useless geek trivial: 1234567890 day

Let .... Ma = 800 ;;; Bullet mass in grains Mv = 2895 ;;; Muzzle velocity in ft/s Dis = (2500*3) ;;; 2500 yards is = 7500 feet distance SG = 32.174 ;;; rough idea of earth standard gravity in ft/s Grav = SG * 2 ;;; convenience for: 64.34800 gip = 7000 ;;; number of grains in a U.S. pound (Ma * (Mv * Mv)) / (Grav * gip) = 14885.1779 foot pounds out the muzzle ;;; that is: ;;; ((800*(2895*2895)) / ((32.174*2)*7000) = 14885.1779 If the bullet maintained a constant speed of 2895 ft/s, it would hit the target 2500 yards away in approximately 3 seconds and hurt like a son of a gun. Now to try and figure out how much is lost by traveling 2500 yards down range 0/ figure out up how to get energy from muzzle velocity and mass, see above 1/ How do we convert this to foot pounds... see above 2/ figure out what has changed 2500 yards down range we are here 3/ found out about a Ballistic Coefficient (BC), seems it is equal to mass / (drag coefficient * cross sectional area), which is equal to (average density * body length) / drag coefficient. 4/ How do we find the sectional density of our bullet? Cal = 0.50 ;;; any 12.7x99mm bullet is ~1/2 inch in diameter mip = Ma / gip ;;; that is 800 grains / 7000 = 0.114285714 pounds SD = mip / (Cal * Cal) ;;; which is is (800 / 7000) / 0.50 squared ;;; and and should happen to be equal to 0.457142857 f = form factor Cd = Coefficient of Drag for OUR bullet (SD / f) = (mip / (Cd * (Cal * Cal))) 5/ Find out what the heck a drag coefficient is.... along with the bullets form factor. A bullets form factor is Cd of our bullet / Cd of standard 'G' bullet, the Cd of a standard G1 projectile appears to be 0.5190793992194678, but hey... I need to skimp on some of this math. Thank you NASA: "The drag coefficient is a number which aerodynamicists use to model all of the complex dependencies of drag on shape, inclination, and some flow conditions. The drag coefficient Cd is equal to the drag D divided by the quantity: density r times reference area A times one half of the velocity V squared. Cd = D / (.5 * r * V^2 * A)". r = Ma / 3.8976378 ;;; we need to figure out what D and A are: but I don't have the bullet ;;; handy to measure its shape, nor weather conditions at the fire site or ;;; anything.... lol. f = (D / (0.50 * r * (Mv * Mv) * A ) / 0.5190793992194678 BC = SD / f so: (SD / f) = (mip / (Cd * (Cal * Cal))) is: ? But I'll take my best guess.... which is that the bullet would probably drop about 1.5 to 2 inches along the way, and hit the poor guy about 4 seconds later traveling at around 1200 ft/s Which would be: (Ma * (1200*1200) / ((32.174*2) * gip)) = ~2550 foot-pounds But hey.... I suck at math and physics, and I sure ain't been to the marines scout sniper school or anything like that lol. Either way, the guy will have a NICE sized barn door through whatever body part you aim at references: http://en.wikipedia.org/wiki/Standard_gravity http://en.wikipedia.org/wiki/Ballistic_coefficient http://www.shootingsoftware.com/coefficients.htm http://www.grc.nasa.gov/WWW/K-12/airplane/shaped.html