Chapter 4.5 CONTINUED ACCREATION PART II
Dec. 2019, changes still needed here to update to latest principles.
There are many collisions to deal with. Each new set with 1P, or with doublets, etc.. Both hitting them, and being hit by them. Most of my cases till now are on the later, but the former is to be considered also. They are not just reciprocal.
But rather than beat around the bush, I will directly consider how a N accretes from smaller sets.
With regards to the accretion of a Nucleon. Originally, I had a set of two rows of 137 PP turned 90◦ from each other topped with a single PP on each side also turned 90◦ to that face.
But as of 2017 and going over chapter 2-5 in finer detail, I have a problem in getting such a N to form. It appears, though I am still working on the details, that I might end up with, at best, one complete row of P's on one side, but only a half row on the other. This is because in order for the particles to line up at center of mass (com) they must rotate to that point. And scenarios to do this do not allow a second full row. To get that on the second 1/2 row on one side P would have to rotate off the end P of the other row and land on top of an established row, which would prevent center of mass alignment.
The extras must be, on each face, running in opposite directions and would need to be at center of mass in order for the transfer of motion as a unit with the faces. This can only be achieved having only two half faces back to back, then an extra would be able to rotate, or be rotated around, in two axes’. This leaves TWO HALF ROWS WITH AN EXTRA ON EACH FACE OF THOSE ROWS. Total of 140 PP in a Nucleon (2-20-2020 see below in red!). This then should transfer linear and rotational motion as a unit, including re-rotation which reverses direction of unit (N), though it would be complicated and after principles 19.4 and 19.5 eventually.
I now conclude that for a P falling onto the face of a N it would produce rotation (spin) of the N.
Only the hits achieving balance on the extras would directly produce linear motion of a N.
But, I have thought out but need to re-figure, on most or some additional P impacts to N, when spin is stopped a sequence of motion within the P of the N, it ends with an extra being rotated back into N producing linear motion again. So, this needs to definitely get figured out.
Okay here is an example of this. A P overtakes a N hitting on the face. Unison motion is established, the rest goes into torque, (other considerations may need to be considered here too) then falls across face of N this causes rotation of P which gives rotation to face at points of contact. This causes those rods to rotate on the P which cause their backswing to hit the other face. This causes re-rotation of the first face into the other face but only those rods that had been in contact with the P, so the far face rotates on those rods causing all of them to rotate into the first face which also then rotates the extra on that side and the P, as well as the back extra being swept up in this motion. So, the whole N and the P are in a rotation in addition to their linear motion.
When another P (Px) hits, say a direct hit on the extra, all motion goes through the extra into torque, then re-rotation into the face and the rest of the N as linear motion (reversing direction).
BUT with the P across face still it would cause all to collapse to rotational motion again, now with no linear motion. Here I see no way to get the P off or produce linear, except perhaps after a second layer of plastered P’s is formed it might rotate into the extra which would re-rotate and accelerate the N linear, leaving the plastered P’s in its wake. How then to get an accelerated P as a fast electron is another story. Very complicated here.
PP off N Direct Hits also overtaking hits
1. across face (after torque and rotation)
a) Adjacent face
b) Far face
2. Off extra
A) Rotates to midpoint and accelerates
B) Rotates to edge
Then question of two of these things happening at once
The Second Layer overtaking and direct
1. Across extra and plastered P (likely) near and far faces (IF ROTATES AROUND TO FRONT)
Accretion of a Nucleon
Lets look at it in total. To get two ½ faces one needs
1. Squared up particles ending parallel with existing particles.
2. All those particles to squeeze together.
This would leave them all at midpoints to the particles on the other face, and vice versa.
The only way I know of to get (1) is to have an overtaking hit – unison motion – uniform unison motion – scooting. #2 usually occurs if, in fact I think only if, a particle is at com to the pusher particle and is being orbited. This allows it to come up against the fulcrum particle. Only this way will everything be in perfect balance around a midpoint to midpoint connection. Pretty nifty when you consider it has to be perfect this way, and chance by random collision of it happening is infinitely small.
This has to accrete though a 3P set, from a doublet there is no other form to follow. From the 3P however there are several paths. The 4Pb and 4Pub. Also perhaps a double doublet. Lets think about that.
Double Doublet from 3P or 4Pb and 4Pub
Can a 3P go to a double doublet, that is add a P parallel to the pusher P. The answer appears to be no, or I would have done it in the cases last chapter. Unless I made a mistake. From the same reasoning there a 4Pub should not from a particle set on the pusher side, as when it is rotating it causes interactions that preclude the addition of a P with the pusher.
And the 4Pb can not do it because P's rotating on it cannot rotate to COM.
Feb 18, 2020 Okay then, but that means something. Unbalanced sets cannot accrete onto the other face! So two half faces is not possible. This is big trouble. Perhaps I should consider a Nucleon that is not perfectly aligned all round, perhaps there is still a way to produce linear motion from it. That would also give a mechanism for N decay which I do not have right now. But right now this leaves the biggest sets as 4Pb and up to a full face, but no corresponding face. Or a unbalanced 1/2 face. Any other accretions will have to be NOT com accretions.
Specific Cases Studies
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).
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 they go off through space together but with their motion not being as a perpendicular to their position/shape.
Else if the rotation completes on the overtaking side the motions also equalize, but with a compounded vector.
So the energy of particle ensemble is the same in each scenario, but apparent velocity is different.
Hereafter if a 5th particle, E, collides with the 4P Balanced we have these possibilities. IT MAYBE THIS CLASS OF INTERACTIONS IS NOT VERY IMPORTANT AS THEY ARE TRAVELING > c SO WILL OVERTAKE PP FLOW QUICKLY AND BREAKUP ANYHOW.
There is no Rotation, only ABCD moving linearly through space. I have not considered yet hits off A behind CBD area.
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, depending on where D hits A;
a) CBD go off with their velocity. A and E are stopped and torque to each other, then rotate to a doublet.
b) Else A torques back into CBD, it cant torque to both and normally would rebound. However lets look at this more closely.
When A attempts to torque perhaps its outlet for this dilemma might here be accelerating CBD, either accelerating them linear or orbiting them, in which case A would rotate a elliptical path around E. I don't think this could happen though as the angle A rotates on E would not be the same as the angle needed to orbit CBD, which can only be pushed directly across the window of contact. So I would give priority of all A's motion being "drained" out into CBD, accelerating it linearly.
Principle 18.5 A rod torqueing in contrary directions, or torqueing and rotating in contrary directions rebounds. Unless said rod has midpoint contacts in which case it is simpler to for its motion to flow into those rods with such contact in a linear fashion.
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 all of A's motion goes into CBD accelerating it linearly, added to its already present linear motion. E torques and rotates on A forming a doublet eventually.
2. E hits C or D, will say C. All A and C's linear is stopped. BD pulls away. C torques.
A) if torque in one direction re-hits B and A, it accelerates
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. This ends, as an outlet, accelerating BD. The
motion of C is portioned out, I suppose, percentagewise based on the mass it
hits. So 66% to BD, 33% to A. The original linear of A when stopped goes into
torque, then re-hitting BCD accelerating BCD also (as C holds A, A loses all
its speed). The portion to back into C back to BD and A as before. The end
result of all this should be BD flying off with a combined motion, A and C
motionless, E torqueing and rotating on C.
A) if torque in one direction re-hits B and A, it accelerates 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. This ends, as an outlet, accelerating BD. The motion of C is portioned out, I suppose, percentagewise based on the mass it hits. So 66% to BD, 33% to A. The original linear of A when stopped goes into torque, then re-hitting BCD accelerating BCD also (as C holds A, A loses all its speed). The portion to back into C back to BD and A as before. The end result of all this should be BD flying off with a combined motion, A and C motionless, E torqueing and rotating on C.
3. E hits on A. Treat as overtaking hit chapter 2. Rotating toward front of A treat as B) below. Else is toward back, after unison motion established and rotation continues;
A) If E reaches midpoint on overtaking side. E scoots around A until aligned by previous scooting mechanism (principle 16). 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, and continuing around CBD and E until it lands such that its midpoint is between other particles, then it continues to rotate and also orbit other particles, particularly vis principle 17.5. Anyhow there a several nuisances to all this but eventually should end with a particle set BCDE pushed in orbit by A while traveling >c still.
B) If E reaches midpoint on the direct hit side A, A motion is stopped and DBC go off. If need be A will accelerate DBC until their sped is => than A's rotational speed hitting it again. A calculation of this is done in Appendix N.
4. E hits end rod, say C. Treat as overtaking hit chapter 2.After torque if;
A) E rotates and hits CBD before it reaches its midpoint, it hits at the point of achieving unison uniform motion. if
a) Its midpoint lands between particles it will follow principle 17.5, but either way it occurs, one or two particles will receive the rotational speed but immediately transfer it to re-rotation in the other axis whereby they then hit A causing it to re-rotate on its axial plane, accelerating BCD linear. This will all equalize so E ends up traveling with the group, at higher speeds than ABCD originally.
b) else if midpoint does not fall inbetween the particles E continues to rotate around until it does the it acts as in a) or if on direct hit side the like still happens as unison moton is present, but here BCD rotate away from A and then follow a pattern as in 3A.
B) If E rotates to the front side of CBD without UUM, it stops that particle it is rotating on and A also. The other two particles go off and will be accelerated by A also as per Appendix N. E rotates and toques and flows on "C" as C torques and follows and rotates on it, and A does the same on C. At some point, unless E reaches its midpoint, C and A will have opposing motion as and C, torqueing to two rods will rebound, leaving all separated. Else if E reaches its midpoint before that a doublet or even a 3P may be formed.
C) But if in rotating toward the front or back it reaches its midpoint so as to accelerate C. Then C scoots to the back and UUM forms, there rotating A. In re-rotating on E it should be the thing were the backside re-hits C and rotates on it and re-hits the face of CBD therefore re-hitting E etc. Until A slips around E and hit CBD on the front face thereafter rotating until ECBD are together but A has no midpoint to midpoint connection so the set spins. So it has both linear and rotational motion. Just what happens to these type of particles in future collisions is a question, I presume they lose their linear motion at some point and break up, but perhaps they stick around for a longer period and effect the PP Flow dynamics.
. Depending on the angle, both dual vectors may be stopped, so
same as #1. If sharper angle only one is stopped, but the other cant just move the rod and leave motion standing on stopped vector. That stopped vector would torque, but instead as "pulled" by other vector, as of it were an opposite motion, it converges together to one vector. So can treat as case #1,
6. Same as 5.
7. E overtakes A. Same as case #3 I believe, 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.
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.
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.