Air resistance in relation to ballistics

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SUMMARY

The discussion centers on the effects of air resistance on the flight time of a .50 caliber sniper rifle bullet over a distance of two miles. Calculations indicate that, neglecting air resistance, the bullet would take approximately 3.857 seconds to reach the target at an initial velocity of 2,800 feet per second. To achieve a flight time exceeding 5 seconds, the bullet's final velocity would need to drop to 1,040 feet per second. The conversation also highlights the complexity of calculating drag forces on bullets, emphasizing the need for drag coefficients and the limitations of simple kinematic equations in high-speed ballistics.

PREREQUISITES
  • Understanding of basic kinematics and equations of motion
  • Familiarity with the drag equation: F = 1/2 ρ v² A C_d
  • Knowledge of bullet dynamics and terminal velocity concepts
  • Awareness of external ballistics principles
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  • Research the drag coefficient for various bullet shapes and sizes
  • Learn about the impact of air density on projectile motion
  • Study external ballistics calculations for supersonic projectiles
  • Explore methods to calculate wind resistance without calculus
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Ballistics enthusiasts, firearms experts, physics students, and anyone interested in the effects of air resistance on projectile motion.

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Having this discussion on another board, basically stemmed from one member saying that a bullet would be in flight for "over 5 seconds"in response to something else someone said about a 0.50 caliber sniper rifle firing at a target 2 miles away.

I figured that negating air resistance, the time would be 3.857 seconds (using Vf^2 = Vi^2 + 2ad then putting finding average velocity (Vf + Vi)/2 and substituting that into v = d/t) "d" is two miles, or 10,560 feet, "Vi" is 2,800 feet per second, and "a" is -32.15 feet per second.

So assuming that was right, I found that for the time to be "over 5 seconds" (I used 5.5), the final velocity would have to be 1,040 feet per second (using v = d/t, solving for v as average velocity, multiplying Va by two and subtracting initial velocity to find the new Vf) is that correct?

Anyway, none of that really matters, just background, my question is, how do you take wind resistance into account in kinematics, and how high would the wind speed have to be to cause a 63% loss of velocity? I'd preferably like to do this without calculus.

Thanks in advance.
 
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Wind isn't needed to slow the bullet - just it's motion through the air is enough.
The force on an object moving through air (at height speed)
F = 1/2 \rho v 2 A C d

\rho is the density of air
v is the velocity
A is the cross section area
C d depends on the shape, for a bullet it's around 0.3
 
Mythbusters did a "can a falling bullet kill you" episode where they found the terminal velocity of a bullet to be around 100fps. You can plug that into the drag equation to find the drag coefficient (the bullet tumbled in their demo, but at least that gives you an upper bound). I think you'll find that the drag needed to explain the 5.5 second flight time is absurdly high.
 
Bullets from most rifles travel at super sonic speeds. At these speeds, the math to calculate the drag is so complex that tables of coefficients are required to do the ballistics calculations.

http://en.wikipedia.org/wiki/External_ballistics

In the case of 50 caliber sniper rifles, the muzzle velocity is very high, enough that the bullet will still be well above supersonic at impact to improve the accuracy.

russ_watters said:
Mythbusters did a "can a falling bullet kill you" episode where they found the terminal velocity of a bullet to be around 100fps.
Must have been a very small caliber weapon. Hunters sometimes get pelted by tiny birdshot falling from above without much harm, but people have died from falling bullets. I'm pretty sure that a 50 cailiber bullet would have a much higher terminal velocity than 100fps.
 
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