Redone Jan. 23, 2017 I think equation C-11 is still okay.
Following from chapters 2-4, for a rod rotating, it may have 1 arm (if rotating from a fixed endpoint) or 2 arms (if rotating from other than the endpoint). For a rod (A) rotating from its endpoint, the velocity of the rotation is calculable by the circumference the midpoint traces (for any time) (see Appendix D). Each point above and below the midpoint has a value > or < the midpoint value, that is that point travels a distance (circular) > or < the distance the midpoint travels. The velocity that can be imparted to a rod B which is perpendicular and at midpoint to midpoint with A, and “orbited” by A as A rotates on C (Figure 4.4) is such that the value at that point on A equalizes with the value (velocity) at the corresponding contact point on B (by principle 9). As motion travels instantly in the rods, B may be given any amount from A.
This appendix is a calculation of these values, I talk of radius's as the velocities but of course this is to calculate, the velocity is whatever it is and is multiplied by the percentages expressed in the calculations in this chapter. Its very important to realize that momentum is the real calculation needed, velocity without considering the mass involved will gets calculations all messed up, so although I talk only of velocity here, it must be related back correctly to momentum. For example for a rod going 100 miles per hour, another rod at he same speed (velocity) and twice the size and mass would have twice the momentum. But more importantly (in not messing up a calculation) a rod going 100 miles and hour, if simply doubled in mass, would then have 1/2 the velocity, or now only 50 miles per hour. The same holds true for line segments, if you double the length, the linear velocity is cut on half, as velocity is ultimately calculated by area swept.
If the rod was just a line (137.036 rod widths long), and with velocity constant the area swept when the rotation point is the endpoint (pi x rod length2) must be the same as the area swept by both arms when the rotation point is other than the endpoint.
For this to be correct a factor indicating the number of extra revolutions (or parts thereof) must be taken into account (because, of course, the shorter the arms become, the less area swept per turn, so to keep velocity constant they must make more turns). So:
pi x rod length2 = factor x (pi x arm1 2 + pi x arm2 2 ) (C1)
Now the change in factor is a uniform type of change, as the change in rotation point is uniform (continuous), so when the rotation point is known the factor can be determined and vice versa. Also, the factor x points (radii) of average velocity equals a constant, here = 2pi x 68.518 per revolution. This basically holds true if the volume of the rod as a cylinder is considered, using the rotation point along the line on the outside of the cylinder. However an adjustment needs to be made. As mentioned in Appendix D the area swept would be neigh impossible to calculate by area analysis. But by using the mid- mass point it can easily be calculated by determining the radius from the point of rotation to the mid-mass point. The radius of average velocity can be figured by similitude's and therefore (using rod length=137.036):
Rotation Point Factor Radius of Average Velocity
0 (lower side) 1 68.5198 (sq rt of 68.5182 + 1/22)
Midpoint on outside edge 2.0003 34.2545 (sq rt of 34.25092 + 1/22)
radius of average velocity x factor always = 68.5198
factor = 68.5198 / radius of average velocity
This is figured as on a plane surface, but is true for the three dimensional cylinder due to canceling out of similitude’s (see Appendix D).
P-M=68.518 (rod widths) P to mmp A & B (mid-mass point) =68.198
To calculate the average velocity of B we need to find the radius from contact point A-C to the midpoint of B, as in figure C1.
So now I will write about velocity and momentum, because this is what needs to be used to calculate this problem further.
Velocity of a rod going in a linear direction means every point of that rod travels at that given velocity.
Momentum of a rod going in a linear direction means all the volume of the rod is traveling at a given velocity. The momentum of the rod is its mass x its velocity. In this atomist hypothesis the mass is proportional to the volume of the rod. Density is not an issue, as this said mass is infinitely dense. All rods are the same size, shape and volume, therefore the same mass.
There is, in the natural world, no velocity without mass. Momentum is only expressed, i.e. present, when mass "displaces" space. The same of velocity, except it can also travel within mass area, instantly. Momentum displaces space over time, mass x velocity, i.e. area swept by a volume of mass per unit of time.
Transfer of motion rod to rod is as per principle 9, is quite straight forward with linear motion transfers.
For a rotating rod transfer of motion become more of an illusion, as every point on the rotating rod has a different velocity, and one might think that is the velocity for transfer at that point of impact with another rod. But it makes no more sense to add up the velocities at every point for a rotating rod then for a rod in linear motion. The only actual force, or energy as such is the momentum, i.e. area swept by the mass, or mass x average velocity of the rotating rod. For the rotating rod only one point, traveling a circular line represent the average speed of the rod, whereas for linear motion every point "scribing" itself though space represents a vector of average motion.
If we take rod B at the midpoint of the rotating rod A (rotating off C, as in figure C-2) rotating from the endpoint of A, point M on A travels at its velocity and the corresponding point M on B must be equal velocity. However it is B that is orbited by A's rotational motion, not just point M.
The "rotational" momentum of B is its mass x the point of average velocity on B. But to equalize (principle 9) the speed with A the points M must be equal, so that A's rotation is not just cut in half but the rdq (of B) / radius P-M for B means B's velocity is such that there is % change in A's speed at point M to make it all work.. So if A & B's speed are equal at M,
velocity B = velocity M x rdq (B) / radius P-M (C4)
and likewise velocity A = velocity M x rdq (A) / radius P-M (C5)
to equalize velocity B = velocity M x 1/2 x rdq (B) / radius P-M (C6)
and velocity A = velocity M x 1/2 x rdq (A) / radius P-M (C7)
and in this case only, resulting velocity A = velocity B (C8)
(rdq – radius
in question, or radius from point of rotation to the mid-mass point
of rotating rod)
Now for any point of contact A-B, or that is any point of rotation off C so any distance F-M (F stands for any point of rotation other than the end point P) , the thing to consider is the circular "orbit" (motion) given B at transfer is such that the velocity of B gives the value of the velocity at point M a velocity equal to what the velocity of A also imparts to point M.. A must retain more than 1/2 its motion to make point M equalize, except when rotating from the endpoint P, where the area swept by B is the same as the area swept by A, though they are at 90 degrees to each other. Calculating it this way the revolutions of A speeding up as arm length(s) decrease for at the same speed as described previously are such that it accounts naturally for change in speed of revolution as is. The equations I have figured for this are as follows:
(See Figure C-1, F stands for any point of rotation (A on C) other than the end point P)
1. At rotation off endpoint of A
P-M/137.036 x 68.5198 = Velocity imparted to B (C9)
2. At rotation point other than endpoint
(F-M/137.036) x 68.5198 + ([1/2 x 68.5198 - (F-M/137.036) x 68.5198] x F-M/[P-M + F-M]) =
Velocity B (C10)
F-M/137.036) x 68.5198 + ([34.25099 - (F-M/137.036) x 68.5198] x F-M/[68.518 + F-M])
Explaining this reasoning a bit, I reckoned that in equation C9 the velocity imparted to B is 1/2 the original velocity of A. As the distance F-M gets shorter, B will have a smaller orbital value, so > 1/2 the original value (total momentum of rod) remains in A. So and take 1/2 the speed for the radius of F-M as part of the speed of B (as a momentum in B). Retain 1/2 the speed of A in A (as mentioned two sentences back). The remaining speed this being divided up is then, if figured by hand little by little I found to follow a descending order with a limit equal to the term F-M/P-M + F-M. So for example if the distance F-M 34.2509, or 1/4 a rod length, the value of velocity for B becomes 1/4 of the total speed (here figured as proportional to 68.5198, therefore 17.12995 plus 1/3 of the 17.125995term above). If distance F-M is 1/8 a rod length the value for B becomes 8.56497 + 1/5 of 8.56497, etc. Of course the speed left in A is the original speed of A minus the speed calculated for B. At these calculated speeds I found the contact point speeds to be equalized.
Note speeds so calculated are only ratios to known quantities, as are all real measurements.