GoSlash27

The reason that map shows 80 DV to Eve is because it assumes a single burn from LKO to transfer. In that sense, it is true. It only takes 80 additional DV to get from Kerbin escape to Eve transfer, but only if you add those DV while still in LKO (due to the Oberth effect). If you're like me and you transfer burn outside Kerbin SOI, there's no way you can set up an intercept on a measly 80 DV. If that's the case, try this map: Best, -Slashy [edit] Ninja'd by mhoram![/edit]

The thing that jumps out at me is that the small 3 stager is generating more DV on paper than the larger asparagus setup. Unless they are lifting radically different payloads that should not be. A 3 stager can get close to the mass efficiency of an asparagus but it can't quite match it, let alone exceed it by a wide margin. I'm thinkin' MJ is misleading you about how much DV you're actually generating. Best, -Slashy

I've encountered this on Kerbin and Eve. Slope changes don't seem to behave intuitively. Best, -Slashy

Oh, I almost forgot the other half of the problem! Not only do you have to generate the ÃŽâ€V for each phase of the launch, but you also have to do it rapidly enough to not waste it fighting gravity and aerodynamic drag. You can generate 4,500 M/s ÃŽâ€V, but if you do it at less than 1 G, you'll never leave the pad. Likewise, if you generate it at the wrong acceleration for a phase, you'll waste a lot of it fighting gravity and drag instead of establishing your orbit. So to that end, you want enough engine to generate the proper acceleration for each phase of the flight. The important equation here is Ga=T/(Ca*Mt) where Ga= Gs of acceleration, T is the thrust in kilonewtons, Ca is the local acceleration at 1 G, and Mt is the total mass of the vehicle you're moving. Naturally, you'll want to rewrite this as we've done above to show how much thrust you need to generate your necessary acceleration. There's no consensus on this, but the conventional wisdom is you need 2 Gs for the boost phase (pad to gravity turn), 2 Gs tapering down to 1G for the transstage phase (gravity turn to apoapsis), and .5 G or even less from apoapsis to orbit. Best, -Slashy

The equation you really need to plan your launch vehicle is this one: ÃŽâ€V= 9.81Isp*ln(Mw/Md) where Isp is the specific impulse of your engine in seconds, ln denotes a natural logarithm, Mw is the mass of your rocket with fuel and Md is the mass of your rocket without fuel. As you have already surmised, what you can do is determined by how much ÃŽâ€V you can generate. This equation is in the form of "this much rocket will make this much ÃŽâ€V". A much more useful version would say "in order to make this much ÃŽâ€V, you need this much rocket". I will show the derivation of such an equation below. If you want to derive it yourself, don't highlight the following text... Mw/Md is what the rocket surgeons refer to as the "mass ratio", which I shall refer to as Rm. Since the mass of fuel and tanks you need are wholly dependent on the mass ratio, first order of business is to rewrite the main equation to solve for mass ratio. ÃŽâ€V=9.81*Isp*ln(Rm) ÃŽâ€V/(9.81*Isp)=ln(Rm) e^(ÃŽâ€V/9.81*Isp)=Rm Next is to define which parts of our mass define the numerator and denominator of our mass ratio. Mt= Mass of our tanks fully loaded Me=Mass of our engines Mp= Mass of our payload and Rfe= the ratio of of the mass of our full tanks to empty. You would simply take the loaded mass of your tank and divide it by the empty mass. Thankfully, all small and large diameter liquid tanks work out to "9". Expressed this way, Rm=(Mp+Me+Mt)/(Mp+Me+Mt/Rfe) We need to rewrite this to solve for Mt. Rm=(Mp+Me+Mt)/(Mp+Me+Mt/Rfe) Rm(Mp+Me+Mt/Rfe)=Mp+Me+Mt Rm(Mp+Me+Mt/Rfe)-Mp-Me=Mt but we have an "Mt" on the wrong side of the equation, so... Rm(Mp+Me+Mt/Rfe)-Mp-Me=Mt RmMp+RmMe+RmMt/Rfe-Mp-Me=Mt commutative RmMp-Mp+RmMe-Me+RmMt/Rfe=Mt Distributive corollary Mp(Rm-1)+Me(Rm-1)+RmMt/Rfe=Mt Distributive corollary (Mp+Me)(Rm-1)+RmMt/Rfe=Mt rewrite that a bit... (Mp+Me)(Rm-1)+Mt(Rm/Rfe)=Mt and move it back over (Mp+Me)(Rm-1)=Mt-Mt(Rm/Rfe) Reverse distributive (Mp+Me)(Rm-1)=Mt(1-Rm/Rfe) (Mp+Me)(Rm-1)/(1-Rm/Rfe)=Mt We can tidy it up a bit by replacing "1" in the denominator with "Rfe/Rfe" (Mp+Me)(Rm-1)/(Rfe/Rfe-Rm/Rfe)=Mt (Mp+Me)(Rm-1)/((Rfe-Rm)/Rfe)=Mt And since division by a fraction is the same thing as multiplying by it's inverse... (Mp+Me)(Rm-1)*(Rfe/(Rfe-Rm))=Mt so Rfe(Mp+Me)(Rm-1)/(Rfe-Rm)=Mt coupled with the reverse of the ÃŽâ€V equation above, you can substitute it in. Rfe(Mp+Me)(e^(ÃŽâ€V/9.81*Isp)-1)/(Rfe-e^(ÃŽâ€V/9.81*Isp))=Mt You now have the power to determine exactly how much fuel you need to make a rocket generate a needed ÃŽâ€V given the engine specs, tank specs, and payload you need to move. This also applies to staging, where you treat subsequent stages as payload for earlier stages. The design process is to start at the end of your mission and work backwards. I need to put a capsule in orbit that can support a Kerbal and return to Kerbin. That will weigh this much. It needs to have solar panels and the ability to dock, so it needs RCS junk, batteries, etc. And we'll deorbit the heavy stuff and let it burn up so our Kerbal has just the parachute and reentry shield, so add in a decoupler. The whole orbital vehicle will weigh this much. That's payload. So how much to get this from exiting the atmosphere to orbit? That's this much ÃŽâ€V. This engine looks good for that, so plug it in. That creates a stage with such-and-such tanks and such-and-such engines, for a total mass of this much. How much ÃŽâ€V to get it from the gravity turn to space? Should I break it down into multiple stages? Rinse and repeat until you've designed all the way back down to the pad. Remember that saving a little weight at the end saves a whole lot of weight at the beginning, which is critical. Good luck and remember that failed launches are really just discoveries of what doesn't work! -Slashy

But of course... if you're using Protractor/ MechJeb, etc. it's no longer a vanilla installation. I generally break my intercept into 3 phases for that reason; I have no way to accurately plan the entire burn from LKO. I wait for a window that looks close (either proper phase or inclination node alignment), shoot for an escape burn from Kerbin SOI, then perform my intercept burn. Definitely less efficient than a single honk from LKO. Best, -Slashy

Same here. Spreadsheets rawk! I also have a part in there where I reversed the rocket equation and applied it to each engine. I just put in a payload, required DV, number of engines and local gravity and it spits out how much fuel (including the mass of the tank) I would need, the recommended number of engines to make various G levels, total vehicle mass, etc. for each engine. I can tell at a glance which combo would be most efficient for a given stage. Best, -Slashy

*edit* looking at that equation, it wouldn't be able to approximate a DV requirement to land or launch, even from an airless body. It looks like that just gives an orbital velocity for a known altitude from a planetary center.You'd need to know the rotational period of the body, it's radius, it's mass, and your orbital altitude (assuming a prograde orbit) to be able to calculate a DV. It would be your orbital velocity minus the planet's rotational velocity. But once you factor in atmospheric drag (something that cannot be done with calculus), you end up with a completely different number. Anywho... the process would be to calculate orbital velocity at your chosen altitude. Rp+A. Your equation looks fine to do that part. Next, you would work out surface rotation velocity by radius and period Vs= 2pi r(m)/p(sec). Then Vo-Vs would give you your DV. If you were to correct for inclination, you would multiply Vs by the cosine of your ascention, assuming that 0* ascention is aligned with the equator prograde. Gravity losses are generally minor. That'd be dependent on the sine of your pitch vector which applies 1 local G of acceleration downward, regardless of what your engines are generating. Since the bulk of an approach or departure are tangent to the surface, gravity losses don't account for much. Not something that you can compute, since your approach profile determines it. The big killer is atmospheric drag, which cannot be mathematically solved. It must be either simulated or gathered from empirical results. Best, -Slashy

That they are, but unfortunately I have to meet the acceleration requirement using the available real estate. Generally when you break the job down into a small DV budget for a stage, the higher thrust to weight of a lighter engine can override the Isp advantage of a heavier but more efficient engine. I'm not a fan of any of the big radius engines, but in this case I need the kilonewtons. Best, -Slashy

I goofed somewhere and gave it way more DV than it needed for the mission. By the time I was circularized and zeroed out my inclination, it had hardly used any of the fuel in the injection stage. The transstage worked perfectly as designed and there was more boost stage DV than I needed. It'll need some tweaking...

I'm actually doing something like that with my (hopefully) properly designed 20t lifter. The transstage is arrayed around the injection stage asparagus style, but the injection stage is merely cargo. I lose some mass ratio doing that, but I have a self- imposed rule against lighting nuclear engines in Kerbal atmosphere. Ironically enough, all the stages of that booster wound up using the same exact engine because I wanted the thrust to taper down from 1.5G to 1G over the course of the burn. The boost stage looks to be a lot like my knocked- together example. single mainsail surrounded by aerospikes with radial engines. It should maintain a constant 2G throughout the boost phase. We'll see how it goes...

I'm with Taki. If I hit the space bar and something falls off, it's a stage.

http://www.sciencechannel.com/tv-shows/are-we-alone/videos/alien-encounters.htm Has anyone else here been following this series? I just saw it for the first time today and I'm hooked. It's sort of a futurist what- if scenario that covers a lot of scientific ground.

Moonfrog, The error in this train of thought is that more thrust does not equal more DV. Whether you have a single engine or a dozen, they will all yield the same DV for a given quantity of fuel given the same Isp, since the DV is actually a function of the Isp and mass ratio. Arguably, the total DV is actually *reduced* by adding engines, since they are mass that is not fuel. The efficiency is mostly as you say; using engines instead of lifting them as cargo. It allows you to meet your thrust requirement for a given phase of a launch using less total engine mass. Also, in this setup the radial rockets don't have a TWR of 1. They have a TWR appropriate for the tank they're lifting and the phase of the launch. An early stage might accelerate it's tank at 2 Gs, while a late stage might only lift it's tank at .7G. As you say, it depends on the requirement at the time... Best, -Slashy

UmbralRaptor, I knocked together an example of what I'm talking about. It was a rush job and I didn't do any prep work, so please pardon the crudity... Here you see the complete stack. Almost 20T of payload at the top. The guidance/ docking unit below that. Further down is the injection stage surrounded by the transstage in an asparagus arrangement. You can't tell by the pic on the pad, but the injection stage actually has 4 engines while the transstage units have twice the fuel but only the one 48-7S engine per tank. The injection stage does the bulk of the lifting for the upper stages, while the transstage units are just along for the ride, feeding it. Likewise, the boost stage sits below. A single Mainsail surrounded in an asparagus arrangement by 6 Aerospikes. They don't add thrust to the stack, but are merely there to lift their individual tanks. The Mainsail does the heavy work. This results in a remarkably small assembly for such a massive payload. Again, this is a crude example. I can do better with some prep time.

That's just it; When I'm designing upper stages, every Kg counts. I'd much rather not have to lift that additional 50 kilos if I don't have to. I'd prefer a 10Kg tank empty and would really love to have it "small" radius instead of "tiny". Maybe the height of a Z1K battery. That'd be about as close to perfect as I could ever want. Just to clarify, I'm not asking to get rid of any bigger RCS tanks. I'd just like a smaller stackable option. Best, -Slashy

"what you mean is those stages are to be seen as neutral, having the same TWR of the core stage so the total TWR does not change." This. And my reasoning is that if you have radial stages providing as much thrust as the core stage, then you are either over-accelerating off the pad or throttling back. Either way, you're carrying more engine than you need to. Likewise, if you are carrying radial drop tanks, you will either under- accelerate off the pad or else put in a bigger engine and throttle down at the end. Again, more engine than you need. If the system is tailored so that you always have just the right amount of engine then you have maximized your mass ratio, allowing you to get more DV with a lighter stage. And the reason I used a transstage as an example instead of a booster is because saving weight on upper stages is critical while saving weight on boosters isn't. Best, -Slashy

No Sir. In this model they do not contribute to the TWR of the stage as a whole. They merely generate enough thrust to make themselves "transparent" to the rest of the stack. Their job is not to provide additional thrust, but rather additional fuel. Simply carrying the fuel with no engines would incur a weight penalty on the engine that's doing all the work, so they generate just enough thrust to lift themselves so the core engine doesn't have to. -Slashy

I'm just kicking around the concept, so I don't have a craft to show as an example. The thinking is that copying the core booster basically makes the radial stages "boosters", while eliminating their engines entirely essentially makes them drop tanks. My suspicion is that the optimal efficiency lies somewhere between those two points. I think it's where they burn just enough gas to provide their own acceleration. -Slashy

I'm assuming a transstage for the example rather than a boost stage, hence the arbitrary "1G" specification. For a booster stage, you'd use 2G on the central core booster to lift the entire stack and each radial "feeder" would generate 2G, but only for itself. Using completely invented numbers by way of illustration, you might be lifting a 20T payload +50T transstage/ injection assembly off the pad. Your central core booster has 30T of fuel, so it has an engine that generates 200T of thrust. The radial "feeder boosters", only have to lift themselves, so each requires only 60T of thrust. Hope I'm explaining it right! -Slashy

I had an interesting thought earlier and I'd like to share it here and let you folks kick it around... The advantage of asparagus staging in terms of efficiency is that virtually none of the weight involved is "dead" weight. The disadvantage is that it's payload capacity is ultimately limited by how much payload the last remaining stage can accelerate acceptably. So what if you set up the center stage to accelerate the payload at a desired rate, while the radially dependent stages merely have enough thrust to lift themselves? Looked at this way, the radial stages become not "boosters", but actually "massles fuel tanks". They provide additional DV by donating their unburned fuel to the core booster, rather than the usual practice of providing additional thrust. If this line of reasoning is correct, an asparagus booster designed in this fashion should be noticeably more efficient than common asparagus designs. As an example, say you're designing an asparagus booster as a transstage for a 50t payload/injection stage combo. You want to generate 2,000 DV at 1G. So you design the core stage to generate maybe 700DV while lifting 20T + whatever it's fuel is. The radial stages have the same amount of fuel as the core, but only enough thrust to lift themselves. Each stage pair would then be expected to keep the core booster running to generate another 200+450+650 and ultimately fulfill the DV requirement. This should also cut down on the inherent instability problems since the bulk of the thrust is behind the payload instead of spread out. Has anyone experimented along these lines? Best, -Slashy

This is probably a first, but I'd like to request the addition of some smaller RCS tanks. I'm finding that on an average flight with 20t payloads and 2 docking evolutions, I only need maybe 10-15% of the amount of propellant provided in the probe- sized RCS tanks. I realize that I can unload unnecessary propellant before the flight, but that still leaves the tanks themselves, which are fairly chunky. Thanks! -Slashy

With one caveat: Often the most efficient engine (in terms of Isp) isn't actually the engine that will yield the lightest overall stage. A lot of the time you can get better results by choosing a lighter engine with worse Isp. When the additional fuel it needs to generate the same DV weighs less than the additional mass of the more efficient engine, it's better to go with the smaller one. Regards, -Slashy

Think of it in terms of those coin funnels your parents used to let you play with when you were little. http://www.youtube.com/watch?v=B39ep3419ro The gravitational force is lower with distance, which is represented by the slope of the funnel; nearly flat out at the edges and nearly vertical close in. A coin travels very slowly when in a circular orbit far from the center, both in terms of angular velocity and vector velocity. It has very little kinetic energy but a lot of potential energy. It's reversed when the coin is in close to the center. Very high kinetic energy, but nearly depleted potential energy. So indeed, a ship in a high orbit *does* have less velocity than a ship in low orbit. It's moving more slowly. But if you were to force it down to a periapsis matching the orbit of the "faster" ship in low orbit, you would find that the ship in high orbit would whiz right by the one in low orbit before returning to it's apoapsis because it has more energy. Best, -Slashy

It'd depend on what you consider a "landing".