arrives at the supposedly correct answer in the example given. But IMO the wording can lead one to calculate it another way, too.

    Now I KNOW the following math is NOT CORRECT. But the wording says "for every foot... it GAINS A UNIT of weight PLUS one (unit of weight)". So it falls a foot. It gains 500, plus another 500, equaling 1000. Then it falls to two feet. It GAINS a unit PLUS a unit, equaling 2000, and so on for each foot as follows:

    500 lbs falling to 4 feet:
    @ 1 ft: 500 + 500 = 1000
    @ 2 ft: 1000 + 500 + 500 = 2000
    @ 3 ft: 2000 + 500 + 500 = 3000
    @ 4 ft: 3000 + 500 + 500 = 4000.

    After bouncing this off a few folks who aren't in the tree business I have decided that this wording could confuse others and should be a lot more precise, even for something that is supposed to be a quick estimating tool. Either that or I'm being OCD here and this kind of thinking simply qualifies me for the ITCC rules committee... [​IMG]
     
  11. Tony

    Tony Well-Known Member

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    Mossball,

    You can't be on the Rules committee until you put your acronyms in the correct order. CDO not OCD! [​IMG] Don't laugh at me I didn't put the alphabet in that order!

    Tony
     
  12. chris_girard

    chris_girard Well-Known Member

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    For those of you who were wondering about the origin of Blair’s Rule of Thumb. "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 plus one. EXAMPLE: 500 pounds falling four feet will hit the rigging at about 2,500 pounds." (Blair 1995).

    Here’s what Don told me a few years ago: “I learned it originally from L. Dean Stringer, the owner of Utilities Safety Supply in Lee's Summit Missouri. He said that he had been taught the formula from a utility lineman. I was advised to mention the more good than harm caveat years later from Susan Cook, a rep for Yale Cordage years ago.
    Apparently, the formula has been accepted. It is in Yale's newest catalog.” (Blair 2009).
     
  13. mark_sauls

    mark_sauls Member

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    CYA formula is force=10X weight of chunk for negative block rigging. Where this conversation usually breaks down is the weight of the wood. As a profession this a a weak spot in our expertise. I think 500# falling 4' would generate more than 2500# I would estimate 2500-4000 depending on all the variables not specified with a rule of thumb. How many beers did the brakeman have last night? Angle of deflection at the block? Any limbs attached to dampen? How much rope in the system is a big one. The ol static vs dynamic loads. How many of you actually weigh the chunk? Then how do you know it was 500?
     
  14. yoyoman

    yoyoman Well-Known Member

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    This is a good discussion and maybe I can add to understanding the question. I will say that I do not like rules of thumb as they are only an approximation and when dealing with math and science that is almost always wrong, huge assumptions are made but it can be better than nothing.
    So your question is in regards to the accumulated force of gravity. Gravity is simply a constant force of acceleration. With a falling object that force can be accumulated and saved. You too can apply a constant force of acceleration just like gravity. And if an object is allowed to move freely without friction the characteristics will be the same as your falling log. Now lay your 500 pound log on a frozen frictionless lake. If you apply 500 pounds of force for 1 foot you have accumulated 500 foot pounds of force. Another foot you have accumulated another 500 pounds of force, after 4 feet you have accumulated 2000 pounds of force. That log now contains all 2000 pounds of your force stored within it. If your log impacts an immovable object, all 2000 pounds of force will be exerted at one moment. Because the object is being supported by the ice we do not need to add the +1. This immovable object is similar to a static rope with no stretch. If you were able to get to the other side of your log, and apply 500 pounds of force in the opposite direction for 4 feet, you could bring that log to a complete stop with no more than 500 pounds of force. This would be equivalent to a very dynamic rope that stretches the equal distance of the distance traveled.
    So you see that rule of thumb is complete dependent on the absorption, elasticity or lack thereof in each rope.
    The horizontal example makes the vertical a little easier for me to understand.
     
  15. Daniel

    Daniel Well-Known Member

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    Blair's formula does not take into account stretch in the rope or allowing a piece to run... don't kick me.. after all I heard it on Treebuzz... one other thing that is important on big limbs is the wind resistance.. I saw that on a video...
     
  16. Daniel

    Daniel Well-Known Member

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    whoops,
    I didn't read yoyo's post til just now..
    pretty much "what he said"
     
  17. yoyoman

    yoyoman Well-Known Member

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    [ QUOTE ]
    ..... one other thing that is important on big limbs is the wind resistance.. I saw that on a video...

    [/ QUOTE ]

    Indulge me by allowing me to apply my example to Daniel's comment.
    You're pushing on that log with 500 pounds of force, in other words, if you put a scale on the side of the log it would read 500 pounds. Let's say you push for 50 feet, all the while the log is accelerating and at the same rate. Now at the 50 foot mark, someone comes to the other side of the log and pushes against you with the same 500 pounds of force.
    So now the log does not continue to accelerate, nor does it stop. It will simply continue at the same speed forever unless something changes. Just like a falling branch from the tree reaches its terminal velocity when air friction equals the force of gravity.
    Again, severely effecting the rule of thumb.
     
  18. moray

    moray Member

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    [ QUOTE ]
    Here’s what Don told me a few years ago: “I learned it originally from L. Dean Stringer, the owner of Utilities Safety Supply in Lee's Summit Missouri. He said that he had been taught the formula from a utility lineman. I was advised to mention the more good than harm caveat years later from Susan Cook, a rep for Yale Cordage years ago.
    Apparently, the formula has been accepted. It is in Yale's newest catalog.” (Blair 2009).

    [/ QUOTE ]

    The above forensic trail for this ridiculous formula made me smile--I'm sure I could find a better formula under a park bench somewhere...

    The comment about Yale's catalog, on the other hand, is not a smiling matter; if a serious rope manufacturer says it's a good formula then a math-challenged arborist might well take the formula to the bank and never consider the matter further.

    I have been a fan of the Yale catalog since before 2009 and not only never saw the crappy "rule of thumb" there but would have been shocked if I had. The main reason I was a fan is because they publish beautiful charts of rope performance (stress vs. strain curves) and a very intelligent (and correct) discussion of dynamic loading of arborist ropes. The current dynamic loading discussion is a little bit different from the first version I remember seeing but it is about as simple as you can make it without sacrificing accuracy or completeness.

    Two of their rigging rope charts are worth looking at: one for double esterlon and another for polydyne.

    Both ropes have a working load equal to 20% of breaking strength. The charts show that when double esterlon is stressed to 20% of breaking strength it has only absorbed something less than 4% of maximum energy capacity! Polydyne, stressed to 20%, aborbs far more energy, nearly 11% of capacity. In absolute terms, per pound of rope at the working load limit, polydyne absorbs 1040 ft-lb vs only 291 ft-lb for double esterlon. If you are going to drop loads on a snubbed-up rope, or have a ground crew who can't be trusted to let it run, or simply want to drop bigger loads, you might want to choose a rope that can absorb a lot of energy vs one that can absorb very little.

    dynamic loading: http://www.yalecordage.com/arborist-rope/dynamic-energy-in-arborist-rope.html

    double esterlon: http://www.yalecordage.com/arborist-rope/rigger-s-line/double-esterlon.html

    polydyne: http://www.yalecordage.com/arborist-rope/rigger-s-line/polydyne.html
     
  19. Daniel

    Daniel Well-Known Member

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    [ QUOTE ]
    If you are going to drop loads on a snubbed-up rope, or have a ground crew who can't be trusted to let it run, or simply want to drop bigger loads, you might want to choose a rope that can absorb a lot of energy vs one that can absorb very little.

    [/ QUOTE ]

    Pure genius .. I wish I had thought of that!

    ps... thanks Moray
     

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