ramoncreager wrote:

"Don't be so sure that Oswald couldn't have hit President Kennedy with that rifle. I've fired weapons in the army and own a 6.5 mm bolt-action rifle"

Did you fire from a 6.5mm MANNLICHER-CARCANO? The rifle the Italians dubbed "the rifle that never killed anyone" because of its terrible design flaws. I have, and it is damned near impossible to get off three successive shots in the time period noted because of the tendency of the bolt action to stick.

Even with shims corrected, and rifle sights perfecto - or as near as I could get them - I have never been able to get off three *accurate* shots in under 10 sec. (and that was after greasing the bolt action thoroughly)

Let's also not forget the testimony of Oswald's DI, Nelson Delgado, who averred that everyone nicknamed Oswald "Maggie's Drawers" - which means he couldn't shoot worth beans and a fiddle even if his rifle could (and wasn't really a "stage prop" as Patricia Dumais suspected from her correspondence with the National Archives)

From the records available, Oswald's highest rating was achieved on Dec. 20, 1956, not long after entering the Marines. This was 'sharpshooter. Three grades are used in all: expert marksman, sharpshooter, and marksman. The shots made in Dallas - on 22 November, 1963 - if made by one person, on the moving targets specified - would have required a 'master expert marksman' level of skill, or the 'best of the best'.

By the time Oswald was discharged from the Marines, he was barely able to qualify in any defined shooter category - with a 191 score (or Marksman).

Logically, also, bear in mind that regular practice would have been required in the Marines and Oswald deteriorated in skill even in this environment. It is rather illogical, therefore, to suppose that in an environment ('civilian') which required no regular training (and for which no evidence of any extraneous 'practice' exists) Oswald's skills would have significantly improved to the level of being able to make the

shots attributed to him on Nov. 22, 1963.

Apart from this the physics belies that Oswald made the key head kill shot.

treat the head as an initially moving free body (due to the limo's motion). Since the limo speed was at approx. 11 mph, this would be 4.9 m/s. Accordingly, as an inelastic collision, we require (The kinetic energy is not conserved):

total momentum before collision + total momentum after = 0

or (using P for momentum)

P(before) + P(after) = 0

The head (3kg) is moving initially at 4.9 m/s to the RIGHT (vector +)"

-------------------> P (h) = 14. 7 kg m/s

the bullet is directed to the LEFT (c=vector -)

<----------------------------

P(b) = -6 kg m/s

where a bullet speed of 600 m/s is assumed, (The muzzle velocity of the Mannlicher-Carcano is estimated at 2100 f/s or 636 m/s) and bullet mass of 0.01 kg. Thus:

Total momentum before collision: = 14.7 kg m/s + (- 6kg m/s)

" " " " " " " " = 8.7 kg m/s

Now, since the sum of the total momenta (before and after collision) must be zero, this means that we require:

Total momentum After collision = - 8.7 kg m/s

In other words, it is directed toward the LEFT (or the REAR of the limo)

Checking:

P(before) + P(after) = 0

[8.7 kg m/s] + [-8.7 kg m /s] = 0

We know that 1.2 kg (approx.) of mass was lost, 1.0 kg of

blood and tissue, and 0.2 kg mass of bone fragment.

An estimate of the fragment's momentum can be found from

the range of the bone fragment. We can calculate the range using:

R = v(o)^2 sin 2(X)/ g

and thence the velocity v(o):

v(o) = SQRT ( R g/ sin 2X)

where v(o) is the initial velocity and X is the 'launch' angle, g the acceleration of gravity (we use 10 m/s^2 to comport with the other approximations, estimates).

For the range, we adopt 25 feet, or the estimated distance the

Harper bone fragment landed from the limo just after the head

shot. (cf. 1, p198):

Thus, R = 7.6 m.

then v(o) = SQRT (7.6 m * 10 m/s^2/ sin 90)

where X is estimated at 45 deg.

Thus, v(o) = - 8.7 m/s

(since it is directed leftward)

However, this does not take into account the 'launch' is

from a 'moving platform' (i.e. limo) moving at 4.9 m/s.

Thus, the actual fragment velocity is:

-8.7 m/s - (4.9 m/s) = -13.6 m/s

The momentum of the fragment is therefore:

P(f) = 0.2 kg (-13.6 m/s) = -2.7 kg m/s

And the (-) direction denotes it is traveling in the backward motion, verified by Jackie's pursuit of it over the limo boot.