Knotworkn

BRAIDS includes information on the following topics:

  Andean Up Braiding   Turkshead
  Braiding   Single strand 8 plait
  Flat TH Cordage Calc   TH w/Pins and Former
  Finger and  Lucet 

  TH Expansion Calculations

  Headhunter   
  Herringbone  
  Lanyard  
 Mats  
  Pineapple  
   Sennits  




Until I get all of the pages entered and links hooked up the items listed above maybe out of order and have to be accessed sequentially.
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Andean  Braiding

Below is a sample of 16 strand Andean Up braiding.  The loop on the end is 8 strand Andean then the two ends are brought together and the 16 strand Andean is continued.



Start with eight strands and knot them at one end.  Locate the position two inches towards the knot end from the center and place a rubber band around the eight strands at that location.  
Hold the eight strands in your left hand formed as a fist with the thumb up and the knot end down and the rubber band inside the closed fist.

Lay the strands in groups of two each, with two going north, two going south, two going east and two going west.   At the start of the braid the meaning of UPPER and LOWER will not mean anything since it will not be apparent, but as the braiding proceeds it will become apparent. To start off we will just use the terms LEFT and RIGHT as viewed by the person braiding. The terms FRONT and BACK will mean back is away from your body and front will be closest to your body.

Two strands will be  picked up with the first, middle and thumb of the right hand. With the fingers of the right hand are pointing to the left, grasp the LEFT_FRONT STRAND and pinch it between the middle and first finger and then grasp the LEFT _BACK  STRAND and pinch it between the first finger and the thumb. The right hand is then rotated so the palm goes from facing to the left to facing down or CCW (counter clockwise). The twisted pair is moved to the BACK of the left hand and placed under and held down with the little finger on the left hand.

With the fingers of the right hand are pointing to the right, grasp the RIGHT_BACK STRAND and pinch it between the middle and first finger and then grasp the RIGHT_FRONT STRAND and pinch it between the first finger and the thumb. The right hand is then rotated so the palm goes from facing to the right to facing down or CW (clockwise). The twisted pair is moved to the FRONT of the left hand and placed under and held down with the ring finger on the left hand.

Given the preliminary descriptions of how to manipulate the braiding, I will establish some shorthand notation that will make it simpler to describe the movements for Z and S techniques and for sixteen-strand braiding.



Refering  to the figure above: 
First letter indicates front or back
Second letter is left or right
Third letter is lower or Upper

For eight strand braiding (left side figure) there is only one set per side so the left /right designation is dropped.

The ARROW indicates the first strands to start with and the direction they should rotate. The up arrow indicates counterclock wise rotation and a down arrow implies clockwise rotation. The rotation starts as indicated by the arrow and alternates direction as the strands are twisted moving from left to right.

The remaining strands are held down along the side of the braiding while the first set of strands are twisted then while holding them down the strands from along the sides are selected and moved up. The whole assembly is rotated 90 degrees (direction is not important as long as the direction of rotation REMAINS THE SAME for all the braiding).

Repeat the above process using the next set of strands and holding the previous strands down alongside of the braiding.

The braiding that starts with the CCW (counter clockwise) rotation will produce Z twist braiding and for the S twist braiding the rotation must start with a CW (clockwise) rotation. 

Having the information above we know that the eight strand braiding  we prevoously started  will  be  Z twist braiding.  Continue the eight strand braidding  until you have completed three inches of the braiding then secure the braiding with another rubber band around it .
Next you will bring the other eight strands up and lay them so that there are four strands  per compass direction,  4 pointing North, 4 pointing South ,4 pointing East and 4 pointing West.   It will be necessary to untie the knot that was intially placed in the end of the eight strands.  The loop formed with the eight strand braiding will be placed downward into the closed fist of the left hand with the weavers exitting by the thumb and first finger.  The four east strands and four west strands will be held placed along side of the loop to start.  To continue with the z twist braiding the process will be to use the diagram above right, with the exception that the arrow should be shown pointing from the fll to the blU indicators. So you would start with the (front-left-lower) strand and the (back-left-UPPER) strand and you would rotate them to the CCW and place them at the back under the little finger.   Next select the (front-left-UPPER) strand and the (back left lower) strand  and rotate them in a CW manner and place them in the front under the ring finger tip. Now pickup the (front-right-lower) strand  and the (back-right-UPPER) strand and rotate them in the CCW direction and place them to the back under the middle finger.   Fianlly pick up the (front-right UPPER) strand and the (back-right-lower) strand and rotate them in a CW direction and place them in the front under the middle finger tip.   Reach in from the sides and grasp the east and west strands and bring tnem up along the sides of the strands just crossed over and while holding the crossed strands down between the east west strands and the top of the braiding release your left hand fist and rotate the whole group 90 degrees so that the old east-west group take up the location where the north-south strands were and vise versa.  The direction of rotation does not matter as long as it is always the same each time the strands are rotated.

I know this sounds difficult, so spend some time to get it straight in your mind and then it will be simple and second nature to you. By varying the colors of the strands  will produce amazing patterns. I prefer to use three colors but experiment . I recommend that you start out by making eight strand braiding until you perfect the technique.  There is a good book on this topic which is  
"Sling Braiding of the Andes" by  Adele Cahlander it is a paperback book published by Weavers Journal  with ISBN: 0937452033 . It is an expensive if you can find it but she does give an good explaintion of this and other techniques in her book.
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Turkhead Knot Expansion Tutorial 

Concepts that enable direct computation of the dimensions of Turkhead knots that are produced by expanding a Turkhead knot either on the left side or the right side of the Standing Part (SP) are presented below.

The notation used here will be outlined now.
Ps and Bs represents the number of “parts” and “bights" respectively in the base or starting knot.
The terms dpL and dbL represent the amount of increase in the parts and bights for the expansion carried out on the LEFT side of the SP
While dpR and dbR represent the parts and bight increase for expansion on the RIGHT side of the SP.
PfL and BfL represent the number of “parts” and “bights" respectively in the final knot after the expansion on the LEFT.
Similarly PfR and BfR represent the number of “parts” and “bights” respectively in the final knot that was expanded on the RIGHT.

Listed below are some of the equations that are useful.

                       dpR + dpL = 2* Ps    (1)
                       dbR + dbL = 2* Bs    (2)

                       PfR = Ps + dpR          (3)
                       PfL = Ps + dpL          (4)

                       BfR = Bs + dbR          (5)
                       BfL = Bs + dbL          (6)

                       PfR +PfL = 4* Ps      (7)
                       BfR + BfL = 4* Bs     (8)

Below is a drawing that shows the base Turkhead knot drawn as a continuous series of triangles linked together with the width of the base on each of the triangles as wide as the number of “parts” Ps.
The red vertical lines are drawn in spaced in intervals as wide as the number of “bigh
ts” Bs.


 

If the difference between the bights and parts is solved for, (m*Bs – n* Ps)  yields the phase relationship for each of the parts relative to the starting point. When the phase shift is +1 or as shown when the “rising edge” of the waveform intersects a vertical line ONE unit from the START of the RISING WAVEFORM then the criteria is met for expansion on the LEFT side of the SP.

You can observe from the drawing that the brown line running parallel to the black line (representing the base knot) forms two peaks that are numbered and shown enclosed in CIRCLES at the top or the triangles. The brown line represents the line that would be the parallel line on the left of the standing part in the expansion.


Note: In the drawing the brown line is drawn on the right side of the base knot (shown in black) however with my background in electrical and nuclear engineering I just think of lagging and leading phase with the waveforms as they would appear on an oscilloscope running from left to right from zero in a time/space reference. The waveforms could just as easily be drawn running from right to left with the brown line drawn on the left side if that makes it easier to interpret.

With each peak there will be two bights formed in the final knot so for the figure above there would be a total of four additional bights or dbL = 4.

There are three bight groups shown between the start of the waveform and to and including the bight group  at the +1 location.   These three bight groups are shown numbered and enclosed in SQUARES at the bottom of the bight group markers.   As there are two lines at each of these locations , then the total number of additional parts added to the final knot will be  dpL = 6 .

Using equations (4) and (6) we get 

                        PfL = Ps + dpL   and     BfL = Bs + dbL
                        PfL= 7 + 6 =13 and      BfL = 5 + 4 = 9

Which gives the Left Side expansion topology: 
                              7p5bL ---> 13p9b


Having solved for the expansion on the left, it is a simple matter to determine what the topology will be for the expansion on the right side of the standing part using the information about the left and the equations above !!! 

From equations (1) and (2)

            dpR = 2*Ps – dpL which gives dpR = 2(7) – 6 = 8

             dbR = 2*Bs - dbL which gives dbR = 2(5) – 4 = 6

Plug that information into equations (3) and (5)

            PfR = Ps + dpR      and   BfR = Bs + dbR
            PfR = 7 + 8 = 15  and    BfR = 5 + 6 = 11

Which gives the Right Side expansion topology: 
                                7p5bR ---> 15p11b

Alternatively the values for a RIGHT  side expansion can be taken directly from a drawing like the one above by drawing the brown line  in the same parallel manner until the brown line intersect the bight group where the phase shift is negative one (-1). 

Extending the brown line in the figure above you will observe that there will be  an additional peak  and an additional bight group included which means an additional 2 more bights and 2 more parts than was required for the LEFT side expansion yielding  a total of  dbR = 3*2 =6 and dpR= 4*2 =8

These concepts were used to create the mathematical expressions used in the Expand portion of the program shown in the screen capture below. The Expand portion of the program is displayed in the lower left area and  computes the Turkhead expansion dimensions for both the RIGHT and LEFT side expansion using direct calculation based only the  “parts” and “bights”  values  of the starting Turkhead knot.  For  a more detailed description and those  looking for a method for you to do direct calculation here is a section below the table that goes into more detail on the math expressions.

The other portion of the program computes the  "RunList" for the starting Turkhead knot and optimizes for a maximum number of OVER's in the knot unless otherwise requested. Additionally it provides the cability to incorporate any coding sequence desired.


Beneath the screen capture is a table listing many lower order expansions.

                                              7p5b  Screen Capture






Turkhead Expansion Table  


In his book on Turkheads, Tom Hall uses the terminology Method1 (M1)to describe the expansion method where the WE starts to the right and parallel to the SP and Method2 (M2) to describe the expansion method where the WE starts to the left and parallel to the SP. I will use the M1 and M2 designations here to reduce the amount of typiing and to establish a consistency with some of the other literature available as well as using L and R to indicate the type of expansion used or for their resultant component increases.

A method of computing the increases in the number of bights (db) and parts (dp) and the resultant THK dimensions (Pf) and (Bf) for either a lefthand side (M2) or righthand side (M1) or both knowing only the dimensions (Ps) and (Bs) of the starting THK and the pin number sequence for the left or lower bight boundary    PL(m_) .

The PL(m_) pin sequence numbers can be computed as shown in the section titled Turks Head Knots using P ins and a Former or by using the following equation(9) which will require an understanding of modulo arithmatic.

PL(m_) = mod[( 1+ m*Ps),Bs]      (9)

where m={0,1,2, . . . ,Bs-1} which means m sequentially takes on the values from 0 to one less than the number of bights in the base THK. If the values for m and the corresponding value of PL(m_) are written down in a tabular form, then the increase in the number of bights for either right or left side expansion can easily determined.   Using the information about the basic foundation knot in addition the values for m_ the increases in the number of parts can be determined.

The values for the increases in the bights are computed using the following equations:
dbR = 2* m_2        (10)  Means 2 times the value of m when PL(m_) = 2 or the
                                       number of intervals until pin number 2 is reached.
dbL = 2* m_Bs       (11)  Means 2 times the value of m when PL(m_) = Bs or the
                                       number of intervals until pin number Bs is reached.


The values for the increases in the parts are computed using the following equations:
dpR = 2*[ (Ps*m_2 - 1)/ Bs]       (12)
   The phase relationship I spoke of in the
dpL = 2*[ (Ps*m_Bs + 1)/ Bs]    (13)    information above is apparent from 
                                                              equations (12) and (13) if you  
                                                              notice the +1 and -1 terms.

Resolving the final topology it is a simple matter of simply adding the appropriate increases in parts and bights to the base knot topology.

Ps X BsR -->( Ps +dpR) X (Bs + dbR) = PfR X BfR    (14)
Ps X BsL  -->( Ps +dpL) X (Bs + dbL) = PfL X BfL      (15)

This process will be demonstrated by going through an example using a 7p6b TH as shown in the section on the Using Pins and a Former.
Since the pin sequence numbers have been calculated there and we will make the tabular values using equation (9)
                                                       m 0  1  2  3  4  5
                                                 PL(m) 1  2  3  4  5  6
 
Solving for the M1 rightside expansion we note that at a pin sequence value of 2 results from a value of m =1 .
Equation(10) results in a value of               Equation(12) results in a value of
dbR = 2* m_2                                             dpR = 2*[ (Ps*m_2 - 1)/ Bs]
dbR = 2* 1                                                  dpR = 2*[ (7*1 - 1)/ 6] = 2
dbR = 2                                                       dpR = 2 .

Using equation(14) results in a final M1 rightside topology
Ps X BsR -->( Ps +dpR) X (Bs + dbR) = PfR X BfR
7 pX 6bR -->( 7 + 2) X (6 + 2) = 9p X 8b

Solving for the M2 leftside expansion we note that at a pin sequence value of 6 results from a value of m =5 .
Equation(11) results in a value of               Equation(13) results in a value of
dbL = 2* m_Bs                                            dpL = 2*[ (Ps*m_Bs - 1)/ Bs]
dbL = 2* 5                                                  dpL = 2*[ (7*5 + 1)/ 6] = 6
dbL = 10                                                     dpL = 12 .

Using equation(15) results in a final M2 leftside topology
Ps X BsL -->( Ps +dpL) X (Bs + dbL) = PfL X BfL
7 pX 6bL -->( 7 + 12) X (6 + 10) = 19p X 16b

A simplified model of Modulo arithmatic can be thought of like a wagon wheel with the number of spokes around the periphery equal to the modulus. Once you go around all the spokes in the wheel you start over again with the number one. Given the modulus is 6, then the numbers for the spokes would be 1,2,3,4,5,6 and the number 8 modulo 6 would be the number 2 . (1,2,3,4,5,6 then 1,2). Likewise 11 modulo 7 would be the number 4 (1.2.3.4.5.6.7.then 1,2,3,4).


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Turks Head Knots raised from basic foundation knots


Many TurksHead Knots (THK) can be produced by first tying one of  three basic foundation knots, which are the 2pX3b, 3pX2b, and the 3pX4b THK. Adding additional turns or  AB crossovers will increase the number of bights and  by using a method called laying the tracks and using a "mule" the  THK can be EXPANDED to produce a variety of larger knots. Once you gain experience in tying THK the need to use the mule can be eliminated.

All of the larger knots (in this tutorial) which have an EVEN number of parts will begin with the 2pX3b THK as a foundation knot and all the larger knots that have and ODD number of parts will have a foundation knot that is either 3pX2b or 3pX4b.

The number of times the foundation has to be raised depends ONLY upon the number of parts!  Subtract the number of parts in the foundation knot  from the number of parts in the desired knot and divide that result  by 2 and the answer is the number of times you need to increase the foundation knot. 

 As to the number of extra turns or AB crossovers needed to get to a specific number of bights, that number is the bightsdesired minus the  bightsStart plus the parts in the foundation knot take all of that and divide it by the partsDesired  and then subtract one. 

All that information is to remaind me about the mathmatics involved but I have generated a table that allows you to EASILY choose the final topology and see what the required foundation knot is and to determine how many extra turns or AB crossovers are needed and how many times the foundation knot has to be raised. The table is presented below and covers many of the knots that are tied in hand.

                                                                  PARTS

 2 3 4 6 7 910 
2 3pX2b        
32pX3b   2pX3b       
4
3pX4b  3pX2b      
2pX3b 3pX2b  2pX3b      
 6   3pX4b  3pX2b    
 72pX3b 3pX4b 2pX3b   2pX3b    
 8 3pX2b     3pX4b  3pX2b  
 92pX3b  2pX3b 3pX2b     2pX3b  
 10 3pX4b      3pX4b   
 112pX3b 3pX2b   2pX3b 3pX4b 2pX3b     
 12         
 132pX3b 3pX4b 2pX3b  2pX3b 3pX2b    
 14 3pX2b   3pX2b      
 152pX3b   2pX3b   3pX4b 2pX3b   
 16 3pX4b  3pX4b      
 172pX3b  3pX2b2pX3b  2pX3b  2pX3b 3pX2b  
 18         
 192pX3b 3pX4b 2pX3b  3pX2b 2pX3b   3pX4b   
 20 3pX2b     3pX2b   
 212pX3b  2pX3b 3pX4b      
 22 3pX4b    3pX4b    



An obvious observation from the table is that there are NO elements shown in rows 12 or 18 this is because all THK must satisfy the rule of no common divisor and both of those numbers are divisible by 2, 3 and 4 which are the factors in all of the foundation knots.

We will use the example of tying a 9pX17b THK to demonstrate how to use the table.  See the chart below.

At the intersection of  PARTS column  9  and  Bights row 17 you find that the foundation knot must be a 3pX2b.

EXAMPLE  9P x 17B




Following the blue line diagonally from right to left you see that you  encounter the 3pX2b foundation  three (3) times before you reach the far right side.  This is the number of times the knot will have to be raised. 

Continuing to follow the blue line upward you move only once (1) time before reaching the top most occurance of the 3pX2b foundation.  This is the number of AB crossovers that will be needed. 

To summarize the 3pX2b foundation is first given an AB crossover which increases the knot to a 3pX5b. Next using the "mule" the knot is raised from the 3pX5b to a 5pX9b,  the process is repeated again and the knot becomes a 7pX13b and finally the process is repeated for the final time which raises the knot to a 9pX17b THK

There are some good sites that discuss the use of the mule and increasing the dimensions of THs.   Until I have added the photos and discusssions of those techniques here I would direct you to the following websites for some good examples of how to increase THs.

The Alaska Museum of Fancy Knots website  has a good tutorial on how to make  Nantucket Sailor's Bracelets which demeonstrates the use of the mule on both odd and even part THs and a discussion of AB crossovers. 
   
A tutorial by Bud Brewer on the KHWW website called Nantucket Mat    demonstrates how to increase a TH by performing an AB crossover  and without using the mule.

Since there is a signifigant amount of information on the table above which contains ALL of the information about "EVEN" and "ODD" part THs on the same chart , I have made a separate table for each type .  

The color of the diagonal lines indicates which side of the Standing Part (SP) that the Working End (WE) should take when starting to lay down the tracks.  The mule will split the tracks and will assist you to determine whether to pass over or under the next bight.   Of course the WE changes to the other side of the SP as it makes the down pass and then reverts back to the original side on the next up pass if applicable. 

The diagonal lines represent the INCREASES in the number of leads from a 2 lead  to a 10 lead for the EVEN part TH knots and from a 3 lead to a 9 lead for the ODD part TH knots.

The column to the left of column2 for EVEN PART THs and vertical column3 for ODD PART THs  shows how many initial bights (loops) have to be tied before doing any expansion.    Each time the same foundations typ2 is encounterd  as you move up in column 2 or 3   represents another loop for 2pX3b THs or another AB crossover for the 3part THs..

The table below is for EVEN PART THs and the starting point for these expansions will be the  2pX3b TH  which is a simple overhand knot where the WE is brought up next to the SP. 

NOTICE:  If the 2pX3b is expanded by taking the WE to the left of the SP, a 4pX5b TH will result, however if you expand by taking the WE to the right of the SP the resulting TH will be 4pX7b !!!!!!!




For ODD PART THs  the starting point for the expansions will be either the  3pX2b TH  or the 3pX4b TH.  The diagrams below show the 3pX2b and 3pX4b starting  knot topologies as wll as a diagram of how to perform an AB crossover.










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Turks Head Knots using Pins and a Former

If you set two rows of pins into and at the sides of a  board or at the sides of a  tubular form and assign the pins numbers such that the pin numbers are same on the left as they are on the right, then all you have to do is to determine the number of BIGHTS you want and you will have to insert the same number of pins on each side  of the form as there are bights.  To determine how to tie the Turkshead you will start at Pin 1 and attach the end of the cord to pin 1.  In order to determine which pin the cord should go around will be determined from the number of leads or parts. 

 THE RULE :  is to take the number of LEADS and subtract 2, then divide that answer in half and that is the number of pins to skip over. 

                        P= (Leads -2)*(1/2)

 What to do if the number does not come out a WHOLE NUMBER (you know it has a remainder)???           (This WILL happen for all ODD part TH's !!!)
That is easy just forget about the remainder and skip the WHOLE NUMBER of pins on the leftthand side BUT on the righthand side you will skip the WHOLE NUMBER PLUS ONE more. Thats all there is to it. 

When you are counting up you will have to stay in  modulo(BIGHTS) arithmatic but that will not be a problem because you will only have the same number of pins as there are bights so you will automatically have to go back to pin 1 when you reach the last bight.  We will show how this works with an example.

EXAMPLE  7pX6b TH 
Begin by finding the number of pins to be skipped
 (7-2)/2 =2.5  then WHOLE NUMBER is  2  
and so the number of pins to SKIP on the left will be Pl = 2
and  the number of pins to SKIP on the right will be Pr= Pl+1 = 2+1 =3 

Starting with the lower pin1 (our righthand side) we skip two so we have
1 + 2 +1 = 4  so this is pin4 on the lefthand side (our topside)  from there we go to 4+3+1 =8 - 6 =2.  This pin2 on the bottom side.  Notice since we went higher than the number of bights we sutracted that number from our total to bring us to the proper location. Continuing on we start with pin 2  so
2+2+1 =         pin5  top                  5+3+1=9-6=   pin3  bottom               
3+2+1=          pin6  top                  6+3+1=10-6= pin4  bottom                    
4+2+1= 7-6= pin1  top                  1+3+1=           pin5  bottom               
5+2+1= 8-6= pin2  top                  2+3+1 =          pin6  bottom                 
6+2+1=9-6=  pin3  top                  3+3+1 =7-6=  pin1  bottom               


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Flat TH  Mats Cordage Length Calculator

The Flat TH Cordage Calculator computes the cordage lengths for  Turks head and Multistrand circular mats  and also generates a KNOT TEMPLATE  which can be followed when tying the knot and a MAT TEMPLATE (that  can be printed).   Below is a picture of the calculator



Shown below is a view of one of the templates the Mat Template can generate.


The calculator may be  borrowed for evaluation   (like checking a book out of the library).  

                              Simply click  Mat Calculator    

To purchase your own registered calculator visit Knotworkn Market       

After the file  downloads to your computer press RUN then RUN again and when the registration page  comes up then press the START button.

I have not personally tested this calculator against  many of the mats that are possible. since I personally prefer the cyclindrical form but I developed this calculator as a tool for the many folks that do tie TH mats  at the request of Gordon Perry. 
I put it here for people to use and to evaluate it.  I have a feedback form on my home page and would appreciate any comments that you are willing to provide especially if they could be used to improve the calculator.
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 Single Strand  8 plait braiding

An example of single strand 8 plait braiding is shown in the picture below. 

 To braid a single strand eight plait grommet obtain at least seven feet (210 cm) of cordage and locate the center.  At the center form a circle of roughly 3 inches (7.5cm) diameter taking cordage from both halves equally.  Begin to form a grommet by making four turns around the starting loop with each of the free ends being sure to keep the lay of the grommet correct.


The pictures above show the first two strands of the grommet. 

Shown below  are the third and fourth strands of the grommet laid up




The next steps show how to interlock the cordage and to start the interweave of the grommet. 

The strand coming from the right passes to the left of the leftmost strand and then turns back towards the right crossing in front of the leftmost strand and over ONE additional strand and then passes UNDER itself and the additional strand.  

The leftmost strand after passing to the right of the rightmost strand then turns towards the left passing OVER the rightmost strand and then immediately passing UNDER the TWO strands that were not passed under by the rightmost strand.  This is the same as passing OVER the same two strands that the rightmost strand passed under.

The picture on the left below shows this process clearly .    The photo on the right below shows the same information as the photo to the left but additionally shows how to interweave the strands in order to make the plait.



Each cordage end is woven over TWO then under TWO.  The weaving should pass over and then under the SAME two strands each time. Continue the weaving on around until the area where the cordage first interlocked.


The figure above shows the weaving pattern as the strands pass through the initial interlock for the first time.  The weaving sequence is always the same over 2 then under two.




In the photos above the over 2 under 2 can easily be seen. The awl is pushed under the two strands that need to be passed under and the two strands between the awl  and the working end  are  the two strands that are to be passed over. The interweaving tends to spiral around the grommets as the weaving progresses.


The lobes of the interweaving are apparent in the photo above with three sets of  lobes then a space  followed by another set of three lobes a space and so on .  At this point  all but the last sequence of weaving has been completed.  The rightmost strand has made  two complete rotations of interweaving and so must the leftmost strand so that with the four strands from the grommet and the four fron the interweave we get an eight strand plait.



The photo above shows the end of the interweave and the strands meet at the same location passing under two strands.  All that is required is to go back through and adjust the tension and remove any slack from the plait.



At this point the ends can be lashed and clipped or they can be fed up through the plait and tied into a two strand Matthew Walker knot or a diamond knot.

Below are some  examples of single strand eight plait knots.




This is a collection of the first three knots of the Periodic Table of Knots formed using single strand 8 plaiting.  The knots are connected together using Spanish Ring knots. 

The Periodic Table of Knots is shown below and is one of those items that I do not know from whom I received the photo.


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