Advertisement
Guest User

Untitled

a guest
Oct 4th, 2017
145
0
Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
text 4.09 KB | None | 0 0
  1. >>10710351
  2. As promised in >>10710575 here is some 6th grade math to give us some perspective.
  3.  
  4. Assumptions:
  5. Height of 32nd floor: 380ft
  6. Distance from corner of Mandalay Bay to centroid of venue: 1400ft
  7. Dimensions of venue: 1250ft x 580ft
  8. People in attendance: 22,000, all male of average height and uniformly distributed
  9. Effects of bullet drop and ambient conditions are not significant (bullets behave like laser-beams)
  10.  
  11. Scenario 1:
  12. '''Paddock aims at the center of the venue and lays scunion'''
  13.  
  14. We'll imagine a cone of fire that makes an ellipse where it intersects the ground, like a spotlight. If bullets are flying randomly, the area of that ellipse which contains a person would count toward probability of a hit and empty space that of a miss.
  15.  
  16. Let's assume Paddock can fire continuously within a 15 degree arc in the horizontal and vertical direction (to visualize this, hold your arm out and make evil-eye sign with your hand. Index to little finger is about 15 degrees. This is pretty conservative estimate for automatic fire)
  17.  
  18. ''Finding the end points of our ellipse''
  19. The angle measured from Paddock's direction of aim to the ground would be:
  20. >Tan^-1(1400ft /380ft) = 75 degrees
  21. Therefore the nearest point of impact measured horizontally from the Mandalay Bay is:
  22. >Tan(75-[15/2])*380ft = 920ft
  23. The farthest is:
  24. >Tan(75+[15/2])*380 = 2890ft
  25. Whoops, so this goes about 1000ft beyond our venue according to Google maps. The center of this ellipse would be at 1910ft horizontal feet from Mandalay Bay. That's very close to the distance to the far edge of the venue, so, for the sake of simplicity, let's assume that half of the bullets don't even land in the venue at all and don't hit anyone.
  26.  
  27. The width of our ellipse (lateral distance from point of aim to point of impact) is symmetrical, so we can imagine an isosceles triangle with a height of 1910ft and an angle of 15 degrees we get:
  28. >2 x (tan(15/2) x 1910ft) = 500ft
  29.  
  30. The area of an ellipse is pi times the product of the major and minor radii:
  31. > pi x (2890ft - 910ft)/2 x 500ft/2 = 777550sqft
  32. So this is the area the bullets might land in
  33.  
  34. The density of people within the venue is
  35. > 22,000 people / (1250ft x 580ft) = 0.03 people/sqft (one person per 33sqft)
  36. Only half of our bullets even land in the venue, so we take half that area and multiply that by the density of attendees.
  37. > 777550sqft / 2 x 0.03 people/sqft = 11660 people
  38. If the average male is about 18in from shoulder to shoulder and we assume, looking directly downward, he has a circular profile, the average attendee takes up an area of:
  39. >pi x (1.5ft / 2)^2 = 1.8sqft
  40. Multiplying by the number of people we have:
  41. >1.8sqft x 11660people = 20,990 sqft
  42. This is our occupied area. So if a bullet from the Mandalay could land anywhere within our calculated ellipse we would expect the odds of a hit to be:
  43. >20990sqft / 777550sqft = 0.027
  44. 2.7% of rounds should hit a person under these assumptions.
  45.  
  46. Borrowing our earlier assumptions:
  47. > 600rds/min x 50% firing time x 10 min = 3000rds
  48. > 3000rds x 2.7% = 81 hits
  49.  
  50. Just a wee bit shy of the 500+ casualties.
  51.  
  52. This is a pretty simple model, but we can conclude the one-shooter theory is plausible under the conditions that Paddock spent a full 5 minutes going cyclic, he had excellent (but possible) control of recoil, nearly all hits were fatal, and almost all non-fatal injuries were not related to gunshot wounds
  53.  
  54. Scenario 2:
  55. '''Paddock is a terminator from the future with perfect accuracy'''
  56.  
  57. If Paddock were to perfectly aim each shot and achieve the 3MOA accuracy a Stoner rifle is typically capable of, we would would expect at this distance (485yds) to have a point of impact anywhere within a radius of:
  58. >3MOA x 485yds x 1in/100yds/MOA = 14.6in
  59.  
  60. The distance from center of mass to the shoulder on an average man is about 9in. If we imagine a lethal hit on a man being anywhere within a circle shoulder-to-shoulder centered on the heart, then we would expect the hotel-terminator to hit:
  61.  
  62. >pi x 9in^2 / pi x 14.6in^2 = 0.38
  63.  
  64. 38% of the time. This would translate into about 1140 hits if each shot were perfectly aimed at center of mass.
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement