Trees and Physics

[ddhlakebound]

Greetings all, ddhlakebound suggested I take a look at this thread because I am a physicist to see if I could contribute.

I should perhaps preface any remarks with the fact that I am a home miller (076AV with 52" mill) and know very little about the practicalities of bringing down heavy lumps of wood from a tree. My only direct experience in this regard was some 20 years ago when I was in a hurry and tied up a 20 ft long x 8" branch with a piece of old rope before cutting the branch with a bow saw. What I didn't take into account was the levering action of the length of the branch so the rope snapped and the branch fell - only about 3 ft, flat onto the neighbors wooden shed. The shed made a couple snapping sounds and wobbled a bit but stayed upright. I managed to get the branch off the shed without any further damage. A year or so later we had a big storm and the shed collapsed!

OK - that aside, I can still help with concepts and formulas. To really understand these problems with, diagrams would be useful - you would all get marks docked if you were in my Physics 101 class. Moray's post on scale drawings is a good one but would really benefit from a diagram or two.

Bobemoto, I'm not quite agreeing with the results on your table .
Lets start by seeing if my diagram represents what you are calculating.
For a 100 lb lump (or for that matter any lump) falling freely from rest over any distance S will reach a velocity of SQRT(2 x g x S).
g = 32.2 ft/s/s/ for most placed on the earth.

When S = 12 ft , v = SQRT(2 x 32.2 x S) = 27.8 ft/second

To decelerate this to zero velocity over distance d =1.27 ft requires a deceleration of v^2/(2d) = 28^2/(2x1.27) = 309 ft/s/s

Now here is where you use F = m x a.

The force F, generated on the rope during the deceleration = mass x deceleration = 100 x 309 = 30900 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 959 lbs PLUS the actual 100lb mass of the lump = a total of 1059 lbs (Bobemoto gets 1043 lbs) Bobemoto , are you taking rope/pulley friction - ie non free falling, into account?

I hope this helps. BTW there is some very good physics software out there that simulates these situations including angles and rope elasticity very well, see something like http://www.knowplay.com/science/interactive-physics.html

Like others have said, none of this will predict super accurately what happens in reality but you will gain some idea of what to expect. I believe a good experiment under controlled conditions will also give you other ideas and help build your experience.

Happy to try to answer other questions

Cheers

Sweet!

Thanks for joining us, and I think you've done a great job with your whole post, you've cleared alot of things up, and communicated VERY clearly in doing it.

Now let me give it a try. I'm sure I'll be missing some of the variables, but this first run is basic.

When S = 3 ft , v = SQRT(2 x 32.2 x S) = 13.86 ft/second

To decelerate this to zero velocity over distance d =1 ft requires a deceleration of v^2/(2d) = 13.86^2/(2x1) = 96.05 ft/s/s

To decelerate this to zero velocity over distance d = 2 ft requires a deceleration of v^2/(2d) = 13.86^2/(2x2) = 48.02 ft/s/s

In the 1ft deceleration, the force F, generated on the rope during the deceleration = mass x deceleration = 425 x 96.05 = 40821.25 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 1267.74 lbs PLUS the actual 425lb mass of the lump = a total of 1692.74 lbs

In the 2ft deceleration, the force F, generated on the rope during the deceleration = mass x deceleration = 425 x 48.02 = 20408.5 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 633.8 lbs PLUS the actual 425lb mass of the lump = a total of 1058.8 lbs

Not taking rope/pulley friction - ie non free falling, into account. I wouldn't even know how to begin to calculate/apply those variables....??

Is this correct?

 

[BobL]

Not taking rope/pulley friction - ie non free falling, into account. I wouldn't even know how to begin to calculate/apply those variables....??

You got it!

The example you chose is a good one because it demonstrates that a 100 lb log falling 12 ft and slowing to zero in 1.27 ft (my example) applies close to the same effective force on a rope as a 425 lb log falling 3 ft and stopping to zero in 2 ft (your example).

Rope and pulley friction will vary. In all cases it will reduce the falling velocity which in turn will reduce the effective final force on the rope in the deceleration phase. I've seen arborists wrap rope around other branches to uses as pulleys (I don't know if this is considered OHS appropriate or not) in this case the friction can be very significant.

It sure would have been nice if the guy who cut down my 60 ft gum tree in my back yard had even stopped to think about how much damage he was likely to do by bouncing 50 lb lumps of tree off my brick paving.

Now here's a physics type question for ya. What is the force on a rope while it is being used to lower a 100lb piece at 2 ft/s and at 10ft/s?

Cheers

 

[Pilsnaman]

May be too late but

Sorry I didn't get into this earlier and all this may have already been taken care of, I didn't go through each post word for word probably because I forgot my Adderall today (problem with ADD is you get distracted and can't remember to take your ADD medication).
Anyway, I figured as a mechanical engineer and someone that did tree work for a year this may be a subject I can help with. Plenty of others have already shown how the basic calcs are done so here is a little more info. As has already been stated the base equation used is Force = Mass x acceleration. It should be noted that in dynamics you rarely hear the word deceleration, instead it is called a negative acceleration. So what we care about is the rate at which velocity is changed (in our case reduced) from the time force is applied to the lowering line to the time our log reaches a velocity of 0 ft/sec. What we need is the maximum acceleration during this time to determine the maximum force applied to the rope, tree, and block. Here is where we just don't have enough information to perform the calculations required for this problem. A rope is acting as a spring when it begins to deform (stretch) but like most springs in the real world the rope does not act as a linear spring, the force it applies increases as it elongates. In the end we only have the ability to do basic approximations due to this lack of information. Something I learned a long time ago is books and calculations will only bring you so far, experience is required to bring you the rest of the way. When I did tree work my first thought was to try and calculate this but quickly realized the experienced climbers knew what to use when and how much wood to take safely. This leaves room for human error but thats part of the job, we can't calculate how strong the dead tree we are removing will be.
One other thing, if anyone is confused about something or has any other engineering related question feel free to ask. Not saying I will always know the answer but the sun shines on a dog's a$$ every now and then.
While this post probably doesn't give the answers you guys want, the numerical kind, hopefully it is somewhat insightful.

 

[moray]

Good thread, it is nice to see people doing equations!

The calculations in the last few posts show how to figure the velocity of the falling load at the moment the rope goes taut, and from that and the stopping distance how to calculate the acceleration required to stop the load. Knowing the mass of the dropping load, it is then straightforward to derive the force.

But if you then go climb your tree and drop the 425 lb chunk, expecting, as calculated, that the rope will experience a force of 1692.74 lbs, you are in for a rude surprise. The actual maximum force, which is the thing we really want to know, is twice that, or 3385 lbs. Why? Because the unspoken assumption that the falling load will be stopped by a constant acceleration (96.05 ft/s/s) is false. We are talking about catching the load with a rope, are we not? We are also talking about the uncluttered simple case of the rope alone doing this work--there is no ground person with a PortaWrap letting the load run.

The rope acts like a spring. Over a usefully large range, from zero up to about 1/3 of the breaking strength of the rope, it is a pretty good spring in that there is a very good correspondence between applied force and amount of stretch (a linear relationship).

In the diagram the upper weight is positioned where the rope first goes taut. The rope begins to stretch as the load continues downward, but the further the load travels, the more the rope stretches, and the more upward force the rope applies to the load. When this force exceeds the weight of the load, the load begins to slow down. The force, and therefore the acceleration of the load that is slowing it down, continues to increase in linear fashion until the load has stopped, represented by the lower weight in the diagram. At this point the rope experiences the maximum force and the load experiences maximum acceleration upward. The load will bounce, but because of large heat losses in the rope, the bounce height will be very low compared to drop height.

The triangle diagram to the right of the two weight positions shows how force on the rope (horizontal scale) varies with degree of rope stretch (vertical scale). The maximum force, at the bottom, is twice the average force.

...
What we need is the maximum acceleration during this time to determine the maximum force applied to the rope, tree, and block. Here is where we just don't have enough information to perform the calculations required for this problem. A rope is acting as a spring when it begins to deform (stretch) but like most springs in the real world the rope does not act as a linear spring, the force it applies increases as it elongates. In the end we only have the ability to do basic approximations due to this lack of information.

I have to partly disagree with the above statement. The rope manufacturer's web sites (Samson Rope, for example) do provide enough information to show that the rope is a roughly linear spring, and give the numerical relationship between force and stretch. I think these numbers are plenty good enough to use in our calculations.

 

[ddhlakebound]

You got it!

It sure would have been nice if the guy who cut down my 60 ft gum tree in my back yard had even stopped to think about how much damage he was likely to do by bouncing 50 lb lumps of tree off my brick paving.

Now here's a physics type question for ya. What is the force on a rope while it is being used to lower a 100lb piece at 2 ft/s and at 10ft/s?

Cheers

Thx BobL...

Examples like your 60' gum tree are why were here trying to understand this stuff better. The customer wants "no damage", but not damaging the LZ means we've either got to build some sort of a cushion to absorb the damage, or rope it to the ground. In big wood, weight compounds quickly. A 34" diameter, 18" long piece of red oak weighs in at roughly 576 lbs. A 16" 6 foot log weighs in at 510#.

So anytime that there is no overhead rig point, we're faced with dynamic rigging to lower it. And as has been noted several times in the thread, even if we can calculate all the forces in the system, there's no way to "rate" the tree.

On lowering a 100# piece @ 2fps, or at 10 fps, since the velocity is constant, only the actual weight of the piece would factor into the force on the rope, right?

So on both 2 & 10 fps.....100# on the working leg, and 100#-friction in the rig point on the running leg.

Hope I'm still passing.....:biggrinbounce2:

 

[lxt]

All these charts are wonderful, math seems to be on, however noone has considered how old is the rope? & the cut force placed upon the rope by certain tree`s; example clove hitch while butting down a sycamore log as opposed to butting down a shag bark hickory! would this matter?

I always try to give a best guess & if worried take a smaller peice or use a 5/8 instead of a 1/2 rope. just wondering!!

LXT...............

 

[Joe]

http://www.arboristsite.com/showthread.php?t=7385

The information in this thread may also help. I had fun in it.

Joe

 

[moray]

http://www.arboristsite.com/showthread.php?t=7385

The information in this thread may also help. I had fun in it.

Joe

Good link, this is a fun read!

I have to confess I couldn't keep from smiling while reading many of the posts, but I was openly admiring the posters for trying heroically to come to grips with a tricky topic.

Besides a persistent confusion about the meaning of things (mass, energy, velocity, force, acceleration) the biggest problem seemed to be that very few people understood that the impact force from dropping a load cannot be calculated without knowing the distance used up in stopping it. I especially didn't like the several references to some rule of thumb connecting impact force with weight of object and distance of drop. Such a rule could only work if everyone always used the same type of rope.

But here is a rule of thumb for calculating dynamic loads when dropping chunks of wood on a pulley where the groundie can let it run. For a load, W, that drops a distance, D, and then runs a distance, R, the load on the rope will be W + (D/R) x W.

Two examples: Drop a load 5 feet and let it run another 5 feet. The maximum load on the rope will be twice the weight of the load.
Drop a load 5 feet and let it run 2.5 feet. The maximum load on the rope will be 3 times the weight of the load.

The basic idea here is that if gravity has 10 feet in which to accelerate the load, and you have 5 feet in which to stop it, you have to supply twice the acceleration of gravity, or 2g, to stop it because you have half the distance in which to do it. If you must stop it in 2 feet, then you will need 5g of acceleration to do it. Since you must still support the load when it has stopped moving, you must add W at the end to get the correct value.

This simple method has an optimistic assumption behind it: the groundie is able to supply completely constant acceleration to stop the load. This certainly won't be the case, but it might be pretty close in the case of an experienced operator.

 

[ddhlakebound]

But if you then go climb your tree and drop the 425 lb chunk, expecting, as calculated, that the rope will experience a force of 1692.74 lbs, you are in for a rude surprise. The actual maximum force, which is the thing we really want to know, is twice that, or 3385 lbs. Why? Because the unspoken assumption that the falling load will be stopped by a constant acceleration (96.05 ft/s/s) is false. We are talking about catching the load with a rope, are we not? We are also talking about the uncluttered simple case of the rope alone doing this work--there is no ground person with a PortaWrap letting the load run.

The rope acts like a spring. Over a usefully large range, from zero up to about 1/3 of the breaking strength of the rope, it is a pretty good spring in that there is a very good correspondence between applied force and amount of stretch (a linear relationship).

Moray, I'm glad you brought this up. I hadn't considered that the acceleration wasn't constant. I do think you've made a couple minor errors after some reflection. I may be wrong.

I think that the first error is that you can't just double the final load number, you must remove the static 425 lbs, then double, then add the 425 lbs back in. So we get 1692 - 425 x 2 + 425 = 2959 lbs.

We can double check this by changing the acceleration distance to zero in the equation.

When S = 3 ft , v = SQRT(2 x 32.2 x S) = 13.86 ft/second

To decelerate this to zero velocity over distance d = 0 ft requires a deceleration of v^2/(2d) = 13.86^2/(2x0) = 192.1 ft/s/s

In the 0ft/instant deceleration, the force F, generated on the rope during the deceleration = mass x deceleration = 425 x 192.1 = 81642.5 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 2535.48 lbs PLUS the actual 425lb mass of the lump = a total of 2960.48 lbs

So the absolute peak load possible in dropping a 425 lb block 3 feet, then stopping it instantly is 2960.48 lbs.

And since the rope we are using does have some stretch, we won't be stopping the load instantaneously. So now I guess we need to apply some other equation or algorithm to determine the actual peak load, which will be somewhere between 1692 and 2960 lbs.

Any way you look at it, that's an awful lot of weight generated dropping 425# a mere 3 feet.

 

[moray]

We can double check this by changing the acceleration distance to zero in the equation.
...
So the absolute peak load possible in dropping a 425 lb block 3 feet, then stopping it instantly is 2960.48 lbs.

ddhlakebound, good reply--I'm glad you're still trying to wrap your mind around this. It is also good you are trying to attack this from various angles, like a true mathematician!

Before I show how I would prefer to understand and solve this problem, let me first comment on your second sentence quoted above. To stop something instantly as you suggest means instantly changing its velocity from some positive number to zero. To change the velocity of something is to accelerate it. To do it in zero time requires an infinite acceleration. Since F=M x A, your proposed situation implies infinite force! I am not sure what you intended here, but this cannot be it!

How would I solve this? We know that when the 425# block has finally stopped, it has dropped 4 feet. During the last foot, the rope was stretching and catching it. We know the rope is like a spring, and that the distance it stretches is proportional to the force applied to it. This force reaches its maximum value when the block is at the 4-foot position, and it is zero at the 3-foot position. At intermediate positions the force is intermediate also. The graph of this relation is a right triangle (see my diagram in post #44).

When the block has descended the full 4 feet, it has given up 4ft x 425# of energy, or 1700 ft-lbs. Where does this energy go? Into stretching the rope. The equation connecting energy and force is E = F x D, or energy equals force times distance. The rope absorbs 1700 ft-lbs of energy while stretching one foot. The average force on the rope during its one-foot stretch will be 1700 lbs. That gives the right answer of 1700 ft-lbs. But we know force starts at zero and increases linearly to some maximum value. This means the maximum must be 3400 lbs.

Another way to say this to say the area of my triangle represents the energy stored in the rope. The only way for this area to be 1700 ft-lbs is for the bottom leg of the triangle, the maximum force, to be 3400 lbs.

And just to make sure you really have a headache, try this experiment: Take the very same rope and the same 425# block of your hypthetical problem. Only this time use 50 feet of rope. Drop the block 50 feet. The rope goes taut and catches the block. Maximum force on the rope? 3400 lbs! (Only assumption here, not quite true in your diagram, is that the length of freefall and length of rope are the same...)

 

[Joe]

Actually, the correct and simplified procedure for figuring the force or tension in a line from a falling anything on a rope is outlined in post #53 of the A.S. link a few posts back. The metric units for distance and acceleration are used there. There are also a few physics links in that thread which one can use to reference introductory concepts applied in that post to calculate the final force. It's there but one needs to spend the time with it.

Sherrill, a sponsor of this site, sells rigging software which can show how loads affect different rope constructions when rigging. It's on page 47 of the 2007 master catalog. It will help.

Here's another link which shows how to figure fall factors into the anything falling while tied to a rope problem.

http://student.kuleuven.be/~m9916724/physics/physics.htm

Joe

 

[moray]

Actually, the correct and simplified procedure for figuring the force or tension in a line [...] It's there but one needs to spend the time with it.

Sorry, I can't let this go by. There are as many correct ways of solving a problem as you can dream up, and the more the better! In fact, WAY better. I like your other comment about spending time with it. To me this means running through the problem various ways, changing parameters, checking the relationship of various quantities, etc. In other words, solve it and related problems many times in many different ways and pretty soon you will truly understand it.

While there are usually many different ways to solve a problem, that doesn't mean they are all equally good. In regard to the problem at hand, even though it can be solved using F = M x A and a little knowledge about springs, it is much cleaner and easier to use the energy relations, as I did above. But do it both ways! The more the better. The results should be the same. Knowing how to attack the problem by more than one route just deepens one's understanding.

Now I'm sure you didn't mean to attach any particular significance to the word "procedure" in your post, but it brought up bad memories of high school math classes where a poor student would ask a good student "What's the formula for such and so?" The useful but unwelcome answer would be "You need to actually understand this stuff so you don't have to memorize some formula or procedure. How will you know which formula to use and when?" Who would you trust to calculate some rigging problem for you--some one who had memorized some formulas or procedures but didn't really understand the physics, or someone who really understood the subject but had memorized nothing?

 

[Pilsnaman]

Just a little note

Moray, I am sorry to say that the rope does not act as an ideal spring just many springs don't. Where on the rope companies' websites do you see such information given (like a spring constant perhaps)? Going to the Sampson rope website and looking at the Arbor-Plex they only give information on the elongation percentage with respect to breaking strength percentage. Now take those three data sets and plot them, do you see a straight line? If you answered yes then you did something wrong because that is the only way this is linear. In reality the relationship between elongation and force is not linear but parabolic or exponential. What you are thinking of is Hooke’s Law of elasticity, which you can look up on-line as the idea that the force applied by a spring is linear to the spring’s deformation. The problem is that this approximation is only applicable to certain kinds of material and synthetics (like rope made of polyester and ployolefin) are not applicable. This is reaffirmed by the following calculations I made:

Lets say we have a 200 lbm (pound mass) piece of wood that falls 15 ft, at which time the rope is held tight. The block is located at the same point as the wood’s starting point and 50 ft from the ground. Assume the wood is dropped perfectly vertical and the rope is tied off to the base of the tree and is perfectly vertical. Also, lets assume there is no friction in the block, that the rope going through the block will have no effect on its elongation, and the rope’s weight is negligible compared with the wood.
The point when all slack has been removed from the rope will be 0 ft. According to the Sampson webpage, ½” Pro Master has an average strength of 6300 lbf (pounds force) and shows a table of percent elongation with respect to percent breaking strength. Using this table you can find the following:
x0 = 0 ft; x1 = -1.3 ft; x2 = -2.08 ft; x3 = -2.54 ft
F0 = 0 lbf; F1 = 630 lbf; F2 = 1260 lbf; F3 = 1890 lbf

Where ‘x’ is the distance that the rope has stretched and ‘F’ is the force applied by the rope.
Now we know all know the equation F = m * a, and solving for acceleration gives us a = F/m. we can now solve for ‘a’ at the four points above to determine the acceleration of the mass. Here ‘F’ is the force of the rope in addition to the force of gravity. Note that the gravitational constant must be multiplied to convert units and a positive acceleration will be in the upward direction.
a = F/m
a1 = (630 lbf*200lbf / 200 lbm) * (32.17 lbm*ft/lbf*sec^2) = 69.17 ft/sec^2
a0 = -32.17 ft/sec^2 ; a1 = 69.17 ft/sec^2 ; a2 = 170.5 ft/sec^2 ; a3 = 271.84 ft/sec^2

Using Excel we can plot these accelerations with respect to ‘x’ and have the software extrapolate an equation for this. Acceleration of the wood piece as a result of the rope elongation can be represented as,
a(x) = 35.71x^2 + 26.86x + 31.23

While an exponential equation may also work it does not allow for negative values which is a problem so I went with the parabolic one. Either way this is not a linear spring and the calculations are much more difficult then most are making them out to be.

 

[ddhlakebound]

ddhlakebound, good reply--I'm glad you're still trying to wrap your mind around this. It is also good you are trying to attack this from various angles, like a true mathematician!

Thanks.

Before I show how I would prefer to understand and solve this problem, let me first comment on your second sentence quoted above. To stop something instantly as you suggest means instantly changing its velocity from some positive number to zero. To change the velocity of something is to accelerate it. To do it in zero time requires an infinite acceleration. Since F=M x A, your proposed situation implies infinite force! I am not sure what you intended here, but this cannot be it!

You are correct, I didn't mean to say that we could stop the load "instantaneously". I did mean that *if* we could set up the system to be totally static, so that when the load reached 3 ft, it stops there, with *nearly* zero stretch, that the absolute peak load (425#, 3ft drop) would be 2960 lbs.

How would I solve this? We know that when the 425# block has finally stopped, it has dropped 4 feet. During the last foot, the rope was stretching and catching it. We know the rope is like a spring, and that the distance it stretches is proportional to the force applied to it. This force reaches its maximum value when the block is at the 4-foot position, and it is zero at the 3-foot position. At intermediate positions the force is intermediate also. The graph of this relation is a right triangle (see my diagram in post #44).

This is the part I'm not very clear on. It seems to me, and again, I could easily be wrong, that the more you stretch the rope, the more difficult it becomes to stretch it more, so this doesn't seem to be a linear relationship, but rather an upward curve, from zero stretch, to the breaking point, where no more stretch is possible.

When the block has descended the full 4 feet, it has given up 4ft x 425# of energy, or 1700 ft-lbs. Where does this energy go? Into stretching the rope. The equation connecting energy and force is E = F x D, or energy equals force times distance. The rope absorbs 1700 ft-lbs of energy while stretching one foot. The average force on the rope during its one-foot stretch will be 1700 lbs. That gives the right answer of 1700 ft-lbs. But we know force starts at zero and increases linearly to some maximum value. This means the maximum must be 3400 lbs.

This doesn't seem logical, even though the avg works out perfectly. When the rope catches the load at 3 ft, and begins to stretch, it's consuming energy to stretch. That energy is no longer available for "weight" when we finally bring the load to a stop, so everything which has been consumed in stretching the rope would not be felt when the load comes to rest.

Another way to say this to say the area of my triangle represents the energy stored in the rope. The only way for this area to be 1700 ft-lbs is for the bottom leg of the triangle, the maximum force, to be 3400 lbs.

True if the stretching relationship is linear. How can we determine that it is?

And just to make sure you really have a headache, try this experiment: Take the very same rope and the same 425# block of your hypthetical problem. Only this time use 50 feet of rope. Drop the block 50 feet. The rope goes taut and catches the block. Maximum force on the rope? 3400 lbs! (Only assumption here, not quite true in your diagram, is that the length of freefall and length of rope are the same...)

This does not jive at all. Lets drop the variables into the formulas.

When S = 50 ft , v = SQRT(2 x 32.2 x S) = 56.75 ft/second

To decelerate this to zero velocity over distance d = 0 ft requires a deceleration of v^2/(2d) = 56.75^2/(2x0) = 3220 ft/s/s

In the 0ft/instant deceleration, the force F, generated on the rope during the deceleration = mass x deceleration = 425 x 3220 = 1368500 lbs ft/s/s which is you divide by g = 32.2 ft/s/s you get a weight equivalent of 42500 lbs PLUS the actual 425lb mass of the lump = a total of 42925 lbs

You would have to get up around 15+ feet of stretch from a 50' rope to get the peak load down to anything reasonably manageable. Basically, dropping 425 lbs 50 feet onto any arborist rope will assuredly result in a broken rope, failed rigging, or failed rig point, unless it is allowed to run a whole whole lot.

Maybe we need the physics professor to clear some more stuff up...:help:

 

[moray]

Moray, I am sorry to say that the rope does not act as an ideal spring just many springs don't. [...]
Either way this is not a linear spring and the calculations are much more difficult then most are making them out to be.

This is an excellent post, Pilsnaman! Thanks for checking on me.

First off, I completely agree that my assertion that the rope acts like a spring is just an approximation. As you point out, even a real spring only behaves approximately like an ideal spring, but the numerical error in treating it as ideal is usually small. In the case of rope, it certainly shows spring-like behavior in the sense that it takes an increase in force to produce an increase in stretch. But how linear is this relationship?

I put together the chart below from the Samson Web site. I have looked at these numbers many times because I own 5 of the ropes and have been interested in the others. For each rope, the three numeric columns represent, in order, the % rope stretch at 10%, 20%, and 30% of ultimate break strength. They are, effectively, 3 data points on the curve of stretch vs. force. None of them will plot as a straight line (though Amsteel-Blue and Tree-Master are pretty close). An ideal spring must, of course, plot as a straight line.

One thing jumps out at me. In every case, the first number is larger, often much larger, than it should be. If the spring behavior is linear, the first number should be 1/3 of the 3rd number, but it is always more than that, and in some cases more than 1/2. (Arbor-Plex, by the way, your choice of rope, is by far the worst rope of the lot in terms of linearity.) I think the explanation for this is that up to about 10% maximum load or so the rope fibers are stretching (they are loaded, after all), but the rope is stretching as well. By this I mean the weave is tightening and aligning more with the tension, spaces between the yarns are closing up, and all the slack and slop in the construction is being eliminated. This causes a lengthening of the rope quite separate from the stretching of individual fibers. Once this initial construction slack is largely gone, the rope behaves from then on (higher loads) in a more linear fashion, as the table shows, because from then on almost all of the stretch comes from stretching individual fibers.

If a particular rope actually behaved like a spring, the linear relation between force and stretch means the difference between columns 1 and 2 must equal the difference between columns 2 and 3. Each increment of force must produce the same increment of stretch.

Look at the first 4 ropes. In all of them the second increment is smaller than the first. The rope is growing stiffer under more tension. In the next group of 5, the second increment is larger than the first. The rope is growing less stiff under more tension! The two groups deviate from linearity in opposite directions! This is why I chose to split the difference and treat them all as roughly linear.

The two figures below show the stretch characteristic of first an ideal rope that behaves like a perfect spring, and second a close approximation of a real rope. The horizontal axis in each case represents force as a percent of ultimate break strength, and the vertical axis represents amount of stretch in arbitrary units. In the second figure I took pains to make the stretch at the 10% load equal to half the stretch at the 30% load--somewhat worse than the behavior of most of the real ropes.

What can we conclude? When I look at the two figures I don't see a huge amount of difference, so I am inclined to continue thinking of the rope as a spring. Because, in every case, the rope is much softer at the beginning of its stretch, we need to allow for that in any calculations we make. The area in the upper triangle and the area in the lower "triangle" (with curved hypotenuse) represent the work (energy) done to stretch the rope. Since the area of the lower figure is clearly less than that of upper figure--the ideal case--less energy has been stored or consumed by the real rope. We must stretch it still more by applying more force in order to store as much energy as the ideal rope. The peak force when the real rope catches a load will thus be somewhat higher than what we calculate for an ideal rope. It looks to me if we allowed for the peak force to be 25% greater than calculated, we would be in the ballpark.

Once again, excellent discussion!

 

[moray]

Whew!

I did mean that *if* we could set up the system to be totally static, so that when the load reached 3 ft, it stops there, with *nearly* zero stretch, that the absolute peak load (425#, 3ft drop) would be 2960 lbs.

How fast are you stopping it with "nearly" zero stretch? The governing equation is F=m*a. If a is huge, the F will be correspondingly huge. Inescapable. There is no such thing as an absolute peak load, to put it bluntly.

This is the part I'm not very clear on. It seems to me, and again, I could easily be wrong, that the more you stretch the rope, the more difficult it becomes to stretch it more, so this doesn't seem to be a linear relationship, but rather an upward curve, from zero stretch, to the breaking point, where no more stretch is possible.

You are absolutely right about the rope. I think you are confusing the word constant and the word linear. The relationship is linear if each equal increment of force causes equal increments of stretch. If you keep adding weight to the rope, 10 lbs at a time, and each time the rope stretches one more inch, you have a linear relationship.

This doesn't seem logical, even though the avg works out perfectly. When the rope catches the load at 3 ft, and begins to stretch, it's consuming energy to stretch. That energy is no longer available for "weight" when we finally bring the load to a stop, so everything which has been consumed in stretching the rope would not be felt when the load comes to rest.

Don't mix up weight and energy. They may be related in a particular situation, and it is enormously useful to understand what the relationship may be, but they are different things. The phrase "energy no longer available for weight" seems to imply you expect one to be converted into the other. The motion of the load goes to zero when the rope is fully stretched, all the energy has been transferred to the rope, the force on the rope reaches maximum, and the accleration of the load also reaches maximum.

True if the stretching relationship is linear. How can we determine that it is?

See my long post above as regards linearity. Others may disagree, but I think it is a useful and easy approximation, and we can add a fudge factor to allow for the softness of the rope during early stages of stretch.

This does not jive at all. [...]
a total of 42925 lbs

You would have to get up around 15+ feet of stretch from a 50' rope to get the peak load down to anything reasonably manageable. Basically, dropping 425 lbs 50 feet onto any arborist rope will assuredly result in a broken rope, failed rigging, or failed rig point, unless it is allowed to run a whole whole lot.

I was sure this would get your attention! This idea, known as Fall Factor, is well-known to rock climbers. It basically says, for a given falling mass, it is the ratio of fall distance to length of rope catching the load that matters. Distance of fall does not matter. So, in our case, dropping 425# 3 feet and catching it with 3 feet of rope is exactly the same, as far as the rope is concerned, as dropping it 50 feet and catching it with 50 feet of rope.

And you are right about the 15+ feet of stretch! Your rope from your example stretched 1 foot from the 3-foot fall; it will stretch 16.67 feet from the 50-foot fall.

This is terrific stuff, but you have to live with it for awhile before it clicks.

 

[Pilsnaman]

The fact is we only have three data points for each rope type, not nearly enough to make any conclusion. That being said, we can not assume this to be linear because the change is negligible in comparison. Take a look at the graph below of the acceleration with respect to distance, this for Pro-Master, and note that I have included the trend line. You are correct in saying the acceleration will have a more drastic change at the beginning of the stretch but what happens near the rope's breaking strength? This data is not given by the manufacturer as they only go up to 30%, and we both know people hear bring their ropes past this. We also know that Hooke's Law is not normally applicable to synthetic springs, like the rope we are evaluating. There is a reason for this, synthetics tend to have spring characteristics that are too far from being linear. The simple comparisons you did are not enough to assume linearity, we are not talking high school physics here.

All this aside here is my engineering evaluation of this system... The only way to correctly identify the forces seen at different parts of a lowering system is to use measuring tools and do tests. This is because of the large number of unknown variables seen in the real world. Just to name a few we have the friction at the block, the angle of the rope going from the block to where it is held/tied off, the angle from the block to where the wood first takes up rope slack (no way the wood falls exactly vertical and/or in a vertical line with the block), the amount the tree gives with this load, how much the groundy lets the rope run...
I just don't see how we can assume all of these variables are negligible, especially ones like angle and tree movement.
On a final note, most of the guys on here don't like when John Doe homeowner or landscaper tries to tell them they don't do something right. That's because a professional has knowledge beyond that of a non-pro. As someone with a Mechanical Engineering degree from Virginia Tech and that does mechanical engineering as a job I have spent years working on things like dynamics, statics, mechanics of materials, vibrations, and deformation. When it comes to something involving mechanical engineering I have more knowledge and expertise then say a software developer. I am not trying to brag or say I am smarter then anyone here, there are A LOT of things I am ignorant on but that is why we have professionals. This just happens to be the subject I am a professional in.
P.S. Sorry for the large image but things were not coming through clear when scaled down.

 

[ddhlakebound]

How fast are you stopping it with "nearly" zero stretch? The governing equation is F=m*a. If a is huge, the F will be correspondingly huge. Inescapable. There is no such thing as an absolute peak load, to put it bluntly.

You are absolutely right about the rope. I think you are confusing the word constant and the word linear. The relationship is linear if each equal increment of force causes equal increments of stretch. If you keep adding weight to the rope, 10 lbs at a time, and each time the rope stretches one more inch, you have a linear relationship.

Don't mix up weight and energy. They may be related in a particular situation, and it is enormously useful to understand what the relationship may be, but they are different things. The phrase "energy no longer available for weight" seems to imply you expect one to be converted into the other. The motion of the load goes to zero when the rope is fully stretched, all the energy has been transferred to the rope, the force on the rope reaches maximum, and the accleration of the load also reaches maximum.

See my long post above as regards linearity. Others may disagree, but I think it is a useful and easy approximation, and we can add a fudge factor to allow for the softness of the rope during early stages of stretch.

OK, I finally understand that there's no absolute peak load. Had to run a couple fractional equations to get it.

I didn't mix up weight and energy. Thats why weight was in quotes. When we bring the load to a stop, the kinetic energy is felt as lbf....basically simulated weight.

I too disagree that the rope works in any linear fashion as a spring. I would like the link to where you got those stretch/load figures, I did a quick search and couldn't find them. But based on the limited info from 3 points, it shows to me that the relationship is not linear.

I was sure this would get your attention! This idea, known as Fall Factor, is well-known to rock climbers. It basically says, for a given falling mass, it is the ratio of fall distance to length of rope catching the load that matters. Distance of fall does not matter. So, in our case, dropping 425# 3 feet and catching it with 3 feet of rope is exactly the same, as far as the rope is concerned, as dropping it 50 feet and catching it with 50 feet of rope.

And you are right about the 15+ feet of stretch! Your rope from your example stretched 1 foot from the 3-foot fall; it will stretch 16.67 feet from the 50-foot fall.

This is terrific stuff, but you have to live with it for awhile before it clicks.

Fall factor has alot more to consider than the distance of the fall. It also includes the TOTAL amount of rope in the system. So to figure the fall factor, we also need to know how high up the rig point is.

Rockers use fall factors because they climb on dynamic rope. In the trees most of the rope we use is much more static, so trying to compare the two is definately apple and oranges.

Now, we are not going to get 1 foot of stretch from a 3 foot rope, and we are not going to get 15 feet of stretch from a 50 foot rope. So a large percentage of the distance needed to stop the load from either fall comes from letting the rope run. Is it easier to let it run one foot or fifteen? So how is it the same on the rope?

The mass and the distance of the fall determine how much kinetic energy we must dissipate to stop a load. The distance we stop it in determines how much force we apply to the rope. The distance of the fall matters very very much, no matter how you look at it. More distance = more energy. For it to matter less, you'd need some very tall trees, and a whole lotta rope in the system, to make your ratio's work out.

I'd be very interested to see you successfully drop 425 lbs 50', then catch it on a 1/2 inch arborist rope. Just make sure if you try it that nobody is near the dropzone, cause I bet the rope, the rigging, or the rig point breaks.

Even discussing it muddies the waters, clouding what we're trying to learn here.

 

[moray]

The fact is we only have three data points for each rope type, not nearly enough to make any conclusion. That being said, we can not assume this to be linear because the change is negligible in comparison. Take a look at the graph below of the acceleration with respect to distance, this for Pro-Master, and note that I have included the trend line.

I, too, am saddened by the fact we have only 3 data points per rope. I guess we have to make the best of it. Sorry, I don't understand your second sentence. I like your chart with the best-fit second-degree polynomial. I show it again, below, this time with two straight lines fitted by hand. This shows how we can do a pretty good job fitting the high-load part of the curve with one straight line, and the low-rope-stiffness, low-load part of the curve with another. But for someone just trying to get a general sense of how all this works, I would suggest a single straight line from the y-intercept to the furthermost data point.

You are correct in saying the acceleration will have a more drastic change at the beginning of the stretch but what happens near the rope's breaking strength? This data is not given by the manufacturer as they only go up to 30%, and we both know people hear bring their ropes past this. We also know that Hooke's Law is not normally applicable to synthetic springs, like the rope we are evaluating. There is a reason for this, synthetics tend to have spring characteristics that are too far from being linear. The simple comparisons you did are not enough to assume linearity, we are not talking high school physics here.

The Samson data actually show the opposite: there is relatively little change in acceleration (equivalently, force) at the beginning of the stretch. Who cares what happens near the rope's breaking strength? No one in their right mind operates their rope at the edge of failure, and I have no idea how to calculate what happens in that region. You keep raising the notion that synthetics are not true Hookeian materials, but nowhere do I claim such a thing nor rely on it. My claim is that if you look at Samson's data sheets, and assume a smooth curve connects the (admittedly sparse) 3 data points, you can clearly fit a straight line to them without too much error. For an even better fit, you can model the low-load and high-load sections with two separate lines. It makes no difference whether the underlying rope material is springlike or not--we have to live with real rope behavior, and the Samson data, sparse as they are, tell us something about that. (But see the citation at the bottom of this post for some really good rope data.)

All this aside here is my engineering evaluation of this system... The only way to correctly identify the forces seen at different parts of a lowering system is to use measuring tools and do tests. This is because of the large number of unknown variables seen in the real world. Just to name a few we have the friction at the block, the angle of the rope going from the block to where it is held/tied off, the angle from the block to where the wood first takes up rope slack (no way the wood falls exactly vertical and/or in a vertical line with the block), the amount the tree gives with this load, how much the groundy lets the rope run...
I just don't see how we can assume all of these variables are negligible, especially ones like angle and tree movement.

To take your last sentence first, who is assuming anything is negligible? This whole thread has been almost exclusively devoted to the simple case of a rope catching a vertically dropped load. No angles, no run-out, no wood, no friction. It might be useful to get a good handle on this simple case before we take on a blizzard of additional factors. As to the utility of measuring tools and tests, I heartily agree (see citation below). But if ddhlakebound wants to develop a mathematical sense for the behavior of a rope catching a load, he, like the rest of us, is going have to live by his wits, tying together the limited data out there with solid math, plausible assumptions, and good approximations.

And then we come to the following amusing and regrettable passage:

On a final note, most of the guys on here don't like when John Doe homeowner or landscaper tries to tell them they don't do something right. That's because a professional has knowledge beyond that of a non-pro. As someone with a Mechanical Engineering degree from Virginia Tech and that does mechanical engineering as a job I have spent years working on things like dynamics, statics, mechanics of materials, vibrations, and deformation. When it comes to something involving mechanical engineering I have more knowledge and expertise then say a software developer. I am not trying to brag or say I am smarter then anyone here, there are A LOT of things I am ignorant on but that is why we have professionals. This just happens to be the subject I am a professional in.

Since, no matter how hard I tried, I just could not find anything constructive or useful in this paragraph, I was at first inclined to ignore it. Maybe I should have. But then I decided I would at least wonder out loud why anyone would interrupt a perfectly good and lively discussion of complex ideas with a sexist and self-inflating paragraph from a poorly written job application?! And why spoil even that by including a completely gratuitous and invidious comparison with a total stranger? Employer to Applicant: "Don't call us, we'll call you. And while you're waiting to hear from us, learn some manners!"


But back to the subject and back to the world of facts and reason. Here is a link to some excellent real-world testing of various synthetic ropes: http://www.xmission.com/~tmoyer/testing/Qualifying_a_Rescue_Rope.pdf
I have run into other papers by the author, who works in search and rescue for the Salt Lake City Sheriff's Office, and does some very nice experiments in his spare time. Every time I read one of his pieces I come away impressed by the quality of both his experiments and his analyses. In this paper there is an excellent chart on page 3 that plots force vs stretch for a number of ropes with something like 15 data points each. All of the plots, save one, look just about as linear as real-world data ever get. The one exception is only mildly curved. Again, this bears out my contention that over reasonable load ranges a rope will behave approximately like a spring.

 

[moray]

good questions

I didn't mix up weight and energy. Thats why weight was in quotes.

I saw the quotes. Point taken. By the way, please don't take anything I say personally. I sincerely admire your desire to develop a mathematical understanding of this stuff. If I sound abrupt or severe at times, that is just my crappy style, nothing personal at all. I want to be helpful, and I love a good technical discussion...

When we bring the load to a stop, the kinetic energy is felt as lbf....basically simulated weight.

This is the kind of expression that makes me wince a little, even though I know what you mean. The kinetic energy is spent by stretching the rope. The rope resists this stretching because it takes force to stretch the rope. When the load has been brought to a full stop, the kinetic energy is all gone and the rope is exerting the maximum force against the load. I think this force is the "simulated weight" you speak of. Right?

I too disagree that the rope works in any linear fashion as a spring. I would like the link to where you got those stretch/load figures, I did a quick search and couldn't find them. But based on the limited info from 3 points, it shows to me that the relationship is not linear.

Follow the link in my previous post. There are some beautiful data sets there. I wish the manufacturers of rigging rope would publish nice big tables of rope-stretch tests so we could make more accurate calculations of various rigging scenarios.

Fall factor has alot more to consider than the distance of the fall. It also includes the TOTAL amount of rope in the system. So to figure the fall factor, we also need to know how high up the rig point is.

I thought I defined it correctly. It is the ratio of the distance the object falls to the length of rope that will arrest the fall. If the rope goes over a biner at a protection point and down to a belayer, that is part of the rope length that must be counted. But this complicates things a bit, because the load on that leg of the rope will be significantly less than the leg attached to the falling load (climber). In my comment about fall factor, I thought it was clear I was comparing a 3-ft fall arrested by a 3-foot rope to a 50-ft fall arrested by a 50-ft rope. In other words, no block, no biners, no belayer, just a knot attaching the rope to some theoretical anchor. In both cases, obviously, the fall factor is 1.

Rockers use fall factors because they climb on dynamic rope. In the trees most of the rope we use is much more static, so trying to compare the two is definately apple and oranges.

Easy to say, but is either statement true? I think climbers use the fall factor concept because it beautifully captures the physics of falling on a rope in a very simple and useful form. As to the second statement, why should it matter how stretchy the rope is? If you look at the math behind the fall factor concept, the stretchiness of the rope doesn't enter into it.

Now, we are not going to get 1 foot of stretch from a 3 foot rope, and we are not going to get 15 feet of stretch from a 50 foot rope. So a large percentage of the distance needed to stop the load from either fall comes from letting the rope run. Is it easier to let it run one foot or fifteen? So how is it the same on the rope?

Wait a minute. I never claimed this was a realistic rope. I merely took your original example, for which I had calculated the maximum rope tension at 3400 lbs and extrapolated it to a 50-ft fall on the same rope. If you prefer a more realistic example, then drop 20 lbs first 10 feet on 10 feet of arborist rope, and then 50 feet on 50 feet of the same rope. The maximum tension on the rope, whatever that turns out to be, will be the same in both cases because the fall factor is the same.

The mass and the distance of the fall determine how much kinetic energy we must dissipate to stop a load. The distance we stop it in determines how much force we apply to the rope. The distance of the fall matters very very much, no matter how you look at it. More distance = more energy.

Perfect! Can't argue with this.

I'd be very interested to see you successfully drop 425 lbs 50', then catch it on a 1/2 inch arborist rope. Just make sure if you try it that nobody is near the dropzone, cause I bet the rope, the rigging, or the rig point breaks.

Even discussing it muddies the waters, clouding what we're trying to learn here.

I can see why discussing a highly unrealistic example may not appeal to you, but it doesn't affect the math. There have been plenty of times in the past when I have struggled with some mathematical or physical problem that suddenly became clear when I picked highly extreme or unrealistic examples to run the math on. But if you prefer to use nothing but highly realistic examples from here on out, I promise I won't stop talking to you.