Ballistics differential equations

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SUMMARY

The discussion centers on the application of differential equations to model the drag forces acting on bullets. The key equation presented is a=(D*V^2*B*A)/(2*M), where D represents air density, V is velocity, B is the ballistic coefficient, A is the cross-sectional area, and M is mass. The user seeks to graph acceleration versus time and subsequently integrate to obtain velocity versus time and displacement versus time. The integration process is outlined, leading to the equation -1/v = (DBA/2M)t + C, where C is the constant of integration.

PREREQUISITES
  • Understanding of basic physics concepts related to drag forces.
  • Familiarity with differential equations and integration techniques.
  • Knowledge of ballistic coefficients and their significance in projectile motion.
  • Ability to interpret and manipulate mathematical equations for graphing purposes.
NEXT STEPS
  • Research methods for graphing differential equations using tools like MATLAB or Python's Matplotlib.
  • Study the integration of multi-variable equations in calculus.
  • Explore the impact of air density on bullet trajectory and performance.
  • Learn about numerical methods for solving differential equations in physics simulations.
USEFUL FOR

Students in physics or engineering, mathematicians interested in applied mathematics, and anyone involved in ballistics or projectile motion analysis.

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i have wandered past my math background into drag equations on bullets. i found the equation that relates acceleration and speed of the bullet. I do not know how to integrate multi-varible equations. How to i get the equation so it can be graphed on an acceleration vs time? I am thinking from there i can integrate again to velocity vs time then again to displacement vs time. I am not really interested in learning how to do this right now, since i will learn it in class later, i just need the answer to complete my program. The equation i have is
a=(D*V^2*B*A)/(2*M)

D=air density
V=velocity
B=ballistic coefficient
A=cross sectional area
M=mass
Thanks for your time
 
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So you have
[tex]\frac{dv}{dt}= \frac{DBA}{2M}v^2[/tex]

That separates as
[tex]\frac{dv}{v^2}= \frac{DBA}{2M}dt[/tex]

which can be integrated to
[tex]-\frac{1}{v}= \frac{DBA}{2M}t+ C[/tex]
where C is a constant of integration.
 

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