[Note: For this chapter also it might be useful to have several full sized pencils handy to help visualize what is being described.]
In Chapter 2 it was calculated that when two particles collide they would, after a process of torque and rotation, move off in unison motion crossed at midpoints.
What happens when these crossed particles (2P) collide with a single particle (1P)?
Rod B is Perpendicular to the page (C may be perpendicular or parallel or in between)
There are four basic collisions to consider:
Direct hits (C at A-B's direction of motion) Overtaking hits (C overtakes A-B)
1. C hits on B 3. C hits on B
2. C hits on A 4. C hits on A
As in chapter 2 these collisions are all off center – off center and we can follow a sequence of events:
1. The particles hit (off center to off center) causing impediment or unison motions to happen.
2. Particles Torque.
3. Rotation occurs.
This would be just as Chapter 2, except we have 3 particles, the first in contact with the second, which is hit (or hits) a third.
It is important to work from all the previous principles.
1. Direct hit C on B
2. All the linear motion of B and C goes into torque and rotation of each. While torquing they are both stationary in space, but A has a linear motion to be conserved. This all instantly (see Appendix B) goes into B speeding its torque.
3. Torque pulls B into A.
a) B re-torques on A, this all being in the first instant, and being still at mid-mass points accelerates A by principle 9. In essence B-A reverse directions instantly in space.
b) C is now following AB and either does not keep up, or overtakes AB, treat as overtaking hit case #3.
1. Direct hit, C hits on A
2. C & A’s linear motion stops, going into torque of each. While C & A torque B moves off with its linear motion.
3. C & A torque with their respective velocities.
4a. When C reaches its perpendicular each rotate on each other, they will torque also to keep up with change in relative positions. When one particle reaches its midpoint first all motion goes into the other participle causes it to rotate faster, and no torque occurring thereafter. From hence that particle also reaches its midpoint, and accelerates the other as per principle 9, forming a new doublet.
4b. Sometimes, unless midpoint contact is reached first, A or C will re-hit B (even allowing that A-C have no linear motion, A’s rotation “catches up” to B), but it will be off center – off center and unaligned. Since A would have to torque on B, but also then C, which it can’t simultaneously, it rebounds (principle 18). This causes all 3 particles to go in separate ways.
1. Overtaking hit, C hits on B, treat as an overtaking hit, Chapter 2.
2. Excess motion goes into C testing on B, bringing that together with the unison linear motions as occurred in last chapter.
3. Then excess goes into torque of C on B. and then C rotates on B.
4. When C reaches its midpoint
a) As C rotates toward the front the overtaking motion is impended and diminished, going into faster rotation.
b) So if C rotates to the front, treat then as a direct hit case #1.
c) If C reaches its midpoint while rotating toward the front, first the test vector and the unison vector combine, but they are the same direction so all goes to re-rotation of B on C. This impends C's overtaking motion turning that all back into B's re-rotation.
d) here as B rotates on C it slides past it to. If it slides off it continues with a virtual rotation which should hit A again sometime and they should end up as a doublet again.
e) Or if it hits A while still rotating on C it torques to both and rebound and all three go separate ways.
f) else if it reaches its midpoint first it causes C to re-rotates on it but B should slip away and all three separate.
Principle 16.5- /Alignment
If on the overtaking side when C reaches its midpoint, C would cause B to re-rotate on C. But first a "virtual" resultant between the linear motion released from the rotational and the already present unisonotion occurs (principle 15.8). This resultant causes a new unison motion, allowing a point outward on C to contact B, but only though a small distance. However each, if possible, rotation point would be such that the next new resultant and new unison motion allows more distance to be spanned. So even though C would press on B before all motions is used up, the mechanics are there so that C should literally "scoot" around B at high speed and uses all motion up, until parallel with A. The time it takes to do so means BA travel a certain distance and C uses real unison motion, as above mechanics, for that portion of motion as necessary.
6. When aligned, C's unison motion is again the same as B and all its excess goes into B causes it to re-rotate around C, and "orbiting" A in said fashion as follows.
7. Unison Rotation-Orbiting:
As said above, from here C’s excess now can cause B to rotate as it is “squared up” to A, so A is “orbited” (see next paragraph). A is accelerated to an equal velocity (area swept) of the “force” of rotation of B, at B’s midpoint (see Chapter 4 and Appendix C).
Here it might be supposed that A, accelerated by B at midpoint is like Principle 9, that is accelerated linear. But that was if B was traveling linear, here it has a rotational motion. This can, I postulate, be considered is a “quality” of motion that can pass through the window of contact of two rods, as linear motion does, but gives a rotational motion to the particle on the other side. So for orbiting that might leave us with the notion that rotational motion in the rod is like linear motion with a continually changing angle of direction., in essence the same as the discussion on infinite numbers in Appendix I. Here every rotation instant has a linear motion, the next instant a certain different one, but there are no two successive angles so as to say it changes from this “angle to that” only the rate of change over any interval of rotation, determined by the rotation precession of one rod on the other. So that at a 90 degree turn there is a precise angle change determined not by speeds, but by fixed geometry. This angle change is “known” in any instant, such that it is passed across the window of contact, causing as it where a rotation/orbiting of the next rod as opposed to linear acceleration. That is unless acted upon by another force/impact it continues as is “imprinted” in the motion of the rod (however as it precesses, that is a continually changing imprinting also!).
BUT I don’t think so, else if force lets go rods would be going at last linear, rather than an imprinted rotational etc. which is needed in light particle perhaps and just makes more sense to me that this rotational motion is a quality in itself, no really able to be conceptualized, but real, both rotational off a point and free flowing rotation of an orbital motion from push. So the tendency I would say if an orbited P hits another is for the vector to the window of contact to become straight lines (shortest distance between two points). It is only when there is a force of rotation off another point directly or to a secondary P also, that rotational or curvilinear quality’s to motion are maintained over linear motion vectors. Also see step 7 below. As principles
17.1 A rod rotating off of another develops a curvilinear motion quality in itself, not a continuously changing linear motion.
17.2 This can be transferred to other rods as long as the initial rotation point on a real rod is present.
17.3 When rods are free from pressing on points on other rods, they may continue to have curvilinear motion from within themselves, in total off of “virtual" rotation points, but further impacts will wipe away the curvilinear motion to the extent any impact causes movement across the window of contact, same occurs there as for linear motion vectors.
See also principle 18.4 in chapter 4.
So A is given an “orbit” with a radius equal to the distance from point of contact on B-A to contact point B-C and the velocity of the orbit, as calculated in Chapter 4 and Appendix C.
8. Declining Orbital Values
Is not the rod orbited put in motion so that no longer pressed on. Likewise with a impact B on A by principle 9 is B no longer pressing on A when there speeds equalize? At impact motion is held, looking for a release (principle 15.1.7). It is the motions within the rod in response to this that determine the next motion of the rod. Once they are both moving in unison there is none of this. For example if a doublet is moving through space and for argument the forward rod (B) is moved downward in the same plane of contact, A does not begin to rotate, it has no other speed than the speed of unison motion, so there is no further occurrences. Rotation and orbiting would continue as started until AC are in contact, in principle.
But as rod rotates on rod, it both precesses, and rotates faster as distance from rotation point to endpoint shortens. This then does cause a continual testing but only an amount above the unison motion present. This then taken from B and imparted to A at a new orbital path and speed. This would combine with the unison orbital motion and speed to re-form a resultant, just as linear motions can be combined. So I am in a dilemma, it would seem to "orbitdown" one would have to combine the original orbit with a change > the actual path B is rotating to get A to be in that same path, yet have decreased speed in the new orbit.. My solution is this:
As to the orbited rod (A), B I think undergoes a free fall until the next point of rotation on C. When it does impact the next point then it undergoes a change in orbital direction (lesser orbit) and speed. A is in the way of this, but more importantly A's orbit pulls it away from allowing B to transfer its new position and speed across the point of contact. but it is pressing on A and causes as it where infinite friction on the point of contact holding A from the back and causing all A's orbital motion to flip and go into B, which re-rotates on C REFORMING A NEW ORBITAL MOTION INTO A. But this must not violate any causes where rebounds occurs by resorting to this principle also, they should be different scenarios.
9. Final Rotation of 3P
When B has rotated to the point that A comes in contact with C then all 3 rotate in unison (no progression on rod) as the rotation of B goes to A to C back to B. So B-A-C rotate simultaneously around the point of contact B-C. This is in addition to the unison linear motion of B-A-C .
In Chapter 4 it will be shown how when a particle D is added on the other side of A, A-C-D are in balance with B and this rotation can “revert” back to a linear motion.
1. Overtaking hit C on A. Treat as Chapter 2, an overtaking hit.
2. Excess motion goes into torque C on A.
3. When C is perpendicular, C tests on A, bringing together that and the unison linear motions as occurs in the last chapter, and then C rotates on A.
4a) When C reaches its midpoint, if on direct hit side, C's excess will go to A causing it to re-rotate (again they are already perpendicular, so no torque). But it will rotate into B, accelerating B in a cruvilinear fashion as in case #3, until B is brought up against C as in case #3 sequence 8.
A Digression for Some More Mechanics.
Rods C & B (imagined) are perpendicular to page and to Rod A (shown) for all Figures 3.2-3.4. C and B are at there own midpoints in this figures.
In figure 3.2 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 3.3) then A doesn’t rotate on C as each rotation on each side of C is balanced and A would accelerates C linearly as per Principle 9, 12 & 15. But any linear of A also presses on B and so all force instead becomes a force of rotation around B, accelerating C in a curvilinear motion as per Principle 15.1.6, 17.1 and 17.2.
Principle 17.4 A rod must be in balance (reciprocal symmetry) to the totality of its contact points with other rods to have and/or to pass on linear motion.
A ↓ midpoint
If C is on the opposite side of the midpoint of (A) from B (Fig. 3.4), and assuming B and C are at their midpoints, then the rotation of A around B & C, are in contrary directions, and A cannot rotate both simultaneously. Here I used to say they rebounded, then I thought they pushed linear , but would now prefer to go with the arm with the greatest force orbiting the other particle, as in 3.3. In principle if the rods B and C achieved a equidistant position from the midpoint of A, then A should, due to perfectly reciprocal rotational force, drive B&C linear.
Now considering a question. Is the force on the rods C and B reciprocal, as the distance between them is the same, though the distance each to end of rod is what is different. But does the increase in rotation make up for that. The answer is no. If B and C are at endpoints of A, yes of course reciprocal. But as you move one of them inward on A, the rotation off that contact point increases, therefore the reciprocity of force is no longer, though the reciprocal distances between them is always the same. So indeed, and much to my surprise, the SHORTER arm gives more force at any given point in relation to the other rod.
But the force at any point is not so much the issue as the total momentum of the arm. It is that that drives the rotation, so despite the point issue that arm that is longer should be the one that is rotating. Also, the force at any point is really the whole momentum of the rod as motion within the rod is instantaneous. So really it is only the velocity of the orbited rod to get it out of the way that determines the impetus it receives from the pusher rod. But my figuring was correct because equidistant points from the rotation point have equal speeds. And any two rods like in figure 3-4 are reciprocal in distance from each other (obviously), therefore equal distance from their reciprocal rotation points. But the shorter arm has a faster rotation due to conservation of momentum (see Appendix C) therefore that is why the point on that arm has a greater value than on the longer arm. But that I conclude is not the issue, the longer arm in total has more force so rotation would not under any circumstances depart from it. Therefore, there WILL be a scenario that when the arms become equal a linear motion will establish, this should have significant repercussions for the mechanics henceforth.
What is one to say in the case of figure 3.4. The longer arm drives rotation, a equidistant postions from the modpoint of A are achieved a a linear motion is established.
A ↓ midpoint
So if the ABC case as above (figure 3.4) causes linear acceleration from the dual rotation, then it would be by tension as the forces are not reciprocal or balanced
lanced, or if part linear and part rotation still the forces are unbalanced. Because of that it would seem more reasonable to assume, as I had come to believe, that the dominant force of rotation "wins out" and rearranges the motion of rotation to itself. This can be seen in one sense if the excess in one rotation induces orbiting, the leftover when pressing on each contact point again produces another excess, until all is "flipped" in that direction.
Principle 17.5 For a system of 3 particles, ABC, with B and C on opposite sides of A's midpoint, and B and C being at their midpoints, rotation occurs as per arm with the greatest total force, orbiting the other particle.
4b) If, on overtaking side C’s excess linear motion goes to A, which rotates toward and around C. Then two possibilities can happen, depending on where C is on A. A can rotate completely around C and hit B. Or in rotating the backside of A hits B. Either
way it should torque again to B and then treat as in case #1, with eventually the doublet re-forming, but with an excess speed over the speed of light. C remains traveling at the speed of light as a single particle.
Principle 17.7 A rotating of a rod continues in rotation until in balance.
Principle 17.8 A rod in balance proceeds in a linear motion until it is put out of balance.
Principle 17.9 Linear and curvilinear motions can be simultaneous in the same rod depending on how impacts are effective in relation to any unison motions that occur with other rods.
See also Principle 18.5 in Chapter 4.5 and Principle 19.6 Appendix A.
And Some More Mechanics Again
As in above mechanics section, but what if the particles B and C are not at midpoints to A, or vice versa. If one P, say B, is stationary and hit by A perfectly perpendicular, it simply rotates on A as per principle 13 and/or 15.1.2. If both B and C are hit by A, A will attempt to rotate by the same notion as principle 17.5. it rotates off then the rod producing the arm with greatest total force, but as it cannot orbit the rod it is not rotating on, but instead motion is forced into that rod, re-rotating it around A.