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Forces involved with snatching using a block


Badgerado
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I am interested in the physics involved with snatching wood, the shockload in my experience feels like more when the lead is longer i.e. half hitch is further away from the block..

I have been told however that in theory the load should be exactly the same unless you take the half hitch above the halfway point of the bit your topping down..

The way I see it that allows it to drop more before you're able to 'catch' it on the rope and let it run, there are other factors involved of course, can people elaborate on this for me (we are talking about a piece that is let run by the groundy not a snuffed off system)..

 

This was the only literature I found on it (not that I looked too hard)

 

http://www.hse.gov.uk/treework/articles/rigging-research-3.pdf

 

Hopefully my lame drawing clarifies what I'm describing

image.jpg.632df6057316ef0a1ef78d42d925f7fb.jpg

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providing you don't tie that hitch higher than the centroid then in fact the higher it is the lower the force as there is more active rope in the system, if you're truly interested then here's the link to the full paper: RR668: Evaluation of current rigging and dismantling practices used in arboriculture

 

Thanks for the reading reference Loler..

So I understand that the longer the active rope in a system that is snubbed off does reduce shock loading.. However when 'letting it run' is it to do with the rope already moving by the time the full load of the centroid comes on that makes it smoother and feel less when your half hitch is closer to the block?

 

Regardless I will be giving that study another read, it's a very well put together piece of work

 

Thanks again, Jack

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Hi

 

The second (RH) one exerts a greater load in terms for force * distance because it is further away from the fulcrum - hinge.

 

The hinge of the cut being the fulcrum, the tree being the lever & the height of the hitch being the distance

 

https://en.wikipedia.org/wiki/Mechanical_advantage explains further

 

Thats how I understand it anyway.

 

N

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That is incorrect N. The force acts through the centre of gravity of the piece, not the butt end, so, as long as it is tied below the centre of gravity, then where is irrelevant as the distance from the hinge (fulcrum point) to the COG remains constant.

Hope that makes sense :)

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Pete,

 

It will actually be a vector force, acting through the centre of gravity (not necessarily the centre of the tree)

 

This force changes as it pivots on the hinge as the tree falls.

 

 

N

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That is incorrect N. The force acts through the centre of gravity of the piece, not the butt end, so, as long as it is tied below the centre of gravity, then where is irrelevant as the distance from the hinge (fulcrum point) to the COG remains constant.

Hope that makes sense :)

 

I agee that the COG is important but the entire weight of the pice acts through the half hitch attachment point. Anything elevated has 'Potential Energy'. he energy is released when it falls. The energy is equal to mass x distance fallen. See attached modification of original diagram. The distance fallen is A to B plus A to C. A to C is really equal to A to B. So the distance is twice A to B. The energy released is mass x distance, but in this example we are assuming mass is the same whatever the ehights. So basically the amount of energy released by the fall is proportional to distance A to B. The groundie and the capstan and the tree have to absorb it as frictional heat, vibration, muscular effort and occasionally the groundie being moved some distance.

 

The shock and how hard it will be to control how it is let run is proportional to the distance between the block and the half hitch. Te groundie can't do anything to slow it down until the rope goes taut. So the OP is right.

image.jpg.5b3c8b3bffe5b0fa33f6d4f8e34e84d5.jpg

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