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Rope Bend Forces

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TheTreeSpyder

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Inside the secret hidden microcosm of a splice, knot, braid etc. the same forces are going on as in a much larger and viewable rig, under the same loads and conditions and the same laws of Nature.

Yet another calculator for pulley forces, by spread of the rope angle to pulley and load.


 
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TheTreeSpyder

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No magic, each block calculated the same, just by the spread of lines to it, and their carried loads.

1147.7 load on block at the preset example of 1000# flowing through a 110 angle. Alter the angle and load and the calculator instantly gives answers to that loading.

You want to know forces of lower pulley, keep same load and change angle of spread input, to what you propose the angle would be on lower. The 2 pulleys are separate entitys, that know not of the other, only of the tensioned lines of load to themselves, and the angle of spread(gotta get MB's attention somehow).

The example is from somewhar's else, the 2 pulleys seem to confuse, but it is all the same; calculate 1 pulley at a time. The sling holding the block only resists inline forces. Cosine of the rope angle from sling angle(half the spread) X potential Load (2x load because 2 legs on pulley).


Next Questions: Why are we only needing to do half the work and calculate the adjacent leg with cosine and not the opposite leg with sine??
 

darkstar

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This gives me some insight when im telling my crew that redirects are cutting down on pulling power.
 
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TheTreeSpyder

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That would be a multiple faceted problem, like which of the 3 positions redirect occupies, on the load, on your work input or on the anchor/pivot.

This is basically with redirect/pulley on anchor; but we could invert picture and anchor both ends and have load on hanging off top pulley. Then, the loading showed on the pulley in calculator simply would create the load in the white box you type in. In other words, reversed is simply reversed!

Redirects on anchor cut down on power of your lift(fighting more friction and elastic length), but also when holding or lowering then they'd cut down on the load's pull. There would be some power loss to elasticity from the rope length increase and the pulley friction. Whether that helps or hinders depends on if the load is the initiating force change or you.

Also, if the redirect adjusted to a better angle of pull for you, then the net change from the redirect would be better angle advantage less the loss of friction and elasticity (which this calculator leaves out). During the range of movemeant, that angle could get better or worse etc.

So, to me ya have to have imagery/ be familiar with the angles, and when 1 change in degree means a lot of change(or nominal) on the inline or perpendicular axis, and how that effects everything. Playing with the numbers and studying the changes will show what ranges give more dynamic changes. Then, how much time and effort does that change take/ is it worth it; also forming good habits of setting up to nearest maximum as ya go. In a nut shell that is it, for everything!
 

gf beranek

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Neat little program, Spidy. Simple and accurate. I worked rigging for years when I first started in the business not even knowing the load forces I was generating. Though I did learn early on from the old timers to double up on the strength ratings of the slings and blocks. Sure why not? Duh....
 
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Koa Man

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All I ever relied on when rigging is looking at the setup and deciding how big a piece it could safely handle. I preferred to err on the safe side. In 24 years none of my rigging ever failed, but I am not crazy like some of you guys. All this math is way over my head.
 
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TheTreeSpyder

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2 legs of pull on pulley, so pulley will all ways have 2x Load AT it; this is the potential, that may pull ON it.

The cosine range is from Zer0 to 1. Multiply the Cosine X Potential (2x Load) to find pull actually ON pulley. We are just finding the percentage of the potential applied. The Cosine scale is just the degree of change.

The cosine of Zer0 degrees is 1; so all the potential force (2x Load) is on the pulley at inline for our 2/1 setups. The cosine of 90 (each leg offset 90 from center is 180, flat line, the other extreme from Zer0) is Zer0. So, at flatline, there is Zer0 pull on pulley, line passes across thru pulley, not down on it.

Feel is great, but understanding sum nums can back up feel, expand it and double check it etc.!
 

Altissimus

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Thanks Spidey...You know your shit!...along the same lines (bad joke bad) maybe you would care to comment on the smaller scale rigging we call climblines , many climbers have gone frome double line to single line...the single line set-up INCREASES load at the tie in point all climbers should understand the differences...
 

PCTREE

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cool thread. good to have some understanding of the forces before you get killed. I have only once in 20 years broken out a crotch which I had half expected so everyone was clear.

hey Spidy did I read right that you are no longer climbing or am I dreaming???? Your agile mind would qualify you as a good tester for my ascender.....
 

darkstar

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2 legs of pull on pulley, so pulley will all ways have 2x Load AT it; this is the potential, that may pull ON it.

The cosine range is from Zer0 to 1. Multiply the Cosine X Potential (2x Load) to find pull actually ON pulley. We are just finding the percentage of the potential applied. The Cosine scale is just the degree of change.

The cosine of Zer0 degrees is 1; so all the potential force (2x Load) is on the pulley at inline for our 2/1 setups. The cosine of 90 (each leg offset 90 from center is 180, flat line, the other extreme from Zer0) is Zer0. So, at flatline, there is Zer0 pull on pulley, line passes across thru pulley, not down on it.

Feel is great, but understanding sum nums can back up feel, expand it and double check it etc.!
I love that Spidy , when you pointed out that 180 is flat line passing straight through the pulley it just clicked.
So now how can i determine the cosine of varying degrees?
Can you give a few examples?
Wow the wider the atotal angle is , its so obvious the less load on the pulley.
But then is the load still constant on the attachement point for the pully?


One more thing , when you have a rope attached to hard between two fixed points and snug and you pull straight down in the middle say 100 pounds .
How much force is now on each attachment point.It seems like it alot ?
I often will tie a tree off static and pull pretty hard to tighten it up .
But then ill slide a pulley up the rope and tie on end off to a stump or another tree , i seem to generate alot of pulling force this way and of course redirect as i choose.
 
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TheTreeSpyder

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You have to use chart to look up cosine. You take the cosine of half the spread, this gives the angle of each leg from centerline/inline.

You don't need to calculate the sine, cuz that is the across force passing thru not onto the pulley. The across force of leg A is balanced out by across force of leg B; so the sine forces pull on each other and not on pulley.

So, we just concern ourselves with the inline forces of the total (2x Load, because 2 loaded legs to pulley). Cosine, gives the percentage of the total pulling on puley, rest is pulling accross.

i'm at werk but will look at making examples later.

To the last paragraph. You then jsut run the calculator in reverse. If you set up anl;angle and load normally, and read the pull on pulley, then if the end was anchored and you placed the load calculated on pulley on pulley or pulled with hand, and it pulled at the stated pounds by the calculator, and held the angle of spread shown, you'd then create the force shown on the load white input window.

In other words there is no magic. If you set load to 1000, and angle to 170 and get 174.3. then if laod end was anchored, and you pulled or set 174.3 on pulley bend point, and it held at 170 degrees spread, then 1000#'s of force would be the line tension! backwiereds is jsut backwierds! But it is stillthe same loading/ actually the same balance point between forces, looked at forwards or backwords.

This leveraging is only so powerful near flatline. If you raise tension like that and then quickly snatch line through frictions, we call it taking a purchase (from the tensioned side to the untensioned side). this increases the rope tension (or moves load). It is called swigging or sweating a purchase.

Gotta run, errr gimp...
 
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moray

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Alter the angle and load and the calculator instantly gives answers to that loading...
What a cute little calculator! How do you make it work right there in the middle of a post? Wizardry!

You know I love your posts, Spydie, and your mathematical instincts are dead on. Reading your posts used to make me want to pull my hair out, if I had any, but now they take no effort at all. But most people are never gonna calculate any of this stuff, and for them I would leave them with a couple of rules of thumb.

1. The absolute worst-case load on a pulley is twice the tension in the line. This applies to the case where the angle in your diagram is zero.
2. When the angle is 120 degrees, the load on the pulley is equal to the tension.

Anyone who has no sense about this stuff, or who doesn't want to figure it out, could stay safe by always assuming the pulley load is twice the tension.
 
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TheTreeSpyder

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Uhhhhhhhhm the Flash thing is builted in by our Host!​

The points about 1x and 2x are apt, thanx.

But, this example goes on beyond that, you inverse it for forces on a speed line (kinda) and also swigging/ sweating purchase from a line. In these cases, we look at the calcs from the perspective of what was the load we placed on the end, to be generated by the number calculated on the pulley etc.

Also, as we go to larger systems, and these base numbers become part of a string of multipliers, then we knead more accuracy IMLHO. So, 2x isn't too bad pulled from end by load. But, 6x in center then multiplied by impacts, then leveraged by something else becomes a stranger beast, that it is best to have the basics down.

The ropes only resist in 1 direction, on 1 axis (inline). So, we only worry about the inline forces given by cosine of the angle of deflection (from inline). So, is less complicated a starting point than trees and hinges, that resist (therefore empower) forces on across axes, so then we have to use not only cosine, but also sine and length to view their secrets!

Playing with the nums can be kinda fun, but the biggest thing is getting a feel for the patterns-and fer dat i think ya kinda gotta pay your dues and do your homewerk!, and when changes of inputs can have the most dynamic effect, and is that effect standing on your side or against ye!

Orrrrrrrr sumetine like dat, i think!​
 
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TheTreeSpyder

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i'm not/can't climb; sum daze hard to walk:cry:....

The inverse of this stuff is speedlining and sweating lines etc. Whereby, the input is at the bend to increase line tension, stretch or move load etc. not the load on the end of the line to calculate tensions by.

When force is applied to a line perpendiculairly/sideways/across, the line must be already tensioned to give resistance to the bend, to empower the leveraging.

For this we take 1/(cosine of half the angle spread) X Half the pull at bend (because that load or your hand pull has 2 legs of support, 1 on eiter side of the bend/pulley). From this perspective, we can't just use the 2x or 1x marks suggested as rules of thumb (which i use sometimes myself on other stuff).

This is where the real powers are seen; mistakes or magic can be maid near flatline.
 

darkstar

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Just one quick question.
We tie both ends of a rope ,creating a taunt line or slack line , like what a tight roper walks across.
Pull down in the middle 100 pounds
How much force are we generating on each tie point?
 

lumberjack

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What's the angle on the rope with the person's weight? Without the weight?

That can be used for an approximation.
 
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TheTreeSpyder

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Just one quick question.
We tie both ends of a rope ,creating a taunt line or slack line , like what a tight roper walks across.
Pull down in the middle 100 pounds
How much force are we generating on each tie point?


you have 2 supports, so each carrying half the weight, so werk with 50# per leg. If support was directly overhead, it would be 50# line tension(50 each leg). As each leg spreads wider than inline, we have to take 1\cosine of that deflection; or in other words 1/cosine of half the spread angle X 50#(half the wieght).

i'm gonna make another calulator for this..

Now, the stretch of line and slip of hitch will provide some relief of forces, and angle falls, but it will settle at a balance point of that 1/2 weight on such and such an angle, gives such and such a force.

Same for sweating line to tighten etc. Then have to multiply if impacting with the bending force. The initial line tension doesn't matter, except to make line bend less, to give more leveraging. Just like if a wrench bent sum, it would give less leverage than non-bending. The power of the leveraging is in the resistance to bending.
 

darkstar

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Wow Spidy, i always heard when you pull in the middle of a taunt line you mutiply the force
I never could figure why but I have noticed the knots getting overly tight.
But alas its wrong ? Pulling down in the middle does not multiply force at all , it cuts it in half on the tie in points.
Correct ?
 
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TheTreeSpyder

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Ummm depends on the rope angle. If it is in a giant U (inline, close to Zer0 deflection/ Zer0 spread lines)when you pull down with 100# it would be like 50# line tension. But, if 100# on a line so tight it only bends to 170 degrees under 100# load, would hit 575# line tension. More if greater angle/flatter, less if more bent. then if that 100# impacted to line with 500# force, it would then raise forces appropriately.

edit, pic below shows a lot of leverage applied to perpendicular device; whenter that device is arm or rope. To get feel for this, imagine yourself in pic...
 

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moray

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Wow Spidy, i always heard when you pull in the middle of a taunt line you mutiply the force ...
All this business of sines and cosines is for geeks like me. There is a shortcut that works for everyone and requires nothing but a decent set of eyes.

Say you hang 100# in the middle of a horizontal taut line. The line is 100 feet long. You notice the load has pulled the center point down about 10 feet. You now know everything you need to figure the line tension. The 50 feet of line on each side of the load is 5 times longer than the drop. The tension in each leg is also 5 times greater than the load it supports (half the 100-lb. total, or 50 lbs. for each leg). The tension is thus 250 lb.

If the 100-lb. load had only caused a drop of 3 feet, the length ratio, as above, would be 50/3 = 16.6, and the tension would be 833 lbs.

It is a lot simpler to eyeball the lengths involved rather than mess around with angles and trigonometric functions like sine and cosine!
 
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