Re: starship-design: FTL Navigation

[STAR1SHIP@aol.com]

In a message dated 8/25/01 9:19:48 PM Pacific Daylight Time, 
toxicroach@swbell.net writes:

> Sweet god.  You people actually know what you're talking about.

ToxicRoach,
One of us knows what he is talking about at least :-)

>  
>  Ok, here's a question--- how much energy would an antimatter (hydrogen
>  atoms) collision yield per mole of anti and reg matter?  How much AM would
>  an AM-powered ship (60000 tons and/or applicable measurement) need to go
>  from Earth to Mars (at the most effiecent time) in, say, 3 months? 

Well when anti matter meets matter all mass is converted to energy. So I have 
done the atomic matter converted to energy calculations so will adapt them 
easily to antimatter requirements and then multiply them by two for 
antimatter considerations.

To simplify the complex rocket equation formulas by removing the energy 
requirements for land launches and landings, I will consider only a lunar 
orbit launch to Mars orbit launch as the radiation prohibits an earth launch 
and Mars landing for environmental concerns and the same for a return trip 
considering an end payload mass of 60,000 tons, accelerating at 1 g 1/2 way 
to Mars and then turn around 180 degrees and decelerate at one g to Mars 
orbit and the same for the return trip which makes the round trip journey 
take less than three months and provides the comfortable 1 g artificial 
gravity field for the travelers. I will time the lunar launch and Mars launch 
times to arrive a Mars and back to Lunar orbit at their minimum distance 
apart. As the distance is away from the sun to Mars and towards the sun's 
gravity field on the return trip the energy required cancels so is not 
considered though a normal 50% fuel surplus for safety reasons missing the 
launch windows or the unlikely event an emergency earth or Mars landing is 
needed so will be provided for dramatic effects useful in Sci-Fiction 
dramatic plots.

Distance from Earth  Note (10 to the 6th power is expressed as 10^6 or E6)
        Minimum (10^6 km)           54.5  = 54.5E6 kilometers
        Maximum (10^6 km)          401.3  = 401.3E6 kilometers

Total distance round trip = 2 time 54.5E6 km = 109E6 km
converting 109E6km to light years so I can use the Java applet calculator to 
figure trip time found at:
http://ucsu.colorado.edu/~obrian/applets/Rocket/Voyage.html

 I divide 109E6 by the distance light travels in a year which is equal to 
2.998E8 m/sec times the numer of seconds in a year from below (3.1557600E7 
sec)
therefore;
109E6 km / (2.998E8 x m / sec) x 3.1557600E7 sec
109E6 km / 9.461E15 meters as the seconds cancel out     
109E6 km / 9.461E12 km to cancel the km's from the nominator and denominator 
109E6/9.461E12 gives the distance in light years (ly)
109/9.461E6 ly  by canceling  E6 from the nominator and denominator then 
dividing
11.521E-6 ly equal to
.000011521 ly round trip. 

To adjust for the java applet calculator I need the one way distance so 
divide by tow and enter 
Trip length: 5.7605E-6 light years.
Acceleration: 1.0 g. 

and the applet returns one way time of

Time on earth: 0.004726034935456117 years multiplied by 365.25 days to get
1.72618426017534673425 days
Time on ship: 0.004726030251834143 years.
1.72618254948242073075 days so the relativistic time dilation effects are 
insignifigant but not zero.

1.72618254948242073075 times 2 for round trip time is equal to:
3.45 days rounded off.

Bon Voyage!

I am ready to calculate your energy requirements to make the specified 
journey in the given time arriving at end of trip with 60,000 ton payload.

This is much easier to do but is a chain calculation with some handy 
shortcuts and approximations to save me from using the calculus for 
integration.

As you figure the energy required to raise 60,000 tons of payload by 
propelling 1/4 of the 6,000,000 tons of propellant to send the payload max 
velocity at  1/2 distance to Mars under 1 g acceleration and then multiply it 
by four and then calculate the amount of propellant mass to be converted to 
energy and then divide by two calculate each mass of matter and antimatter to 
convert to energy.

To determine first the max velocity for 1 forth the trip required I use 
V=AT 
for energy required.

Energykenetic(Ek) = Mass(M) times velocity(v) / 2

>From Newtons third law of equal action and reaction I use conservation of 
momentum general rocket equation MrVr = MpVp with subscripts r for rocket and 
p for propellant to determine how much propellant I need to accelerate to 
what velocity to get the payload to the required velocity for each trip phase 
instead of the special rocket equations requiring integration as the exhaust 
velocity is measured from a point fixed in space twixt the exhaust port and 
exhaust and not the same as the exhaust velocity measured relative to the 
rocket needed in the special rocket equations from Newtons second law of f=ma 
as it is not needed to get the answers to your questions. 

v=1g time 3.25 days/4 
v= (9.8 m/sec^2) times 3.25 times 24 hours times 60 minutes times 60 
seconds/4 to get:
v=(9.8 m  times 280800 sec ) / 4 times second times second
v= (9.8 times 280800 /4) m/sec
v=687960 m/sec
v=687.960 km/sec


Beginning with a 6,060,000 ton craft with common 1/100 ratio of payload to 
propellant I do an approximate simple integration and say that approx 1/2 of 
the mass is ejected in the first 1/4 distance traveled 1/4 the mass the 
second 1/4 distance 1/8 the third forth of the distance and 1/16 the final 
distance knowing that 1/2 plus 1/4 plus 1/8 +1/16 does not equal to one but 
it is well within the 50% saftey margin of the extra fuel carried.

For the first quarter distance from,

MrVr=MpVp
Vp=MrVr/Mr 
3,060,000 tons times 687.960 km/sec= 3,000,000 times Vp

Vp=(3,060,000 tons times 687.960 km/sec)/ 3,000,000 tons
Vp=701719.2 km/sec average propellant velocity

Ek=MV/2
Ek=(3000000 tons times 701719.2 km/sec ) / 2
Ek=1500000 tons times 701719.2 km/sec
Ek= 1052578800000 km tons/sec
Ek= 1.0525788E12 km tons/sec

Conveting matter to energy from Einstein's E=MC^2
M=E/C^2
M= (1.0525788E12 km tons/sec) / (2.998E5  km / sec)^2
M= (1.0525788E12 km tons/sec) / 8.988004E10 km / sec
M= 1.0525788E12 tons/ 8.988004E10
M= 0.117109293676326801812727275154751E2
M=11.711 tons of the 300000 tons converted to energy or 
M= 5.855 tons of antimatter mixed with 5.855 tons of matter.

For the second phase
Vp=(1,530,000 tons times 687.960 km/sec)/ 1,500,000 tons
Vp=701719.2 km/sec average propellant velocity

Ek=MV/2
Ek=(1500000 tons times 701719.2 km/sec ) / 2
Ek=750000 tons times 701719.2 km/sec
Ek= 526289400000 km tons/sec
Ek= 5.26289400000E11 km tons/sec

Conveting matter to energy from Einstein's E=MC^2
M=E/C^2
M= (5.262894E11 km tons/sec) / (2.998E5  km / sec)^2
M= (5.262894E11 km tons/sec) / 8.988004E10 km / sec
M= 5.262894E11E11 tons/ 8.988004E10
M= 0.585546468381634009063636375773754 E1
M= 5.855 tons of the 150000 tons of propellant converted to energy or 
M= 2.927 tons of antimatter mixed with 2.927 tons of matter.

Since I noticed that this was 1/2 the first stage energy requirement value 
then I can say 

For the third distance or the first accleration of the trip home requires
2.927/2 or 1.464 tons of matter of the 750,000 tons of propellant converted 
to energy.

For the final deceleration returning to lunar orbit
.732 tons of the remaining 375,000 tons of propellant are converted to energy.

Mtotal=M1 + M2 + M3 + M4
Mtotal= 5.855 + 2.927 + 1.464 + .732

Recalling that 1/2 + 1/4 + 1/8 +1/16 does not equal to 1
then 
5.855 times 2 =  11.77 tons of matter converted to the energy required to go 
to Mars at 1 g and back in 3.45 days without breaking into my 50% reserve 
fuel carried.

5.885 tons of anti matter would needed to be carried. But for practical 
reasons, and since the ship began with 6,060,000 tons I would carry 100 tons 
of convertable fuel in case I decided to make a side trip to say Pluto. I 
will let you do do that math :-)

Use E = MC^2
or 11.77 tons times C^2 to calculate the energy used for the Mars trip- I 
recommend you calculate horsepower with miles per second instead of metric 
units.
  
Tom
To see animated atomic engine to do the above see link:
http://members.aol.com/tjac780754/Page1.html
To see the patent pending on the above rocket engine see:
http://members.aol.com/tjac780754/indexb.htm

  



>  >  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