Aerial Ladder as High Directional or Crane

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Originally Posted by MG3610

a little late here...but...

So if I get this, as soon as you hang a pulley and use the aerial as a high overhead COD, youve doubles the load because (if I think this correctly) the load exerts its exact weight on one side of the pulley, and your haul force exerts equal, or maybe a tad more due to inedfficiency, force on the other side of the pulley.

Right?

If this is the case, a rescuer, baslet, and load can easily add up to 700lbs, so youre looking at a 1400lb load with no shocking. Whats the answer here to avoid this dangerous predicament, if my assumption is correct?

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This is where the ROPE guys lose me. I use COD every day,I just use wire rope instead of kermantle. A Lineal load applied to a vertical structure(Ladder,wrecker boom,etc.)ASSUMING that it is vertical loading is the weight of the object being lifted,PERIOD. Minus input forces if the MA is low and any swing or shock forces.Our Tower is set up to do these kinds of lifts,if I were doing it with our old Ladder,I'd bridle the tip so both sides were loaded evenly.If you use a 3 to 1 ma,it will take 100# input to raise a 300# load. If you go 4 to 1 Ma it is easier on the directional and the patient,but it takes more rope. 75# input to raise 300#. Or is there something about rope rigging that changes the physics I use everyday? T.C.

Did some searching, and answered my own question.

Rescue 101 -

If I read your post correct your school of thought is an incomplete depiction of what a MA is doing throughout the system. N2DFIRE explained the accurately, but here is another view.

1. A pulley is a simple machine that produces a 2:1 advantage (this is in a vacuum and under the correct conditions.)

2. The pulley provides this advantage whether attached to the load or when used as a COD. That is ALL pulleys is a system are multiplying forces. Your input to MA theory is correct, but your only considering the total output of the MA and not its impact on the anchor. Use the T method to trace forces to see what the anchor is actually seeing.

With that said, I'll use an example you have used. A 4:1 is commonly used attached to the end of a high-point. When used vertically it would be a 4:1. While the load is seeing a 4:1 MA, the anchor is seeing a 5:1. (Remember the pulley has no idea what end is moving the load). Therefore your anchor end is seeing on additional unit of tension (whatever that is).

This additional load is minimal, but many times it is not accounted for when teams rig an MA to an anchor to haul a load. Obviously the higher the load weight the more impact this additional unit can make.

The ideal way to use an aerial as a high-point is to run the rope up the bed of the ladder and over a pulley. This is why many manufacturers are making pulley plates that mount to the ladder. By doing this it eliminates additional multiplication of forces, eliminates the potential of torquing the ladder, and allows the ladder itself to absorb forces inline (the resultant force will point more toward the ladder vs perpendicular to the ladder). Maybe Eric Ulner will chime in on the effects of resultant forces on the ladder. I know works with Reed and would imagine they use these calculations frequently with the AV

Of course there are other methods....

Mike - If I had to go through all of that to stabilize the aerial I would devise another plan. Part of TRT is efficiency and good use of resources. Tying back a ladder, in my opinion, is neither of those. After 21 years I have learned to never say never, but let's say it would be a last resort and I would have to be lifting more than a rescuer and victim. :)

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Originally Posted by jmatthe2

Rescue 101 -

If I read your post correct your school of thought is an incomplete depiction of what a MA is doing throughout the system. N2DFIRE explained the accurately, but here is another view.

1. A pulley is a simple machine that produces a 2:1 advantage (this is in a vacuum and under the correct conditions.)

2. The pulley provides this advantage whether attached to the load or when used as a COD. That is ALL pulleys is a system are multiplying forces. Your input to MA theory is correct, but your only considering the total output of the MA and not its impact on the anchor. Use the T method to trace forces to see what the anchor is actually seeing.

With that said, I'll use an example you have used. A 4:1 is commonly used attached to the end of a high-point. When used vertically it would be a 4:1. While the load is seeing a 4:1 MA, the anchor is seeing a 5:1. (Remember the pulley has no idea what end is moving the load). Therefore your anchor end is seeing on additional unit of tension (whatever that is).

This additional load is minimal, but many times it is not accounted for when teams rig an MA to an anchor to haul a load. Obviously the higher the load weight the more impact this additional unit can make.

The ideal way to use an aerial as a high-point is to run the rope up the bed of the ladder and over a pulley. This is why many manufacturers are making pulley plates that mount to the ladder. By doing this it eliminates additional multiplication of forces, eliminates the potential of torquing the ladder, and allows the ladder itself to absorb forces inline (the resultant force will point more toward the ladder vs perpendicular to the ladder). Maybe Eric Ulner will chime in on the effects of resultant forces on the ladder. I know works with Reed and would imagine they use these calculations frequently with the AV

Of course there are other methods....

Mike - If I had to go through all of that to stabilize the aerial I would devise another plan. Part of TRT is efficiency and good use of resources. Tying back a ladder, in my opinion, is neither of those. After 21 years I have learned to never say never, but let's say it would be a last resort and I would have to be lifting more than a rescuer and victim. :)

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Better try again, A pulley is ONLY 2:1 if it is ON the load(Load lines),IE one end deadlined, then run thru a pulley(on the load)then back to the winch(driven end,be it winch or man).Otherwise, it is merely a COD. MA ONLY occurs in lines to load. In MY world if you snatch off a deadman (tree) you have the weight of the vehicle(3000#) on BOTH sides of that block(and the tree). If you anchor one end and put the block on the vehicle NOW you have a 2:1 with 1500# on each leg. I understand MA VERY well,I use it just about every day. Where I get confused is in the ROPE world where apparently everything I've ever been taught(or learned)is different. The one CONSTANT is MA is determined by lines to LOAD.If the line go to a high directional or multiple anchor points they are just COD UNLESS they go to load. T.C.

OK - correct me if I'm wrong. (I never was good at physics.)

A single pulley, hung from your platform. a 300lb load on the end. To lift the load, there will be roughly 300lbs on each side of the rope, and 600lbs on the anchor, right? The pulley used in this fashon acts as a force multiplier, and no MA is achieved, correct?

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Originally Posted by Rescue101

Better try again, A pulley is ONLY 2:1 if it is ON the load(Load lines),IE one end deadlined, then run thru a pulley(on the load)then back to the winch(driven end,be it winch or man).Otherwise, it is merely a COD. MA ONLY occurs in lines to load. In MY world if you snatch off a deadman (tree) you have the weight of the vehicle(3000#) on BOTH sides of that block(and the tree). If you anchor one end and put the block on the vehicle NOW you have a 2:1 with 1500# on each leg. I understand MA VERY well,I use it just about every day. Where I get confused is in the ROPE world where apparently everything I've ever been taught(or learned)is different. The one CONSTANT is MA is determined by lines to LOAD.If the line go to a high directional or multiple anchor points they are just COD UNLESS they go to load. T.C.

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Hey 101,
Based on this I think we all talking Apples n Apples we're just looking at them from different view points.

Let's stick with your example(s) of the 3000# car.

You are correct that if you have the car directly behind your wrecker and you run your cable down to the car, thru a snatch block, and back to your wrecker that each section of line is only carrying 1500#'s thus the 2:1 advantage. But the MA didn't make the car lighter, it just divided the work from one run of cable to two. The hook on the snatch block is still carrying the whole 3000# "load" of the car.

Now let's say the car is sitting beside your wrecker and you run your line out to a (really big) tree and thru a snatch block on the tree and then back to the 3000# car beside you. As you pointed out above; there is no MA to this system and as you begin to pull you now have 3000#'s of force throughout the cable from the car all the way to the winch. The physics in this example work the same as above - it's just the loads that have changed. Instead of having 1500#'s per line off the pulley, you now have 3000#'s on each line. So now the hook on the snatch block (and thus the tree / anchor) is seeing 6000#'s of force.

Translate that from the horizontal plane to vertical:
You are standing on the ground beside a 250# load with a pulley as a high COD on an aerial tip. One end of the rope is on the load, the other in your hands. As you begin to apply 250#'s of force into one side of the system, the load then resists with an equal 250#'s on the other side - thus the aerial tip is now supporting 500#'s of total downward pulling force (load).

This is where the concern of overloading aerials from use as a high directional comes in. The closer the angle of your pulling forces comes to 180 - the greater the doubling force on the anchor for the pulley.

These same basic principles would still apply if you were using a "complex" MA system to raise the load. The total net force on the ladder would be the weight of the load plus the input force required to lift it.

250# load with a 4:1 MA = 250# load + 63.5# input force = 313.5# tip load.

The solution to this is (as others have stated) to pull along the direction of the ladder down toward the turn table. This in turn decreases the angle between the ropes from the 180 degree position to something closer to 90 degrees and in turn reduces the overall loading force on the aerial tip.

Without having to learn complex trigonometry, vector analysis, etc. - for the "average" line firefighter a good set of general rules for aerials as High Directionals are:

1) The force on the ladder tip will ALWAYS be more than just the load
2) If you're pulling from ground to tip and back to ground (close to a 180 degree bend around the pulley) then 1/2 tip load capacity should be your MAX load weight.
3) If you're pulling from ground to tip and back down along the ladder bed (closer to a 90 degree bend) then 2/3 tip load capacity should be your MAX load weight.

Caveat to rules 2 & 3 - the angle and extension of your aerial device may affect it's tip load limits - also it WILL affect the angle of the rope around the pulley so you should take this into consideration when making your estimates on lifting capacity at every incident.

Bear in mind that these are just generalized rules and that you should have someone "do the math" for your device(s) and have them verified by the Mfg prior to making any lifts.

Hope this clears the confusion and gets us all back on the same page. As I said before I think we're all understanding how the system(s) perform - I think we're just looking at them from different points of view or looking at the force(s) on different components of the system.

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Originally Posted by MEAN15

OK - correct me if I'm wrong. (I never was good at physics.)

A single pulley, hung from your platform. a 300lb load on the end. To lift the load, there will be roughly 300lbs on each side of the rope, and 600lbs on the anchor, right? The pulley used in this fashon acts as a force multiplier, and no MA is achieved, correct?

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Yes sir - that is correct - provided you are pulling directly back down to the ground.

The factor of "Force Multiplication" is related to the angle of wrap around the pulley. The closer to 180 degrees -the closer to double the force.

Also - all of this is theoretical load since there are various losses in a real world system that would in turn further increase the anchor (aerial device) load.

Edit - sorry so slow to reply to this one - I was writing the "book" of a post above *LOL*

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Originally Posted by MEAN15

OK - correct me if I'm wrong. (I never was good at physics.)

A single pulley, hung from your platform. a 300lb load on the end. To lift the load, there will be roughly 300lbs on each side of the rope, and 600lbs on the anchor, right? The pulley used in this fashon acts as a force multiplier, and no MA is achieved, correct?

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Correct........T.C.

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Originally Posted by N2DFire

Hey 101,
Based on this I think we all talking Apples n Apples we're just looking at them from different view points.

Let's stick with your example(s) of the 3000# car.

You are correct that if you have the car directly behind your wrecker and you run your cable down to the car, thru a snatch block, and back to your wrecker that each section of line is only carrying 1500#'s thus the 2:1 advantage. But the MA didn't make the car lighter, it just divided the work from one run of cable to two. The hook on the snatch block is still carrying the whole 3000# "load" of the car.

Now let's say the car is sitting beside your wrecker and you run your line out to a (really big) tree and thru a snatch block on the tree and then back to the 3000# car beside you. As you pointed out above; there is no MA to this system and as you begin to pull you now have 3000#'s of force throughout the cable from the car all the way to the winch. The physics in this example work the same as above - it's just the loads that have changed. Instead of having 1500#'s per line off the pulley, you now have 3000#'s on each line. So now the hook on the snatch block (and thus the tree / anchor) is seeing 6000#'s of force.

Translate that from the horizontal plane to vertical:
You are standing on the ground beside a 250# load with a pulley as a high COD on an aerial tip. One end of the rope is on the load, the other in your hands. As you begin to apply 250#'s of force into one side of the system, the load then resists with an equal 250#'s on the other side - thus the aerial tip is now supporting 500#'s of total downward pulling force (load).

This is where the concern of overloading aerials from use as a high directional comes in. The closer the angle of your pulling forces comes to 180 - the greater the doubling force on the anchor for the pulley.

These same basic principles would still apply if you were using a "complex" MA system to raise the load. The total net force on the ladder would be the weight of the load plus the input force required to lift it.

250# load with a 4:1 MA = 250# load + 63.5# input force = 313.5# tip load.

The solution to this is (as others have stated) to pull along the direction of the ladder down toward the turn table. This in turn decreases the angle between the ropes from the 180 degree position to something closer to 90 degrees and in turn reduces the overall loading force on the aerial tip.

Without having to learn complex trigonometry, vector analysis, etc. - for the "average" line firefighter a good set of general rules for aerials as High Directionals are:

1) The force on the ladder tip will ALWAYS be more than just the load
2) If you're pulling from ground to tip and back to ground (close to a 180 degree bend around the pulley) then 1/2 tip load capacity should be your MAX load weight.
3) If you're pulling from ground to tip and back down along the ladder bed (closer to a 90 degree bend) then 2/3 tip load capacity should be your MAX load weight.

Caveat to rules 2 & 3 - the angle and extension of your aerial device may affect it's tip load limits - also it WILL affect the angle of the rope around the pulley so you should take this into consideration when making your estimates on lifting capacity at every incident.

Bear in mind that these are just generalized rules and that you should have someone "do the math" for your device(s) and have them verified by the Mfg prior to making any lifts.

Hope this clears the confusion and gets us all back on the same page. As I said before I think we're all understanding how the system(s) perform - I think we're just looking at them from different points of view or looking at the force(s) on different components of the system.

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In purist terms,correct. THIS is where it gets interesting with the ladder. If you run the RUNNING line over the rung and down the top of the ladder(walking side) NOW you've lessened the ultimate load as opposed to pulling vertically. See WHY I get confused? In my world the load is the constant and the only thing you can do to change it is apply MA or DECREASE resistance. As I said earlier,our Platform is DESIGNED and rated to do this type of work(within operational limits) so I never spent much time thinking about it. But it certainly has been an educational thread. Thank you all, T.C.

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Originally Posted by Rescue101

See WHY I get confused?

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Trust me brother you are NOT the only one. ;)

About 95% of the time I have to sketch the system out on paper so I can "see" what the forces are doing before I really understand it myself.

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Originally Posted by Rescue101

But it certainly has been an educational thread. Thanks you all, T.C.

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I have recently began taking some training in the realm of "Big Rig Rescue" and all I can say is my hat is off to you guys in the recovery industry. The system physics may work the same but you guys are some real wizards when it comes to force / load calculations & rigging.

In reading some of the various threads on here regarding Heavy Rescue / Recovery - I have learned a lot from your post - so let's just say I'm finally getting to pay a little back for a change. ;)

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Originally Posted by Rescue101

But it certainly has been an educational thread. Thanks you all, T.C.

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I have to agree.... flashbacks to 9th grade physics.

Good stuff.

Am I safe in saying....If the load indicator on the platform is still in the green for load limit and there is no side loading we should be good? So we don't have to rack our brains on the MA and F, etc, etc. I'll agree though I totally draw things out many times.

Attached is the Newton's 2nd Law analysis of three MA systems for the idealized case of
* no acceleration of the load (load is already moving at constant speed)
* direct downward pull
* pulley's with 100% efficiency

Note that the force on the tip of the ladder is a maximum of twice the weight of the load for a COD, and asymptotically approaches a minimum of the weight of the load as the MA is gradually increased.

If the pull is directed along the length of the ladder instead of directly downwards, the force on the tip of the ladder will have a component along the length of the ladder. This will decrease the moment exerted about the base of the ladder (a good thing).

Also note that the forces can momentarily exceed many times the weight of the load during while it is accelerating, e.g. being moved up to a constant speed from rest. This is one of the reasons we require a minimum 10:1 SSSF in rope rescue.

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Originally Posted by jmatthe2

Rescue 101 -

... Maybe Eric Ulner will chime in on the effects of resultant forces on the ladder. I know works with Reed and would imagine they use these calculations frequently with the AV

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Finally acknowledging your nod, jmatthe2...

Posts by N2DFire and servantleader have it pretty much covered. Happy 4th.

Sorry to drag up an old thread but there is some great info in here. Pat Rhodes did a great video on this very subject you can find it here http://www.rescueresponse.com/store/...e-rigging.html