Forces in rigging

[Dan@JBT]

I have been putting together a rudimentary forces calculator for my own use as a trainer I would like someone to check my formulas, and suggest other variables that I may have overlooked, and as I'm sure you would do anyway critique it in general.

I am aware of the many limitations and inaccuracies of a calculator... this is only for training and expressing the importance of letting the rope run etc.

the variables that are taken into consideration are:

mass of log

dropping distance (x the force of gravity)

rope length

rope stretch

branch flex (movement at the high point)

run in the rigging line

the formulas are:

force = mass x acceleration

impact force = mass x acceleration / breaking distance (assuming breaking is even)

so for example a 300kg log dropping 1m and the breaking distance is the sum of limb flex 10cm and rope stretch 45cm (15m rope in the system x 3%rope strech) and the run the log is given through friction device

so if the log isn't let run the impact force would be

300kg x 1m x 9.8/(.1+.45)= 5349N
and at the block it would be 5349x1.66=8879N

if the log is let run even 50cm the impact force would be
300kg x 1m x 9.8/(.1+.45+.5)= 2802N
and at the block it would be 2802x1.66=4651N



any feedback would be appreciated.
thanks.

Dan

 

[imagineero]

You've got some reading to do mate ;-)

http://www.hse.gov.uk/research/rrpdf/rr668.pdf

 

[Stayalert]

braking is way more better than breaking.....just sayin'

 

[yoyoman]

I have been putting together a rudimentary forces .......

rope stretch
.....
impact force = mass x acceleration / breaking distance (assuming breaking is even)

Dan

G'day mate,
Obviously a topic with so many variables to consider.

In your computations that you might consider, right next to rope stretch, is rope elasticity. An ideal rope would absorb the impact over a given length bringing the object to a stop without bounce.
Consider a rope that not only has stretch but elasticity to reverse the course of the falling object.
If an object comes to rest after traveling 3' and then stops, it is only half the force of an object that travels 3', and in the process reverses course. In other words, it hits the end of the rope, travels 1.5 feet down, stops, and reverses course to almost the same speed in that same 1.5 feet but now upwards, same 3' distance traveled. It bounced. Double the force. This is why some test results can be so much higher than just the numbers of a given fall.

 

[PassionForTrees]

I dont know about all that, but a lot of people dont get it when it comes to chunking pieces down about 4 feet to a dead stop.
If you have a 200lb piece drop 4 feet. there are formulas already in place to accurately measure load forces. like such
200 x 4 = 800 + 200 + 1000lbs. So the weight of the piece times the drop distance then add the weight of the piece again gives you the accurate total.
But when you are talking stretch and elasticity??? all I know is it's like guessing then, because there are cycles to failure we dont know when that rope will see it's last run! Keep thinking and improving our knowledge is great, there is so much to learn!

 

[jefflovstrom]

I dont know about all that, but a lot of people dont get it when it comes to chunking pieces down about 4 feet to a dead stop. But when you are talking stretch and elasticity??? all I know is it's like guessing then, because there are cycles to failure we dont know when that rope will see it's last run! Keep thinking and improving our knowledge is great, there is so much to learn![/QUOTE)

Maybe you need an operation's guy,,,,
Jeff

 

[yoyoman]

.....gives you the accurate total.
....

I knew the rule of thumb formula would come up.
Good rule of thumb but it is a rule of thumb, with all the accuracy of a rule of thumb, taking all the variable conditions and coming up with an answer that fits somewhere in all of those variables. In other words, as long as you are using generally accepted rigging practices, with the tools most often used and in a manner that they are most often used, you have yourself a pretty good ballpark number.

 

[PassionForTrees]

I knew the rule of thumb formula would come up.
Good rule of thumb but it is a rule of thumb, with all the accuracy of a rule of thumb, taking all the variable conditions and coming up with an answer that fits somewhere in all of those variables. In other words, as long as you are using generally accepted rigging practices, with the tools most often used and in a manner that they are most often used, you have yourself a pretty good ballpark number.

Well, of course, shouldnt you be using accepted rigging practices and the right tools in the manner they are supposed to be used. I have seen the dynamiter ( dont know if it is spelled right ) used on similar and various pieces and they all very accurate using that formula. problem is most people do not know this. and they are rigging like 600 lbs pieces letting it drop 4 feet on average to a dead stop using a half inch line!!! do the math.... 600 x 4 = 3000lbs!!! holy ####e! we never knew this before. most half inch lines have a 6-7thousand lbs breaking strength, and we are to be working in the 10% of that breaking strength 10 % of 6-7000 is only 6 - 700 lbs and this piece weighs 3000lbs. Cycles to failure comes alot quicker on this, along with trunk / Spar / Crotch failure too! :msp_sneaky:

 

[yoyoman]

Well, of course, shouldnt you be using accepted rigging practices and the right tools in the manner they are supposed to be used. I have seen the dynamiter ( dont know if it is spelled right ) ....

Yes I agree you should. Is rigging set at 15% WLL?

Some interesting quotes about this rule of thumb.
Thanks imagineero for the suggested reading, took me a while but I enjoyed it.

"
The origin of this rule is not stated, and Blair warns that it does not mirror the actual physics equations that are required to assess peak loads. It is also unclear from what point the distance of fall is to be measured.

.... Uncertainty about how to measure distance of fall may have led to the assumption that a log of four feet in length would fall a distance of four feet, thus generating a peak force five times its weight (according to Blair’s rule of thumb). This may be an underestimation, as the distance, from the initial position of the centroid to its position after the fall, is likely to be significantly greater than the length of a typical log. ...

The limitations of such rules of thumb become obvious when they are applied to safety assessments in scenarios that do not necessarily match those from which the rules were derived. Shock loading may result in greater force magnification than letting a log run. Stronger rope will have greater stiffness, and its use will increase peak forces when a log is snatched (i.e. not slowly decelerated). Shorter rope length, as the stem is topped down, may also result in greater peak forces if it is not compensated for by shorter sections. At the same time, a short, thick stem will damp the peak force less than a high, slender one.
...
...For stiffer ropes or other fall factors, the deviations from the linear behaviour would make this rule of thumb less reliable than it was observed to be in the present field studies. A force factor of 10, as recommended by Blair (1995), would have exaggerated the forces by a factor of between 40 and 120%


The rule of thumb proposed by Blair (1995) for peak forces in rigging, is that, for each foot the log flies, it gains a unit of weight plus one (cf section 8.1.3). This rule of thumb produced force estimations way beyond those actually recorded when the true distance of fall was considered. ....
Evidently, neither the distance of fall, the dimensions of the log, nor its mass alone, can serve as a basis for reliable force estimation.
...."

 

[Zale]

Is that you Smokey?

 

[imagineero]

Some interesting quotes about this rule of thumb.
Thanks imagineero for the suggested reading, took me a while but I enjoyed it.

I found it to be a bit academic, and not exactly compelling reading. The math went way over my head, but the data and explanations were worth it. There's an often misunderstood interplay between breaking strain, WLL, shock force, doubling of forces at pulley and CTF. My simple interpretation of it is here;

http://www.arboristsite.com/arborist-101/212035.htm

Shaun

 

[Dan@JBT]

You've got some reading to do mate ;-)

http://www.hse.gov.uk/research/rrpdf/rr668.pdf

all 370 pages printed... I'm going to read it cover to cover... when I find the time...

 

[imagineero]

all 370 pages printed... I'm going to read it cover to cover... when I find the time...

Stock the fridge with beer first, the magic learning juice will expand the mind!

 

[yoyoman]

I found it to be a bit academic, and not exactly compelling reading. The math went way over my head, but the data and explanations were worth it. There's an often misunderstood interplay between breaking strain, WLL, shock force, doubling of forces at pulley and CTF. My simple interpretation of it is here;

http://www.arboristsite.com/arborist-101/212035.htm

Shaun

Very well written! Your knowledge of the subject is very impressive and I enjoy reading and learning from your depth of understanding. There is a lot to consider with many variables and complicated scenarios. When it comes to the big math I just ask my son who is a math major.
I don't know as much about it as I would like, I just know you can't solve "it" with one simple rule of thumb.
Good job putting "rule of thumb" and a very complicated procedure in perspective.
Thanks for taking the time. One day soon I want to come "down under" and climb those giant eucalyptus you have there.
Cheers mate.

 

[miko0618]

even if you could get everything else dead on, you wouldn't know the actual weight of the piece. so your going through all of this to find specific calculations with an estimated weight as a foundation. seems foolish. also your teaching it so it should be dumbed down and exaggerated so if they make an error in calculation everything will hold.

 

[pdqdl]

G'day mate,
Obviously a topic with so many variables to consider.

In your computations that you might consider, right next to rope stretch, is rope elasticity. An ideal rope would absorb the impact over a given length bringing the object to a stop without bounce.
Consider a rope that not only has stretch but elasticity to reverse the course of the falling object.
If an object comes to rest after traveling 3' and then stops, it is only half the force of an object that travels 3', and in the process reverses course. In other words, it hits the end of the rope, travels 1.5 feet down, stops, and reverses course to almost the same speed in that same 1.5 feet but now upwards, same 3' distance traveled. It bounced. Double the force. This is why some test results can be so much higher than just the numbers of a given fall.

That is just plain wrong. You do not double the force applied to the rope by giving it stretch.

The force applied to the rope is not doubled, it is stored. When the energy in the log (imparted by gravity) has all the energy sucked out of it by friction and the rope, the stored energy in the rope is applied to the log, and voila! You have bouncing log.

Force=mass x acceleration...true, but the -acceleration applied to the log is exceptionally hard to calculate. The different forces mentioned by others in this thread do not come together in a consistent way, and most have failed to factor in friction during the free fall and loaded-fall. If you are attempting to calculate the force absorbed by the TieInPoint, you must also know the vectors of the load as it is applied, the mass of the TIP itself, and the wind-resistance of the TIP.

Quite frankly, I consider the problem incalculable. It is impossible to measure the variables required to come up with an accurate calculation, unless you do repeated drops of the same log, and then calculate backwards. That, of course, defeats the purpose of doing the calculation.

Much simpler: Know the weight of the wood you are cutting off, know the rated strength of the equipment you are using. Calculate the total momentum-force applied to the equipment at the moment that the rope draws tight. If it exceeds the safe working load for the equipment, don't do it. This is a pretty safe method, since all rigging will have some damping effect, reducing the applied force to less than the rated strength, and ...all is well.

 

[yoyoman]

That is just plain wrong. You do not double the force applied to the rope by giving it stretch.

.....

Much simpler: Know the weight of the wood you are cutting off, know the rated strength of the equipment you are using. Calculate the total momentum-force applied to the equipment at the moment that the rope draws tight. If it exceeds the safe working load for the equipment, don't do it. This is a pretty safe method, since all rigging will have some damping effect, reducing the applied force to less than the rated strength, and ...all is well.

Good points, I just want to understand what you say is wrong.

What I said was that if a falling object comes to a stop in 3', that acceleration will be half of the same object that comes to a stop and also reverses course to the same speed in 3' (1.5' down, 1.5' up) of travel. (example, the wrong kind of rope is used and is not allowed to run, it bounces)

 

[yoyoman]

I have been putting together a rudimentary forces calculator for my own use as a trainer I would like someone to check my formulas, and suggest other variables that I may have overlooked, and as I'm sure you would do anyway critique it in general.
..............

any feedback would be appreciated.
thanks.

Dan

Hi Dan,
Here are some numbers to back you up.
Again, this does not count if a different rope is used and the log bounces.

 

[pdqdl]

Good points, I just want to understand what you say is wrong.

What I said was that if a falling object comes to a stop in 3', that acceleration will be half of the same object that comes to a stop and also reverses course to the same speed in 3' (1.5' down, 1.5' up) of travel. (example, the wrong kind of rope is used and is not allowed to run, it bounces)

Again, that just isn't true.

Consider a different example of similar physical properties: You are in a car, doing 60 mph. You slam on the brakes and stop in 100 feet, or...same speed, same car, but you hit the end of a huge tow strap that stretches down to a stop in 100 feet. You will naturally be snatched backwards at the end of tow strap "stop".

Which method of stopping has the greatest acceleration force? You just went from 60mph to zero in 100ft. The deceleration is essentially the same, although braking tends to stop fastest in the beginning and then fade, whereas the tow strap will stop with the greatest rate at the end.

As I hope you understand, the stopping forces are not different, they just dissipate the energy differently. When the brakes stop the car, the tires burn rubber, the brakes heat up, and the suspension bounces down in the front. Energy is converted to heat. This is what happens to your "run to a stop" log scenario.

When the tow strap stops the car, some of the energy is converted to heating the strap, but most of the energy is used to stretch the strap. When all the energy of the car is spent, the strap yanks the car backwards because the strap has stored the momentum of the car, rather than dissipating it as heat.

The bouncing log acceleration is really two different acceleration calculations. The first occurs when the log is slowing down, and the second occurs when the rope yanks it back. Since the rope cannot possibly store more energy than the initial fall, the second acceleration is smaller than the first.

If you still don't see that bouncing backwards does not double the acceleration force, imagine that the tow strap ruptures, right as the car comes to a stop.

 

[pdqdl]

Hi Dan,
Here are some numbers to back you up.
Again, this does not count if a different rope is used and the log bounces.

View attachment 316195

According to this reasoning, you could create a perpetual motion machine by dropping weights and capturing the excess force when a rebound occurs. It just does not work that way.

The total energy of any falling object scenario is the mass of the object and how far it falls. End of story. After that, you are only calculating where the total energy is being applied. If rebounding doubled the force applied, you would be creating more energy than you started with.