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jt512


Oct 9, 2008, 12:45 PM
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Re: [cracklover] tension [In reply to]
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cracklover wrote:
jt512 wrote:
robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
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

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

While it is true that for newtonian physics to work, a frame of reference is not required...

Huh?

In reply to:
...that doesn't mean that one cannot have a frame of reference and speak intelligently about the motions of objects within that frame. This is the point which JT is unwilling to acknowledge.

In order to speak intelligently about Newtonian physics you have to understand that your observations are dependent on your frame of reference. My original statement was in response to a couple of people who seemed to think that the tension in their rope depended on their rope's velocity. Failure to understand frame of reference misled them into thinking rather unintelligently about the behavior of objects.

Jay


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Oct 9, 2008, 1:07 PM
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Re: [jt512] tension [In reply to]
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jt512 wrote:
cracklover wrote:
While it is true that for newtonian physics to work, a frame of reference is not required...

Huh?

I mean that choosing one frame of reference versus another doesn't matter. The physics works out the same.

In reply to:
In reply to:
...that doesn't mean that one cannot have a frame of reference and speak intelligently about the motions of objects within that frame. This is the point which JT is unwilling to acknowledge.

In order to speak intelligently about Newtonian physics you have to understand that your observations are dependent on your frame of reference. My original statement was in response to a couple of people who seemed to think that the tension in their rope depended on their rope's velocity.

Who the fuck knows what the OP was thinking? So far as I can tell, he hadn't actually thought about the problems he wanted answered well enough to even generate a question that made sense.

In reply to:
Failure to understand frame of reference misled them into thinking rather unintelligently about the behavior of objects.

Jay

I suppose you could be right. But when I read:
In reply to:
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 lack of understanding about the relevance of a frame of reference certainly doesn't shine through to me.

But probably this is not the post you're referring to? Could you please direct me to the posts of the "people who seemed to think that the tension in their rope depended on their rope's velocity"?

Thanks!

GO


onceahardman


Oct 9, 2008, 3:16 PM
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Re: [cracklover] tension [In reply to]
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Umm, this might be my fault, when I posted this on page 2 of this thread:

In reply to:
mtnrock, at great risk of furthering misunderstanding, I'm going to try.

The situation is static. Nothing is moving. Climber of weight X is hanging from a normal slingshot-type belay.

Since he is not moving, the tension in the rope between the climber and the 'biners on the toprope anchor is also X. X=X, no motion.

Between the toprope anchor 'biner and the belayer, you also have tension = X, since the rope is not moving.

You now have TWO downward tension vectors, each = X. 2X going downward. The anchor is not moving, so there must be an UPWARD tension vector on the anchor of 2X, or else the anchor would have to be accelerating.

Does that help?

Awkwardly stated, but an effort at simplifying things for mtnrock, by showing a static situation (which is, of course, equivalent to a constant velocity situation, with frictionless ropes and biners, etc.


jt512


Oct 9, 2008, 3:59 PM
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Re: [cracklover] tension [In reply to]
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cracklover wrote:
Could you please direct me to the posts of the "people who seemed to think that the tension in their rope depended on their rope's velocity"?

It was the one that "onceahardman" reposted above and mtnrock's punctuation-free response:

mtnrock wrote:
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


Jay


(This post was edited by jt512 on Oct 9, 2008, 4:00 PM)


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Oct 10, 2008, 7:33 AM
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Re: [jt512] tension [In reply to]
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jt512 wrote:
cracklover wrote:
Could you please direct me to the posts of the "people who seemed to think that the tension in their rope depended on their rope's velocity"?

It was the one that "onceahardman" reposted above and mtnrock's punctuation-free response:

mtnrock wrote:
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


Jay

I see. Yech, what a mess. Anywey, I think we're all on the same page now.

Just for shits and giggles, mtnrock, have we answered your question? If so, do you think you could explain, in your own language, what the forces are for a slingshot belay? This means it's a toprope fall in which the belayer is at the bottom, the rope runs up through the anchor and back down to the climber.

Assume there was no slack when the climber falls. Assume that the climber weighs 150kg.

The three forces I'm looking for are: the peak force on the belayer, on the climber, and on the anchor.
For extra credit, discuss how the frame of reference informs your answer.

Okay gang, let's see how well we did in informing mtnrock.

Good luck!

GO


knudenoggin


Oct 12, 2008, 11:10 PM
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Re: tension [In reply to]
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I'm curious if this physics is for rockclimbers only,
or if it applies to mundane things like barbell weights
and cord & 'biners and less frictive sheaves like pulleys?

For beside me are four 25# weights, in the following configuration:

two wgt.s are tied together and to a 7mm line that runs up through
a pretty low-friction (relative 'biners, anyway) pulley wheel;

which line then runs down to tie to a 'biner;

through which runs quarter-inch solid-braid nylon line tied,
at each end, to each of the other two 25# wgt.s, resp..

Now, hoisting up one of these last 25pounders, to give it a short
drop on a sling-shot-through 'biner belay by the other 25pounder,
which 'biner anchor cord itself is sling-shot through the pulley to
the 50# belay, and . . .

what has all this preceding discussion said will happen?
(aside from showing posters' excessive love the quote function!)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - -

As I read it, the dropped (a few inches) 25# generates > (2*25#),
coupled with 4/3(25#) on the flip side (or raising that 25# belay),
producing > 10/3 (25#) = 111.111111... # of tension fed into
the pulley wheel, belayed by a mere 50#, which should thus go
halfway into orbit.
(But my faith in this is such that I've not alerted NASA, or local
air-traffic control, or even a ceiling repair guy just yet.)

Is that right?

*kN*


jdefazio


Oct 13, 2008, 5:29 AM
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Re: [knudenoggin] tension [In reply to]
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Majid-ize a schematic please.


jt512


Oct 13, 2008, 7:25 AM
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Re: [knudenoggin] tension [In reply to]
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knudenoggin wrote:
I'm curious if this physics is for rockclimbers only,
or if it applies to mundane things like barbell weights
and cord & 'biners and less frictive sheaves like pulleys?

For beside me are four 25# weights, in the following configuration:

two wgt.s are tied together and to a 7mm line that runs up through
a pretty low-friction (relative 'biners, anyway) pulley wheel;

which line then runs down to tie to a 'biner;

through which runs quarter-inch solid-braid nylon line tied,
at each end, to each of the other two 25# wgt.s, resp..

Now, hoisting up one of these last 25pounders, to give it a short
drop on a sling-shot-through 'biner belay by the other 25pounder,
which 'biner anchor cord itself is sling-shot through the pulley to
the 50# belay, and . . .

what has all this preceding discussion said will happen?
(aside from showing posters' excessive love the quote function!)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - -

As I read it, the dropped (a few inches) 25# generates > (2*25#),
coupled with 4/3(25#) on the flip side (or raising that 25# belay),
producing > 10/3 (25#) = 111.111111... # of tension fed into
the pulley wheel, belayed by a mere 50#, which should thus go
halfway into orbit.
(But my faith in this is such that I've not alerted NASA, or local
air-traffic control, or even a ceiling repair guy just yet.)

Is that right?

*kN*

If you want to have any chance of getting a correct answer, you're going to have provide a diagram of your setup.

And how come everybody's posts except your go all the way across the text field?

Jay


mtnrock


Oct 13, 2008, 7:43 AM
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In reply to:
My original statement was in response to a couple of people who seemed to think that the tension in their rope depended on their rope's velocity.

it doesn't depend on its velocity and i said it didn't but it does depend on its acceleration.


sungam


Oct 13, 2008, 7:51 AM
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Re: [jt512] tension [In reply to]
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jt512 wrote:
And how come everybody's posts except your go all the way across the text field?

Jay
He
likes
the
return
button.


knudenoggin


Oct 13, 2008, 12:29 PM
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Re: [jt512] tension [In reply to]
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jt512 wrote:
If you want to have any chance of getting a correct answer,
you're going to have provide a diagram of your setup.
C'mon, it's quite simple: a compound sling-shot, with 25# being dropped and
25# on the *belayer's* side of this, cord run through a 'biner;
this anchor 'biner is supported itself by line running up/down through
a pulley wheel, which has much less friction than a 'biner, and its *belay*
side had the 50# wgt. In crude jargon, then, its 2:1 hauling on 2:1,
theoretical MA. The drop is something >FF0, say, 3". And I guess that
the dropping side had about 1' to 'biner, and 3' down to belay wgt.; the
pulley was about 6' above its "belay" wgt., with about 1.5' down to the 'biner.
So, the idea was that the lower, 'biner system was generating the 10/3W
(W = 25#), and delivering this force into the upper and less frictive system,
which should then generate 4/3(10/3)W on the 50# anchor wg.t.

In reply to:
And how come everybody's posts except yours go all the way across the text field?
Damn good whine: how come, indeed!
Well, I don't like reading wiiiiiiiiidddddddddeeeeee lines any more than you
like a lack of punctuation. And why keep quoting texts that are already presented
above? Most folks aren't going to MajiDelete things into oblivion: what's the
point of repeating 20-50 lines just to add another line or two?

*kN*


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Oct 13, 2008, 12:33 PM
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Re: [jt512] tension [In reply to]
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Isn't it kinda silly to make up a configuration completely different from the one described and then suggest that the same analysis ought to apply?

In any case, that 111 pound result is (in the correct configuration not employed here) merely an instantaneous peak load, rather than, as implied by the orbit allusions, a constant upward force, so neither ICBM's, jets, nor ceiling repair personnel will have to be scrambled in any case.

But you knew that.

Edit: I instantaneously added a "tan" to the former "instaneous" to repair the the gap alluded to by KnudeNoggin in a post that follows.


(This post was edited by rgold on Oct 14, 2008, 12:29 AM)


jmvc


Oct 13, 2008, 1:08 PM
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Re: [Lazlo] tension [In reply to]
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Lazlo wrote:
When you say tension, do you mean force?

I would have thought so.

http://www.thefreedictionary.com/tension


(This post was edited by jmvc on Oct 13, 2008, 1:08 PM)


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Oct 13, 2008, 2:07 PM
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Re: [knudenoggin] tension [In reply to]
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knude, the 50# weight won't move. This diagram represents an approximation of the system at the point of peak force:



GO


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Oct 13, 2008, 7:32 PM
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Wait a minute, I just re-read your post (for like the fourth time). I think this is what you're saying, right KN?



GO


knudenoggin


Oct 13, 2008, 7:33 PM
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rgold wrote:
Isn't it kinda silly to make up a configuration completely different from the one described and then suggest that the same analysis ought to apply?
It would be, yes; but how is what I described that? --a sling-shot system,
compounded simply to highlight/amplify the effects I questioned.
AND it is a system that folks can readily DO, here in The Lab, after all.
But, okay, your turn: how can we demonstrate the theory?
What will show this peak force?
(Because the above experiment with weights showed no hint of it
(to the untrained eye?)!)

In reply to:
that 111 pound result is ... merely an instaneous peak load,
Which, to my quick look, seemed to be missing something, like one syllable.
But age-old [ca. '95] wisdom of Ken Cline yet retrievable has it also!
In reply to:
The short answer is that fall force is not linear with free-fall
distance or rope characteristics, and top roping on static rope won't
result in equipment failure and death. A 200 pound slingshot top
roper will load the anchor with about 660 pounds of force in a fall
with zero slack, regardless of whether the rope is dynamic, static,
bungee, or steel cable. As you add free-fall, the force felt by the
anchor starts to depend on the type of rope, but the transition isn't
all that abrupt for the types of rope that climbers use.

That sure seems like a heckuva load to not show effects, as I had tried to do.
The 25# dropped even some slight distance didn't budge its belay,
nor did the anchoring line budge the 50#.

In reply to:
But you knew that.
Nope, I'd be knewdNoggin then.
Rather, I've been finding that friction is huge in such often hyped pulley
systems (such as using pure rope, nothing metal), with behaviors quite
different from what is written, which is that near-theoretical MA obtains:
it's nowhere close.

*kN*


knudenoggin


Oct 13, 2008, 8:10 PM
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cracklover wrote:
knude, the 50# weight won't move. This diagram represents an approximation of the system at the point of peak force:

[IMG]http://i38.tinypic.com/11jc4du.jpg[/IMG]

GO
Thanks for a quite nice image!
But the system I described was compound: replace, in your image, the 25#wgt.
with the 'biner, and a line run though IT anchored w/25# and the other
end has the 25# wgt. that is dropped. I.e., where you show 25# dropping
to generate double its mass in force would be a 'biner fielding such generated
force amplified to the 10/3 (25#) value.
THAT is what didn't budge the 50# (or anything).

----

But I don't follow your annotations!?

How do you figure "5 lb (friction)" ? The rule-of-thumb efficiency of 60-66%
would hold that the 50# force was reduced to 30-33 lb.s.
(Or are you showing my pulley vs. generic 'biner and assuming that it
has a high, 90% efficiency? But then, still, my next question ... :-)

And what is the red, up-arrowed "5 lb.s" at the line under the 50 lb. wgt.?
I'd think that if you were taking off 5 lb. at top, you'd show 45 lb. here, maybe?!
Which would not raise the 50#.

*kN*


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Oct 14, 2008, 1:21 AM
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Re: [knudenoggin] tension [In reply to]
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knudenoggin wrote:
rgold wrote:
Isn't it kinda silly to make up a configuration completely different from the one described and then suggest that the same analysis ought to apply?
It would be, yes; but how is what I described that? --a sling-shot system,
compounded simply to highlight/amplify the effects I questioned.
AND it is a system that folks can readily DO, here in The Lab, after all.
But, okay, your turn: how can we demonstrate the theory?

It's very late, I have much work to do and a lot of climbing coming up---it's Gunks Reunion Week, so I ain't promising any theoretical treatises.

But looky here Knude, my original comments referred to a weight applied to a rope fastened to a fixed anchor point, but you've got a moveable anchor point; the 25 lb counterweight. That anchor point will start moving up as soon as the rope tension from the drop overcomes carabiner friction, so there will be less stretch in the rope than you'd get with a fixed anchor point and so less tension to transmit to the next part of the system.

Off the top of my very sleepy head, I'd say that if the rope to the counterweight starts sliding when the tension on the dropped weight side is 4/3 X 25, then the total peak load on the lower structure will just be about 7/3 X 25 or about 67 lbs, which won't budge the 50 lb load with the same friction to overcome on the top biner. This seems to be what you are observing, which of course does not mean that what I've just said is right.


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Oct 14, 2008, 8:46 AM
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knudenoggin wrote:
cracklover wrote:
knude, the 50# weight won't move. This diagram represents an approximation of the system at the point of peak force:



GO
Thanks for a quite nice image!

No prob. But (as I said before) after re-reading your post again, I realized that you had two 25 pound weights, not one. Thus the second diagram (see a few posts up).

In reply to:
But the system I described was compound: replace, in your image, the 25#wgt.
with the 'biner, and a line run though IT anchored w/25# and the other
end has the 25# wgt. that is dropped. I.e., where you show 25# dropping
to generate double its mass in force would be a 'biner fielding such generated
force amplified to the 10/3 (25#) value.

Look up in the thread. I posted what I think is exactly that. I'm happy to plot out the forces for you, but only after you've verified that the image I posted jives with the question you're asking. I don't want to answer the wrong question a second time.

In reply to:
But I don't follow your annotations!?

Well, maybe there's something to be gained by looking at that first diagram - even though it's not the question you asked. So sure, let's walk through it.

In reply to:
How do you figure "5 lb (friction)" ? The rule-of-thumb efficiency of 60-66%
would hold that the 50# force was reduced to 30-33 lb.s.
(Or are you showing my pulley vs. generic 'biner and assuming that it
has a high, 90% efficiency? But then, still, my next question ... :-)

Yes, the top thing that looks kinda like a pulley is... a pulley. And yes, I'm assuming 90% efficiency, or 10% friction (5 lb).

In reply to:
And what is the red, up-arrowed "5 lb.s" at the line under the 50 lb. wgt.?

That is the force the ground is exerting to hold up the 50 lb weight, while the tension in the line from the weight to the pulley is holding the other 45 lb.

In reply to:
I'd think that if you were taking off 5 lb. at top, you'd show 45 lb. here, maybe?!
Which would not raise the 50#.

*kN*

Not exactly, but I think you get the idea.

GO


knudenoggin


Oct 14, 2008, 10:44 AM
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cracklover wrote:
In reply to:
Thanks for a quite nice image!

No prob. But (as I said before) after re-reading your post again, I realized that you had two 25 pound weights, not one. Thus the second diagram (see a few posts up)
...

Look up in the thread. I posted what I think is exactly that. I'm happy to plot out the forces for you, but only after you've verified that the image I posted jives with the question you're asking.
BINGO, well done, again!
Yes, 25# dropped (or, as in OP, even just *released*) onto 'biner system,
which delivers force to pulley'd one w/50# belay wgt..

-----------

> The usual Physics 101 explanation involves a vertical spring with a mass at its bottom.

And so I guess my problem is that I substituted for this mass the lower, 2nd sling-shot (2:1 TMA) system,
a sheave with so-calculated FORCE, not mass?!
And the theory I'm slow to understand applies to the force by mass in this lower system,
but then the upper one has acting on IT not a falling mass, directly, but a force resulting
from that?! --to which there is no doubling effect? (although a suspended mass STILL exists)

-----

Now, to the first & direct point, without complication, when I dropped (just a few inches, but SOMEthing more
than mere release) the upper 25# wgt. onto the >> 'BINER-SHEAVED sling-shot with equal, 25# belay wgt. <<
there was no movement of that BELAY/anchor wgt. (or 50# one) (just a standing to attention from its leaning
at, oh, maybe 85deg angle against something).

Which to my naive eye, didn't seem to show the delivery of the predicted peak force of 4/3 25# (but which the
aspect of instantaneity might dismiss?). And one should also say that it was more than 4/3(25#) given
the few-inches drop, but maybe then the factors of rope stretch go some ways to mitigate any rise?

*kN*


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Oct 14, 2008, 11:44 AM
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Knude, the gizmo you have at the bottom is called an Atwood Machine and is famous in elementary physics. You might google around a bit and see if you can find any treatment that describes what happens if the rope stretches; the usual case involves the use of an inelastic string.


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Oct 14, 2008, 1:17 PM
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To be honest, Knude, I think the amount of time of the peak force is so short that we need to think about the problem a little differently. Basically, the tiny amount of energy involved is mostly converted to heat by your cords, rather than converting it to work done in lifting objects.

Look at it this way - what if you had a rubber band attached to a five lb weight on the floor. Attach the top of the rubber band to a lever, and drop a 1000 lb weight on the other end of the lever. If the lever can only go up two inches, and the rubber band can easily stretch that much, then it doesn't matter that an enormous amount of force is involved - all of the energy is easily turned to heat by stretching the rubber band two inches.

Now your cord isn't quite that stretchy, but I think showing a simple diagram of the forces, discounting the energy your cord can convert to heat, would be meaningless.

GO


Partner cracklover


Oct 14, 2008, 1:20 PM
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Re: [cracklover] tension [In reply to]
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Oops, that's what I get for doing real work and then coming back to my post. It appears that RGold explained it away in two words.

Ah well. Have fun, Rich! I'd see you at the Gunks, maybe, but I'm no longer local.

Cheers!

GO

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