The Other Single Bullet Theory

[Charles Collins]

Where does your paper mention a 2 degree downward angle? Your charts indicate a slight upward angle. 

In any event it doesn’t matter much.  A fall of 5 feet requires .56 seconds so the average speed will be 250/.56=447 fps.  And let’s say it leaves the asphalt at 600 fps and when it strikes the manhole it is down to around 300 fps. At 600 fps leaving the asphalt it has (600/2000)^2=0.09=9% of the energy it had before hitting the asphalt.  So it lost 91% of its 1860 J=1693J. in striking the asphalt.  That still results in a lot of energy transferred to the asphalt.

Where does your paper mention a 2 degree downward angle? Your charts indicate a slight upward angle.

The below image is from the report. I posted it here in my first reply to your response. I pointed out the -2-degrees angle in my reply. I drew a red arrow on the graph to make it easy to find.




These graphs are designed to aid in sighting in guns and their sights, and for calculating hold-overs. They are not really made for ricochets. This is an improvised attempt at using one of the graphs for the purpose of demonstrating the idea of the other single bullet theory. I agree that if you are only looking at the graph itself it appears level, but that is not the angle that was input. This is apparently a limitation of the graph. Also, I may need to adjust for only a 5-foot drop, the jury is still out on that. I appreciate your input. I am still experimenting with actual results of actual ricochets.

You are still ignoring the fact that the vertical component of the direction (-2-degrees) gives the ricocheted fragment a “running start” compared to dropping a bullet from the same height.

[Andrew Mason]

Where does your paper mention a 2 degree downward angle? Your charts indicate a slight upward angle.

The below image is from the report. I posted it here in my first reply to your response. I pointed out the -2-degrees angle in my reply. I drew a red arrow on the graph to make it easy to find.




These graphs are designed to aid in sighting in guns and their sights, and for calculating hold-overs. They are not really made for ricochets. This is an improvised attempt at using one of the graphs for the purpose of demonstrating the idea of the other single bullet theory. I agree that if you are only looking at the graph itself it appears level, but that is not the angle that was input. This is apparently a limitation of the graph. Also, I may need to adjust for only a 5-foot drop, the jury is still out on that. I appreciate your input. I am still experimenting with actual results of actual ricochets.

You are still ignoring the fact that the vertical component of the direction (-2-degrees) gives the ricocheted fragment a “running start” compared to dropping a bullet from the same height.

I am not ignoring that at all.  That just gives it the constant rate of drop of v₀sin(2) resulting in a drop of 8.7 feet over 250 feet. The additional drop of 5 feet requires t=sqrt(2h/g)=(2x5/32)^(1/2)=0.56 seconds. That is what limits the speed of the fragment.

[Charles Collins]

I am not ignoring that at all.  That just gives it the constant rate of drop of v₀sin(2) resulting in a drop of 8.7 feet over 250 feet. The additional drop of 5 feet requires t=sqrt(2h/g)=(2x5/32)^(1/2)=0.56 seconds. That is what limits the speed of the fragment.

Sorry, but I disagree. The dropped bullet accelerates from zero velocity without any other significant force other than gravity. The ricochet has a "running start" due to its momentum. If its velocity is 1500-fps, then I believe that 2/90 = .0222222 X 1500 = 33.33 fps vertically is its "running start."

[Andrew Mason]

Sorry, but I disagree. The dropped bullet accelerates from zero velocity without any other significant force other than gravity. The ricochet has a "running start" due to its momentum. If its velocity is 1500-fps, then I believe that 2/90 = .0222222 X 1500 = 33.33 fps vertically is its "running start."

A projectile follows the following path defined by its horizontal (x) and vertical coordinates (y):

x=x₀ + v₀cos(θ)t
y= y₀ + v₀sin(θ)t - (1/2)gt²

See: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5

If you apply that to your situation,
x=x₀=0;
(1)  v₀cos(2)t=250 ft.

y=y₀=0;
v₀sin(2)t=250 sin(2)=8.7 ft
(2)  (1/2)gt²=16t²=5 ft

This last equation (2) means t= √(5/16)=0.56 seconds
Substituting into (1) gives v₀=250/(.56x.9994)=447 fps

[Charles Collins]

A projectile follows the following path defined by its horizontal (x) and vertical coordinates (y):

x=x₀ + v₀cos(θ)t
y= y₀ + v₀sin(θ)t - (1/2)gt²

See: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5

If you apply that to your situation,
x=x₀=0;
(1)  v₀cos(2)t=250 ft.

y=y₀=0;
v₀sin(2)t=250 sin(2)=8.7 ft
(2)  (1/2)gt²=16t²=5 ft

This last equation (2) means t= √(5/16)=0.56 seconds
Substituting into (1) gives v₀=250/(.56x.9994)=447 fps

That’s a very interesting and useful link you posted Andrew. Thank you. At first glance, it appears that the different calculators are intended for level or upwards directions. However, I tried a -2-degree launch angle in this one and a shorter time interval than you keep trying to use and got some results that appear to be “in the ballpark” of what is needed for the situation in Dealey Plaza. And hopefully, the numerical results in this image illustrate what I have been trying to say.



Although I did indicate the ballistic graphs in my report were not intended to be totally accurate (only to show the concepts), I think the difference between what you are calculating and what I am coming up with is larger than it might should be. So, I think that one of us is doing something wrong. I want the report to be honest and reasonably accurate. So your input is valuable to me and appreciated. I plan to double check the elevations I used and look for any errors on my part. Please let me know what you think about the above calculator results. Thanks again.

[Charles Collins]

A projectile follows the following path defined by its horizontal (x) and vertical coordinates (y):

x=x₀ + v₀cos(θ)t
y= y₀ + v₀sin(θ)t - (1/2)gt²

See: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5

If you apply that to your situation,
x=x₀=0;
(1)  v₀cos(2)t=250 ft.

y=y₀=0;
v₀sin(2)t=250 sin(2)=8.7 ft
(2)  (1/2)gt²=16t²=5 ft

This last equation (2) means t= √(5/16)=0.56 seconds
Substituting into (1) gives v₀=250/(.56x.9994)=447 fps

I did find that in reviewing the elevations in Dealey Plaza and using interpolation between two measured points’ elevations, instead of just the closest point, the difference in elevation between the bullet strike and the manhole cover is actually closer to 12.7-feet instead of the 13.7-feet originally indicated in my report.

I also want to point out that the equations and calculators on the webpage that you linked to do not include anything for air resistance. You still have not addressed the air resistance that the tumbling fragment would have. You did address the air resistance of the bullet before it struck Elm Street, but nothing at all afterwards. Here’s a blurb from the webpage you linked to and that contains the equations.

“ The motion of an object under the influence of gravity is determined completely by the acceleration of gravity, its launch speed, and launch angle provided air friction is negligible.”

I think that the air resistance of a tumbling fragment is significant, not negligible. Even you acknowledged that the air resistance is “huge”. We need to take the air resistance into consideration. I think that the ballistic calculator that I used in my report allows this. I used a very low ballistic coefficient input into that calculator to account for high air resistance. That is where I believe the difference in our results lies.

[Andrew Mason]

I did find that in reviewing the elevations in Dealey Plaza and using interpolation between two measured points’ elevations, instead of just the closest point, the difference in elevation between the bullet strike and the manhole cover is actually closer to 12.7-feet instead of the 13.7-feet originally indicated in my report.

I also want to point out that the equations and calculators on the webpage that you linked to do not include anything for air resistance. You still have not addressed the air resistance that the tumbling fragment would have. You did address the air resistance of the bullet before it struck Elm Street, but nothing at all afterwards. Here’s a blurb from the webpage you linked to and that contains the equations.

“ The motion of an object under the influence of gravity is determined completely by the acceleration of gravity, its launch speed, and launch angle provided air friction is negligible.”

I think that the air resistance of a tumbling fragment is significant, not negligible. Even you acknowledged that the air resistance is “huge”. We need to take the air resistance into consideration. I think that the ballistic calculator that I used in my report allows this. I used a very low ballistic coefficient input into that calculator to account for high air resistance. That is where I believe the difference in our results lies.

Air resistance in the direction of travel is huge. So air resistance to horizontal movement is significant especially at high speeds around 2000 fps. Gravity is not in the direction of travel.  There is no difference, as you acknowledge, in time to drop a given distance for a bullet fired from a gun and a bullet that is just dropped.  Air resistance for the dropped bullet is negligible over a drop distance that we are talking about here. That is why air resistance does not affect the time required for a bullet to drop such a short distance.