Flight Centrifuge

The flight centrifuge

A flight centrifuge was designed  to simulate the  flight in a space vehicle. The arm which is $d=40 ft$, rotates about axis A-A. The cockpit rotates itself about the axis C-C. Inside the cockpit the seat rotates about the axis B-B.  Motors control these rotations and different situations of flight are simulated.  The distances are given in the figure.

1. What is the acceleration of the pilot head about the three given axes of rotation A-A, B-B, C-C in this flight centrifuge.

2. If $\omega_1 = 2 rad/s$, $\omega_2 = 3 rad/s$, $\omega_3 = 4 rad/s$, $\dot \omega_ 1 = 3 rad/s^2$, $\dot \omega_2 = 2 rad/s^2$, and $\dot\omega_3 = 4 rad/s^2$, what is the torque of the seat about the CM of the man?  The weight of the man is 800 N and its radii of gyration in the seat are: $k_x = 600mm$, $k_y = 500mm$, and $k_z =150mm$.

Flight Centrifuge


1) For the flight centrifuge if $\phi$ is the angle of rotation with respect to CC axis, $\theta$ is the angle of rotation with respect to BB axis and $\psi$ is the angle of rotation with respect to AA axis then the correspinding angular speeds are

$\omega_1=\omega_1 =d(\phi)/dt$

$\omega_2 =d(\theta)/dt$


and the corresponding vector has the components

$\omega = (\omega1, \omega2, \omega3)$


The accelerations are

$\epsilon_1= d(\omega1)/dt = d^2(\phi)/dt^2$

$\epsilon2 =d(\omega2)/dt = d^2(\theta)/dt^2$

$\epsilon3=d(\omega3)/dt =d^2(\psi)/dt^2$

the corresponding vector has the components

$\epsilon = (\epsilon_1, \epsilon_2, \epsilon_3)$

The acceleration of the head is then

$a = a(radial) + a(tangential) = \omega \times (\omega \times R) + \epsilon \times R$

where $\times$ is the cross product.

$R = (3, 3, 40)$   since $\psi$ is the angle with respect to CC axis (as given in text)

$\omega \times R = \begin{vmatrix}

i & j & k\\

\omega_1 & \omega_2 & \omega_3\\

3 &  3& 40

\end{vmatrix} = i(40\omega_2 -3\omega_3) +j(3\omega_3 -40\omega_1)+k(3\omega_1 -3\omega_2)$

$\omega \times (\omega \times R) = \begin{vmatrix} i &  j& k\\  \omega_1 & \omega_2 &\omega_3 \\  40\omega_2-3\omega_3 & 3\omega_3-40\omega_ 3&\omega_1-3\omega_2  \end{vmatrix}=$

$=i*[(3\omega_1-3\omega_2)*\omega_2 -\omega_3*(3\omega_3-40\omega_1)] +…$

$\epsilon \times R = \begin{vmatrix}

i & j & k\\

\epsilon_1 &\epsilon_2  &\epsilon_3 \\

3 & 3 & 40

\end{vmatrix} = i*(40*\epsilon_2 -3*\epsilon_3) +j*(3\epsilon_3 -40\epsilon_1) +k*(3\epsilon_1-3\epsilon_2)$


2) To compute the torque one needs to compute the moments of inertia with respect to axes CC, BB , AA and the other components of the inertia tensor

$I(B) =I_{xy} = mx^2 = G/g*k_x^2 =800/9.8*0.6^2 =29.39 kg*m^2$

$I(C)=I_{yy} =my^2 =800/9.8*0.5^2 =20.4 kg*m^2$

$I(A) =I_{zz} =m*z^2 = 800/9.8*0.15^2 =1.837 kg*m^2$

$I_{xy}= I_{yx} = m*(xy) = G/g*k_x*k_y = 800/9.8*0.6*0.5 =24.49 kg*m^2$

$I_{xz} =I_{zy}= m*(xz) =800/9.8*0.6*0.15 =7.35 kg*m^2$

$I_{yz} =m*(yz) =800/9.8*0.5*0.15 =6.12 kg*m^2$

Thus the inertia matrix is

$[I] = \begin{vmatrix}

Ixx &Ixy  &Ixz \\

Iyx & Iyy &Iyz \\

Izx &Izy  &Izz


The torque is

$T = [I]*\epsilon + \omega \times ( [I]*\omega )$

where $\epsilon = (\epsilon_1, \epsilon_2, \epsilon_3)$

and $\omega =(\omega_1, \omega_2, \omega_3)$

(see Wiki)