GoSlash27

RIC, Certainly no apology needed. I was the one who wandered off into the weeds and I think this gave me some food for thought on a problem I've been wrestling with. Best, -Slashy

RIC, Oh, sure it would be. Engine A consumes yada percentage of the fuel at such-and-such an Isp. Engine b consumes yada percentage of the fuel at such-and-such an Isp. Therefore the combined Isp is weighted proportionally. This is actually what you're doing when you weight by F/Isp. And yeah... definitely in the weeds. My point was just that either description would be technically correct, so Pincushion's speculation is accurate although we think of it as weighting by thrust. And of course reconstituting mass flow rate by Isp and thrust is convenient because those figures are more readily available. Best, -Slashy

RIC, The process to weight them would be different, so you couldn't just plug in Rf in place of T. When you invoke "F/Isp" in the denominator you are talking about mass flow rate at that point. Likewise, when you invoke F in the numerator that's really the same thing as "RfIsp"*. I would expect a proper weighting based on mass flow rate to look something like (Rf1Isp1+Rf2Isp2+...+ RfnIspn)/(Rf1+Rf2+...+Rfn) although I haven't actually verified that tl/dr: Aside from it's use in the rocket equation, Isp can be readily converted to exhaust velocity by multiplying by 9.81. This exhaust velocity is convertible to thrust when the mass flow rate is given or mass flow rate when the thrust is given. *not exactly, but it's proportional and that's what matters Best, -Slashy

At the risk of getting too far into the weeds, either approach would technically work since the mass flow rate is proportional to thrust. Thrust is a much more convenient value for the calculation though. Best, -Slashy

Yeah we know, but still. Others may run into this problem and need suggestions, so may as well post them. Now in the spirit of your OP, you could try Tom Kelly's suggestion: https://youtu.be/rZNHktj14H8?t=103 I hear Bolivia is really nice this time of year. Best, -Slashy

Smashing Kirby, you can always temporarily arrest the spin by hitting the period then comma. The jump into time warp stops the tumbling, but I'm sure it will build back up over time. Best, -Slashy

Serious answer to a not- serious question: Take a satellite contract in the Kerbin SoI. They'll front you more than enough money to build it and then reward you handsomely when it's done. Problem solved Best, -Slashy

liebniz, You could orbit 2 sats over Kerbin to run as backwards clock hands. One would have an orbital period of 3 hours (1,577,000m), while the other would have an orbital period of 51m 25.7s (347,900m) You could then set them in a constellation of 6 geosynchronous sats (2,869,000m) and voila! A giant celestial clock! The geosynchronous sats would be the clock face. The slow hand would make 1 sweep per day (hour hand) and the fast hand would make 6 sweeps per day (minute hand). Of course, I'm way too lazy to actually do this Best, -Slashy

Yay! Another satisfied customer Best, -Slashy

1b) Designing our Munar ascent stage. Before we begin this, we need to know one more thing about the Mun: What's the surface gravity (g)? g=GM/r^2 The Mun's surface gravity is therefore 1.63m/sec^2 We will pick a reasonable minimum acceleration in "g"s. 1g exactly will no doubt be wasteful of fuel and gobs of acceleration will be difficult to control with precision and sacrifices lightness for moar boosters. We'll go with 1.2g q#1: if I were to build a rocket using just 1 of my proposed engines, how much mass could it lift at 1.2g Munar? a: M= T/(1.2*1.63) where M is in tonnes and T is in kN. A single LV-1R could lift up to 1.02 tonnes total mass on the Mun at a minimum of 1.2G. q#2: How much of this mass would have to be fuel in order to generate 700 m/sec DV? a: Using the reverse rocket equation, we know that our wet-to-dry ratio (which I'll refer to as Rwd) needs to be e^(DV/(9.81Isp)), 1.25. We know that our fully fueled rocket will weigh 1.25 times as much full as it does when empty. Converting this to fuel percentage, we subtract 1 and then divide it into the fully fueled portion. .25/1.25= 1/5= 20%. We know that our fuel tanks weigh 1/8 as much as the fuel they contain, so fuel tanks account for 2.5% of our mass. q#3: How much of this theoretical lifter would be payload? We take our total mass and then subtract the mass of the engine, fuel, and tanks. Mp= M-Me-Mf-Mt 1.02-.02-1.02(.2+.025)= .77 tonnes. A theoretical rocket using one of these engines could place .77 tonnes into Munar orbit. We can repeat this process with several different candidate engines. We need to know how much payload we actually need to place into orbit. We'll need a lander can, docking port, RCS quads, perhaps a battery and solar panels. Say that weighs about a tonne. This is just a little more than 1 of these engines can handle, but 2 can do it no problem. So now we design it for real. Rwd= 1.25 Referring to my "reverse rocket equation", (Rwd-1)(NMe+Mp)/(9-Rwd) will give you the mass of your empty fuel tanks. N= number of engines (2) Me is the mass of your engine (0.02) Mp is the mass of your payload (1) So our tank weighs .03t when empty. Fill it with LF&O and it weighs 9 times as much; .302 tonnes. So that's it. Our 1 tonne ascent module, .302t of fuel tankage, 2 LV-1R engines. This comes out to 1.34t total mass. This whole assembly now becomes payload for the descent stage. - - - Updated - - - So now that I've gone into detail for the ascent stage, I won't bother repeating it for the other stages of the mission. It's the same procedure. Our ascent stage (plus decoupler) is payload for the descent stage. It needs to make 1150 m/sec at a minimum of 1g Munar with a payload of 1.4tonnes. A single 48-7S will do this with a total mass of 2.51 tonnes. The Munar lander is, in turn, payload for the Munar transstage. In a parallel fashion, you have to work out your trip home using the CM with reentry capsule. These are both payloads for the TMI stage, which is payload for the upper transstage, which is payload for the booster. In working out the transstage, we again use the vis-viva to get us an apoapsis at the Mun's altitude. One important thing to note is the period of this orbit, because we actually have to arrive at the Mun. Likewise, we need to know the Mun's orbital velocity around Kerbin so we can work out how much to lead it by. This is all in the "Orbmech" reference. Happily, it works out that the Mun is ideally situated when you see it rise above Kerbin's surface. And you've already worked out what your velocity will be when it's time to cut engines. Apologies for the wall o' text, but this should be enough to get you going. A couple more points to add that should prove helpful: If you know your current mass, engine Isp, and fuel/ o2, then it's a simple matter to compute your remaining DV to make sure you're not in trouble. each 90 units of fuel and 110 units in the tank *of the current stage* is a tonne. current mass/(current mass-fuel mass) = Rwd 9.81xIspxln(Rwd)= DV You want to keep track of this throughout the mission. You probably want to get into the habit of doing an "around the horn" check before each burn in order to be safe. If you know your current mass, engine thrust, and DV of the burn, you can compute the burn time. Handy to figure this out ahead of each burn. Good luck! -Slashy

AFA how, Stage 1 is planning out the mission. It is assumed that you know what the aero losses are during launch to LKO so you have that part of the DV budget already figured out. 1a) Planning the launch to Munar orbit from the surface. We'll pick an altitude arbitrarily.. say 10km. We already know the Mun's mass and sea level radius. From this, we can compute what our orbital velocity would be at the surface. v= sqrt(GM/r) where M is the Mun's mass in grams and r is the Mun's radius in meters. G is the universal gravitational constant; 6.67408 × 10-11 Plugging in the Mun's mass and radius gives us 571 m/sec. Extending the apoapsis to 10km using the vis-viva (step by step process in the "orbmech" link I posted) shows that you need another 7 m/sec. Circularizing at 10 km requires yet another 7 m/sec. It would therefore take an absolute minimum of 585 m/sec DV to get from the Munar surface to orbit and you'd want to pad that figure to account for gravity losses, cosine losses, and imperfect flight path. Probably 20% will be fine for launch, but you'll want to be extra generous for landing DV, perhaps even doubling it. So we'll design our Munar lander for 700 m/sec DV for launch and 1,150 m/sec DV for landing. We know from the vis-viva some important figures we'll want to remember for future use: Orbital velocity= 557 m/sec. Velocity to set periapsis to sea level= 550 m/sec Velocity to set apoapsis to 10km during launch= 577m/sec

Der Anfang, You can do most of it on paper and don't need to use maneuver nodes at all, but there are some situations where you'll have to refer to the map at least temporarily. For example, you need to verify your periapsis when you cross SoI boundaries to ensure that you're not going to smack into the surface or miss the atmosphere on the way back. For safety's sake you need to at least verify in map mode that your burn has gone as predicted. Also it is the only way you can see your ship information tab. You need that from time to time during the mission. For mission planning you use the vis-viva equation https://en.wikipedia.org/wiki/Vis-viva_equation For stage design you'd use the reverse Tsiolkovsky rocket equation http://forum.kerbalspaceprogram.com/threads/102809-The-reverse-rocket-equation-explained This resource is excellent: http://www.braeunig.us/space/orbmech.htm Best, -Slashy

Brainlord, The procedural fairings are a bit buggy, I'm afraid. Generally if you carry the fairing forward a little before closing it it will work better. You will see the "close" option text turn green when it's happy. Good luck! -Slashy

Scotius, I'd imagine that would be necessary for the catalytic cracking and fuel distillation processes. I don't think either would work without centrifugal force. Of course in KSP there is no need for that. Best, -Slashy

Red shirt, I like your attitude! Remember that you are operating in Munar gravity and vacuum, so you need far less engine and fuel than testing on Kerbin would suggest. We did it back then without the aid of mods as a matter of course. In fact, I planned my whole mission and designed the vehicles using nothing more than pen/ paper and a Pickett No. 120 sliderule. It's just as feasible now as it was back then, so don't give up hope! I look forward to seeing your mission report. Best, -Slashy

Nothalogh, On the return vessel. Like the Apollo, there was a separate CM module. Unlike the Apollo program, there was no docking capability in .18 demo. This meant 2 "Mercury" sized launches instead of 1 "Saturn" sized launch. I did a munar orbit rendezvous and EVA to the CM for the trip home. Jeb made it home safely and had a tickertape parade, followed by my immediate purchase of .24 Beta. Best, -Slashy

My current project is pretty mundane... I'm trying to create a universal stack modeler to go incorporate into my stage designer spreadsheet. The design and fuel flow logic are giving me fits Best, -Slashy

Delta-Cheese, Your sink rate is your vertical velocity towards the surface (or in other words how fast you're falling). It's in that half- moon looking gauge to the right of the altimeter. If you're falling too fast, parts get explodey on impact. Best, -Slashy

Red Shirt (love the name BTW) You probably saw my first Mun lander from back in the .18 demo. Yes absolutely you can still do it using only RCS thrusters. All the parts stats and celestial body parameters are the same as they were back then. It's not the most efficient way to do it now, but it was at the time. Best, -Slashy - - - Updated - - - All, 4 RCS quads would be adequate for a 1t payload on the Mun. No biggy, it's just not a very efficient way to go about it. Best, -Slashy

Alshain, You're very welcome. Give 'em Hell!, -Slashy

Alshain, The Isp would be a constant for identical engines. You're doubling the thrust and flow rate, but exhaust velocity remains constant. Best, -Slashy

Jack Bush, If you burn more horizontal, it will extend your apopasis further prograde. This gives you more spacing in time while still adding velocity without raising it. Check it out: http://wikisend.com/download/969842/SRBLifter.craft Download and install this simple rocket and observe the path it follows in the first stage. Don't engage SAS or touch anything, just launch it watch what it does. This rocket has been set up to execute a near- ideal gravity turn without any input from the user. http://s52.photobucket.com/user/GoSlash27/slideshow/KSP/GravTurnBoosterDemo If you mimic this flight path during your launch, it will drop you right into orbit without overshooting your altitude or passing your apoapsis. Good luck! -Slashy

Delta-Cheese, Here's the "reverse gravity turn" technique that Alshain's referring to: http://forum.kerbalspaceprogram.com/threads/104638-Reverse-gravity-turn-landing-technique-for-airless-bodies Still the easiest way I've found to place a lander right where you want it without mods. 2km/sec DV is more than plenty to accomplish the mission. Good luck! -Slashy

It does model that... sorta. But it's a very small amount of drag and it disappears rapidly as you approach Mach 1. For best results you always want to have the nose as clean as possible. You can compensate for the stability issues with fins, but you really want to avoid adding drag wherever you can when building spaceplanes. That's the whole key to their performance. Best, -Slashy

I concur with RIC. A draggy nose is a recipe for all the control issues you're having. See if you can extend the procedural fairing and give it a more gradual ogive. As for the control surfaces, I really can't tell by that picture. Cool concept, BTW Best, -Slashy