The Other Single Bullet Theory

[Charles Collins]

"If you read the report you might remember..."

I know.

Confusing answer.

Regardless, why don't you just answer my question?

You need to consider re-wording your question. It is a loaded question as it is.

But I can probably save you some time and effort by telling you that my report on “the other single bullet theory” is about where that bullet went. The Roselle/Scearce idea is about when the shot occurred. Two different subjects.

I have given you an answer in the form of a range of time when there was a clear path for the bullet to cross Elm Street. Any time within that range is okay with me. I do believe that the Roselle/Scearce time is within the range that I already indicated.

[Charles Collins]

The running start is what reduces the fall due to gravity from 13.7 feet to 5 feet (or 4).  You can't count it twice.  The rate of vertical descent increases that vertical descent speed (v₀sin(2)) by gt where g=32 ft/sec² and t=time in seconds. The time, t, is horizontal distance/horizontal speed.

I am not counting anything twice. If the fragment already has a vertical velocity, gravity adds to it.

Regardless, the difference in our thinking is your lack of applying the air resistance to the fragment after it ricochets off of Elm Street.

We already established that simple geometry and a 2-degree downward ricochet angle reduces the amount of drop necessary to  approximately 47.64” (12.7’-8.73’). Your claim that the fragment has to be going at a drastically reduced speed to have the time necessary to drop that far does seem to be correct. However, your claim that the necessary velocity and energy loss would have to be due to the collision with the pavement on Elm Street is not correct. The reason it is not correct is that you have not taken into account the air resistance force on the fragment after it ricochets.

I believe that I have figured out a few things about the ballistic calculators. And I have found a different ballistic calculator that shows more information. Here are a couple of images that show how the air resistance affects things.


First, here is a table showing how quickly the velocity and energy values are reduced. The distance from the mark on Elm Street to the manhole cover we have been using using is 250’ (~83.33-yards). Based on that, and looking at the table, we can see that after a distance of only 16-yards the velocity is already reduced by about half of the initial 1500 fps. And the elapsed time is only about 0.05-seconds. Wow, that is due to the huge (your word) effect of the air resistance particularly during the trans-sonic stage. And this already solves the “mystery” of a lot of our differences.




Next are some graphs. I have determined that the “level” line at the top of the top left graph represents the line of sight. I have set the height of the “rifle” sights at zero above the bore. Therefore that line represents sighting through the bore of a rifle aimed at 2-degrees below level (the same as the direction of the ricochet off of Elm Street). So the necessary drop is ~47.64”. The ballistic calculator indicates a drop of ~46.8”. Therefore we are within an inch and that should be close enough for the purposes of this discussion.




I hope this helps you understand the importance of accounting for the air resistance. And that the needed reduction in the velocity and energy of the fragment is not totally due to a loss at the time of the collision with the pavement on Elm Street. If this were taking place in a vacuum, yes; but not in the atmosphere in Dealey Plaza on 11/22/63.

[Andrew Mason]

I am not counting anything twice. If the fragment already has a vertical velocity, gravity adds to it.

Yes.  But the time it takes to add a distance h is ½gt² so for an additional 4 feet, t=√(2x4/32)=.5 seconds.  That is how long it takes for the fragment to travel from the place where it strikes the asphalt to where it strikes the manhole (in your scenario). So its average speed over that 250 feet is 250/.5=500 fps. 

Regardless, the difference in our thinking is your lack of applying the air resistance to the fragment after it ricochets off of Elm Street.

We already established that simple geometry and a 2-degree downward ricochet angle reduces the amount of drop necessary to  approximately 47.64” (12.7’-8.73’). Your claim that the fragment has to be going at a drastically reduced speed to have the time necessary to drop that far does seem to be correct. However, your claim that the necessary velocity and energy loss would have to be due to the collision with the pavement on Elm Street is not correct. The reason it is not correct is that you have not taken into account the air resistance force on the fragment after it ricochets.


I believe that I have figured out a few things about the ballistic calculators. And I have found a different ballistic calculator that shows more information. Here are a couple of images that show how the air resistance affects things.

First, here is a table showing how quickly the velocity and energy values are reduced. The distance from the mark on Elm Street to the manhole cover we have been using using is 250’ (~83.33-yards). Based on that, and looking at the table, we can see that after a distance of only 16-yards the velocity is already reduced by about half of the initial 1500 fps. And the elapsed time is only about 0.05-seconds. Wow, that is due to the huge (your word) effect of the air resistance particularly during the trans-sonic stage. And this already solves the “mystery” of a lot of our differences.



Next are some graphs. I have determined that the “level” line at the top of the top left graph represents the line of sight. I have set the height of the “rifle” sights at zero above the bore. Therefore that line represents sighting through the bore of a rifle aimed at 2-degrees below level (the same as the direction of the ricochet off of Elm Street). So the necessary drop is ~47.64”. The ballistic calculator indicates a drop of ~46.8”. Therefore we are within an inch and that should be close enough for the purposes of this discussion.



I hope this helps you understand the importance of accounting for the air resistance. And that the needed reduction in the velocity and energy of the fragment is not totally due to a loss at the time of the collision with the pavement on Elm Street. If this were taking place in a vacuum, yes; but not in the atmosphere in Dealey Plaza on 11/22/63.

If the fragment:

• started out at 1500 fps, meaning the bullet would have lost only 44% of its incident energy of 1860 J (assuming the bullet struck asphalt at 2000 fps) - which still works out to 800 Joules - and
• averaged 500 fps over that 250 ft distance, and
• was able to drop a further 6.8 feet over the next 141 feet

what do you say was its speed over the last 141 ft distance? You have it starting out at about 170 fps based on your chart.  The drop of 6.8 feet occurs with the fragment starting out on a horizontal or low upward angle.  If it was a horizontal initial trajectory, that means it took  t=√(2x6.8/32)=.65 seconds to travel that 141 feet so its average speed was 141/.65=216 fps.  Pretty hard to average 216 fps when you start out at 170 fps and then keep slowing down.

[Tom Sorensen]

Only one of you guys can be right, but both could be wrong. The abundance of assumptions is staggering. Fascinating to watch two LNs battle it out!

[Tom Graves]

Only one of you guys can be right, but both could be wrong. The abundance of assumptions is staggering. Fascinating to watch two LNs battle it out!

It's almost as much fun as watching two zombified-by-KGB*-disinformation JFKA CT's squabble over their tinfoil-hat theories!

*Today's SVR and FSB

[Charles Collins]

Yes.  But the time it takes to add a distance h is ½gt² so for an additional 4 feet, t=√(2x4/32)=.5 seconds.  That is how long it takes for the fragment to travel from the place where it strikes the asphalt to where it strikes the manhole (in your scenario). So its average speed over that 250 feet is 250/.5=500 fps. 
If the fragment:

• started out at 1500 fps, meaning the bullet would have lost only 44% of its incident energy of 1860 J (assuming the bullet struck asphalt at 2000 fps) - which still works out to 800 Joules - and
• averaged 500 fps over that 250 ft distance, and
• was able to drop a further 6.8 feet over the next 141 feet

what do you say was its speed over the last 141 ft distance? You have it starting out at about 170 fps based on your chart.  The drop of 6.8 feet occurs with the fragment starting out on a horizontal or low upward angle.  If it was a horizontal initial trajectory, that means it took  t=√(2x6.8/32)=.65 seconds to travel that 141 feet so its average speed was 141/.65=216 fps.  Pretty hard to average 216 fps when you start out at 170 fps and then keep slowing down.

Now, you seem to be discussing the second ricochet (from the manhole cover apron to the curb on the south side of Main Street). We will never know the specific velocities and angles for certain. So, as I said before we have to assume some reasonable numbers. The 1500-fps velocity figure for the first ricochet is what seemed to me to be a reasonable guesstimate before I started my own experimentation. My experiments with 30-degrees ricochets have them retaining a much higher percentage of their velocities than that, closer to 95%. So this might surprise you but when we increase the velocity of the first ricochet to 1900-fps we get these results. These are within a reasonable range for this discussion, especially since we are not using exact measurements and only reasonable assumptions, etc.







For the second ricochet, if we assume an angle of positive 1-degree, simple geometry tells us the drop needs to increase by ~29.5” from ~81.6” to a total of ~111.1”. And if we a assume a velocity of ~187.8-fps, the fragment fragmented again and is now more aerodynamic, we get these results:






Again, well within a range suitable for this discussion.