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
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.
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
Non-Inertial Frame 3 Centrifugal Force
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
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)
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.
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
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.
PE 1PE 2
CF Vectors thru 720 degrees
Top View
Pendulum Centrifugal Force Vectors
KE 1KE 2
CF 1 +CF 2
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
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
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
Energy Calculation
System Energy = 77,840 joules
Time for One Oscillation = 2.006 sec
77,840/2.006 = 38,803 watts (38.8 kw)