The Other Single Bullet Theory

[Tom Graves]

If you read the report you might remember this image showing a clear path for the bullet/fragment(s) at about Z133. This appears to be true for a period of time from shortly before to shortly after Z133.

No need to be snide, Chuck.

Why don't you like Roselle's and Scearce's hypothetical "Z-124"?

[Charles Collins]

No need to be snide, Chuck.

Why don't you like Roselle's and Scearce's hypothetical "Z-124"?

I wasn’t being snide. Whatever gave you that idea?

I didn’t say anything about the Roselle and Scearce idea. Why do you think I don’t like it?

[Andrew Mason]

A total velocity of 1500 fps with a downward component of 0.035 X 1500 = 52.5 fps. In 0.2419 seconds it would drop approximately the required 12.7-feet. The vertical component of the drag force would be acting to slow this descent while gravity would be acting to speed it up. Since the vertical component of the drag force would be stronger than gravity, the net effect would be deceleration of the descent. But, the point I have been making is that since there is a vertical velocity to begin with, the time required to accelerate from zero velocity (of a dropped bullet) is eliminated. Here is an image that shows the dropped bullet’s much smaller velocity and distance at the same 0.2419 seconds elapsed time. That is the running start I have been describing.

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.

[Tom Graves]

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[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.