## Monday, May 4, 2015

### How I Became A JFK Assassination Researcher - While Doing Solar Physics

On occasion, including at astronomy and astrophysics lectures I've given over the years, people ask me how I became interested in the Kennedy assassination. Usually this has been after seeing a letter to the Editor or newspaper article I've published on it, or after a radio interview (including several over VOB in Barbados). The answer is fairly straightforward, and the only reason it would "blow" minds is that the culture of specialist isolation remains too pervasive among academicians. It is as if they believe one can't "walk and chew gum" - academically - at the same time. But the proper apportioning of time does allow the curious researcher to pursue his/her primary specialist interest while also pursuing a less prominent interest.

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: 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 = H-CG) and we assume a shot directed as shown and let R = 0.6m  for example.(The 'torso' can actually be modeled as a solid, uniform mass, cylinder 'block' centered at x and symmetrical about both sides of the axis H-CG. )

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[3r2 + l2]

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 = dL/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-m2)  »  3150 rad/s2

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/s2 (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:

w2 = w(i)2 + 2aq

whence:

q=  [w2 - w(i)2]/ 2a

q = [0 – 1.042]/ 2(-2.36 rad/s)

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

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

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)

Which (from direct estimates using the Z-film)  appears to be about one third to one half  the actual displacement. Still, one would expect some degree of error.  For example, 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.