It is true, the ‘Jet Effect’ does not have to push the head straight back. With the side of the head exploding outwards, the ‘Jet Effect’ would push the head to the left. But what I am interested in the effect it would have on pushing the head to the rear. The ‘Jet Effect’ theory was developed to come up with a possible reason why the head started moving backwards.

**Question: What was the change of speed, and the resulting speed of the head, rearward, as a result of the ‘Jet Effect’? **

For instant, you could say:

+ The initial speed of the head forward was: 2.0 mph

+ When the head exploded, it added 2.5 mph motion to the rear

+ Resulting in a motion to the rear of 2.5 – 2.0 or 0.5 mph to the rear

Basically, I am interested in your estimate of the rearward component of the velocity vector.

Do you mean rearward momentum of the head and body relative to the body or to the car? There is a difference because JFK was turned to the left and leaning to his left and forward. Also, the head was leaning forward more than the body. I don't see the head being driven back relative to the body. It was driven to the body left which was to the car rear-left and this caused the whole body to move rear-left.

This is going to be a rough estimate, but here goes:

The head and body pivot. The head pivots on the neck. The upper body, being seated, pivots from the seat. So we really need to find the angular speed imparted to the head by the bullet and the angular speed of the head and body from the jet effect. The angular speed (ω) of a body can be determined from the angular impulse or angular momentum (

**L**) and moment of inertia (

**I**) of the body by:

** L=Iω**. We are going to use the moment of inertia of a sphere for the head pivoting about the neck and the moment of inertia of a cylinder pivoting on the seat for the head/torso.

My estimates of the motion of the head due to the bullet imparting momentum due to its impact to the head:

1. JFK was turned 15 degrees to the left. He was leaning left at an angle of 10 degrees and his head was leaning forward 40 degrees. I will assume the head mass is 5 kg. which appears to be the average mass of a human head (google).

2. The bullet was travelling at a right to left angle of 5 degrees and a downward angle of arctan(60/275)=12 degrees relative to the horizontal. The car was on a downward 3 degree slope. So the downward angle relative to the car was 9 degrees.

3. The 10 gram bullet entered the head travelling 1900 fps or 580 m/s, passed through the head and exited at 360 fps or 110 m/s and did not change direction. So the change in momentum was .01 x 470 =4.7 kg m/s which imparted a forward momentum of 4.7 kg m/s to the head at an angle of 5 degrees right to left and 9 degrees down.

4. The bullet struck above the centre of mass of the head. I would put the centre of mass of the head halfway between the ears, so the bullet path was about and inch or 2.5 cm above the centre of mass.

5. The bullet struck JFK just before z313 was exposed but after the end of exposure of z312. There is 30 ms. between frames where film is not being exposed. I will assume that it struck 30 ms. before the exposure of z313 began.

6. So 4.7 kg m/s impulse to the 5 kg head would result in a 5 degree right to left and 9 degree downward velocity of just under 1 m/s: 940 mm/s or .94 mm/ms. So in the 30 ms before exposure of z313 began, the centre of mass of the head would have moved 30x.94= 28 mm or just over an inch in a direction 5 degrees to the right and 9 degrees down, assuming it was free to pivot from the neck.

7. Average speed through the head would have been (580+110)/2=345 m/s. Assuming a path of 20 cm or .2 m, the transit time through the head, assuming uniform stopping force was applied to the head would have been t=d/v=.2/345=.0006 s = .6 ms. So we don't really have to take into account the fact that this impulse took some time to impart momentum.

8. Since the head pivots from the neck and since the bullet struck above the centre of mass, one has to determine the angular impulse to the head. The head centre of mass (middle of the ear to base of the neck) is about 17 cm above the neck, so the angular impulse is r x p=.17 x 4.7= .8 kg m^2/sec which is the angular momentum of the head. To determine the angular speed, we divide by the moment of inertia of the head. Assuming it to be approximately a 5 kg sphere of radius 10 cm rotating about a pivot 17 cm. from the centre of mass, the moment of inertia is .02 + 5x(.17)^2=.16 kg m^2. So the angular speed would be .8/.16=5 radians/sec or about 280 degrees per second or .28 degrees per millisecond.

**In the space of 30 ms (the time between end of frame 312 and beginning of frame 313) the head would pivot 9 degrees forward**.

9. There is a limit to how far forward the head can move because the chin cannot go past the chest.

Here are my estimates of the motion of the head and body due to the jet-effect:

1. The jet effect did not begin until the bullet exited the head. At that point there was no longer any forward momentum being transferred to the head from the bullet.

2. Frame z313 appears to show a streak from the motion of a piece of matter, possibly a skull fragment moving at an upward angle of about 65 degrees above car horizontal. It appears to be about 1 metre long and it starts about a metre from the head. This gives us information not only about the speed of the ejected matter but also the timing of when it started. Since the exposure is 25 ms duration, 1 m of travel in 25 ms is 40 m/s. Since it had already travelled 1 m before z313 began to be exposed, this means that the head erupted about 25 ms. before z313 was exposed or about 5 ms. after the end of exposure of z312. Since z312 does not provide any indication of the head being hit, this suggests that the bullet hit less than 5 ms after the exposure of z312 ended.

3. There appears to be a substantial amount of brain, skull bone and blood erupting from the head wound. Doctors reported that a substantial portion of the brain was missing. If the mass of matter exploding from the head was only 300 g, travelling initially at 40 m/s, the total momentum would be 12 kg m/s or almost 3 times the momentum of the incoming bullet. But it does not all explode in the same direction. It forms what appears to be a hemispheric cloud around the exit wound. So the average would be about cos 45 x total momentum = 70% of 12 kg m/s = 8.4 kg m/s/

4. So the head and the body to which it is attached would, under those assumptions, recoil to the body left with 8.4 kg m/s of momentum. Unlike the forward momentum imparted by the bullet, however, there is nothing really to stop the upper body from moving to the left so it keeps going.

5. If the torso received an impulse of 8.4 kg m/s applied to the right side of the upper skull, which would be a distance of approximately 90 cm above the seat or pivot point of the torso, the angular impulse, r x p would be .9 x 8.4 kg m^2/s = 7.5 kg m^2/s.

[Correction: I was using the wrong axis of rotation for the torso. The moment of inertia of a cylinder of length Len, mass M and radius R about an axis at one end is: MR^2/4 + M(Len)^2/3. If the centre of mass of the torso is midway, 45 cm above the seat (Len=.9), and the torso is approximately a solid cylinder of mass=40 kg and radius 20 cm (moment of inertia, I = MR^2/4+M(Len)^2/3= 40x(.2)^2/4+40x(.9)^2/3=.4+10.8=11.2 kg m^2/s). So the angular speed of the torso would be: L/I=7.5/11.2=.7 radians/sec or 40 degrees/second

**In 55 ms or one frame, it would rotate a bit more than 2 degrees to the body left**.