Author Topic: There can be no symmetric midline (TIB) single loop.  (Read 3690 times)

X1

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There can be no symmetric midline (TIB) single loop.
« on: June 13, 2013, 12:47:20 PM »
  ANY mid-line (TIB) eyeknot can be tied by the same general method : bights are reeved through other bights, until, at the very end, there is one final bight which is reeved through another, and becomes the eye of the loop - or, alternatively, the whole nub is reeved through a final bight, which shrinks and becomes a collar.
  Now, to retain a perfect symmetry, there is only one way to follow : Form and reeve two symmetrically located bights at each stage, that is, tie a double loop - but to tie a single loop, we have, at some point, to choose which of the two bights of the still symmetric TIB knot we will reeve though the other, and that choice, inevitably, destroys ( "breaks" ) the perfect symmetry. The "last" two symmetrically located bights become one eye and one collar - and the symmetry preserved till this stage disappears.
  Ashley states that " ...there seems no end to the number of knots of this nature [ double loops tiable-in-the-bight ] that are possible." Almost all possible sequences of forming symmetric bights and reeving them through other symmetric bights, produce stable, secure double loops, with more or less communicating bights. As with every practical knot, the task is to tie loops which consumes the less material, are adequately secure, and their particular tying sequence can be memorized easily by the average knot tyer.
  Therefore, we have to live with this limitation of the topology of our world : The two legs of our single TIB loop will not follow symmetric paths inside the knot s nub, so our loop, like ANY possible TIB single loop, will be loaded differently from its two ends... People do not seem to realize that the Alpine Butterfly loop is not a symmetric knot - although, at light loadings, the difference seems irrelevant, at heavy loadings - near, say, 33 % or even 50% of the strength of the rope - the difference becomes quite pronounced.
   So, the task of a "symmetric" single midline loop becomes more modest ; we try to entangle and secure each eye leg / end segment in the best possible way, so the loop will not be deformed much if it is loaded by its one or its other side. That is what I, too, had tried to do with the pet Loop, presented recently. My understanding is that the way this has been achieved in the pet Loop is quite satisfactory, given the simplicity of the tying method of the knot as TIB ( the same as the Span loop ) - and, because of the much better way, relatively to the Span loop, its tail is secured, when this eyeknot is loaded from its one side as a post-eye-tiable ( PET) bowline-like loop, it is preferable from the Span loop, indeed.

X1

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Re: There can be no symmetric midline (TIB) single loop.
« Reply #1 on: June 13, 2013, 03:09:19 PM »
   The interested reader can well pose the following two reasonable objections :

1. Imagine a vertical mid-line bight/loop, placed perpendicularly to the horizontal line where it is formed, and bights at the left and at the right of it, which can wrap the two legs of this bight as half hitches. Would nt the TIB eyeknot, that is going to be generated this way, be an axially symmetric knot, in relation to the vertical central axis of the loop ?

    No... because the two bights of any pair of symmetrical bights that is going to be wrapped around the two eye legs of tis bight/loop, can not encircle those legs and be symmetrical at the same time - the two symmetrical bights at the left and at the right side of the central bight/loop are independent, they can not be entangled with each other. So, the main bight/loop should be reeved through one, and one only, of them at each time, therefore each one of them would be placed "above" or "below" the other, in relation to the vertical axis. That is a cause of asymmetry, with the direct consequence that the main bight/loop would behave differently if it is hanged by the one or the other end.

2. What about the double / retraced fig.8 loop, or fig.9 loop, etc., where the knot is formed "with a bight", "in the end" ? Are nt those TIB loops not symmetric ?

   No... because each line that follows its twin partner along their common path inside the knot s nub, is located, in relation to this partner, "above", "below", at the side, etc., in different parts of the knot. Even if the symmetric stopper-turned-into-an-eyeknot is carefully dressed, as it should be, the perfect symmetry can not be attained. However, if the two lines of the double line that forms the loop are twisted around each other ( like they would have been in a 2-stand laid rope ), the asymmetry will get less and less noticeable and important, as the coils will be more in number and less widely spaced along their double-helical path. But such multi-twisted segments can not be parts of a practical / easy tiable knot.
   ( I should also point out that the eyeknots formed by retracing a symmetric stopper, are not post-eye-tiable, by definition : the stopper - as a whole, but also each individual link of it - is topologically equivalent to the overhand knot, the fig.8, knot, the fig.9 knot, etc, but not to the unknot - otherwise it could have not remained tied around and within itself. Although the present thread is about TIB single loops, in general, what interests me most are the TIB AND PET loops, where this almost-symmetry of those double line / retraced symmetric stoppers does not apply and so it can not be utilized. )
« Last Edit: June 13, 2013, 03:25:56 PM by X1 »

Dan_Lehman

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Re: There can be no symmetric midline (TIB) single loop.
« Reply #2 on: June 15, 2013, 04:52:47 AM »
   The interested reader can well pose the following two reasonable objections :

1. Imagine a vertical mid-line bight/loop, placed perpendicularly to the horizontal line where it is formed, and bights at the left and at the right of it, which can wrap the two legs of this bight as half hitches. Would nt the TIB eyeknot, that is going to be generated this way, be an axially symmetric knot, in relation to the vertical central axis of the loop ?

I disagree : while unwieldy, this objeciton that you have
found --in general-- solves the riddle against your assertion
--there ARE symmetric TIB eyeknots.  (E.g., tie the two
formed-in-SParts bights together with a reef knot
or Ashley's #1452 ; the TIB eye is beside each of
the respective bight SParts equally, symmetrically,
and the joining knot is symmetric.  Then another, simpler
solution comes to mind : sort of embedding the single-eye
aspect in forming #1408/1452 (that poor symmetry of
the zeppelin is less accommodating, for it wants
tails going opposite directions, and here I'm sort of making
the single eye be in the position of adjacent tails!).


--dl*
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X1

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Re: There can be no symmetric midline (TIB) single loop.
« Reply #3 on: June 15, 2013, 09:54:48 AM »
--there ARE symmetric TIB eyeknots.

  Noope ! Do not confuse a triple loop knot with a single loop knot, because the two side-loops have short eye legs !  :)  If we accept, besides the main single loop that is tightened by the pulling oft its eye legs, any number of pairs of not-pulled-by-anything bights/eyes coming out just a little bid of the knot s nub, and be stabilized there only by inner friction ( the ingenious side-bights embracing-each-other-in a side-Reef  "counter-example"... ), then, of course, anything goes ! If there were no friction, how would you stabilize those bights ? By toggles ?  :)  They would be swollen into the depth of the nub, and consumed by the standing ends and/or the eye legs of the main single loop in no time.
   A single loop TIB knot is meant to be a four limbs knot - not a four limbs and/plus who-knows-how-many pairs of hanging here and there non-functional bights/eyes ! Show me the (symmetric TIB single loop) knot !
   
« Last Edit: June 15, 2013, 09:57:43 AM by X1 »

X1

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Re: There can be no symmetric midline (TIB) single loop.
« Reply #4 on: June 15, 2013, 12:31:14 PM »
   If we are willing to accept non-functional elements in our knots ( bights, or even mote complex structures, like overhand knots, etc. ), and seek for what is most important, structural-under-loading symmetry, not shape symmetry, then things change...
   See the attached picture for such a load-symmetric single TIB loop. It is essentially an Englishman s knot ( ABoK#1022), when it is loaded from the one or the other side of the main line. The only difference from the ABoK#1022 is the one - different each time - additional overhand knot ( the "above", during loading by the one side s standing end, or the "below", during loading by the other side s standing end - "above" or" below" in relation to the central overhand knot in a vertically placed loop ). This additional third overhand knot remains tightened, but it is not squeezed onto the central overhand knot - so we can say that it is non-functional.
 

X1

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Re: There can be no symmetric midline (TIB) single loop.
« Reply #5 on: June 15, 2013, 01:46:15 PM »
   Is there any difference between the pairs of non-functional bights of DL counter-example, and the non functional overhand knot in the knot shown in the previous post ? I believe there is. If we imagine that these elements - and only these, in the whole knot - are made from a very slippery material, in the case of DL knot, the moment the knot will be loaded the pairs of bights will be

...swollen into the depth of the nub, and consumed by the standing ends and/or the eye legs of the main single loop in no time.

   So, although those elements are not functioning as the two standing parts and the two eye legs of the main bight/eye, they have to induce and absorb enough friction forces for their own sake, in order to be able to retain their place in the knot and their shape - otherwise they will degenerate into straightened lines and they will be untied altogether.
  On the contrary, in the case of the knot presented in the previous post, if the additional element ( the "above" or the "below" overhand knot, in this case ) will not induce or absorb any friction forces, especially then, nothing will change ! The symmetry of the knot, regarding loading, would be sustained, and the non functional overhand knot will remain knotted, in its place.
( Notice that the three overhand knots can well be adjacent to each other. During loading from the one or the other side, only two of them will be squeezed upon each other, the third not playing any role at all. I have shown them more spaced than necessary, because there have been complains about my pictures recently, by people claiming they have not being clear enough. )
 

Dan_Lehman

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Re: There can be no symmetric midline (TIB) single loop.
« Reply #6 on: June 17, 2013, 06:25:18 AM »
--there ARE symmetric TIB eyeknots.
Do not confuse a triple loop knot with a single loop knot,
because the two side-loops have short eye legs !  :)
If we accept, besides the main single loop that is tightened by the pulling oft its eye legs,
any number of pairs of not-pulled-by-anything bights/eyes coming out just a little bit
of the knot's nub, and be stabilized there only by inner friction ( the ingenious side-bights
embracing-each-other-in a side-Reef  "counter-example"... ), then, of course, anything goes !

Touche', yes, some *ends* of my *singlEye* knot would
be bights, as you point out.  Okay, you have tougher demands;
but the possibility of tying with bights qua ends is interesting
--although in this case one would be just solving a problem,
and otherwise have an awkwardly large knot.

Quote
If there were no friction, how would you stabilize those bights ? By toggles ?  :)

If there were no friciton, how would you ... knot?!   ::)


--dl*
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X1

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Re: There can be no symmetric midline (TIB) single loop.
« Reply #7 on: June 17, 2013, 12:45:08 PM »
If there were no friciton, how would you ... knot?!   ::).

   Elementary, my dear dL... :)   By using the pair of the eye legs ( of the single loop s main bight ) as a "pivot" through the two parts of the knot. This pivot does not need friction to remain in place, because it is loaded by both sides. In any properly called "single loop knot", we should keep in mind that there are four, not two limbs : the two standing ends and the two eye legs. Therefore, we have a two-parts knot. If we imagine all of the limbs loaded, we can see that the knot could remain knotted, indeed, even if there were no friction.   
    I had to use this simple "thought experiment" in order to clarify what I had meant by this "non-functioning elements" neologism - because any segment simply penetrating and/or interwoven within the knot s nub does affect its form, by its mere spatial presence within the knot, by the bulk of its body ( so, it makes a difference, regarding the shape and place of the functioning elements as well ). Those pairs of bights you had tied to each other with the reef or the ABoK#1452 ( in fact, with whatever other bend you can think of  :) ), do alter the flow of the tensile forces within the functioning elements, although their presence is not necessary for the knot - i.e., if and while the knot is loaded by its four limbs, it will remain knotted, even in their absence, albeit in a very different form.     
   Let me try again, using a second mental picture, which could possibly clarify the first one I have used previously.
   Imagine there is no inherent friction whatsoever in any segment ( "functioning" or not ) of the rope on which a single TIB eyeknot is tied on. If you load ALL its four limbs ( the two standing ends and the two eye legs ), pulling them from whatever direction you wish, can you untie the knot ? No. What you will be left with will be two still entangled segments, which can not but remain entangled, not because of the geometry, but because of the topology. However you manipulate the ends of those segments, without altering the topology of the whole knot ( i.e., without re-tuckings of these ends through the knot ), you will not be able to disentangle them. This will tell you that one, at least, of the two segments is not TIB  : what was TIB was the initial whole knot, but now you load/pull the ends of its four loaded limbs, one, at least, of its two parts can not become both topologically equivalent to the unknot, so you can not disentangle them from each other without altering the topology of the knot. Therefore, you do not need friction to be able to "knot" a four ends tangle of two ropes, simply because it is not but a non-TIB bend of those ropes.
   I will attempt a definition of the "functional" and "non-functional" elements of a knot:  If all its limbs are loaded by ( say, just to simplify the mental picture ) an equal load/force, and we have reached a mechanical equilibrium, the "non-functional" elements would be the elements whose rope length would have been consumed, they would have been swallowed inside the still knotted tangle, where they would have been almost straightened out. ( This "almost" means that what is left of a long "non-functional " bight, like the one of the two that are tied by the Reef or the ABoK#1452 of the suggested counter-example, can still remain as an open, widely curved rope segment, but not the convoluted within its "twin" double line of the initial "non-functional" bight. )

the possibility of tying with bights qua ends is interesting--although in this case one would be just solving a problem, and otherwise have an awkwardly large knot.

 
   You may mean that you can tie two "Gordian" knots, each with the one of two ( or four, six, etc.) symmetric bights and the one end of the two ends of the whole knot, which do not need friction... and so they are non- "functional", although they are not condemned to disappear in the absence of friction forces. Firstly, this knot would be awkwardly complex, as you say - there are no "simple" Gordian" knots that can be used as parts of a practical knot. However, I believe that such a knot can not even exist and be symmetric, because any connecting segments ( the "pivot" of the main bight, or other bights extending from those Gordian parts and encircling the pivot and/or each other, would "break": the symmetry : one of them would still have to be "above" or "below" the other, etc.

 
   
« Last Edit: June 17, 2013, 01:16:37 PM by X1 »