The Other Single Bullet Theory

[Charles Collins]

Here’s an image showing the drag coefficients of different shaped objects. A bullet is designed to be streamlined to a certain extent. You can see for yourself by comparing the drag coefficients of the different shaped objects.




An irregular shaped fragment, that is also tumbling, from the impact with Elm Street would have an extremely high drag coefficient due to turbulence. The drag would be magnified as the fragment transitioned from supersonic to subsonic velocity due to naturally increased turbulence in the trans-sonic velocity range. The net effect would be similar to deploying a relatively large parachute (think fighter jet airplane landing then slowing down with the assistance of parachutes).

A large drag coefficient means a low ballistic coefficient. I used a very low ballistic coefficient of 0.00405 in the online ballistic calculator. There is no way we can know exactly what the values actually were in Dealey Plaza. I indicated that we have to assume some reasonable values. I came up with what seems reasonable values to me, just to show that this idea is feasible.

[Andrew Mason]

Andrew, you cannot ignore the air resistance. Here’s a little blurb that might surprise you.

“The invention of the ballistic pendulum by the English ballistician Benjamin Robins in 1740 had led to the astounding discovery (at that time) that the drag force on a bullet was many times more powerful than the force due to gravity, and that it changed markedly with bullet velocity.”

No surprise at all. Drag is huge.  FBI expert Robert Frazier used 1 fps per foot so over 165 feet of travel bullet speed would go from 2165 fps to 2000 fps. That is a deceleration of roughly 1fps/(1/2000s) = 2000 fps². Gravity is only 32 fps².

The drag is a function of v² where v is the velocity in the direction of motion through the air.  That does not affect the time a bullet takes to drop a short distance due to gravity because it's falling speed is so much slower. That's all I am saying.

In the time it takes the bullet to travel 200 feet (.1 sec at average 2000 fps)
 the bullet will be falling at 3.2 fps and will have dropped .16 ft.  So in determining falling time you can ignore air resistance.

[Charles Collins]

No surprise at all. Drag is huge.  FBI expert Robert Frazier used 1 fps per foot so over 165 feet of travel bullet speed would go from 2165 fps to 2000 fps. That is a deceleration of roughly 1fps/(1/2000s) = 2000 fps². Gravity is only 32 fps².

The drag is a function of v² where v is the velocity in the direction of motion through the air.  That does not affect the time a bullet takes to drop a short distance due to gravity because it's falling speed is so much slower. That's all I am saying.

In the time it takes the bullet to travel 200 feet (.1 sec at average 2000 fps)
 the bullet will be falling at 3.2 fps and will have dropped .16 ft.  So in determining falling time you can ignore air resistance.

No, that is not all you were trying to say. You were trying to calculate the energy transferred to the asphalt without taking into consideration the energy lost due to drag.

[Andrew Mason]

No, that is not all you were trying to say. You were trying to calculate the energy transferred to the asphalt without taking into consideration the energy lost due to drag.

No. I am just pointing out that if you want the bullet fragment, after hitting the asphalt, to fall 13.7 feet vertically over a horizontal range of 250 feet, (and then fall another 6.8 feet over the next 141 ft) it can’t be travelling much more than 300 fps after hitting the asphalt.

I took into account the loss of energy due to air resistance.  The muzzle speed was 2165 fps or 660 m/s so the 10g.bullet initially had mv^2/2 energy=(.005(660)^2)=2,178 J.   Just before hitting the road it had slowed to 2000 fps due to air resistance.  So its energy on impact with the road was .005(610)^2=1,860 J.  Upon hitting the road, the bullet fragmented but all fragments kept moving slowly enough to fall 13.7 feet over a range of 250 feet which means it had an average speed of 270 fps.

Even if it was moving at 440 fps (134 m/s) after hitting the road and slowed to 100 fps just before hitting the manhole (but then it could not have hit the road near Tague 141 ft away) that means it had retained only (134/610)^2=0.05=5% of its original energy. So it deposited 95% of its 1860 Joules of energy (1767 J) in colliding with the road. I am saying it was a bit more than that - about 1800 J.

[Charles Collins]

No. I am just pointing out that if you want the bullet fragment, after hitting the asphalt, to fall 13.7 feet vertically over a horizontal range of 250 feet, (and then fall another 6.8 feet over the next 141 ft) it can’t be travelling much more than 300 fps after hitting the asphalt.

I took into account the loss of energy due to air resistance.  The muzzle speed was 2165 fps or 660 m/s so the 10g.bullet initially had mv^2/2 energy=(.005(660)^2)=2,178 J.   Just before hitting the road it had slowed to 2000 fps due to air resistance.  So its energy on impact with the road was .005(610)^2=1,860 J.  Upon hitting the road, the bullet fragmented but all fragments kept moving slowly enough to fall 13.7 feet over a range of 250 feet which means it had an average speed of 270 fps.

Even if it was moving at 440 fps (134 m/s) after hitting the road and slowed to 100 fps just before hitting the manhole (but then it could not have hit the road near Tague 141 ft away) that means it had retained only (134/610)^2=0.05=5% of its original energy. So it deposited 95% of its 1860 Joules of energy (1767 J) in colliding with the road. I am saying it was a bit more than that - about 1800 J.

Andrew, you are trying to over-simplify more than one aspect. And you are trying to compare apples to oranges. Your methods do not work for this application. I understand the physics behind firing a bullet level and dropping a bullet straight down and they both hit a level surface at the same time. You are trying to specify a slower velocity for the fragment to give it enough time to drop 13.7-feet. However that is simply not the situation we are faced with in this case.

First of all, as I have already explained, the fragment leaves the surface of Elm Street at a negative 2-degrees angle (not level). Therefore, at a distance of 250-feet it is going to be at an elevation that is 8.7-feet lower than it was at the Elm Street mark without any help from gravity due to simple geometry. Also, unlike the dropped bullet, due to the downward component of the direction (-2-degrees) of the fragment's velocity when it leaves the Elm Street mark the fragment is not starting from zero vertical velocity.

So the fragment only needs to drop approximately 5-feet due to gravity (not 13.7-feet). Plus it has a "running start" compared to a dropped bullet. You can compute all of this if you wish. However, I will likely continue to believe the ballistic calculators.