WJETech (we push from the inside)
Tecnologías William Jhon
Elliott Somerville E.I.R.L.
Rut: 76722619-5
Español
Version 3 of the
The Fluid Space Drive V3
(FSD) is a method of propulsion in microgravity environments that has no need
to expel propellant to generate thrust.
It does this by
propelling a mass (M2) inside the FSD from one end of the spacecraft to the opposite end but modifying the mass’s
drag coefficient decreasing it’s velocity when traveling to the rear end of the
spacecraft (M1)
What follows below is a
step by step description of the method and the corresponding calculations (you
can see a simple “tongue in cheek” description of the method here).
First the FSD’s dimensions (for this simulation):
Fig 1
Fig 1 shows a pressurized structure (M1) or
spacecraft in a micro-gravity environment, inside the spacecraft is a 100k mass
(M2) we call Ram Mass Assembly or RMA.
The RMA has freedom to move across the length
(not breadth) of the spacecraft.
The pressurized
structure (mass M1) is 30m long with a radius of 1,5 meters, the skin is 6mm
aluminum (more or less Lunar Lander) (1), M1’s mass will be
808,02 k (2)
Fig 2
Fig 2 illustrates the
RMA (M2), it has a series of flaps attached to servo motors, when the flaps are
closed they form the shape of an open cylinder that functions as an air brake
increasing the RMS’s drag coefficient therefore slowing it’s relative velocity
inside the pressurized structure (M1).
Note: The illustration
does not show the ideal aerodynamically shape for low drag, see note 3 for a better design
|
Recap |
|
M1 (pressured structure) mass = 808.02 k |
|
M2 (RMS) mass = 100 k. |
Cycle 1 (forward
ramming piston is activated)
Fig 3
The forward ramming
piston is activated with a 400n force giving the RMA (M2) a push (F1) in the –X direction (Fig x) accelerating it in the –X direction
(a2) with to 4 m/s (formula 1)
Formula 1
The opposite force (F2) gives the pressurized structure an acceleration in the +X
direction (a1) of 0,495 m/s (formula
2).
Formula 2
Fig 4
As mass M1 is traveling
unhindered (4) in the +X at 0.495 m/s (V1) it’s velocity will remain constant until
another force is applied.
Mass M2 is traveling in
a pressurized environment, the fluid (air) exerts a force (drag) as it
encounters the air brake decreasing M2’s –X velocity.
The drag force can be
calculated by using the drag equation (formula 1).
Fd = 0.5 * p * v² * A * Cd
Fd=
drag force.
P=
mass density of fluid (At 20 °C and 101.325 kPa, dry air has a density of
1.2041 kg/m3) in this example we will use 1.2 kg/m3 (x)
V=
Velocity in m/s
A
= surface of the 1m radius air brake (3.1415 m2)
Cd = drag coefficient
for an open cone 2.2
Formula 3
Therefore using the
above data we get:
Formula 4
The 66.350n force
pushing in the +X direction will change M2’s velocity by 0.6635 (formula 5)
resulting in a final velocity (for this time interval) of 3.3364 m/s (formula
6)
Formula 5
Formula 6
This calculation is to
be repeated every time interval until M2 collides with M1’s inner rear hull
with a final velocity of 1.4071398
|
Column A = time interval (1 second) Column B = total displacement of mass M1 at time interval. Column C = total displacement of mass M2 at time interval. Column D = Velocity of mass M1 at time interval. Column E = Velocity of mass M2 at time interval. Column F = Drag on M2 at time interval.
Drag change (M2)
Velocity change (M2) |
Fig 5
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For M1 (spaceship) |
|
|
Elapsed time |
11 s |
|
Distance travelled |
5.44 m |
Fig 6
Recap
After the 400n force is exerted, the elements do the
following:
M1 travels in the +X direction at 0.495 m/s constant
velocity (formula 2).
M2 travels with an initial velocity of 4 m/s but is slowed by
the drag generated by the air brake, it travels for 11 seconds until it
collides with M1’s inner rear hull with a final velocity of 1.325 (Fig 5)
End of cycle 1
Cycle 2: M2 collides with M1’s rear (-X) inner hull.
Fig 7
The instant M2 contacts M1 an electromagnet (not
shown) is activated to prevent any “bounce” insuring the collision is
inelastic.
Therefore we have two masses (M1 and M2) that
have an inelastic collision, to find the final velocity we use the formula:
Formula 7
Where M1 = mass of body
1
M2 = mass of body 2
V1 = Initial velocity of body 1
V2 = Initial velocity of body 2
v = Final velocity of both the bodies
The final velocity for inelastic collision is
given by:
Formula 8
Formula 9
Therefore M1 and M2 are traveling in the +X
direction at 0.294591 m/s after inelastic collision 1
For simplicity we permit masses M1 and M2 to
travel together at 0.294591 m/s for 1 time interval (1 second), traveling at
this time a distance of 5.7399 meters in 12 seconds the +X direction.
End of cycle 3
|
For M1 (spaceship) |
|
|
Total elapsed time |
12 s |
|
Distance travelled |
5.7399 m |
Cycle 4: The rear ramming piston is activated with a 200n
force.
Fig 8
The rear ramming piston
is activated with a 200n force, this gives the RMS (M2) a push (F5) in the +X direction (Fig 8) accelerating it in the +X direction
(a2) with to 2 m/s (formula 10)
Formula 10
As M2 was traveling with an initial velocity of
0.2945 m/s when the acceleration is added it has a final velocity of 2.2945 m/s
The opposite force (F6) gives the pressurized structure (M1) an acceleration (velocity
change) of 0.24752 m/s in the -X direction (formula 11).
Formula 11
As M1 had an previous
velocity of 0.294591 m/s (formula 9) it’s present velocity (formula 12) is
0.047075 (V1 + V2 = 0.294591 - 0.24752 = 0.047071)
Formula 12
As M2 travels inside M1
(fig 9) its velocity is decreased by drag, but because the air brake is
deactivated (flaps in open position) its drag coefficient is only 0.3, M2 “catches
up” to M1’s inner hull in time interval 29 (29 seconds from start of cycle)
with a velocity of 1,6 m/s (M2 velocity of 1.7 – M1 velocity of 0.004)
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Column A = time interval (1 second) Column B = total displacement of mass M1 at time interval. Column C = total displacement of mass M2 at time interval. Column D = Velocity of mass M1 at time interval. Column E = Velocity of mass M2 at time interval. Column F = Drag on M2 at time interval.
Drag change (M2) Velocity
change (M2) |
Fig 9
Fig 10
After 17 second of
travel (29s from start of cycle fig 3) in the +X mass M2 “caches up” with mass
M1 colliding with M1’s forward inner hull (Fig 11)
Fig 11
The instant M2 contacts M1 an electromagnet (not
shown) is activated to prevent any “bounce” insuring the collision is
inelastic.
Therefore we have two masses (M1 and M2) that have
an inelastic collision, to find the final velocity we use formulas 13 and 14:
Formula 13
Where M1 = mass of body
1
M2 = mass of body 2
V1 = Initial velocity of body 1
V2 = Initial velocity of body 2
v = Final velocity of both the bodies
The final velocity for inelastic collision is
given by:
Formula 14
Formula 15
For simplicity we permit M1 and M2 to travel
together for one time interval (1s), it will advance an additional 0.229931m
With this action in time
interval 30 (seconds) M2 is in its original position (attached to M1’s forward
inner hull) and the propulsion cycle is complete, therefore we have that M1
(the spaceship) has traveled a total of 6.75m in 30s.
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Parameters at end of cycle For M1
(spaceship) |
|
|
Total elapsed time |
30 s |
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Distance travelled |
6.77 m |
With the above
parameters we can calculate how much M1 (the spaceship) has accelerated using
formula 16
Formula 16
The spacecraft has an
acceleration of 0.015044m/s.
At first glance that
does not seem like much, but as the spacecraft will be able to accelerate
continuously as long as electrical power is applied it will soon become the
fastest machine humanity has ever constructed.
Therefore, if we start
with 0 velocity:
In one minute it will
accelerate to 0.9 m/s
In one hour it will
accelerate to 54 m/s
In one day it will
accelerate to 1,296 m/s
In a week it will
accelerate to 9,072 m/s (in two weeks it will surpass Voyager II’s present
velocity of 15,400 m/s)
In one month it will
accelerate to 38,880 m/s
In one year it will
accelerate to 473,040 m/s
In one decade it will
accelerate to 4,730,400 m/s
Notes (NOT COMPLETED
YET)
Note 1: thickness and material of the pressured
structures “skin” (
Note 2: Calculating M1’s mass
M1 length (h) = 30 m
M1 radius = 1.5 m
Thickness of aluminum
skin = 0.6 mm = 0.0006 m
Volume of M1 = pi * r *
r * h = 3.141592654 * 1.5 * 1.5 * 30 = 212.0575
Note 3 RMS design
Note x: Superior
materials are available for the pressurized structures skin but the
characteristics of aluminum are well known and adequate for this document.
Before we chant “Alpha
Centauri here we come” conceder what percentage of the force excreted by the
air drag on M2 is transmitted to the spaceship’s hull (nullifying the effect)
Let us include a factor pt
(momentum transmitted) if pt=1 we will understand that all the force slowing M2
is transmitted to the spaceship’s hull, and if pt=0 no force is transmitted to
the spacecraft.
The true pt value has to
be determined by experimentation (just as the drag coefficient of various
shapes is deduced by experiments), if pt has any value that is less than 1,
then the method described in this document works as a practical method of
propulsion in microgravity.
Assessment 1, decreasing
M2 velocity by direct contact with the spacecraft (fig 7)
Fig 7
If M2 is slowed by the
friction created by direct contact with the spacecraft’s inner hull (fig 7),
all the force (F4) that is slowing M2 will be transmitted (F5) to the
spacecraft (M1) and regardless of M2’s final velocity we can assume the
spaceship (M1) will not gain any acceleration (therefore the pt factor is 1)
Assessment 2, decreasing
M2’s velocity by air brake when diameter of spaceship is almost identical to
air brake’s diameter (tight fit) (fig 8).
Fig 8
If the air brake’s diameter is a tight fit with
the spacecraft’s diameter, it will act like a piston in a cylinder compressing the
air in the –X side and come to a full stop before reaching the –X inner hull,
but it is a logical assumption that the spacecraft will not gain acceleration
in any direction.
Assessment 3, decreasing M2’s velocity by air
brake when diameter of spaceship larger than the air brake’s diameter by a
factor of X2 or more (fig 8).
Fig 8
If we increase the spacecraft’s diameter with
relation to the air brakes, there will be sufficient space for the air to flow
around M2 without greatly affecting the spaceships velocity, we can assume
that the pt factor is less than one, and if we continue increasing the
spaceship’s diameter (fig 9) the pt factor will decrease.
