## 2 First Order Theory of Propulsion Generation by Means of Rotation Energy
To give the answer in Figure 2, the centrifugal force \(F_{u}\) is represented in two orthogonal components:  
\[F_{ux}=F_{u}\text{cos }\varphi_{x}\text{ (effective centrifugal force component, see below)} \tag{4}\]  
\[F_{uy}=F_{u}\text{sin }\varphi_{x}\text{ (noneffective centrifugal force component, see below)} \tag{5}\]  
\[\varphi_{x}=\omega t\pm mod\ 2\pi\text{ instantaneous phase angle, }t=0\ :\varphi_{x}=0\] \[t=T\ :\varphi_{x}=2\pi\]  
Both components describe strictly harmonic forces having zero averages over each rotation period, i.e.,  
\[F_{x}=\frac{1}{2\pi}\int\limits_{0}^{2\pi}F_{ux}d\varphi=0\] \[F_{y}=\frac{1}{2\pi}\int\limits_{0}^{2\pi}F_{uy}d\varphi=0\]  
Our approach to solve the problem how to obtain a non-zero resulting translatoric propulsion of the rotating disc: (partial) rectification of centrifugal force must be realized. To explain the need for force rectification, let us assume here that wanted propulsion should appear in the x-direction. Consequently, we have to modulate unbalance during each rotation in the following way (Figure 3):  
Figure 1:0 Components of centrifugal force vector. \(F_{u}\) instantaneous centrifugal vector. \(F_{ux}\) effective x-component of \(F_{u}\). \(F_{uy}\) noneffective-component of \(F_{u}\). \(\varphi\) instantaneous rotation angle of \(F_{u}\) against \(x\). In an azimuthal angle region \(2\alpha_{x}\) highly symmetric to the x-axis, i.e., the wanted propulsion direction, we generate non-zero unbalance. Otherwise, outside of this region, i.e., in the remaining region \(2\pi-2\alpha_{x}\), this unbalance is always compensated. Thus, we have to switch on and off the rotor unbalance within time period \(\tau\), which is only a small fraction of cycle time \(T\)  
Figure 1:1  
To achieve this, unbalances are selectively generated, for example, by rotor-fixed actuators, so that the resulting pulsating centrifugal forces cause a directed propulsion of the rotor mass and, thus, the flying object via the rotor bearings. If we succeed in managing this ultrafast modulation due to rotation, we obtain a train of force impulses, each pulse of duration \(\tau\) and separated by cycle time \(T\) from each other. This impulse train will result in a significant change of the translational state of rotor mass \(M\): According to [PERSON], force \(F\) acting on mass point \(M\) equals the first derivate of its translational impulse \(P=M_{v}\) (\(v\) = translational velocity), thus  
\[ \begin{array}{c}\frac{dP}{dt}=F(t)\\ \Delta P=M\Delta v=F(t)\Delta t\end{array}  \tag{6}\]  
The resulting impulse change during each rotation follows from  
\[\Delta P=\int\limits_{-\frac{\tau}{2}}^{\frac{\tau}{2}}F(t)dt=2\int\limits_{0 }^{\frac{\tau}{2}}F(t)dt\]  
We obtain the effective impulse component (x-component) of centrifugal force:  
\[F(t)=F_{ux}(t)=F_{u}\cos\omega t\]  
\[\Delta P_{x}=\int\limits_{0}^{T}F_{u}\cos\omega tdt=\int\limits_{-\frac{\tau} {2}}^{\frac{\tau}{2}}F_{u}\cos\omega tdt\]  
\[\Delta P_{x}=2\int\limits_{0}^{\frac{\tau}{2}}F_{u}\cos\omega tdt=2\bigg{(} \frac{F_{u}}{\omega}\bigg{)}\sin\frac{\omega\tau}{2} \tag{7}\]  
Otherwise, due to the (ineffective) y-component of centrifugal force, zero average of impulse change is confirmed:  
\[F_{uy}=F_{u}\sin\varphi_{x}\]  
Figure 3: Active angle interval \(2\alpha_{x}\) symmetric to thrust direction \(x\). \[\Delta P_{y}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}F_{uy}d\varphi=0 \tag{8}\]  
Let us assume the angle of unbalance activity (Figure 3)  
\[\alpha_{x}\leq\frac{\pi}{2},\ \ \sin\!\alpha_{x}\leq 1\]  
we can rewrite (7) to obtain the non-zero force magnitude in the x-direction  
\[F_{x}=4\pi m_{u}Rf^{2}\sin\alpha_{x}=m_{u}R\omega^{2}\frac{\sin\alpha_{x}}{\pi} \tag{9}\]  
Based on Equations (7)-(9) above we conclude in summary:  
The magnitude of propulsion force depends linearly on rotor radius \(R\) and on unbalance mass \(m_{u}\) and increases with the square of rotational frequency \(f\).