CascadeClimber

That Aztech thing looks familiar. Where have I seen it before... Oh yeah: And the climber's version:

I have a 95 4Runner that I bought with 37,000 and now has 120,000. It has been a great, reliable car. Repairs are much less frequent that my previous Big-3 cars (including a Jeep CJ7 and a Fiero), but Toyota parts and repair work are expensive. I was also concerned about the head gasket issue, which affects the 4Runners through 1995, but I was told by Michael's Toyota in Bellevue that the factory has agreed to cover head gasket repairs of affected vehicles regardless of age or mileage. As to price, I got so sick of the inflated prices in the Puget Sound area that I ended up buying my 4Runner in Bend, Oregon. I think I bought it in 1998 and paid $17,500. Similar vehicles here were in the $25k range.

Hackmeister, you're working in a heterogenous environment, which is the exception, not the norm. You can't plug you Mac into a Windows 2000 Domain and logon without some configuration work. Sure, you can probably pick up an IP address from a DHCP or BOOTP server and load web pages, but that won't get you access to permission-restricted network resources. On an average day my laptop is attached to 4-6 networks (wireless and wired, Win2k and WinNT domains), without rebooting or any other such hassle. Thankfully, all OSes have gotten more robust and user friendly in the last few years. I was a Mac evangelist for years. I gave up not because of anything Apple did (though they did and do lots of stupid things), but because of Office 4.2 for the Mac, which was Windows code running in an interpreter. It crashed the machine over and over and I couldn't get any work done. Damn MS. I understand that the latest version of Mac Office is nice, but what happens when you need to use an Access database? As with most things in life, the bottom line here is this: Opinions are like assholes: Everyone has one, and everyone thinks their's is the only one that doesn't stink. -L Oh yeah, I agree 100% on the mega/giga/terra/google-hertz overhype. Unless you are doing major number crunching, as in video rendering, you don't need the latest, greatest, most fastest wonder widget with the highest number. And, a 1Ghz CISC (Intel et al) processor is not as fast (in most cases) as a 1Ghz RISC (Mac) processor. Now, isn't there some ice to climb by now?!?

Look at me, I figured out the quote thing. Dell has good product. Not without problems, but good. Easier to deal with than the HP-Compaq monster, and much more likely to exist in three years than Gateway and Micron. I buy Dell for my clients, I buy Dell for my business, and I buy Dell for home. But If I didn't have to interact with the rest of the electronic world, I'd own a Mac. Wintel may not have the best product(s) in a pure sense, but they are, sadly, the best solution in most situations today. -L

Try to find one (new or used) with a decent warranty. Otherwise you'll be SOL when (not if) it breaks. My clients have a lot of laptops and the ones that get moved around a lot have little things go wrong with them. There is risk in spending too little. You don't need a Mercedes, but you don't want a Yugo, either. There should be some kicking sales post-Christmas. -L

Why can my small brain not discover how to "reply with quote"??? Don't situate the belay directly below the line of climbing for the next pitch. Regardless, a leader fall onto the belay is going to be highly unpleasant for everyone involved, regardless of position, rope system, how much Smoker-custom-roast-espresso you had that morning, or how many hail-Marys you said the night before. Perhaps the best position is with your head tucked between your knees so you can kiss your ass goodbye. So leaders, get a damn screw in to protect the belay on multipitch ice routes. Can it please get cold now so we can stop talking about ice and start climbing it?!?!?!?!?!

So getting back to the original issue, if the force introduced into the system by the falling climber is constant, who can tell us where the additional compensating force comes from? Bueller? Anyone?

Overall it is 1-1, but the effect on either load alone is 0.5-1, I think. Did this thread have something to do with climbing??

If L1<>L2, then 2-1, I think. If L1=L2, then 1-1??

But what about the friction in the system, and the force of gravity on the anchor points and the rope, and hydrogen bonds between the sheaves and the rope, and what about the moon and its gravitational effect and...ouch. I think I just hurt myself.

The first looks like a Z-pulley, 3-1. The second looks like a form of a C-on-a-Z, 6-1. Do I pass?

Loren, if I only had a nickel for every time my physics prof. answered questions like this by saying "Draw the Free-Body Diagram" ! I was trying to do this last night. I gave up and watched The Daily Show instead. Maybe I'll break out my old notes and textbooks and try again. It gets complicated when you toss in the frictional forces...

It depends entirely on the position of the belayer. In the best case scenario, which is not difficult to approximate in the field, almost 0 force/energy is applied to/absorbed by the belayer, it is simply passed through to the anchor. All you have to do is get into a position that aligns the force of the fall with the anchor and remove slack in the system that connects the belayer to the anchor. I try to lean against the anchor and face perpendicular to the fall-force vector (facing 90 degrees from the direction of pull in the case of a fall). So in Alex's example, the spiral fracture occurred because the belayer was looking up at the leader and was spun around by the fall. But if he was oriented differently, with no slack, the force would not have spun him and he would have felt much less force overall. Running the rope through the anchor simply makes predicting the direction of the fall-force vector, and consequently the best body position, easier -L

If the rope doesn't go through the anchor can't we assume that some of the force is absorbed by the belayer and transfered into the ground, muscles, breaking bones, before it gets to the ancor. It seem like with all the details of newtonian physics some of the more intangible forces have been left out. I wish I had a nickel for every time a prof said, "But for now let's ignore those forces." You are right. A dynamic rope absorbs energy in a fall, as does friction at each point where the rope runs over a surface, including the friction of the kern against the mantle as the kern stretches, and the friction of the individual kern fibers on eachother. It turns out that calculating some of these quantities is very difficult. The rope for example, would presumably act like a spring. But a spring resists pull/absorbs energy at a constant rate (the 'spring constant'), and ropes don't. The complication of the minutae is why they are often left out of calculations and why much of the data about impact forces, energy absorbtion, and ultimate strength of climbing gear are determined empirically- through testing. It is easier, faster, and more accurate to do it that way. Can be think of the belayer like a really big screamer? Yep. Especially if they end up with a compound spiral tib-fib fracture. That would make a screamer out of me. Hey, this is the most I've used that dusty old ME degree in ten years! -L

Hey Lyle, how can I get my hands on one of those? Can I be Steve Berard? Or can I just buy one? -Loren

The article is excellent. But it doesn't discuss the situation in question here. The closest illustration is #3, where the force generated by the fall is 5.7kN, but the top draw experiences 8.9kN. This illustrates the additive effect I was pointing out. But what is doesn't do is show that the force on the belay without the gear would be equal to the force of the fall. If the rope is run from the belayer to the belay anchor to the climber and the leader falls with no other protection in place (or the second falls at all) the resultant impact force on the belay anchor is significantly higher (double, ignoring friction) than if the rope is not run through the belay anchor. So if I am belaying on a shitty swedgie anchor, or bad screws, I don't run the rope through the belay, inorder to minimize potential impact force on it. If the anchor is bomber, I do, because it is more comfortable and the force of a fall on me is less, due to friction. However, if the force is high enough, I could do a face-plant into the rock or ice which would make me even uglier than I already am! -L

The force on the draw, and thus the anchor is STILL 1060, or thereabouts. Agree? No. The force on the anchor is not the same. It can't be if the sum of all the forces equals 0. The point I was trying to make, perhaps in an unclear manner, is that running the rope from the belayer through the belay anchor will result in the (near) doubling of the force on the belay anchor if the leader falls with no gear in above the belay (or if the gear fails), or if the second falls. A fall of either type tranmits twice the force to the anchor, less frictional forces, as it does if the rope is not run from the belayer to the anchor to the climber. So if I mistrust the belay anchor, I don't run the rope through it when I am bringing up a second or belaying a leader. Maybe this is better settled over a beer...

Here, from the Beal web site: In case of a fall, the anchor sustains a double load: that of the climber and that of the belayer. These two forces combine in a pulley effect. If the rope does not absorb enough energy, the load on the anchor point will be very high, with increased risk of failure… Rope with a low impact force is the key to limiting the consequences of the pulley effect. -L

Alex- You are right, the sum of the forces in x, y, and z planes must be zero, or some part of the system must be in motion. Where you are mistaken is in your assertion that the force on the anchor is constant. It is not. The constant in the system is the force related to the fall of the leader or second. Surely you can see that the y-vector downward force created by the fall cannot change as a result of the anchor system. F=MA, the mass of the climber times (in the case of a fall) the acceleration of gravity, 9.8m/s^2. What you are saying is that the force of the falling leader is different if the rope runs through the anchor. The orientation of the rope has nothing to do with the mass of the climber and the acceleration of gravity. It is counter-intuitive in this presentation, but if you think of it as a C-pulley, with the anchor as the victim, you know that the force required to hold the c-pulley in a static position is 1/2 the weight (in newtons or pounds-force) of the victim (ignoring friction). I'll go dig through my Statics book to find a reference if this still seems wrong to you. -L

What Forrest said. To look at it another way, it is a C-pulley system. 2-1 mechanical advantage, except it is a 2-1 mechanical disadvantage in this scenario. -L

Because to keep the system static, the anchor must absorb the force of the falling leader plus the equal force of the stationary belayer (less static and dynamic friction). This is the best pic I could find quickly: If you remove the rope from the quickdraw, the equal and opposite force on the anchor is only that of the falling leader. This also applies to belaying a second through the anchor. -Loren

It is worth noting that running the rope through the belay can effectively double the impact force on the belay in such a situation. If the belay failed you would end up with a much bigger problem than a tib-fib fracture. Sometimes I run the rope through the belay, and sometimes not. It depends on my opinion about the strength of the anchor system and the potential for a factor-2 fall. -Loren

<spray> Ice was found, fear was felt, fun was had, ice was climbed. </spray> -Loren

Looks good, Tim & Jon. The threaded mode is nice.