A Non Fiction Trilogy



     So now continuing from the 4P particle produced in case #10 & #13 of the last chapter. Each of these ensembles is different. Case #10  has a lopsided arrangement, I will call it 4P Unbalanced  (4P ub) . Case #13 is in balance, I will call it 4P Balanced (4P b). 

  Overview [this probably needs updating]

        If a 5th particle is added, unbalanced with 4 one side and 1 the other still, then there is rotation again as in the 3p case, but only now the particles are most likely not in a linear motion of c, but other values < and > c. This could continue to 6P, 7P, 8P and up with the same type of results. Furthermore, the same can happen on the other side, so a 2x2 set, then 3x2, 3x3, 4x3 etc. Eventually both sides are filled up.

          Before each side is filled up, PP hitting from certain directions can “breakup” this proto-nucleon group.   But once the faces (rows) are complete all motions occurring when a PP hits a proto-nucleon are transferred within the proto-nucleon, equalizing the velocities of all proto-nucleon particles.

         If a PP hits a proto-nucleon face it may rotate perpendicular to the rods of that face.   If that PP is then across all the rods of that face they rotate around the PP, causing the other face to rotate in unison also. When the proto-nucleon has rotated so the PP is at a midpoint center of mass all rotational motion goes to linear again as per principle 14. I call this PP an “extra”. An extra also would align on the other face, so both faces obtain an extra, forming a nucleon.

         Thereafter any PP hitting each face either hits the extra & face, or incompletely hits the face only, and “rebounds” away. Else a PP rotates the face of the proto-nucleon as above, however as this PP can never align at a center of mass position it will eventually get thrown off whereas the extra eventually stays put. A small percentage rotates the extra until at center of mass and transfers motion to the nucleon, which rotates back around the PP and if also reaching center of mass will revert to linear motion accelerating that PP. This sort of thing is how the nucleon gets accelerated and decelerated.

          This leaves a nucleon with two rows of particles, 137 in each row, or more likely 135 as the last one would be a tight fit, one row turned perpendicular to the other, and the two extra as described above. Now that meant the N was composed of 135 x 2 +2 or 272 particles. When accelerated by one PP particle the total mass would then be 273 PP.


Specifics for the Particle ensemble from Case #13, Chapter 4.

        The 4P Balanced particle should be traveling at the speed of light plus whatever rotational value the 3P particle had plus whatever value > c the 4th particle added to the system. If, when ABCD come together and rotational motion reverts to linear, CBD being in balance on A, it is on the direct hit side then the original linear motions and the new linear acceleration are contrary motions and they equalize to a single resultant motion as per principle 19.2. Thereafter go off through space together but with their motion being not as a perpendicular to their position/shape.

        Else if the rotation completes on the overtaking side the motions also equalize, but with a combined/compounded vector.

        So either way the energy of particle ensemble is the same, but apparent velocity is different.





Further Collisions

       Hereafter if a 5th particle, E, collides with the 4P Balanced we have these possibilities.


Chart 4.5-1

There is no Rotation, only ABCD moving linearly through space.

If ABCD is one vector. For each Direct Hit

  1. E hits on A

  2. E hits on C or D

For each overtaking hit

3. E hits on A

4. E hits on C or D

ABCD is a compounded vector.   For each Direct Hit

  5. E hits on A

  6. E hits on C or D

For each overtaking hit

7. E hits on A

8. E hits on C or D


1.       E hits A directly, CBD go off with their velocity. A and E are stopped and torque to each other, then rotate to a doublet.


2.       E hits C or D, will say C.   All A and C's linear is stopped. BD pulls away. C torques, re-hits B and A, accelerating them both, and D gets accelerated also. All acceleration on A causes re-rotation of A which off C and D is still balanced during instant and reverts to linear acceleration of all. But this causes all of C to drain into A (as is propelling A back toward itself), This prevents the acceleration of BD in the other 90o direction from occurring.

          Here it all becomes a jumbled mess.  A cannot go linear because C is stationary. A rotates off C and D getting nowhere, so all the motion of ABCD drains through A into B which shoots off, and hits E, torqueing slightly and rotating slightly to its mid-mass point. Some depends on where E is, but a most likely scenario is for when B rotates on E its arm re-hits A, torqueing and rotating on it, then accelerating it and then when A re-rotates it rotates off C and D and rebounds, B following it and forming a high speed doublet. E is left to form a slower doublet with C, and D is left stationary in space.



3.       E hits on A. Treat as overtaking hit chapter 2.


A)      If E reaches midpoint on overtaking side. E scoots around A until aligned by previous scooting mechanism.   Here, something interesting happens, amazing how different scenarios can develop over such silly simple things. E’s excess will cause A to rotate back on it, but when it reaches a point its backside will re-hit ABC and re-rotate on that again. Then the process will re-peat until at some point A’s rotation on E will clear CBD and go around A then re-hit CBD from the other side, have dual rotation and rebound.

           However, if A should reach its midpoint such it accelerates E. A will re-hit CBD and rotate around them as many times as needed until E hits D, and held by same causing acceleration of D to B and C which slide until the circular motion of E is free to transfer to all such that they all can circularly rotate freely together. As the circular motion is transferred it combines with the sideways linear to hit A causing all to go rotating off of A, creating a loop. So end up with a lopsided particle CBDE – being pushed in circular by A rotating off of C and motion going through E at its midpoint to form loop with rest. middle-3 unbalanced


B)      If E hits reaches midpoint on the direct hit side DBC go off and E and A form doublet.


4.      E hits End rod, say C. Treat as overtaking hit chapter 2. If

A)      E rotates and hits CBD before it reaches its midpoint, it ends up with dual rotation and rebounds.

B)      Likewise if E hits on the front side of CBD as in A), it also rebounds.

C)      But (and perhaps this sort of thing should be considered in other cases too) if in rotating toward the front or back it reaches its midpoint so as to accelerate C. Then C in re-rotating on E and re-hits BD and A. This causes acceleration of CBD and A (two direction acceleration). Over the instant C and A’s vectors come together so that BD would pull away, but that would cause All C’s to go the A which, over the instant would still cause A to, though its faux rotation, accelerate CBD and itself faster in a linear direction, the same as its original. E is left with just enough motion to keep up and CBDA slide past it.


5.               Both dual vectors should be stopped, so same as #1.


6.               E hits, say C, C’s vector toward E torques, but continues as linear along other vector, attempting to accelerate   BD. All A’s motion is stopped. However when C torques it runs up a big phone bill, oops. When C torques it would re-hit BD and A, but BD are already moving away, so all goes to A.   A rotates on C, but accelerates against CBD linear. All motion drains from C, C can’t rotate with ABD. All this has occurred within an instant according to priorities and "inferred motions".   BD slide off of A as it rotates with their linear plus a rotational motion, causing them to loop through space thereafter. A rotates on C as well as E, but they all should break up when midpoints are reached (NOT FIGURED).


7.       E   overtakes A.   Same as case #3, except results will be particles ACBD still have a compounded motion but E does not. So again, the energy levels are the same as in 5, but technically that would not be apparent just by comparing their motions though visible space, if one could do that, which one cannot.

8.       Same as case #4, but with caveats as in #7.


        So to the extent this is all correct one is left with a problem, no clear accretion toward a Nucleon like particle yet. the only thing to do is build upon the particle from Case #3A.

Specifics for the 4P Unbalanced Particle ensemble from Case #10, Chapter 4.

          So here the 4P ub has, as from chapter 4, a linear velocity of c, and a rotational velocity also, just like the 3P particle. 


      Chart 4.5-2

DIRECT HITS           The linear motion of a single particle (E) is directed at the linear motion of a 4P ub.

            The long arm of A is rotating away from its linear motion (toward back).

1         E hits off the long arm of A rotating toward back.

2         E hits off the short arm of A rotating toward front.

3         E hits off C.

4         E hits off B.

5         E hits off D

            The long arm of A is rotating toward its linear motion (toward front).

6         E hits off long arm of A rotating toward front.

7         E hits off the short arm of A rotating toward back.

8         E hits off C.

9         E hits off B.

10       E hits off D.

OVERTAKING HITS            (E overtakes ABCD)

The long arm of A is rotating toward back

11     E hits off the long arm of A rotating toward back.

12     E hits off the short arm of A rotating toward the front.

13     E hits off C.

14     E hits off B.

15     E hits off D.

 The long arm of A is rotating toward the front.

16     E hits off the long arm of A rotating toward the front.

17     E hits off the short arm of A rotating toward the back.

18     E hits off C.

19     E hits off B.

20     E hits off D. 

1-10  Similar to the Direct hit cases chapter 4.

12.  Like case 10 chapter 4

A) E hits on A, treat as overtaking hit chapter 2. That is:

1. E's motion swings to unison with A per principles 15.4 - 15.6.

2. E torques and follows also per principles 15.2 and 18.2.

3. A) E rotates per principle 13 and 15.6. Because It loses its above mechanism for following the rotation of A. E instead will have a virtual rotation as A pulls away, but this virtual rotation should be enough to still cause E to hit A, but off center, so E again would torque, and also follow A around. This all happens in the first instance, and repeats for every other "instance". MATH NEEDS TO BE DONE HERE AS TO IF TO FAST CAUSE REBOUND ETC Therefore E rotates A while following its rotation around and also has a unison linear motion the same as ABC, until E reaches its midpoint.

   B) Once rotation begins and follows as outlined in 3A here, if, and likely, E rotates to direct hit side it is actually impeding the rotation of A. When this happens E is at top of A.  A rotates on E. All rotation ABCD goes into A's rotation on E, as it sweeps up past ABCD. Depending on position E and A may reach midpoints and go off as a doublet with speed > c. Or more likely A rotates all around E, each torqueing and rotating on each others, and then A re-hits BCD and rebounds, leaving E  space with virtual rotation and linear speed same as BCD's linear, and BCD continue with linear speed at c.


C) But if by chance E is on the overtaking side relative to A's rotation when it reached midpoint, it accelerates A to re-rotate back around E. But it can’t as that motion runs into BCD. This causes their rotation to rotate off of D, moving the end of the arm A away from E so that E's release linear has a amount that re-hits A above midpoint, and off balance some, causing the process of torque and follow, and rotation downward until the Particle E is parallel with BCD. Then CBADE synchronizes, even though the feedback off B to C to D to A if necessary, until EDCB are aligned. That is EDABC can re-rotate off of E without rebound because rotation accelerates B at midpoint which rotates back though C and D equalizes the system, until all four particles BCDE are in line and touching for the ensemble ABCDE which moves off at c and with a rotational velocity also. So 1-middle-2 unbalanced   

16. As in case #13 chapter 4. Same as that complicated scenario except when finished ABCDE are a unbalanced set and rotation occurs off of the longer arm from D, the other side re-rotating through B locks all together and allows the rotation to occur off of D. So ABCDE are traveling at c with a large rotational momentum.  So 1-middle-2 unbalanced  

Other cases not worked out yet, but should be similar to chapter 4 cases. 



    Now without going into further details, will leave that for others in the future I hope. Proceeding from the sets ABCDE above, 3 cases overall from both charts all ended as lopsided partices. As one goes with another particle F, then G  etc. they should in alternaive sequences either fill up to a balance particle or move to unbalanced particles of one type or the other, totally unbalanced, or partially unbalanced, up until the point of reaching the set that is all filled out.  
    So it appears if I follow that scenario, though detials might prove otherwise, that

with 2 particle, crossed and balanced (no rotation)
with 3 particles, middle particle-1 beside it unbalanced (PARTICLE A FOR ALL THESE IS PERPENDICULAR TO THEM)
with 4      "       ,   1-m-1 Balanced
with 5      "       ,   1-m-2, and m-3 unbalanced
with 6      "       ,   2-m-2 Balanced, 1-m-3 and m-4
with 7 particles  2-m-3, 1-middle-4, and m-5
with 8 particles   3-m-3 Balanced , 2-m-4 and a 1-m-5, and m-6
with 9  a 3-m-4 2-m-5, 1-m-6, and m-7
with 10  4-m-4 Balanced, 3-m-5, 2-m-6, 1-m-7, m-8
with 11 4-m-5 , 3-m-6, 2-m-7, 1-m-8, m-9
with 12 5-m-5 Balanced, 4-m-6, 3-m-7, 2-m-8, 1-m-9, m- 10
with 13  5-m-6, 4-m-7, 3-m-8, 2-m-9, 1-m-10, and m-11
with 14 6-m-6 Balanced, 5-m-7, 4-m-8, 3-m-9, 2-m-10, 1-m-11, m-12
with 15 6-m-7, 5-m-8, 4-m-9, 3-m-10, 2-m-11, 1-m-12 and m-13
with 16  7-m-7 Balanced, 6-m-8, 5-m-9, 4-m-10, 3-m-11, 2-m-12, 1-m-13, m-14 
etc up to 135 (assuming 136-137 are to tight to fit) plus particle A = 136

Now the question is how to get the row on the other side!   

>Chapter 5 Elements