... question 1 is meaningless without angles. Also, sort of meaningless in general...

question 2 also meaningless without specific dimensions.

Majid, man, this isn't your day.

Every thing is right there

may be you should respond after your PMS is over

How is the tree the "redirect" anchor? If you know anything about physics, the greater percentage of load is on the tree, as pictured.

As for #2... does it really matter when there is a tree falling on top of you?

No, I get it... the problem is: the. tree. broke. Now it's falling. Down. Onto the climber. Look at the picture. Maybe the rope will act as a pendulum and swing the climber out of the way of the falling tree...

I honestly don't know why I am keeping this going. I guess I am just getting sick of these asinine TR tests that make absolutely no sense whatsoever.

Tie the rope to the tree, assuming no ginsu edge for the rope to run over.

few feet away from edge

...in the back some distance away...

A-Less than 18 KN

B-18 KN plus an additional (40%+) but no more than 36 KN

C- up to 2X of the original belay anchor

D-18 KN plus an additional 10 % = 19.8 KN

E-Same as main anchor or 18 KN

Is this 30 feet or three feet? What does "few" mean here? Not that it matters but the question looks ill-conceived without sticking in real numbers even if they don't matter in the end.





Do you mean directly in line with the climb and 3000 feet behind the tree or at a 90 degree angle (rope redirection) but six inches from the tree? Again, the question looks pretty lame without this information. It makes it look like you are not competent even if the numbers don't affect the answer.


Why would anyone ever say that an anchor SHOULD be less than any specific value. That's just stupid. Also, zero is less than that and zero is not really acceptable, is it?


Completely unnecessary for a TR anchor.


What the hell does "up to" mean? 0.00001kn qualifies as being "up to" 2x of the original belay anchor. This is a gibberish answer without giving both ends of the range. Or maybe zero, as the assumed lower end of an "up to" range is what you meant like in the "lower than 18kn..." answer. That's pretty dumb.

BTW, I'd like to mail you up to 2 million dollars. What is your address.


At least this is a reasonable choice since it's not a range from zero to 2000000kn or anything dumb like that. Still, This is getting tiring.


Ah. Less gibberish.

My guesses are:

1. If the rock is close to the tree and when the tree falls off the cliff such that the climber will get dropped less than 1 foot from the event, the tree is pointless as part of the anchor.

or

2. If the rock is far enough from the tree that when the tree falls, the climber hits the ground regardless of where he/she is on the climb then the tree needs to BE the anchor and the rock is useless. It must hold the full force and the anchor must be appropriate for any TR situation. Of course if this is the real question then the force the tree must hold will be more than the rock must hold based on the angle of the rope through the tree anchor carabiners. So what is that angle?

or

3. If the rock is far enough to the side but close enough to the tree that the climber won't hit the ground when the tree fails but will swing to the side and hit the death-sticks that protrude from the side of the climb then the tree again must be the anchor and the rock is pointless.

or

4. If the rock is lined up with the tree and the climber then what is the tree use for again? Maybe keeping the rope from rubbing on the edge of the cliff then it needs to hold some reasonable TR fall. If the edge is sharp then we're back to the tree needing to hold the full force and the rock is no longer useful again.

On the other hand, another possible answer is that this is a deep-water free-solo type of climb and as long as the tree won't hit the climber, the anchors are not necessary. The climber simply falls into the water then swims to shore at which point eh tree falls on him there.

Dave

If I were a mod I'd consider to move all this bull$hit of "TR tests" from the beginners forums before someone takes them seriously. Unfortunately this self-appointed "rope master" can create quite a bit of confusion.

Yes you hit the nail on the head about the idiot being a self-appointed rope master.

ciao
Santana

If I were a mod I'd consider to move all this bull$hit of "TR tests" from the beginners forums before someone takes them seriously. Unfortunately this self-appointed "rope master" can create quite a bit of confusion.

With 90º angle you get 40%+ forces?? Check your basic trigonometry:

F_anchor = 2xFxcos(45) = 1.05xF --> an extra 5%, not 40% (and that would be without friction in the biner)

Stretch of 40%?? That's a bungee, not a climbing rope.

Top ropes generating 18kN???

Please....

If I were a mod I'd consider to move all this bull$hit of "TR tests" from the beginners forums before someone takes them seriously. Unfortunately this self-appointed "rope master" can create quite a bit of confusion.

With 90º angle you get 40%+ forces?? Check your basic trigonometry:

F_anchor = 2xFxcos(45) = 1.05xF --> an extra 5%, not 40% (and that would be without friction in the biner)

Stretch of 40%?? That's a bungee, not a climbing rope.

Top ropes generating 18kN???

Please....

note; most dynamic rope elongations are between 30 to 38% and yes they act like a bongee cord.

GOT IT?

1st tree 100% of the load.
2nd tree 100% of the load
3rd tree 52% of the load.

most dynamic rope elongations are between 30 to 38% and yes they act like a bongee cord.

GOT IT?


http://mtntools.com/...e/03DynamicRope.html

How much does the first anchor (red color) gets ?

100% of the load which is 12 KN ?

is this right ?

I did not specify what the FF was or how much FF will creates during a top rope and you started coming with bunch of #a and nonsense

The way you drew it, yes. I don't care what the load is. I am giving you the %age of load on the anchor based on the angle of the rope attached to the anchor. For an angle ≤90º, the load on the anchor is 100%. When the angle is >90º there is a reduction factor of the load on the anchor point, as some force is redirected to another point. I approximated the angle you drew to mean that there would be about 52% of the force on that anchor, about 130º.

Well,The anchor with red rope (redirect) does not get 100% but sees 200% or twice as much. This is for the fact that you have load on one end and another load on the opposite side or the other end of the rope.

if you build your redirect to handel 12 KN, it will fail cause at 6 KN load, it is already receives 12 KN or twice as much.

So the point of this thread is . . . what?

To waste our Time... oh yeah, and confuse beginners.

To waste our Time... oh yeah, and confuse beginners.

If I were a mod I'd consider to move all this bull$hit of "TR tests" from the beginners forums before someone takes them seriously. Unfortunately this self-appointed "rope master" can create quite a bit of confusion.

With 90º angle you get 40%+ forces?? Check your basic trigonometry:

F_anchor = 2xFxcos(45) = 1.05xF --> an extra 5%, not 40% (and that would be without friction in the biner)

Stretch of 40%?? That's a bungee, not a climbing rope.

Top ropes generating 18kN???

Please....

While I agree with your general sentiments re: majid, something's awry with your math on the forces generated by redirects. Using your math, cos(45) = sqrt(2)/2, so 2*F*sqrt(2)/2 = sqrt(2)*F ~= 1.4*F

Other than that, though, right on. This thread borders on silly.

Damn.... you're right (damn calculator in radians!).
An apology is due: My apologies, Majid, the extra force is 40% as you correctly stated.
I stand by everything else I said.
Thanks glytch.