The Governing Equations of the Typhoon Submachine Gun

A few months ago I downloaded Crysis 3 off the XBLA because it was on sale for cheap. One of the new weapons in the game is a toy called the “Typhoon.” I will let the video speak for itself:

This weapon is referred to technically as a “stacked projectile weapon”, where basically, the magazine *is* the weapon. The technology has been around for a while, pioneered by an Australian company called Metal Storm (and humorously, were nearly stolen away by the Red Chinese.) I’ll let this promotional video explain it for you:

How Feasible is the Typhoon as an Infantry Weapon?

I wondered after beating the game just how practical my favorite weapon would be in reality. My biggest concerns were that it would be too heavy, or that the recoil would be completely unmanageable. In the game, your character has many abilities, including super-strength. However, the Typhoon is wielded by normal human enemies. The weapon would have to have recoil manageable enough for a regular soldier to handle, and not weigh so much as to be unwieldy and burdensome.

Like I said in my other post of what chemical engineers do, our first order of business is to write and solve the equations of mass, energy, and momentum balance. Those results are presented in this blog post.. The result is quite similar to the rocket equation, which makes sense – the Typhoon is basically a handheld rocket motor that exhausts lead bullets instead of gases.

For the next article on this subject, I’ll go into some specific numbers and possibly some multiobjective optimization results, depending on how much time I have next week.

Mass balance equation

\frac{d}{dt}(M_{empty} + (m_b + m_p)n) = -R(m_b + m_p)

\frac{dn}{dt} = -R

Momentum balance equation

\frac{d}{dt}(M_{total}v_r) = R(v_b - v_r)(m_b + m_p)

\frac{d}{dt}([ M_{empty} + (m_b + m_p)n]v_r) = R(v_b - v_r)(m_b + m_p)

\frac{dv_r}{dt}=\frac{Rv_b(m_b + m_p)}{ M_{empty} + (m_b + m_p)n}

Energy balance equation

\frac{d}{dt}(\frac{1}{2}M_{total}v_r^2+nm_pE_p)= - \frac{1}{2}Rm_b(v_b - v_r)^2 - \frac{1}{2}Rm_p(v_b - v_r)^2

After about a page of derivation, the energy balance reveals:

v_b = \sqrt{\frac{2m_p}{m_b + m_p}E_p}

We can substitute this result for v_b into our momentum balance differential equation. This equation can subsequently be integrated using u-substitution to yield our final result:

v_r(t) = v_b \ln(\frac{M_{empty} + (m_b + m_p)N_{fm}}{M_{empty}+(m_b+m_p)(N_{fm}-Rt)})

Nomenclature

M_{empty} Mass of the Typhoon when it has an empty magazine.

m_b – Mass of a single bullet

m_p – Mass of propellant used to fire a single bullet

R – The firing rate

M_{total} Total mass of human, empty Typhoon, and current magazine weight.

v_b – Velocity of a single bullet.

v_r – Recoil velocity

n – Number of bullets remaining in magazine

N_{fm} Total number of bullets in a full magazine.

E_p Energy content of the propellant, in Joules per kilogram

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s