[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, rebound and continue moving as single particles or move off in unison motion crossed at midpoints.
What happens when these crossed particles (2P) collide with a single particle (1P)?
Rod A 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)(AB's direction of motion) Overtaking hits (C overtakes A-B)
1. C hits on A 3. C hits on A
2. C hits on B 4. C hits on B
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 particularly conservation of motion and that one rod can only accelerate another from its midpoint.
1. Direct hit C on A
2. All the linear motion of A and C goes into torque and rotation of each. While torquing they are both stationary in space, but B has a linear motion to be conserved. This all instantly (see Appendix B) goes into A speeding its torque.
3. All torque will pull A into B in which case A re-torques and re-rotates B, in an instant, and therefore is still at mid-mass points and accelerates A by principle 9. In essence A-B reverse directions instantly in space.
4. 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 B
2. C & B’s linear motion stops, going into torque of each. While C & B torque A moves off with its linear motion.
3. C & B torque with their respective velocities. One will complete its torquing before the other, (for example B) with all its velocity going into rotation, but the torque of C still occurring causes B to torque and rotate simultaneously.
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 reached 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.
4b. Sometimes, unless midpoint contact is reached first, B or C will re-hit A (even allowing that B-C have no linear motion, B’s rotation “catches up” to A), but it will be off center – off center and unaligned. Since B would have to torque or rotate on A, but also then C, which it can’t simultaneously, it rebounds (see principle 18). This causes all 3 particles to go in separate ways.
1. Overtaking hit, C hits on A, treat as an overtaking hit, Chapter 2.
2. Excess motion goes into torque of C on A.
3. When C is perpendicular, C tests on A, bringing together the unison linear motions as occurred in last chapter, and then C rotates on A.
4. When C reaches its midpoint
a) If on the direct hit side of A, C releases its rotational motion to linear motion which causes A to rotate on C (they are already perpendicular so no torque first). A rotates on C about a quarter rotation and re-hits B. A torques to B, rotates slightly and is at mid-point. It then imparts its excess motion to B by principle 9, leaving A and B still in unison motion, but a new one which is of the old unison motion and 1/2 the excess motion, as a compounded motion now. This means in future impacts there is more to consider (see Appendix A, Section 4).
A and B will still be a pair (doublet), but with a change in direction and real and also apparent velocity, always moving off away from C.
If on the overtaking side when C reaches it's midpoint, the testing of C on A would cause A to re-rotate on C. But first a "virtual" resultant between the test linear and the already present Unison motion linear occurs (as in chapter 2 dynamics) . This resultant would, if carried out, cause a new unison motion, allowing a point outward on C to contact A, 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 A before all motions used up, the mechanics are there so that C should literally "scoot" around A at high speed and use all motion up, until parallel with B. The time it takes to do so means AB travel a certain distance and C uses real unison motion, as above mechanics, for that portion of motion as necessary.
5. When aligned, C's unison motion is again the same as A and all its excess goes into A causes it to re-rotate around C, and "orbiting" B in said fashion as follows.
6. Unison Rotation-Orbiting:
As said above, from here C’s excess now can cause A to rotate as it is “squared up” to B so B is “orbited” (see next paragraph). B is accelerated to an equal velocity (area swept) of the “force” of rotation of A, at A’s midpoint (see Chapter 4 and Appendix C).
Here it might be supposed that B, accelerated by A at midpoint is like Principle 9, that is accelerated linear. But that was if A was traveling linear, here it has a rotational motion. If (as stated) B is accelerated with a rotational motion is this not a violation of the reasoning behind Principle 14? For one thing, if it was then the hypothesis is unworkable. But as no angle can be assigned to the rotational motion anyway, indeed the slope of a circle goes from infinity to 0 as you work back to a single point. This then can 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. Stating it as a Principle:
Principle 17: A rotating rod passes a rotational motion to any rod positioned at its midpoint on the side of direction of the rotation. This rotation remains as an “orbital” quality until acted on by another impact.
So B is given an “orbit” with a radius equal to the distance from point of contact on A-B to contact point A-C and the velocity of the orbit, as calculated in Chapter 4 and Appendix C.
7. Declining Orbital Values
As A rotates on C, the length of its arm shortens, increasing the speed of revolution of A (but not its net velocity-area swept) But B has a set circumstance and velocity of orbit for each instance. But as A precesses on C it would be displaced outward past the midpoint on B causing as it would a rotation on B re-hitting its midpoint and orbiting B linear as it were again. But as these two orbits are off separate mechanisms, the later within B itself, intersects under the other orbit and puts a stop causing as it were a hold and feedback of motion from B to A, increasing A's rotation on B. This all happens "virtually" in a instance within the rods, the result is the orbiting of B occurs with appropriate diminution of speed as the rods C and B become closer together from the rotation of A around C.
8. Final Rotation of 3P
When A has rotated to the point that B comes in contact with C then all 3 rotate in unison (no progression on rod) as the rotation of A goes to B to C back to A. So A-B-C rotate simultaneously around the point of contact A-C. This is in addition to the unison linear motion of A-B-C .
In Chapter 4 it will be shown how when a particle D is added on the other side of B, B-C-D are in balance with A and this rotation can “revert” back to a linear motion.
1. Overtaking hit C on B. Treat as Chapter 2, an overtaking hit.
2. Excess motion goes into torque C on B.
3. When C is perpendicular, C tests on B, bringing together that and the the unison linear motions as occurs in the last chapter, and then C rotates on B.
4a) When C reaches its midpoint, if on direct hit side, C's excess will go to B causing it to re-rotate (again they are already perpendicular, so no torque). But it will rotate into A, accelerating A (as they have mid-point to mid-point contact). This will cause an increase in A's linear (taken from the rotation/excess) speed equal to enough that it keeps pace with B's rotation. Thereafter A moves off over the speed of light now, and B rotates on C until at midpoint and; 1. if on the direct hit side treat as per principle 15.7. 2. if on the overtaking side treat as per principle 9 and 15.4.1.
4b) If, on overtaking side C’s excess linear motion goes to B, which rotates toward and around C. Then two possibilities can happen, depending on where C is on B. B can rotate completely around C and hit A. Or in rotating the backside of B hits A, and re-rotates A, either going all around A, or again its back side hitting C and re-rotating it. Thereafter the same process occurs again, till at some point B is left rotating around C and hitting A. Either way at some point B is trying to rotate A and B both off-center contacts, and can't so it rebounds as per principle 18 which is
Principle 18: A rod rotating in contrary directions always rebounds. A rod torquing in contrary directions, or torquing and rotating in contrary directions also rebounds, as well as a rod in contrary linear motions.
This leaves B moving off from and A and C. A and C are separated but moving in the same general direction.