A scissors
is made of two rods attached by one of its ends. While the scissors closes, the other two end began to move one against
another. The point of contact, called "wire scissors' sliding the wire to
the tip.
By opening
and closing, it is remarkable that the wire slides from one point to another
stem growing fast enough that in a given moment it's been impossible to
accompany him with his eyes.
In example,
I'll use a very large scissor, with 300.000 kilometers each rod. Positioned at
90º, as a Cartesian plan. Between the two rods is positioned a sphere of
insignificant size.
Then we
have a sphere resting in right angle between the two rods. We started to close
the scissors at 300,000 kilometers per second on each stem. They approaches
faster because each one goes against the other and doubles the final speed.
The final
positioning of scissors would be 45º degrees as each rod is going toward each
other.
To get the
time it takes to close the scissors we just need to use the angular velocity of
Newton:
W = A / T
where:
W = angular
velocity (km \ s);
A = angle
(degrees);
T = time
(s).
then we
have:
600 km \ s
= 45 / T
T = 45/600
T = 0.075
s.
We now know
that the scissors will close 45º at 0.075 seconds and the sphere will travel a
distance of 300,000 kilometers at this time. We now calculate the speed like a
straight. Because the wire will go through an entire rod.
V = D / T
V = 300,000
km / s 0.075.
V =
4,000,000 km \ s
where:
V =
velocity (km \ s);
D =
distance (km);
T = time
(s).
Thus, this
sphere can achieve in just 300,000 kilometers a speed of 4,000,000 kilometers
per second, well, much more than light itself.
In another
example:
Let's use a 50 meters rod.
Pi = 50
meters x 160 meters (rounded)
112 500 000 rpm / 60 = 1,875,000 revolutions
per second
1,875,000 x
160 = 300,000,000 m \ s (angular velocity\W)
This is the
limit for a rod of 50 meters, 112.5 million rpm (revolutions per minute).
Now let's
see how the wire will produce with a max angular speed for a rod of 50 meters.
W = A / T
600,000,000
m \ s = 45 / T
T = 45/600
000 000 m \ s
T = 0.00000002 s
V = D / T
V = 25 m
/ 0.00000002 s
V =
1,250,000,000 m \ s
A 50 meters rod produce 4x the speed of light and
is "only" needed an engine with power of 112.5 million rpm.
Starting
from the beginning of the two first Newton equations, to construct a final
equation for rods and scissors respecting relativity:
W = A / T
V = D / T
Then we
replace T:
V = D / (W
/ W)
or
V = D / W 1
x / A
V x A = W x
D
W, which was
within this equation can be replace by this:
wheel
diameter x pi = length of the circumference x (Rpm/60) = m\s x 3.6 =
Km \ h
or
W = [D x π x (rpm/60)]
The result
was:
V x A = D /
2 x [D x π x (rpm/60)]
D / 2 -
half because it is for a half rod and not the entire one.
More
detailed yet:
rpm/60 = H
(hertz), thus:
Π x D x H
<= 300,000,000 m \ s
Minor equal
to 300,000,000 m \ s because we do not want to exceed the speed of light, it is
not permitted
to physics laws.
Then
D x π x H <= C
soon,
(D x π x H) / C = 0
We return
to the equation:
V x A = D / 2 x π x D x H / C
However, as
we have two rods moving toward one another, W is doubled.
2 x V x A x C = D x
(π x D x H) ²
Every
scissors or similar mechanism must obey this mathematical rule or law:
D x (π x D x H) ² / 2 x V x A x C = 0
Where,
π =
pi;
D =
diameter;
H = Hertz;
V =
Velocity of wire;
A = Angle;
C = speed
of light.
Thus, the
speed of the wire scissors, does not exceed the speed of light and to take away
the actual proof, we use the second example to see how it changes the whole outcome
when entering the line of relativity.
We use a
rod of 50 meters at a frequency of 1,875,000 Hz, which is the maximum that the
rod supports.
D x (π x D x H) ² / 2 x V x A x C = 0
50 x (50 x
3.14 x 1.875 million) ² / 2 x V x 45 x 300 000 000 = 0
V = 160 475
260 m \ s or 577710937.5 km \ h
This will
be the maximum speed that the wire will shear obeying Einstein's relativity.
But there
is a problem. It was to run the wire to 1.25 billion m \ s according to Newton
and naturally, but relativity will stop the wire to 160 million m \ s. As
Einstein himself said, space-time is
what distorts.
Suppose the following
will occur in the simulation of this event:
As was to
run the wire to 1.25 billion meters per second and can only go to 160 mi, rods
and wire will not distort. I think that at nearly 90% of the length of the rod
will be formed beads / circles, by themselves, and return to 70% extension of
the rod to then, follow the natural sense. That's because relativity
"stop" the wire, in fact she needs to distort the rod to turn this
event in the limit of speed. I call it the time-warp, what happens is that the
scissors will distort 1 billion kilometers and wonder, the space in front of
the sphere will goes backward ? behind the sphere that is being pushed?
I think
this will happen in the simulation, and this is what I'm producing at the
moment.
As this
simple example, where 3 is the bending moment:
Under construction.
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