# 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

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
M=E/C^2
therefore replacing E with calculated E(kinetic) of 1/2(100 tons times
c^2/100)
therefore
M= [1/2(100 tons times c^2/100)]/C^2 and clearing first then second
parenthesis
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
speed.
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

Tom
Profession- High Speed, High Energy Physics

```