The Other Single Bullet Theory

Users Currently Browsing This Topic:
0 Members

Author Topic: The Other Single Bullet Theory  (Read 18363 times)

Offline Tom Sorensen

  • Full Member
  • ***
  • Posts: 134
Re: The Other Single Bullet Theory
« Reply #35 on: March 12, 2025, 06:20:03 PM »
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!

Online Tom Graves

  • Hero Member
  • *****
  • Posts: 3495
Re: The Other Single Bullet Theory
« Reply #36 on: March 12, 2025, 06:49:29 PM »
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

Online Charles Collins

  • Hero Member
  • *****
  • Posts: 4402
Re: The Other Single Bullet Theory
« Reply #37 on: March 12, 2025, 07:57:49 PM »
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.