We use Seagrave tiller trucks with a 250# tip load for a high point. That tip load is for any extension / elevation combo. So it makes sense that this is for full extension straight off the side of the truck. It is a number based off of the operating pressures of the lifting cylinders.
Seagrave says the max operating pressure of those cylinders is 2500 psi. So, we are able to judge how close we are to our ladder's weight limits by keeping an eye on the cylinder pressure gauge. I'm not sure about other manufacturer's operating pressures.
We have a 60 ton rotating boom wrecker that does not have that pressure gauge and instead have to rely on weight estimates and a capacity chart on the wrecker. I've wondered why they don't have pressure gauge.
A LOT of wreckers DO,the gauges are in with the operating handles. Any that I'm aware of will do their rated capacity retracted and something in the 50% area extended. T.C.
so after much trial and error we've decided on a method and used it quite a bit in training now and are content with what it offers yet still provides a solution to our needs. Attached is a training document i put out showing 2 of the 3 configurations we are going to employ.
We went with 3 configurations:
1) fixed crane - litter attached to tip, full aerial function with package attached
2) reeved crane - ropes operated from turntable, over tip, dual tensioned mains, no traditional belay
3) high directional - fixed pulley at tip, no aerial function once operation has begun)
I received the blessing from the manufacturer on the first 2 configurations and i'm working on the 3rd with them, specifically looking at increasing tip loads if you eliminate the egress section of the aerial which carries the least rating in many positions. I'll post more once i complete that component.
Let me know what you think as i'm sure you all will...
Looks good, you've put a lot of research into it and got the manufacture involved. Like how the one person load on the non-fixed systems has been identified. How fast is it to set up? They look very intensive.
If possible, tag lines will help a lot in all the set-ups.
---section removed--- figured out the 3:1 system on the aerial.
How big is your bay? It looked like the one picture was an indoor training tower, or is that an optical illusion?
recently we had an evaluation involving a subject in a tank where the fixed crane evolution was selected as most effective. i'd say we had the stokes ready to raise within 10 minutes of setting the parking brake on the aerial.
tag lines are used on all setups, i guess i must have forgot that part. one thing i learned is to place the tagline on the end of the stokes that will keep tagline tender in visual contact with the aerial operator since they can serve both as spotter and tagline operator.
those pictures were taken at our indoor training facility. looks big but you don't have easy movement of the aerial in there. i should have taken a picture of the whole scene as i shoe-horned the truck in there between other rigs and had to be pretty careful not to take out hvac ducting and lights with the stick...
Mike, would love to see that if you ever get a chance.
Originally Posted by stickboy42
If I remember from when this started I was thinking that the fixed crane method was being used with a longer attachment, creating problems with keeping the high point above the pt. The short throw of the aztecs and the ability of the rescuer to adjust them would eliminate the problems I was envisioning. I also like the dual 3:1's (reeved crane) a lot the more I looked at it.
Interesting question I have. After reading here about useing the aerial as a hd if you attach a 3/1 or 4/1 M A to the tip of your 250# rated tip are you not setting the tip up for overload and or failure.
WHAT'S the LOAD? If it DOES NOT exceed 250# then NO you aren't. T.C.
Originally Posted by rescueraver
Rescueraver I see where your line of thought is coming from but, I think you are slightly left of center in your thinking. You have to remember by using a mechanical advantage system all that you are doing is lessening the input force used to move an object. 1:1 = 250#s force to move 250#s, 3:1 = 84.3#s force to move 250#s, 4:1 = 62.5#s to move 250#s, so on and so forth. Higher the MA, the lesser the input, the easier/less work involved in moving said object. The load does not change, just the input force, the load is almost always a fixed weight. Not many ways to lighten a load, some, but not many. You are still moving the same weight. In actuality you are lessening the strain on the structural members of the aerial since you will be pulling against the structure with less force to move the same weight. There will always be the same weight hanging from the aerial, just less force in certain directions when the system is being raised. Pure physics, don't let the idea of MA confuse you.
Great info. Noticed from the pics that you appear to be using Pierce's Lyfe Pulley Rescue System. Any experience in your testing using the system with the MA sewn off the tip in either a 2:1 or 4:1 configuration and anchored to the bottom square rung. I did some training with a local department with this system, neither they or I had much experience utilizing it, so it was a teach and learn for all of us. After researching the system, I've found that it can be rigged two ways, one supporting a maximum load of 500# with the main line anchored at the base of the aerial, and the other supporting a maximum load of 250# with the main line anchored to a separate anchor independant of the apparatus. Both ways can be utilized as a COD/High point or a crane type set-up. Correct me if I am wrong, however if the aerial has a 1000# tip load, wouldn't you be better off rigging the main 4 rungs back from the tip with a sling or webbing and maintaining the full use of the 1000# tip load?
Sorry I'm "coming late to the party" on this one but as I read through the post's here I think we're mixing apples & oranges in our discussion or we're overlooking some principles of physics. Also variations on the location of our working end, running end, and anchor points will result in drastically different scenarios.
Calculating the actual forces of the various way you should use & rig an aerial really gets into some pretty moderate physics and vector analysis so some of my examples will be greatly over simplified.
That being said, the first and main concern for placing any type of load on an aerial device is Dynamic Loading (often referred to as Shock Loading). Anyone who's ever had any training on working from aerial devices has surely been told not to routinely "jump on" the ladder when leaving the structure / working area (especially if the tip is unsupported). A 250# geared up firefighter that jumps / falls 6" onto an aerial device will generate an impact load of over 1,400# - the same physics apply if you "drop" a 250# load (combined load of Pt, stokes, rigging, etc) from the same 6" (say while trying to get them off a ledge or parapet).
Suddenly the tip load, design safety factor, and age/condition of the device are VERY important)
It should be noted that swinging a suspended load can also create shock loading of rotational forces which not only affect the aerial itself but also the torque box - after all the aerial is just a big lever and 250# of force on the tip of a 100' device is equal to 25,000lb/ft of torque at the base.
Now lets assume that we make every single move flawlessly and create absolutely Zero shock loads whatsoever and look at rigging.
A 250# load connected directly to the tip of the aerial device creates a 250# "load" or downward force on the tip - easy.
Assume you have the same load in a 1:1 mechanical advantage (MA) system (which is actually a misnomer since there is no advantage to the system rather just a change of direction) anchored at the tip of the aerial. You now have 250# downward force on the load side of the pulley plus an equal & opposite 250# downward force on the working side of the pulley which in turn translates to a 500# load on the tip of the aerial device where the pulley is anchored.
The only benefit of a true MA system (in terms of tip load only) is that you reduce the anchor force by reducing the required input force to hold/raise the loading force.
Or in math terms
A 2:1 MA system would still have the same 250# load but only require a 125# input force on the working end thus translating into a 375# load on the tip.
A 4:1 MA system would then be 250# load + 62.5# input force = 312.5# tip load
For the purist in the crowd these examples assume all force directions are parallel to one another and are therefore the maximum forces applied to the anchor for the given loads/forces. Once you begin pulling at angles, the anchor (tip) loading can decrease further based on the pulling geometry.
Now all this isn't meant to say you can't or shouldn't use an aerial device as any type of high point (or even a "crane" of sorts by rotating a suspended load) - but in addition to getting the device designers blessing on your intended usage, you should also request them to provide you with Working Load Limits (WLL) for those usages as well as yourself being able to adequately estimate your real life loads and stay within the given limits.
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.
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?
I keep hearing from administration that it is an OSHA violation if you use the ladder as "a crane". Does anyone know what standard they are referring to? I cannot find which one they are basing this off of. I know one of the standards says something about using machinery and Im guessing this is what they are basing it off of. I want to be able to take something in writing and discuss the issue.
This is a great topic because it seems to have so many "what if's" and unknowns. I'm here in the firehouse working this easter reading through these posts and as always my creative and sometimes left-field juices get flowing.
The greatest focus and rightfully so seems to be on the tip load. So here comes a crazy idea and I'll again this is just an idea and no force loads have been measured with that being said.....
Why not connect a rope to the tip of the ladder and run it back to an anchor opposite the direction of the ladder. Once the load has been applied to the main line tension the line attached to tip going to the opposite anchor. In essence your back tying the anchor hoping to equalize some of that force. What are your thoughts? Do you think this will help the tip loading problem. I must also point out that the line used to back tie the overhead anchor (ladder tip) cannot come in contact with the ladder, if it did the load force would be placed back on the ladder defeating the purpose of back tying.
Over at another site(VES) there was a story about a dept using a stick to aid in lifting a large man out of a bath tub in a house. They cut a hole in the roof, etc and used the pulley as the one pictured on the end of the stick on page one as a COD and had their system down on the ground.. They were getting harrassed on overloading, etc but after contacting Pierce they were told that if they were right above it and no side load and the load indicator was still in the green. GO for it. The stick doesn't know where that force is comming from, just that it has weight pulling down on it.
Note: this was a tip rated 750lbs 105' stick.
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.
Originally Posted by MG3610
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. :)
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.
Originally Posted by jmatthe2
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?
Originally Posted by Rescue101
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.
Yes sir - that is correct - provided you are pulling directly back down to the ground.
Originally Posted by MEAN15
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*
Originally Posted by MEAN15
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.
Originally Posted by N2DFire
Trust me brother you are NOT the only one. ;)
Originally Posted by Rescue101
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.
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.
Originally Posted by Rescue101
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. ;)