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Re: starship-design: Re: FTL travel
In a message dated 3/26/00 1:21:20 PM Pacific Standard Time, stevev@efn.org
writes:
> STAR1SHIP@aol.com writes:
> > Taking for example the general equation MVof exhaust=MVof payload.
> > to accelerate a given payload mass of 5 tons twice light speed requires
I
>
> > accelerate 100 tons of exhaust to 1/10 light speed (at 1 g
acceleration).
> > Using the equation E(kinetic)=1/2 MV we get the energy required.
> > using the equation E=Mc^2 we get the mass that needs to be converted
> > providing the energy.
>
> That equation isn't relativistically correct.
I believe I now understand your use of term relativisic velocity. All
velocities are relativistic in that they do change from the perspective of
the observer. Relativistic velocities is a term used loosely in an
enginneering sense to describe relativistic effects that are more noticable
at velocities near light speed, usually .8c or better when the effects become
significant. Below that speed the effects are insignifigant but not zero.
This has nothing to do with velocity relativistic described by Einstein.
I previously pointed out why and how the relativistic rocket equation MV
times gamma of payload=MV times gamma of exhaust converts to the real
equation MV of exhaust= MV of payload by canceling the gamma factor out.
> In the situation where your 5 ton payload _instantaneously_ reacts 100
> tons of fuel which is ejected in one direction at 0.1 c, then we would
> have:
>
> Mp * Vp / sqrt(1 - Vp^2) = Mf * Vf / sqrt(1 - Vf^2)
First my engine does not use instanteous acceleration. It ejects propellant
gradually or I would have to use the particle accelerator equation.
Einstein found and saw the problem with relativistic equations containing a
sqrt term as some root solutions from square roots gave invalid results of
imaginary solutions so corrected his relativistic equations to eliminate that
problem. I suspect you learned that equation in historical sequence without
examining and using his later corrections.
> Solving for Vp we would get about 0.89 c. And this is still
> oversimplified.
As I stated before velocity relativistic is always going to calculate to
below light speed. When the rocket exceeds light speed relativistic it
becomes unobservable because the observer light is limited to light speed.
You have to transform from velocity relativisatic back to velocity real to
get the actual velocity of the ship. What part of Velocity real = distance
divided by time dilated did you not understand?
>
> If you carry 100 tons of fuel and accelerate at 1 g the resulting
> velocity is much lower because you aren't just pushing 5 tons of payload
> but all the remaining fuel at that point in the trip.
The energy calculation E=1/2 MV determines the energy required and the mass
conversion formulae M=E/c^2 tells you the amount of mass needed to get the
energy. The variables you mention are not part of the given problem or
equation solution. To prove this to your own satisfaction using identity
proofs instead of using twice light speed and 5 tons use the 100 ton and 1/10
light speed with your MV portion of your equation. The different values you
will calculate prove the equation violates conservation of momentum. as .89c
you first calculated does not equal something else and P=MV always and P is
always conserved in valid equations not just arbitarialy selected root
solutions of square root terms.
>
> If your fuel is capable of reacting to produce exhaust products that
> travel at 0.1 c, then you're also converting a nonnegligible amount of
> fuel mass into energy, and that needs to be accounted for too.
It is nonnegligible, but treated engineering wise as insignifigant but not
zero. The equation is thus simpified and when reduced to practice the rule of
thumb is to carry more fuel than you calculate needed.
>
> > Machines like partical accelerators that use the general instantaneous
> > acceleration equation of E'=1/2 MV gamma, reguire the Energy
relativistic
>
> > equation to be used for the energy is delivered intantaneously (small
> > increment of time)from the relativistic(observer or rest) frame instead
> of
> > the frame of moving(inertial) object and is therefore limited to below
> light
> > speed relativistic so that momentum is conserved.
> The same laws of physics apply to rockets as to particles in particle
> accelerators.
I agree completely.
>You don't get to pick and choose which laws of physics
> you want to apply in which situations.
You need to understand clearly the difference between math and physics in
order to chose (pick)the correct equations for any given problem. Your
mangled choices to support your erronous light speed limit beliefs while
completely ignoring all other facts and wwworking equations claiming them not
valid when cleary shown otherwise is not acceptable.
You can repeat making unsubstaniated claims (arrogance) a thousand times and
they do not become true.
Albert Einstein paraphrased partial quote (1955)--
It is possible a machine other than a particle accelerator can be found to
send objects faster than light.
End guote--
I found the machine predicted(atomic rocket)- end discussion.....unless you
change from your unproven, unsubstatiated claim or provide from a credible
source the proof we have little to talk about unless you can suggest
something.
I have no desire to teach the mistaught and misbehaving ;=)>
I have the patent rights to a star ship and you just are not going to get to
heaven(s) unless I decide you are behaving.
Best Regards,
Tom
<A HREF="http://members.aol.com/tjac780754/indexb.htm">Plasma Rocket Engine</
A>