1
$\begingroup$

I am designing a man-"portable" 30mm sniper rifle. Although it may be classified as a anti-materiel cannon.

I have the mass of the round in total, and the mass of the projectile leaving the barrel, as well as the pressure inside the barrel. I have the weapon's weight as well.

Here is my source : http://www.pmp.co.za/index.php?page=mediumcalibre6

Now, my problem.

  1. How do I calculate the force that this bullet imparts upon the rifle?

  2. If I use a spring-damper system to reduce the force on the shooter, how do I actually determine the force on the shooter?

If you do not find the need to supply me with a direct answer, which is understandable, could you please redirect me to a website/book that might help me?

$\endgroup$
1
  • $\begingroup$ You have the pressure in the barrel during the moment the round leaves the barrel? If not, what physical situation does your pressure pertain to? $\endgroup$ Mar 17, 2014 at 23:51

1 Answer 1

1
$\begingroup$

My initial guess for modelling the forces involved in firing a rifle would be to model the rifle, bullet and pressurized gas as a piston in which the gas would experience ideal adiabatic expansion until the bullet leaves the barrel. The force exerted by the pressure of the gas on the barrel in the radial direction will cancel out. The pressure does exerts a force on the riffle and the bullet, both proportional to the cross sectional area, $A$, of the barrel (the base of the cylindrical volume of the gas in the barrel). These forces will contribute to the acceleration of both objects, which will result in a increase of the enclosed volume of the gas and therefore a drop in pressure. When assuming the only force involved is due to pressure, the following differential equations can be composed: $$ p=p_0\left(1+\frac{A}{V_0}(x_b-x_r)\right)^{-\gamma}, $$ $$ \frac{\partial^2}{\partial t^2}x_b=\frac{pA}{m_b}=\frac{p_0A}{m_b}\left(1+\frac{A}{V_0}(x_b-x_r)\right)^{-\gamma}, $$ $$ \frac{\partial^2}{\partial t^2}x_r=-\frac{pA}{m_r}=-\frac{p_0A}{m_r}\left(1+\frac{A}{V_0}(x_b-x_r)\right)^{-\gamma}, $$ where $x_b$ and $x_r$ are the positions of the bullet and rifle, $m_b$ and $m_r$ the masses of the bullet and rifle, $p$ is the pressure of the gas, $p_0$ is the starting pressure of the gas, $V_0$ is the starting volume of the pressurized gas and $\gamma$ is the adiabatic index.

This model does has quite some simplifications, such as no friction between barrel and bullet, the air in front of the bullet in the barrel has no influence and no high pressure gas can escape between the bullet and the barrel.

However even this simplified model might be to complex and unnecessary. Because by knowing the muzzle velocity and using the conservation of momentum you that you would have to dissipate the same impulse as how much the bullet has gained, but distribute it over a longer period to reduce the the average force.

Ideally you would want some kind of friction between the moving and static part of the rifle. Because this would mean that there is a constant force acting between the two part and thus on you and therefore minimizing the maximum force experienced and the distance needed to transfer the impulse of the moving part to the user. However friction often means wear, so other systems might be a better option if smallest distance is not that important.

To answer your second problem I would have to simplify it. Say that after firing a bullet of mass $m_b$ at velocity $v_b$, the barrel (and other parts of the rifle moving along with it) with mass $m_r$ should have an equal momentum but in the opposite direction as that of the bullet. From this you can find the initial speed, $v_r$, of the moving part of the rifle: $$ v_r=v_b\frac{m_b}{m_r} $$ After this the spring and the damper will exert a force on this moving part, however the same force will also be exerted onto you: $$ F=m_r\ddot{x}=-kx-c\dot{x} $$ This equation has a known solution and by using the initial conditions $x(t=0)=0$ and $\dot{x}(t=0)=v_r$ you can find the equation of motion and thus the exerted force.

$\endgroup$
1
  • $\begingroup$ This is beautiful answer, thank you. I will post my findings as soon as possible for those who want to see the final result. $\endgroup$
    – 22134484
    Mar 18, 2014 at 6:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.