Re: starship-design: Inspiration inside!

[STAR1SHIP@aol.com]

Check my SR and GR math please.

Note: Comments not needed- solve problem below the dotted line only.

Considering atomic rockets theoretical max velocity.

One relies on momentum to carry the rocket beyond c.
I conserve momentum as momenum(P) = MV and as each action has an equal and
opposite reaction, In a rocket(r) the momentum of the rocket in one direction
is always = to the momentum of the exhaust(e) in the other direction at any
given time.
is expressed in a general equation as MrVr=MeVe conserving momentum.
 
Example.
A 100 ton ship accelerates at one g by providing the energy from converting 
1/2
ton of 95 tons exhaust to the energy required to accelerate the 94.5 tons of
exhaust out the exhaust port at 1/10 c. This exhaust momentum propels the
remaining 5 ton ship to 1.89 C speed in the opposite direction.
 
Showing:
MrVr=MeVe 
5 tons times Vr = 94.5 tons times 1/10 C
Vr=.945 c/5
Vr= 1.89C
As 1/2 ton mass converts to energy it leaves the exhaust mass so the figure of
95 tons changes to 94.5 tons giving a rocket velocity of 1.89 C conserving
momentum.  

See Working diagram Animation
 http://members.aol.com/tjac780754/Page1.html

An ordinary chemical rocket can control by throttle an exhaust flow rate to
give 1 g acceleration for several hours based on tests of rockets reaching an
average 15g for many minutes such as "Hellious".

At less than 1 year = 356 days time 24 hours =8544 hours of 1 g acceleration
are needed to exceed c.

Versus present chemical rockets, 8,544/3(several) times more power is needed 
so
that acceleration time can be measured at greater than 356 days to exceed c.

The chemical Vs atomic power to mass unit ratio based on atomic bomb and
reactor test measurements are best "guesstamated" at 1,000,000 to 100,000,000
times the power possible with chemical rockets so atomic rockets of good 
design
can maintain 1 g acceleration for 351 to 35100 years.

Dotted line---------------Presented for your Solution---------------------


PROBLEM BACKGROUND:
In the problem I give, I state the acceleration to be a constant 1 g.
This is the acceleration of the object with respect to the object as measured
and calculated aboard the object by dropping an object on the rocket ship deck
and timing the fall time to calculate the acceleration rate at a fixed time 
and
by reading his watch at the beginning of the ships acceleration,the rocketman
then subtract the time difference and calculates his velocity at any time by
knowing the current time.

I did not give the relativistic acceleration
that changes with an arbitrary external observers frame of reference from any
external observation point with respect to the ship(wrt) as the external frame
of reference can be of any velocity including zero(at rest) or position 
outside
the ship giving different velocity and acceleration measurements, for in
relativity there is no "absolute reference frame" from which all measurements
are taken.

Sticky Note :-):

Stick to the given problem and focus and concentrate: In the reference frame 
of
the moving object the local(wrt ship) acceleration determines the final
velocity and is given as 1 g. Do not mix frames of reference.
Note: In the given problem, A hypothetical uniform(linear with slope zero)
gravitational field of 1 g of unlimited size does not exist in nature. In
practice a small uniform gravitational field of 1 g exists near the earth's
surface but becomes significantly nonuniform(nonlinear) the further away
from the earth's surface  as the gravitation field strength varies
inversely proportional with the square of the distance between the two masses.

To construct an artificial
uniform gravitational field of size sufficient to contain the distance
traveled by the object "feeling or experiencing " the 1 g acceleration, it
is necessary to accelerate a rocket at a local rate of 1 g for the duration of
the flight as "acceleration producing artificial gravity is indistinguishable
from gravity produced from mass."(Einstein)

So you can see the difference between actual acceleration and relativistic
acceleration I will temporarily modify the given problem and change
 1.9 C to 9/10 C to conform to common
preconceived notions of a light speed limit and belief systems so that you can
calculate both actual velocity(Vreal) wrt the moving object and relativistic
velocity(Vrel.) relative to an arbitrary observer and compare the results of
Special Relativity(SR)
or General Relativity(GR) versus(vs) Newtonian Mechanics(NM).

---------------------------------------SOLVE------------------------------
------------

What is the time(t) required for an object(p) accelerating(a) at one 
gravity(g)
to reach a velocity(v) of 9/10 C when free falling in an hypothetical uniform
gravitational field of 1 g of unlimited size or artificial uniform
gravitational field of  Rocket accelerating in deep space at a constant 1 g?
(Neglecting air resistance, nearby gravitational masses such as planets, stars
and external observers)
------------------------------------Defining
Parameters--------------------------------
Initial conditions:
The notation: (:=) represents (is defined as equal to)
                    (^) represents exponent such as x^2 for x squared.
                    (Xp) represents the variable X with the subscript p
V:= velocity in units of lightspeed (C) = 9/10 C expressed as warpspeed point 
9
C:= light speed in meters(m)/sec(s) or m/s =  2.998 time 10^8 m/sec 
g:=9.8 m/s per sec = 9.8 m/s^2
T:=time(t) 
---------------------------------------converting to algebraic short
hand------------------------- 
Expressing the word problem  in standard algebraic shorthand, 
from the large set of equations in the field of physics formula, I select the
general formula Velocity is equal to Acceleration times Time expressed as 
v=at.
>From the this I derive the special equation to solve the initial condition
problem.
  
Vp=gt  using the constants 
g=9.8 m/sec^2  and 
C=2.998 time 10^8 m/sec  to calculate 
1 year in seconds = 60 seconds times 60 minutes times 24 hours times 365.25
days=31557600 seconds 
t=Vp/g to calculate T as T=t

calculating (hint...................
>From the original problem,
if 1.9C=(1.8 years)g then g=1.9C/1.8years
and  .9C=gx and 
x=.9C/g then x= .9C(1.8 years)/1.9C therefore
x = 0.85 years trunctuated to two signifigant digits = T giving the NM,SR and
GR solution

I am eagarly wating for your NM, SR and GR solution . . . . . . . . . . .