The Other Single Bullet Theory

[Charles Collins]

I have created a report for another Single Bullet Theory. I figured one SBT wasn't enough. 

You can find the report at this link:

https://sites.google.com/view/theothersbt/home

Please review this report if you are interested and let me know what you think. Thanks!

[Andrew Mason]

I have created a report for another Single Bullet Theory. I figured one SBT wasn't enough. 

You can find the report at this link:

https://sites.google.com/view/theothersbt/home

Please review this report if you are interested and let me know what you think. Thanks!

You have put a good effort into your paper, Charles.  While your theory that a wildly missed shot is better explained by accidental firing, the suggestion that Oswald would have been so careless as to fire accidentally seems just as highly improbable as an aimed shot that missed the car.

I do have a few questions, though. 

1. You seem to equate velocity with energy as in: "The impact of the first ricochet (off of Elm Street) would have reduced the velocity considerably. We have to assume some values that seem reasonable. If we assume it lost about 31% of its velocity/energy,".  Energy varies as the square of velocity so if the bullet loses 31% of its speed (retaining 69%) it ends up with .69^2=48% of its original energy, not 69%.

2. You say "the end point of the calculated trajectory as the bullet mark on the manhole cover, the graph shows an elevation drop of about 13.7-feet in about 83.3-yards (similar to what actually exists in Dealey Plaza). One of the main points that needs to be understood about the graphs is that the trajectories are essentially horizontal (level) for the first half of the distance, and slowly curve downward over the course of the second half.".

A fragment is still a projectile and follows the laws of physics. Neglecting air forces, any object drops at g=9.8 m/s^2 (32 f/s^2).  So if it starts with zero vertical velocity, the distance, h, that it falls is proportional to time^2:  h=gt^2/2.  So t=√2h/√g.  And since horizontal speed does not decrease much, time is proportional to distance travelled (t=d/v).

If the bullet hit the pavement and where you suggest and a fragment deflected at a 3 degree angle above the road surface so it started perfectly horizontal, in order to drop 13.7 feet (to the level of the manhole cover 83.3 yards or 250 feet) away, 2h/g=2x13.7/32=.856. So t=√.856=.925 seconds. So it would take .925 seconds to travel 250 feet. This means the fragment was travelling at an average speed of 250/.925 = 270 fps.  So that fragment has lost 1-270/2165 = 87.5% of its initial speed, and 98% of its initial energy.  If the bullet lost 98% of its energy in hitting the asphalt - ie. 600 [edit: this is wrong.  2000 fps is 600 m/s which works works out to 1800 J.] joules of energy would be deposited in a tiny area of the asphalt - there would have to be a significant movement of asphalt around that area.

[Charles Collins]

You have put a good effort into your paper, Charles.  While your theory that a wildly missed shot is better explained by accidental firing, the suggestion that Oswald would have been so careless as to fire accidentally seems just as highly improbable as an aimed shot that missed the car.

I do have a few questions, though. 

1. You seem to equate velocity with energy as in: "The impact of the first ricochet (off of Elm Street) would have reduced the velocity considerably. We have to assume some values that seem reasonable. If we assume it lost about 31% of its velocity/energy,".  Energy varies as the square of velocity so if the bullet loses 31% of its speed (retaining 69%) it ends up with .69^2=48% of its original energy, not 69%.

2. You say "the end point of the calculated trajectory as the bullet mark on the manhole cover, the graph shows an elevation drop of about 13.7-feet in about 83.3-yards (similar to what actually exists in Dealey Plaza). One of the main points that needs to be understood about the graphs is that the trajectories are essentially horizontal (level) for the first half of the distance, and slowly curve downward over the course of the second half.".

A fragment is still a projectile and follows the laws of physics. Neglecting air forces, any object drops at g=9.8 m/s^2 (32 f/s^2).  So if it starts with zero vertical velocity, the distance, h, that it falls is proportional to time^2:  h=gt^2/2.  So t=√2h/√g.  And since horizontal speed does not decrease much, time is proportional to distance travelled (t=d/v).

If the bullet hit the pavement and where you suggest and a fragment deflected at a 3 degree angle above the road surface so it started perfectly horizontal, in order to drop 13.7 feet (to the level of the manhole cover 83.3 yards or 250 feet) away, 2h/g=2x13.7/32=.856. So t=√.856=.925 seconds. So it would take .925 seconds to travel 250 feet. This means the fragment was travelling at an average speed of 250/.925 = 270 fps.  So that fragment has lost 1-270/2165 = 87.5% of its initial speed, and 98% of its initial energy.  If the bullet lost 98% of its energy in hitting the asphalt - ie. 600 joules of energy would be deposited in a tiny area of the asphalt - there would have to be a significant movement of asphalt around that area.

Thanks for the reply Andrew.

You have put a good effort into your paper, Charles.  While your theory that a wildly missed shot is better explained by accidental firing, the suggestion that Oswald would have been so careless as to fire accidentally seems just as highly improbable as an aimed shot that missed the car.

An accidental discharge can happen to anyone at any time. Even to highly trained, safety conscious professionals. We acquired a .45 caliber hole in our dining room table, chair, and floor when a gunsmith and safety instructor that we had known over 50-years had an accidental discharge. My idea that an accidental discharge happened to LHO could have been if the barrel of the rifle hit the box on the window sill. He most likely didn’t have an opportunity to actually practice his shots while in the sniper’s nest. And he most likely was looking through the scope as he began to try to track the moving target. If his finger was on the trigger the sudden unexpected collision with the box could cause the accidental discharge.


1. You seem to equate velocity with energy as in: "The impact of the first ricochet (off of Elm Street) would have reduced the velocity considerably. We have to assume some values that seem reasonable. If we assume it lost about 31% of its velocity/energy,".  Energy varies as the square of velocity so if the bullet loses 31% of its speed (retaining 69%) it ends up with .69^2=48% of its original energy, not 69%.

Thanks I will revise the velocity/energy wording.



A fragment is still a projectile and follows the laws of physics. Neglecting air forces, any object drops at g=9.8 m/s^2 (32 f/s^2).  So if it starts with zero vertical velocity, the distance, h, that it falls is proportional to time^2:  h=gt^2/2.  So t=√2h/√g.  And since horizontal speed does not decrease much, time is proportional to distance travelled (t=d/v).

If the bullet hit the pavement and where you suggest and a fragment deflected at a 3 degree angle above the road surface so it started perfectly horizontal, in order to drop 13.7 feet (to the level of the manhole cover 83.3 yards or 250 feet) away, 2h/g=2x13.7/32=.856. So t=√.856=.925 seconds. So it would take .925 seconds to travel 250 feet. This means the fragment was travelling at an average speed of 250/.925 = 270 fps.  So that fragment has lost 1-270/2165 = 87.5% of its initial speed, and 98% of its initial energy.  If the bullet lost 98% of its energy in hitting the asphalt - ie. 600 joules of energy would be deposited in a tiny area of the asphalt - there would have to be a significant movement of asphalt around that area.

I said essentially level not level. I show the bullet fragment leaving Elm Street at a -2-degrees angle. This is one-degree above the road surface which slopes down at a -3-degrees angle. Also I have the fragment weighing 120-grains instead of the 161-grains it had as a whole bullet. The ballistic coefficient is a measure of efficiency the projectile has while in flight. The drastically lowered BC causes air friction to become a much larger factor in slowing the velocity over time and distance. I have circled the velocities and energy of the fragment beginning at the Elm Street mark and ending at the manhole cover for you. I believe that a significant amount of energy would have been lost due the fragmentation of the bullet when it hit Elm Street. And the amount of asphalt moved would depend on its hardness. Elm Street asphalt is relatively hard I do believe.

[Andrew Mason]

I believe that a significant amount of energy would have been lost due the fragmentation of the bullet when it hit Elm Street. And the amount of asphalt moved would depend on its hardness. Elm Street asphalt is relatively hard I do believe.

The hardness of the street determines whether, or how deep, the bullet penetrates. Asphalt, being a composite material, contains hard and soft particles. So the best way to determine what happens to asphalt struck by a 6.5 mm jacketed 10 gram bullet at 600 m/s striking asphalt at a 30 degree angle above horizontal would be experiment.

If it loses 98% of its energy in hitting the asphalt (which would be the case if the bullet fragments deflected from the asphalt were travelling at 270 fps, which is the case if a fragment dropped far enough to hit the manhole 250 distant) then it deposits all 98% of its incident energy in colliding with the asphalt: 1800 Joules.  1 gram of TNT releases 4184 Joules of energy when it explodes.  So this would be like about half a gram of TNT being detonated in the asphalt surface. [Note in my earlier post I said 600 Joules - I was thinking of speed not energy - since corrected]

[Charles Collins]

The hardness of the street determines whether, or how deep, the bullet penetrates. Asphalt, being a composite material, contains hard and soft particles. So the best way to determine what happens to asphalt struck by a 6.5 mm jacketed 10 gram bullet at 600 m/s striking asphalt at a 30 degree angle above horizontal would be experiment.

If it loses 98% of its energy in hitting the asphalt (which would be the case if the bullet fragments deflected from the asphalt were travelling at 270 fps, which is the case if a fragment dropped far enough to hit the manhole 250 distant) then it deposits all 98% of its incident energy in colliding with the asphalt: 1800 Joules.  1 gram of TNT releases 4184 Joules of energy when it explodes.  So this would be like about half a gram of TNT being detonated in the asphalt surface. [Note in my earlier post I said 600 Joules - I was thinking of speed not energy - since corrected]

Are you trying to calculate this in a vacuum? Or do you try to account for air resistance (drag)?

[Charles Collins]

The hardness of the street determines whether, or how deep, the bullet penetrates. Asphalt, being a composite material, contains hard and soft particles. So the best way to determine what happens to asphalt struck by a 6.5 mm jacketed 10 gram bullet at 600 m/s striking asphalt at a 30 degree angle above horizontal would be experiment.

If it loses 98% of its energy in hitting the asphalt (which would be the case if the bullet fragments deflected from the asphalt were travelling at 270 fps, which is the case if a fragment dropped far enough to hit the manhole 250 distant) then it deposits all 98% of its incident energy in colliding with the asphalt: 1800 Joules.  1 gram of TNT releases 4184 Joules of energy when it explodes.  So this would be like about half a gram of TNT being detonated in the asphalt surface. [Note in my earlier post I said 600 Joules - I was thinking of speed not energy - since corrected]

Robert Frazier testified that the energy of the Carcano bullet as it leaves the muzzle is calculated to be about 1676 foot-pounds on average. This online calculator appears to agree with Frazier’s figure. It also agrees with the shooterscalculator.com figure for a 120-grain projectile at 1500 fps.



This indicates to me that that particular fragment still contains about 36% of the initial energy of the bullet. There are obviously other fragments that make up the other 41-grains of that bullet that also still contain some of the initial energy. A significant amount of energy is required to fragment a 161-grain full metal jacketed bullet. So how much energy is transferred to the asphalt surface depends on the angle, the hardnesses of the two colliding bodies, etc.

[Andrew Mason]

Are you trying to calculate this in a vacuum? Or do you try to account for air resistance (drag)?

Air resistance is not a significant factor in determining how much energy was lost in striking the asphalt if you want it to drop to the manhole and then to Main St. near Tague.

The air does not affect how long it takes to drop.  So to drop 13.7 feet to strike the manhole will take just under a second.  So the average speed has to be 270 fps.   Now that could be because it started out faster than 270 fps and was movingly slower when it hit the manhole.

But you also want it to go another 141 feet and drop another 6.8 ft to the level of Main St.  That takes .65 seconds so its average speed is over that distance is 141/.65=216 fps.

 So before it hit the manhole it was going at least 216 fps and was travelling an average of 270 fps from where it hit the road. Assuming roughly constant resistance force v_avg= (v_f+ v_i)/2.  So if v_f=216 and v_avg=270 fps v_i=324 fps.  That is the maximum speed after hitting the road.

That means it retained only 324/2165=15% of it initial speed and .15²=2.25% of its energy. So that means it lost 97.75% of its muzzle energy in hitting the road.

[Charles Collins]

Air resistance is not a significant factor in determining how much energy was lost in striking the asphalt if you want it to drop to the manhole and then to Main St. near Tague.

The air does not affect how long it takes to drop.  So to drop 13.7 feet to strike the manhole will take just under a second.  So the average speed has to be 270 fps.   Now that could be because it started out faster than 270 fps and was movingly slower when it hit the manhole.

But you also want it to go another 141 feet and drop another 6.8 ft to the level of Main St.  That takes .65 seconds so its average speed is over that distance is 141/.65=216 fps.

 So before it hit the manhole it was going at least 216 fps and was travelling an average of 270 fps from where it hit the road. Assuming roughly constant resistance force v_avg= (v_f+ v_i)/2.  So if v_f=216 and v_avg=270 fps v_i=324 fps.  That is the maximum speed after hitting the road.

That means it retained only 324/2165=15% of it initial speed and .15²=2.25% of its energy. So that means it lost 97.75% of its muzzle energy in hitting the road.

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.”

https://www.sierrabullets.com/exterior-ballistics/2-0-the-ballistic-coefficient/