Letters to the Editor
droogoy
Published Letters: 590 Editor's Choice: 9
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Entry-exit wounds
[Read the article: From Dallas to Baghdad]
[Read more letters about this article: Here]vidor1 wrote:
"False. There was a small entry wound in the back of the skull and a big one in the right front."
El wrongo. The small entry wound was in the right front, the massive exit wound was in the rear. (As the Parkland drs. noted). The path of "cavitation" with the wound width increasing, was toward the REAR of the skull not the front. Also, the way the brain fragments blew out disclosed that a "dum-dum" bullet was responsible, not your garden variety 6.5mm Mannlicher-Carcano fare.
The photos you (and other referred to) had clearly been doctored and as David Lifton also noted ('Best Evidence') there were clear indications the autopsy itself was "doctored" when at Bethesda. The best policy would have been to allow the Parkland drs. tend to the matter, and not Humes and Boswell (the Bethesda autopsists) who had only limited experience with gunshot wounds.
The Harper bone fragment itself was later idenitfied as a piece of the rear skull. Or "occipital region" bone. In my earlier post, I noted Dr. Charles Baxter read that the "temporal and parietal bones" were missing. I forgot to mention this had been an error, since he was reading from his own script which clearly said temporal and occipital bones missing. Evidently, like most drs., he couldn't read his own hand writing. (The script was entered as the WC testimony evidence-material so superseded what he actually said)
The report of the "occipital bone" missing would comport with the statements of Drs. Carrico and Crenshaw (in his book) and also with the finding of the occipital bone fragment - which had clearly been dislodged.
By virtue of the laws of conservation of linear momentum in physics (see my earlier analysis) a rear bone CANNOT be dislodged or displaced by a rear shot! It can only occur via the forward momentum from a FRONTAL shot. Hence, to blow out the occpital bone (which, incidentally, is also clearly seen flying over the limo boot in the Nix film) the shot must have come from the FRONT.
This was surely the kill shot, but Oswald was not alleged at any time to be in the front of the limo as it progressed down Elm St. toward the Triple Overpass. Hence, it follows that Oswald could not have fired the kill shot, even if he had been part of the plan.
We now come to the central issue of wound location and placement. If one accepts the autopsy report (written) as accurate (**) it is placed:
"in the upper right posterior thorax just above the upper border of the scapula there is a 7 x 4 millimeter oval wound. This wound is measured to be 14 cm from the tip of the right acromion process and 14 cm below the tip of the right mastoid process."
** HSCA, Vol. 7, pp. 228-29, 223, 218,
! ! Neck
! x !_______
!Shoulder
C * !
The above simplified view (rear view) shows how far off this places the correct bullet entry site (C)from the site which is not correct (x) - but which was referenced(WCT, p73) by the Warren Commission Report. The actual bullet wound in the back can be seen clearly in photos (ibid.). It can also be referenced with respect to the bullet wound in the back, and the puncture made by the bullet in the jacket JFK was wearing ('Killing of a President', p78).
This wound placement - as described in the autopsy report reference(above) and shown in the photographs, is also clearly deduced by reference to the detailed anatomical diagrams in Gray's Anatomy (8, cf. Mastoid process, p. 31 -top, p. 216; and Acromion Process, p. 88, p.91).
The Mastoid Process is the concave bone structure just behind the ear. The Acromion Process is the peak or summit of the triangular shaped bone overlying the scapula, and is "flattened from behind outwards, directed at first a little upwards, ,so as to overhang the glenoid cavity. Its upper surface, directed upwards, backwards and outwards, is convex,
rough and gives attachment to some fibers of the deltoid."
All of the above, mean that the position reported by the Warren
Commission, is at the very least erroneous and quite
possibly prevarication. Certainly, the misplacement of this
crucial back wound undermines the entire underpinning of the single bullet theory.
Worse, it discloses that the Bethesda autopsists themselves may have been complicit, i.e. lied, to make the single bullet theory work. This claim is reinforced by the fact that on their Autopsy Body Chart, they did correctly identify the location - not in the base of the neck (as their WC
testimony averred) but six inches lower (as depicted in the diagram above). Thus, either they falsified the facts in their testimony, or the Warren Commission itself did, to make it compatible with their single bullet-lone nut nonsense.
Again, repeating, there is NO evidence of a bullet wound in the back of the HEAD. The only evidence for a rear wound is in the upper back, as described above. The kill shot had to have come from the front, in order to blow out the occipital bone and propel it 25' behind the limo where W. Harper found it.
A simple torque model can disclose the physics. We use a torque or moment about the center of gravity, acting
perpendicular to moment arm as shown below:
H 0 <------ bullet (e.g. from GK)-produces force F on H
!
!R
!
!
x CG
In the above, H denotes head, and CG is the center of gravity of the body (near point of contact with limo seat). R is the 'moment arm' (R = XO) and we assume a shot is placed at O and let R = 0.6m for ex.(The 'torso' can actually be modeled as a solid, uniform mass, cylinder 'block' centered at x and symmetrical about both sides of the axis XO. )
For a bullet of mass m= 0.01 kg, traveling at 545 m/s (see above) initially and decelerated to 445 m/s in 0.0004 s, we have a force of F = dp/dt = m(dv/dt) = 0.01 kg( 545 m/s - 445 m/s)/ 0.0004 s = 2500 N force exerted at O.
Assume the very reasonable estimate of 20 cm (about 8 inches) for the bullet path thru the head. At the velocities noted, this implies an average of 495 m/s for which the 'head transit' time is:
dt = 0.20 m/ 495 m/s = 0.0004 s
This would produce a torque (couple)T = F x R about 1500 N-m
at O. Thus a shot from the front (depicted) would produce a torque of 1500 N-m that would displace the head O toward H by some angular amount.
To find it we need the moment of inertia (I) of the 'torso' and from the cylindrical model we find: I = M/12[3r^2 + l^2] where M is the cylinder mass (say 50 kg), r is the cylinder radius (shoulder to center of torso dist. = 0.25 m), l =R,the cyl.length, say 0.6 m). Now, the force of the bullet F impacting at moment arm distance R from CG exerts an external
torque on the torso such that:
T = F x R = dL/dt = I(@)
where dL/dt is the rate of change of the angular momentum of the body, I is as defined above, and @ is the angular acceleration @ = dw/dt or rate of change of angular speed). On substitution (say using the above reasonable values) we can solve for @:
@ = T/I = 1500 N-m/ (2.281 kg-m^2) = 657.6 rad/s^2
This is the angular acceleration under impulse. From this the angular velocity (arising in same time) can be obtained as:
and w = @t = 657.6 rad/s^2 (0.0004s) = 0.263 rad/s
Under conditions of uniform deceleration, we have:
@(u) = (w(f) - w(i))/ t
where t is the elapsed time over Z-frames 314-321, with a mean
film advance rate of 18 frames per second. (t = 0.44s). The
initial angular velocity is known from the impulse calculations
(0.263 rad/s) and the final angular velocity w(f) = 0.
@(u) = (0 - 0.263 rad/s)/ 0.44s = - 0.598 rad/s
Again, under conditions of uniform motion we have:
w^2 = w(i)^2 + 2@ (theta)
whence:
theta = [w^2 - w(i)^2]/ 2@
theta = [0 - 0.263^2]/ 2(-0.598 rad/s)
= 0.05783 rad
This must now be added to the angle made under impulse:
theta (total) = theta (impulse) + theta (uniform)
0.0001 rad + 0.05783 rad = 0.05793 rad
theta (deg) = 0.05793 rad (57.3 deg/rad) = 3.3 degrees
This is the total amount of angular motion backwards undergone by JFK's body with respect to a central axis, as a result of being struck by the frontal shot.