CHAPTER 2 INITIAL COLLISIONS OF PRIMARY PARTICLES
[Note: For this chapter it
might be useful to have a couple of full size pencils handy to help
visualize what is being described.]
Section 1: Infinite Quantities and Rod Motion Mechanics
What length is this rod? What units should we use? Let’s consider using a dimension-less quantity, that is a ratio: width to length. Taking the width to be = 1 part, then the length to width ratio will = 1:?. The value for length is of no real concern yet, but in Chapter 5 it is postulated to be 137.
If two cylindrical particles A & B are brought together with their tops being both in the same plane, there is then perfect alignment. For 2 particles traveling randomly in space what is the chance they would collide in perfect alignment? Near zero. Why? By Principle 10 there are an infinite number of misaligned positions; therefore A-B can “never” be perfectly aligned by random collision. However if never then that denies that the contact occurs across given points, but what might be supposed is for any given alignment the chance of such through random collision is 1/infinity.
Another way to state the same is for any difference in alignment, that difference is infinitely sub-dividable, so perfect alignment is infinitely impossible (or not possible, except for 1/infinity).
Consider the motion of a rod (B) as perpendicular to its ends. If it hypothetically hits a rod A (stationary) at the midpoint of each rod (that 1/infinity chance), and perpendicular to A for discussion sake, then;
Principle 12: the speed of
motion times the quantity of rod above and below the point of contact
being equal, there would be no preference for rotation to occur at
either arm so consider rod A accelerated in a linear
direction as per Principal 9.
Principle 13: for any speed
of motion times the quantity of rod above or below the point of
contact, this being unequal, the rod will rotate in the direction of
the longer arm.
motion of the rod so involved goes to a rotational quality. This
rotational motion is as “natural” as linear and itself can be
transferred rod to rod (see Case #3, Chapter 3). Next:
Principle 14: Rotational motion continues until the rod rotating reaches its mid-point and is at balance, and then linear motion is re-formed.
When the rotating rod, progressing around rod A reaches its midpoint, its arms being balanced, it returns to a linear motion again, as per Principle 12 and 14, and so accelerates A (now in a linear direction that is perpendicular from the ends of B at the point where it has reached its midpoint).
the rods may be aligned (perpendicular to each other) at their
midpoints by the process of rotation as this passes one point across
another, which is a different sense from that of random possibly in
Transfer of Motion Rod to Rod
Any tangent drawn to a perfect circle contacts the circle but no area can be described for the point of contact. As, for any chord drawn parallel to the tangent, consider the arc between the chord and tangent. For any chord drawn further toward the tangent (still parallel) this arc only gets smaller but an arc remains ad infinitum. Such a point on a circle is of a mathematical point quality (indefinable by area). Therefore the point of contact of two curvilinear figures is a point of “zero” area, to which motion must pass through this “window” to be transferred rod to rod (See Appendix B for more information on this).
If the rods are traveling though space and collide, by principle 10 the rod (B) would be slanted relative to the hit rod (A).
A force of motion to torque the rod B perpendicular with rod A would occur, until the perpendicular position is reached.
This torque occurs as a motion in space that takes time, and before any other acceleration can happen, but occurs at the same time a unison motion component occurs between A and B, only the excess over this goes into torque.
At one time I considered that this motion across the mathematical point(s) needed to be in a straight line. Then I considered any angle is possible across the window of contact, short of parallel to the rods. Now I'm not sure.
But, taking a step forward, consider this; the force/motion toward/across the window of contact means that only if it is constrained along the short axis will it torque. The point of contact on a rod is determined by their axis’s. With a cube impacting a cube (face to face), all that contact permits free flow of motion per its quality. However, I still think on impact it would cause rotation of both bodies around their mutual center of mass, as expressed in chapter 1. If two spheres impacted I think only a straight line transfer of motion would be possible, always at center of mass so always impelling the forward mass straight ahead.
A rod is somewhat in between that. Along the long axis there is free motion along many vectors (1 dimension) to that axis. The short axis has only one vector possible (one point). Neither are in 2 dimensions like the cube face. So I think this is what happens; the motion from the impacting rod has its maximum force per “area” along the one point (the apex of a one-dimensional line cutting a circular cross section around the short axis). So the lines of force congeal into that vector at the crossing point. In the impacted rod the opposite occurs, rather than maximum force we need to consider the least resistance which is the 1-dimension motion vectors radiating along the long axis. So, motion goes with greatest force at the least resistance when the rods are perpendicular to each other, and this “requirement” to the window of contact is what produces torque in the impacting rod (though infinite friction).
These constraints relate to the window of contact only, the motion within the rods is another matter covered below.
When on impact the rod is held I am saying that the motion goes toward the contact point, but really can say instead that motion moves forward in parallel with that plane, and any imbalance causes rotation due to predominate force on arms. Then there is a linear test reason in all cases, as needed under "unison motion" below. And it is then okay to say that infinite pressure causes the motion to go across without motion toward the contact point, because rotation breaks from it. Then, when and if, motion goes across contact point to the other rod, THEN it spreads out and non-parallel symmetry is okay. This might be because “before” going across the whole rod is in motion, so pressure exists, in going into new rod that motion is “free” from pressure in first instance and so can just spread out (radiate) into new rod. Hence the force then on the longer arm is predominate and proceed as in 15.1.5 below. Further torque is still okay because it is the “shape” of the window that cause the trans-lateral pressure and torque, regardless of the way the motion goes to or across the window. But when this motion comes across it radiates/spreads out to the full area. For if it filled across in a straight line and dragged the particle into motion by infinite friction between mass points here also, then the notion of rotation of the forward rod has no basis. Instead I see this scenario
1. Motion within a rod without contact is as is. This is primary motion type.
2. Motion on impact within that rod is forward in parallel vectors aligned parallel to the line running from the contact point and perpendicular with respect to the line on the long axis emanating from the contact point. This then if reciprocal around the contact point results in a linear push. This allows for a linear testing at each point of rotation. If linear motion crosses the point of contact when in balance it is due to infinite friction between the parallel vectors. Or if unbalanced it results in a rotational push/motion. This is the secondary motion type.
3. Motion goes with greatest force at the least resistance when the rods are perpendicular to each other, and this “requirement” to the window of contact produces torque in the impacting rod (though infinite friction). This is the tertiary type of motion.
4. After crossing the window of contact the motion fans out/radiates. This is the fourth type/level of motion, with the resulting pressures taking that rod “back” to a level one motion.
5. The vectors of motion then if symmetrical to a circular cross section that is perpendicular to the ends of the rod, and drawn from the radiation point, will result in linear motion. Else if there is an imbalance the rod rotates forward in relation to the average motion vector, "pivoting" virtually from the radiation point (though it is on the backside to the direction of rotation).
6. This causes a back swing which hits the first particle and causes motion to reverse to that point, leading to a change of rotation onto the first rod.
So, these four “levels” of motion are related to certain circumstances; free movement, impacts, the window of contact, filling new mass area, completion of the process.
7 . Any type of impact that presents a situation where motion cannot be expressed in space as it had been, results in transfers of motion, torque, rotation or rebound, or vibration as a last resort, to satisfy the conservation of motion principle #4. This occurs along the lines of the other principles so mentioned here and elsewhere (in the case of rebound and vibration).
Now there are fundamentally two types of collisions, direct hits and overtaking hits, gone over in detail in Section 2. Suffice to say as an introduction, on direct hits all (absolute) motion is stopped, goes into torque per principle 15.1.3, then rotation per principle 13, then back to linear motion per principle 14. For an overtaking hit there is a unison motion that is achieved by the overtaking particle (B) with the overtaken particle (A), thereafter any excess above the unison motion goes into torque and then rotation, The unison motion is initially following the the same vector of motion B had traveling to meet A, therefore in theory B would slide across A relative to it.
The excess motion, after torque, means any excess motion as rotation is in a perpendicular to
In the case of B sliding by A due to the vectors at angles to each other, this unison motion establishes and occurs while B is torqueing.
Then after rotation the added problem is with the direction of the linear motion of A and B different, the rotation causes an additional displacement of B relative to A that is a change from the established unison motion at the first instance. That is the rod B itself has changed position, so that its motion though space, unless the same as A, intersects A’s linear motion at different spots, even though B’s linear motion is in the same direction to the horizon as when established in the first instance. This can be seen most clearly when B comes to the top of A, it is now clearly moving up and off of A. Conversely if rotating to the bottom of A, all its linear motion is suppressed.
HOWEVER, this doesn't occur because in the first instance (after torque is complete) as B begins to rotate it "tests" the window of contact as it were with linear motion which combines with the linear unison motion and because it diminishes toward the linear motion of A that would , as it where, increase rotation and the testing cycle occurs again. So all B's linear overtaking motion moves toward a equal plane (but not line) of direction as A's linear motion, increasing the amount of rotation possible as it does. This change of direction of B's linear motion is instantaneous as all occurs within the rods mass.
After which B actually begins rotating on A, with a unison motion component now in the same plane as A's linear motion. Now each point of rotation causes a testing which also vectorizes with the unison motion of B.
If the long arm of rotation is toward the back of A (behind the direction of A's motion) it causes an increase in rotation and more vectorization, to a limit of what is necessary for a new unison motion keeping up with A. All these times then B is sliding relative to A as well as rotating. Only when B rides across the point in the back of A that is behind its vector of motion, does the unison motion of B comes in equal to that of A, having been drawn down from the original , and each step of thereafter, of the unison motion B had.
After which A and B move off with unison speed and direction, and with B rotating on A. Now each point of rotation causes a testing but the rotational motions "linear test" does not vectorize with the unison motion thence (see more on testing vectors after principles). So, several principles are needed here:
15.2 A linear "testing" occurs at each point of rotation
of a rod rotating on another rod. For overtaking hits this
instantaneously forms a resultant with its unison overtaking motion.
15.3 If this secondary linear motion can be expressed, in the rod or by
accelerating another rod, then the resultant with the unison linear
motions are compounded (two simultaneous linear motions occurring within rod).
15.4 If this secondary linear motion is impended, as it were, before (and
causing) rotation, the two linear motions merge to the resultant,
forming one linear motion.
Principle 15.5 If, in the resultant impending motion of the overtaking rod, motion can be lost from the unison motion component to an increase in rotation, to make the resultant not impending, this occurs.
Principle 15.6 If the resultant increases the overtaking motion, this can come from the testing vector of the rotational motion until impediment is reached.
Principle 15.7 Uniform Unison Motion-If in the
resultant impending motion of the overtaking rod, and motion must be
taken from the rotational component, this does not occur, unless the
rod is at midpoint balance, but unison linear motion toward a horizon
remains as it is, likewise the rotational motion remains at the same
As in the case described before Principle 15.2
More on Testing Vectors
During 15.2, 15.4 and 15.5 and as described in paragraphs before that, the value of the overtaking vector is diminishing. In 15.6 when in the same plane it is increasing, so frees up rotational (as it is pressing on the other particle) testing to extend to linear motion. But with unison it is neither, presses but no extension; zero value. That is why it remains as is although testing occurs. When midpoint is reached then it is of course active as it crosses the window.
A. You have to have uniform unison motion with rotation or the 3P rotation case would be sliding all over the place, etc.
B. You have to have overtaking vector swing to A or cannot get accretion.
C. So there is the question of rotating to a midpoint and issue #2 below. How to say that it is different than uniform unison rotation.
There are 5 stages
1. Pulling the overtaking vectors into plane of forward rod. This is longer to shorter vectors, diminishing.
2. Accounting for the overtaking vectors in this plane as rod rotates. Here the vector increases, if rod is rotating toward the
back. Diminishes to zero (increasing rotation) if rod B is rotating toward the front of A.
3. Locking into uniform unison motion and maintaining it when back rod crosses behind A’s vector of motion.
4. Transfer of motion when reaching midpoint, and firstly vector addition at instant that happens which induces scooting.
5. Final transfer of motion when alignment achieved.
Vector addition scenario for each case.
In #1, the swing of the overtaking motion onto a plane with A.
1. As in the description in the Unison Motion section and associated principles, the collision causes an overtaking motion to establish, the rest of what was a linear motion flips to the window of contact. So, before rotation begins in can combine, with itself sort of, after it was separated into two parts. However, torque does occur before these vectors combine and rotate, but the scenario holds, though delayed for the torqueing.
2. But for rotation itself there is only a faux testing vector, held, impended as it were, only if a gap is formed is it a real vector as explained below in (a.) and others. So (below) simultaneously it is pressing from rotation but also forming a gap and closing that gap by extension of a test vector. A strange but reasonable application of the concept of instantaneousness.
So on #2 where B is rotating toward the back, there are 7 steps
a. B's motion is stopped by A, excess goes to rotation.
b. The unison motion vector is too little as B rotates.
c. It would not release from rotation alone.
d. But the rotation presses B on A anyhow so,
e. A test vector releases any motion needed to hit A.
f. This is combined with the over-taking vector.
g. This should be by principle 15.3 I believe.
h. Because of a. and b. the test vector had extension and f. was possible. That is the test vector only becomes a real vector if a gap is to be filled.
i. This all happens as a continuum, there is no first moment, in essence it begins to happen while still in the instant before
#3 on locked unison motion
Here as each point of rotation is reached, the testing is held but not impended or freed up, so the “test vector” is in essence zero. And everything as far as overtaking motion stays the same.
Even more on same.
Sequence 1: Sequence at impact
1. Stage 1: Impact, lines of force parallel and forward. Contact will hold back rods motion.
2. When parallel vectors can't cross into window per principle 15.1.3. Then lines of force become simultaneously parallel and to point of impact causing torque to occurs by principle 15.1.3 again, until rods are perpendicular to each other (I think this must occur before vector can formed across window as without window being breeched there is no constrained vector). This would also mean: Principle 15.8: Rotation can only occur on the perpendicular, as lines of force cannot initiate rotation until stage two motion is reached.
3. Window constrains vector (after torque) to a perpendicular vector, again now only parallel and forward.
4. Stage 2: (After torque) “testing” vector extents to common contact point. That vector is real and needs to be combined, unless there is unison motion which causes vector not to press/impact causing it to remain as stage 1.
5. Stage 3: Vector crosses common point into the other rod only if the back rod is in balance (so there is linear motion rather than rotation).
Sequence 2: Sequence once testing vector is stage 2.
As in principles 15.2 – 15.7
Proof motion swings in line with window
At first it does as resultant of overtaking and testing vectors always leaves excess, so always impeded, and rotation cannot occur until that impeded resultant vector is crossing the window at the perpendicular. With unison motion no stage 2 impediment so it can test and rotate without combining with the unison overtaking vector.
But when the resultant is brought perpendicular it is not aligned with the motion of A but only in the same perpendicular plane. Then each new rotation point and testing draws vector to that new perpendicular by the same reasoning as above. When at back point it becomes and locks into uniform unison motion with B, so stage 1 only and vectors changes stop.
Section 2: Collisions
of Two Particles
Now, in more detail,
to consider how two particles might normally collide, that is off
center to each other. There are two basic ways these two particles
might collide, via a direct hit, or by one overtaking another.
1. B could overtake A from
2. Also with many “slants”
relative to A.
3. To the extent B has the velocity to overtake A, there is a unison motion component of B, following its original motion vector, which would slide B across A.
5. After torque is complete,
6. The excess
velocity over the unison motion goes into rotation of B on A, on the
perpendicular cross section of A.
7. If the long arm of rotation is toward the back of A, as B rotates on A, while they both maintain a unison linear motion
per principle 15.6, if B passes behind A's vector of motion, uniform unison motion is established toward the same horizon as per principle 15.7.
8. When B reaches its midpoint, linear motion is renewed per principle 14, and B would accelerates A, such that A rotates back around B (principle 5.1.6), while they are still in unison motion as in (7). However this does not happen as B's overtaking/unison vector must combine with this new vector of acceleration.
9. Here then if on the direct hit side those two vectors merge to a resultant which causes a direct hit with A and all motion both is stopped, treat as a direct hit.
10. Else if on the overtaking side the vectors merge and cause scooting of B behind A as per principle 16 next chapter. From there B accelerates A to re-rotate on it.
When A reaches its midpoint A and B are at midpoint to midpoint
a) If this is on the "overtaking side" of B then that rotational motion in A reverts to linear. This combines with the overtaking vector as per principle 16 (next chapter) and B scoots around A until behind A's motion and the vectors are in the same direction. Then A accelerates B as per principle 9. And they go off in unison with a equal linear motions. This means doublets are formed here and the motion of the same will be in the same direction as the primary particle flow. Their finial speeds depend of course on the initial speed of A and B. For B to have overtaken A of course its speed was > A. There is also the consideration of a particle passing in front of another before being overtaken (side swiped) , in which cases B may have a speed equal to or less than A. Such interaction I leave to others to figure for now.
b) If on the direct hit side there is a problem with the transfer of linear motion from A to B. It is contrary to the unison linear motion of both B and A such that the rods would be having two opposite (not overlapping here) vectors of motion. In this case of contrary linear motions in A and B t causes the contrary motion to be draw together with the unison motion as one vector. However not as standard vector analysis either, but the drawn vector is in the spot as triangulated, but with the length of each vector added (see Appendix A, Section 4 Contrary Motion Vectors, especially principle 19.2). So the Rods A and B go off together with the same speed and direction, but the vector of motion relative to the ends of the rods is not perpendicular.
Proceeding from 5. If the long arm of rotation is toward the front of A, then B's unison motion, indeed all B's linear motion, diminishes to zero, When and if B rotates in front of A treat as a direct hit.
If it reaches its midpoint before then it should, after accelerating A and vector additions, reverse its rotation back the other way, it should as vectors of testing is added each rotation point, reach a spot where the unison motion and added vectors converge to a perpendicular. After that scooting occurs as above and treat as 10 above.
With a direct hit A-B, by Principle 10, both A and B will be off midpoints. All linear motion of each goes to torque and then rotation of each other.
One PP (A) will reach its midpoint before the other (according to their relative velocity’s and initial positions) where it will then transfer all its velocity into the (added) rotation of the other PP (A becomes stationary).
When that PP reaches its midpoint (both PP’s then are at a midpoint – midpoint contact) it will move in a linear direction moving each equal to ½ the velocity of its last rotational speed (by Principle 9). This then will result in two particles moving off in unison but randomly in any direction.
* (except if one got in to more extended case studies of reverse overtaking from PP flow)