Safe Rig?

[NZcaver]

Because the deviation acts only as a change of direction for the force, I contend that the main anchor always remains at 100% but the deviation anchor force varies with the angle as per Tim's diagram.

Ian - sorry, the wording in my previous post about the main and deviation anchors "sharing" the load was probably a little ambiguous.

It's not stated in Tim's diagram, but it's reasonable to assume the tension on each side of the pulley sheave always stays equalized. This means your main anchor should always see about the same force regardless of the angle of deviation - assuming no friction at the deviation point (and no knot of course, that would make it a Y-hang). Just like you said in your post, if the deviation point is directly behind the anchor point (away from the load), the anchor will still see about the same force - just in the opposite direction.

Say your load is a big caver with gear, weighing about 220lbs total. You rig a 45 degree redirect over the hole. Once the caver is on-rope and below the redirect, your main rope anchor point sees about 1kN of force (the same as if there was no redirect). But your redirect anchor point (ignoring friction) should see an additional ~770N of force, or about 170lbs.

I could be wrong, but that's my understanding (edited to correct even more ambiguity).

[Dwight Livingston]

Because the deviation acts only as a change of direction for the force, I contend that the main anchor always remains at 100% but the deviation anchor force varies with the angle as per Tim's diagram.

Yeah, I agree.

Dwight

[Scott McCrea]

If you have a 45 degree Y with a pulley deviation, where the deviation leg sees 77% of the load and the main leg receives 100% (according to Ian). But, if you replace the pulley with a knot (maybe a double fig 8), the two legs would get 77% each. What gives?

[Mike Rz]

Ian writes:"What's the difference between a plain old rope anchored to a tree and flung down a shaft, bent at 45 degrees at the lip, and the same rope held at 45 degrees by a deviation?

Absolutely nothing

Ian writes: D'you mean to say that in each case the tree bears only 77% of the load? Assuming no friction at the lip or deviation. No way - all the load is on the tree in the first example... so it must also be 100% in the second. "

Not at all. In the first case you have forgot the earth. Even if there is no friction, the earth is is still providing a normal (perpendicular to the surface) force at the 45 degree lip. Thats what makes the cases identical. If you don't believe me, stick your finger between the weighted rope and the earth. Then explain where the squashed finger came from. The conclusion/assumption that the rope has 100% of the weight is wrong.

Remember:
F=m*A
and
"you cannot push a rope

other than that, it is all sines and cosines

Mike the mechanical engineer

PS: there nothing wrong with Tim's diagram. If you draw free body diagrams (FBD's), all the forces balance. [/quote]

[Dwight Livingston]

If you have a 45 degree Y with a pulley deviation, where the deviation leg sees 77% of the load and the main leg receives 100% (according to Ian). But, if you replace the pulley with a knot (maybe a double fig 8), the two legs would get 77% each. What gives?

They wouldn't both have 77%. The main line would have more than 77%. If you look at the 60 degree example, both would have 100%.

Dwight

[Scott McCrea]

If you have a 45 degree Y with a pulley deviation, where the deviation leg sees 77% of the load and the main leg receives 100% (according to Ian). But, if you replace the pulley with a knot (maybe a double fig 8), the two legs would get 77% each. What gives?

They wouldn't both have 77%. The main line would have more than 77%. If you look at the 60 degree example, both would have 100%.

I didn't make my little example clear enough. And it's wrong anyway. So, never mind.

But, Mike Rz's post cleared things up a bit for me. Thanks!

If I understand correctly, the only time a main line anchor will see 100% of the load is at 0 degrees deviation. At other angles, the deviation is sharing the load and the main line anchor's load is less than 100%. Right? If so, a deviation that creates an angle greater than 60 degrees needs a stronger anchor than the main would.

[Dwight Livingston]

For a nice demo of y-hang loads, check out http://www.phy.syr.edu/courses/mra/devl ... demo1.html This is for knotted y-hangs, not deviations. You can select the "knot" in the diagram and drag it around and see the loads change. To see it in percent, change the load to 100N.

Dwight

[NZcaver]

If I understand correctly, the only time a main line anchor will see 100% of the load is at 0 degrees deviation. At other angles, the deviation is sharing the load and the main line anchor's load is less than 100%. Right? If so, a deviation that creates an angle greater than 60 degrees needs a stronger anchor than the main would.

Yes and no. Here's how I understand it (and maybe Mike can chime in again on this).

The only time the main anchor sees 100% of the force generated by the load is when there's no deviation. When a deviation is added, things get a little more complicated. Don't get confused by reading about knotted Y-hangs. You can't directly use the same theory, because a simple knotted Y-hang shares the load force (generally equalized) between 2 anchor points. Whereas a deviation pulley (or carabiner) is free to equalize itself (and therefore the forces) between the main anchor point and the load. See the difference?

Technically speaking, the load itself is always constant regardless of any deviation angle. It's the total force applied by that load that will change. In our example, the force on the main anchor always stays about the same, equalized by the deviating pulley or carabiner. However, as the angle of deviation increases, the force on the deviation anchor will also increase - as per Tim's diagram. This is a form of mechanical advantage, acting on the deviation anchor. So yes, if your angle of deviation is more than 60 degrees, the deviation anchor will experience a higher force than the main anchor. But that doesn't mean the main anchor experiences any less force - it still sees the same force applied by the load that it did with no deviation in place.

[Edit] I modified Tim's diagram a little to reflect the forces involved with rigging deviations. Hopefully this helps unravel the mystery... assuming I'm correct, and the percentage figures in Tim's original diagram were correct (someone please check).



Fun!

[Mike Rz]

I think I like lurking better.......

I mis-spoke (sort of) in my previous post. When I said: "The conclusion/assumption that the rope has 100% of the weight is wrong."

I should have said:
The conclusion/assumption that the rope has 100% of the generated loads is wrong. NZcaver said it better.

The main anchor experiences a force equal to the hanging load (minus friction) and directed along the angle of the rope. The deviation anchor experiences a force which, depending on its angle, can be greater or less than the hanging load.

Tim's diagram has one mistake. The 130 degree angle should be 120 degrees (not that it matters that much).

And I think it a safe rig (if the anchors are sufficient for the deviations)

[Tim White]

Tim's diagram has one mistake. The 130 degree angle should be 120 degrees (not that it matters that much).

Yep, 120 degrees, not 130 degrees. Typo. Sorry, I was slamming out the diagram in a rush.

[LifeOnALine]

Sheesh this is turning into a clusterbuck...


Here - play with this...

http://www.lifeonaline.com/toys/deviation.php

~DM~

[Scott McCrea]

OK, I get it now. Thanks y'all!

Fun toy, Dave!

[ian mckenzie]

So, back to the original post - is this a safe rig?

What I've heard is: Main anchor-wise, yes, as the deviations cause no more force on the main anchor than if you had just flung the rope straight over the lip. BUT, you must ensure your deviation anchor(s) are capable of any force multipliers that the angle in the rope might exert on them. I should think that an oval carabiner on some knotted webbing round a substantial tree would be fine.

But ditch the safety line, and the pulleys - just use a carabiner and pass it as for a normal deviation.

This has been interesting for me - normally, mid-pitch deviations are very low-angle, and have very little force exerted on them, so cavers often get away with light cord looped round tiny nubbins. But, as the deflection angle increases (e.g. at the top of a pitch) you have to think about it. Last month I was on a trip where the rigger did indeed use a small nubbin for a big-angle deviation; I never thought anything about it at the time, but I'm thinking about it now...

[NZcaver]

Here - play with this...

http://www.lifeonaline.com/toys/deviation.php

Dave - that's great!

And to think, I wasted all that time and hurt my brain making those posts. I searched the internet for a vectored force calculator like yours, and nothing quite fit the bill. Did you just make that little gizmo? Will it be around in perpetuity (or at least for a while)?

Thanks!

[NZcaver]

So, back to the original post - is this a safe rig?

Assuming the anchors are sound, Landon's rigging will probably be fine. It's good to keep the angle of deviation in mind, at least as a rule of thumb. Something like, say, trying to keep it to 60 degrees or less.