Gravity machine

BERT’S PROJECTS: THE GRAVITY MACHINE

Thin Wire
Mass m
Length l
Mass m’=m
GRAVITY TORSION BALANCE ASSEMBLY

The torsion balance behaves
as a harmonic oscillator with
a natural frequency T

T=𝟐𝛑
𝐈
𝐊
I=
𝒎𝑳 𝟐
𝟐
T=𝟐𝝅
𝒎𝑳 𝟐
𝟐𝑲
I is the moment of inertia
K is the torsion constant
By measuring the period of
oscillation, we determine K the
torsion coefficient of the balance
Equation(1)

The two larges spheres can
rotate in the same plane as
the two small spheres
around the same axisLarge Mass M, M’

Top view: The large spheres are
equidistant to the small spheres. All
gravitation effects are cancelled out.
GRAVITY

We rotate the two larges
spheres clockwise… …
GRAVITY

Bringing them as close as possible to the
small spheres m and m’. The distance
between the large and small spheres is r.
Top view:
GRAVITY

The force due to gravity
between the large and
small sphere is F=
𝑮𝒎𝑴
𝒓 𝟐
Top view:
GRAVITY

small sphere is F=
𝑮𝒎𝑴
𝒓 𝟐 Top view:
This force applies a
torque to the torsion
balance equal to 𝑳
𝑮𝒎𝑴
𝒓 𝟐

small sphere is F=
𝑮𝒎𝑴
𝒓 𝟐
𝑮𝒎𝑴
𝒓 𝟐
Note: This is INCORRECT as the
force is not perpendicular to the
torsion balance. It is easy to
correct using simple
trigonometry. These calculations
have been omitted for clarity in
this presentation.

small sphere is F=
𝑮𝒎𝑴
𝒓 𝟐 Top view:
𝑮𝒎𝑴
𝒓 𝟐
The torque rotates de
torsion balance by an
angle θ :
𝒌𝜽 = 𝑳
𝑮𝒎𝑴
𝒓 𝟐

Top view:
2θ
By using a laser
and a mirror at the
center of the
torsion balance, it
is possible to
measure very small
variations of 𝜽.
The torque rotates de
angle θ :
𝒌𝜽 = 𝑳
𝑮𝒎𝑴
𝒓 𝟐
small sphere is F=
𝑮𝒎𝑴
𝒓 𝟐
𝑮𝒎𝑴
𝒓 𝟐

Top view:
2θ
By using a laser
and a mirror at the
center of the
torsion balance, it
is possible to
measure very small
variations of 𝜽.
The torque rotates the
angle θ and
Eq.(2) 𝒌𝜽 = 𝑳
𝑮𝒎𝑴
𝒓 𝟐
And
Eq.(1)
By replacing K from (2)
into (1), we obtain:
𝐺=
𝟐𝛑 𝟐 𝐋𝐫 𝟐
𝐌𝐓 𝟐 𝛉
T=𝟐𝝅
𝒎𝑳 𝟐
𝟐𝑲

Top view:
We can obtain a second
measurement of θ by
rotating the large spheres
anticlockwise
2θ’
GRAVITY

Cavendish did this experiment in 1797 and obtained a
value for G of 6.74 10 -11 S.I. This is within 1% of current
value of : 6.67428 10 -11 S.I.!
IT MUST BE
EASY THEN!!
GRAVITY

Gravity is an astonishingly
weak force. EXAMPLE
IT MUST BE
EASY THEN??
STEEL BALL
1M DIAMETER
STEEL BALL
1CM DIAMETER
F

weak force. EXAMPLE
IT MUST BE
EASY THEN??
STEEL BALL
1M DIAMETER
STEEL BALL
1CM DIAMETER
F
Mass 4,084kG or 9,001lbs!! Mass 4gr

weak force. EXAMPLE
IT MUST BE
EASY THEN??
STEEL BALL
1M DIAMETER
STEEL BALL
1CM DIAMETER
F=????
Almost 4 tons of steel,
just about the mass of 2
large SUV.!!!

weak force. EXAMPLE
IT MUST BE
EASY THEN??
STEEL BALL
1M DIAMETER
STEEL BALL
1CM DIAMETER
The force of gravity between the 2 spheres is F= 4.2 10 -9 N But what does it mean?

weak force. EXAMPLE
F= 4.2 10 -9 N is the weight of a steel ball with a diameter of 0.2mm!!
0.2mm is the average thickness of a human hair. Such a ball would
NOT be visible with the naked eye! The force of gravity is
RIDICULOUSLY WEAK indeed.
Cavendish measuring G within 1% with the limited resources of its
time was a truly gifted experimental scientist. Or maybe, a lucky one!
IT MUST BE EASY THEN??
I THINK NOT!

GRAVITY MACHINE
The balance must
be EXTREMELY
sensitive. It must
be located inside a
rigid enclosure.

GRAVITY MACHINE
Polycarbonate Window
for Laser beam
Adjustable Tube supporting
the balance. By changing
the length, we can adjust the
TORSION COEFFICIENT
Angular offset
adjustment of
the balance
Rigid Frame
supporting the
balance at
waist level
Side View
Alignment of Balance
and Turntable by Laser
3” PVC PIPE

Handle for moving
the mass on the
turntable beam
Large Mass
Support # 1
ANGLE
digital
readout

GRAVITY MACHINE
Remote control
for internal lights
Green Laser
Red Laser
Small air pump to give an
impulse to the balance
Small Mass 42.4gr1/8” Carbon Rod
Light Weight Hub
Front Coated Mirror (from an SLR)
Front Coated Mirror
(Telescope)
Aluminum Hub for
¼” Carbon Rod
TORSION BALANCE

GRAVITY MACHINE – M
The best material for the
large mass would be:
Material Density
Platinum 21.5
Uranium 20.2
Gold 19.3
Mercury 13.6
Lead 11.4
Silver 10.5
Copper 9.0
Iron 7.9
The force due to gravity between
the large and small sphere is
F=
𝑮𝒎𝑴
𝒓 𝟐 .To maximize F we
need M as large as possible r as
small as possible. We need the
largest mass in the smallest
volume. Therefore, we need to
use a material with largest
possible DENSITY.

The best materials for the
Material Density
Platinum 21.5
Uranium 20.2
Gold 19.3
Mercury 13.6
Lead 11.4
Silver 10.5
Copper 9.0
Iron 7.9
The cost for most of these heavy
metals is ASTRONOMICAL as they
are very rare. This is no wonder as
no element with an atomic number
above IRON can be synthetized in
stars through fusion processes.
They can only be synthetized by
converting potential gravitational
energy in a Nova or Super Nova
event. They are indeed
ASTROMICALLY expensive…

GRAVITY MACHINE – COST $$$$$
Grapefruit O.D.= 10 cm bowling ball O.D.= 22 cm
Metal Density(kg/m3)
Mass
(kg)
Price USD
Mass
(kg)
Price USD Metal Trading Price
Steel 7800 4.1 $1.55 43.5 $16.53 Steel $0.38
lead 11400 6.0 $14.33 72.6 $174.30 Lead $2.40
gold 19300 10.1 $417 355.35 123.0 $5 077 962.52 Gold $41 300.00
uranium 20200 10.6 $877.87 128.7 $10 680.99 Uranium $83.00
platinum 21500 11.3 $493 072.97 137.0 $5 999 218.79 Platinum $43 800.00
The metal prices below are TRADING PRICES
for bulk material and NOT retail prices. A quick
trip to Home Depot will show you that retail price
for Steel IS NOT $0.38. Prices for Uranium is for
refined Uranium (99.3%U 238 , 0.7%U 235 )
American Forces shot 1,000 metric tons of DUF (depleted Uranium) in three weeks during the invasion of Iraq, mainly 30mm
rounds (PGU-14/B) from the evil GAU8 Avenger Gatling gun of the A-10 Thunderbolt and 120mm APFSDS (Armor Piercing Fin
Stabilized Discarding Sabot) from the M1A1 and M1A2 Abrams battle tank. Maybe, I could get a good deal there…. Maybe not!

The best materials for the
Material Density
Platinum 21.5
Uranium 20.2
Gold 19.3
Mercury 13.6
Lead 11.4
Silver 10.5
Copper 9.0
Iron 7.9
Lead is the only heavy metal that is
somewhat affordable. Yet, it is still
expensive and, unfortunately, it is
TOXIC. As a result, we decided
against melting and casting large
quantity of lead. We chose the least
expensive material in this list: IRON.
Pure IRON is rare. However, low
carbon alloys (MILD STEEL) is
widely used in construction.

SHAPE
The most efficient packaging of
a material is the SPHERE. In
addition, the gravitational field
around a sphere of mass M is
equivalent to the gravitational
field of an infinitesimal point of
mass M, greatly simplifying the
calculations.
However, a large sphere of steel is a
very uncommon object. Getting one
made would be excessively expensive.
On the other hand, large round bars
of steel are quite common and can be
purchased on the recycling market.
F=
𝑮𝒎𝑴
𝒓 𝟐 is only valid
if M and m are solid spheres

GRAVITY MACHINE – GRAVITATIONAL FIELD OF A CYLINDER
GRAVITATIONAL FIELD
gc(r) is the gravitational field of a
cylinder of mass M at a distance r
from the axis of cylinder in a plane
perpendicular to the main axis and
equidistant to both extremities of the
cylinder. The gravitational force on
mass m (sphere) is:
F= gc (r) m
®
d
Main axis
Mass m
Mass M
R
2L

GRAVITATIONAL FIELD
We were unable to obtain an analytical
solution to the integrals* for gc(r) nor
were we able to find such a solution in
published articles*. Instead, we
developed a simulation in Excel by
modeling the cylinder as an assembly of
rods of negligible diameter.
®
d
Main axis
Mass m
Mass M
R
2L
*I strongly believe that a solution exists. I am just
to old for this. It’s just a matter of finding the
magic substitution… as it often is with integrals.

In the iso-thickness model, the
cylinder is cut into N shells of equal
thickness. Each shell is then
separated into T sectors. At the center
of each sector, a rod is located with a
mass equal to the mass of the sector.
In this particular example N=5 and
T=24, the cylinder is replaced by a
concentric distribution of 120 rods.
Note: This a scan
from a hardcopy

GRAVITY MACHINE – ISO-MASS MODEL
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
-1.1 -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1
n=1
n=2
n=3
n=4
n=5
N=5
T=12
60 Rods
R=1
In addition to the iso-thickness model, we studied
the iso-mass model. In the iso-mass model, the
cylinder is split in N concentric shells of equal mass
and each shell is split in T sectors. Each sector is then
replaced by a rod of constant mass m.
𝒎 =
𝑴
𝑵𝑻
with m = Mass of the cylinder
The external radius of the shell(n) is:
𝒓 𝒏 = 𝑹
𝒏
𝑵
And the Cartesian coordinates of the rod(n,t) are:
𝑥 𝑛, 𝑡 =
𝑅
2
𝑛
𝑁
+
𝑛 − 1
𝑁
cos(
2𝜋𝑡
𝑇
)
𝑦 𝑛, 𝑡 =
𝑅
2
𝑛
𝑁
+
𝑛 − 1
𝑁
sin(
2𝜋𝑡
𝑇
)
We used the Iso-thickness model in our simulation.

GRAVITATIONAL FIELD
The gravitational field of Rodn,t at the
location of mass m is:
𝑔(n,t) =
𝟐𝑮 𝑴(𝒏,𝒕)
𝒙 𝑳 𝟐+𝒙(𝒏,𝒕) 𝟐
The total gravitational field is the vector
summation of each individual rod
contribution.
r
Main axis
Mass m
Mass M
R
2L
Xn,t
of mass Mn,tRodn,t

GRAVITATIONAL FIELD
In this experiment, the mass (m) is very
close to the surface of the cylinder. To
ensure a good computation of the
gravitational field, it is critical to use as
many rods as possible. We used up to
1 million rods!Yes, in Excel!
However, the results showed that 10,000
rods is sufficient.
d
Main axis
Mass m
Mass M
R
2L
of mass Mn,tRodn,t
Xn,t

GRAVITY MACHINE – ON THE ISSUE OF SHAPES (1)
QUESTION
Given a material with a density ρ, what
shape would maximize the strength of the
gravitational field.
QUESTION
The answer is obvious: THE SOLID
SPHERE, OF COURSE! … But is it?
The force due to gravity between a
large and small sphere is
F=
𝑮𝒎𝑴
𝒓 𝟐
To maximize F we need M as large
as possible and r as small as
possible. Assuming m is very small,
the force will be maximum when m
is on the surface of M, at a distance
r equal to the radius of M.

Max g on the surface of the sphere
g =
𝟒
𝟑
𝝅𝝆𝑮𝑹
g =
𝟒
𝟑
πρGR3/X2 at a distance X
M
m
Solid Sphere of diameter R and density ρ
F The volume of a sphere is V=
𝟒
𝟑
𝝅𝑹3
Therefore, to double the value of gmax, we need
to increase the mass 8 times! But if we double
the density of the material, we double gmax.

M
Let’s replace the mass M with a very tall
cylinder.
If H=10D=20R, the mass of the cylinder is
M = 20ρπR3
This is 15 times the mass of the sphere.
What would be the maximum
value of the gravitational field?
Solid Cylinder of
diameter R and
density ρ with
H>>R=20R
H>>R

If H>>R, The
gravitational field in the
middle section of the
cylinder and close to its
surface is perpendicular to
the surface of the cylinder.
Therefore, the flux of the
gravitational field across
the surface S is:
Flux=2πXhg
m
Solid Cylinder of diameter R
and density ρ
F
H>>R
S
h
X

By applying the GAUSS
Theorem, the flux of the
gravitational field across a
surface is equal to 4πGM
Where M is the mass
included in the surface S.
M = ρπR2h
4πGρπR2h =2πXhg
g=2πρGR2/X
g=2πρGR if X=R
m
Solid Cylinder of diameter R
and density ρ
F
H>>R
S
h
X

SPHERE
The gravitational field on the yellow
dashed line, in close proximity of the
surface is: gs=4/3πρGR3/X2. The
maximum value on the surface is:
gs=4/3πρGR or gs/gc=2/3
Gs = 67% of gc
If H=10D=20R, the mass of the
cylinder is M = 20ρπR3
mm
M
15M
XX
2R
2R
TALL CYLINDER
The gravitational field on the yellow
dashed line, in close proximity of
the surface is: gc=2πρGR2/X. The
maximum value on the surface is:
gc=2πρGR and gc/gs = 3/2
gc = 150% of gs
If H=10D=20R, the mass of the
cylinder is M = 20ρπR3
H=20R

0%
50%
100%
150%
200%
250%
300%
350%
400%
450%
500%
0
500
1000
1500
2000
2500
3000
3500
4000
4500
1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0
X/R
gs/G
gc/G
gc/gs
Example for a steel round
bar of 6.75inch diameter
R= 0.0857m
ρ= 7800kg/m3
gs is proportional to 1/X2
gc is proportional to 1/X
gc/gs is proportional to X
Sphere of radius R
Tall cylinder of radius R
H=20R Cylinder over
sphere ratio

Gravitational Field Modelling
The torsion balance must be
extremely sensitive. Therefore, it
must be protected from any air
current by an enclosure. In our
design, the torsion balance is within
a 3” PVC pipe. The outside diameter
of the pipe is 3-1/2”. The minimum
distance between the small mass m
and the surface of the large mass M
is 1.75” or 44.45mm
3” PVC Pipe
Small Mass M on
Torsion Balance
m
1.75” / 44.5mm

Our earlier field calculations using the Gauss Theorem are
only valid for very tall cylinders. We did extensive modelling
using our Excel simulator for cylinders of various H/R ratio.
We found that H/R=4 provided a good compromise of
reasonable total mass while still behaving like a tall cylinder
(i.e. decay in 1/X) within 50mm from the surface of the
cylinder. Using cylinders instead of spheres is very attractive
as round bars of large diameter are commonly used in the oil
and gas industry and can be purchased in the surplus or
recycling market.
Hard copy of an Excel Simulation

Example of field calculation using
our Excel simulator. The two curves
represent the fields of a cylinder
and sphere of equal radius with
H/R=4.5. The ratio of gc/gs on the
plot clearly indicates that gc decays
in 1/X. The vertical dashed lines
represent various size of PVC pipe.
Ratio gc/gs
Note: Scan of a
hardcopy as all
data was lost.

GRAVITY MACHINE
The large masses in our
gravity experiment have a
mass of 50.2kg. Each are
cut from a round bar of 6-
3/4” diameter. They were
purchased from a Houston
surplus metal supplier.
Nevertheless, steel is quite
expensive and the total
cost was $300.

GRAVITY MACHINE
All the elements of the
machine have been
fabricated (and currently
stored in the master
bedroom…) It only
requires assembly and
testing with various small
masses and fishing lines
for the balance. The only
thing missing is time…

Gravity machine

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