Homepage Forums Dense Plasma Focus (DPF) Science and Applications How fast could FF get us to Mars?

This topic contains 5 replies, has 6 voices, and was last updated by  vlad 4 years, 8 months ago.

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  • #1568

    dennisp
    Member

    FF of course is already a rocket. An old forum post claimed exhaust velocity of around 3% lightspeed, giving us a very efficient rocket.

    Without adding extra reaction mass, though, I’d think thrust would be quite low. Maybe too low to be optimal for a fast Mars trip, I don’t know.

    So how fast could FF get us to Mars, if we used the plasma jet directly? And if that would be slow, how much could we improve matters with extra reaction mass?

    #13203

    FF of course is already a rocket. An old forum post claimed exhaust velocity of around 3% lightspeed, giving us a very efficient rocket.

    Without adding extra reaction mass, though, Iโ€™d think thrust would be quite low. Maybe too low to be optimal for a fast Mars trip, I donโ€™t know.

    So how fast could FF get us to Mars, if we used the plasma jet directly? And if that would be slow, how much could we improve matters with extra reaction mass?

    I can’t give you a straight answer but I can give you clues. One device of the size shown by LPP focus fusion is not enough, so you would need several or a much bigger machine. Maybe It’s possible to built several units in the same shell?

    The trust depends of how much and how fast you let the gases out of the nozzle. The connection between speed and velocity is according to square root of the temperature. You can get much more thrust if with more mass and lower temperature. First you can use the surrounding air compress it, heat it with the alfa particles from fusion and let it out. Higher up there are no air, so then you can use gas you brought with you, to heat and let out. You can not bring unlimited amount of gases, but surely enough to get out of earth gravitation. Other could as you know.

    Finely the alfa particles are the only to give thrust. There energy is about 3/2*KB*T and equals to m*v*v/2.
    KB*T should equal 7Mev according to physics table of binding energy and m = 4*1,6*10 exp(-27). This gives v=58 milj m/s
    But that’s very theoretical so lets say efficiency is 50 % gives v=40milj m/s.

    The rocket can never reach that speed and it’s probably relatively heavy and the mass flow tiny. But it will get close the velocity out of the nozzle, even if it takes times, since there are almost no resistance in space.

    #13205

    vansig
    Member

    it turns out the fusion reaction is more energetic, p + 11B –> 3 He + 8.7 MeV; but for Focus Fusion, i expect the exit beam should consist of around 600 keV particles, or all the exhaust along with some unreacted fuel, in a tightly focused beam, before electrical conversion.

    so i get a lower overall velocity in my estimates.
    600 keV = 9.61305939 ร— 10^-14 joule, m[alpha] = 4.0026u = 6.6465 ร— 10^-27 kg, E = 1/2 mv^2; so velocity of the beam, v = sqrt(2E/m) is about 5.378 x10^6 m/s, or 1.8% of c.

    we should keep in mind that fusion yield is not 100% of available fuel, (it might be 20%), and that 80% or more of the exit beam’s kinetic energy needs to be tapped to keep the reactor running for the next pulse, and that the entire process generates a good amount of waste heat to be carried off (up to half of all energy converted).

    Even so, rate of fuel consumption is still extremely small. So overall thrust is small.

    #13247

    Vulvox
    Member

    How would the thrust compare with what a VASIMIR rocket propulsion system can develop?

    #13308

    GullyFoyle
    Member

    I came across an Older Business Plan of Lerner’s today. It has a section on Space Propulsion.

    #13317

    vlad
    Member

    Formula that links rocket engine power, thrust and its specific impulse is more than simple ๐Ÿ™‚
    P = I / 2
    where P is specific power of rocket engine (in the sense of 1 Watt per 1 N of thrust), without loss of course
    and I is a specific impulse of rocket engine, in m/s

    specific impulse of the rocket engine in m/s is equal to the speed of exaust velocity
    hence for specific impulse of 0.03c (9E6 m/s) and power of 5MWt the thrust would be… about 1 Newton
    not too much I believe ๐Ÿ™‚

    To travel to Mars you definetely should “exchange specific impulse to thrust”
    You can (very) roughly estimate optimal specific impulse for some space route as… say equal or – much better – twice to the deltaV of this route.
    So if your estimate deltaV for the route from Earth low orbit to Mars low orbit and back to… say 40 km/s (for fast piloted mission with flight time measured by weeks not months) then the optimal specific impulse would start from 4000s, and 8000s would quite nice.

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