Results of tree-pulling (SIM) a possible fraud?

anonymoustreelover

Results of tree-pulling (SIM) a possible fraud?

I would like to ask the following to Detter, Brudi and Bischoff (2005) to clarify a doubt we have:

Quote (Bond 2006):
“Detter et al. (2005) reproduced a graph of t/R ratio by radius that summarizes close to 5,000 individual tree investigations (Figure 3). A full 45 percent of this large number of trees “in parks and along roadsides” had a t/R ratio less than 0.3.”

Is that not wrong?, since:

The degree of hollowness diagram, as published by Detter et al (Bond 2006, figure 3) might simply be a mathematical tramp?
As understood in their workshops, the formulae it uses takes in account the modulus of elasticity (MOE ) as given by Wessolly and assuming an ellipse shaped trunk at the height of the elastometer measurements at about 15cm from the ground.
While this is probably seldom the case in their nearly 5000 trees?
In reality, at the measuring points at the trunk base usually there are wide root flares and a irregular outer cross-section, thus not being a nice ellipse at all. So the cross-section modulus is in reality lower than the one the ellipse formula assumes in the SIM method! And the real fibre-elongation measurements might indicate a lower MOE than the high Wessolly values, even in sound trees, not?

So, if we're not wrong, the result is that the SIM method uses the measured stiffness of the tree (high fibre-elongation and irregular outer geometry ) and compares it with what it likes better (a high Wessolly MOE and a perfectly ellipse shaped geometry), though it is not correct.
Hence, the calculations, wrongly, suppose then a lower stiffness “than what should be” and assume the trunk is hollow or has a defect…

In other word, and correct me if we're wrong, but saying that these results (the t/R to R diagram) “show how all these hollow trees still are standing” seems completely false!
Most of them might have not a single mechanical defect, and in the case there is a defect, its mechanical “weakness” could have been highly exaggerated by the calculations!

The answer to the above quote, if I am correct would be thus that the statement “A full 45 percent of this large number of trees in parks and along roadsides” had a t/R ratio less than 0.3” is FALSE?
I suppose that you never thought about this and that you published unintentionally results that seem completely wrong if my simple analysis is correct (if not, it could be regarded as scientific fraud, which nobody wants and surely not you).

It would be great to have an answer on this, since many more will come up with the same questions!
Or better would be, I only suggest, if the authors Detter et al. could demonstrate scientifically that their t/R diagram is no fraud, in a peer-reviewed journal?(but please, preferably no ISA one, since the ISA supports you already with a stunning ease)

Greetings ,
Anonymous for obvious reasons.
(don't search for me, my job is done here, now it is up to you and the arb-scene)

References:
Bond, Jerry. 2006. FOUNDATIONS OF TREE RISK ANALYSIS: Use of the t/R ratio to Evaluate Trunk Failure Potential. Arborist News. ISA.

Figure t/R by R for 4807 standing trees (Detter et al. 2005):
Detter, Andreas, Erk Brudi, and Frank Bischoff. 2005. Statics Integrated Methods: Results from Pulling ests in Past Decades. Barcelona. ISA Spain.



resuming:
1. The MOE for each tree-species of the Stuttgart Strength Catalogue might be higher than the one found in a standing tree. So it is easy to be mislead by the lower MOE found in a standing tree.
In this way, even completely defect-free cross-sections can have a lower MOE than the one used in the treepulling SIM and the calculations would tell us that there is a hollow while the trunk in reality is sound and full.

2. The way of measuring the geometry of the pulled tree (elastomethod) can result in a deformed vision of the relative stiffness of a cross-section, since a very irregular stem base, for example, is assumed to be an ellipse. So the calculations will assume a much bigger cross-section modulus ( a full ellipse) than the stem base, with its root buttresses and spacing in between.
And the combination of comparing MOE’s and the practical impossibility of registering the real cross-section modulus will lead to a, theoretically, very low stiffness and thus, according to this method, a very low t/R, FALSELY.

3. the load-bearing geometry or cross-section modulus of this irregular cross-section of a trunk will be overestimated when the ellipse formula is employed in the elastomethod. The combination of this methodological error (also an understandable limit in the real field-work) and taking as a standard a higher MOE than the real MOE of this tree, leads to a “hollow trunk and very thin residual wall”, although the trunk is completely full or almost never as hollow as calculated by the Elastomethod (SIM tree-pulling).
If so, the t/R over R of Detter et al 2005 would be completely false!

Eric Frei

Interesting indeed.

I believe this is the publication you speak of? (PDF attached below)

So lets get something real straight here.

In the diagram below the t/R ratios weren't ever proven by other devices such as core samples, resistograph, picus or dissection? If that's the case then they're theoretical t/R ratios. Also without hard evidence what happens about some decay where wood is present but has rot, could be soft or hard and brittle skewing results?

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Attached Files

bond_tR_tree_risk_assessment.pdf (1.63 MB, 159 views)

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Tim Craig

If you apply a pull force of 100kg at 12 metres up a stem what will the force be at the base of the tree?

I reckon it'd be about 1200Kns...is this right?

Can any one explain this with a diagram?

The reason I ask is because I was trying to work out the potential force on the base of the tree when trying to pull it over with a 4x4 winch.

Eric Frei

moment = force x perpendicular distance from pivot.
? = 100kg x 12m

The rope is not pulling at 90 degrees to the stem so not all of your 100kg of applied force is being used.

But in essence you are on the right track. You are applying a moment to the tree (rotational force) on a hinge (scarf/backcut). Just do the sums on what force is really applied on the horizontal vector, from this handy vector tool it appears to be 71kg

So the correct answer would be 71 x 12m = 852kn

If I'm wrong I'd happily like to be shown where.

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Tim Craig

The figures you come up with look good, thanks for the explanation.

Eric Frei

Error, decimal point issue. It would be more like 8.52kn

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moray

You are raising the interesting issues that face anyone trying to pull over a tree, but with your permission, let me rephrase the question. Since the kg is a unit of mass, not force, we really need to specify the pull force in newtons. The gravitational force on a kg of mass at sea level is about 9.8N, so let's say you apply a horizontal 980N force 12 meters above the hinge. The resulting torque about the hinge will be 11,760 newton-meter. The bull rope will undoubtedly not be horizontal, so you must pull harder than 980N to get the 980N you want at the 12 meter level. If the rope angle is 45 degrees from horizontal, for example, you must increase the pull by a factor of about 1.41 (square root of 2): it will talke 1386N rope tension to produce a horizontal component of 980N.

Let's assume you apply 1386N at 45 degrees and, after bending slightly, the the tree stabilizes. The hinge was more than strong enough to resist the torque.

Here's an interesting variation that illustrates a point no one ever mentions. Imagine everything is as described above, but the tree is rooted in 3 meters of soil on a huge barge, and the barge is floating freely in a calm lake. You stand on shore and apply 1386N to your 45-degree bull rope. What happens?

Therrin

*Therrin scratches his head in confused wonderment*

Gee, you're right! Noone ever mentions the effects of pulling on a tree which is rooted to a barge on a calm lake.

Odd...

Lets see, so then you'd have to factor in a couple things here. A couple things which I happen to have an uncanny knowledge of. For example;

You'd have to consider buoyancy. Archimedes principle says that any object immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced. So your displacement volume (aka "D") determines the buoyant force. Buoyancy is expressed by the ratio W/D; weight of the vehicle (aka "W") to the weight of displaced water.
Buoyancy in this case would be greatly affected by the loading of the tree in 3 meters of soil, on the barge. We would also have to consider the water density, but you've mentioned it's a "lake", so I'll take that to mean fresh water. Keeping in mind that a cubic foot of fresh water is right around 62.4 lbs.

Also, as your tree grows and flourishes, unless it grows perfectly symmetrical, it would set up an unequal weight distribution along the longitudinal axis which would cause the barge to have an up or down angle from the horizontal. Hopefully, said barge would have a trim system in place to counteract this effect, which would transfer weight or ballast forward or aft, port or starboard in order to remedy this effect.

Having your tree rooted to a barge on a lake brings up the topic of stability. Stability is the property of a body that causes it, when disturbed from a condition of equilibrium, to develop forces that tend to restore it to its original condition. Equilibrium is a state of balance between opposing forces which may exist in three states: Stable, Neutral, and Unstable.
Still with me?
Okay, say for example, if when a 45 degree angle is put on your tree/barge... forces are set up which act to reduce the angle, your tree/barge is stable. Neutral equilibrium exists when the tree/barge remains in its displaced position at 45 degrees after a force that displaced it is removed; and Unstable equilibrium exists when the tree/barge continues movement after a slight displacement.
Stability on your tree/barge deal is intimately related to the center of buoyancy as well as the center of gravity. The center of buoyancy is the geometric center of volume of the displaced water. The center of gravity is the effective center of mass. Let's call these two centers B and G, respectively.
When your barge-with-tree rig is in stable equilibrium, its center of buoyancy and center of gravity are in the same vertical line.
I'ma throw another curveball at you, because there's some lingo that's gotta be covered. That's right, here it is... Metacenter. Oh yeah, you like that?
This is the point of intersection of a vertical line through the center of buoyancy on the tree/barge rig while floating upright and a vertical line through the new center of buoyancy when it is inclined a small amount. (indicated by letter "M")
When you pull your 45 degree angle on the tree, the center of buoyancy moves from B to B1 because the volume of displaced water at the left of G has been decreased while the volume of displaced water is increased.
The center of buoyancy, being at the center of gravity of the displaced water, moves to point B1 and a vertical line through this point passes G and intersects the original vertical at M. The distance GM is called the metacentric height. This stuff is really indicative of a fundamental law of stability. Okay, when M is above G, the metacentric height is positive and the barge is stable because a moment arm, GZ, has been set up which tends to return the barge to its original position. It's pretty obvious that if M is located below G, the moment arm would tend to increase the inclination. In this case, the metacentric height would be negative, and the barge would be considered unstable.
What I'm getting at is that when the barge heels over, there is a shift in the position of the buoyancy center because of the volume shape change below the water line. I'm pretty sure that the shape of the hull, the trim and ballasting systems which are either present or not, and the overall displacement and centers of buoyancy and gravity all need to be defined before you can decide what exactly will happen when you try to pull your tree/barge over at a 45 degree angle.
(afterall, what if it's just a tiny tree on a huge barge, the tree would snap, of course, and NOTHING would happen to the barge....go figure) I'll draw the picture of what I mentioned above showing roughly how I'm picturing the size of the tree and barge that you asked about.
Then again, it might just stay at 45degrees for as long as you maintain the pull, and then right itself when you let go. (But the barge would probably sink) *shrugs*

Can I get more data, please?

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Therrin

Awww...no witty replies? My feelings are hurt.
You touch on my knowledge base aquired while building a personal submersible, and then you dont respond.

what gives....

moray

Good catch, Therrin, that was a rather dopey scenario; here in Maine the lakes are never calm.

But kudos on taking the physics way beyond what I intended.

Your comments on stability, said much better than I could have, are at the heart of what I was after. More narrowly, your concept of equilibrium is what I wanted to illustrate.

Take our tree again, this time rooted in the ground. When we pull on the bull rope, we have applied an external force to the tree. If it bends slightly and achieves a state of equilibrium, then we know it satisfies TWO conditions: the sum of all the force vectors acting on the tree equals zero, and all of the torques acting on the tree sum to zero as well (you do have to specify a center of rotation to describe a torque. If there is no net torque about some arbitrary center, say the center of mass, or the hinge, then there is no net torque about any center).

The point of the barge example was to show that all torques can be in balance but the vectors may not. When we pull on the barge tree, the barge will tip very slightly, and as you have described so well, the greater bouyancy resulting at the end that dips will produce the necessary restoring torque. Presto, rotational stability! Because there is a downward component of force due to our pull, the barge as a whole will sink deeper into the lake until the extra buoyancy from the increased displacement exactly equals the downward force. The barge both sinks a little and dips a little to reach equilibrium.

Ah, but we don't have translational stability, not at first. The horizontal 980N we are applying at a height of 12 meters is not resisted by some equal countervailing horizontal force. Consequently, the barge starts to move towards us. After a period of acceleration it reaches a steady velocity such that the drag force from the water is 980N. We have equilibrium again, as all the forces and all the torques are in balance. This example shows that the typical tree-pulling scenario doesn't create JUST a torque. There is a horizontal component of force that the roots must withstand.

Another common tree world scenario, rigging down heavy branches using a natural crotch, also presents some interesting aspects. Let's say you have two wraps around a branch right up against the branch collar, and you are lowering something that weighs a ton. Any decent branch will support the ton, and two or three wraps will give you plenty of friction to control the lowering. No worries. But all that friction isn't exactly free. Friction is a force and it has a direction. The rope feels it as a force opposing the motion of the rope, and the limb feels it as an equal force in the same direction as the motion. More precisely, the limb experiences a very large torque trying to twist the limb off. I raise this as a matter of interest only--it may be that no limb would ever fail in this mode because it would always bend down and break first. But it seems like a good idea to be mindful of all the forces and all the torques present in any given situation.

Great stuff, Therrin, I enjoyed reading your lengthy disquisition.

Eric Frei

Also we tend to have a hinge in the tree we are pulling, but I have seen some dudes just pull them over without cutting.

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Therrin

Glad to FINALLY be of some help with my lengthy dissertations.