There IS some analysis of the missile parts by Commander Donalson on his website, although that page is currently unavailable for some reason..
I'll post a link in case it reappears, and post the pertinant section of the analysis.
From Radar Analysis:
Evidence of a Missile
A Sweep by Sweep analysis of the radar data clearly depicts a plume of very high velocity metal exploding out of the aircraft's right side at approximately 20:31:13. The lead ejecta almost certainly has to be a missile body! With the aircraft on a heading of 071 degrees True, this radar contact separated laterally from the aircraft track on a ballistic trajectory of 190 degrees True, traveling about 3,200 feet in less than 7 seconds. This 119 degree change of direction and high velocity could not possibly have been the result of a Center Wing Tank explosion.
This is not theory or speculation, but hard physical evidence. It takes almost 2 seconds for a bullet fired from a military sniper rifle to travel that distance. Only a missile body, with it's hundreds of thousands of foot pounds (ft/lbs)* of kinetic energy would have the inertia to perform this way on radar. It only takes 6,000 ft/lbs of energy for a 50 caliber heavy machine gun bullet to punch through 3/8 inch thick steel plate. Boeing 747's are constructed of various aluminum alloys and rarely in thickness' exceeding 1/2 inch.
With this much energy, the missile body would slice through the aircraft "like a bullet through a tin can", just as I stated in my April 24, 1997 letter to the Wall Street Journal, in response to Chairman Hall's "It wasn't a missile" letter to the same paper. It is startling to note also that the trajectory of the ejecta (missile body & other parts) is on a direct line from an unidentified boat, 2.9 nautical miles to the Northeast of Flight 800 when it exploded.
* [ 15lb missile body @ 1500 ft/sec, apply 1/2 MV squared = 524,720 ft/lbs of energy ]
He states that the ejected object traveled at 3200 feet in about 7 seconds, yet he states the speed as 1500 feet/sec. That's obviously wrong. He also neglects to convert the weight of the missle body from pounds to the unit of mass in the foot-pound-second system, which is a slug.
So the velocity should be;
v = 3200/7 = 457 feet/sec
The equation for kinetic energy is;
E = (m * v2)/2, where;
E = Energy (foot pounds)
m = mass (slugs)
v = velocity (feet/sec)
A slug = 32.17 pounds, so we'll add that to the equation to convert from pounds to slugs..
E = (m * v2)/64.34
Solving for E, we have;
E = (15 * 4572)/64.34 = 48,690 foot pounds
I'd say that value would ALSO suggest a missile, as its energy is still extremely high..
As the muzzle energy of a .50 BMG rifle round is a little over 12,000 foot pounds, I'd say that 48,690 foot pounds is A LOT of energy. In comparison, a .308 rifle bullet has a muzzle energy of about 2520 foot pounds, and IT can pass through a telephone pole or a tree....