2) Newton's Unfinished Theorem, The Physics of Inertial Propulsion Drive, (Section 2) by ggutsche1

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									    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:

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       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

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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:




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   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:

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 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:


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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:




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      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
								
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