is tension on a rope doubled in a top rope system when its on a redirect belay or a direct from the anchor belay.

it seems like it should but im draawing a blank if not please say why

You need to explain it better.
But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through.
Korekt me if ayem rong, but I do not b-leave aye am.

Yer knot Rong. Yer Gamsum.

Eye can stil b rong.

Sorry for hi-jacking, OP. I'll play nice now.

OP'z prob'ly thinkin' "Crap, the two guyz I didn' wan' respondin' responded!"

That’s poor logic. In fact a fall straight onto an anchor produces more force on the anchor then when the rope runs through the anchor to the belayer. When you fall the belayer is lifted off the ground which increases the fall distance and decreases the peak load on the quickdraw / cam / anchor.

Belaying straight off the anchor for a second will produce more force on the anchor then doing a redirectional into your harness.

So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

Belaying straight off the anchor for a second will produce more force on the anchor then doing a redirectional into your harness.

jt512 has given a clear and correct explanation.

A redirected top-rope anchor has to be able to withstand far more than triple body weight to be even remotely safe, since most belayed falls with have small but non-zero fall factors.

is tension on a rope doubled in a top rope system when its on a redirect belay or a direct from the anchor belay.

it seems like it should but im draawing a blank if not please say why

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight — that is, a little more than triple the climber's weight.

Jay

When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.

where does the 2/3 come from? is this an empirical value derived from research somewhere (in which case i'd like to read the link if you remember where to find it)? or is the 2/3 a value that can be determined by math and physics? in that case would you mind posting the math or PMing me?

also, why is the climber's force double the static weight? clearly the climber must realistically undergo some acceleration and therefore an increase in force, but why is it double? is this value research based or math based?

In his paper, Richard refers to it as a "convention" to account for the reduction in force on the belayer's side of the rope due to friction between the rope and the carabiner. I presume that it is based on empirical data.


The peak force of twice the climber's weight for a fall-factor-0 fall is a consequence of using Hooke's Law to describe rope stretch. As you'll see, Richard provides two separate derivations of this result.

Jay

Colatown, the 2/3 value for carabiner efficiency is an empirically derived value that has been stated over and over in various places, and now has the status of "folklore." I don't know the original reference. I have seen 1/2 stated, but much less often. The fact is that the number will be different for different rope-carabiner combinations and probably, like other friction coefficients, has a sliding and a static value.

The fact that the peak tension in a rope with applied weight W is 2W is a simple conservation of energy consequence of Hooke's Law, viewed as a reasonable approximation to the behavior of a climbing rope. (Yes, there is some acceleration involved, since a climber who weights a rope will drop as the rope stretches.) The result assumes that the climber's weight is instantaneously applied to the climbing rope with no slack in the rope. Using various strategies to ease the weight onto the rope could, in theory, keep the peak rope tension no greater than the climber's weight, but this would not apply to catching a top-rope fall.

Whether the real result is more or less than 2W would be a matter for experiment. I'd guess a little less than 2W because of the internal damping effects of rope construction, but given the manifold uncertainties and variations in real-life practice, 2W is a very reasonable working estimate.

Apparently, I derived this result in the .pdf jt512 refers to, though I'd have to check back to be sure.

If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers, or an estimation that knowing the real figures will not contribute in any meaningul way to practical behavior in the field.

If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers...

So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

haha this post got so off topic i just had a simple question on tention and it went into all this stuff i can't really say it was simple because i can't word anything well but how can i put a diagram so i can show what im talking about more clearly.

I claim that a toprope anchor should be built to withstand a force of at least 3000 lb.

Stop it. You're kill me.

Jay

Damn Jay, that's right up there with "all your base are belong to us."

yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

thats not what im saying in both situations there is no acceleration so T doesn't increase

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay

Yes, especially if I (first) stood in front of the parked car and then in front of the car going 30 mi/hr. I'm not sure, though, that I'm willing to do that experiment. Would you do it for me?

r.c

Yes, especially if I (first) stood in front of the parked car and then in front of the car going 30 mi/hr. I'm not sure, though, that I'm willing to do that experiment. Would you do it for me?

r.c

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

ok wait cause your guys are so set on how i word something. that's not what the point i was trying to get across. i was trying to show if there is no acceleration tension does not increase.

ok wait cause your guys are so set on how i word something. that's not what the point i was trying to get across. i was trying to show if there is no acceleration tension does not increase. and it really seems people are just on this site to piss people off

na just saying all you do is insult people

Not true--I do that and a whole lot more. You'll see.

Curt

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt

It would probably help to have a real physicist to weigh in, instead of two guys who sometimes play one on the internet; but all the real physicists I know want to get paid to answer questions like this. But, that said, I claim that since velocity is relative then so must be momentum, since they are proportional; so a momentum of 0 must also be an artifact of having chosen a special frame of reference.

Jay

What you're actually saying is that both velocity and momentum have no generally accepted meanings in newtonian physics--and hence, can not be used to differentiate between stationary and moving objects. Obviously, I think that's silly.

Curt

So, are you saying that Newtonian physics does not recognize the relativity of motion?

Jay

I am saying (in yet another way) that the first derivative of displacement, with respect to time, is velocity--and even mathematically it is clear that an object moving at a constant velocity and a stationary one can easily be differentiated. By the way, in the first post of yours that I replied to, you fixed the frame of reference by saying that one comparative case involved "zero velocity."

Curt

Well, I don't know how to explain that there is no difference between zero velocity and non-zero velocity without using the phrase "zero velocity." I suppose that if I tried hard enough, I could do it, but it would have made the post even more abstract.

Jay

I will now happily wait for rgold to arbitrate.

Curt

But let's not complicate things...

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.
Jay

^^^repeat ad-infinitum.

Curt is correct here regarding the semantics of Jay's statement. The rest is like arguing over who is better looking, since the answer depends on who you ask (i.e. what reference frame they are in)...

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Fine. Then I'm better looking too.

Curt

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

Some of you physics guys have lost the big picture... Come on: this is high school stuff...

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay,

The problem is that you said the two cases are "indistinguishable" and that simply isn't true. Again, if I plot position versus time for an object moving at constant velocity I will get a straight line. If I plot position versus time for a stationary object, I will get a point. If I now decide to change my frame of reference and "sit" on the object moving at constant velocity, I may indeed opt to say that the "stationary" object is the one in motion, relative to me. It doesn't matter--because if I then create the same two graphs of position versus time they are merely reversed and I can still easily differentiate one graph from the other.

Curt

You may be able to differentiate one graph from the other, but it's just begging the question. Those two graphs will completely disagree as to which object is the stationary one. How much more indistinguishable do you want than that?

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

I have a what now?

/still hasn't finished college

From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

I have a what now?

/still hasn't finished college

BTW, just to stir the pot, both your sailor and your landlubber are rong.
The cannon ball is staying still (duh) and the other things are moving around it.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay

Well, you're close, right? And it is a degree in physics--isn't it?

Curt

But, they will always be different--and hence not "indistinguishable."

Curt

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay

The question implies that Newtonian velocity is an absolute quantity, which it isn't; it is a relative quantity.

Jay