Saturday, September 20, 2025

Nuclear Fusion Propulsion Space Craft



Interstellar Space travel




Sean Kennedy

  • February 12, 2012

  • InfiniteVerse Incorporated

  • New Space Propulsion System

A. The area labeled “A” consists of an electromagnetic flow around the exterior part of the craft allowing deuterium or tritium to smash into each other.  The core chamber will be the central area of energy use  in the rear of the craft, but the laser carrying this will only act as a way to create plasma from hydrogen fusion, or you can use a laser in the reaction chamber, but they just could cause problems  because it is too much energy concentrated in one area at one time.

B. This area labeled “B” consists of the living quarters, computer center, storage, research, and other normal functions such as recreation and fitness

C. This area is the water chamber. This will take up 78% of the entire ship, and heated water will make an earth like atmosphere like a bio dome to maintain normal life for all. In here we will conduct electricity for the living quarters as well as act as a solvent and coolant for the Hydrogen reaction chamber and core. It will be filtrated through an area which will be a vacuum chamber to keep optimal coolant conditions

D. This is the area for a hydrogen fusion catalyst or feeder for the reaction to take place. This will be monitored through fiber-optic wiring with the use of graphene to better the electric current to make quick and efficient computer updates to the mother board . Compartmentalization of fuel into small rooms will keep risk low, energy production efficient, and you can control how much energy you want to use and create, but honestly i don't think this technology could be mainstreamed for many years

E. The actual reaction chamber for Hydrogen fusion is found on what’s labeled “E.” From there we can take the energy  creation for the core.  Hydrogen fusion in theory will keep a usable magnetosphere deflecting cosmic rays and space particles.  The hydrogen fusion will turn into plasma once it is hit by the laser beam. The collision will be like a jump start of energy to get the craft going.

F. The core is found right in front of the thruster. This area will be monitored by cutting edge computer technology and complementary human/robotic analysts. The relation between minimizing input and maximizing output will be monitored at all times. This plasma made from light and hydrogen fusion.

G. The thruster is the area where the energy is expelled into a combustion system which will concentrate the energy to one small place.



Why Chose It?

  • In the end, this is the most efficient way to maximize input to output ratio into 4 basic functions. You have electromagnetism to create the levitation with the help of a superconducting compound. The electro-magnetic wave will send the atoms at each other around the exterior particle accelerator of the craft which will kick start the craft. Unfortunately, creating such conditions is extremely hard with little to show of it as of right now. The water helps regulate atmospheric composition for the living quarters, and make electricity for the living quarters and computer system while shielding the astronauts from harmful radiation. The catalyst will help promote and compartmentalize the energy flow through computers all the way to the core of the propulsion system through a web of fiber-optic web design to maximize efficiency of energy output/input for reaction and intelligence purposes. The reaction chambers will be kept at absolute zero temperature to minimize risk of melt-down, and the temperature of the area should be able to be changed if humans need to intervene in the area. In the end, this is a ground breaking idea. The craft energy/mass will match that of our earth’s core creating an earth-like gravitational field.




What Should Be Done With This Craft:


First of all, Earth's crust is has only got so many natural resources, and the further we dig down, the more trouble it may cause us, so why not search the solar system for natural resources that can help us improve that problem by extracting from places and objects that are abundant in elements and resources that we need on earth such as H20, helium-3, lithium, iron, Uranium, as well as many other things that can prove to be useful in the future.  At the same time, it will be useful for researchers that can now get to places much faster than ever before, but will it be safe for humans to travel that fast.  i am not sure on how to stop the effects of inertia, but creating a magnetosphere may help in stopping this problem.  Will the craft burn up when going so fast because we'll have the energy output of a downgraded sized star?  Will anyone be willing to take a step in a direction that will allow us to perform interplanetary travel within weeks?  I haven't done much calculations on how much energy from fusion a 3 foot by 3 foot star would create to make the craft go at an ever accelerating rate, and the outer limits of the Earth's gravitational field?  Could we use an electro-magnetic track at a gradual upscale to get the craft propelled into space, avoiding any means of combustion here on Earth.  Maybe make most of the track underground for those people that think electromagnetic radiation is more deadly than that of which you get from your cell phone or microwave because of the speed at which you go?  The speed of an object sometimes causes more radiation than the mass of the object, look at the early big-bang.  Some might want to create this craft in space for it may be too harmful to make it on Earth's surface?  Ideas are only ideas until you make them happen?


1. Fusion Reaction Energy


The most practical fusion reaction using hydrogen isotopes is deuterium–tritium fusion:


D + T \rightarrow He^4 + n + 17.6 \, MeV


Energy released per reaction:


E = 17.6 \, MeV


Convert to joules:


1\, eV = 1.602 \times 10^{-19} \, J


So


E = 17.6 \times 10^6 \times 1.602 \times 10^{-19}


E = 2.82 \times 10^{-12} \, J


per fusion reaction.


2. Energy per kilogram of fuel


For deuterium–tritium fuel:


E_{kg} \approx 3.4 \times 10^{14} \, J/kg


Compare that to chemical rockets:

Fuel

Energy per kg

Chemical rocket

~10⁷ J/kg

Fusion

~10¹⁴ J/kg



Fusion has 10 million times more energy density.


3. Rocket Thrust Equation


Your spacecraft thrust comes from Newton’s rocket equation:


F = \dot{m} v_e


Where

  • F = thrust

  • \dot{m} = mass flow rate

  • v_e = exhaust velocity


If fusion plasma leaves the nozzle at:


v_e = 100,000 \, m/s


and mass flow is


\dot{m} = 0.01 \, kg/s


Then


F = 0.01 \times 100000


F = 1000 \, N


About the thrust of a small jet engine.


4. Specific Impulse (Efficiency)


Rocket efficiency is measured by specific impulse.


I_{sp} = \frac{v_e}{g_0}


Where


g_0 = 9.81\, m/s^2


If exhaust velocity is


v_e = 100,000 \, m/s


I_{sp} = \frac{100000}{9.81}


I_{sp} \approx 10,200 \, s


That’s 20× better than chemical rockets.


5. Particle Accelerator Energy


To initiate fusion collisions, particles must reach high kinetic energy.


Kinetic energy equation:


E_k = \frac{1}{2}mv^2


or in accelerator physics:


E = qV


Where

  • q = particle charge

  • V = accelerator voltage


For a proton accelerated through 1 million volts:


E = (1.6\times10^{-19})(10^6)


E = 1.6\times10^{-13} J


Many billions of these particles must collide each second to maintain fusion.


6. Electrolysis Fuel Production


Water electrolysis reaction:


2H_2O \rightarrow 2H_2 + O_2


Energy required:


237 \, kJ/mol


Hydrogen produced can be used as:

  • fusion fuel (if converted to deuterium/tritium)

  • plasma propellant.


7. Rocket Velocity (Tsiolkovsky Equation)


Maximum spacecraft velocity:


\Delta v = v_e \ln \left(\frac{m_0}{m_f}\right)


Where

  • m_0 = initial mass

  • m_f = final mass


Example:


v_e = 100000 \, m/s


Mass ratio:


m_0/m_f = 5


Then


\Delta v = 100000 \ln(5)


\Delta v \approx 160,900 \, m/s


≈ 161 km/s


This is fast enough for rapid interplanetary travel.


8. Power of the Engine


Power output:


P = \frac{1}{2} \dot{m} v_e^2


If


\dot{m} = 0.01 kg/s


v_e = 100000 m/s


Then


P = 0.5 \times 0.01 \times (10^5)^2


P = 50,000,000 W


P = 50 \, MW


About the output of a small power plant.


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