Aerospace Reactionless Propulsion and Earth Gravity Generator

Aerospace Reactionless Propulsion

And

Earth Gravity Generator

by

Elijah Hawk

Art by

PROLOGUE

These mathematical models are presented here as theoretical conceptsthat may, or not, represent actual workable mechanizes. According tothe present, well established view of the existing laws of physics, they willnot work. It is the view of this author that the existing laws of physics,which are based on the the Inertial Frame of Reference, need to be modifiedin order to reflect the dynamics within the Non-inertial Frame of Reference.It is to this end that this work is hereby presented for others to evaluate.

Chapter 1

Chapter 2

Chapter 3

Distinction

Inertial Frame of Reference vsNon-inertial Frame of Reference

Mathematical ModelReactionless Propulsion

Mathematical ModelEarth Gravity Generator

“The only way of discovering the limitsof the possible is to venture a little waypast them into the impossible.”

Arthur C. Clarke (Clarke's second law)

“In order to do the impossible, you must see the invisible”

David Murdock

“Conventional wisdom leads to stagnation.Unconventional wisdom leads to advancement.”

Elijah Hawk

Chapter 1

Inertial Frame of Reference vsNon-inertial Frame of Reference

Inertial Frame of Reference

In physics, an inertial frame of reference (also inertial reference frame or inertial frame or Galilean reference frame or inertial space) is a frame of reference that describes time and space homogeneously, isotropically, and in a time-independent manner.

Landau, L. D.; Lifshitz, E. M. (1960). Mechanics. Pergamon Press. pp. 4–6.


Inertial Frame 1 Earth Gravity


Inertial Frame 2 Rocket Acceleration

Rotational Frame of Reference

Non-Inertial Frame 3 Centrifugal Force

3000 ft 3000 ft

1 rpm

1g1g

12,000 ft @ ½ rpm

48,000 ft @ ¼ rpm

Centrifugal Force (Rotating Reference Frame)

In classical mechanics, the centrifugal force is an outward force which arises when describing the motion of objects in a rotating reference frame. Because a rotating frame is an example of a non-inertial reference frame, Newton's laws of motion do not accurately describe the dynamics within the rotating frame. (John Robert Taylor)

Einstein's Principle of Equivalence

The equivalence principle was properly introduced by Albert Einstein in 1907, when he observed that the acceleration of bodies towards the center of the Earth at a rate of 1g (g = 9.81 m/s2 being a standard reference of gravitational acceleration at the Earth's surface) is equivalent to the acceleration of an inertially moving body that would be observed on a rocket in free space being accelerated at a rate of 1g. Einstein stated it thus:

“We assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system”.

—Einstein, 1907

The Hawk Principle of Equivalence

All Three Frames of Reference Affect Mass Proportionately the Same

Inertial Frame 1 Earth Gravity

Inertial Frame 2 Rocket Acceleration


The Hawk Equivalency

Newton's Laws/Inertial Frames

The laws of Newtonian mechanics do not always hold in their simplest form.... Newton's laws hold in their simplest form only in a family of reference frames, called inertial frames. This fact represents the essence of the Galilean principle of relativity: ”The laws of mechanics have the same form in all inertial frames”.

Milutin Blagojević: Gravitation and Gauge Symmetries, p. 4

The Laws of Physics Vary

Physical laws take the same form in all inertial frames. By contrast, in a non-inertial reference frame the laws of physics vary depending on the acceleration of that frame with respect to an inertial frame, and the usual physical forces must be supplemented by fictional forces.

Milton A. Rothman (1989). . Courier Dover Publications. p. 23

Sidney Borowitz & Lawrence A. Bornstein (1968). A Commentary View of Physics

Spiral Galaxies

Modified Newtonian Dynamics (MOND) is a hypothesis advanced by Mordehai Milgrom (Milgrom, 1993) in order to explain the anomalous rotation of spiral galaxies. Many such galaxies do not appear to obey Newton's law of gravitational attraction....

EarthTech International Website http://earthtech.org/mond/ Harold Puthoff, Ph.D

http://earthtech.org/mond/

Inertial vs Non-Inertial Frames of Reference

Inertial

Frame of Reference

Non-Inertial

(Rotating)

Frame of Reference

vs

Newton's Laws Fully ApplyLaws of Physics Well Established

Newton's Laws do not Necessarily ApplyLaws of Physics VaryLaws of Physics not Well Established (Yet)

(Non-Rotating)

Chapter 2

Mathematical ModelReactionless Propulsion

Pendulum Definitions

1) Displacement: At any moment, the distance of bob from mean position. It is a vector quantity.

2) Amplitude: Maximum displacement on either side of the mean position.

3) Vibration: Motion from the mean position to one extreme, then to the other extreme and then back to the mean position. (Time Period = “T”)

4) Oscillation: Motion from one extreme to the other extreme. One Oscillation is half Vibration.



3000 ft 3000 ft

1 rpm

1g1g

12,000 ft @ ½ rpm

48,000 ft @ ¼ rpm


ROOM

TETHER

TEST STAND

PENDULUM

1g

T = 2(pi) Lg

KE 1

KE 2

PE 1

PE 2

KE 1

PE 2KE 2

EDGE VIEW

KE 1PE 1KE 2PE 2 PE 2

TOP VIEW

CF=0

CF=0

CF=0

CF=MAXCF=MAX

Pendulum Motion in Rotation

Plot of Pendulum CF Vectors (Oscillation only)

Note: There are two centrifugal forces superimposed along the pendulum arm. One from the rotation about the spin axis. The other from the pendulumoscillation only as shown in the “Top View” sketch above.

General Pendulum Formulas

2

Vibrations/Revolution Options

The Hawk Anomaly

PE 1PE 2

CF Vectors thru 720 degrees

Top View

Pendulum Centrifugal Force Vectors

KE 1KE 2

CF 1 +CF 2

Calculations 1

Calculations 2

Given:

Determine GravitySpin Radius=25 cm(Spin Diameter=50 cm)

STEP 2:

Centrifuge Gravity Formula

F=5.59 X 10 DN -6 2

5.59 X 10 (50 cm) (1000 rpm) =279.5 g’s2-6

279.5 X 9.8 m/sec = 2739.1 m/sec2 2

Calculations 3

Given:

Determine Pendulum LengthSpin Radius=25 cmGravity 2739.1 m/sec

STEP 3:

2

Pendulum Formula

T=2(pi)Lg

L=g

4(pi)

2739.1 X .0144 = 1 m

0.12 sec (For One Vibration)(@ 1000 rpm)

Spin Axis/Pendulum Length = 1:4 (Constant)

T2

2 4 X 9.869

Calculations 4

Given:

Determine Pendulum “h”Length = 1 mDisplacement = 0.1 m

STEP 4:

Pythagorean Theorem

L - (L - D ) = h2

1 - (1 - .1 ) = 5.013 mm

L

Dh

LFormula for Angle

= ASIN ( )DL

ASIN ( ) = 5.74 Degrees.11

2

2 2

Calculations 5

Given:

Determine System EnergyMass = 25 kgGravity = 2739.1 m/sec“h” = 5.013 mm

STEP 5:

Formula for Energy:

P.E. = K.E.P.E. = mgh

P.E. = 25 X 2739.1 X .005013 = 343.25 Joules

K.E. = 1/2 mv

2

2

V = = 5.24 m/sec (Max. Pen. Velocity)343.2525 2( )

Calculations 6

Given:

Determine Centripital Force @ K.E. Max.Mass = 25 kgMax. Pen. Velocity = 5.24 m/secPen. Length = 1 m

STEP 6:

Formula for Centripital Force:

CF = 2mv

R

25 X (5.24) 1 m

= 686.44 N (154.32 LBf)2

g

g

FulcrumFulcrum

PE 1

PE 2

KE 2KE 1

SP

IN A

XIS

Side View

Work/Energy Relationship

W = fa

f

Energy Calculation

Energy = 343.25 joules

Time for One Oscillation = 0.06 sec

343.25/0.06 = 5720.8 watts

5720.8/686.44 = 8.3 watts/Newton

5720.8/154.32 = 37.1 watts/lbs

37.1 @ 50% eff = 74.2 watts/lbs

37.1 X 2.2 @ 50% eff = 163.2 watts/kg

Alternate (Real World) Comparison

Bell Jet Ranger Helicopter Performing 1g work against Gravity

Weight = 2500 lbs+/-

Horse Power to Hover = 250 hp +/-

250/2500 = 0.1 hp/lbs

0.1 X 745 = 74.5 watts/lbs

74.5 X 2.2 163.9 watts/kg

(compare results to previous page)

K.E. Max. CF @ RPM

RPM Newtons Lbf

1000 686 154

1500 1545 347

2000 2746 617

2500 4291 965

3000 6178 1389

4000 10984 2469

5000 17162 3858

6000 24714 5556

ConstantsPendulum Length = 1mSpin Radius = 25 cmAmplitude = 100 mmMass = 25 kg

Continuous 1g Space Travel

Destination Time MPH @ Mid Point

Moon 3.5 Hrs 136,947 MPH

Mars 2.08 Days 1,956,445 MPH

Jupiter 5.88 Days 5,540,258 MPH

Saturn 8.38 Days 7,897,326 MPH

Uranus 12.23 Days 11,523,886 MPH

Neptune 15.46 Days 14,567,166 MPH

Pluto 17.17 Days 16,748,180 MPH

Chapter 3

Mathematical ModelEarth Gravity Generator

g g

Fulcrum

Fulcrum

PE 1 PE 2

KE

2

KE

1

SPINAXIS

Earth Gravity

Earth Gravity Generator

AC Output

Generator

CF CF

Note:System at EquilibriumRPM. (CF = g)

Motor

Hawk's Anomaly

Calculations 1Determine Equilibrium RPM

Where CF = g

Given: Bob Mass = 250 kg Spin Radius = 1 m

Calculations 2

Calculations 3

Calculations 4

Calculations 5

Energy Calculation

System Energy = 77,840 joules

Time for One Oscillation = 2.006 sec

77,840/2.006 = 38,803 watts (38.8 kw)

Q.E.D.

Aerospace Reactionless Propulsion and Earth Gravity Generator

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