A Non Fiction Trilogy

CHAPTER 4   CONTINUED ACCRETION PART I 



            In Chapter 3, Case #3 (see also Figure 4.4) it was outlined how 3 primary particles might accrete.   Along with unison linear motion through space this triplet always has a unison rotational motion. Now for some more mechanics.



Rods C & B (imagined) are perpendicular to page and Rod A (shown) for all Figures 4.1- 4.4


  

                                                                              
  Figure 4.1



            In FIG. 4.1 Any force of rotation off of B presses on C, therefore all rotation occurs off of C.

            If C is at the midpoint of A (Figure 4.2) then A doesn’t rotate on C as each rotation on each side of C is balanced and A accelerates C linearly as per Principle 9, 12 & 15. But any linear of A presses on B and so all force becomes a force of rotation around B, accelerating C in a curvilinear motion as per Principle 17.


                                                                            

                                               

                                                                                                                    

        If C is on the opposite side of the midpoint of (A) from B (FIG. 4.3) then the rotation of A around B & C, are in contrary directions, and A cannot rotate both simultaneously but rebounds, as per Principle 18.

 

                                             

                                                                                                                                     

  

  3P-1P Collisions


            Applying these mathematical considerations to the 3P rotating case

 

 

 

Figure 4.4


                                                                                                          

           

           


Continued Accretion


         The 3P particle with rotating and linear motion (at the speed of light) from chapter 3 is the key. From this particle "accretion' can continue toward a Nucleon. If, with similar processes as described in chapters 2 and 3 a fourth particle is added, making three in a row and one perpendicular at the midpoints, this 4P case would then be balanced and all the rotational motion would, under the right circumstances, become linear motion, accelerating all four particles to a unison linear speed. This speed however would now be > c. It could maybe be accelerated and decelerated much as miniature nucleon.


         For 3P-1P collisions the number of cases to consider is necessarily greater than those considered in 2P-1P collisions. They are:



DIRECT HITS           The linear motion of a single particle (D) is directed at the linear motion of a 3P triplet (ABC).


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


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

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

3         D hits off C.

4         D hits off B.


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


5         D hits off long arm of A rotating toward front.

6         D hits off the short arm of A rotating toward back.

7         D hits off C.

8         D hits off B.


OVERTAKING HITS            (D overtakes ABC)


The long arm of A is rotating toward back


9       D hits off the long arm of A rotating toward back.

10     D hits off the short arm of A rotating toward the front.

11     D hits off C.

12     D hits off B.


 The long arm of A is rotating toward the front.


13     D hits off the long arm of A rotating toward the front.

14     D hits off the short arm of A rotating toward the back.

15     D hits off C.

16     D hits off B.

 



 

Case by Case from Chart 4.1


1. A) When D hits A D's linear and A's linear are stopped. Except for a portion of A's that continues linear, to the extent that the rotation of A is away from its original linear motion, it allows that portion to be freed up at first instance.


B) B & C 's linear is stopped and goes into rotation of A-B-C through A. Except that portion as in A) that so remains linear. Also as B&C increase A's rotation, that portion in A is increased that remains linear. This in turn increases the B & C linear that remains, and all repeats until a balance is struck with D.


C) Then as D torques it should also have or release (from torque) a linear motion, which, during each "test” across the window of contact would cause it to literally follows A's rotation.


Principle 18.2 When a rod torquing loses any contact it reverts to linear motion.


    That is torque is only maintained by the vectors attempting to cross the window of contact perpendicular to the two rods. When there is no contact that force is not constrained and the forces within the rod produce the linear motion it was trying before to achieve.

    From here it seems the majority of the time ABC would slide past D. leaving ABC moving at a much slower linear speed than c. But with a larger rotational velocity than previous.


D) But it is more complicated than that because it will depend to on D's velocity (how quickly it torques) and where on the arm of A it hits. And rather ABC do slide past, or rather D begins to rotate and what might happen there. NOT FIGURED YET


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2. A) All motion of D stops and D torques. All motion of A stops and goes into torque. All motion of BC is stopped and goes into A, speeding A's torque. So BC are left motionless, and AD are torquing to each other.


B) After AD torque they rotate as per chapter two direct hits. In the process of rotation one or the other should re-hit BC and rebound. So AD should end up, one of them. with a linear speed c, the other freestanding in space rotating.


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3. A) D hits C, both lose their linear motion, and C loses its orbital motion. However when C torques it re-hits A and B and accelerating them both but their vectors compound, so B runs into A  accelerating  it, with all establishing a equilibrium of motion, moving away from CD. So A-B go off with a speed > c.

B) As AB move off B hits D. A continues off with its linear motion. B and D torque. D is torquing to B and C at the same time, so it rebounds as per principle 18. This rebounding results in a linear motion traveling perpendicular to the two points of contact DC and DB. When B torque is released it reverts to linear motion (principle 18.2) and follows (its speed being greater in most cases, but perhaps not always) D re-hits it, treat like overtaking hit chapter 2. C is left stationary in space.

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4. A) When D hits B. B and D's linear motion stops, except for a small amount as in 1A).D and B torque. When B torques it re-hits C, accelerating C linear, which vectorizes with C's current linear motion. A goes off at c. C goes off > c. Because of this resultant in C it does not re-hit D.


B) B still has orbital rotation around its virtual midpoint contact to ABC. So it rotates thus as D torques and follows (per principle 18.2), D with a "virtual" rotation, which is really a continuously changing linear motion.


C) Then after torque D rotates, the orbital rotation of B continues and does not interfere with D's rotation. When D reaches its midpoint, it transfers linear motion to B, causing it to re-rotate around D. Also B's orbital rotation is stopped by a stationary D and all goes into increasing its rotational velocity. After which CD move off crossed when at midpoints as per principle 9.


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5. A) All of D and A's motions are stopped and they torque to each other. BC go off at c with their orbital rotation continuing around the virtual contact point ABC.


B) AD proceed as an direct hit chapter 2, with finial speed > c because of A's rotational velocity added to things.


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6. A) D hits A and all A's and D's linear motion is stopped except for a portion that is free do to A's rotation toward back. BC move off at c with their orbital rotation continuing around the virtual contact point ABC.


B) A continues to rotate around its virtual contact point ABC, while also torquing. So it rotates thus as D torques and follows (per principle 18.2), D with a "virtual" rotation, which is really a continuously changing linear motion. Both have the small portion linear motion as mentioned in 6A). As D torques and follows A however, its two linear motions are contrary and form a resultant (see Appendix A Principle 19.2). This further slows A motion and increases its torque, also the freeing up more "following" linear in D which forms a bigger resultant and all repeats itself until all A linear is stopped and all D is moved into only the "following" linear. Thereafter A and D finish there torques and proceed to rotate. It D reaches its midpoint first A's virtual rotation get changed to "real" re-rotation on D as it is accelerated by D. This happens rather D accelerated from the front or back relative to A's rotation. It simply switches its virtual rotation to off of D because it’s a new and real rotation point. So from the backside it actually flips rotation to a reverse direction. These are added to the rotation from the torque motion forming a resultant rotational speed.



Principle 18.4 A rod in virtual rotation, when given a real rotation point will always re-form a rotation off that point, regardless where that point is on the rod.



Thereafter a doublet is formed and it moves off with whatever the total motion is.


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7. A) D hits C. All of D and C's linear is stopped and goes to torque. All of A's linear goes to rotation off C, then to increasing proportionately the Rotation ABC. B goes forward with its linear motion, which moves it outward on A. This causes A to be re-rotate on B, but on C also so it rebounds. B continues with linear motion (c) and virtual orbital motion. C has only a virtual motion and torque, which continues as D is torquing and following C's motion.


B) B re-hits D most likely (although it depends a little on their relative velocities and initial positions rather D's torque may move it out of the way; (NOT FIGURED)). B and D toque to each other. C rotates virtually then comes back and re-hits D. D can't torque to both B and C so rebounds. B follows (NOT FIGURED) as torque is released. C follows also as its torque is released buts with virtual orbiting also.

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8. A) D hits B. All of A's rotation and linear motion are stopped going to torque. C and B's rotation drain into A and are stopped likewise. B's linear is stopped likewise. All A's motion goes into B's torque. C goes off at c and as case #7 hits D. D can’t torque both B and C so rebounds. C follows but B follows much faster and re-hits D, treat as an overtaking hit chapter 2. C follows and depending on D's speed may or may not re-hit it (NOT FIGURED). A is motionless in space.


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9. A) Treat as overtaking hit chapter 2. All rotation ABC is stopped. A and D torque. If in torquing A re-hits BC it is torquing both them and D so rebounds with speed of its linear motion plus speed of all the rotational motion that was in ABC. BC continue on "together" at their linear speed of c.


B) D follow behind A most likely. It depends on its original speed, as to if all its torque is released or not (NOT FIGURED FROM HERE).



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


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


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


3. A) D rotates per principle 13 and 15.6. Because It loses its above mechanism for following the rotation of A. D instead will have a virtual rotation as A pulls away, but this virtual rotation should be enough to still cause D to hit A, but off center, so D 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 D rotates A while following its rotation around and also has a unison linear motion the same as ABC, until D reaches its midpoint.


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


C) But if by chance D is on the overtaking side relative to A's rotation when it reached midpoint, it accelerates A to re-rotate back around D. But it can’t as that motion runs into BC. This cause their rotation to rotate off of C, moving the end of the arm A away from D so that D'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 D is parallel with BC. Then CBAD synchronizes, even though the feedback off B to C to A if necessary, until DCB are aligned. That is DABC can re-rotate off of D without rebound because rotation accelerates B at midpoint which rotates back though C and equalizes the system, until all three particles BCD are in line and touching for the ensemble ABCD which moves off at c and with a rotational velocity also.       



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11. D hits C. All rotation ABC is stopped. Here things get kind of stuffed up, like in my cabin. All rotation goes into C (easiest outlet) and C and D torque, but C re-torques to B and A, then accelerates both A and B by principle 9 because it has a midpoint alignment to both, though with B a whole line of contact. Also, as A and B are perpendicular to C, this causes a dual motion (compounded motion) to occur. However, when C pushes A it causes A to rotate on C back to B and then to C and as C can't rotate so the whole process “starts over”.

     So by principle 5, the only outlet for all rotational motion to go is into accelerating B. Here B has that linear acceleration and its original linear motion, which come together as a compounded resultant. From there B accelerates A, by principle 9, such that the result is A moving off >c, and B sliding down it and away with a diminished compounded motion. C is left moving at c and D overtaking it from the get go as per overtaking hit chapter 2, resulting in a doublet forming.


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12. D hits B. All rotation ABC is stopped. Then thing happen almost as in as in #11 but in reverse. Here all rotational motion ends up in B. B torques to D but re-torques to C and A both, so rebounds (first toward D actually, then back between AC). Here the motion ends up again split in this case between A and C. Both A and C get accelerated immediately until their speeds are such that B traveling in a diagonal without pressing C and A. Note: first draw the vector of B rebound together with its original linear motion, then split it between C and A, but any acceleration of C must come from the compounded motion component that is in that direction. Likewise in #11 above).

     This will leave A going off > c and BC sliding past it in unison motion > c. D, depending on its velocity may or may not keep up. If it does it would do the torque and follow thing after B, then rotates on B. NOT FIGURED FROM HERE.



13. D hits A D establishes unison motion sliding across A. Excess over that goes to torque on A and following A’s rotation. Now a lot may depend on where in A’s rotation the first impact is, and where in that rotation each subsequent event occurs. But I will follow this scenario. When D begins to rotate, its testing and unison motion merge to the plane of that testing motion. From there D may continue to torque, follow and rotate due to the factors mentioned in previous cases. Each rotation point changes the unison motion if it is getting closer to the linear motion of A, but here there is no merging to that point because D is not pointed in that plane. However, as A rotates it will pass a point where it is perpendicular to its linear motion. With D following through this point, the vectorization of D’s unison motion will be in a plane equal to A’s linear motion and be locked in after that in D on that plane but not line of A’s motion.



A DIGRESSION in one rotation the chance of ending on any point for any instance in time is 1/∞

so to speak. The chance of any point on the rotating rod to land on that same point on the other rod is then 1/∞2. But if the rods are also sliding past each other what chance is there then? Figure if the slide is over an area equal to the circumference of the rod during one revolution then that should make the chance 2/∞2. This is leaving out any figuring for time.



Back to reality! There are so many options here, but I do have a goal, that is this should be the case that accretes a larger particle, which can be built on to make a nucleon, if possible. But I need to get rod D in the front and squared up and then re-rotate ABC to D.


       Consider this, if in the unison motion of D, one first has the Unison motion at impact, that is still in the linear direction of D’s motion before impact. Then after torque and follow, that unison linear motion can vecotize with the testing motion that occurs before rotation. But it occurs to me these two vectors may resolve across an arc that intersects the linear motion of ABC. This would depend in large upon the angle in which D impacts A, and the rotation of D on A so the "testing" for each point of rotation is in direction on the other side of the plane of ABC linear motion from the original overtaking motion vector of D. But in the case where it does, the linear motion of ABC should act as a limit to the resolving if the two vectors as per previous ideas and principles. In that case once the resultant is equal to the linear motion ABC that gets locked into D, allowing it to rotate freely on A by the torque and follow from the consistently open window of contact presented by A’s rotation, as previously expounded on in other cases. But only up until the point that D rotates to the front of A’s rotation. Then all rotation ABC is stopped, A tries to re-rotate on D, but immediately re-rotates on C again. But there is a continuity here, in that (explained below) here A can rotate off D pushing both BC forward and then immediately upward around D. See D’s first rotation block of A occurs, of course, just after it has rotated to the “top” of A. If it were to block A on the top A would be able to rotate up past BC and around D. But it is after the top, though no point can be ascribed to it, it still is a continuity such that the point of rotation up and also the point re-hitting BC are such that they are adjacent in style, though not in measure, and the rotation of ABC can turn up and around D. 3-1-18 bit more complicated still So because his is in "instant type pattern" A can push "under" BC causing BC to rotate but in a direction away from D . All torques and rotates with following, both D and A and BC.  D may reach its midpoint sometime and then BC continue to rotates around A as A rotates (scoots) around D as itself is rotated by A! Until at some point BC are in parallel to D then rotation occurs locked into that plane and brings BC into contact with D so that they all comes together. Once that is all so the sides around the midpoint contact A-B are balanced and all motion reverts to linear.

      Why do I think ABC re-rotate off D when it should rotate off of C also, and be dual rotation and rebound? This is because of the feedback from B to C to A. When A rotates off of C it goes through B to A and back to C. This locks them together as it were. Even when A rotates off of D this mechanism holds then together even while being rotated off D, a continuous interplay allowing the rotation of D to override the rotation off of C.



SO THERE ARE OTHER SENERAIOS here that have not been done, and I am a little confused why they seem to be more possibilities based on position here but not in the other cases, but cannot do now.


14. A) D hits on A, all rotation ABC is stopped, D forms unison motion the rest goes to torque on A.

     B) Depending on angles and slants when A torques to D it [1] May re-hit BC  [2] May torque away from BC.

          [1] If it re-hits BC, it can’t torque to D and rotate C at the same time so it rebounds away from BC. But this repeats the same process over again. So the question becomes, as in case #11 what becomes of this pent up motion when the rod in question cannot freely rebound. Does it move thought free space in first available opening to same, or does it move into another rod that can move? It's kind of a case by case basis. Here I see no way it can accelerate B as an outlet, because even the slightest acceleration of B brings it out of contact.

         In case #11 it was different circumstance with C accelerating two particles at once, but one creating a loop-back, but if all acceleration was feed into other (B) it was free to move. C there technically would have accelerated in the direction of C also, but the feedback though A keep it from doing so, allowing an instantaneous removal of all excess motion from C to B.

        Here in case #14 the easiest thing is for A to form a continuous loop of pent up motion, or for A to shoot off at the first free space in between the two forces of rotation and torque with all the rotational motion absorbed within itself. This is what I suppose to happen.

        So attempting to make a principle for this:



Principle 18.6 If a rod cannot freely rebound in space so has pent up motion. This motion is released in any way that is easiest (per principle 5), without violating other principles, such as principle of contact for transfer of motion, considering the nature of the pent up motion and what avenue for escape it has to other rods, or by traveling in any direction in free space, with preference to the former. The last resort is the motion remains pent up in a continual loop until further collisions of the pile of rods frees up things.


        So A shoots of with a rebounded linear motion, and its original linear motion, which form a resultant and which should by the forces of circumstance above merge to one motion, but only so much so that A has a motion enough to follow BC off though space sliding across them, then away. BC continue off at c. D follows A as part its torque is freed, then torques with the rest, but A is sliding past it also, so after A is away, D either hits BC (UNDONE) or it reaches rotation and when A slides off is left rotating virtually in space as well as having a linear speed from its unison overtaking component, gradually traveling away from BC area, unless it hits them in doing so (UNDONE).



          [2] Else B torques away from BC. A and D torque each other, and then rotate each other, D or A re-hits BC (UNDONE FROM HERE)



15. A hits on C, all rotation ABC is stopped. C and D torque, with all rotational velocity ABC going into C first. As in case #11 C in torquing re-hits A and B accelerating them each linearly. All of C’s original linear motion get drawn together toward B as the vectors form a resultant, But also the same for B. so they form a pair moving off together with a new compounded linear motion. Any that went to A causes A to rotate on C through B back to C as case #11, the easiest outlet is all the motion to go into accelerating B. So A is left moving at c, as BC move off > c. D follows BC if it has the velocity to do so. UNDONE HERE.



16. D hits on B. D forms a unison motion and torques to B as it follows it,. All rotation ABC continues. It is somewhat sililar to case #12, but her D should never achieve a perfect unison motion with ABC.

       ABC will slide past D, although it could start to rotate and other things happen, but should not accrete larger particle (NOT AT ALL DONE HERE)



Conclusion

Much to work with here as far as a system goes, only the case #10 & 13 accretes to a larger particle which I will follow up on in Chapter 4.5


>Chapter 4.5 Continued Accretion Part II