Is my simplified bullet drop equation accurate for long-range shooting?

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SUMMARY

The simplified bullet drop equation presented in the discussion is accurate for calculating the drop of a bullet over long distances. The equation derived is d_{drop}=\frac{gd_{target}^{2}}{2v_{bullet}^2}, where g represents gravitational acceleration, d_{target} is the distance to the target, and v_{bullet} is the bullet's velocity. For a rifle with a muzzle velocity of 2850 feet per second, the shooter must aim approximately 55 feet above a target located one mile away to compensate for bullet drop. This model effectively illustrates the relationship between distance, bullet velocity, and gravitational effects on trajectory.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion.
  • Familiarity with gravitational acceleration (g = 32.2 ft/s²).
  • Knowledge of bullet velocity and its measurement in feet per second.
  • Basic algebra skills for manipulating equations.
NEXT STEPS
  • Research the effects of air resistance on bullet trajectory.
  • Learn about advanced ballistic calculations using software like Hornady Ballistics Calculator.
  • Study the impact of altitude and temperature on bullet drop.
  • Explore the use of ballistic coefficients in long-range shooting.
USEFUL FOR

This discussion is beneficial for long-range shooters, ballistics enthusiasts, and anyone interested in understanding the physics of projectile motion and bullet drop calculations.

warfreak131
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I was wondering if you guys could see if my equation on bullet drop is correct. I know there are more factors than what I am presenting, but this is only a simplified version...

The distance an object with 0 initial vertical velocity falls is d_{drop}=\frac{1}{2}gt^{2}, and t=t_{flight}, the total time the bullet is in flight.

The velocity of the bullet is v_{bullet}=\frac{d}{t} and in this case, d=d_{target}, and t=t_{flight}=\frac{d_{target}}{v_{bullet}}.

Combine these two and you get: d_{drop}=\frac{gd_{target}^{2}}{2v_{bullet}^2}

Is this correct?
 
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Correct.
 
cool, so just a fun fact using this model, if you had to hit a target a mile away with a rifle that shoots at 2850 feet per second from the muzzle (assuming velocity stays constant along the path), you would have to shoot about 55 feet above the target
 

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