Ricardo_Montalban Posted July 30, 2004 Share Posted July 30, 2004 !! whoo hoo!! my first unless i'm missing something, that site only describes forces required to pull the weight. i didn't see anything about forces of the weight and hauling the weight against the anchor/s. not saying you're wrong, just still not convinced. Quote Link to comment Share on other sites More sharing options...
cj001f Posted July 30, 2004 Share Posted July 30, 2004 unless i'm missing something, that site only describes forces required to pull the weight. i didn't see anything about forces of the weight and hauling the weight against the anchor/s. not saying you're wrong, just still not convinced. Forces at each anchor are equal & opposite to the one applied on the rope. To find the total force on the anchor, sum the forces on each anchor. Quote Link to comment Share on other sites More sharing options...
erden Posted July 30, 2004 Share Posted July 30, 2004 Force is never multiplied on the anchor with a block and tackle (which is what a Z haul is, essentially See Here That URL is very easy to misread. "Does this change anything? Not really." is bogus, because the system applies twice the force on the ceiling. The statement there only applies to the tension on the rope. When one builds an anchor, one best make sure that it is strong enough to carry the weight of the load, and one's own weight. In the case of a z-pulley, if the traditional set up is used, you will apply 1/3 of the load weight on the rope, and you will be pulling away from the crevasse. At the anchor, there will be a force equal to 2/3 of the load. Bottom line is: based on the number of pulleys involved, if you pull toward the crevasse for some weird reason, you will add the pull that you apply to the load already on the anchor. This is the case in big wall anchors and hauling for example, that is why I suggested that the system must support your weight as well... It does not make sense to argue over hypotheticals and generalizations, best to do the physics on actual set ups, hence my default example of the z-pulley. Erden. Quote Link to comment Share on other sites More sharing options...
snoboy Posted July 30, 2004 Share Posted July 30, 2004 In the last example cited on cj's link, the force on the anchor (ceiling) would be 125 lbs so yes, it would be higher, but it is a sum, not a multilple. Maybe the easiest way to explain this is that each rope that goes to the load (to a moving point) shares an equal portion of the load. In a z-drag, there are three ropes sharing the load - one goes to the rescuer's hand, the other two go to the ratchet/pulley on the anchor. The total load is divided equally between those three, so the anchor holds only 2/3 of the load while the rescuer is hauling. When the rescuer releases his pull, the load is fully on the ratchet, and the anchor supports 100% of the climber's weight. Of course these loads are theoretical, and will be increased by friction. Note in the first example on cj's link, that the anchor is actually taking 200lbs! That would be the classic redirected belay setup on a multipitch. In the second one it is taking 150... In each case the pulley to the right does nothing except change the direction of pull, and increase the load on the anchor... Quote Link to comment Share on other sites More sharing options...
erden Posted July 30, 2004 Share Posted July 30, 2004 When the rescuer releases his pull, the load is fully on the ratchet, and the anchor supports 100% of the climber's weight. Of course these loads are theoretical, and will be increased by friction. Good catch - a necessary clarification to what I said! E. Quote Link to comment Share on other sites More sharing options...
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