o->
						   /
		      o                          o
	       ^      !                         ip1
	       o             /
			    o
 o
   \            <--o->          <--O->               o
		 ip1             ip2               ip1

 \                  o                  ^
    o               !                  o
   ip1

The particles shown above represent those in the volume of a sphere with
radius Rv that one collector ip2 (O) particle occupies, showing smaller but
more ip1 o particles. All move in random direction. Let's consider ip2 moving
in 2 directions only in space and consider the ip1 particle moving in these 2
same directions. We considered that they must be alligned before coalescing
can take place and we gave this probability as asr1 and asr2 explained in the
main text. We have the alignment but will they touch so we can apply the Collision
Efficiency expression? Let the arrows denote slow -> and fast --> vector velocities.
The table below should show the possibilities.

case ip1     ip2    Collide?
 1   <--     <-      no           The 6 cases --> -->, -> ->, <-- <--, <- <-, --> <--,
 2   <-      <--     yes          and -> <- were not included because so unlikely that
 3   ->      <--     yes          the 2 particles would ever be exactly the same speed.
 4   -->     <-      yes          Rather, one would move relatively faster --> than the
 5   -->     ->      yes          other -> at some reference point. Also, the 4 vectors
 6   ->      -->     no           moving away from each other <- ->, etc. need not to be
				  considered.

The relative vector velocities -> means that it is moving slower than the other particle vector
-->. Notice, regardless of the vector speed, they collide if heading towards each other in
case 3 & 4. So we use the 4 yes cases to multiply by the independent joint probabilities
asr1 x asr2 by. Now since this is for 1 ip2 particle, we have np2 # of ip2 particles so then
for total np2's, we must also include np2 as a factor.

Since for 1 ip2 particle, we have a number of ip1 particles, as seen above, as the greater
number of smaller particles feed the larger. How many more np1 particles than np2? We have
np1/np2 times as many. Putting all this together, we have npar = 4 x asr1 x asr2 x np2 np1/np2 =
4 x asr1 x asr2 x np1 before applying the Collision Efficiency expression.