You guys bring up some excelent questions. Could any explain to me why having documented evidence of the structural integeraty of the tree was not need when doing an investigation? That was the closest educated guess from where the rigging and cuts were made. According to the guy on the rope the load did not run. the calculations were entered into a program which gave us the forces applied to the tree. the point I am trying to get across is how much will any given tree withstand before failure? If anyone has some answers for this let me know where you can find the specifis data and research.
Side loading trees when rigging
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At best it is a guessing game you come to learn over time spent working in trees. Green wood will hold only 70% of what kiln dried holds under ideal conditions and can be as low as 30%. Dead wood is to be trusted at 30% to ZERO of good wood load, take your hammer up with you and tap the tree to sound it if you don't have a meter. Your question is a good one for a carpenter or mechanical engineer to answer on loads to wood but out in the field is much different. Go out to a treeline and you will see the trees have trunks that are oval not round and this is because they are growing to withstand wind forces upon them. Side loads would be different here because of shape.
What will it hold also depends on are the growth rings tight or wide. Most important is what kind of tree are you tying into, hardwood or softwood. There is no difinitive answer to the question because there are so many variables to consider, a rough estimate is what you come up with and your life depends on making the right desicision, a tough call.
You want to know something about bending moment, shear, deflection, stress and elastic theory.
At best it is a guessing game you come to learn over time spent working in trees. Green wood will hold only 70% of what kiln dried holds under ideal conditions and can be as low as 30%. Dead wood is to be trusted at 30% to ZERO of good wood load, take your hammer up with you and tap the tree to sound it if you don't have a meter. Your question is a good one for a carpenter or mechanical engineer to answer on loads to wood but out in the field is much different. Go out to a treeline and you will see the trees have trunks that are oval not round and this is because they are growing to withstand wind forces upon them. Side loads would be different here because of shape.
What will it hold also depends on are the growth rings tight or wide. Most important is what kind of tree are you tying into, hardwood or softwood. There is no difinitive answer to the question because there are so many variables to consider, a rough estimate is what you come up with and your life depends on making the right desicision, a tough call.
You want to know something about bending moment, shear, deflection, stress and elastic theory.
Kneejerk Bombas
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Another thing to throw into the mix is that no arborist ropes can hold 85,000#s.
The tree failed long before 85,000 lb.s.
The tree failed long before 85,000 lb.s.
Tom Dunlap
Addicted to ArboristSite
You need to do some research on wood strength information. Foresters have all of that information already. The Forest Products Lab in Madison, WI would be the first stop.
I agree Mike, in order to have a rope that could withstand 85k, we'd be stealing the anchor rope off one of the ore boats up in Duluth/Superior
Tom
I agree Mike, in order to have a rope that could withstand 85k, we'd be stealing the anchor rope off one of the ore boats up in Duluth/Superior
Tom
murphy4trees
Addicted to ArboristSite
8,500 lbs. sounds more like it for 450 lbs. dropping 9'.
Daniel
Daniel
Stumper
One Man Band
Hmmm.... Geofore, the bending resistance of green wood is lower than that of dry wood but the resistance to breakage?. Green wood can flex much further. It may even take a 'set' (permanent deformation) but the increased flexibility makes it more difficult to break (speaking in generalities).
I have been assuming(perhaps incorrectly) that the 85000 lb figure was supposed to be the stress at the location of failure. Dropping 450 lbs 9' won't yield 85000 lbs of force BUT if the force generated is transfered to the end of a lever (treetop) them the force at the opposite end of the lever could reach that quite conceivably. introducing leverage makes it possible to generate tremendous forces without breaking a rope of much lower tensile strength.
No offense intended but leaving aside the fact that a failure occurred, this whole scenario gives the impression of some very imprudent rigging.
I have been assuming(perhaps incorrectly) that the 85000 lb figure was supposed to be the stress at the location of failure. Dropping 450 lbs 9' won't yield 85000 lbs of force BUT if the force generated is transfered to the end of a lever (treetop) them the force at the opposite end of the lever could reach that quite conceivably. introducing leverage makes it possible to generate tremendous forces without breaking a rope of much lower tensile strength.
No offense intended but leaving aside the fact that a failure occurred, this whole scenario gives the impression of some very imprudent rigging.
imprudent rigging
If this guy is rigging for the first time he will make a mistake or two. Green wood does have flexability but a sudden load may not give it time to flex that is where you stress ease it down. The TIP may be able to handle a load slowly but not suddenly. I'm sure that is why we have safe working load vs tensile strength on our equipment. You can expect it to handle the SWL all the time but not the sudden leap to max or over max load before failure. Guessing what wood will hold is at best tricky. A furmula that will always work for wood loads will not work as a hard and fast rule all the time because wood varies from tree to tree.
I know from experience the trees I pulled down today would break in half when the roots broke loose or they would break just because one part of the tree can't go that fast and stay in one piece. The kind of thing that happens when you knock down a tall smokestack made of bricks. I've done that before also. Had the trees been green instead of dead they would fall down in one piece because they could flex, bend as they fall. Dead and punky they can't stand the strain and break in two or more pieces on the way down.
It may not be imprudent rigging as much as not taking up the slack beforehand. The way it was rigged may have been right but letting the piece free fall 9' may have been where the whole thing went wrong. It would be like tying a blake and not dressing it. The knot was tied right but not dressed, the result would be a fall before sudden impact. Don't take up the slack and things go wrong. Would it have held if it were loading the rigging at 450# per second or 8,500# instantly? My money is on the slow loading vs instantanious load.
If this guy is rigging for the first time he will make a mistake or two. Green wood does have flexability but a sudden load may not give it time to flex that is where you stress ease it down. The TIP may be able to handle a load slowly but not suddenly. I'm sure that is why we have safe working load vs tensile strength on our equipment. You can expect it to handle the SWL all the time but not the sudden leap to max or over max load before failure. Guessing what wood will hold is at best tricky. A furmula that will always work for wood loads will not work as a hard and fast rule all the time because wood varies from tree to tree.
I know from experience the trees I pulled down today would break in half when the roots broke loose or they would break just because one part of the tree can't go that fast and stay in one piece. The kind of thing that happens when you knock down a tall smokestack made of bricks. I've done that before also. Had the trees been green instead of dead they would fall down in one piece because they could flex, bend as they fall. Dead and punky they can't stand the strain and break in two or more pieces on the way down.
It may not be imprudent rigging as much as not taking up the slack beforehand. The way it was rigged may have been right but letting the piece free fall 9' may have been where the whole thing went wrong. It would be like tying a blake and not dressing it. The knot was tied right but not dressed, the result would be a fall before sudden impact. Don't take up the slack and things go wrong. Would it have held if it were loading the rigging at 450# per second or 8,500# instantly? My money is on the slow loading vs instantanious load.
How about a drawing? For i have had trouble getting the picture from the start and like to understand these things and their impact and lessons on all of our work.
The tree could have been dissected with a chainsaw to come to the same conclusion-no flaws present in the wood.
Anyone can make an educated guess. What one uses to make that guess is what is important.
You weren't on the job when this occurred?
This would mean the information given is second hand knowledge and it isn't clear how far the load fell and how much it weighed. It is possible to know how far the friction device was from the tree. This doesn't mean if it were located on the same tree it wouldn't have broke.
How can we know the information was interpreted correctly? Is it a certainty pressure is what caused the tree to break?
Perhaps this is a question being asked to the posters here. A few of the posters have already started working with what they know and are giving freely. They don't mind helping, they're proving it, and these guys are alright.
Joe
There is enough data to support the idea side loading trees can cause structural failure in tree stems. Because of this fact
one also needs to be aware when a friction device is used to lower a load, one is side loading the tree no matter where that friction device is located. It's common sense when one takes large pieces, there is more of a chance for failure.
When a fricton device is anchored away from
the tree being rigged with the load hanging there and no motion, the block will point away from the stem, not down it's length. This is enough to show there is a side loading of the stem. The block points in the direction of the resultant tensions in the lowering line. The vertical component of this resultant force is what people will be thinking about when rigging.
Because it is there doesn't mean there is great danger of stem failure. If a large enough piece is removed, the stem will fail no matter where the friction device is located.
The late Dr. Peter Donzelli (The Art and Science of Practical Rigging videos and workbook) had also addressed the issue of sling tensions from the resultant tensions of lowering lines. It can be shown through trigonometry along with empirical data, the actual tension of the rigging block sling is lowered by an increase in the angle made by the lowering line when the friction device is located away from the tree being rigged. What Dr. Donzelli warns us about is a couple is made on the stem when the sling angle points away from it.
Mike Maas guys his trees. His solution to prevent side loading is sound and for some a common practice. If one is afraid of side loading a tree when rigging, guy it.
Joe
one also needs to be aware when a friction device is used to lower a load, one is side loading the tree no matter where that friction device is located. It's common sense when one takes large pieces, there is more of a chance for failure.
When a fricton device is anchored away from
the tree being rigged with the load hanging there and no motion, the block will point away from the stem, not down it's length. This is enough to show there is a side loading of the stem. The block points in the direction of the resultant tensions in the lowering line. The vertical component of this resultant force is what people will be thinking about when rigging.
Because it is there doesn't mean there is great danger of stem failure. If a large enough piece is removed, the stem will fail no matter where the friction device is located.
The late Dr. Peter Donzelli
(The Art and Science of Practical Rigging videos and workbook) had also addressed the issue of sling tensions from the resultant tensions of lowering lines. It can be shown through trigonometry along with empirical data, the actual tension of the rigging block sling is lowered by an increase in the angle made by the lowering line when the friction device is located away from the tree being rigged. What Dr. Donzelli warns us about is a couple is made on the stem when the sling angle points away from it.Mike Maas guys his trees. His solution to prevent side loading is sound and for some a common practice. If one is afraid of side loading a tree when rigging, guy it.
Joe
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Sometimes we redirect off the base of the host tree with a block (so that the stress of the load/impact comes down columnar strength of the spar), to another anchor with the Porty ready. This keeps the Porty out of the way of the loads. And it is alot easier to pretighten across that horizontal waist level line, than vertically, for the so challenged especially!!
i believe in guying trees, even if Mike does!!! Also i believe in rigging tops in such questionable conditions (as i think we are speaking), perpendicular to the 'floppy' axis of the spar. What i mean is like perpendicular to it's lean, instead of just rigged of the 'normal' side of it's lean. If i am redirecting the rigging line around for gathered strength/ friction of several Upper Supports, i set the force of the pulls likewise, unless i just can't, or am looking to turn that 'floppiness' around to aid instead of hinder, and use it like a giant spring!
i think that the sideloading could be more severe if the pull was directed horizontally across or aslightly higher during the final snap on the spar's movement, to another redirect, from high angle support.
i believe in guying trees, even if Mike does!!! Also i believe in rigging tops in such questionable conditions (as i think we are speaking), perpendicular to the 'floppy' axis of the spar. What i mean is like perpendicular to it's lean, instead of just rigged of the 'normal' side of it's lean. If i am redirecting the rigging line around for gathered strength/ friction of several Upper Supports, i set the force of the pulls likewise, unless i just can't, or am looking to turn that 'floppiness' around to aid instead of hinder, and use it like a giant spring!
i think that the sideloading could be more severe if the pull was directed horizontally across or aslightly higher during the final snap on the spar's movement, to another redirect, from high angle support.
treeclimber165
Member A.K.A Skwerl
I guess I knew this, but never really thought about it. I just assumed everyone knew this. It's really obvious if you are roping down limbs with your lowering line rigged through two separate crotches in a canopy. The higher crotch (with the tighter angle on the rope) will take more of the weight. If you think about it, a rope passing through a block at a 180* angle will have to hold 2X the weight of the limb being rigged (not counting impact loads). But a rope passing through a block in a straight line won't put any tension on the block at all. So therefore a rope passing through a block at a 90* angle will put a load on the block equal to the weight of the limb being rigged.
In this crude drawing, the higher fork on the right will take about 2/3 of the total load (4/3 the weight of the limb). The lower fork on the left will only take about 1/3 (2/3 the weight of the limb).
i disect it like this whenst putting these rigs together:
Actual load percentages will depend on angles of lines on supports, nos. of supports, degree of support(running/hanging) and friction i think.
Pulleys placing more load on overhead support anchors of the system, at same angles, as you can't beat the force of the load on the support system, you can only beat the reudced load on the control line, this is reduced by friction, which pulleys don't do. This is the only way i can see to reduce the total load on supports, from the force presented.
After that it goes to spreading that presented total load over multiple supports, at X angles. The final fine tuning would be reducing load force by reducing impacting(speed and drop) and/or possibly running the load (so that it isn't all being supported). i think that those are the total variables to watch, after that, reduction in load mass input into the system would be final.
Of course; prestretching the lines to reduce impacting, 2/1, adding an extra line, leg on load etc. all plays into all that folding out of it. But i find that the above factors to be all inclusive of the diagnosis of forces and strategies presented in that day's puzzle.
Actual load percentages will depend on angles of lines on supports, nos. of supports, degree of support(running/hanging) and friction i think.
Pulleys placing more load on overhead support anchors of the system, at same angles, as you can't beat the force of the load on the support system, you can only beat the reudced load on the control line, this is reduced by friction, which pulleys don't do. This is the only way i can see to reduce the total load on supports, from the force presented.
After that it goes to spreading that presented total load over multiple supports, at X angles. The final fine tuning would be reducing load force by reducing impacting(speed and drop) and/or possibly running the load (so that it isn't all being supported). i think that those are the total variables to watch, after that, reduction in load mass input into the system would be final.
Of course; prestretching the lines to reduce impacting, 2/1, adding an extra line, leg on load etc. all plays into all that folding out of it. But i find that the above factors to be all inclusive of the diagnosis of forces and strategies presented in that day's puzzle.
Mahk
ArboristSite Member
B.S.This is why Mark is correct about his statement that there will be 1.41 times the load on a sling with a 90° angle made with the rope by the block.
How one starts is sum(add) tensions from the vertical and horizontal components of each leg of line.
T=tension of the line.
sum=add
x=vertical component
y=horizontal component
The vertical components of each leg are:
(sum)[Tcos 180° + Tcos 270°] = T(-1)+0=-1
(sum)[Tsin 180° + Tsin 270°] = 0 + T(-1)=-1
T can represent any line tension which is =
to the weight of the load. I chose 1 for simplicity. It needs to be understood T is the same for both vertical and horizontal components in this example.
What we're interested in knowing is the tension of the sling attached to the lowering block. This means we need to know the resultant tension along with the angle the resultant tensions make with the rigging block sling.
To do this we use the pythagorean theorem
to sum the 2 components we found earlier.
R=resultant force or in this case tension.
^2=the power of 2 or squared
sqrt=square root
The formula we use to find the resultant tension is R^2=x^2+y^2 and R=sqrt(x^2+y^2) Therefore:
R^2=(-1)^2+(-1)^2
The square of a negative # is a positive #.
R^2=1+1
R^2=2
Take the square root of both sides of the equation to get:
sqrt(R^2)=sqrt(2)
The square root of a square, like that of R, is simply R. Therefore
R=sqrt(2) and sqrt(2)=1.41
Therefore R-the resultant force or tension- is 1.41. This is the tension of the sling attached to the rigging block.
To find the angle the sling makes due to the tensions in the line, take the inverse tangent of the vertical components divided by the horizontal components.
tan^(-1)[(-1)/(-1)]=45°+180°=225° which is a 3rd quadrant angle because both x and y components are negative. Otherwise, a rope which makes a 90° angle through a block will make a resultant angle of 45° which is an equal distance between the 2 legs.
Watch making these kind of statements till they're correctly understood.Joe
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Kneejerk Bombas
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Originally posted by Joe
I like your style... but disagree. Anyone who has used 2 blocks in a tree knows this.
Do you carry a laptop into the tree with you and do the math before each cut? Common sense and good intellectual intuition trump a math whiz in a tree anytime.
:Eye::Eye:
This would be ideal. I read an article in TCI magazine titled "Opposing Pendulums".
In this article a guy put together 1 or more removals where he did use a laptop computer to figure line tensions, weights and other junk. I can't state for fact it was before each cut, but he used one to do the job. Imagine the stuff one could learn doing this daily for awhile.
Well, I really am trying to make a point. A guy can get himself into trouble if the limits are pushed based on this type of thinking. There is an approach one can use to come to correct conclusions. Correct conclusions = more efficient use of resources. The math in that specific post is a tool that will help overcome the incorrect conclusion of the sling tension "if" one decides to believe it to give a correct conclusion. I believe what I learned through the math in that post to be true. So do many other people. There may be a time when some person did something based on their intuition without support of the math that gets them into serious trouble, like when underestimating sling tensions based on intuition. One needs to be careful with what they know or don't know. Safety can be thrown into this argument.
:angel:
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It's sort of refreshing and conjures up old memories eh?
I found about the same resultant tension. A chart in the book On Rope labels the resultant as 185% of the load.
Well, it's really 76.5 lbs. On Rope states it's 77% of the load which would make it 77 lbs.
When natural crotch rigging these figures will change dramatically.I'll buy this statement.
Joe
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Well, wee kinda ain't in the tree now!
i think that the math in the tree is kinda extreme for my taste, i generally am interested in tendencies and ranges for competant aproximations and understanding what kind of support, steering and brakeforce is needed for a particular path and how the line can be laced to what available points to maximize the grace-full overpowering of a giant, and the lessons learned there in. These approximations are then worked with a 8-10+/1 Safety Factor.
But, on the ground i think that we could investigate the numbers to make sure that all on occasions they backup the men-tall images that we guide our decisions by. Joe leaves me behind on the numbers a lot(which i used to think i was gifted at!!), but then he is whom i will turn to ask something before running my mouth! (Well sometimes); even if i end up not agreeing, i still listen and try to digest all that he says!
If you start with a pulley support directly at the center of the clock, and the tie off back to the load itsef; the Support Load is = to the load (all supports terminate at the load and support only). Detatch the control line from the load and hold it @6o'clock (next to the load). The Support Load is now 2x load! The support, now has 2 pulls on it, the load line and the control line pulling down, with a pulley (no friction), the control line must match the load; so the load is the force pulling on the OverHead Support, with 2/1 leverage! As the angle between the load and the control line opens up from 6o'clock to 12o'clock; that plley in the center of the line goes from bearing 2xload to 0#, depending on the angle of bend around the pulley!
These figures are for non frictional supports, slide friction in, and the control line load lessens, thereby reducing Support Load, control line load can be reduced to 0 with total brake force of friction = to load; thereby reducing Support System Load to = Load
i think that the math in the tree is kinda extreme for my taste, i generally am interested in tendencies and ranges for competant aproximations and understanding what kind of support, steering and brakeforce is needed for a particular path and how the line can be laced to what available points to maximize the grace-full overpowering of a giant, and the lessons learned there in. These approximations are then worked with a 8-10+/1 Safety Factor.
But, on the ground i think that we could investigate the numbers to make sure that all on occasions they backup the men-tall images that we guide our decisions by. Joe leaves me behind on the numbers a lot(which i used to think i was gifted at!!), but then he is whom i will turn to ask something before running my mouth! (Well sometimes); even if i end up not agreeing, i still listen and try to digest all that he says!
If you start with a pulley support directly at the center of the clock, and the tie off back to the load itsef; the Support Load is = to the load (all supports terminate at the load and support only). Detatch the control line from the load and hold it @6o'clock (next to the load). The Support Load is now 2x load! The support, now has 2 pulls on it, the load line and the control line pulling down, with a pulley (no friction), the control line must match the load; so the load is the force pulling on the OverHead Support, with 2/1 leverage! As the angle between the load and the control line opens up from 6o'clock to 12o'clock; that plley in the center of the line goes from bearing 2xload to 0#, depending on the angle of bend around the pulley!
These figures are for non frictional supports, slide friction in, and the control line load lessens, thereby reducing Support Load, control line load can be reduced to 0 with total brake force of friction = to load; thereby reducing Support System Load to = Load
This is doing homework. We don't do homework in the tree. We adjust what we learned on the ground in the tree. It looks like the Spyder made a bigger web to capture more informaton.
It's funny a TreeSpyder would climb a tree, make a web, then disassemble that support I thought Spyders would use to make a web to capture food. I think we have encountered the 1st TreeSpyder that eats the actual tree itself.
Joe
