Patent Publication Number: US-2012024633-A1

Title: Gyromotor

Description:
The U.S. patent application claims the priority of U.S. Provisional Application No. 61/400,613 filed on Jul. 30, 2010. 
    
    
     ORIGIN OF THE INVENTION 
     The invention was made by John M. Vranish as President of Vranish Innovative Technologies LLC and may be used by John M. Vranish and Vranish Innovative Technologies LLC without the payment of any royalties therein or therefore. The work was done by John M. Vranish on his own time and at his own expense. 
     BACKGROUND OF THE INVENTION 
     There is a large and growing presence of objects in earth orbit associated with human activity. There is need to maneuver these objects and to have ready access to them and this requires a practical transportation system that works in earth orbit where vacuum and micro gravity conditions prevail. This, in turn, suggests an action and reaction motor is required that runs on renewable energy. 
     Rockets are the means currently employed and these are severely limited in their usefulness. The prime means of earth orbit maneuver is hydrazine rocket motors, a World War 2 era propulsion technique that powered the Me  163 . Komet. Hydrazine rockets run out of fuel, are corrosive and volatile and lack capability for precision control. Ion engines are emerging as a more modern alternative, but, these also run out of fuel. Ion engines, in their present stage of development, are too low in thrust for practical earth orbit operations because activities would take too long. 
     A propulsion means is needed that provides a safe, useful level of thrust and that runs on renewable energy without emitting a plume. Three (3) approaches were tried with three different approaches to the physics of propulsion and all three are in different stages of development. Gyromotor is the latest evolution of one of these approaches and has reached the point where it needs patent protection. Any plume-less action and reaction motor is subject to skepticism and controversy and Gyromotor is no exception. The skeptics worry that inertial activities confined to a closed system cannot affect activities outside said closed system. John M. Vranish respects these arguments, takes them seriously and addresses them in the specification of this patent application. Experiment will settle the issue. In the mean time this patent application establishes the origin of the John M. Vranish Gyromotor concept. 
     FIELD OF THE INVENTION 
     The present invention relates generally to action and reaction propulsion motors and more particularly to action and reaction propulsion systems that utilize gyroscopes. The present invention relates generally to gyroscope systems and more particularly to gyroscope systems used in propulsion applications. The present invention relates particularly to electromechanical and motion control systems. 
     DESCRIPTION OF THE PRIOR ART 
     
         
         1. Hydrazine rockets are currently used in earth orbit space operations and have been so for many years. These are chemical action and reaction engines that emit plumes and provide linear motion. As a practical matter, hydrazine is hard to resupply in space and is non-renewable. 
         2. Ion Engines are being developed but, are not yet in extensive use. These are electromagnetic action and reaction engines that emit plumes and provide linear motion. As a practical matter, the ions in the plume are also non-renewable in space. Ion Thrusters are being pursued in many forms.
       a. Electrostatic   b. Electromagnetic Lorentz Force   c. Hall Effect   
     
         3. CGM (Control Gyroscope Moment) systems use gyroscopes for attitude control (Single Gimble, Dual Gimble, Variable Speed). These are action and reaction motors that do their rotation work without emitting a plume but, cannot provide linear motion. They can be supplied in space using electrical energy from the sun via solar panel. 
         4. There have been attempts to use gyroscopes to provide both rotary and linear motion without emitting a plume.
       a. The Generation of a Unidirectional Force [Bruce E. DePalma—Simularity Institute 1974] “The mechanical generation of a unidirectional force, is shown to be a consequence of the variable inertial property of matter.” (Gyroscopes are used.) [prior art]   b. John M. Vranish Abandoned patent application [prior art]. This reached the Preliminary application stage before it went abandoned.   
     
       
    
     SUMMARY OF THE INVENTION 
     Gyromotor is a type of action and reaction motor which generates thrust without plume ejection. Whereas rockets react equal and opposite to ejected mass momentum, Gyromotor cycles gyroscopes, each mounted on the end of a moment arm, in a back and forth rowing motion to drive a spacecraft, without external mass ejection analogous to rowing a boat. Gyroscope inertial properties are configured to provide maximum resistance torque during the drive stroke and reconfigured to provide minimum torque resistance during the return stroke. The gyroscopes are turned by a moment arm so the torque resistance provides a useful linear force component to drive the spacecraft, with said linear force greater during the drive stroke than the return stroke, analogous to an oar in water during the drive stroke and in air during the return stroke. The space craft moves in reaction to the net linear forces and momentum is conserved. The torques and forces are pseudo and are generated by change of gyroscope spin axis during said moment arm rotation, similar to centrifugal and coriolis effect pseudo forces. The Gyromotor Invention will provide maximum resistance torque and resistance linear force during the Drive Stroke because the gyroscopes are each spinning and are oriented such that the spin axis of each is perpendicular to the axis of moment arm rotation. Maximum net action and reaction force, then depends on minimizing resistance torque and linear reaction force during the return stroke. 
     Two (2) methods of reconfiguring the gyroscopes to provide minimum torque and linear force during the return stroke are considered. In one method, the spin of each gyroscope is reduced to zero, prior to the Return Stroke, so gyroscope orientation doesn&#39;t matter. In an alternate method, the spin axis of each gyroscope is redirected prior to the Return Stroke such that each is parallel to the angular direction of rotation. Thus, there is no change in direction of gyroscope spin during return, with no gyroscope torque resistance and no reactive linear force even though the gyroscopes are spinning. 
     Electro-mechanical devices and systems essential to performing the Gyromotor functions are described. These include a system for rotating said moment arms, a system for spinning gyroscopes, a system for cancelling gyroscope precession in the preferred embodiment and a method for cancelling the effects of changing the orientation of each spinning gyroscope in said alternate method. Also included in this description are representative form, fit and function numbers to provide expected performance information and construction and operating particulars needed to achieve said performance. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
       A more complete appreciation of the invention and man of its attendant advantages will be readily appreciated as the same becomes better understood by reference to the following detailed description when considered in connection with the accompanying drawings wherein: 
         FIGS. 1   a  and  1   b  illustrate the base components of a Gyromotor and show how said base components move during said Gyromotor drive and return strokes:  FIGS. 1   a  and  1   b  also show the inertial reaction force difference between said drive stroke and said return stroke. 
         FIGS. 2   a  and  2   b  show how a pair of spinning gyroscopes, on the end of a turning moment arm, creates an inertial reaction force with a linear component useful for powering a vehicle.  FIG. 2   a  shows how said inertial reaction force is physically created by using a turning moment arm to turn said pair of spinning gyroscopes affixed to end of said moment arm.  FIG. 2   b  interprets the actions and results of  FIG. 2   a  as a lever arm functionally equivalent diagram. 
         FIG. 3   a  details how Gyromotor applies directional inertial force to said vehicle during said Drive Stroke and  FIG. 3   b  details how the Gyromotor removes said directional inertial force during said Return Stroke. 
         FIG. 4  shows how gyroscopes can be configured in pairs to cancel precession, while adding said directional inertial force of each. 
         FIG. 5   a  illustrates a configuration whereby a motor gear drive can operate through an idler gear to rotate said gyroscopes on the end of a moment arm.  FIG. 5   b  illustrates an alternate configuration whereby said motor gear drive can operate through an idler gear to rotate said gyroscopes on the end of a moment arm. 
         FIG. 6   a  illustrates a configuration whereby idler gears can be configured in coaxial pairs to independently spin the gyroscope pairs and rotate the moment arm on which the gyroscopes are mounted from a top view perspective.  FIG. 6   b  illustrates the configuration introduced in  FIG. 6   a  from a side section view perspective. 
         FIG. 7  illustrates a configuration whereby one pair of motor gear drives can operate on a pair of idler gear arrangements, according to  FIG. 6   a , to provide and control gyroscope spin, while a second pair of motor gear drives can operate on a second pair of idler gear arrangements, to provide and control moment arm rotation also according to  FIG. 6   a , with gyroscope spin and moment arm rotation functions independent. 
         FIG. 8   a  illustrates the Alternate Drive Cycle Drive Stroke wherein said directions of gyroscope spin are oriented to maximize inertial drive force on said vehicle.  FIG. 8   b  illustrates said Alternate Drive Cycle Return Stroke wherein said directions of gyroscope spin are re-oriented to minimize inertial drive force on said vehicle. 
         FIG. 9   a  illustrates the inertial torque reacted to said vehicle by changing the spin direction of said gyroscopes at the end of said Alternate Drive Cycle Drive Stroke, preparatory to said Alternate Drive Cycle Return Stroke.  FIG. 9   b  illustrates the inertial torque reacted to said vehicle by changing the spin direction of said gyroscopes at the end of said Alternate Drive Cycle Return Stroke, preparatory to said Alternate Drive Cycle Drive Stroke.  FIG. 9   a  and  FIG. 9   b  together show the net torque reacted to said vehicle by changing spin direction to be zero over each complete Alternate Drive and Return Cycle. 
         FIG. 10   a  Illustrates the mechanical parts used to perform said Alternate Drive Cycle Stroke and their arrangement. A top down section view of said mechanical parts is presented in the region where they interface with said gyroscope pair. 
         FIG. 10   b  Illustrates said mechanical parts used to perform said Alternate Drive Cycle Stroke and their arrangement. A side section view of said mechanical parts is presented in the region where they interface with said gyroscope pair. 
         FIG. 11  Illustrates said mechanical parts used to perform said Alternate Drive Cycle Stroke and their arrangement. A side section view is presented in the region where they interface with said Motor Gear Drives. 
     
    
    
     DETAILED DESCRIPTION OF PREFERRED EMBODIMENT 
     In accordance with the present invention, a Gyromotor includes: 1. a Gyroscope Arm System, 2. A Motor Control System to control and motivate said Gyroscope Arm System, 3. a Housing for 1 and 2. The Gyroscope Arm System includes a Left Arm System and a Right Arm System n which each of the two arm systems contains a pair of co-axial gyroscopes mounted on the end of a moment arm. The Motor Control System includes a Left Motor Control System and a Right Motor Control System. The Left Motor Control System rotates the Left Moment Arm and attached gyroscopes and, independently, spins the Left Arm System gyroscopes at angular velocities equal and opposite to each other. The Right Motor Control System rotates the Right Moment Arm and attached gyroscopes and, independently, spins the Right Arm System gyroscopes at angular velocities equal and opposite to each other. For linear travel, the Left and Right Arm Systems are rotated towards and away from each other in a coordinated back and forth rowing motion. The gyroscopes are spinning during the Drive Stroke and are not spinning during the Return Stroke, with the spin axis of each gyroscope oriented perpendicular to the axis of its moment arm rotation. The Left Motor Control System contains a motor and gear system and controller and the Right Motor Control System contains a mirror image motor and gear system and controller. The Housing contains said Gyroscope Arm and Motor Control Systems. The preferred embodiment is configured and operated according to  FIGS. 1   a  and  1   b.    
     I. GYROMOTOR DRIVE METHOD (FIGS.  1   a ,  1   b ) 
     Two (2) moment arms are counter-rotated back and forth in opposition to each other in a cyclic manner as per  FIGS. 1   a ,  1   b . Each moment arm has two (2) gyroscopes mounted on its end with the spin axis of each oriented in the direction of gyroscope instantaneous velocity. Normally this would produce no net motion as per the Zero-Sum nature of Newton&#39;s Laws of Motion. It would move in the +X direction during the Drive Stroke and return the same amount in the −X direction during the return stroke. But the gyroscopes during the Drive Stroke are at full spin during the Drive Stroke and are without spin during the Return Stroke and this difference in gyroscopic spin upsets Zero-Sum in favor of the Drive Stroke. We will now show why this is so. 
     A. Drive Stroke 
     During the Drive Stroke, the gyroscopes are spinning and oriented as shown in  FIG. 1   a . The gyroscope pair attached to the left moment arm is labeled  1   a  and is rotated on the end of the left moment arm by a motor and gear system labeled  2   a . The gyroscope pair attached to the right moment arm is labeled  1   b  and is rotated on the end of the right moment arm by a motor and gear system labeled  2   b . The angular momentum vector of a set of coaxial spinning gyroscopes is labeled +{right arrow over (L)} and −{right arrow over (L)} in  FIG. 1   a.    
     We know a spinning gyroscope has an angular momentum vector of 
       mR 2 ω S {right arrow over (a)} ωs ={right arrow over (L)}  (I1)[1][2]
 
     And, when {right arrow over (L)} is changed with respect to time a torque {right arrow over (τ)} is generated such as: 
         d{right arrow over (L)}/dt={right arrow over (τ)}=d ( mR   2 {right arrow over (ω)} S )/dt=mR 2   d{right arrow over (ω)}   S   /dt   (I2)[3]
 
     For each gyroscope pair, the torque generated by turning the +{right arrow over (L)}, and −{right arrow over (L)}, vectors add. 
     For the gyroscopic orientation shown in the Drive Stroke ( FIG. 1   a ), the direction of the angular momentum vector changes, even though all angular speeds remain constant, such that: 
       d{right arrow over (ω)} S /dt=ω S d({right arrow over (a)} ωS )/dt=ω S ω R {right arrow over (a)} ωR   (I3)
 
       And 
         mR   2 ω S ω R   {right arrow over (a)}   ωR ={right arrow over (τ)}(where:  W   R =forced angular velocity of rotation)  (I4)
 
     This torque must be provided by the motor and gear systems labeled  2   a  and  2   b  in  FIGS. 1   a ,  1   b ,  2   a.  
 
B. Return Stroke ( FIG. 1   b )
 
     For the Return Stroke the spin is zero so {right arrow over (L)}=0 and: 
     
       
         
           
             
               
                 
                   
                     
                        
                       
                         L 
                         → 
                       
                     
                     
                        
                       t 
                     
                   
                   = 
                   
                     
                       τ 
                       → 
                     
                     = 
                     0 
                   
                 
               
               
                 
                   ( 
                   
                     I 
                      
                     
                         
                     
                      
                     5 
                   
                   ) 
                 
               
             
           
         
       
     
     C. Torque to Force 
     The torque produced in turning the gyroscopes (labeled  1   a   1  and  1   a   2  in  FIG. 2   a ) is given in eq. (I4) above and must be provided by the motor and gear system labeled  2   a . Considering the Left Arm System ( 1   a ), 
       Σ{right arrow over (M)}=0  (I6)
 
     But, each motor and gear system supplying the torque and the gyroscope pair reacting the torque are separated by a moment arm R T  (labeled  1   a   3 ) so a force {right arrow over (F)} O  must be induced on the end of that moment arm such that: 
       P O R T =T R (gyroscope reaction torque)=i t (motor input torque)  (I7)
 
     This force {right arrow over (F)} O  must be reacted with an equal and opposite {right arrow over (F)} O  exerted by each motor and gear system on the Housing (or Drive Vehicle) labeled  3  in  FIGS. 1   a ,  1   b  and  2   a . We begin by discussing the circumstances of 
     The forces and torques for an inertial lever arm terminated by a gyroscope must obey: 
       ΣF X =0  (I8)
 
       And: 
       ΣM Z =0  (I9)
 
     (The forces in Y and Z are always self cancelling by the symmetric construction technique of using two (2) counter-rotating sets of identical Drives.) 
     This relationship between system geometry and forces and torques can be interpreted in a lever arm equivalent diagram as shown in  FIG. 2   b  where: 
       1   a   1 =the front gyroscope and  1   a   2 =rear gyroscope of gyroscope pair  1   a.      1   a   3 =Moment arm length R T .
   2   a =Motor Gear System for Left Arm System.
 
F G =Gyroscopic force opposing turning.
 
R G =Distance between gyroscope spin axis and radius of gyration.
 
T G =Gyroscope torque opposing turning.
 
T O =Torque provided by motor gear system.
 
F R =Force equivalent response to T O .
 
T R =Torque from motor gear system being reacted into Housing labeled  3 .
 
The Right Arm System ( 1   b ) mirrors the Left Arm System and each adds thrust in the X direction.
 
       FIGS. 3   a  and  3   b  show the Drive Stroke and Return Stroke for both Moment Arm Systems  1   a  and  1   b.    
     Where: 
       1   b   1 =the front gyroscope and  1   b   2 =rear gyroscope of gyroscope pair lb.
   1   b   3 =Moment arm length R T .
   2   b =Motor Gear System for Right Arm System.
 
     And remaining construction and operation details of Right Moment Arm System replicate and mirror those of the Left Arm System. Similarly a lever arm equivalent diagram can be set up for the Right Moment Arm System that mirrors that shown in  FIG. 2   b.    
     In  FIG. 4  the advantages of mounting gyroscopes in back to back co-axial pairs can be seen. The co-axial pair arrangement with the gyroscopes spinning in equal and opposite angular velocities, allows the torque induced reactive forces to add while the precessions cancel. The arrangement also allows the moment arm to operate on the exact center of the spinning gyroscopes. 
     II. GYROMOTOR EFFECT AS COMPARED TO ZERO-SUM 
     The concept of an action and reaction motor in which no plume is ejected is counter-intuitive and is considered by many to be impossible. These concerns will now be addressed. We begin by considering the Vehicle (labeled  3 ) as a Space Craft operating in earth orbit. Returning to  FIGS. 1   a ,  1   b ,  2   a ,  2   b , and  3 , we note that the F O  produced by each Drive Motor on the Space Craft serves to “push off” against the Space Craft while F O  on the end of each moment arm “pushes off” against a separate inertial body (the spinning set of gyroscopes). So, we get a transfer of force and momentum to the Space Craft with respect to its external environment even as we see an equal and opposite transfer of force and momentum to the gyroscopes. An observer external to the Space Craft would see the Space Craft move in one direction and the gyroscopes move in the opposite direction according to conservation of momentum. The force produced by the gyroscopes rotating on the end of a moment arm acting over the Drive Stroke time has the dimensions of momentum and acts as a rocket plume with mass, velocity and momentum. The Space Craft would react equal and opposite to the gyroscope momentum. During the Return Stroke, gyroscope spin is off and the force produced by gyroscopes rotating on the end of a moment arm is zero. Thus, the momentum of the returning gyroscopes is zero and the Space Craft does not react. The Space Craft would experience a net momentum in the Direction of the Drive Stroke similar to rowing a boat. An observer inside the Space Craft would not notice a difference between the Drive and Return Strokes. In both strokes the Space Craft would seem to be stationary and the gyroscopes would move the same distance with respect to the Space Craft and at the same angular velocities. Newton&#39;s Laws seen by an observer inside the Space Craft-Gyroscope structure would seem unaffected (Zero-Sum), except for the force measured between gyro arms and Space Craft housing. To an observer outside the Space Craft, Newton&#39;s Laws would be satisfied by the Space Craft motion in reaction to the net reaction force between the gyro arms and the Space Craft housing. Newton&#39;s Laws would be obeyed but, they would be Zero-Sum in a different sense. The Space Craft would move with respect to the external observer. The speed of the gyro arms would appear slightly slower during the Drive Stroke and slightly faster during the Return Stroke. In this sense, the external observer would also see Zero-Sum. But, Space Craft motion would continue and that is what matters most. 
     E. Governing Equations of Cycle Drive and Return Strokes. 
     Because a torque is added to the ends of the moment arm R T  during the Drive Stroke but, is absent during the Return Stroke, the net Drive Force remains to drive the Space Craft in return. This net drive force will now be determined. 
     Equal and opposite torque operating on opposite ends of a moment arm is mathematically equivalent to equal and opposite forces operating perpendicular to the moment arm such that: 
       {right arrow over (τ)}={right arrow over (F)}X( R   T )= F ( R   T ) sin θ{right arrow over (a)} X   +F ( R   T ) cos θ{right arrow over (a)} Y   (I10)
 
     The Y components cancel each other and we are left with 
     
       
         
           
             
               
                 
                   
                     F 
                     X 
                   
                   = 
                   
                     
                       
                         τ 
                         → 
                       
                       
                         R 
                         T 
                       
                     
                      
                     sin 
                      
                     
                         
                     
                      
                     θ 
                   
                 
               
               
                 
                   ( 
                   
                     I 
                      
                     
                         
                     
                      
                     11 
                   
                   ) 
                 
               
             
           
         
       
     
     {right arrow over (τ)} is constant when {right arrow over (ω)} R  and {right arrow over (ω)} S  are constant. When the Spin Axis is oriented in the direction of tangential instantaneous velocity the torque generated at each gyroscope is: 
     
       
         
           
             
               
                 
                   
                     τ 
                     → 
                   
                   = 
                   
                     
                       
                          
                         
                           L 
                           → 
                         
                       
                       
                          
                         t 
                       
                     
                     = 
                     
                       
                         I 
                          
                         
                           
                              
                             
                               
                                 ω 
                                 → 
                               
                               S 
                             
                           
                           
                              
                             t 
                           
                         
                       
                       = 
                       
                         
                           
                             mR 
                             2 
                           
                            
                           
                             ω 
                             S 
                           
                            
                           
                             
                                
                               
                                 
                                   a 
                                   → 
                                 
                                 ωS 
                               
                             
                             
                                
                               t 
                             
                           
                         
                         = 
                         
                           
                             mR 
                             2 
                           
                            
                           
                             ω 
                             S 
                           
                            
                           
                             ω 
                             R 
                           
                            
                           
                             
                               a 
                               → 
                             
                             
                               ω 
                                
                               
                                   
                               
                                
                               R 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     I 
                      
                     
                         
                     
                      
                     12 
                   
                   ) 
                 
               
             
           
         
       
     
     R=R G  (gyroscope radius of gyration)
 
For a gyroscope set on the end of each of two (2) oars we have a torque of 2{right arrow over (τ)} and a linear force of:
 
                     F   X     =           τ   →       R   T          sin                 θ     =       2      m            V   .     →     X       =       2        mR   2          ω   S          ω   R        sin                 θ                     a   →     X         R   T                   (     I                 13     )               We know ∫ t1   t2 2m{dot over ({right arrow over (V)} Xdt= 2 mV X{right arrow over (a)}   X (momentum in X)  (I14)[4]
 
     We also know: 
     
       
         
           
             
               
                 
                   
                      
                     t 
                   
                   = 
                   
                     
                        
                       θ 
                     
                     
                       ω 
                       R 
                     
                   
                 
               
               
                 
                   ( 
                   
                     I 
                      
                     
                         
                     
                      
                     15 
                   
                   ) 
                 
               
             
           
         
       
     
     So: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             ∫ 
                             
                               θ 
                                
                               
                                   
                               
                                
                               1 
                             
                             
                               θ 
                                
                               
                                   
                               
                                
                               2 
                             
                           
                            
                           
                             
                               
                                 2 
                                  
                                 
                                   mR 
                                   2 
                                 
                                  
                                 
                                   ω 
                                   S 
                                 
                                  
                                 
                                   ω 
                                   R 
                                 
                                  
                                 sin 
                                  
                                 
                                     
                                 
                                  
                                 θ 
                               
                               
                                 R 
                                 T 
                               
                             
                              
                             
                               ( 
                               
                                   
                               
                                
                               
                                 
                                    
                                   θ 
                                 
                                 
                                   ω 
                                   R 
                                 
                               
                               ) 
                             
                           
                         
                         = 
                         
                           
                             2 
                              
                             
                               mR 
                               2 
                             
                              
                             
                               
                                 ω 
                                 S 
                               
                                
                               
                                 ( 
                                 
                                   
                                     
                                       - 
                                       cos 
                                     
                                      
                                     
                                         
                                     
                                      
                                     θ2 
                                   
                                   + 
                                   
                                     cos 
                                      
                                     
                                         
                                     
                                      
                                     θ1 
                                   
                                 
                                 ) 
                               
                             
                           
                           
                             R 
                             T 
                           
                         
                       
                     
                   
                   
                     
                       
                         = 
                         
                           2 
                            
                           
                             mV 
                             X 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     I 
                      
                     
                         
                     
                      
                     16 
                   
                   ) 
                 
               
             
           
         
       
     
     With an X direction momentum from the Inertial Oars provided to the Boat of: 
     
       
         
           
             
               
                 
                   
                     
                       2 
                        
                       
                         mR 
                         2 
                       
                        
                       
                         
                           ω 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               
                                 - 
                                 cos 
                               
                                
                               
                                   
                               
                                
                               θ2 
                             
                             + 
                             
                               cos 
                                
                               
                                   
                               
                                
                               θ1 
                             
                           
                           ) 
                         
                       
                     
                     
                       R 
                       T 
                     
                   
                   = 
                   
                     2 
                      
                     
                       mV 
                       X 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         for 
                          
                         
                             
                         
                          
                         each 
                          
                         
                             
                         
                          
                         cycle 
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     I 
                      
                     
                         
                     
                      
                     17 
                   
                   ) 
                 
               
             
           
         
       
     
     By conservation of momentum the Boat acquires an X direction velocity of 
     
       
         
           
             
               
                 
                   
                     V 
                     Boat 
                   
                   = 
                   
                     
                       V 
                       X 
                     
                      
                     
                       
                         2 
                          
                         m 
                       
                       
                         m 
                         sc 
                       
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         for 
                          
                         
                             
                         
                          
                         the 
                          
                         
                             
                         
                          
                         Drive 
                          
                         
                             
                         
                          
                         Stroke 
                          
                         
                             
                         
                          
                         of 
                          
                         
                             
                         
                          
                         each 
                          
                         
                             
                         
                          
                         cycle 
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     I 
                      
                     
                         
                     
                      
                     18 
                   
                   ) 
                 
               
             
           
         
       
     
     The Zero-Sum inertial stalemate has been broken by changing inertial mass properties and conditions have been created to drive a Space Craft using internal inertial means only. 
     F. Back to Back Gyroscope Pairs 
     The gyroscopes are operated in back to back counter rotating pairs as per  FIG. 4 . This is done for several reasons. The Gyromotor requires the gyroscopes be operated with forced torque applied. This, in turn, means the individual gyroscopes seek to perform precession. The back to back arrangement, sharing the same spin axis and counter-rotating at the same angular speeds means that precession effects are self-cancelling and not a factor in Gyromotor performance. Also, construction and operation is simplified and form, fit, function is improved. 
       {right arrow over (τ)}={right arrow over (Ω)} P   ×{right arrow over (L)}   G   (I19)[3][5]
 
     Where: 
     {right arrow over (τ)}=torque
 
{right arrow over (Ω)} P =angular velocity of precession
 
{right arrow over (L)} G =angular momentum of gyroscope
 
     In  FIG. 4  we see that when two (2) gyroscopes share a common spin center and counter-rotate back to back, their natural angular velocities of precession oppose each other and cancel. 
     Thus, in the back to back configuration net: 
       Σ{right arrow over (Ω)} P =0.  (I20)
 
     The torque from gyroscope  1   a   1  and the torque from gyroscope  1   a   2  add. The bevel gear drive  1   a   31  causes the gyroscopes to counter-rotate at equal and opposite speeds. The forces generated by turning the gyroscopes acts at R G  as shown in  FIG. 4 . 
     II. TOWARDS A PRACTICAL GYROMOTOR 
     A Gyromotor can be constructed according to  FIGS. 4 ,  5   a ,  5   b ,  6   a ,  6   b  its Drive Method can be applied according to  FIGS. 3   a ,  3   b . The construction methods shown in  FIGS. 6   a ,  6   b ,  7  enable the gyroscopes and moment arm to be cycled while the electric motors that power them remain stationary in the Space Craft housing. This reduces un-sprung weight, enables faster cycle times and eliminates the danger of failure from electrical cable and connection problems. These motors are operated in pairs to operate a single moment arm and pair of gyroscopes. This arrangement enables the gyroscopes to be spun up or spun down independent of moment arm rotation. The gyroscopes on the end of each moment arm are positioned and operated in counter-rotating pairs, sharing the same spin axis according to  FIG. 4 . This cancels out gyroscope precession and simplifies construction. 
     A construction method is illustrated in  FIGS. 5   a ,  5   b ,  6   a ,  6   b  and  7 .  FIGS. 5   a  and  5   b  show alternate arrangements in which electric motors, that drive the Gyromotor arms, can be located in the Space Craft housing and can drive the gyroscope mechanisms through gearing without disturbing the force and torque balance needed to drive the Space Craft with reaction forces. All the example positions as per  FIGS. 5   a  and  5   b , leave a reaction force operating on the Space Craft. This is because one force of a chain of forces and reaction forces reacts against gyroscope inertia separate from the Space Craft structure. This uncovers and isolates an equal and opposite force which operates on the Space Craft housing.  FIG. 6 , shows a mechanical structure which can transform the forces into the appropriate gyroscope and arm motions.  FIG. 7  shows how the features shown in  FIGS. 5   a ,  5   b  and  6  work together as a Gyromotor drive system. 
     We choose one (1) 4490 . . . B Micromo dc servo motor, 11,000 rpm, 390.533 oz-in. stall torque, with a gear box of 40/1 to perform the Drive and Stroke rotation. This provides: 
     
       
         
           
             
               
                 
                   
                     
                       ( 
                       
                         390.553 
                          
                         
                             
                         
                          
                         
                           oz 
                           · 
                           in 
                           · 
                           40 
                         
                       
                       ) 
                     
                     / 
                     
                       ( 
                       
                         16 
                          
                         
                           
                             oz 
                             lb 
                           
                           · 
                           12 
                         
                          
                         
                           in 
                           ft 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     81.3610416666667 
                      
                     
                         
                     
                      
                     
                       ft 
                       · 
                       lbs 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       torque 
                       ) 
                     
                   
                 
               
               
                 
                   
                     ( 
                     
                       II 
                        
                       
                           
                       
                        
                       1 
                     
                     ) 
                   
                    
                   
                     [ 
                     6 
                     ] 
                   
                 
               
             
           
         
       
     
     At a speed of: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             11 
                             , 
                             000 
                              
                             
                                 
                             
                              
                             
                               
                                 ( 
                                 
                                   rev 
                                   min 
                                 
                                 ) 
                               
                               · 
                               2 
                             
                              
                             π 
                              
                             
                                 
                             
                              
                             
                               ( 
                               
                                 rad 
                                 rev 
                               
                               ) 
                             
                           
                           
                             ( 
                             
                               
                                 40 
                                 · 
                                 60 
                               
                                
                               
                                   
                               
                                
                               
                                 ( 
                                 
                                   sec 
                                   min 
                                 
                                 ) 
                               
                             
                             ) 
                           
                         
                         = 
                         
                           
                             ω 
                             R 
                           
                            
                           
                               
                           
                            
                           
                             ( 
                             
                               rad 
                               sec 
                             
                             ) 
                           
                            
                           
                               
                           
                            
                           
                             ( 
                             
                               available 
                                
                               
                                   
                               
                                
                               rotation 
                                
                               
                                   
                               
                                
                               speed 
                             
                             ) 
                           
                         
                       
                     
                   
                   
                     
                       
                         = 
                         
                           28.7979326579064 
                            
                           
                               
                           
                            
                           
                             rad 
                             sec 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     II 
                      
                     
                         
                     
                      
                     2 
                   
                   ) 
                 
               
             
           
         
       
     
     We stay with the same motor for gyroscope spin up and spin down and reserve judgment on the gear box for the moment. 
     We select gyroscopes with flywheels of 0.5 ft radius, weighing 5 lbs and spinning at 600 rpm=20π rad/sec. We select a moment arm of 0.75 ft. and rotate it at 15 (rad/sec). 
     600 rpm means #4490 . . . B can support a spin up MA of: 
     
       
         
           
             
               
                 
                   
                     
                       11 
                       , 
                       000 
                        
                       
                           
                       
                        
                       rpm 
                     
                     
                       600 
                        
                       
                           
                       
                        
                       rpm 
                     
                   
                   = 
                   
                     18.3333333333333 
                     = 
                     MA 
                   
                 
               
               
                 
                   ( 
                   
                     II 
                      
                     
                         
                     
                      
                     3 
                   
                   ) 
                 
               
             
           
         
       
     
     We use 15=MA to be conservative. 
     
       
         
           
             
               
                 390.533 
                 
                   16 
                   · 
                   12 
                 
               
               · 
               15 
             
             = 
             
               30.510390625 
                
               
                   
               
                
               
                 ft 
                 · 
                 
                   lb 
                    
                   
                     ( 
                     
                       
                         
                           available 
                            
                           
                               
                           
                            
                           spin 
                            
                           
                               
                           
                            
                           up 
                            
                           
                               
                           
                            
                           torque 
                            
                           
                               
                           
                            
                           from 
                            
                           
                               
                           
                            
                           electric 
                            
                           
                               
                           
                            
                           motor 
                         
                         &amp; 
                       
                        
                       
                           
                       
                        
                       gear 
                        
                       
                           
                       
                        
                       box 
                     
                     ) 
                   
                 
               
             
           
         
       
     
     This Spin up torque provides a spin up angular acceleration (d{right arrow over (ω)} S /dt) as per: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           τ 
                           → 
                         
                         = 
                         
                           
                             mR 
                             2 
                           
                            
                           
                             
                                
                               
                                 
                                   ω 
                                   → 
                                 
                                 S 
                               
                             
                             
                                
                               t 
                             
                           
                         
                       
                     
                   
                   
                     
                       
                         = 
                         
                           
                             
                               5 
                                
                               
                                   
                               
                                
                               lb 
                             
                             
                               32.2 
                                
                               
                                   
                               
                                
                               
                                 ( 
                                 
                                   ft 
                                   / 
                                   
                                     sec 
                                     2 
                                   
                                 
                                 ) 
                               
                             
                           
                            
                           
                             
                               ( 
                               
                                 0.5 
                                  
                                 
                                     
                                 
                                  
                                 ft 
                               
                               ) 
                             
                             2 
                           
                            
                           
                             
                                
                               
                                 
                                   ω 
                                   → 
                                 
                                 S 
                               
                             
                             
                                
                               t 
                             
                           
                         
                       
                     
                   
                   
                     
                       
                         = 
                         
                           30.510390625 
                            
                           
                               
                           
                            
                           
                             ft 
                             · 
                             lb 
                           
                         
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     II4 
                     ) 
                   
                    
                   
                       
                   
                 
               
             
             
               
                 
                   
                     
                       
                         
                           
                              
                             
                               
                                 ω 
                                 → 
                               
                               S 
                             
                           
                           
                              
                             t 
                           
                         
                         = 
                         
                           128.8 
                            
                           
                               
                           
                            
                           
                             
                               ft 
                               
                                 sec 
                                 2 
                               
                             
                             · 
                             
                               
                                 30.510390625 
                                  
                                 
                                     
                                 
                                  
                                 
                                   ft 
                                   · 
                                   lbs 
                                 
                               
                               
                                 5 
                                  
                                 
                                     
                                 
                                  
                                 lbs 
                               
                             
                           
                         
                       
                     
                   
                   
                     
                       
                         = 
                         
                           785.9476625 
                            
                           
                               
                           
                            
                           
                             rad 
                             
                               sec 
                               2 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     II 
                      
                     
                         
                     
                      
                     5 
                   
                   ) 
                 
               
             
           
         
       
     
     Which requires a time from zero to 600 rpm of: 
     
       
         
           
             
               
                 
                   
                     
                       20 
                        
                       
                           
                       
                        
                       π 
                        
                       
                           
                       
                        
                       
                         ( 
                         
                           rad 
                           / 
                           sec 
                         
                         ) 
                       
                     
                     
                       785.9476625 
                        
                       
                           
                       
                        
                       
                         ( 
                         
                           rad 
                           / 
                           
                             sec 
                             2 
                           
                         
                         ) 
                       
                     
                   
                   = 
                   
                     t 
                     = 
                     
                       0.0799440676138 
                        
                       
                           
                       
                        
                       sec 
                     
                   
                 
               
               
                 
                   ( 
                   
                     II 
                      
                     
                         
                     
                      
                     6 
                   
                   ) 
                 
               
             
           
         
       
     
     With the knowledge our motor gear box combinations can meet our arbitrary design requirements, we calculate an estimated performance. 
     
       
         
           
             
               
                 
                   
                     
                       2 
                        
                       
                         mR 
                         2 
                       
                        
                       
                         
                           ω 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               
                                 - 
                                 cos 
                               
                                
                               
                                   
                               
                                
                               θ 
                                
                               
                                   
                               
                                
                               2 
                             
                             + 
                             
                               cos 
                                
                               
                                   
                               
                                
                               θ 
                                
                               
                                   
                               
                                
                               1 
                             
                           
                           ) 
                         
                       
                     
                     LT 
                   
                   = 
                   
                     2 
                      
                     
                       mV 
                       X 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         for 
                          
                         
                             
                         
                          
                         each 
                          
                         
                             
                         
                          
                         cycle 
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     I 
                      
                     
                         
                     
                      
                     16 
                   
                   ) 
                 
               
             
           
         
       
     
     We estimate that start up from rest to full rotation speed takes (π/4)rad as does slow down from full rotation speed to stop. We also choose LT=0.75 ft and, conservatively say 1 cycle per second can be performed. Thus we have: 
     
       
         
           
             
               
                 
                   
                     
                       
                         2 
                         · 
                         
                           ( 
                           
                             5 
                             32.2 
                           
                           ) 
                         
                         · 
                         
                           
                             ( 
                             0.5 
                             ) 
                           
                           2 
                         
                         · 
                         20 
                       
                        
                       
                         π 
                         · 
                         
                           ( 
                           
                             2 
                             
                               2 
                             
                           
                           ) 
                         
                       
                     
                     0.75 
                   
                   = 
                   
                     2 
                      
                     
                       mV 
                       X 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         lb 
                         · 
                         sec 
                       
                       ) 
                     
                      
                     
                       ( 
                       
                         per 
                          
                         
                             
                         
                          
                         cycle 
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     II 
                      
                     
                         
                     
                      
                     7 
                   
                   ) 
                 
               
             
           
         
       
     
     And: 
     
       
         
           
             
               
                 
                   
                     
                       27.5955461997414 
                       0.75 
                     
                     = 
                     
                       36.7940615996552 
                        
                       
                           
                       
                        
                       
                         lb 
                         · 
                         
                           sec 
                           sec 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     ( 
                     
                       momentum 
                        
                       
                           
                       
                        
                       transfer 
                        
                       
                           
                       
                        
                       rate 
                     
                     ) 
                   
                 
               
               
                 
                   ( 
                   
                     II 
                      
                     
                         
                     
                      
                     8 
                   
                   ) 
                 
               
             
           
         
       
     
     This means our 10 lbs of gyroscope rest mass (2 arms with 5 lbs of gyroscopes each) would acquire a speed increase on a per cycle basis of: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           36.7940615996552 
                            
                           
                               
                           
                            
                           
                             lb 
                             · 
                             sec 
                           
                         
                         
                           10 
                            
                           
                               
                           
                            
                           lb 
                         
                       
                       · 
                       32.2 
                     
                      
                     
                         
                     
                      
                     
                       ft 
                       
                         sec 
                         2 
                       
                     
                   
                   = 
                   
                     118.47687835089 
                      
                     
                         
                     
                      
                     
                       ft 
                       / 
                       sec 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         for 
                          
                         
                             
                         
                          
                         a 
                          
                         
                             
                         
                          
                         10 
                          
                         
                             
                         
                          
                         lb 
                          
                         
                             
                         
                          
                         payload 
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     II 
                      
                     
                         
                     
                      
                     9 
                   
                   ) 
                 
               
             
           
         
       
     
     For a 2,000 lb space craft this equates to: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             10 
                              
                             
                                 
                             
                              
                             lb 
                           
                           
                             2 
                             , 
                             000 
                              
                             
                                 
                             
                              
                             lb 
                           
                         
                         · 
                         118.47687835089 
                       
                        
                       
                           
                       
                        
                       
                         ft 
                         sec 
                       
                     
                     = 
                     
                       0.5923843917545 
                        
                       
                           
                       
                        
                       
                         ft 
                         sec 
                       
                     
                   
                    
                   
                       
                   
                    
                   
                     
 
                   
                    
                   
                       
                   
                    
                   
                     ( 
                     
                       for 
                        
                       
                           
                       
                        
                       a 
                        
                       
                           
                       
                        
                       single 
                        
                       
                           
                       
                        
                       cycle 
                     
                     ) 
                   
                 
               
               
                 
                   ( 
                   
                     II 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
               
             
           
         
       
     
     With a cycle rate of one (1) cycle per second, within 20 sec the 2,000 lb object will acquire a speed of 11.84768783509 ft/sec. These speed values are encouraging. 
     III. PROTOTYPE ESTIMATED PERFORMANCE AND FORM, FIT, FUNCTION 
     0.75 ft moment arm
 
0.5 ft radius of gyration
 
1.25 ft radius for gyro ring mounted on a moment arm=2.5 ft dia foot print.
 
[5 ft diameter for two (2) Drivers]
 
[Height &gt;1.5 ft+Electric Motor]
 
5 lb gyro ring weight
 
Micromo Brushless DC Servomotor 4490 . . . B, 1.732 in. dia, 3.543 in. length wt 750 g [750 grams=1.65346696638658 lbs. If we double the size to include the gear box, we get approx 6.5 lbs of motor and gear box weight for one (1) Drive Arm system and approximately 13 lbs. for the entire system.]
 
     These are rough estimates but, the values are encouraging especially for a motor to drive a Space Craft of 2,000 lbs. 
     IV. AN ALTERNATE DRIVE METHOD (FIGS.  8   a ,  8   b ,  9   a ,  9   b ,  10   a ,  10   b ,  11 ) 
     The Return Stroke can also produce zero torque if the gyroscope spin axis vectors do not change direction during the return stroke as shown in  FIG. 8   b . In this instance (d{right arrow over (a)} ωS /dt)=0 so: 
     
       
         
           
             
               
                 
                   
                     τ 
                     → 
                   
                   = 
                   
                     
                       
                          
                         
                           L 
                           → 
                         
                       
                       
                          
                         t 
                       
                     
                     = 
                     
                       
                         I 
                          
                         
                           
                              
                             
                               
                                 ω 
                                 → 
                               
                               S 
                             
                           
                           
                              
                             t 
                           
                         
                       
                       = 
                       
                         
                           
                             mR 
                             2 
                           
                            
                           
                             ω 
                             S 
                           
                            
                           
                             
                                
                               
                                 
                                   a 
                                   → 
                                 
                                 
                                   ω 
                                    
                                   
                                       
                                   
                                    
                                   S 
                                 
                               
                             
                             
                                
                               t 
                             
                           
                         
                         = 
                         
                           0 
                            
                           
                               
                           
                            
                           
                             ( 
                             
                               
                                 because 
                                  
                                 
                                     
                                 
                                  
                                 
                                   
                                      
                                     
                                       
                                         a 
                                         → 
                                       
                                       
                                         ω 
                                          
                                         
                                             
                                         
                                          
                                         S 
                                       
                                     
                                   
                                   
                                      
                                     t 
                                   
                                 
                               
                               = 
                               0 
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     IV 
                      
                     
                         
                     
                      
                     1 
                   
                   ) 
                 
               
             
           
         
       
     
     This leaves the problem of switching the orientation of the gyroscopes at the end of the Return Stroke so torque can be generated during the Drive Stroke and switching gyroscope orientation again before the Return Stroke. Each orientation switch produces a torque as shown in  FIG. 9   a . But, as also shown in  FIG. 9   b , the orientation switches can be performed so as make the switching torques self-cancelling as per: 
       Σ{right arrow over (τ)} SH =0  (IV2)
 
     The orientation switching method appears a viable alternative to spinning down and spinning up the gyroscopes. 
       FIGS. 10   a ,  10   b ,  11  illustrate a mechanical arrangement capable of performing the Alternate Drive Method. This arrangement is an extension of the arrangement shown in  FIGS. 6   a ,  6   b  and  7  in which an additional co-axial geared shaft system is added ( 2   b   23  and  1   b   4  in  FIG. 11 ) and an additional set of bearings is added to enable  1   b   4  to rotate inside  2   b   21 . A third Motor and Gear System per gyroscope pair is also required along with added capability for the control system. 
     V. SUMMARY AND CONCLUSIONS 
     1. The Gyromotor concept appears to work. It seems possible to generate useful reaction thrust from a motor that performs an internal cycle to generate external thrust and/or force and that uses renewable energy. It seems possible to do so by changing the inertial properties of parts internal to the motor while leaving the rest mass of each unchanged. This, in turn, seems possible to accomplish by using gyroscopes in novel ways. Newton&#39;s Laws of Motion seem not to be violated.
 
2. The construction of a practical Gyromotor seems straight forward and well within current art.
 
3. The performance and thrust to weight of a Gyromotor seems useful for applications in micro-gravity, such as low earth orbit and space beyond. The form, fit, function factors also seem favorable. The thrust to weight is not sufficient to provide lift-off against earth gravity.
 
4. Gyromotor presents an important opportunity to further performing useful work in low earth orbit and space beyond and this Gyromotor paper establishes a preliminary and tenuous level of credibility.
 
5. The technical community needs to prove out the concept up or down. They could start with a credible simulation and from there move to hardware and developments as results determine.