Rotational Energy Based Propulsion System

Abstract

This application is for specifying the basic physics behind a propulsion system based upon using rotational energy in place of kinetic energy to induce a new kind of orbit that resembles levitation. Evidence from an experiment is presented showing that the author was able to measure a small upwards force on a spinning gyroscope in a vacuum to justify the physics. A plan for building a flying craft using this mechanism is shown where the upwards force opposing gravity is from the centrifugal forces of rapidly rotating flywheels that rotate at speeds comparable to orbital velocity speeds. Rotational energy can be translated into kinetic energy by selectively exposing small portions of the spinning flywheels to the atmosphere, which allows for the craft to accelerate in any direction.

Claims

1. A new mechanism for achieving levitation by using rotational energy is explained. This mechanism can be used to provide structural support for structures without adding additional weight to the structure (unlike a support column).

2. If there is a mechanism that allows for the flywheel to interact with the atmosphere (such as a small exposure) then the strong drag forces between the flywheel and the atmosphere can be used to generate a thrust in any direction coplanar with the flywheel.

3. To go up (or down) the craft would need to increase (or decrease) the total rotational velocities of both flywheels. This would transition the craft into an oscillating elliptical orbit in which it would either ascent or descent using the same orbital dynamics equations as a normal elliptical orbit.

Description

BRIEF DESCRIPTION OF FIGURES

[0011] FIG. 1 explains the source of repulsive forces between two rotating rings. In the limit that one mass is far larger than the other, the equations simplify.

[0012] FIG. 2 shows data collected during our experiment which showed a decrease in the measured weight of a gyroscope while spinning in a vacuum chamber. As the gyroscope slowed down, the weight returned to normal.

[0013] FIG. 3 shows the similarity in orbital trajectories between an object in an elliptical orbit and another object with the same amount of rotational energy as the first object's kinetic energy. These trajectories illustrate how an orbit that looks like levitation operates using the same physics as a normal orbit.

[0014] FIG. 4 shows the cross-sectional cut through view of a craft with two flywheels spinning in opposite directions (so that total angular momentum is zero), and how this configuration allows the craft to levitate and also accelerate in any direction by opening shutters that result in frictional forces between the flywheels and the atmosphere.

BEST MODE FOR CARRYING OUT THE INVENTION

[0015] The first issue in building the invention is to design a flywheel that can tolerate the extreme rotation velocities required for achieving a new positive force upwards. There is a density to tensile stress ratio requirement that must be met in order to meet these velocities. Currently only the carbon nanotube materials seem to be beyond this threshold, however we anticipate further developments in materials engineering will provide additional options.

[0016] The craft requires two flywheels rotating in opposite directions. This neutralizes the total angular momentum, which allows the craft to be rotated. This also allows the rotational velocity of the flywheels to increase or decrease without inducing unwanted rotation of the craft.

[0017] The flywheels must be sheltered from the surrounding environment in a vacuum chamber, however there must also be some mechanism for translating the flywheel's rotational energy into kinetic energy in some direction so that the craft can maneuver.

[0018] The mechanisms for powering the flywheels and for allowing controlled interaction with the surrounding environment are not developed in this patent.

DETAILED DESCRIPTION

[0019] The first iteration of the invention is simply a rapidly rotating gyroscope on scale, within a vacuum chamber. The purpose of this was to measure the lift force and verify that the physics were correct. This has already been accomplished.

[0020] For the following equations refer to FIG. 1 for visual variable definitions. Consider a cylindrically symmetric rigid body of density ?, at distance z from the center of the dominant body M, and rotating at angular velocity ? about the {circumflex over (z)} axis. For an arbitrary differential mass dm in this body, moving at velocity {right arrow over (v)}=wr{circumflex over (?)}, the force acting on it is given by the sum of centrifugal and gravitational forces.

[00001]$\stackrel{.fwdarw.}{\mathrm{dF}}=\mathrm{dm}\left[\frac{v^2}{r}\hat{r}+\left(\frac{v^2}{z}-\frac{\mathrm{GM}}{z^2}\right)\hat{Z}\right],\mathrm{dm}={?}_{\left(r,z\right)}\mathrm{rd}?\mathrm{dr}\mathrm{dz}$

[0021] Since the body is cylindrically symmetric, the integral over the radial {circumflex over (r)} component evaluates to zero, leaving only the components along the {circumflex over (z)} axis.

[00002]$\begin{array}{cc}\stackrel{.fwdarw.}{F}=?\stackrel{.fwdarw.}{\mathrm{dF}}=2???\left[\frac{{?}^2{r}^2}{z}-\frac{\mathrm{GM}}{z^2}\right]\hat{z}{?}_{\left(r,z\right)}r\mathrm{dr}\mathrm{dz}& \left(1\right)\end{array}$

[0022] Consider the case where the overall size of the rigid body is very small compared to r<<R, and R is the object's distance from the origin. Further, assume the body has an internal and external radius of r.sub.1 and r.sub.2, then the density function is:

[00003]${?}_{\left(r,z\right)}=\{\begin{array}{cc}?h{?}_{\left(z-R\right)},& r_1?r?r_2\\ 0,& \mathrm{else}\end{array}$

[0023] Where h<<R is the thickness of the body along the {circumflex over (z)} axis, and ?.sub.(z-R) is the Dirac delta function. Then equation (1) simplifies:

[00004]$\begin{array}{cc}\stackrel{.fwdarw.}{F}=?h2?{?}_{r_1}^{r_2}\left[\frac{{?}^2{r}^2}{R}-\frac{\mathrm{GM}}{R^2}\right]r\mathrm{dr}\hat{z}=?h{?\left[\frac{{?}^2{r}^4}{2R}-\frac{{\mathrm{GMr}}^2}{R^2}\right]}_{r_1}^{r_2}\hat{z}& \left(2\right)\end{array}$$m=?h?\left(r_2^2-r_1^2\right),\stackrel{.fwdarw.}{F}=m\left[\frac{{?}^2}{2R}\left(r_2^2+r_1^2\right)-\frac{\mathrm{GM}}{R^2}\right]\hat{z}$

[0024] The positive term in equation (2) is the lift force. Notice that if we set r.sub.2=r.sub.1 such that all the mass were to be concentrated in a rotating ring, then we can determine the angular velocity required to counteract gravity and produce a stationary orbit.

[00005]$\begin{array}{cc}\mathrm{Assume}\begin{array}{c}.Math.\stackrel{.fwdarw.}{F}.Math.=0\\ r_2=r_1=r\end{array},\frac{{?}^{2}2r^2}{2R}=\frac{\mathrm{GM}}{R^2},?r=\sqrt{\frac{\mathrm{GM}}{R}}& \left(3\right)\end{array}$

[0025] Equation (3) is the same equation for the circular orbital velocity of a point particle about the dominant body M. Therefore, just as a point particle will remain in a stable circular orbit at the orbital velocity, so too will a rotating ring of matter remain in orbit, if it is rotating such that each part of it is moving at the same orbital velocity. For the case where r<<R, this kind of orbit could look like levitation since the ring may seem to remain stationary.

[0026] The next iteration of the invention would be to use an industrial flywheel that is already designed to rotate in a vacuum, and have that on a scale appropriate to measure the change in weight as a function of energy stored in the flywheel. The inventor has not had the resources to do this yet.

[0027] The first development of a craft that would be maneuverable would be to develop a submersible craft, similar to a submarine, but disk shaped. The additional buoyancy forces would supplement the deficient lift force as we work through the engineering issues of actually building it.

[0028] Further iterations of the invention would require building a specialized flywheel out of carbon nanotubes (or other material with sufficient tensile strength to density ratio) that could achieve rotational velocities sufficient to generate a net lift force capable of lifting the craft into the air. Many of the same principles for maneuvering in water would also apply to air. Here is the explanation for why something like carbon nanotubes are required:

[0029] For an arbitrary differential volume dV, the total force acting along the z axis is given by the centripetal force and the gravitational force on this element.

[00006]${\mathrm{dF}}_z=\left[{?}_E^{2}z-\frac{{\mathrm{GM}}_E}{z^2}\right]?\left(r,z\right)dV=\left[\frac{{?}_D^2{r}^2}{z}-\frac{{\mathrm{GM}}_E}{z^2}\right]?\left(r,z\right)\mathrm{rd}?\mathrm{dr}\mathrm{dz}$

[0030] Integrating this over the volume of our body will give the general formula for calculating the lift of a cylindrically symmetric rigid body rotating at angular velocity ? about the z axis.

[00007]$F_z=2???\left[\frac{{?}^2{r}^2}{z}-\frac{{\mathrm{GM}}_E}{z^2}\right]?\left(r,z\right)r\mathrm{dr}\mathrm{dz}$

[0031] For the case where our rotating body has a constant density, and an overall height h<<z such that the integral only needs to consider r, and our disk has an internal and external radius of r1, and r2, then the equation can be solved:

[00008]$F_z=2?h?{?}_{r_1}^{r_2}\left[\frac{{?}^2{r}^2}{z}-\frac{{\mathrm{GM}}_E}{z^2}\right]r\mathrm{dr}=2?h{?\left[\frac{{?}^2{r}^4}{4z}-\frac{{\mathrm{GM}}_E{r}^2}{2z^2}\right]}_{r_1}^{r_2}=2?h?\left(r_2^2-r_1^2\right)\left[\frac{{?}^2\left(r_2^4-r_1^4\right)}{4z\left(r_2^2-r_1^2\right)}-\frac{{\mathrm{GM}}_E}{2z^2}\right]$$\mathrm{Note}:m=?2?h\left(r_2^2-r_1^2\right)$$\mathrm{Lift}\mathrm{Force}:F_z=m\left[\frac{{?}^2}{4z}\left(r_2^2+r_1^2\right)-\frac{{\mathrm{GM}}_E}{2z^2}\right]$$\mathrm{Condition}\mathrm{for}\mathrm{Levetation}:?=\sqrt{\frac{2{\mathrm{GM}}_E}{z\left(r_2^2+r_1^2\right)}}$

[0032] As an example, lets assume our flywheel is a uniform disk of inner radius of 1 m, and outer radius 2 m, then in order to levitate, the disk will need to spin at:

[00009]$?=\sqrt{\frac{2{\mathrm{GM}}_E}{z\left(r_2^2+r_1^2\right)}}=\sqrt{\frac{2*6.67*{10}^{-11}*5.97*{10}^{24}}{6.37*{10}^6*\left(4+1\right)}}\frac{\mathrm{rad}}{s}=5.*{10}^3\frac{\mathrm{rad}}{s}$

[0033] At such a rotation, the rim of our flywheel would be moving at 12,900 m/s. It is extremely fast, and would require some advanced materials to avoid explosion. The maximum hoop stress before exceeding the tensile stress for a uniform disk flywheel is given by:

[00010]$\frac{{?}_t}{?}=r^2{?}^2?1.*{10}^8\frac{m^2}{s^2}$

[0034] For our 2 m radius flywheel, it would need to be constructed of a material with an ultimate tensile strength to density ratio of less than that maximum value. Graphene is one material that may work, with a maximum tensile strength of 50000 MPa and a density of 1 g/cm.sup.3, which works out to

[00011]$\mathrm{Graphene}:\frac{{?}_t}{?}=\frac{5*{10}^{10}\mathrm{Pa}}{1000\mathrm{kg}/m^3}=5*{10}^7\frac{m^2}{s^2}$

[0035] This value will not quite be sufficient for levitation; however carbon nanotubes have a theoretical tensile strength of up to 300 GPa with a corresponding density of around 0.55 g/cm.sup.3. If we could build our flywheel out of carbon nanotubes material, we have a tensile strength to density ratio of:

[00012]$\mathrm{Theoretical}\mathrm{Maximum}\mathrm{of}\mathrm{Carbon}\mathrm{Nanotubes}:$$\frac{{?}_t}{?}?\frac{3*{10}^{14}\mathrm{Pa}}{550\mathrm{kg}/m^3}=5.5*{10}^{11}\frac{m^2}{s^2}$$\mathrm{Current}\mathrm{Maximum}\mathrm{of}\mathrm{Carbon}\mathrm{Nanotubes}:$$\frac{{?}_t}{?}?\frac{6.3*{10}^{13}\mathrm{Pa}}{550\mathrm{kg}/m^3}=1.1*{10}^{11}\frac{m^2}{s^2}$

[0036] These values greatly exceed the minimum required value by a safe margin, so carbon nanotubes would be strong enough to levitate as a flywheel near the Earth's surface.

[0037] The final craft would consist of at least two sets of flywheels rotating in opposite directions such that the total angular momentum from each flywheel would sum to zero. The craft would require a mechanism for translating the rotational energy of the wheel into kinetic energy by some sort of interaction with the surrounding fluid or gas. This mechanism would be developed as a separate patent later.

DETAILED DESCRIPTION OF THE DRAWINGS

[0038] FIG. 1

[00013]$\mathrm{Conservation}\mathrm{of}\mathrm{Momentum}:$${\stackrel{.fwdarw.}{P}}_0=0={\stackrel{.fwdarw.}{P}}_1+{\stackrel{.fwdarw.}{P}}_2=m_1{\stackrel{.fwdarw.}{v}}_1+m_2{\stackrel{.fwdarw.}{v}}_2$$\mathrm{Conservation}\mathrm{of}\mathrm{Angular}\mathrm{Momentum}:$${\stackrel{.fwdarw.}{L}}_0=0={\stackrel{.fwdarw.}{L}}_1+{\stackrel{.fwdarw.}{L}}_2=m_1{\stackrel{.fwdarw.}{d}}_1?\left({\stackrel{.fwdarw.}{?}}_1?{\stackrel{.fwdarw.}{r}}_1\right)+m_2{\stackrel{.fwdarw.}{d}}_2\left({\stackrel{.fwdarw.}{?}}_2?{\stackrel{.fwdarw.}{r}}_2\right)$

[0039] Now we shall consider the forces acting on both rings. There are two forces, the gravitational force pulling the objects together, and a centrifugal force pushing them apart. The centrifugal term has a component along the z axis, and is given by:

[00014]${\stackrel{.fwdarw.}{F}}_{\mathrm{cf}}=\left[m_1.Math.{\stackrel{.fwdarw.}{d}}_1+\stackrel{.fwdarw.}{r_1}.Math.{{?}_{\mathrm{com}}}^2\cos ?\right]\hat{z}=\frac{{m_1\left(r_1{?}_1\right)}^2}{\sqrt{d_1^2+r_1^2}}\cos ?\hat{z}={m_1\left(r_1{?}_1\right)}^2\frac{d_1}{d_1^2+r_1^2}\hat{Z}$$\left(\mathrm{using}\sqrt{d_1^2+r_1^2}{?}_{\mathrm{com}}=r_1{?}_1\right)$${\stackrel{.fwdarw.}{F}}_{m_1}=\left[-\frac{{\mathrm{Gm}}_1{m}_2}{{\left(d_1+d_2\right)}^2}+{m_1\left(r_1{?}_1\right)}^2\frac{d_1}{d_1^2+r_1^2}\right]\hat{Z}$${\stackrel{.fwdarw.}{F}}_{m_2}=-\left[-\frac{{\mathrm{Gm}}_1{m}_2}{{\left(d_1+d_2\right)}^2}+{m_2\left(r_2{?}_2\right)}^2\frac{d_1}{d_2^2+r_2^2}\right]\hat{Z}$

[0040] There are two important limits for these equations, the first is when the masses and rotations are approximately equal, and the other is when one mass M is far larger than the other mass m. in these equations, R is the distance between the masses to match normal portrayal of the force of gravity. We also assume that R>>r which significantly simplifies the equations. These equations reduce to the normal point mass version when r?=0.

[00015]$\begin{array}{cc}\underset{\underset{R?r}{m_1?m_2}}{\lim }{\stackrel{.fwdarw.}{F}}_{m_1}=-{\stackrel{.fwdarw.}{F}}_{m_2}=\left[-\frac{{\mathrm{Gm}}^2}{R^2}+\frac{2{m\left(r?\right)}^2}{R}\right]\hat{Z}& \left(1\right)\end{array}$$\begin{array}{cc}\underset{\underset{R?r}{M?m}}{\lim }{\stackrel{.fwdarw.}{F}}_m=-{\stackrel{.fwdarw.}{F}}_M=\left[-\frac{\mathrm{GMm}}{R^2}+\frac{{m\left(r?\right)}^2}{R}\right]\hat{Z}& \left(2\right)\end{array}$

[0041] FIG. 2

[0042] Sample of experimental data showing a decrease in effective weight in proportion to the rotational energy. Vacuum is achieved at approximately 400 seconds, and therefore fluid dynamics effects are not responsible for this weight reduction. See our attached paper (Evidence of Lift Force on Gyroscope) for more details.

[0043] FIG. 3

[0044] With the right choice of coordinate systems, the equations of motion are equivalent. First we consider the elliptical orbit in polar coordinates. The equations for the force on the elliptical orbit in these coordinates is given by:

[00016]$\stackrel{.fwdarw.}{F}\left(\stackrel{.fwdarw.}{r}\right)=m\stackrel{.Math.}{\stackrel{.fwdarw.}{r}}=m\left[\left(\stackrel{.Math.}{r}-r{\stackrel{.}{?}}^2\right)\hat{r}+\left(\frac{1}{r}\frac{d\left(r^2?\right)}{\mathrm{dt}}\right)\widehat{?}\right],$$F\left(r\right)=-\frac{\mathrm{GMm}}{r^2}=m\left(\stackrel{.Math.}{r}-r{\stackrel{.}{?}}^2\right),$$0=\frac{1}{r}\frac{d\left(r^2?\right)}{\mathrm{dt}}$

[0045] Since there is no force acting along the angular direction, we know that the second term is equal to zero. This is equivalent to the conservation of angular momentum. Meanwhile, if we rearrange the force equation for the force along the radial direction, this gives exactly the same equation as (2):

[00017]$m\stackrel{.Math.}{r}=-\frac{\mathrm{GMm}}{r^2}+\frac{{\mathrm{mr}}^2{\stackrel{.}{?}}^2}{r}=-\frac{\mathrm{GMm}}{r^2}+\frac{{\left({\mathrm{mr}}^2\stackrel{.}{?}\right)}^2}{{\mathrm{mr}}^3}=-\frac{\mathrm{GMm}}{r^2}+\frac{L_1^2}{{\mathrm{mr}}^3},$$\mathrm{where}\mathrm{Angular}\mathrm{Momentum}.Math.L_1.Math.={\mathrm{mr}}^2\stackrel{.}{?}$

[0046] Where the angular momentum is given by L. Compare that equation to the equation for motion of the spinning orbit, which is given by (2):

[00018]$m\stackrel{.Math.}{z}=-\frac{\mathrm{GMm}}{z^2}+\frac{{m\left(r?\right)}^2}{z}=-\frac{\mathrm{GMm}}{z^2}+\frac{{\left({\mathrm{mzr}}_o?\right)}^2}{{\mathrm{mz}}^3}=-\frac{\mathrm{GMm}}{z^2}+\frac{L_2^2}{{\mathrm{mz}}^3},$$\mathrm{where}\mathrm{Angular}\mathrm{Momentum}.Math.L_2.Math.={\mathrm{mzr}}_o?$

[0047] These equations are not just the same form, but are exactly the same as long as the elliptical orbit and the rotational orbit have the same initial conditions. In the figure above, consider the point at which the orbits are starting at the lowest altitude R.sub.min and the same initial speed V.sub.max. At that point we can calculate the angular momentum and energy of the system:

[00019]$\mathrm{Same}\mathrm{initial}\mathrm{conditions},$$r=R_{\min }=z,r{\stackrel{.}{?}}_o=V_{\max }=r_o{?}_o$$\mathrm{Same}\mathrm{angular}\mathrm{momentum},$$.Math.L_1.Math.={\mathrm{mr}}^2{\stackrel{.}{?}}_o={\mathrm{mR}}_{\min }{V}_{\max }=\mathrm{mz}{r}_o{?}_o=.Math.L_2.Math.$$\mathrm{Same}\mathrm{energy},$$E_1=-\frac{\mathrm{GMm}}{r}+\frac{1}{2}{m\left(r{\stackrel{.}{?}}_o\right)}^2=\frac{\mathrm{GMm}}{R_{\min }}+\frac{1}{2}{mV}_{\max }^2=\frac{\mathrm{GMm}}{z}+\frac{1}{2}{m\left(r_o{?}_o\right)}^2=E_2$

[0048] Therefore, both of these orbits have the exact same equations of motion for the coordinate that is equivalent to the distance from the center of dominant mass M.

FIG. 4

[0049] Diagram of final craft with two oppositely rotating disks. To interact with the atmosphere through drag, there is a retractable area of the shell that allows for contact between the spinning disks and the atmosphere. Thrust is induced in one direction by exposing opposite sides of both disks to the atmosphere. The resultant drag acts like a drag racer's wheel on the asphalt.