Re: starship-design: FTL Navigation

[STAR1SHIP@aol.com]

In a message dated 8/25/01 10:08:57 AM Pacific Daylight Time, STAR1SHIP 
writes:


>  In a message dated 8/24/01 1:50:08 PM Pacific Daylight Time, lparker@
> cacaphony.net writes:
>  
>  << I haven't worked this acceleration to light speed problem since I was
>   sixteen...and I'm not going to say how long ago that was.
>  
>  Sorry about the delay in response Lee, had to downgrade my aol mailreader 
from 7.0 to 5.0 to get it to work with this mailing list so my messages get 
through readable.
>  
>  I did the acceleration to light speed at 1 g  in 1966 in high school 
physics and it was verified independently later by another published in 
Scientific American. When I patented my rocket engine in 1988, it never 
occurred to me I would have to go back and relearn special relativity to 
respond to the barrage of critism form those unable to calculate it's max 
velocity from the engine parameters. Instead of skeptics focusing on the 
engine, the focus is on relativity which is boring to me as I am more 
interested in discussing the engine for practical star travel as even were a 
light speed limit applicable to my rocket and it's max velocity be sublight 
speed near C it will do for star travel today and allow one to journey to the 
stars within the lifetimes of the astronaughts and myself.
>   
>   
>   Nevertheless, I think even Starfish made a mistake. Look carefully at your
>   seconds units and your canceling and try again....
>   
>  StarShip often makes mistakes in math so do check my math with a preferred 
MathCAD program but at times do it manually- Like so:
>  ------------------------
>  The original math.
>  
>  c=at        therefore 
>  t=c/a       and replacing parameters with given values
>  t= (2.998E8 m/sec)/(9.8 m/sec^2) inverting parenthesized denominator and 
multiplying gives
>  t=2.998E8 s^2 / 9.8 s  by simultaneously canceling the m in nominator and 
denominator
>  t=2.998E8 sec / 9.8    by canceling the s in the nominator and denominator
>  t=3.059E7 sec to reach c at 1 g.
>  
>  One year in sec = 365.25 days x 24 hours x 60 minutes x 60 seconds or 
31557600 seconds: therefore;
>   3.059E7/3.156E7 = .969 years
>  .969 years times 365.25 days =  353.93 days given some small rounding 
errors as the best measurements of 1 g and c give slightly more than 355 days 
to reach c velocity.
>   
>  checking my calculation math by solving for a original variable given the 
calculated variable.
>  
>  c=at
>  t=3.059E7 sec to reach c at 1 g.
>  a=9.8 m/sec^2
>  c= 3.059E7 sec x 9.8 m/sec^2
>  c= (2.9782E8 x m x sec)/(sec x sec) seconds cancel in denominator and 
noninator cancel leaving a second in denominator 
>  c= 2.9782E8 m/sec approximately =  the beginning 2.998E8 m/sec
>  The apprximation is due to rounding errors introduced giving accuracy to 
two signifigant digits as accleration was to two signifigant digits.
>  
>  The math checks okay the original equation stands as is.
>  To see a second math check done by computer with MathCAD visit.
>  http://members.aol.com/tjac780754/indexC.htm
>  
>  Tom
>  
>  Lee
>    >>
>  
>  Sorry about the delay in response Lee, had to downgrade my aol mailreader 
> from 7.0 to 5.0 to get it to work with this mailing list so my messages get 
> through readable.
>  
>