Blocks sharing loads through differing angles

F

Frank1

Hi,

Can anyone help me / refresh my aging memory?

Recently, been dismantling a large, spreading, ash tree (not much height, lots of long slender branches as spread). Many of the long slender branches had to be lowered / rigged, to avoid property below.

Set up lowering system through two blocks: one at a static high point in the centre of the crown and the other lower down, placed upon the branches that were been removed (or on adjacent limbs, as appropriate). Consequently, I ended up with an angle of approximately 45 degrees from my top block running to my bottom block.

In terms of my question: when the load is applied the two blocks and anchors are sharing the load? Because of the 45 degree angle, is the top block receiving 75 % and the lower 25%?

I remember a thread about this and some work from Laz1? But cannot find it – any help would be greatly appreciated. I was interested in the pure physics rather than all the variables and pros and cons of different systems / strengths of split tails and blocks etc.

Frank 1
 
Depends on your set up but the total load = line pull x the angle factor multiplier which is 1.84 for 45o.

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http://science.howstuffworks.com/pulley.htm
 

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Kevin, I think Frank is talking about redirects. Not mechanical advantage. Or, does the same principle apply for redirects? And for that I don't know the answer. I would, however, like to know that answer to that myself.

Frank. Do you know how to use "Paint" to create an illustration? That could be helpful too.
 
The Block tension will be the same multiple of the line tension, for a given angle; whether that block is mounted on anchor, effort input or work output position in the scenario. Just like electricity doesn't know/ care what color wire it is conducted thru.

The total tension per leg will be the same, just the angle of that leg's of pull is then divided between inline or sideways force from that total leg/ line tension. We only recognize the inline forces of the angle as pulling force on the pulley/ block. the sideways tension is mirrored by the opposite leg to oppose without pulling on block, as direction of the angle is mediated to shift the pulley

A picture would be nice. i think with as wide an angle as you propose (from leg to leg on lower pulley); so if top angle is 45, bottom is 135? Then, the lower pulley would have leverage over the line. For, at 120 we'd hit break even, whereby the line tension would equal the block tension. Bent more closed than 120, the line tension takes leverage over the block (line tension is less than block up to 2x power). But flatter than 120, the block has leverage over the line tension(the block pull is less than, so has leverage over the line tension). So mostly, you have an extra pull point less than 1x, and a main support slightly less than 2x for more rope to run a different path. But, the lower angle of pull will be horizontal(which would be across a vertical spar). Plus, an extra sweat/ swig point in between, where the sideways pull leverage on a point is very high/ very much less than the line tension it induces(but line must be tensioned, for it is the resistance to bend that gives the leverage).

In Kevin's picture, we have a 2x lift; with 1.5x support load. If we inverted the system, we'd have a 3x lift (but not be able to use bodyweight as a force, but could leg lift force instead of arm) and .66x load on support. Increasing sheaves; increases power, but also decreases support load when effort input moves the opposite direction of the work output. But, when we inverted the system to make effort input move the same direction as work output; we create a condition of more power with added sheaves, but more support loading too!
 
I think the enclosed website may be helpful. It uses the head to tail method for finding a resultant vector or the resultant force. The excercises do require one uses a protractor and a ruler to find lengths along with the angles made with the vectors. (Homework!)

http://www.mathwarehouse.com/vectors/resultant-vector.php

This next site shows more than 2 vectors being added using the head to tail method. It also shows some math involved
with finding these forces.

http://www.1728.com/vectutor.htm

If one also does a google search using "resultant vectors head to tail" then this topic can be futher researched.

Joe
 
Frank1 I removed a ash tree very similar to your description on tues. About 70-75 ft tall with very long leaders. I also used two pulleys. I guess really 1 block and 1 pulley to be correct. The question was asked at one ppoint about wether my block was higher or was the pulley higher and I stated I wanted them at nearly the same hieght b/c I felt this was the best way to even out the force. I have always been under the understanding that when using a pulley at a lower point was more for redirect than force distribution. I am of course talking about a pretty significant hieght difference.
 
In simple terms, the greater the angle of the rope, the less force there is on the anchor. A 45* angle is 1.84 times the load. This would equate to a <9% reduction at the block and anchor. A 90* angle is 1.41 times the load. This equates to almost 30% reduction in force. A significant # when rigging hazardous wood.
The load reaches 1X at a 120* angle.
 
i think these numbers are all good guides, but then we have to recognize too that this changes the angle of pull on the mount/ support. Whereby; on a vertical support/spar at some point 2x load inline on spar is better than 1.4x non-inline.

This effect can be buffered by an imbalance of weight to opposite side in tree, that now becomes ballast to your loading angle. So, like a chess game, it can pay to plan moves ahead, as to what you are giving away and opening up at each move.

Also, as we move from static load considerations, to dynamic, there is more line in an angle to ground, than a straight line, so an angle has more elasticity in the system. This would be more pretensioning to the same tension, but more buffering on the drop/down side.

That changes for a 2/1 or more support systems, whereby there is less static load than 1/1 as Kevin shows. But on the dynamic side though, even though a 2/1 has more line in the system, there is now less elasticity. This is because there is less tension per leg(as Kevin shows), which then then yields less elastic dampening. The higher the MA, the less the static support load, but the more dynamic load (mostly). This (d)effect is seen more with efficient systems, where friction is helping less/ dampening is more on own to dissipate the input forces and transmit the rest.
 
Hello all,

Thank you for the responses: cannot get enough of the “how stuff works web site”.
Not sure if I’ve been clear enough, so further explanation and illustration attached as requested.

A (also refer to illustration): long slender branch. If I snatch it ‘off itself’ substantial force is on the one block / anchor point, with possible risk of branch failure?

B (also refer to illustration): As above scenario, but include another block at high point in centre of crown (substantial anchor point). When branch is removed / snatched the higher block / anchor bears / shares the load, dependant upon angle / vector formed between the blocks (or block and pulley, depending upon your terminology)? Thus, reduced risk of branch failure?

To reiterate from my original post, I used the set up in scenario B. And would admit that I set this up through experience and ease of working. I credit myself with a reasonable level of knowledge, and understanding, but in this scenario I could not get my brain around the “basic concepts” of the loads and forces.

Anyway must get back to the how stuff works site. I’ve a few “basic” questions about this particle generator they are about to switch on in Switzerland! Don’t worry, I’ll post them to Steven Hawkins’s site.


Kind regards
 

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Scenario B, is like they are saying. At fully closed angle between the legs of pull (upside down U/ Zer0 deflection from inline)on any pulley point, you'd have a full 2x load (both legs of tension added together to be supported in any pulley scenario) on block statically(stir in movement potential, friction in movemeant decreases that loading on hold and lower, increases force on compression/ lift x impact of change). The legs of pull in scenario B on the upper pulley are at almost that 2x potential on the block. Thus the 1.84 x line tension multiplier given. Then that multiplied force exerts on the support, at whatever length /angle multiplier, of the block force.

Conversely (from upside down U, or really any U of inline/pairallell legs of pull)at flat line, a block would have a lot/ almost infinite leverage over the stiff(loaded) line (as opposed to the opposite case of the upside down U where line has leverage over the block instead). We've gone from block loaded 2x line tension (taking on the full force of both pulls on it) to almost no pull/ theoretical Zer0 multiplier (flatline), where the legs of pull sidewards force act as ballast/equal and opposite each other and thus 2 legs of pull give less than 2x loading on block/bend). The lower pulley is between almost a straight angle between the legs of force. But the scale of change is not even from Upside down U to flatline. At 120, the block loading will equal the line tension; so small changes in degrees near flatline(that goes from 1x to theoretical infinite in 60 degrees between 180 and 120), give more of a jump in change of force(than degrees off of a perfect 2x gives change to ratio, which goes from 2x to 1x leverage in 120 degrees). i'm thinking the lower pulley has 2 lines of tension on itso 2x potential loading from those legs of pull like the upside down U, then less the non-inline/ perpendicular parts of those pulls. So then that is further away from 2x, being on the down side of the 120 point where block tension = line tension. So, the lower 135deg (180-45?)angle is like 75% of line tension on block(less than the 2 total pulls of line tension on the block). Then that multiplied force and it's median angle in relation to the branch angle x the length of the branch support...

The flat line bending (as used to advantage in Brion Toss's swashbuckling tales of sweating /swigging) for leveraging a line only happens on loaded lines, and more so on more loaded lines (another multiplier in the string). Rope is flexible and therefore must be tensioned to resist bending across it's length. This resistance to bending (perpendicular force applied to device) is what gives leverage. In non-flexables (wood, steel) length is yet another multiplier, but not in flexible rope(except to alter angle to more relaxed-closed..or not).
 
Hey all,

Yes I still lurk around from time to time.

Thought this was a neat thread.

I hope that this spreadsheet is openable as it was made on openoffice. If it does, change the yellowed numbers to adjust the results.

If it doesn't work, let me know and I can post some screenshots.

Spydey, please check my math. I started with heights and distances because I feel tree guys are better at estimating that than angles

Dave Spencer
 

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Let me clarify what I have done here. Of course Spydey caught me on something I didn't think anyone would notice :) Thanks for keeping me honest.

The math I have done here assumes that there is no pulley sheave that the rope is going around. This would slightly affect the angle of the rope therefore the results.

My goal in this is the same as my old forces on a speedline calculator. Pete Donzelli chastised me on trying to make a calculator that would predict exact forces in a system with so many variables that are not impossible but prohibitive to account for; Things like rope stretch, flex in wood, wind affecting loading and resulting forces.

My goal with both calculators was never to have a treeworker measure angles, plug it into a calculator and be able to say this will impart 1321.3972 lbs of force. My goal was to give people a general understanding on how forces change when you change the system. I can't count how many people thanked me for the forces on a speedline calculator because they said they did not understand how much the angle affects the loading. But by just playing around with some numbers, they could visualize the exponential increase.


Dave Spencer
 
[ QUOTE ]
Hi Dave :)

Thanks for working up this formula.

Tom

[/ QUOTE ]

Yes, thanks Dave.
It's very generous of you to take the time to create these excel spread sheets with formulas. This and your 'forces on a speedline' spreadsheet are great tools and fun to play around with.
 
you know i thought on the out set of this thread it was getting too complicated, but after looking at norm hall's chart i get it and have actually started to think more and implement the angles on a more regular basis. i have even referred some people to this thread. thanks for making me more aware.
 
The spreadsheet adjusts for me thanx; even using Excel_2003.


Force = Power x Distance. The sling holding the block, only resists distance from force on the inline, not any perpendicular axis. Because inline is the only axis resisted against distance change from force, that is the only power applied to block sling.

So therefore we only deal with the inline forces to that sling on each of 2 legs, not the perpendicular in calculating block forces. The perpendicular forces are each others equal and opposite/inline to each other. But the sling holding the block is the equal and opposite of the inline forces. Thus we can take cosine(for adjacent/inline leg) of the angle of deflection(hypontoneuse) from inline on the sling, for sum total of each of the 2 leg's pull on sling/block.

Thus, for a 300# load, we have 2 legs of 300# tension on the block all ways, no matter what angle, we have 600# total force. It is just that part of the 600 total resists itself, and part of it resists the sling. Half the spread of the legs is the sling angle, because both legs of tension 'push' the other way equally, sling is center median/ nominal/ inline.

Inline and perpendicular force makes a right angle. The hypotenuse of the geometry is the actual path of the conducted force. The adjacent is the inline part of the hypotenuse total force of that leg and the opposite is the perpendicular force. How much of the hypotenuse force is inline is cosine of the angle, how much is not inline is the cosine of the angle. Thus, the cosine of angle from center/inline X line tension, tells how much of the 300# of that leg of tension is pulling on the sling. Or line tension X 2 X cosine of half of spread. The sine of the half the spread is the force pulling on the opposite leg, not the pulley.
 
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