How to Land a Tree Flat in 3/4 Rotation: Solving the Equations

[coolgen10]

TL;DR Summary

Arborist Tree Problem - How much to cut for a perfect landing

Can anyone tell me how to solve this problem?

I have the stem of a tree that is X feet tall. It's just a cylinder as the top and all branches have been cut off.
I want to cut off the top portion such that when it falls, it will do precisely a 3/4 rotation and land perfectly flat.
What fraction Y of X should I cut from the top?

From experience, I know that Y should be approximately 1/5, but how do we solve this?

We can assume the tree is a perfect cylinder, it is compromised Y% from the top such that the top will fall and rotate around the point of compromise until it is horizontal with that point. At that point, the top cylinder will break off and the entire piece will continue to free fall to the ground, maintaining whatever rotational momentum that was created during the first part of the fall. At this point, it will need to complete 1 more half rotation to land flat.

To see a real live example of this, look here:


I do not believe the mass of the tree or height of the tree is needed to solve this.

First we need to know how fast the top of the tree is moving when it reaches horizontal. This would be the same a solving for a cylindar just falling down. How do I do that? Would it just be 50% of the speed of the entire piece Y falling its height Y?
Once we know the speed of the tip, we can calculate the speed of rotation so we know how long it will take to complete another 1/2 rotation.
Once we know that time, we can than calculate how high the piece will need to be relative to it's length so that it's remaining fall will take exactly that amount of time.

Equations for any or all parts of this would be appreciated.

 

[jedishrfu]

If the tree section rotates initially then it will continue to rotate during the fall to earth. One can see that in Scottish caber toss competitions.

 

[Baluncore]

We can assume the tree is a perfect cylinder, ...

The cut at height complicates the way the cut part releases from the remaining trunk. I believe we must assume the tree has zero diameter at the hinge, to eliminate that complexity. The hinge must operate until the cut part has rotated through 90°, at which point the hinge breaks and the simulation math changes.

Will the cut part actively lift off the remaining trunk as it passes through 90°?
There are two possible transitions to free fall. The end of the cut part may rest on the top of the trunk for some time, or the cut part is immediately free-falling, independent of the trunk. To identify when separation occurs, we must study the acceleration and rotation of the cut part. The complexity there comes because the centre of mass is not accelerating at g, until the hinge has released and is free. Before release, the point accelerating at g will be elsewhere, probably at 1/3 of the length of the cut part.

 

[coolgen10]

We must assume the cut breaks completely free at 90 degrees. So first how do we solve for the speed of the log at the 90 degree point? The video is a good example.

 

[berkeman]

We must assume the cut breaks completely free at 90 degrees. So first how do we solve for the speed of the log at the 90 degree point? The video is a good example.

I couldn't see the cuts from the video. How deep do you usually cut the first "V" notch cut? And how close to the "V" do you get with the backside felling cut? It seems like there is resistance to the fell of the upper tree part from the web between the final cut and the "V", no? So it's not like a freefall situation with a simple fulcrum point under the felling tree piece...

https://deerassociation.com/safely-drop-large-trees-chainsaw/

 

[Baluncore]

We must assume the cut breaks completely free at 90 degrees.

The hinge may break by then, but when will the end of the cut part actively lift off the remaining trunk. That is the critical "zero force" transition in the model when the mathematics changes.

So first how do we solve for the speed of the log at the 90 degree point?

I think that is a simple falling column or tower, that sets the rotation rate.

 

[Baluncore]

So it's not like a freefall situation with a simple fulcrum point under the felling tree piece...

Correct. But assume the tree has zero diameter, which will reduce the complexity of the hinge geometry.

 

[berkeman]

Correct. But assume the tree has zero diameter, which will reduce the complexity of the hinge geometry.

But if the OP miscalculates this, the people standing in the fell zone will die while they are taking their selfies...

 

[jack action]

Given the tree is cut at a height $h$ from the ground, the log above being of length $L$, once the log is horizontal, the downward velocity $v$ will be related to its angular velocity $\omega$ with:
$$v = \omega \frac{L}{2}$$
At that point, the amount of energy in the log will be equivalent to the potential energy gain from the height already fallen, i.e. $mg\frac{L}{2}$. This energy will be converted to kinetic energy, such that:
$$mg\frac{L}{2} = \frac{1}{2}I\omega^2 \color{red}{\cancel{+ \frac{1}{2}mv^2}}$$
Where $I = \frac{1}{3}mL^2$ for a cylinder rotating about its end. Then:
$$mg\frac{L}{2} = \frac{1}{2}\left(\frac{1}{3}mL^2\right)\left(\frac{2v}{L}\right)^2 \color{red}{\cancel{+ \frac{1}{2}mv^2}}$$
Or:
$$ \color{red}{\cancel{\frac{3}{7}}} \color{green}{\frac{3}{4}}gL = v^2$$
Now we know the velocity when the log is horizontal.

At that point, we know that the log will continue to accelerate during the fall of height $h$. This height will be equivalent to:
$$h = vt + \frac{1}{2}gt^2$$
Where $t$ is the time taken before reaching the ground. But the log will also rotate at a rate of $\omega$ as it falls. We want the log to do half of a rotation at that point, i.e. $\pi$ radian. So:
$$\pi = \omega t = \frac{2v}{L}t$$
Putting the 2 equations together, we get:
$$h = v\left(\frac{\pi L}{2v}\right) + \frac{1}{2}g\left(\frac{\pi L}{2v}\right)^2$$
$$h = \frac{\pi}{2}L + \frac{1}{2}g\frac{\pi^2 L^2}{4\left(\color{red}{\cancel{\frac{3}{7}}} \color{green}{\frac{3}{4}}gL\right)}$$
Or:
$$\frac{h}{L} = \frac{\pi}{2} +\color{red}{\cancel{\frac{7}{24}}} \color{green}{\frac{1}{6}}\pi^2 = \color{red}{\cancel{4.45}} \color{green}{3.22}$$
So the length of the log $L$ with respect to the total length of the tree ($h+L$) is $\color{red}{\cancel{\frac{1}{5.45}}} \color{green}{\frac{1}{4.22}}$, which is pretty close to your evaluation of $\frac{1}{5}$.

I think it's right ... or maybe I made a wrong assumption and just got lucky.

Edit: see post #16.

 

[berkeman]

LOL, after I watched the OP's video, this advertisement showed up in my CNN news feed. Cookies are everywhere...

 

[coolgen10]

Your analysis looks correct to me. Thanks for writing it all out. I cut trees for fun, am a Mechanical Engineer by college degree, and worked as a physicist as an intern, but design IT systems IRL.

 

[jedishrfu]

Cutting trees is a dangerous occupation. As a homeowner, I’ve run into a couple of instances where I thought the cut branch would do one thing but instead did something entirely different.

One time while cutting a branch while standing on a ladder with a pole saw, the branch broke and instead of just falling, it fell and rotated right toward me nearly knocking me off the ladder. It was due to the cut being angled.

There was a recent video online where a couple of folks decided to push a dead tree down. As the tree started to fall it slid back towards one fellow who jumped in the air only to get whacked by the tree as the base suddenly rotated upward catching him in mid jump and tossing him even higher in the air like a playground teeter-totter.

 

[Baluncore]

Tree felling is a profession, not a hobby. Mistakes will be made while learning the art. Take precautions and minimise the risk.

First understand the Dunning-Kruger effect, then decide if you will cut the tree yourself.
https://en.wikipedia.org/wiki/Dunning–Kruger_effect

 

[coolgen10]

Tree felling is a profession, not a hobby. Mistakes will be made while learning the art. Take precautions and minimise the risk.

First understand the Dunning-Kruger effect, then decide if you will cut the tree yourself.
https://en.wikipedia.org/wiki/Dunning–Kruger_effect

No kidding. I would not recommend it without proper gear and training. I have both. As the saying goes in the industry. Go low and slow when learning something new. Always tie in twice before cutting. And please don’t use a ladder. There lots to learn, but having a good understanding of the physics behind behind it all is one of the keys to succeeding in the profession.

 

[coolgen10]

Cutting trees is a dangerous occupation. As a homeowner, I’ve run into a couple of instances where I thought the cut branch would do one thing but instead did something entirely different.

One time while cutting a branch while standing on a ladder with a pole saw, the branch broke and instead of just falling, it fell and rotated right toward me nearly knocking me off the ladder. It was due to the cut being angled.

There was a recent video online where a couple of folks decided to push a dead tree down. As the tree started to fall it slid back towards one fellow who jumped in the air only to get whacked by the tree as the base suddenly rotated upward catching him in mid jump and tossing him even higher in the air like a playground teeter-totter.

Please don’t use a ladder. It’s a recipe for disaster. If you can’t reach it, call a professional. If it’s bigger than 5” in diameter or near property, call a professional. As you’ve said there are many videos on the internet of stupid homeowners.
On the other hand there are also many videos of professionals doing it right with the proper safety gear, tools, and training.

 

[TSny]

$$mg\frac{L}{2} = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$$
Where $I = \frac{1}{3}mL^2$ for a cylinder rotating about its end.

If you use the moment of inertia about the end ($\frac 1 3 mL^2$), then $\frac 1 2 I \omega^2$ would be the total kinetic energy. You would not include the additional term $\frac 1 2 m v^2$.

Alternately, you could use the moment of inertia about the center of mass ($\frac 1 {12} m L^2$). Then the total kinetic energy would include the additional term $\frac 1 2 m v^2$.

Otherwise, your analysis looks good to me.

 

[Baluncore]

What proportion of the length should you avoid cutting ?
If the cut top only rotated 90° after the hinge released at horizontal, it would land on its head. Apart from rotational inertia, it might then fall in any direction.

 

[256bits]

I had a maple dying.
Called in a 'professional'.
He used a ladder with a chain saw to cut the limbs off.

 

[jedishrfu]

We had an agile professional that danced from limb to limb with his chain saw cutting dead branches. It was quite scary to watch.

Another team came in and used mountaineering rope techniques to cut branches and slowly lower them to the ground. They were really professional and fast.

 

[jack action]

If you use the moment of inertia about the end ($\frac 1 3 mL^2$), then $\frac 1 2 I \omega^2$ would be the total kinetic energy. You would not include the additional term $\frac 1 2 m v^2$.

Alternately, you could use the moment of inertia about the center of mass ($\frac 1 {12} m L^2$). Then the total kinetic energy would include the additional term $\frac 1 2 m v^2$.

Otherwise, your analysis looks good to me.

I knew something was wrong because the answer did not make sense to me. Now it does.

The true energy equation should have been:
$$mg\frac{L}{2} (1-x) = \frac{1}{2}I\omega^2$$
Where $x$ is the amount of energy required to break the tree as it falls (the cracking noise we hear).
Which would have led to a smaller $v$, thus:
$$\frac{h}{L} = \frac{\pi}{2} +\frac{1}{6(1-x)}\pi^2 $$
If we assume that the experience from the OP requires $\frac{h}{L} = 4$, then $x=0.323$, or about one-third of the energy is used to break the log from the tree during the fall. With the number I previously found, we would have to add energy to the system which wasn't right.