# John2020

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1. ## Reactionless Drive and Newton's 3rd law

Hi Ghideon! Maybe I was a little bit misunderstood. My work as also what I briefly stated above is that Newton's laws of motion always hold, however the construction above and the maths that backed it show that there is a particular case where Newton's 3rd law for isolated systems appears to be incomplete. In other words, according to my view Newton didn't actually consider the case of the induced internal forces that are direction of translation screw rotation depended (not collinear) that means whatever the induced forces (action and reaction) will evolve in the same direction (since the translation screw rotates only in one or the other direction and never both at the same moment). In case the moderators of this forum allow me to share the link to my work there you may find more details (along with the implications of these findings in special relativity and Lorentz transformations) about the concept which is based on two basic findings: a) The induced internal reaction force is mathematically proven to have the same direction with the induced internal action force b) From a purely mathematical perspective, a second interpretation of the rocket equation is also possible. An internal force causes a mass transfer between two points inside the system, resulting in a change of system’s effective inertia (reduction) after a time interval. In order this to work, you cannot use collinear forces (as seen in Fig.1 - Lower) but induced. The paper develops this argument as follow: "The rate of mass ejection corresponds to a mass transfer from the system’s interior to a point away from it (rocket) after a time interval, without affecting its center of mass. From a purely mathematical perspective, a second interpretation is also possible. An internal force causes a mass transfer between two points inside the system, resulting in a change of system’s effective inertia (reduction) after a time interval.". These two views can be described with the same rocket equation. Seeing the reactionless drive from the point of an external observer and have no knowledge of what is going on inside, its motion can be justified just by a change of its inertia (caused by the internal mass transfer) or better reduction of its inertia (gain of inertia would never have as result, motion). And about your question:"Is the rig you have provided the only way to break the laws and build a reactionless drive?". The answer is: The construction (linear actuator) in Fig.1 - Upper is the simplest Reactionless Drive ever as also serves as the basic mechanical equivalent of all Reactionless Drives. If this may work, it is very easy (it is a little bit tricky actually) to develop a pure electromagnetic device (no need of superconductors and exotic material. Just known applied electromagnetism) without moving parts. Any reactionless drive should have the following ingredients: 1.Translation mechanism (e.g. translation screw) 2.Induced internal forces (without translation mechanism there are no induced forces) 3.A part of the system or all the parts of the system should be coupled on the translation mechanism Hint: The inner functionality of Fig.1-Upper resembles the electromagnetic induction and Lenz law (check the equations above).
2. ## Reactionless Drive and Newton's 3rd law

@swansont May I share the link to my paper over a post and through my profile? It will help the readers of this thread to understand better the whole idea.
3. ## Reactionless Drive and Newton's 3rd law

r_A and r_R are the position vectors of the action and reaction torques and the τ_A and τ_R are the corresponding action and reaction torques.
4. ## Reactionless Drive and Newton's 3rd law

Hi everybody! First of all I have to make clear that I am not a physicist, however as many other science fiction enthusiasts, I have an interest in alternative propulsion methods (reactionless drive etc). My theory is published in a non-arxiv.org repository and essentially explores the feasibility of constructing a reactionless drive (an initial mechanical prototype). As we know Newton's 3rd law forbids the construction of a device that may move by means of internal forces, something that all propellantless propulsion/reactionless drive theories tend to ignore. The conclusion of my research can be briefly stated as follow: A reactionless drive may become feasible if and only if Newton's 3rd law is not the whole story that means it may probably be incomplete. Fig.1 - Proof of concept. Action--reaction forces in isolated systems. Upper: Induced internal forces. Redeployment of the center of mass and acceleration of the system. Lower: Collinear internal forces. No change in the center of mass and no acceleration of the system. Attention! The induced internal reaction force was not deliberately inverted (same direction with the induced internal action force) but as a consequence of the mathematical proof. The mathematical description assumes the reactionless drive (see Fig.1.Upper) is internally powered (motor and power supply on board (appears hidden in Fig.1 - Upper)) as also there is no dissipation of energy due to friction (ideal machine). Starting from the conservation of angular momentum, the net external and internal torques in FIG.1 - Upper, are $\sum \vec{\tau}_{ext} = \vec{0}$ $\overbrace{\vec{F}_{A}}^{\longrightarrow} + \overbrace{\vec{F}_{R}}^{\longleftarrow} = \vec{0}$ $\vec{r}_{A} \neq \vec{r}_{R} \Rightarrow \vec{\tau}_{A} \neq \vec{0} \text{ and } \vec{\tau}_{R} \neq \vec{0}$ $\sum \vec{\tau}_{int} + \vec{\tau}_{A} + \vec{\tau}_{R} =\vec{0}$ $\sum \vec{\tau}_{int} + \left(\vec{r}_{A} \times \vec{F}_{A} \right) + \left(\vec{r}_{R} \times \vec{F}_{R} \right) = \vec{0}$ $\sum \vec{\tau}_{int} + \left(\vec{r}_{A} \times \vec{F}_{A} \right) + \left(\vec{r}_{R} \times \left(-\vec{F}_{A} \right)\right) = \vec{0}$ $\overbrace{\sum \vec{\tau}_{int}}^{\curvearrowleft} + \overbrace{\left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}^{\curvearrowright} = \vec{0}$ An ideal translation mechanism (translation screw in FIG.1 - Upper) can maintain, amplify or reduce the magnitude of the input force by delivering the same amount of energy (no energy dissipation through friction) entering the system (energy conservation). At this point, developing a general expression for the net induced force requires the introduction of the dimensionless factor $$n_{T}$$ (ideal mechanical advantage) along with a definition of the net induced torque. Hence, $n_{T} = \frac{|\vec{\omega} \times \left(\vec{r}_{A} - \vec{r}_{R}\right)|}{|\vec{u}_{T}|} = \frac{\left(2 \pi |\vec{r}_{A} - \vec{r}_{R}|\right)}{|\vec{l}_{T}|}$ $\overbrace{n_{T} \sum \vec{\tau}_{int}}^{\curvearrowleft} + \overbrace{n_{T} \left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}^{\curvearrowright} = \vec{0}$ $\sum \vec{\tau}_{T} = n_{T} \sum \vec{\tau}_{int}$ $\overbrace{\sum \vec{\tau}_{T}}^{\curvearrowleft} + \overbrace{n_{T}\left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}^{\curvearrowright} = \vec{0}$ Dividing the above by the position-vector magnitude $$|\vec{r}_\mathrm{A} - \vec{r}_\mathrm{R}|$$ yields $\overbrace{\frac{\sum \vec{\tau}_{T}}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}}^{\longleftarrow} + \overbrace{\frac{n_{T} \left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}}^{\longrightarrow} = \vec{0}$ $\sum \vec{F}_{T} = \frac{\sum \vec{\tau}_{T}}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}$ Due to conservation of energy, in an ideal machine (FIG.1- Upper), the power output equals to the power input: $P_{T} = P_{A} \Rightarrow \frac{|\sum \vec{F}_{T}|}{|\vec{F}_{A}|} = \frac{|\vec{\omega} \times \left(\vec{r}_{A} - \vec{r}_{R}\right)|}{|\vec{u}_{T}|} = n_{T}$ The above shows when the $$\vec{F}_\mathrm{A}$$ is constant then, the angular velocity of the translation screw $$\vec{\omega}$$, the induced force $$\vec{F}_{T}$$ and the translational velocity $$\vec{u}_{T}$$ of mass $$m_{T}$$ are also constant. The force $$\vec{F}_T$$ can be also written as $\sum \vec{F}_{T} = - \frac{n_{T} \left(\left(\vec{r}_{A} \times \vec{F}_{A} \right) + \left(\vec{r}_{R} \times \vec{F}_{R} \right) \right)}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}$ $\sum \vec{F}_{T} = - n_{T} \left(\vec{F}_{A_T} + \vec{F}_{R_T} \right) = \text{const.}$ Nevertheless, the above equations address just the motion of mass $$m_{T}$$. By expanding the ideal mechanical advantage to include varying angular and translational velocities, a new net force expression is derived that may apply for the motion of the system as a whole. Thus, $d \vec{\omega}\begin{cases} = \vec{0} &:\quad \vec{F}_{A} = \text{const.} \text{ and } \vec{F}_{R} = \text{const.} \\ \neq \vec{0} &:\quad \vec{F}_{A} \neq \text{const.} \text{ and } \vec{F}_{R} \neq \text{const.} \end{cases} \\ n_{r} = \frac{|\mathrm{d} {\omega} \times \left(\vec{r}_{A} - \vec{r}_{R}\right)|}{|\mathrm{d} \vec{u}_{T}|}, \\ \sum \vec{F}_{ind} = - \frac{n_{r} \left(\left(\vec{r}_{A} \times \vec{F}_{A} \right) + \left(\vec{r}_{R} \times \vec{F}_{R} \right) \right)}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}, \\ \sum \vec{F}_{ind} \propto d \vec{\omega}, \\ \sum \vec{F}_{ind} = - n_r \left(\vec{F}_{A_I} + \vec{F}_{R_I} \right),\\ d \vec{\omega} \begin{cases} = \vec{0} \Rightarrow \vec{d} \vec{u}_{T} = \vec{0} &:\quad - n_{r} \left( \vec{F}_{A_I} + \vec{F}_{{R_I}} \right) =\vec{0}, \\ \neq \vec{0} \Rightarrow \vec{d} \vec{u}_{T} \neq \vec{0} &:\quad - n_{r} \left( \vec{F}_{A_I} + \vec{F}_{{R_I}} \right) \neq \vec{0}. \end{cases}$ Applying the mass transfer mechanism (FIG.1 - Upper), yields $\mathrm{d} \vec{\omega} \neq \vec{0} \Rightarrow {d} \vec{u}_{T} = \vec{u}_{rel} \neq \vec{0}, \\ \frac{\mathrm{d} \vec{p}}{\mathrm{d}t} = \sum \vec{F}_{ind} = - n_{r}\left(\vec{F}_{A_I} + \vec{F}_{R_I} \right) \neq \vec{0},\\ \frac{\mathrm{d} \vec{p}}{\mathrm{d}t} = -n_{r} \vec{u}_{rel} \frac{\mathrm{d} m}{\mathrm{d}t} \neq \vec{0} \Rightarrow \vec{a} \neq \vec{0}, \\$ I will be very happy if you can help me evaluate the above idea!
5. ## Testing latex

Some further tests:
6. ## Testing latex

Some further tests: $\sum \vec{F}_{T} = \frac{\sum \vec{\tau}_{T}}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}$
7. ## Testing latex

$\frac{\Delta y}{\Delta x} = \frac{f(x+h)- f(x)}{h}$ $\Delta y$ $y = \int f(x) dx$
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