# Reaction Forces and Apparent Thrust in Dual Oscillating Control Moment Gyroscopes
## 2 Materials and Methods
### Dynamic Equilibrium: Newton's Laws
#### 2.6.1 Approximate Model: Piecewise-Constant Angular Velocity  
In this case, by virtue of Equation (11), and in conjunction with \(\ddot{\theta}=\dot{\omega}=0\), Equation (12) becomes  
\[\overrightarrow{F}_{\rm excite}=\Bigl{(}-F_{g}+2m\omega^{2}r\sin\theta \Bigr{)}\overrightarrow{j}\,, \tag{13}\]  
where \(\omega=\omega_{\rm servo}=\dot{\theta}\) represents the speed of the servomotor and \(r\) is the eccentricity of each rotating mass. Then, the torque equilibrium by the front force (\(F_{\rm front}\)) and the rear force (\(F_{\rm rear}\)) gives  
\[-F_{\rm front}l_{1}+F_{\rm rear}l_{2}+\tau_{\rm cart}=0\,, \tag{14}\]  
while the force equilibrium in the vertical \(y\)-direction implies  
\[F_{\rm front}+F_{\rm rear}+F_{\rm excite}=0\,. \tag{15}\]  
When solving the linear system of Equations (14) and (15), the analytical expressions for the two reaction forces become  
\[F_{\rm front}=-\frac{F_{\rm excite}l_{2}+\tau_{\rm int}}{l_{1}+l_{2}}\,, \tag{16}\]  
and  
\[F_{\rm rear}=-\frac{F_{\rm excite}l_{1}-\tau_{\rm int}}{l_{1}+l_{2}}\,. \tag{17}\]  
and  
\[F_{\text{rear}}=\frac{l_{1}}{\underbrace{l_{1}+l_{2}}_{\text{Static term}}F_{g}}+ \underbrace{\left[\frac{l_{1}}{l_{1}+l_{2}}\left(-2m\omega^{2}r\sin\theta \right)\right]}_{\text{Dean-drive term}}-\underbrace{\frac{l_{1}+l_{2}}{L_{ \text{rotor}}\omega\sin\theta}}_{\text{Gyroscopic term}}\,. \tag{19}\]  
Before continuing, it is worth mentioning that usual contra-rotating inertial drives (called Dean drives) operate at constant angular velocities. Due to this fact, in this paper, we call the first term in Equations (18) and (19) the \"Dean-drive term\", because it refers only to centripetal forces produced by the rotation of the out-of-balance concentrated mass \(2m\). One may observe the following in Equations (18) and (19):  
* Both reaction (ground) forces at the front and rear ends of the cart are influenced by the inertial forces (\(2m\omega^{2}r\sin\theta\)) as well as the gyroscopic term (\(L_{\text{rotor}}\omega\sin\theta\)).
* While in the Dean drive the excitation force is distributed proportionally to the lever lengths (\(l_{1}\) and \(l_{2}\)), as shown by the second terms inside the square brackets, the _asymmetry_ due to the internal torque (directed to the negative side of the \(x\) axis)--imposed by the operation of the dual gyroscopes--results in an _equal differentiation_ of the reaction forces at the front (with the \(+\) sign) and rear (with the \(-\) sign) supports. In other words, what is lost at the front support is gained at the rear support, and vice versa.
* When \(\sin\theta>0\) (i.e., the axles are beyond the horizontal level), if we isolate the out-of-balance mass (Dean drive term), then it relieves both the front and rear reaction forces. Nevertheless, the additional gyroscopic term operates as follows. The front reaction force \(F_{\text{front}}\) further _decreases_ while the (previously decreased by the Dean-drive term) rear force \(F_{\text{rear}}\)_increases_. The latter finding is in accordance with the experimental results reported by the creators of the prototypes [22; 23; 24; 25], and thus the above discussion demystifies them.
* The algebraic sum of the two reaction forces in Equations (18) and (19) causes the _deletion_ of the gyroscopic term, and thus it is the same term as that in the Dean drive.
* Given \(\omega=\omega_{\text{servo}}\), the difference between the Dean drive effect and the gyroscopic effect highly depends on the spin (\(\omega_{\text{rotor}}\)) and thus the associated angular momentum \(L_{\text{rotor}}\).