I am not going without a window as the Mercury astronauts demanded one and received it and I will have a radar at sub light speeds. I do not know why I would even need a radar as I would use simple line of sight navigation at sub light speeds and navigate with my clock and 3 axis accelerometer at superlight speed to plot my position in space and time. Any communication with external rockets would establish their position relative to mine.

I am on your side so why fight as the problems atomic technology has caused in the past can be solved with atomic technology in the future. I just use algebra and any high school graduate should be able to follow my math or at least read my bold calculations and the seven simple equations [Eq. 1-7] as windows scientific calculator has the natural log number function LN like so:

At one billion dollars/engine I can deliver 80,000 tons of payload to Mars for $6.25 / ton in 1.73 days and as each engine can be used 100 times reducing the cost to 6.25 cents/pound, or I can deliver 8,000,000 tons of Earth's radioactive metals waste from dismantled nuclear reactors and decommissioned atomic bombs to deep space low solar orbit for 6.25 cents per pound. This puts radioactive metals out of the hands of terrorists who will given time convert it with public domain atomic bomb plans to atomic bombs and blow up reactors bypassing the safety systems preventing a China syndrome effect greatly amplifying the destructive effects of even small atomic bombs. It is equally likely they will detonate a small atomic bomb in a atomic waste dump site making it airborne fallout.

Would you like to calculate yours our your children's number years remaining of surviving here. Would you like to calculate how long of time it will take them if you leave the hundreds of thousands of tons of toxic metals on earth if you do not let me get rid of the stuff?

My guess is you would not last very long unless you let me get rich as a toxic garbage man so I can fund my trip to the stars when the job is done. I will fight tooth and nail - Resistance is futile.

Drat-These fellow's below also have a plasma engine able to get to Mars in a few days but I can find no details <_< - they must keep it very secret however you are lucky you have me to talk to. h34r:

The final velocity of a rocket propelled projectile depends upon how much
propellant you have and how fast the exhaust velocity is. The formula for
determining the speed of a rocket is called the "dynamic" ROCKET EQUATIONS:

^denotes exponet like x squared = x^2

[Eq. 1] Vf = Ve * LN((1/(1-u)) where Vf=final velocity, Ve=exhaust velocity of 81840 miles/sec
u=propellant fraction, c = light speed 186,000 miles/sec
LN:=Natural Log: that is log e, where e is the natural constant that is approximately 2.718281828. So x=log y <=> e^x=y. do not confuse with exhaust e

(Eq. 2) Vf=Ve[ln(Minitial/Mfinal)]

(Eq. 3) Vf = Ve * LN(1/(1-u)) ---> u = 1- 1/EXP(Vf/Ve)

Of rocket
90.227 % Propellant = 902.727 tons propellant
6.000 % Structure = 60 tons structure
3.773 % Payload = 37 tons payload
u = .90227 = 902.727 tons of propellant / 1000 tons of rocket
Ve= 81840 miles/sec therefore:

Vf = (81840 miles/sec ) (LN(1/(1-.090227))
=(.44 times 186,000 miles/sec ) (LN 10.)
=(81840 miles/sec)(2.3255467)
Vf =190,322.7 miles/sec = 1.023 C or warp speed 1.02 arriving with 37 tons of payload and 60 tons of structure.

checking my results of equation one with equation 3

Vf=Ve[ln(Minitial/Mfinal)]
(190,322.7 miles/sec)/(81840 miles/sec) = ln(1000 ton / 97tons)
2.3255 = ln 10.10.30927
2.325546 ~ =2.33304 rounding to 3 signifigant digits
2.33=2.33 so answer checks correct with either equation

A round trip to Mars with the same engine at a constant 1 g acceleration to midpoint and a constant 1 g decelleration to stop at Mars requires an engine capable of substaining 1 g the entire trip time of 1.73 days and so an engine capable of obtaining a velocity = 1g times 1.73 days as

[Eq. 4] V=AT therefore a

Vmars in miles / sec = 1 g times 1.73 days
= [(32.2 feet/sec*sec)/(5820 feet/miles)](1.73 days)(24 hours/day)(60 minutes/hour)(60 seconds/min)
= (0.005532646 miles/sec*sec)(149472 seconds)
Vmars = 826.975670 miles/sec = 0.004446105 C

Vf/Ve=[ln(Minitial/Mfinal)
(826.97567 miles/sec)/(81840 miles/sec) = ln (Minitial / (Mintial-902.727 tons)
0.01010478 = ln (Minitial) / (Mintial-902.727 tons)
1.01015601 = (Minitial) / (Mintial-902.727 tons)
1.01015601(Mintial-902.727 tons)= (Minitial)
1.01015601Mintial -911.895105 =Minital subtracting Mi both sides
0.01015601Mintial -911.89510=0 adding constant both sides
0.01015601Mintial = 911.89510
Minitial=911.89510 / 0.0101560
Mintial = 89,788.7 tons for the Mars rocket

Propellant = 902.727 tons propellant with Ve=of 81840 miles/sec and Vf= 826.975 miles/sec
Structure + payload = 88,886 tons as 80,000 tons is payload and 8,886 tons is structure
propellant fraction (u)= 902.727 tons / 89,788.7 tons = 0.0100539 ~= 1/100

[Eq. 5] E=MC^2
M=E/C^2 tells one what energy that some mass converts to.

[Eq. 6] Ekinetic = (MV^2)/2 so I need to convert mass to energy to propell 902.727 tons of propellant to 81840 miles/sec
Ek=(902.727 tons )(81840 miles/sec)(81840 miles/sec)/2
Ek=6046271901331.2/2 tons-mile^2/sec^2
= 3023135950665.6 tons-mile^2/sec^2 therefore

M = (3023135950665.6 tons-mile^2/sec^2) / C^2 as C= 186,000 miles /sec
M = (3023135950665.6 tons-mile^2/sec^2) / 34596000000 miles^2/sec^2
M = 87.3839736 tons of propellant mass coverted to energy ~= one part of propellant in 10 parts converted to energy

[Eq .7] Power=weight x distance traveled/time (P=WD/t)
distance traveled=AT^2/2 = (32.2 feet / sec^2 )(149472s)(149472s)/2 = (32.2 feet / sec^2 )(2.9929 days)/2
____Horsepower can be determined by dividing the result of this equation by 550 foot-pounds per second.
____Example 1: A rocket totals 89,788.7 tons, If the rocketman flies his constant 1 g rocket past Mars a distance of 359,704,248,422.4 feet vertical feet in 1.73 days, his horsepower can be calculated as:

Power = 179577400 pounds times 359704248422.4 feet / 149472 serconds
= 64594753700648693760ft-pounds /149472 sec
=1305683087416087.7619663648124191 ft-lb/sec
= (1305683087416087.7619663648124191 ft-lb/sec) / (550 ft-lb/sec) dividing by one horsepower
= 2,373,969,249,847 horsepower ~=2.4 trillon horsepower or 1,770,981 gigawatts for the light speed or Mars misson with approximately a thousand ton engine

Eq. 1-7 is an incomplete set of Rocket Science equations.

As I am ejecting the propellant at .44 C a small amount of relativistic effects limit the exhaust momentum transferred to the rocket estimated at 5% loss and as I convert 1 propellant part in 10 to energy than that mass is not considered in the propellant momentum estimated at 10% loss so the final mass and velocity calculations above are estimated at 15% high until I recalculate with relativity equations and propellant mass lost converted to energy so the above are to be considered ball park calculations until that time.

In any case, simply decreasing the payload mass a relative small amount estimated at 15%, I can achieve the stated velocites as calculated above for both the light speed and Mars trip.

I may have missed some previous repliers questions they really want answers to. If so please ask them now and thank you for your patience. :unsure:

__________________
Thomas Hulon Jackson
Scientist/Engineer/Technician