Aerial Rescue

[TLHamel]

Norm, yes the moving block has two legs but shouldn't the reaction force still be 3:1? I know that it's not, but why? If a 5:1 fiddle block is used to pull a tree over, the moving block has 5 legs and the anchored block has 4 legs; reaction force is 4:1. So take the same fiddle block and rig it opposite. It's now "rigged to disadvantage", 4 legs on the moving block and 5 legs on the anchor. 4:1 output, but 5:1 on the reaction force... right??? Or is it not possible to have a reaction force greater than output force? Anyway, when lifting a victim, the RADS setup in my photo increases TIP load, but doesn't yield a greater MA than just pulling up on the tail by hand.

 

[Raven]

Hey Mark B., no disrespect meant, just trying to think of ways for the ordinary groundie to do these rescues. I think these written words can sometimes be a little confusing, pictures sure help, thanks Taylor.

I think we all agree that there's more than one way to skin a cat, as long as you've practiced first.

Peace

 

[Mark_B.]

Thanks Raven,

no offence taken, guess I was just feeling fragile. I get days like that.

M.

 

[Norm_Hall]

[ QUOTE ]
If a 5:1 fiddle block is used to pull a tree over, the moving block has 5 legs and the anchored block has 4 legs; reaction force is 4:1. So take the same fiddle block and rig it opposite. It's now "rigged to disadvantage", 4 legs on the moving block and 5 legs on the anchor. 4:1 output, but 5:1 on the reaction force... right???

[/ QUOTE ]

To calculate the MA, count the parts of rope at the traveling block. Presuming all parts are parallel that is your MA. In the attachment(typical Z rig), the MA is 3 to 1. The end of the rope you are pulling on is also counted. So, in the 5 to 1 set up, if the traveling block has 5 parts of rope, including the end your pulling on, the MA is 5 to 1. If you reverse the blocks, the end you pull on is now on the stationary or anchored end, and the MA is 4 to 1.

 

[Mahk_Adams]

???? Norm wrote:

[ QUOTE ]
Taylor, count the parts of rope 'at the traveling block' (the grigri) (2), that's your MA,(2X) if the ropes are parallel.


[/ QUOTE ]


but then:


[ QUOTE ]
In the attachment(typical Z rig), the MA is 3 to 1.

[/ QUOTE ]


???



And aren't you and Taylor saying the same thing about the fiddle blocks?




Taylor I think you are close in your description of the GG setup. The reaction force can be higher than the MA, and that situation is called 'rigged to disadvantage'.

But I don't understand why you got 2X's at the TIP. I think it's a double whip tackle and should be 1.5X's the load. Try replacing the GG with a pulley and see what you get.

 

[TLHamel]

My head hurts. I believe I am not seeing the forest for the trees!

Norm, I understand the moving block concept in a simple MA system. Moving = output force. Stationary(anchor) = reaction force. The Grigri moves(output 2:1), so I figured if I had 3 legs opposite it, I should have 3x reaction force. Obviously, this is incorrect.

Mark, my first measurement was over 2x with the Grigri installed so I switched for a pulley. I read 2x with the pulley. The scale could be goofy and I am lifting a light weight so the loss to friction may be high. Anyway, forget that. I am lifting 100% of the load, obviously. During the lift, the static leg(green arrows in attachment) must hold 50% of the load. The other leg comming out of the Grigri holds 50% and I need to pull(down in this case) on the third leg with 50% of the load force. 50+50+50=150%=1.5x the load at TIP.

Works for me!

Goodnight everyone!

 

[Mark Chisholm]

I agree T. If you want a load to move upward, you need to apply more force than you would to keep it static. That is why there is a definite increas on the TIP.