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. 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, this image illustrates 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.