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Re: starship-design: FTL Navigation
In a message dated 8/26/01 12:14:39 PM Pacific Daylight Time,
toxicroach@swbell.net writes:
> Subj: RE: starship-design: FTL Navigation
> Date: 8/26/01 12:14:39 PM Pacific Daylight Time
> From: toxicroach@swbell.net (Toxic Roach)
> To: STAR1SHIP@aol.com
>
> Thanks very much. I actually followed that. That will be very very
> useful.
> I'll be printing a hardcopy. That must have been a pile of work. Thanks
> VERY much.
A pile of work and with one significant error. I used for kinetic energy the
equation:
Ek=MV/2 arriving at the 11.77 tons of mass needed to convert to the energy
required. I failed to show my units cancellation so did not catch my error
until after posting at this step
M= (5.262894E11 km tons/sec) / 8.988004E10 km / sec
This equation should of been;
Ek=(MV^2) /2
giving the total mass converted to be:
11.77^2 tons = 138.5 tons
of propellant required converted to energy or 69.3 tons of antimatter mixed
with the propellant and require me to carry 50% more for a safety margin. I
carried 100 tons for side trips, I could of made the journey successfully
back to earth by turning of the engine near mid one way trip and coasting
part way
each direction. The only problem would be explaining to the pretty copilot
navigator's mother that I ran out of fuel causing horse around in zero g and
to be late a couple of days late bringing her home and hope "Mom" understands
(as Scotty would say) I am a rocket scientist and not a Genius who does
"enjoy my work" and hope she buys the out of gas excuse and I will promise to
carry 1000 tons reserve fuel my next trip.
Curious about the horsepower generated by the engine, I will
Use E = MC^2
or 138.5 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.
Checking Encarta Ency. for horse power (hp) definition
One horsepower was originally defined as the amount of power required to
lift 33,000 pounds 1 foot in 1 minute, or 550 foot-pounds per second.
I need to convert 138.5 tons times (186,000 miles/sec)^2 to units of feet,
pounds, and seconds to get E in horse power.
138.5 tons times 2000 lbs/ton = 277000 lbs
(186,000 mile/sec) times 5280 = 982080000 feet/sec
therefore:
E=MC^2
E=277000 lbs times 982080000^2 feet/sec
E=267161272012800000000000 foot-pounds per sec
E=2.671612720128 E23 hp
E=2.67E23 horse power
Tom
> -----Original Message-----
> From: owner-starship-design@lists.uoregon.edu
> [mailto:owner-starship-design@lists.uoregon.edu]On Behalf Of
> STAR1SHIP@aol.com
> Sent: Sunday, August 26, 2001 9:23 AM
> To: toxicroach@swbell.net
> Cc: starship-design@lists.uoregon.edu
> Subject: Re: starship-design: FTL Navigation
>
>
> 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
>
>