Begin with a diagram, modeling the hull as a circular cylinder and the con as a rectangular prism.

Use Eqs. 1 to determine the moments of inertia of the con about its center of gravity:

(Note, dm = ρ c dy dz)

Using a similar derivation, we get

These expressions for the moment of inertia could also be found from Appendix B, or a table of inertial properties.

Equations 2 give the products of inertia of the con about its center of gravity:

(Note, dm = ρ a dx dy)

Using a similar derivation, we get

     I_y'z' = I_x'z' = 0

Problem Simplification- Skew View

Problem Simplification - Side View

The expressions for the products of inertia could have been determined simply by noting that the product of inertia about a plane of symmetry is always zero.

Use the parallel axis theorem (Eqs. 3) to transfer the moment of inertia of the con about its center of gravity to the center of gravity of the sub:

     I_xx = I_x'x' + m_c (d_y² + d_z²)
          = 1/12 m_c(a² + b²) + m_c (d_y² + d_z²)

     I_yy = 1/12 m_c(a² + c²) + m_c (d_x² + d_z²)

     I_zz = 1/12 m_c(b² + c²) + m_c (d_x² + d_y²)

Use the parallel plane theorem (Eqs. 4) to transfer the product of inertia of the con about its center of gravity to the center of gravity of the sub:

     I_xy = I_x'y' + m_rd_xd_y

     I_yz = m_cd_yd_z

     I_xz = 0

Use Eq. 8 to express the angular momentum of the con about its center of gravity (Eq. 5) in scalar form:

     H_cg = H_xi + H_yj + H_zk

Here, we have

     H_x = -I_xxω_x

     H_y = I_yyω_y

     H_z = -I_yzω_y

Express the position and velocity of the con in rectangular components, and determine the moment of the linear momentum of the con about the center of gravity of the sub:

     ρ_con = d_yj + d_zk

     v_con = v_xj + v_yk

Use Eq. 7 to determine the angular momentum of the con about the center of gravity of the sub:

     H_A = ρ_con × m_cv_con + H_cg  

          = (-I_xxω_x - d_zv_y)i + (-I_yyω_y - d_zv_x)j
                     + (-I_yzω_y - d_yv_x)k