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Hey! Recently, I've been messing around with using Ion engines powered by fuel cells. As it happens, the combined ISP of ion engines powered exclusively by fuel cells is a pretty respectable 1293 seconds. One large fuel cell can power two ion engines, so I built an all-purpose crew vehicle that uses 16 large fuel cells and 32 ion engines, allowing all 32 of them to be run at full throttle for as long as one's heart desires, and at any distance or occlusion to the Sun. It also comes with RCS thrusters and a forward mounted docking port. The command module has the standard decoupler, heatshield, and parachutes for return. When placed in LKO using its launch vehicle, it has 4,100+ m/s of delta-v (vac) and an initial Kerbin TWR of 0.25. When flying, keep in mind that the delta-v value that KSP will display on your staging diagram will be less than your true delta-v, because the game does not factor the consumption of lf/ox into the calculation. The ratio of rate of xenon to liquid fuel consumption is roughly 50.08, so the orbiter has 22,800 xenon to 450 lf (ratio of 50.66). This means that all three fuel levels stay roughly equal as a percentage of their initial value throughout the duration of the mission. Because the Ion + fuel cell combined specific impulse is 1293 s, you can estimate your remaining delta-v in m/s at any point using this formula: Delta-v = 9.81 * 1293 * ln ( m / 18.851 ) where "m" is your current mass in t. The launch vehicle comes with the standard launch escape system (LES), with action groups: ABORT: activate LES, separate command pod AG1: activate LES, separate LES (use prior to orbital insertion, or shortly after abort) AG2: toggle fuel cells AG3: deploy parachutes Craft file: https://kerbalx.com/Jamie_Logan/Ranger-FCI-13 Gallery: Hope you liked it! Check this out:

Let me start with the following disclamer. The results I present are for an ideal ship weighing approximately 335 t with 39t of fuel (Xenon or Argon). The ship was designed to shuttle 200t to mars and bring 20t back. The fuel that is provided is ideal fuel and assumes that burns are instantaneous. In reality more fuel would need to be brought along. The drive is assumed to have a nuclear powersupply, no such power supply exists AND if it did the cooling systems for the power supply would be much greater in mass than the mass provided for engines, power-supply, and radiative cooling. The vessel produces in excess of 0.0105 acceleration and has a total dV in the 8.7 km/sec range. The first set up assumes that the burn angle occurs around the periapsis of an opposing exit vector. So if a burn span is 270' then the burn begins 45' after the exit vector and continues until 45' bfore the exit vector, resting as it travels through the apoapsis. The first set up gets the ship to Earths average hill sphere with zero velocity relative to the Earth. SpanBurn TBurn (sec)(cumm.) T (to escape) Dv (m/s) Waste(m/s) 360 ' = 2π 604413 >7 days 6633 3422 270 = 3π/2 524607 >7 days 5726 2514 240 = 4π/3 452699 >13 days 4917 1705 180 = π 364657 >60 days 3937 725 120 = 2π/3 324279 >200 days 3491 280 90 = π/2 313283 >400 days 3370 159 60 = π/3 306224 ~1000 days 3293 81 Burning to Mars. Adding the 6471769 J/kg required to reach Mars at its apogee. SpanBurn TBurn (sec)(cumm.) T (to escape) Dv (m/s) Waste(m/s) 360 800005 >9 days 8902 5018 270 719818 >8 days 7964 4080 240 634915 >13 days 6863 3099 180 527567 >60 days 5759 1875 120 491702 >200 days 5355 1471 90 483796 >400 days 5265 1381 60 476635 ~1000 days 5185 1301 Note that some waste in the final burn to mars is unavoidable, because once spefici orbital energy >=0 one cannot 'kick' past the periapsis and so once the final burn that has SME_>0 has also be a parto of the burn that goes to mars despite the fact that most of the burn occurs at low velocity (and thus low dE/dt during the burn). There are several ways to improve this. 1. Burn down 5 to 10' more than prograde to keep Pe low (from 270 to to 200) 2. On the last jump, jump to 270 and also burn down at 30' 3. Switch the ION drive to burn at a lower ISP and higher thrust. 4. Have an auxillary chemical based engine on board. The dV difference between escape, and mars intercept is only about 400 at burn from 6531 km and so a reasonably small fuel tank with powerful engine would suffice, the fuel tank can be ditched after the burn, it will go into a circumsolar orbit. Lessons to be learned: In a burn to Mars from Earth orbit, you do not want to use a spiral (360 burn span) orbit. It wastes both fuel and time. Since a Mars mission would be at least 90 days the loss of 5 days getting a better fuel economy is worth it. This means that blocking off 120' of the orbit as a no burn zone is a reasonable move. For a space tug that has no real cost of time except for capital cost of inventory. blocking the burn from a half to 2/3rds of the orbit is a reasonably good choice.

Before I start I need to define my terms F = force (measured in Newtons, N) M = mass (as in vessel mass or fuel mass) M* = mass ejection rate (M/sec) kW = 1000 watts = energy production rate or 1000 joules per second. Power = Power available (Kw) Power efficiency = Power not wasted Heating = Power wasted Irradiance = Solar output measured at 1 AU , 1361 Watts per sq. meter Panel efficiency = percent of irradiance converted to electrical power ISP = Fthrust / M* this value is 9.802 x ISPg (sec) which I will not use because basically it adds a term to every equation not needed, particularly for a craft that is interplanetary. ISPg is best applied to chemical thrusters, when you don't really care about where ejection force power comes from, you can wave your hands about bond energies, etc. Given ISP: M* = Fthrust / ISP ION drive equation is F = ( 2 * Power efficiency * Power ) / ISP Given: ION drives cannot shift ISP more than a 4 fold range Given: HiPep produces 0.67 N (rated) at 39.3 kW at 9620s (94300 v) between .75 and .8 efficiency. The form factor is .31 x .91 x .1? = Assuming density of 2 and a fill volume of 0.1 the Mthruster = ~5 kg. Thus one can estimate the following: N/area = 2.37N/m2 F/power input (N/kW) = 0.67/39.3 = 0.017 N/kw at 93000v. For the sake of simplification I will use 0.015 N/Kw at 100000v. Thruster modification cost for on the fly down-ratable ISP (ISPdr; arbitrary guess) = k * SQRT(100000/ISPdr). The cost is paid assuming that the thruster is built for 100,000v (10202 s), and therefore added cost is to allow it to achieve lower ISP during the trip. This allows a significant burn say at Perigee allowing a kick then a higher ISP on a second pass and then most efficient ISP to reach Mars. While these numbers appear not to have much value . . .we then can add solar panels Maximum panel efficiency is 40% lets assume that this applies to non-atmospheric, then we can apply the following: Irradiance (kW/m2) * Panel efficiency = Output power per sq. Meter 1.361 * .4 = 0.544 kW/m2 DC. If we assume this is 24V we have 22Amp so we need 10 gauge wire to feed the power center of the space craft. We are going to take a base assumption that 1 square meter of panel weighs a kilogram. Therefore the space craft has a minimum of 1 * thruster weight. So lets build an ion driven space craft, to keep things simple we will choose a spacecraft that with 1 meter squared solar panel 0.544 kW (panel output) x 0.015 N/kW (thruster input) = 0.00816 N. This then consumes M* of 0.00816N/100000v = 8.16E-8 kg/sec. We have alternatives. We have a theoretical weight per thruster of 5kg per .67N = 7.5 kg/n * 0.00816 = 0.0612 kg With our panel we have 1.0612 (Assume the microelectronics the thruster is glued to the back of the panel) ISP Thrust (N) M*x10E-9 duration/kgfuel 100000 0.00816 81.6 141d 50000 0.01632 326.4 35d 25000 0.03216 1305.6 9d 12500 0.06432 5222.4 53h 6250 0.12864 20889.6 13h As of yet we do not know our dV. Two reasons for this I haven't added any fuel mass. Nor have I added any structure and control mass. Our theoretical acceleration is very close to thrust since solar panel of 1 kg dominates the mass. Thus we can readily determine our theoretical maximum for each ISP. In this case I am going to add fuels and mass of 1.2 and tabulate the accelerations. To do this I am going to assume that for every kg of fuel added 0.2 kg of structure is added. With our theoretical ion drive we have then 1.0612 + Fuel*0.2 = dry mass, starting mass = dry mass + fuel mass + tunability cost mass. The equation is ISP * ln (full mass/dry mass) ISP (ISPgo) 0.1 kg 0.5 kg 1 kg 2.0 kg 4.0 kg 100k m/s (10ks) 8845 C 35808 58383 86235 114713 50k (5.1k) 4325 D 17581 B 28753 42622 56896 25k (2.55k) 2143 F 8725 C 14288 B- 21210 B+ 28353 12.5k (1275) 1058 F 4317 D 7082 C- 10534 C+ 14100 B- 6.25k (637) 519 F 2127 F 3498 D- 5218 D+ 7009 C- 3.125 (318) 253 F 1042 F 1719 F 2575 F 3472 D- The first gets our craft from earth LEO to Mars LEO and back, but over a period of 200 days and of course the spacecraft does not carry any payload to speak of, the second gets us close to Mars, the third gets us to a geosynchronous orbit (or so). With 5 times as much fuel the third ISP gets our craft to Mars. In the second fuel column the starting mass is one/third fuel, and even with that mass the ship is unable to return from Mars with a decent ISPgo of 1275 sec (almost 3 times the SSME ISP). With 1 kg of fuel it makes it back to Earth's sphere of influence (however with any additional payload, its stuck at Mars). One of the things that I did was under-engineered a decoupler on the Gas tanks so that tanks are used serially and then discarded, this gets rid of 10% of the structural cost, but backbone cost still remain 10% of the tanks weight, this improves the ISP for heavier fuel loads a little. I have assigned a grade based upon the usefulness of each reaching Mars assuming it uses 1 ISP and does not refuel along the way. But we can pretty much assume that a ship that cannot even break earth’s orbit or reach L2 refueling or extra-lunar refuelling point is of little value and can be ignored. Consider than in a manned Mars mission at least 1/3 of the ships dry mass will be devoted to survival and other related activities. So at about 4500 dV even refueling will not help with humans onboard. The problem with more fuel loads, are two fold, the through put of the thruster (electrode wear and tear) and breaking Earth's orbit. So lets look at the (get the hell away from earth) accelerations. Lets keep in mind that unless we like leaving our astronauts for months in LEO and LMO for no particular reason, we need to be shooting for accelerations of at least 1mN per starting mass second. Lazily (and to control column widths) I will use the mm but you should assume that is mm/sec2. Parenthesized numbers assume a humanized payload addition of 30% , the second number describes how many time the vessel would need to be refuelled to go to Mars enter its low orbit and return. ISP 0.1 (HPL) kg 0.5 kg 1.0 kg 2.0 kg 4.0 kg 100k m/s 6.9 mm (5.51, 1) C+ 4.9 (4.1, 0), C 3.6 (3.2) D 2.4 (2.1) D- 50k 13.8 (12.9, 2+) B+ 9.8 (8.3, 0) B- 7.2 (6.3) C+ 4.7 (4.3) C 2.8 (2.6) D- 25k 20 (17, 0+*) B 14.4 (12.8) B+ 9.4 (8.7) C+ 5.6 (5.3) C 12.5k 39 (33, 3*) C- 29 (25.4, 1*) A- 18 (17*,0+) B 11 (10.5*,0+) B- 6.25k 38 (37*,1+) C+ 22 (21*,0+) B 3.12k 44(42*, 2+) C * if one pays the thruster cost here then the acceleration only need be used in LEO, and scale ISP to the dV need to alter a profile, possible to reduce the refuels. Based on this the very best choice here is a ship that can achieve 25000 to 100000 ISP (2750 to 10202 ISP) and keeps the humanized portion of the payload below 0.3 dry mass. About the same amount of fuel as solar panel and the bare minimum of structure. I can Imagine a ball, with two bars sticking out and solar panels. Though I think probably given we need to add structural weight, 1.5x fuel might be better. So if this is the case suppose for a multiyear trip how much weight. Lets say 10 tons per astronaut, sounds pretty small. So our scale is 10000/0.3 = 33333, for 2 nauts that is 66666 How does this equate: 66666 meters of solar panel at 66 tonnes producing maximally 36 mW of Power. I should add that if one uses my 10 by 100 meter solar panel (KSP creation) that is 66 Solar panels. The thrusters 11.5 tonnes produce 0.543 to 4.3 kN of thrust (double the PB-ION) and also produce 8.125 MW of waste heat, that needs to be dissipated. The thruster footprint at 0.1M height is 153 square meters (a circle of radius 6.9 meters radius). Saturn V rocket has a radius of 5.1 meters. Our payload in LEO weighs 190 tonnes. Saturn V = 140 tonnes to LEO. Now judging by the structure, it has to be assembled in space.

First off, let me just say that from a couple year ago when I first started building parts this group has come a l o n g w a y in terms of improvements. What I have done: We need however some just basic glossary of terms somewhere, and the term defined like raycasttransformed = (this is what it does, this is how to employ it). . . . . . alphabetized and stickied. It would be nice to see an index at the begin than a headed paragraph on each keyword. The KSP specific area that hung me up were these terms and thrustVectorTransformName, animationName, and raycastTransformName. Built a new, from scratch, ION drive complete with emmissive - would absolutely not have been possible without the tutorials. Built a new, from scratch, Solar panel (10 meters by 100 meters, no joke), extendable non tracking (on purpose) solar panel. - animation would not have been possible without tutorials and searches here Built from scratch a new set of lattice elements (including 5 meter long aluminun poles, corners, T, 30 angles, pole anchors. So here is the basic issue. Im trying to control the power input for my ion drive. So basically this is it, I want to burn 70 units of electricity a second (For my panels im equating a unit to a kilowatt, the ion drive needs to burn 70 units per 0.001 thrust the xenon gas used then needs to be used accordingly to the ISP which is 9000. How to do? Im not exactly sure how electric charge is managed in an ion engine.