Unidirectional motion using rotating masses along figure-eight-shaped trajectories

Abstract

This paper revisits the notorious Dean’s drive, aiming at causing unidirectional motion when attached to an object at rest (inertial propulsion). Instead of the usual circular path along which lumped eccentric masses move, a novel figure-eight-shaped curve is proposed. It is generally shown that the rotational kinetic energy accumulated in the rotating masses produces an equivalent linear ‘initial’ velocity to the object, and therefore, it further works like an oscillating projectile within the gravitational field. Conditions for the variation of the angular velocity, in order to achieve a ‘very short-time hovering’, are derived. The study does not consider any elastodynamic effects and reduces only to rigid-body analysis. Further improvements such as the spinning of the entire mechanism about its axis of symmetry are discussed.

Appendices

Appendix 1: Balancing of lateral inertial forces when the object does not rotate around the z axis

In order to get a better idea about the inertial forces induced by the rotating masses, in this appendix we will consider that the object does not rotate around the z axis, i.e., ω _z  = 0. In this case, Eqs. (4) and (5) simplify to:

$$\{ x_{a} ,y_{a} ,z_{a} \} = \{ - r\cos \omega t,\, - (r\sin \omega t \; \cos \omega t + R\sin \omega t),\,\,( - r\sin^{2} \omega t + R\cos \omega t)\}^{T}$$

(A-1)

and

$$\{ x_{b} ,y_{b} ,z_{b} \} = \{ r\cos \omega t,\,\,\,( - r\sin \omega t\cos \omega t + R\sin \omega t),\,\, - (r\sin^{2} \omega t + R\cos \omega t)\}^{T}$$

(A-2)

Therefore, the center of mass of the couple (a, b) will be at the position:

$$x_{m} = {{(x_{a} + x_{b} )} \mathord{\left/ {\vphantom {{(x_{a} + x_{b} )} 2}} \right. \kern-0pt} 2} \equiv 0$$

(A-3)

$$y_{m} = {{(y_{a} + y_{b} )} \mathord{\left/ {\vphantom {{(y_{a} + y_{b} )} 2}} \right. \kern-0pt} 2} = - \frac{r}{2}\sin 2\omega t$$

(A-4)

$$z_{m} = {{(z_{a} + z_{b} )} \mathord{\left/ {\vphantom {{(z_{a} + z_{b} )} 2}} \right. \kern-0pt} 2} = ( - r\sin^{2} \omega t)$$

(A-5)

If m is the mass of either of particles ‘a’ or ‘b’, by virtue of (A-3), (A-4) and (A-5), the components of the resultant inertial force are given below:

$$F_{x} = - 2m\ddot{x}_{m} \equiv 0$$

(A-6)

$$F_{y} = - 2m\ddot{y}_{m} = -4m \omega^{2} r \sin2\omega t$$

(A-7)

$$F_{z} = - 2m\ddot{z}_{m} = 4m \omega ^{2} r \cos 2 \omega t$$

(A-8)

The balancing of inertial force in the y-direction can be achieved using a second couple of masses (a′, b′) of the same characteristics and the same initial configuration, which rotates in the opposite direction at an angular velocity $\omega^{{\prime }} = - \omega$ (i.e., $\omega_{a}^{{\prime }} = - \omega_{a} = -\omega$ and $\omega_{b}^{{\prime }} = - \omega_{b} = \omega$). In such a case, concerning the ensemble of the two couples, (a, b) and (a′, b′), the total F _x component remains equal to zero, the F _y component becomes zero as an algebraic sum of two opposite quantities [$\sum {F_{y} } = - 4m\omega^{2} r(\sin 2\omega t + \sin 2\omega^{{\prime }} t) \equiv 0$], whereas the F _z component becomes double than what it was when only the couple (a, b) existed.

In summary, the synergy of the couples (a, b) and (a′, b′) leads to the following resultant forces:

$$\begin{gathered} \sum {F_{x} } = 0 \hfill \\ \sum {F_{y} } = 0 \hfill \\ \sum {F_{z} } = 8m\omega^{2} r\cos 2\omega t \hfill \\ \end{gathered}$$

(A-9)

Appendix 2: Rotation about the z axis

Taking into consideration that $\omega^{{\prime }} = - \omega$, the center of mass of the two masses (a′, b′) mentioned in Appendix 1 lies at the position:

$$\begin{gathered} x_{m}^{{\prime }} = {{(x_{a}^{{\prime }} + x_{b}^{{\prime }} )} \mathord{\left/ {\vphantom {{(x_{a}^{{\prime }} + x_{b}^{{\prime }} )} 2}} \right. \kern-0pt} 2} \equiv 0 \hfill \\ y_{m}^{{\prime }} = {{(y_{a}^{{\prime }} + y_{b}^{{\prime }} )} \mathord{\left/ {\vphantom {{(y_{a}^{{\prime }} + y_{b}^{{\prime }} )} 2}} \right. \kern-0pt} 2} = - \frac{r}{2}\sin 2\omega^{\prime}t = \frac{r}{2}\sin 2\omega t = - y_{m} \hfill \\ z_{m}^{{\prime }} = {{(z_{a}^{{\prime }} + z_{b}^{{\prime }} )} \mathord{\left/ {\vphantom {{(z_{a}^{{\prime }} + z_{b}^{{\prime }} )} 2}} \right. \kern-0pt} 2} = - r\sin^{2} \omega^{\prime}t = - r\sin^{2} \omega t = z_{m} \hfill \\ \end{gathered}$$

(B-1)

Combining (A-4) with (B-1), the center of mass of the above-mentioned four masses (a,b) and (a′, b′), lies at the position:

$$\begin{gathered} X_{m} = {{(x_{a} + x_{b} + x_{a}^{{\prime }} + x_{b}^{{\prime }} )} \mathord{\left/ {\vphantom {{(x_{a} + x_{b} + x_{a}^{{\prime }} + x_{b}^{{\prime }} )} 4}} \right. \kern-0pt} 4} \equiv 0 \hfill \\ Y_{m} = {{(y_{a} + y_{b} + y_{a}^{{\prime }} + y_{b}^{{\prime }} )} \mathord{\left/ {\vphantom {{(y_{a} + y_{b} + y_{a}^{{\prime }} + y_{b}^{{\prime }} )} 4}} \right. \kern-0pt} 4} \equiv 0 \hfill \\ Z_{m} = {{(z_{a} + z_{b} + z_{b}^{{\prime }} + z_{b}^{{\prime }} )} \mathord{\left/ {\vphantom {{(z_{a} + z_{b} + z_{b}^{{\prime }} + z_{b}^{{\prime }} )} 4}} \right. \kern-0pt} 4} = - r\sin^{2} \omega^{\prime}t \hfill \\ \end{gathered}$$

(B-2)

Eq. (B-2) suggests that in the above perfect balance, the center of mass of the two devices (i.e., of the four masses) does not laterally move (X _m  = Y _m  = 0). Therefore, if we now consider the rotation ω _z around the z axis, it is evident that this rotation will not further induce additional inertial forces. In other words, the object on which the two devices are attached will perform a purely translational motion in the z-direction.