Author Topic: Symmetry and the Jammed Bend  (Read 3990 times)

DDK

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Symmetry and the Jammed Bend
« on: November 16, 2011, 05:44:00 AM »
While looking at the jamming characteristics of differently dressed Figure 8 Bends (which I'll discuss on another board), I came to see that one technique for loosening the jammed bends was not available to me.  That technique would be the splaying open of the bend by driving a tapered pin through its center.  Likely, others have observed this phenomena, but, I thought it might be of interest to discuss its relation to symmetry.  This technique is always available for the many bends which have axial inversion symmetry, such as the Smith/Hunter's Bend, but, not necessarily available for those bends which have central (point) inversion symmetry, such as the Figure 8 Bend.

Axial Inversion Symmetry:       R1( x, y, z ) = R2( -x, -y, z ) [ read as the section of the first rope located at point ( x, y, z ) is equivalent (twinned if you like) to the section of the second rope located at point ( -x, -y, z ) ].  The origin ( x, y, z ) = ( 0, 0, 0 ) is not completely defined because z = 0 is undefined.  There is an axis of symmetry.

Central Inversion Symmetry:    R1( x, y, z ) = R2( -x, -y, -z ).  The origin ( x, y, z ) = ( 0, 0, 0 ) is completely defined.  There is only a point of symmetry.

The implications of these symmetries are (theoretically) that no rope can reside on the axis or point of symmetry (for a bend, i.e. two ropes).  Twinned sections of the two ropes can reside on either side of the axis or point, but, not occupy the symmetry location and still satisfy the above equations.  Another way to say this is that if one of the ropes is residing on the symmetry location, the twinned part of the other rope would also have to reside on that same spot, which, of course, can't be so.

This means that for the axial inversion symmetry bends, there is always a STRAIGHT LINE FREE (void) OF ROPE right through its center, suitable for driving a tapered pin.  This is not necessarily true for central inversion symmetry and is not true for the Figure 8 Bend.  For bends with central inversion symmetry, there is only one point which must be free of rope.

DDK

xarax

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Re: Symmetry and the Jammed Bend
« Reply #1 on: November 16, 2011, 01:07:59 PM »
technique for loosening the jammed bends

   It would be nice if we could somehow classify all  the possible techniques for loosening jammed bends. ( loosening and untying, not cutting with a sword, as the warrior s infamous method of "untying" the Gordian knot...  :))

splaying open of the bend by driving a tapered pin through its centre. 

   0. If one would choose to drive a tapered pin through a knot s nub to loose a jammed bend, the worst  point he could find to drive it would be the centre of the knot !  He should rather let it through some other off centre path, where he could push or pull a rope strand that happens to be less tensioned than its surrounding strands. To destroy a symmetric thing, the most difficult way is to try to separate into two symmetric parts. On the contrary, you should try to isolate an off-centre element, try to remove it from its original place, and so destroy the general equilibrium of the compound.
 
   1. When you are talking about the axial symmetry of a bend, I guess you mean the original axial symmetry of a 2D representation of this bend. As this axis is not occupied by any rope, you say that this would also be the case in the symmetrically tightened knot. That, independently of any subsequent capsizing, the path of this axis would remain free in the tightened knot as it were in the lose knot, for a pin to penetrate through the knot s nub and reach the centre. I can not see how we can prove this assertion...When we tighten the knot, the rope strands can become twisted in unpredictable ways, and this formelly free axial path to the centre can now be obstructed.
   2. Even if this path to the centre remains free of any rope strands in the tightened knot as it were in the loose knot, the final orientation of the axis of symmetry, in relation with the axis of loading of the jammed bend, is of great importance. Because the axis of symmetry does not have to remain at a 45 degrees angle in relation with the standing ends, as it was in the 2D symmetric diagram. For an effective push-pull of the adjacent strands by a pin, the orientation of the needle should allow you to push-pull the tensiomed rope strands in a perpendicular, not a parallel direction.
   3. Even if we wish and we can drive a tapered pin through the centre following the free path alongside the axis of symmetry of the original loose knot. this would be probably only one of the many ways we can access the centre, and probably not the most easy one. See the attached picture of the Water X bend (1). There is an obvious wide open easy path to the centre of the knot, which is different/perpendicular from/to the axis of symmetry. And how on earth are you going to loosen a jammed Double line Double overhand bend ? ( See the second attached picture).
   4. I am not convinced about the usefulness of examine 2D representations of knots. On the contrary, I think that it is time to move forward of this stage -which can not even help us to enumerate the many possible simple knots. That is why I try to confront with the peculiarities of the knots directly in their 3D tightened form, however more difficult that might be. So, for me a symmetric bend is a bend where, in the dressed/tightened form, the two links of the bend are symmetric to each other, I.e., each one is a symmetric transformation of the other.

1.  http://igkt.net/sm/index.php?topic=2154
« Last Edit: November 16, 2011, 01:16:17 PM by xarax »
This is not a knot.

DDK

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Re: Symmetry and the Jammed Bend
« Reply #2 on: November 16, 2011, 03:25:15 PM »
. . .
   1. When you are talking about the axial symmetry of a bend, I guess you mean the original axial symmetry of a 2D representation of this bend. As this axis is not occupied by any rope, you say that this would also be the case in the symmetrically tightened knot. That, independently of any subsequent capsizing, the path of this axis would remain free in the tightened knot as it were in the lose knot, for a pin to penetrate through the knot s nub and reach the centre. I can not see how we can prove this assertion...When we tighten the knot, the rope strands can become twisted in unpredictable ways, and this formelly free axial path to the centre can now be obstructed.
 . . . 
   4. I am not convinced about the usefulness of examine 2D representations of knots. On the contrary, I think that it is time to move forward of this stage -which can not even help us to enumerate the many possible simple knots. That is why I try to confront with the peculiarities of the knots directly in their 3D tightened form, however more difficult that might be. So, for me a symmetric bend is a bend where, in the dressed/tightened form, the two links of the bend are symmetric to each other, I.e., each one is a symmetric transformation of the other.
 . . .

It is indeed the 3-dimensional bend that I am talking about.  Yes, in practical situations, asymmetric tightening might (always to some degree) break the axial inversion symmetry, thus, the use of my caveat of "theoretical" in the OP.  However, the breaking of symmetry (the distortion/wiggle of the symmetry axis) is often not severe enough (or even noticeable) to prevent the use of a tapered pin.

To see that it is the symmetry axis in the 3D Bend to which I am referring, consider the Smith/Hunter's Bend.  If one examines this bend near its center, one finds three "layers" of pairs of twinned rope, each rope in contact with its twin.  If one takes a section of rope and its twinned section and imagines the point between them and then repeats this for the other two pairs, the axis of symmetry can be found by connecting these collinear points.  A blunted toothpick could be used to examine and skewer your bend.

The proof of this 3D assertion was given in the OP.  Notice that the symmetry axis is (0, 0, z).  That is, x = 0, y = 0 and z to vary.  If we replace any of these symmetry points, (0, 0, z), in our symmetry equation R1( x, y, z ) = R2( -x, -y, z ), we come up with the contradiction  R1( 0, 0, z ) = R2( 0, 0, z )  which states that sections of both ropes are occupying the same space. Thus, neither rope can occupy any of these symmetry locations which collectively produce a continuous line through the bend.

DDK

edit:  The toothpick suggestion I offered brought up another question.  If one places the toothpick along the symmetry axis, what would happen to the toothpick when the bend is tightened?  I did this for the Smith/Hunter's Bend, tightening to the point of making it difficult to untie without tools and found the toothpick was unbroken but had highly compressed at points between the twinned ropes.  Not exactly a parlor trick, but, interesting.
« Last Edit: November 16, 2011, 03:50:45 PM by DDK »

xarax

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Re: Symmetry and the Jammed Bend
« Reply #3 on: November 16, 2011, 04:32:20 PM »
It is indeed the 3-dimensional bend that I am talking about.
The proof of this 3D assertion ...

   In a 3D axially symmetric knot, it is a tautology  that a part of a rope can not occupy a point on the axis of symmetry,  i.e., be at the same place two times !  :) You do not need any "proof" for this... I was talking about the knots derived by tighteming an axialy symmetric (γ symmetric, as Miles calls them) 2D representation of those knots on an square lattice.
This is not a knot.

DDK

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Re: Symmetry and the Jammed Bend
« Reply #4 on: November 17, 2011, 04:11:44 PM »
   0. If one would choose to drive a tapered pin through a knot s nub to loose a jammed bend, the worst  point he could find to drive it would be the centre of the knot !  He should rather let it through some other off centre path, where he could push or pull a rope strand that happens to be less tensioned than its surrounding strands. To destroy a symmetric thing, the most difficult way is to try to separate into two symmetric parts. On the contrary, you should try to isolate an off-centre element, try to remove it from its original place, and so destroy the general equilibrium of the compound. . . .
edit:  The toothpick suggestion I offered brought up another question.  If one places the toothpick along the symmetry axis, what would happen to the toothpick when the bend is tightened?  I did this for the Smith/Hunter's Bend, tightening to the point of making it difficult to untie without tools and found the toothpick was unbroken but had highly compressed at points between the twinned ropes.  Not exactly a parlor trick, but, interesting.

Given the compression I found on the toothpick, and then, after splaying open a jammed Smith/Hunter's Bend through its center, I have a new appreciation of your suggestion of attacking an off-center element of the bend.  Due to the high compression at the core of a bend, driving a pin through the center may be just too damaging to the rope you are trying to save by not cutting.

DDK

Dan_Lehman

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Re: Symmetry and the Jammed Bend
« Reply #5 on: November 25, 2011, 04:08:30 AM »
While looking at the jamming characteristics of differently dressed Figure 8 Bends (which I'll discuss on another board), I came to see that one technique for loosening the jammed bends was not available to me.  That technique would be the splaying open of the bend by driving a tapered pin through its center.  Likely, others have observed this phenomena, but, I thought it might be of interest to discuss its relation to symmetry.  This technique is always available for the many bends which have axial inversion symmetry, such as the Smith/Hunter's Bend, but, not necessarily available for those bends which have central (point) inversion symmetry, such as the Figure 8 Bend.
...

??  I find a center point in a Fig.8 bend .

As for driving a spike through a jammed knot,
good luck w/that in many cases --it's not what
I'd expect to be a usually available/effective
means of loosening a knot.

--dl*
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DDK

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Re: Symmetry and the Jammed Bend
« Reply #6 on: December 01, 2011, 08:58:49 PM »
??  I find a center point in a Fig.8 bend .

As for driving a spike through a jammed knot,
good luck w/that in many cases --it's not what
I'd expect to be a usually available/effective
means of loosening a knot.

--dl*
====

I do not believe I understand your comment about the Figure 8 Bend having a center point.  I mention in the OP (theoretically) that bends with central inversion symmetry ( like the Figure 8 ) have only a center point about which they are symmetric and not a symmetry axis.  Are we not in agreement?  I go on to mention that no rope (again, in the theoretically "ideal" symmetric structure) can occupy these symmetry locations.  So, the center point for the Figure 8 Bend should exist, be well defined and be free of rope.

I would like to note a practical consideration which may be related to your comment about a center.  Its relevance is speculation on my part.  I have more recently noticed that the Figure 8 Bend does offer an approximate "line" through which one might drive a spike and which does go through its center.  In that respect, the Figure 8 Bend was not a very good example on my part for discussing the distinction between axial inversion and central inversion symmetry as it relates to the prospect of available "lines" for the driving of spikes.  It is still true, however, that this "line" (or any line one might find) is not a symmetry axis in the Figure 8 Bend which was part of the distinction that was being made and, also, that central inversion symmetry does not guarantee the existence of such a line as does axial inversion symmetry.

I would second your skepticism regarding the efficacy of driving a spike through the center of a bend.

DDK

Dan_Lehman

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Re: Symmetry and the Jammed Bend
« Reply #7 on: December 03, 2011, 08:11:22 PM »
I do not believe I understand your comment about the Figure 8 Bend having a center point.

Because, I think, I'm only partly understanding the issue
(and thus making only partial sense).  I see the Fig.8 knot
as occupying the center point, and thought that was what
you were pointing to; but with the end-2-end knot, that
point is vacant, passed on either side by the two ends ... .

Quote
... central inversion symmetry does not guarantee the existence of such a line as does axial inversion symmetry.

?!  Really?  Doesn't CIS guarantee that any angle of approach to
and thus through the point is as good as another (and make it
unlikely that all can be unavailable), as opposed to having only *AN*
axis defined?!

--dl*
====

DDK

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Re: Symmetry and the Jammed Bend
« Reply #8 on: December 04, 2011, 07:33:57 PM »
@dl (and with comments more generally)

I think I see where some of my usage of "having a center" may have been confusing.  To have a center is to have an exactly defined point about which the structure is symmetric.  It is completely separate from the issue of whether the point is occupied or vacant (with rope in our case).  Thus, the Figure 8 Knot and the Figure 8 Bend both have center points as do all CIS structures.   As dl mentions, in the Knot the center point is occupied and in the Bend, the center point is vacant.  As an aside, it is interesting to note that all CIS knots (i.e. one rope) will have occupied center points and all CIS bends (end-to-end joining, i.e. two ropes) will have vacant center points.

In the OP, the reason the occupancy of a point or set of points was discussed is that (in theory) a collection of let's say vacant continuous collinear points (a vacant "line") might be of interest if they represent a path that a spike could take without "obstruction" in the spreading open of a bend.   An obstruction in this case would refer to rope which would need to be one could say pierced for a spike to progress through the knot.  Clearly, there are other important practical considerations which have been mentioned and I am generally in agreement with the lack of practicality of splaying open a bend through its center.

In CIS bends, there is only a single point which we can be assured is vacant of rope.   It is true that if one can find a vacant "half-line" from the exterior of the bend to its center, than the other side's "half-line" will also be vacant - thus a complete vacant "line".  But the existence of just such a "half-line" is not required for CIS.  As a thought experiment, imagine leaving the working ends of lets say the Zeppelin Bend rather long.  It is possible to then imagine wrapping and tucking these ends so as to obstruct any vacant "lines" which might exist while still maintaining CIS.

In axial inversion symmetry (AIS) bends, it is possible to perform the same thought experiment and obstruct all the available vacant "lines" except for one, the symmetry axis.  The symmetry axis can not be obstructed and one still maintain AIS.  Thus, there must always be at least one vacant "line" in AIS bends.

DDK