To redirect or not

[Climber020]

I have this ash about 12" dbh or so, around 75 feet tall, a back lean of about 15 degrees, and an 8 foot S curve at the base of the tree. The tree is over others tree and a fence. It is one the end of woods and most of the weight of the foliage adds an addition 10 degrees of lean from about 2/3 the way on up. The area where it needs to be felled is directly on the opposite side of the lean. I would usually just set up a pulley system on a tree in the direction I would like it to fall. But being there is such a lean and the S curve makes it a little more complicated in my opinion. I have guessed that with only 150' rigging rope I will not be able to get far enought away from the tree to get the leverege that I would need.This would create a large angle in the pull rope I would amagine the it would put addition stress on the bend at the bottom of the tree. So the answer I came up was to put a floating croatch with a rescue pully above the tree that I will put my fiddle block and portarap on. I believe this would accomplish more leverage for pulling the tree back and over. And along with this I am hoping to lessen the stress that would be put on the S curve and direct it more to the notch. Would this be a longical assumption or will there still be a large amount of stree be put on curve. Also when notching it would you notch it above the curve of below it. Below I can do on the ground but the bend ends a little bit above 8' which means climbing up and notching it.
Your advice is very much apprectiated.

 

[glens]

If you attached the pull line 50' above the point where you make the cut to a point 109' away at that same height (the cut), you'll have a straight-line length of 120' between the two attachment points. That sounds pretty doable with a 150' rope to me.

The angle between the rope (as a straight line) and the ground would be 25°. That seems plenty flat to be safe.

For every 100# of tension in the line you'd have 42# of downward force and 91# of lateral force at the top.

Think you could manage 5 or 6 hundred pounds of pull? That sure ought to get the tree over nicely. You probably wouldn't even need to leave the ground for that job.

Make the back cut perpendicular to the grain at that point, even if it ultimately goes uphill a bit, and keep a couple of wedges snugged in tight as you go as backup/assistance.

 

[Climber020]

I only had about 80 feet or so room to work with with. But I set it up as I described and just walked the tree right on over. Landed perfectly in between two others trees. Only problem we had was resetting the fiddle block. It was the first time I had used it. Had to reset if 3 times with the hook on a pole saw. But it was a little difficult to do like this. After getting back to the shop I put a micro pulley below the french prussik and attached a small diameter rope to the carabiner and put it up over a branch to pull the top block back up the tree. Worked quite well. But thanks for the advice glen. But may I ask where you learned the calculation as as per force and relationship of the angles to the rigging. I just bought The Art and Science of Practical Rigging hoping to expand my knowledge. Any other good rigging book that you know of? Ty

 

[glens]

It's too simple. Literally. Remember your geometry classes from grade school?

Really, this can be too simplistic but usually it's close enough. All you need is a calculator which has a square root function unless you want to also determine what the angles will be. In that case you'll need one with trig functions: sine, cosine, and tangent.

Knowing any two sides and one angle of a triangle you can readily fill in all the other blanks. When the one known angle is 90 degrees figuring the length of the last side is a real piece of cake.

Force vectors in a right triangle are easy. They're directly proportional to the lengths of the sides. Draw a triangle with one square corner which represents the intersection of the vertical and horizontal sides. Take the length of each of those sides and multiply it by itself, then add the two figures together. Find the square root of the result and you'll have the length of the third, long side. The relationships between the length of the sides correlates to the relationships between the forces present.

If you've got a vertical side of 3 and the long side is 5, take 5 times itself and subtract 3 times itself. That's 25 minus 9, equals 16. What number times itself equals 16? Why 4 does! Remember that if you ever need to produce a square corner. A triangle with side length proportions of 3, 4, and 5 will do it for you.

Let's expand that triangle by a factor of 20. Now the sides will be 60, 80, and 100. If you had a rope attached 60 feet up a tree on one end and 80 feet from that tree on the other end, you'd have 100 feet of rope (not counting the sag!) between the points. For every 5 pounds you pull on the low end of the rope, there will be 3 pounds of downward force (vertically) and 4 pounds of sideways force (horizontally) at the high end. Likewise, there will be 3 pounds upward and 4 pounds lateral on the low end for every 5 pounds of tension in the rope.

Now you've got two examples. You should be able to "see" how it works. If not, holler.

 

[Climber020]

The first part of that I do remember, was a math wizard, and the second part makes complete sense. I have a few more questions now to hopefully make complete sense of falling in general. The first is how to find the maximum angle that the pull rope and the ground make to generate enough force for pulling of a desired tree. I know that the smaller the angle the easier it is being there is more lateral force and less vertical force. But when in a tight situation where you need to know what you need inorder to pull the tree over. The second question is calculating the force required to pull over the tree. Not enough can result in a catastraphy as well as too much can lead to barber chairing the tree. I usually will just look at the tree and decided how many pullies to set up if any. My thought is there but I would like to be more precise on my measurements then just my guess taking in concideration the size and lean of the tree. Both questions are tied together being the smaller the angle the less force need to be applied to the tree in order to fall it being you will have more force working in your favor and not being put back down on the tree. Thanks again.

 

[glens]

[ QUOTE ]
... how to find the maximum angle that the pull rope and the ground make to generate enough force for pulling of a desired tree. I know that the smaller the angle the easier it is being there is more lateral force
and less vertical force. But when in a tight situation where you need to know what you need in order to pull the tree over.

[/ QUOTE ]
The steepest you can have the rope? That's a loaded question. What is the situation? Assuming you're talking about an un-redirect pull it'll obviously be less than 45°. Precisely, the closest you can have that pull anchor would be wherever 45° comes to on the ground plus the remaining distance to the top of the tree from your upper attachment.

Let's just pull a figure out of the air and guess the highest you'll attach the rope to the tree to be at 2/3 height (because it's not a spar and the higher stuff wouldn't be strong enough to use). A triangle with sides of 2 and 3 would come to 34° at the bottom, but you really don't need to concern yourself with the exact figure of the angle. One thing you'll know is that for every unit of lateral force there'll be 2/3 of that downward. The ratio of the sides of that "triangle" would be 2/3/3.6 so that's the way the forces would be present within it.

That was fun but doesn't really answer your question. You talk about tight spots. The tightest spot in my opinion would be a spar that'll just fit into the space you have to lay it; where you could get by with 45° if you redirected at the tail. Those sides are 1/1/1.4

You could use a steeper rope if you redirected the tail, but why not lay the rope as flat as you can and just go the 45°? You could also redirect or pull from a higher point at the tail. If you do that, though, just make sure that the lateral and downward (if you redirect down; upward if you pull from there) forces that'll be present there can be accommodated.

[ QUOTE ]
... calculating the force required to pull over the tree. Not enough can result in a catastraphy as well as too much can lead to barber chairing the tree.

[/ QUOTE ]
That's an interesting one. To be honest, I've never really thought about it. But I'll go out on a limb and give it a go

I'm thinking it would be something like this:

Let's consider a 6,000 lb spar ('cause that'll be easier to envision) that's 40 ft long and is leaning 8 ft out away from the pull. That's 40² - 8² = 39.2² Maybe we don't need all that info. We can probably use a ratio of the side lengths to the weight and figure something like X/6000 = 8/40 so X=1200. Does that seem right? Would it take 1200 lb of initial lateral force at the top to get that thing upright? Yeah, why not...

Let's guess further and say that the inverse (1/X) of the percentage up the stem you attach your line would be the multiplier for the initial lateral force needed. Thus if you were at 80% height, it would be 1/0.8 = 1.25, so 1.25 × 1200 = 1500. Hey, I'm just guessing here, but it doesn't "sound" too far off to me.

Don't forget that in those last two paragraphs we were considering the lateral force, not the force you'd have to apply to the rope. That would be greater, but we already know how to figure that, right?

 

[Climber020]

Ok so if I did my math right you would need about or more (gravity) of 1949 lbs. of force to pull the tree to an upright position. Then you would ony need 150 lbs. of force or less (gravity) to move the top of the spare 1 foot in the intended felling direction.
Gravity will be working against you when you try to get the tree upright but then will begin working in your favor.

 

[Kevin]

20, should be 400-500lbs. to pull that puppy over.

 

[glens]

I can neither confirm nor deny your figure. I don't know what you used as a basis for it.

In the spar above, it would take (if I'm correct, anyway) 1200 lb of force at the top before it would start to move. For every little bit it did move closer to being vertical, there would be correspondingly less force required to move it yet further. That would continue until it was fully vertical when not much more than a notion should get it to where gravity was helping instead of hindering. I should mention that all this ciphering is based on a zero-resistance hinge point. If you've got a 2" tree with 4" hinge, all bets are off

Nevertheless, let's work through your 1949 lb of force, fleshing it out a little to see how far away your anchor might be.

Assuming tied at 80% height (40 × 0.8 = 32). That was 1200 ÷ 0.8 = 1500. You ended up needing 1949. The ratio 1949 ÷ 1500 is the ratio of the hypotenuse to the horizontal distance, or 1.3 ÷ 1

1.3² - 1² = X² so X = 0.83, the relationship of your pull height to its distance. We know the height was 32, so 32 ÷ 0.83 = X ÷ 1

X = 38.6

Since the tree was 8 ft out at the top, 80% of that would be 6.4 ft so subtract that from the 38.6 since the spar is that much closer to your tail anchor at ground level (we're using right triangles, not necessarily actual ones). 32.2 ft away from a 40 ft spar? I'd guess that's not what you started with.

Again, for full height on the spar.

1949 ÷ 1200 = 1.62

1.62² - 1² = 1.27²

40 ÷ 1.27 = X ÷ 1

X = 31.5 ft distant (still need to subtract the 8'!).

What you're shooting for in this case would be an X (height ÷ distance) less than 1, not more than. The first scenario had that, but the second is losing ground.

Plausibly:

Tied at top [40'] going to 50' distant from the base of the tree.

40² + (50 + 8)² = 70.4²

X ÷ 1200 = 70.4 ÷ 58

X = 1456 required force

Tied at 32', 50' distant:

32² + (50 + 6.4)² = 64.8²

X ÷ 1500 = 64.8 ÷ 56.4

X = 1723 required force

Is this any closer to helping?

 

[Climber020]

Yes than you.

 

[glens]

Don't mention it; mental exercise is good too.

Technically, I should have used 39.2 instead of 40, and 31.4 instead of 32 in my calculations, since those figures actually represent the true vertical heights for the two spots on that spar. The resultant force values using those figures are less than mine in the last two examples above, but by less than 10 lb. If the spar were leaning much more it would progressively become more important to figure correctly if you were going to take the time. A little safety margin never hurt anything.