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=𝟐𝛑𝐈

𝐊

I=𝒎𝑳𝟐

𝟐

T=𝟐𝝅𝒎𝑳𝟐

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

𝒓𝟐

Top view:

GRAVITY




𝒓𝟐 Top view:

This force applies a

torque to the torsion

balance equal to 𝑳𝑮𝒎𝑴

𝒓𝟐




𝒓𝟐




𝒓𝟐

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.




𝒓𝟐 Top view:




𝒓𝟐

The torque rotates de

torsion balance by an

angle θ :

𝒌𝜽 = 𝑳𝑮𝒎𝑴

𝒓𝟐

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 = 𝑳𝑮𝒎𝑴

𝒓𝟐




𝒓𝟐




𝒓𝟐

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 = 𝑳𝑮𝒎𝑴

𝒓𝟐

And

Eq.(1)

By replacing K from (2)

into (1), we obtain:

𝐺=𝟐𝛑𝟐𝐋𝐫𝟐

𝐌𝐓𝟐𝛉

T=𝟐𝝅𝒎𝑳𝟐

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. EXAMPLEIT MUST BE

EASY THEN??

STEEL BALL

1M DIAMETER

STEEL BALL

1CM DIAMETER

F



EASY THEN??

STEEL BALL

1M DIAMETER

STEEL BALL

1CM DIAMETER

F

Mass 4,084kG or 9,001lbs!! Mass 4gr



EASY THEN??

STEEL BALL

1M DIAMETER

STEEL BALL

1CM DIAMETER

F=????


Almost 4 tons of steel,

just about the mass of 2

large SUV.!!!



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

𝒓𝟐.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

𝒓𝟐is 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=5T=1260 Rods

R=1

In addition to the iso-thickness model, we studiedthe iso-mass model. In the iso-mass model, thecylinder is split in N concentric shells of equal massand each shell is split in T sectors. Each sector is thenreplaced by a rod of constant mass m.

𝒎 =𝑴

𝑵𝑻with m = Mass of the cylinder

The external radius of the shell(n) is:

𝒓 𝒏 = 𝑹𝒏

𝑵And the Cartesian coordinates of the rod(n,t) are:

𝑥 𝑛, 𝑡 =𝑅

2

𝑛

𝑁+

𝑛 − 1

𝑁cos(

2𝜋𝑡

𝑇)

𝑦 𝑛, 𝑡 =𝑅

2

𝑛

𝑁+

𝑛 − 1

𝑁sin(

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) =𝟐𝑮𝑴(𝒏,𝒕)

𝒙 𝑳𝟐+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

𝒓𝟐

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

𝟑𝝅𝝆𝑮𝑹

g = 𝟒

𝟑πρGR3/X2 at a distance X

M

m

Solid Sphere of diameter R and density ρ

F The volume of a sphere is V=𝟒

𝟑𝝅𝑹3

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

FH>>R

Sh

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

FH>>R

Sh

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 g

s/g

c=2/3

Gs

= 67% of gc

If H=10D=20R, the mass of the

cylinder is M = 20ρπR3


mmM 1

5M

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 g

c/g

s= 3/2

gc

= 150% of gs

If H=10D=20R, the mass of the

cylinder is M = 20ρπR3

H=

20

R


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/g

sis 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/g

s

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…

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Gravity machine