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Re: starship-design: Re: FTL travel

In a message dated 4/3/00 4:00:08 PM Pacific Daylight Time, 
Shealiak@XS4ALL.nl writes:

> Hi Tom,
>  >>  gamma = 1/Sqrt[1-v^2/c^2]
>  >
>  >I can probably show you six or seven gamma formulas from the
>  >equation fields
>  The one on your page <http://members.aol.com/tjac780754/indexC.htm> does
>  contain only v and c as well. (It misses the squareroot though.)
>  Timothy
Hey Timothy,
I do not use the square root one as Einstein instructed it had problems of 
returning imaginary solutions of real numbers and not just imaginary numbers. 
He fixed the problem with a different derivative I was taught in 1968 at the 
FAA Academy.

The field of equations at indexC.htm is a unified field where all equations 
are in the same system of measurements and more important where variables 
like v in different formulas that are not the same are defined and 
subscripted properly. In this manner standardized definitions are made and 
discussions do no start out with "no this means that and not this" As It was 
Einstein's Unified Field Theory that more real work could be accomplished If 
all spoke the same language-in such a manner different scientific disciplines 
like electronics, chemistry, physics, math...etc could have a single source 
field of equations to work from to derive equations to solve unknowns.

If you think indexC is confusing you should see my electronics field of 
equations I will scan in when I find it.

I use the American system of measurements my self so click on Mathcad's units 
of measurement and select those I am most conformable with as I make enough 
errors without adding conversion errors. I could not land on Mars if I mixed 
units of measurements ;)

Checking my math.
if E(kinteic)=1/2MV^2
and E=MC^2

then 1/2MV^2=E(kinetic)=E=MC^2 

The energy required to send 100 tons of propellant at 1/10 light speed is 
said at
E(kinetic)=1/2[100 tons times (c/10)^2] 
E(kinetic)=1/2(100 tons times c^2/100)
To find the amount of Mass to be converted to the required energy
E=MC^2 solving for M is
therefore replacing E with calculated E(kinetic) of 1/2(100 tons times 
M= [1/2(100 tons times c^2/100)]/C^2 and clearing first then second 
M= 100 tons times c^2/200C^2 and canceling C^2 therefore
M=100/200 tons or
M= 1/2 ton of matter to convert to energy to propel 100 tons of propellant to 
1/10 light speed.

>From MeVe=MpVp
since 1/2 ton is converted then only 99.5 tons of propellant reach 1/10 light 
so for a five ton payload the payload velocity calculates to be
Vp=MeVe/Mp placing the calculations in to solve
Vp=99.5 tons times 1/10 C/5 tons solves to
Vp=9.95C/5 solves to
Vp=1.99 C

Profession- High Speed, High Energy Physics