The Other Single Bullet Theory

[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.

[Charles Collins]

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.

Gravity is not in the direction of travel.

What!?     Gravity is downward toward the center of the earth. The ricocheted fragment has a downward component right from the time it leaves the Elm Street surface. That downward component increases over time. Especially in the second half of the calculated trajectory (parabolic curve). So I think that you are just plain wrong.

Some information from “Modern Practical Ballistics” by Arthur J. Pejsa:

Back in the 1730s Benjamin Robbins invented the ballistic pendulum and showed that the drag force (air resistance) is fifty to one hundred times as great as the force of gravity. Additionally, at speeds approaching the speed of sound, there is a rapid and dramatic twofold to threefold increase in air drag; between approximately 1000 and 1200 fps, air drag is found to increase with about the fifth or sixth power of the air speed rather than with the square, or second power as it does at lower speeds. This is the infamous “sonic barrier” we hear of.

I am currently building and experimenting with a DIY ballistic pendulum. So far, ~30-degrees ricochets off of a concrete surface appear to retain the vast majority of their initial velocities. This agrees with the results of the high speed filmed 30-degrees ricochets that I included screenshots from in my report. You can search and find and watch that video on YouTube if you use the information at the bottom of the last screenshot. They show the various penetration depths of the various angle ricochets in ballistic gel.

I intend to continue to study and learn more from Pejsa’s book and my experiments. So far I have to disagree with your assessment and claims based on what I have learned so far. Again, I do appreciate your input it tends to give me incentive to learn more about these things.

[Andrew Mason]

Gravity is not in the direction of travel.

What!?     Gravity is downward toward the center of the earth. The ricocheted fragment has a downward component right from the time it leaves the Elm Street surface. That downward component increases over time. Especially in the second half of the calculated trajectory (parabolic curve). So I think that you are just plain wrong.

Some information from “Modern Practical Ballistics” by Arthur J. Pejsa:

Back in the 1730s Benjamin Robbins invented the ballistic pendulum and showed that the drag force (air resistance) is fifty to one hundred times as great as the force of gravity. Additionally, at speeds approaching the speed of sound, there is a rapid and dramatic twofold to threefold increase in air drag; between approximately 1000 and 1200 fps, air drag is found to increase with about the fifth or sixth power of the air speed rather than with the square, or second power as it does at lower speeds. This is the infamous “sonic barrier” we hear of.

I am currently building and experimenting with a DIY ballistic pendulum. So far, ~30-degrees ricochets off of a concrete surface appear to retain the vast majority of their initial velocities. This agrees with the results of the high speed filmed 30-degrees ricochets that I included screenshots from in my report. You can search and find and watch that video on YouTube if you use the information at the bottom of the last screenshot. They show the various penetration depths of the various angle ricochets in ballistic gel.

I intend to continue to study and learn more from Pejsa’s book and my experiments. So far I have to disagree with your assessment and claims based on what I have learned so far. Again, I do appreciate your input it tends to give me incentive to learn more about these things.

At a 2 degree downward slope, the downward component initially is -.035 times the horizontal component. So even at 2000 fps the downward speed would be no more than -70 fps.  And after .1 second it would be -73.2 fps.  (v_y=v_0y-gt)

 If there is more drag it will take longer to fall a given distance so it would have to be moving horizontally at an even slower speed for it to fall 5 feet (or 4 feet now with your revision) by taking into account vertical drag due to air resistance.

Projectile motion is fairly basic if one does not take into account air resistance. That’s why it is a first year physics subject.  Fluid dynamics is typically a third or fourth year subject as it is very complicated.

[Charles Collins]

At a 2 degree downward slope, the downward component initially is -.035 times the horizontal component. So even at 2000 fps the downward speed would be no more than -70 fps.  And after .1 second it would be -73.2 fps.  (v_y=v_0y-gt)

 If there is more drag it will take longer to fall a given distance so it would have to be moving horizontally at an even slower speed for it to fall 5 feet (or 4 feet now with your revision) by taking into account vertical drag due to air resistance.

Projectile motion is fairly basic if one does not take into account air resistance. That’s why it is a first year physics subject.  Fluid dynamics is typically a third or fourth year subject as it is very complicated.

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.