Anyway, it was in the late fall of 1980 while based in the Physics Dept. of the University of the West Indies (and barely months after my first research paper was published in

__The Journal of the Royal Astronomical Society of Canada__) that a colleague - a visiting E&M professor from University College London - offered a staff room challenge I couldn't resist. He'd been reading the first printing of David Lifton's

*'Best Evidence'*and became fixated on his Fig. 2 (in the photo section), showing JFK's head being blown backwards over frames 312-13 of the Zapruder film.

He asked me whether I believed the displacement of JFK's head could be "quantified". I said I believed it could using some basic parameters and then spent the next few days preparing a working draft of how it might be done using basic mechanics.

I used a torque or moment about the center of
gravity, acting perpendicular to a moment arm R as shown below:

For example, for a bullet of mass 0.016 kg, traveling at
748 m/s (see above) initially and decelerated to 448 m/s in 0.0004 s, will have a
force of

**F = dp/dt****= m (dv/dt**) =
0.016 kg ( 748 m/s - 448
m/s)/ 0.0004 s = 12000 N force exerted at H

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 598 m/s for which the
'head transit' time is:

dt
= 0.20 m/ 598 m/s =
0.00033 s

This would
produce a torque (couple)

**T = F x R**of about 7200 N-m at H. Thus a shot from the front (depicted) would produce a torque of 7200 N-m that would displace the head by some angular amount.
To find it we need the moment of inertia (I) of
the 'torso' and from a simplified 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 cylinder 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**= d

**L**/dt =

**I**

**a**

where dL/dt is the rate of change of the angular
momentum of the body, I is as defined above, and a is the angular
acceleration a = dw/dt or rate of change of angular speed).
On substitution (say using the above
reasonable values) we can solve for a:

a =
T/I = 7200 N-m/ (2.281 kg-m

^{2}) » 3150 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 = at = 3150
rad/s

^{2}(0.00033s) = 1.04 rad/s**Ft**

**_**

**! !**

**! !<----impulse o:p="">**

**! !**

**--! !_________________**

**uniform motion !**

**!------------------------**

The sort of physical situation we have can be
depicted above. The foregoing treatment basically took care of the very brief
(0.00033s) impulse part of the motion. There is still a uniform part left to
reckon (torso motion under uniform deceleration). (N.B. The Dirac Delta
function and Laplace transforms are usually
used for impulse computations, but that is omitted here so as not to overly
complicate the solution).

Under
conditions of uniform deceleration, we have:

**w**

**(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 (1.04
rad/s) and the final angular velocity:

w(f) = 0.

a(u) = (0 –
1.04 rad/s)/ 0.44s = - 2.36 rad/s

Again,
under conditions of uniform motion we have:

w

^{2}= w(i)^{2}+ 2aq
whence:

q= [w

^{2}- w(i)^{2}]/ 2a
q = [0 – 1.04

^{2}]/ 2(-2.36 rad/s)
=
0.229 rad

This
must now be added to the angle made under impulse:

q (total) =
q (impulse) + q (uniform)

=
0.001 rad + 0.229 rad = 0.230 rad

q(deg) = 0.230 rad (57.3 deg/rad) = 13.1 degrees

This
is a smallish angular displacement which the entire torso makes relative to its
'pivot'. This can be translated into linear dimensions by using the fundamental
definition of arc length:

q = s/ R
so s = q x (R) = 0.230 rad (0.6m) = 0.138 m

or
13.8 cm (5.4 inches)

*the entire mass of the torso*is taken into account, as opposed to just the head mass.. In any case, my UCL colleague was impressed with the result and suggested I publish it one day. (I did, some 17 years later in my REAL FAQ on the newsgroup alt.conspiracy.jfk.) In the meantime, he gave me his copy of Lifton's book which commenced my part time research into the physics aspects of the Kennedy assassination - much of which I exposed in Part 5 of my FAQ published on this blog in November, 2013, e.g.

http://brane-space.blogspot.com/2013/11/frequently-asked-questions-on-jfk_15.html

Thus began my path toward being a JFK assassination researcher as well as a solar physics expert in the specialized fields of Poisson limits on flare occurrence, and solar flare prognostication applied to geo-effective effects, including magnetic substorms and sudden ionospheric disturbances.

(Note: Up to the time of working on this displacement problem the only book I'd read on the assassination was Gerald R. Ford's

**. ) Working out this problem and others changed my mind from the lone nut doctrine to the physics disclosing a definite conspiracy at work. In the end, it was**

*'Portrait of an Assassin'***that integrated and drove my JFK research and solar research!**

*physics*The attainment of expertise, or what I consider such, was probably not attained until I wrote my first FAQ for the alt.conspiracy.jfk newsgroup in 1997. By then I had plowed through thousands of released documents and files (on the basis of the

*1992 JFK Records Act*) and read virtually every book on the event I could lay my hands on. In addition, I had become informed on deep politics and made this a specialty area. All this was possible after retiring from full time physics research, teaching. This extra time also enabled me to complete the book I had visualized since beginning my research journey.

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