VIEWS: 226 PAGES: 28 CATEGORY: Science & Technology Papers POSTED ON: 2/3/2012 Public Domain
TECHNOLOGY ADVANCEMENTS USED BY THE PRESENTED INERTIAL PROPULSION DRIVE The preferred technology used for mechanical motivation of the propulsion mechanism is the core-less DC motor or iron-less rotor DC motor, because of their very small electrical inductance, allowing very short drive pulse durations without loss in power and delay in time. A further advantage of the core-less motor is the extremely small mass moment of inertia of the rotor and thereby the very high power density in comparison to motor mass. The DC electrical motor-generator was first introduced by Werner von Siemens in the late 1900 in Germany, for street cars and sub-way-cars. The DC motor-generator can act as both, as a motor and as generator of electricity. This reversibility is also referred to as dynamic regenerative breaking. The technology was improved in great strides in the early 60tis with the advent of the DC motor employing printed circuit rotors and the advent of the power transistor for H-bridge power switches. The advancements in permanent magnet materials for electrical motors delivering very high magnetic field densities together with the advent of tiny field effect power transistor, new battery power technologies, carbon fiber technology and extreme short drive pulse durations allows the DC motor technology to trend to an overall power density to far below one horse power per Kg mass. Some of these advancements are being used in toys, like the rapidly moving RC car toys. For the example inertia drive presented in this publication, mechanical power switches are depicted to illustrate the concept but sensor operated field effect power transistor technologies are of course readily available. Are the presented hardware technologies capable of delivering a self- contained propulsion vehicle capable of defying gravity? No, not yet, but if a model is receiving its energy and its control information through an umbilical cord, such a example could come very close, for a short time run, in power overload condition! Next two pictures, Picture #6 and #7 presents the natural depletion of kinetic energy in a rotational to straight line coupled motion pertaining to the principles set out in picture #2 and #3 applying to the separation of unequal inertial masses by the work of the potential energy contained in an inertial mass moment of a flywheel-rotor. Picture #6 presents the formula for a mutual separation-acceleration between a straight line displacement motion of a rotor- flywheel separating from the inertial mass of a device. Picture #7 presents the stopping- de-acceleration of the straight line displacement rotor- flywheel motion. There are two separating motion and two stopping motion of the rotor- flywheel rotational to straight line displacement coupled motion: ® Page -55- Important to note in picture #6 are the inverse quare root out of three sums wich indicates a compound feedback system. Next presentation is picture #7, the calculation for the rotor-flywheel angular velocity progression pertaining to the rotational to straight line displacement coupled motion progressing from a straight line velocity Vf from a rotational velocity ωb to a momentary stop condition after a 90º turn increasingly progressing onto ωc. The straight line velocity of the isolated system, during this type of motion, is opposing the velocity +Vd due to the straight line displacement energy conserving collision type motion of the flywheel-rotor on proven on page 70 with vectors and is also ® Page -56- applicable if additional energy is induced during this type of motion. Important to note in picture #6 we have a compound feedback system while in the next picture #7 we have a singular feedback system: ® Page -57- ® Page -58- MATHEMATICAL FOOTPRINT SUMMARY UNIFORM VERSUS NON-UNIFORM MASS MOTION FOR STRAIGHT LINE DISPLACEMENT Postulation: The secant line connecting the end-points of a curve represents the average slope of that curve and is the proven average slope of the curve. Thereby: Force, average = mass * Velocity, gain / Time, duration No matter how the mass got there, only the velocity gain and time duration determines the average force. Speed, average = Distance, displacement / Time, duration No matter how the mass got there only the distance displacement and the time duration determines the average Speed. Therefore: The first two formulas combined: Force, average = mass * Velocity, gain * Velocity, average / Distance, displacement Finally: The above formula converted from force to work Work, performed = mass * Velocity, gain * Velocity, average = Kinetic, energy, consumed And: Work, performance = mass * Velocity², gain / 2 = Kinetic, potential, energy Ref: Proven by experiment: Gaspard Coriolis Because real usable: impulse² =mass * 2 * work. performed THEREFORE IMPULSE = mass / (2 * Velocity, gain * Velocity, average) MATH FOOTPRINT FOR CALCULATING THE UNIFORM ROTATIONAL TO STRAIGHT LINE COUPLED MOTION The Continuous Centripetal Force for Rotational motion is: ® Page -59- Force, continuos, centripetal = mass, orbital, motion * Acceleration, centripetal Acceleration, continuous, centripetal =Velocity², tangential / Radius, crank ω = Angular, velocity Velocity, tangential = ω * Radius, orbit, motion Acceleration, continuous, centripetal = (ω * Radius, orbit)² / Radius, crank Acceleration, centripetal =ω² * Radius, orbit Therefore average Force for ¼ turn: Force, average, ¼, turn = mass, flywheel * Radius, orbit * ω² * 2 / π Furthermore by multiplying by time duration we get the impulse: Time, duration,¼, turn = π / 2 * ω Impulse, average,¼, turn = mass, straight line, displacement* Radius, orbit * ω MATH FOOTPRINT FOR CALCULATING THE NON- UNIFORM ROTATIONAL TO STRAIGHT LINE COUPLED MOTION Rotor Angular Velocity at 0º, the start of straight line motion = ω,a Rotor Angular Velocity at 90º, the end of straight line motion = ω,b ω,a > ω,b Average (Mean Value) Angular velocity for the ¼ Rotor turn: ω mean, value = ( ½(ω,a + ω,b )) Squaring the mean angular velocity we get:( ½(ω,a + ω,b ))² Therefore the Average force for non uniform Rotational to Straight line Coupled motion for a drive phase (the drive phase will be presented later) is: Force, average,¼ turn, non, uniform = = mass, straight line, displacement * Radius *( ½(ω,a + ω,b ))² *2 /π Therefore by multiplying Force, average, ¼turn, non, uniform * Time, duration, ( ½(ω,a + ω,b )) turn; we get Impulse: ® Page -60- Time, duration,¼, turn = π / (2( ½(ω,a + ω,b ))) Impulse, average,¼ turn = mass, straight line, displacement * Radius, crank * ½(ω,a+ω,b) Average angular velocity for rest of the ¾ rotor turn is a ω,b progression rising up by the multiplying ω,b with constant C2. Then slowing back down by multiplying ω,bC2 with constant C1 because of the kinetic energy flow into and out of the straight line mass motion. This principle is presented in the next picture #8: ® Page -61- The picture #8 reveals that the drive phase impulse having a larger mechanical energy potential is ( ½(ω,a + ω,b )) average progression and is opposed by one Idle phase impulse having a (½(ω,b +ω,bC2) average progression. The ω,bC2 accounts for the kinetic energy flow from the stopping of the straight line displacing mass during the Idle Phase presented in picture#6. Thereby, the three impulses progressing for each 1/4 turn after the drive phase must be alternately subtracted and added to arrive at the exact resultant internal self contained impulse. The sum of impulses during the idle phase: impulse, idle=mass *radius*(-(½(ω,b+ω,b C2))-(½(ω,b+ω,b C2*C1)+(½(ω,b+ω,b C2²*C1)) Refined to: Impulse, idle=mass *radius*(½ω,b(-1-C2-1-C2C1+1+C2²C1)) Drive phase and Idle Phase will be defined later. The idle phase impulses from 90º to 360º can be algebraically solved to the mathematically exact: Impulse,idle=mass*radius*1/2ω,b(-1-C2-C2C1+C2²C1) C3=(-1-C2-C2C1+C2²C1) The Net Impulse is:Impulse, net= mass, straight line, flywheel, displace * Radius, crank * ½(ω,a+ ω,b(1+C3)) The Self contained interna