truss climber mechanical advantage theory

[brin831]

Guys, few people in the shop got into this discussion today and while trying to prove my point someone brought up a great picture that made me wonder if everyone was crazy ...

https://s-media-cache-ak0.pinimg.com/originals/57/0e/06/570e06b8027cadd279645d200298241b.jpg

example 1 is essentially what you are creating but i've always heard and there is even documentation stating that that the climber system creates a mechanical advantage doubling your lifing load of the motor.

http://www.sgps.net/ground_support_systems2.html

or

https://books.google.com/books?id=hfeXAAAAQBAJ&pg=PA177&lpg=PA177&dq=self+climbing+truss+mechanical+advantage&source=bl&ots=vZD-kkyBAW&sig=9gEFO-7luib-2SbQPnOEsxpu12M&hl=en&sa=X&ved=0ahUKEwjRhtmc85_TAhXDbSYKHU52BsQQ6AEIOjAK#v=onepage&q=self climbing truss mechanical advantage&f=false

page 177

so can someone with more brain power and more eloquent that I please explain how this works and that i'm not crazy after all these years ... at least more than i thought i was.

 

[Van]

Because the hook of the chain motor is attached AND the motor itself is attached, to what you are lifting, you are actually creating what is depicted in example #2 of your first link. If the chain were hooked, lets say, to the base of the column then you would be creating a #1.

 

[RonHebbard]

Because the hook of the chain motor is attached AND the motor itself is attached, to what you are lifting, you are actually creating what is depicted in example #2 of your first link. If the chain were hooked, lets say, to the base of the column then you would be creating a #1.

I'm fully in agreement and posing a question for others as to the loads being applied to the points in the illustrated examples.
If this results in a thread of dissenting posts, I'll leave it to @Van to explain and correct.
Toodleoo!
Ron Hebbard.

 

[Footer]

Think of it like this... with the chain also hooked off to the moving truss every foot of chain pulled in by the motor the truss only raises 6". With that, you have a 2:1 ratio which doubles your lifting capacity of said motor. Each "leg" of the motor is only supporting half the weight.

 

[derekleffew]

Is THAT why 2-ton chain hoist s have a pulley and twice as much chain (and weigh twice as much)?

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[RonHebbard]

Is THAT why 2-ton chain hoist s have a pulley and twice as much chain (and weigh twice as much)?

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And often run faster to match climb rates with other motors.
Toodleoo!
Ron Hebbard

 

[RonHebbard]

I'm fully in agreement and posing a question for others as to the loads being applied to the points in the illustrated examples.
If this results in a thread of dissenting posts, I'll leave it to @Van to explain and correct.
Toodleoo!
Ron Hebbard.

Just to belabor this a little more / further.
If you're sitting astride a beam and pulling up a load on a rope, the load on the beam is your full weight plus the weight of the rope and the load. Purely for example: A 100 pound load equals 100 pounds plus you and the rope.
If you're standing on the ground and throw a rope up and over a beam and pull up the same 100 pound weight, the load on the beam is now the 100 pound weight plus the 100 pounds of downward force you're exerting to lift the 100 pound weight: 200 pounds in round numbers ignoring frictional losses and other factors.
O.K., I've opened the 'worm can', let's see how many wriggle out and who chimes in with what details.
@Van possibly you'll keep an eye on this.
Toodleoo!
Ron Hebbard.

 

[TNasty]

Here's how I always put it:

No matter what, you're always going to do X amount of work (work is calculated by W=FD, W being work, F being force, and D being displacement). Mechanical advantage just enables you to do less "instantaneous work", or in other words, reduces the force needed at the cost of increasing displacement needed for the same effect.

It's just like using a block and tackle in backwoods engineering, each rope in the tackle shares, or mitigates, the tension force to its neighbors.

@RonHebbard said it pretty well. If you're familiar with the "Free Body Diagram" concept, you'll notice that no matter what, you're going to need to get a total force up (in tension) equal to the gravitational force pulling down on what ever it is that you're lifting just to keep it moving at a constant velocity, or to keep it from falling. Note that in order to start moving, you're going to need a force imbalance (upwards from tension). This can be calculated by F=MA, (F = Force in Newtons, M = mass in kilograms, and A = Acceleration in Meters per second per second (or just M/S*)).

Example: Say you have a pulley system where you need to exert 49N of force to counteract gravity's pull (V=0), and you want to accelerate by 1 M/S* for 1 Second to travel at 1 M/S. The target object weighs 10 Kilograms. You'll do 10*1=10N. Now, you need to add the 49N of counteractive force to actually move up.

It's all quite simple if you don't think too hard about it, and just try to envision it.

 

[JChenault]

Worm challenge accepted.

Just to belabor this a little more / further.
If you're sitting astride a beam and pulling up a load on a rope, the load on the beam is your full weight plus the weight of the rope and the load. Purely for example: A 100 pound load equals 100 pounds plus you and the rope.
If you're standing on the ground and throw a rope up and over a beam and pull up the same 100 pound weight, the load on the beam is now the 100 pound weight plus the 100 pounds of downward force you're exerting to lift the 100 pound weight: 200 pounds in round numbers ignoring frictional losses and other factors.
O.K., I've opened the 'worm can', let's see how many wriggle out and who chimes in with what details.
@Van possibly you'll keep an eye on this.
Toodleoo!
Ron Hebbard.

Digging through the worm can
( There is something about this scenario that I am having trouble with)

( For discussion assume the rope has 0 weight, and everything is friction-less)

If I sit on a beam and am holding a 100 lb weight and I weigh 100 lb, the load on the beam is 200 lb ( sounds right to me)
If I weigh nothing, the load on the beam is 100 LB.
If I tie the line to the beam, the load on the beam is 100 LB ( Still seems reasonable)

If I tie two lines from the beam to the load, the total weight on the beam would still be 100 lb. ( seems correct)


If I throw a rope over the beam and hold the weight off the floor, the load on the beam is 200 LB.
If I tie the end of the rope to the deck, the load would still be 200 lb. ( still makes sense)

If I tie the end of the rope to the 100 lb weight the load on the beam would still be 200 LB . But this scenario seems to be the same as simply running two lines from the beam to the load which we have agreed is 100 LB.


What am I missing??

Taking it a step further. If we refer to the diagrams at the top of the post - if you look at figure one - by your analysis, the load on the beam would be 200N ( 100N for the weight, and 100N for the downward force you are exerting on the rope). But if you look at figure two, the load on the beam would be 150N ( 50N for each line going to the weight, and 50 N downward force on the rope). Why does the beam care?

 

[RonHebbard]

Worm challenge accepted.



Digging through the worm can
( There is something about this scenario that I am having trouble with)

( For discussion assume the rope has 0 weight, and everything is friction-less)

If I sit on a beam and am holding a 100 lb weight and I weigh 100 lb, the load on the beam is 200 lb ( sounds right to me)
If I weigh nothing, the load on the beam is 100 LB.
If I tie the line to the beam, the load on the beam is 100 LB ( Still seems reasonable)

If I tie two lines from the beam to the load, the total weight on the beam would still be 100 lb. ( seems correct)


If I throw a rope over the beam and hold the weight off the floor, the load on the beam is 200 LB.
If I tie the end of the rope to the deck, the load would still be 200 lb. ( still makes sense)

If I tie the end of the rope to the 100 lb weight the load on the beam would still be 200 LB . But this scenario seems to be the same as simply running two lines from the beam to the load which we have agreed is 100 LB.


What am I missing??

Taking it a step further. If we refer to the diagrams at the top of the post - if you look at figure one - by your analysis, the load on the beam would be 200N ( 100N for the weight, and 100N for the downward force you are exerting on the rope). But if you look at figure two, the load on the beam would be 150N ( 50N for each line going to the weight, and 50 N downward force on the rope). Why does the beam care?

When you start getting the Fig's (Figures) and Newton's involved, you're going to upset my diabetes. (I'll go back in my hole now.)
Thank you John @JChenault for accepting the "worm can challenge" and furthering this thread in an interesting, and potentially educational, direction. Optimistically we can keep @Van involved here as well.
Toodleoo!
Ron Hebbard.

 

[TNasty]

If I tie the end of the rope to the 100 lb weight the load on the beam would still be 200 LB . But this scenario seems to be the same as simply running two lines from the beam to the load which we have agreed is 100 LB.


What am I missing??

Here's what you're missing. (Pardon the use of Newtons, it's just what I've always used in the world of forces)

If the frame of reference remained the 100N mass, you would notice that the sum of its forces equal zero. If at any moment the forces were not equal (say it did have an excess of 100N in the upward direction when the rope gets attached back to the mass), the target object would start to accelerate. If the object started to accelerate (upwards in this proposed case), an "effect" force (such as a normal force from the ground, or in this case, tension from a wire) would then be released. This would then result in the object accelerating back down until the force reaches equilibrium, but the object will continue to travel to some extent, as its momentum will not be stopped by a mere frame in time where it's traveling at a constant velocity (equilibrium), and then gain momentum and velocity back up.
The reason stuff doesn't bounce endlessly like a poorly scripted video game is due to dampening (shedding momentum into earth) and other forces like drag.

 

[porkchop]

Yes to basically everything that's already been said. Single lift line hoist double over and become a 2:1, double lift line hoists (2-tons in most cases) become 4:1.

I'm really only adding to the conversation to throw out the term "Self-climbing ground support".

 

[brin831]

guys this is has gone exactly the direction i was hoping because last thing i wanted was a discussion about if you don't know the weight of what you are lifting, qualified person, bla bla, that many threads of this nature go to.

that said ... does anyone have a better name than "Self-climbing ground support" ??

also i'd like to say this exact discussion continues in the shop with varying levels of disagreement ... some logical some not so much ... i'll probably share some of the responses so continue on!

 

[porkchop]

also i'd like to say this exact discussion continues in the shop with varying levels of disagreement ... some logical some not so much ... i'll probably share some of the responses so continue on!

If you have the time and space at work to test actual results vs just quote people from the internet then run the hoists for one minute and see how far the truss moved. If you're using 16 foot per minute hoists the truss will have moved 8 feet. It's not like the hoist magically runs at half speed when using it in this way so that 8 feet of unaccounted for chain had to go somewhere. That's what makes 40 foot towers using 8 fpm 2-tons in a 50Hz country so much fun (my thumb hurts just thinking about it).

 

[JonCarter]

Remember your H.S. physics or mechanics class: 1. The load being lifted is the TOTAL WEIGHT of the payload DIVIDED by the number of lines lifting it. 2. The SPEED it moves is the "puller's" speed DIVIDED by the number of lines lifting the load. (All of this is theoretical and must be adjusted for assorted frictions along the way.) Now, the problem amounts to: counting the # of lines involved, adding up the total weight being lifted and figuring friction.

 

[icewolf08]

Now, to add another layer to the discussion:

In the case of these "self climbing ground support" towers, if you are lifting the truss with a 1-ton hoist, how much load can you actually lift on the truss?

If you have 2:1 mechanical advantage can you lift 2-tons or, since it is still a single chain on a hoist with a motor and brake designed for 1-ton, can you only lift 1-ton?

 

[porkchop]

Now, to add another layer to the discussion:

In the case of these "self climbing ground support" towers, if you are lifting the truss with a 1-ton hoist, how much load can you actually lift on the truss?

If you have 2:1 mechanical advantage can you lift 2-tons or, since it is still a single chain on a hoist with a motor and brake designed for 1-ton, can you only lift 1-ton?

Now you're just stirring the pot. The hook is taking half the weight and the motor is taking half the weight, and those individual loads are within what they have been designed for. The load of both is carried by the tower not the hoist. There is certainly an argument about dynamics, but given that CM turns a 1-ton into a 2-ton by doing little more than changing the hook and running the chain back to the hoist body obviously it doesn't bother them.