And we're to have a definitive answer to both concepts by the end of this summer. Is this correct?
Quote from: MP99 on 03/27/2012 10:06 pmQuote from: Star-Drive on 03/27/2012 04:06 amQuote from: MP99 on 03/23/2012 06:01 pmQuote from: mboeller on 03/23/2012 01:03 pmHarry White's and Paul March's PDF at NETS2012 about "Advanced Propulsion Physics: Harnessing the Quantum Vacuum"http://www.lpi.usra.edu/meetings/nets2012/pdf/3082.pdfQuoteHistorical test resultshave yielded thrust levels of between 1000-4000 micro-Newtons, specific force performance of 0.1N/kW,and an equivalent specific impulse of ~1x1012 seconds.Where does that Isp come from?cheers, MartinSee attached slide.Thanks, but I'm not seeing that derivation unless your power source is pure matter/anti-matter with 100% conversion efficiency to usable power. ISTR you mentioning a spacecraft a while ago powered by an H2/O2 power-cell. By retaining the reactants, only the mass of the power output (by E=MC2) goes overboard.If I've remembered that correctly, that seems to be completely the wrong way to analyse the situation. Instead, you are producing power by reacting H2 & O2. To calculate Isp correctly, the reaction product (water) should be sent overboard, and the thrust equated to the rate of consumption / disposal of hydrolox.cheers, MartinMartin: Try to remember that we are NOT talking about rockets in this example, which you are trying to do, though I tried to use a standard rocket parameter to bridge the gap between the two propulsion concepts and to demonstrate the performance enhancements that such a field propulsion device could bring to bear on the tyranny of the rocket equation. Instead we are talking about gravity/inertial (G/I) field propulsion systems that use the ambient G/I field to generate the Mach-Effect (M-E) momentum transfers from the vehicle to the field and thus to the rest of the universe that created this field in the first place. So the G/I field propulsion process does not require the expulsion of mass or E&M radiation away from the vehicle to generate the noted reactive forces, for it directly reacts with the G/I field instead just like a ship uses its propeller to interact with the ocean's water to generate thrust.
Quote from: Star-Drive on 03/27/2012 04:06 amQuote from: MP99 on 03/23/2012 06:01 pmQuote from: mboeller on 03/23/2012 01:03 pmHarry White's and Paul March's PDF at NETS2012 about "Advanced Propulsion Physics: Harnessing the Quantum Vacuum"http://www.lpi.usra.edu/meetings/nets2012/pdf/3082.pdfQuoteHistorical test resultshave yielded thrust levels of between 1000-4000 micro-Newtons, specific force performance of 0.1N/kW,and an equivalent specific impulse of ~1x1012 seconds.Where does that Isp come from?cheers, MartinSee attached slide.Thanks, but I'm not seeing that derivation unless your power source is pure matter/anti-matter with 100% conversion efficiency to usable power. ISTR you mentioning a spacecraft a while ago powered by an H2/O2 power-cell. By retaining the reactants, only the mass of the power output (by E=MC2) goes overboard.If I've remembered that correctly, that seems to be completely the wrong way to analyse the situation. Instead, you are producing power by reacting H2 & O2. To calculate Isp correctly, the reaction product (water) should be sent overboard, and the thrust equated to the rate of consumption / disposal of hydrolox.cheers, Martin
Quote from: MP99 on 03/23/2012 06:01 pmQuote from: mboeller on 03/23/2012 01:03 pmHarry White's and Paul March's PDF at NETS2012 about "Advanced Propulsion Physics: Harnessing the Quantum Vacuum"http://www.lpi.usra.edu/meetings/nets2012/pdf/3082.pdfQuoteHistorical test resultshave yielded thrust levels of between 1000-4000 micro-Newtons, specific force performance of 0.1N/kW,and an equivalent specific impulse of ~1x1012 seconds.Where does that Isp come from?cheers, MartinSee attached slide.
Quote from: mboeller on 03/23/2012 01:03 pmHarry White's and Paul March's PDF at NETS2012 about "Advanced Propulsion Physics: Harnessing the Quantum Vacuum"http://www.lpi.usra.edu/meetings/nets2012/pdf/3082.pdfQuoteHistorical test resultshave yielded thrust levels of between 1000-4000 micro-Newtons, specific force performance of 0.1N/kW,and an equivalent specific impulse of ~1x1012 seconds.Where does that Isp come from?cheers, Martin
Harry White's and Paul March's PDF at NETS2012 about "Advanced Propulsion Physics: Harnessing the Quantum Vacuum"http://www.lpi.usra.edu/meetings/nets2012/pdf/3082.pdf
Historical test resultshave yielded thrust levels of between 1000-4000 micro-Newtons, specific force performance of 0.1N/kW,and an equivalent specific impulse of ~1x1012 seconds.
there are several recent news and discussions at Talk Polywell...
completely clear on the "propellantless" element of the topic title, but you're really missing the point here.Isp relates the consumption of consumables against the amount of impulse generated. If you are producing electricity via H2/O2 in a fuel cell, a kilogram of consumables will be converted to a certain amount of impulse through the thruster. That is clearly the basis on which Isp is calculated, and works irrespective of whether you throw the reactants overboard.
Your argument seems to be that action-reaction by propellant expulsion is always going to be more efficient than action-reaction by other means such as Mach-Woodward (ie. so why bother with Mach-Woodward at all?)Well, you may not always be able to gather propellant mass along the journey if you run out, but you'll probably still be able to gather light energy.
Quote from: MP99 on 04/01/2012 03:01 pmcompletely clear on the "propellantless" element of the topic title, but you're really missing the point here.Isp relates the consumption of consumables against the amount of impulse generated. If you are producing electricity via H2/O2 in a fuel cell, a kilogram of consumables will be converted to a certain amount of impulse through the thruster. That is clearly the basis on which Isp is calculated, and works irrespective of whether you throw the reactants overboard.But dude, as the ESA's SMART-1 probe to the Moon showed, you don't even need all your consumables to even be onboard in the first place. Well, sure the propellant for that mission was onboard the spacecraft, but the energy used to accelerate and eject it wasn't - it was coming from the Sun.So you don't necessarily need your energy to come from an onboard fuel cell, because it could come from solar power even.Your argument seems to be that action-reaction by propellant expulsion is always going to be more efficient than action-reaction by other means such as Mach-Woodward (ie. so why bother with Mach-Woodward at all?)Well, you may not always be able to gather propellant mass along the journey if you run out, but you'll probably still be able to gather light energy.
Of course, you're not limited by chemical energy densities - power it from a solar cell and you can keep going for ever. However, with SEP having such high Isp it will compete quite well for Dawn-like solar powered missions.
Quote from: sanman on 08/17/2012 09:52 pmYour argument seems to be that action-reaction by propellant expulsion is always going to be more efficient than action-reaction by other means such as Mach-Woodward (ie. so why bother with Mach-Woodward at all?)Well, you may not always be able to gather propellant mass along the journey if you run out, but you'll probably still be able to gather light energy.Don't worry about it; it's not true. At least theoretically. His conclusion is only valid for the low-performance thrusters demonstrated to date.WarpStar 1, IIRC, was designed assuming 1 N/W thrusters were available. Last I heard, there was no theoretical reason why this would not be possible, though the new force prediction derivation may have something to say on the feasibility front... anyway, let's assume that number for now.1 N/W, in the context of a conventional rocket, means that mdot*v_exh = 0.5*mdot*v_exh^2 (assuming 100% efficiency). In other words, v_exh = 2 m/s, for a specific impulse of 0.204 seconds.With a Mach-effect thruster capable of 1 N/W, hydrogen/oxygen consumption to produce one newton of thrust with fuel cells at 60% efficiency relative to the HHV would be ~1e-7 kg/s, leading to an effective v_exh based on thrust/mass flow of ~9.5 million m/s, or a specific impulse of about 970,000 seconds. Even using MP99's much more pessimistic power system assumptions, you're still looking at about 464,000 seconds. Assuming you dump the resulting water overboard, which I would not necessarily recommend...In other words, the operating principle of a Mach-effect thruster decouples the specific impulse from the thrust-to-power ratio.
GeeGee just posted on talk-polywell.org that Prof. Woodward book "Making Starships and Stargates: The Science of Interstellar Transport and Absurdly Benign Wormholes" can now be preordered at amazon:http://www.amazon.com/Making-Starships-Stargates-Interstellar-Exploration/dp/1461456223/ref=pd_rhf_gw_p_t_1I have already preordered the book
Well, a cubesat has a max mass of 1.3 kg, and so the acceleration (assuming all the Q thrust bits fit in the 1.3 kg) would be 0.1 mm/s2 or 0.00864 km/s per day. Accounting for gravity and drag losses and etc, it's about 4 km/s to get to the Moon on a low-thrust trajectory, which means roughly 1.25 years. Enough for a powerful demonstration, but not immediately practical.And, that's assuming you can cram enough solar cells in the cubesat package.