# Reaction Forces and Apparent Thrust in Dual Oscillating Control Moment Gyroscopes
## 2 Materials and Methods
### Differential Torques  
Next is the detailed mechanical analysis of differential torques. Note that the ends of the two spindles are pivoted to the cart at a separation distance \(D\) in the \(x\) direction. Quite schematically, Figure 4 illustrates the vertical plane on which the two gyroscopes rotate about the axes (spindles) oriented in the horizontal direction and passing through the points L (left) and R (right), which here are the same as points \(O_{1}\) and \(O_{2}\) illustrated earlier in Figure 2. For the sake of easiness, we assume that the center of gravity G lies in the middle of the segment (LR) and, obviously, it always remains at the same position (actually, it merely oscillates along the \(y\) axis). By construction, the gyroscopes perform rotations of 90 degrees, and thus for the polar angle, we have \(-45^{\circ}\leq\theta\leq 45^{\circ}\). When \(\theta=0^{\circ}\) (horizontal axes), the spins of the gyroscopes are in the _same_\(x\) direction, as shown by the blue-colored arrows in Figure 2a as well as in Figure 4a. In the general case, when setting the symbols \(\overrightarrow{i}\) and \(\overrightarrow{j}\) to represent the unit vectors in the \(x\) and \(y\) directions, respectively, given magnitude of the angular momentum of the rotor \(L_{\text{rotor}}\) for the arbitrary polar angle \(\theta\), we have the following momenta for the left and right gyroscopes:  
\[\text{Point L:}\quad\overrightarrow{L}_{\text{left}}=L_{\text{rotor}}( \cos\theta\overrightarrow{i}+\sin\theta\overrightarrow{j}) \tag{1}\]  
and  
\[\text{Point R:}\quad\overrightarrow{L}_{\text{right}}=L_{\text{rotor}}(\cos \theta\overrightarrow{i}-\sin\theta\overrightarrow{j}) \tag{2}\]  
Therefore, the total (resultant) angular momentum becomes the following (see also Figure 4b):  
\[\overrightarrow{L}_{\text{res}}=\overrightarrow{L}_{\text{left}}+ \overrightarrow{L}_{\text{right}}=2L_{\text{rotor}}\cos\theta\overrightarrow{i }. \tag{3}\]  
As a result, the application of [PERSON]'s second law (in rotation) on the gyroscope results in  
\[\overrightarrow{\tau}_{\text{int}}=\frac{d\overrightarrow{L}_{\text{res}}} {dt}=-\big{(}2L_{\text{rotor}}\dot{\theta}\sin\theta\big{)}\overrightarrow{i }\, \tag{4}\]  
where \(\dot{\theta}=\omega_{\text{servo}}\) is the angular velocity of the servomotor which controls the function \(\theta(t)\). Based on [PERSON]'s third law, the internal torque applied from the gyroscopes to the cart will be the algebraic negative of that in Equation (4), and thus  
\[\overrightarrow{\tau}_{\text{cart}}=-\overrightarrow{\tau}_{\text{int}}= \big{(}2L_{\text{rotor}}\dot{\theta}\sin\theta\big{)}\overrightarrow{i}. \tag{5}\]  
Therefore, the fully controlled motion of the dual gyroscopic system leads to an _internal_ torque \(\overrightarrow{\tau}_{\text{cart}}\) (given by Equation (5)), which is entirely transmitted to the cart. Considering Equation (5), the well-known formula  
\[L_{\text{rotor}}=I_{\text{rotor}}\omega_{\text{rotor}}\, \tag{6}\]  
and also the above definition \(\dot{\theta}=\omega_{\text{servo}}\), we eventually have  
\[\overrightarrow{\tau}_{\text{cart}}=-\overrightarrow{\tau}_{\text{int}}= \big{(}2I_{\text{rotor}}\omega_{\text{rotor}}\omega_{\text{servo}}\sin\theta \big{)}\overrightarrow{i}. \tag{7}\]  
Although the longitudinal vertical plane \(Gyz\), which is located in the middle of the distance \(D\), is a plane of symmetry (geometrically), due to the same chosen direction of the spin vectors, the total change in the total (resultant) angular momentum \(\overrightarrow{L}_{\text{res}}\) is a vector  
\[\overrightarrow{\tau}_{\text{int}}=(\Delta\overrightarrow{L}_{\text{rotor} })_{L}+(\Delta\overrightarrow{L}_{\text{rotor}})_{R}\, \tag{8}\]  
which points in the negative direction of the horizontal \(x\) axis, as shown in Figure 4b.