Value discussion of
Inertial force deviation propulsion system
Let's try to discuss how many thrust we can get when we
made this system .
At first, we show shape of the system clearly.
The acceleration in point Q is the following expression.{Time,x,y,z}
The direction of thrust looks down when we choose direction
of
as plus.
By the way, z component is "centripetal acceleration"
not "centrifugal force" .
We don't multiply "a" by "mass" yet
.
And
shows relativity theory factor.
If we think the velocity of light c as infinite ,=1.
It is just the Newtonian mechanics.
Next we try to discuss with the following assumption.
Velocity of light c=299792458 m/s
Gyro radius@r=1 m
R=1 m
=100,000 rad/sec (15,915 Hz)
=100,000 rad/sec (15,915 Hz)
( If or=-100,000 rad/sec,the direction
of thrust looks up.)
At first, I will try to plot
,
to discuss the distribution of time delay on Q.
If=1.00000000@There are no time delay
on gyro.
Bump of time delay causes inertial force deviation.
The centripetal acceleration of point on gyro should cancel
in Newtonian mechanics.
But, it doesn't cancel in this device.
Next we plot the following expression
to discuss how centripetal acceleration remains behind
.
Roughly, we can understand
that maximum 450 G of residual centripetal acceleration
exists.
We can calculate thrust
by multiplying area of domain applied reddened of this
figure
by line density and radius r of gyro.
For example, if we operate the device with the above-mentioned
condition ,
and this gyro is ring type
and line density of ring is 10Kg/m (mass of ring gyro is
63 kg) ,
we can get roughly 70,000[N](7,000Kgf) thrust.
But, this device is very terrible the machine .
At this condition, acceleration of gyro tip is
3,000,000,000 G !!!
The world of 3000000000 G is the world that weight of hair
becomes 100Kgf.
We need very strong material !
And, the bearing which is usable by 16,000Hzi960,000rpmj
doesn't exist in the world, too.
The bearing of present age usable by 2,000 Hz maximum .
We may need waiting for an age of great-grandchild.