GoSlash27

I posted results on a test of this in the other thread: My DV figures from my test using the Pathfinder: 1) Prograde gravity turn with direct injection: 5230 m/sec 2) Prograde orbit followed by injection: 5,260 m/sec (+30 m/sec) 3) Vertical direct injection: 5,570 m/sec (+340 m/sec) These were launches to hit Munar apoapsis to see which approach took the least DV. The profiles were as follows: TWR for all cases was a maximum of 2G. I held the TWR to the standard for typical prograde profiles* and left it running hot for the vertical ascent at 2G for a reverse terminal velocity fall into space for the vertical. * Prograde profile was gravity kick at 7KM, 68* pitch at 15KM, 45* pitch at 25KM, and 23* pitch at 35KM with a constant, linear reduction throughout. 1 -> 2G acceleration in the vertical boost phase, .75G acceleration in the transstage phase, and .5G in the injection phase. TL/DR: Going vertical is never going to be more efficient regardless of where you're going. If you just want to do it because it's easier, then have at it. Just remember that you will need padding in your budget. The reduced Oberth effect from a vertical launch means that this method will require more and more budget padding the further away your destination. Best, -Slashy

Yours, mine, or both? -Slashy

^ This. In stock form and using stock parts, an Eve SSTO is impossible without cheating or using loopholes in the physics.* You'll definitely need mods to make it happen, and since I've never used mods, I'll leave that advice to others. Best, -Slashy *subject to change when .90 comes out

So since I've outlined the process of reformatting the equation, I won't bother doing it again here. Solving for number of engines: e^(DV/9.82Isp)= Rwd Rwd(Mp+Met)-Mp-Mft ___________________ = N The number of engines Me(1-Rwd) ex. "Mommy, are we there yet"?? I've got a mass-mover made of a Jumbo 64. It's hauling 80 tonnes of junk to monitor the lonely space potato (2,300 m/sec DV). How many engines can we hang on it and still make it? e^(2300/(9.82*800)) = Rwd 1.34= Rwd 1.34(80+4)-80-36 _______________ = 4.50 2.25(1-1.34) We can use 4 engines and still make it. If we try 5, we'll be lost in space.

That's true, but we need the extra variables for what we're doing. If we're running a spreadsheet or (God forbid) doing it by hand, we don't have a hook to populate that. And the correct g0 is, indeed 9.82. All of the equations should reflect that. RIC, Boy, does it ever! I *hate* posting equations in here! Dispatcher, That would be epic! I don't know how to code, though. Maybe someone with those skillz can make it happen someday. Best, -Slashy

So as an example of this form: I have a mass mover consisting of a jumbo 64, a single LV-N and 1.6T of assorted stuff. How much payload can it haul and still make a DV of 6,600 m/sec? e^(DV/9.81Isp)= Rwd. e^(6600/(9.81*800)= 2.319 Rwd(NMe+Met)-NMe-Mft _____________________ = Mp (1-Rwd) 2.319(1*2.25+4)-1*2.25-36 _______________________ = Mp (1-2.319) 14.49-2.25-36/(-1.319) =Mp -23.76/-1.319 = 18.0 tonnes. Removing our 1.6t of gear leaves 16.4 tonnes of payload. Hmm... I'm gonna have to rework this. I must've goofed somewhere, 'cuz that ain't right. I'll figure out what I gacked up and revise it... *edit* Got it. Had a case of number dyslexia The equation is good to go.

So how to the mathing for rearranging the rocket equation: DV=9.82Isp*ln(Rwd) in it's standard form. Since our desired output is hiding inside the "Rwd" part, we need to isolate that. First, we get the "9.81Isp" away from it by dividing both sides. DV/(9.82*Isp)= ln(Rwd) Now, to get inside of a log, we have to raise e (approx 2.718) to it's power. And we can't just do it to one side, so... e^(DV/9.81Isp)= Rwd. Now... Rwd has an alter- ego. Aside from meaning e^(DV/9.82Isp), it also means the total mass of our ship when wet divided by the total mass of our ship when dry. Rwd= (NMe+Mp+Mft)/ (NMe+Mp+Met) where Mft= the mass of our full tanks and Met is the mass of our empty tanks We need to isolate this for "Mp", assuming we already know the rest. Algebra- infested waters ahead... Move the denominator to the left Rwd(NMe+Mp+Met)=NMe+Mp+Mft Distribute on the left RwdNMe+RwdMp+RwdMet=NMe+Mp+Mft Get everything not related to Mp on the left and everything related to Mp on the right. RwdNMe+RwdMet-NMe-Mft=Mp-RwdMp Tidying up a bit on the left and extracting Mp on the right Rwd(NMe+Met)-NMe-Mft = Mp(1-Rwd) And finally... Rwd(NMe+Met)-NMe-Mft _____________________ = Mp (1-Rwd) Now... we could tidy this a bit more, since our full tanks are always a known ratio to our empty tanks, but this should suffice.

Pecan, funny you should mention that. solving for payload is exactly what I'm about to cover in this example, but I don't recommend using it for payload to orbit. The payload is limited by thrust as well as DV, and the rocket equation doesn't account for thrust. You could well end up with the correct DV, but not enough thrust to lift off the pad. I recommend using this equation for ops in zero G where the acceleration isn't so critical. Best, -Slashy

Rearranging the equation means that *anything we use as an input* can instead be used as the output! We've used DV as the output (how much DV will this rocket make) and now we've used Mt as the output (how much tankage do I need for this rocket). Let's take a look at a couple more options. N: How many LV-Ns can I hang on the back of this ship and still make it to Duna and back? Mp: How much mass can my space tug get to Laythe? I imagine there's even more variations that could be useful, but these 2 are questions I've tackled by rearranging the equation. So... how do we do this? Warning: Maths ahead...

Now... I know what you're thinking: Yeah, but I could just go into the VAB and stack tanks on until I have enough and it'll tell me that! Ah, but here's what it won't tell you: If you build a spreadsheet with all the engine types and their values, you could plug in the job you need to do and have it spit out these answers for every engine at the same time. Using these answers, you can tell at a glance which engine will yield the lightest overall stage for your particular job. And knowing the mass of the assembly, it can recommend how many engines you need to use, then recalculate based on the new engine mass. Likewise, knowing the cost of the parts, a spreadsheet can tell you at a glance which option is the least expensive. You could do that and take careful notes in the VAB, but it's gonna take an awfully long time to do it that way. This is just one weapon in the arsenal of what rearranging the rocket equation can do for you...

NOTE: g0 has been corrected to 9.81 in KSP's engine as of version 1.0. Please disregard the "9.82" stuff from these earlier equations. Most of us are familiar with the rocket equation, but don't use it because it doesn't tell us anything we can't find out automatically using a mod. But did you know that you can use the same equation to answer other questions besides "How much DV will this stage make"? Questions that no mod can answer for you, and are incredibly handy to know? In this tutorial, I will introduce the "reverse rocket equation" and show you how to use it for fun and profit. The rocket equation, like any other, is like a machine. You put in your raw materials at one end, pull the lever, and it spits out your desired product. In the case of the rocket equation in normal form, we put in *Mass and number of engines--> *Mass of fuel tanks--> *Mass of fuel and o2--> *Mass of payload--> and finally *Isp of our engines.--> This all gets fed into the machine and we crank it by mathing. Once it's done, it spits out --> DV. Well, like all equations, we can reorganize things a bit and get it to spit out a different output. Suppose, for example, you want to answer this question: If I need to impart a certain amount of DV to a known mass using a known number and type of engine, how much will my loaded fuel tanks weigh? Knowing the answer to that question will save you all sorts of time in the VAB, because it tells you exactly how much rocket you need to do a job. It works like this: Your inputs are -DV Your required DV in m/sec -Isp Your engine(s) specific impulse in seconds -N The number of engines you are using -Me The mass of the engine you are using in tonnes -Mp The mass of the payload you must lift in tonnes and finally -Rfe The ratio of your tank's loaded mass to empty mass. For most liquid fuel tanks, this is 9. For other types, check the wiki. First step is to solve for a placeholder value that we don't want to have to calculate twice; e^[DV/(9.82*Isp)] = Rwd The required wet- to- dry ratio of your rocket (save this value) Now for the rest of the process. (Rwd-1)(Nme+Mp) _______________ = Mt The mass of your required fuel tank in tonnes when empty. (Rfe-Rwd) Mt*Rfe = Mf The mass of your required fuel tank in tonnes when full. Mf+NMe = Ms The mass of your stage in tonnes when full. and finally Ms+Mp = M The mass of your entire stage and payload. Now... e^(yada yada) might look daunting, but it's really no big deal, so don't let that throw you. Using a calculator, it goes like this (data in quotes, keypresses underlined) "delta v" divided by "9.81" divided by "Isp" equals e^x MS In the case of the Windows popup calculator, e^x is hiding under the ln key, so the process is "delta v" divided by "9.81" divided by "Isp" equals Inv e^x MS In a spreadsheet, it might look something like this: Example: I'm using a single 48-7S to inject a 12 tonne payload into orbit. I need to impart 700 M/sec DV. What tanks do I need to accomplish this and how much will it weigh as a payload for the stage that got it there? e^(700/(9.81*350)) = 1.23 (1.23-1)(1*.1+12) _____________ = .23*12.1/7.77 = .358. Our empty fuel tanks will weigh 358 kg (9-1.23) .358*9= 3.22. Our loaded tanks will weigh 3.22 tonnes 3.22+.1= 3.32. Our stage will weigh 3.32 tonnes 3.32+12= 15.3 (rounded). This assembly will present a payload of 15.3 tonnes to the preceding stage. ------------------------------------------------------------------------------------------------ Edit: the useful equation petting zoo. We'll keep everything useful we derive here for safe-keeping. e^[DV/(9.82*Isp)] = Rwd The required wet- to- dry ratio of your rocket (save this value). We use this value in any form where we're looking for an answer involving mass. (Rwd-1)(Nme+Mp) _______________ = Mt The mass of your required fuel tank in tonnes when empty. (Rfe-Rwd) Mt*Rfe = Mf The mass of your required fuel tank in tonnes when full. Mf+NMe = Ms The mass of your stage in tonnes when full. Ms+Mp = M The mass of your entire stage and payload. Rwd(NMe+Met)-NMe-Mft _____________________ = Mp The largest payload mass a rocket can carry and still make it's DV requirement (1-Rwd) Rwd(Mp+Met)-Mp-Mft ___________________ = N The maximum number of engines you can use and still get there. Me(1-Rwd)

Fengist, It really just requires that you get into space, which is easy enough. What I found to be the tricky part is finding a ballistic trajectory that drops the glider in range of KSC. And yes, SS1 has a throttle. Solid fuel, but liquid O2. Best, -Slashy

In theory, direct rendezvous is the most efficient profile because it inserts you exactly into the correct position in the correct orbit immediately. Doing a Hohmann transfer means you have to first 1) establish orbit 2) burn normal/anti-normal to correct your inclination 3) burn prograde/ retrograde to alter your eccentricity for intercept and then 4) do the opposite of 3 to restore eccentricity. 3 and 4 are a waste of fuel because 4 erases the changes made by 3 and both maneuvers cost fuel. 2 is a waste of fuel because it's far more efficient to launch into a perfectly inclined orbit from the outset than it is to do it up there. And also doing it all at once maximizes the Oberth effect. Now... this is in theory. In practice, I can't say because I've never used MechJeb and therefore don't know how that mod does it. I have managed to do it by hand on occasion, and it did save me fuel. I would do it that way every time if I could, but the window to pull it off is small compared to my time to orbit variations and I'm not good enough to pull it off with that sort of precision every time. Best, -Slashy

Outstanding! I love seeing this sort of thing. Now that you've got the first one under your belt, the rest will be easy. Best, -Slashy

^ Pics. There are several reasons why this could happen, and we can usually tell what the problem is if you post pics. Best, -Slashy

Abusing the loopholes in the physics engine. Using control surfaces as wings allows you use lift as thrust and drag as lift. Then attempting to disengage parts while travelling faster than 800 m/sec will result in an imparted acceleration indefinitely. This allows you to do all sorts of crazy stuff without fuel. It's cheaty as hell. Best, -Slashy

No, *heck* no! haha Nobody has done that sort of thing with natural logs since back before the American Revolution. Just plain not needed. Way back when, they'd just look it up in a log table. And never to that many decimal places, since that kind of precision was never needed (something we've forgotten these days). After someone invented the slide rule, they'd just use that. Then calculators, then computers, then spreadsheets. But being an electronics guy, I *can* guesstimate base 10 logs in my head with a fair degree of accuracy. All comms electronics techs and engineers pick up that ability over time. The method to do it isn't the same as what LD describes, but he's talking about a different base. Having said that, your point is absolutely valid. You don't actually have to know the first thing about natural logs to run this equation. All you have to be able to do is find the button marked "ln" on a calculator and push it. Anybody can do that. Best, -Slashy

Love to hear this sort of thing! We're always here to help, so feel free to hit us up. Best, -Slashy

It's worse than you think. I have actually built a successful Kerbin-> Laythe and back manned SSTO. It also did Duna and Eve (yeah, Eve SSTO) on the same trip. With absolutely no fuel expended at all. Combining an infiniglider with Kraken drive completely breaks the game. But that's not what we're talking about here. Best, -Slashy

None of this counters what I've said, so either I gave a misleading impression of what I meant or you misinterpreted what I said. A vertical lift SSTO requires more engines to lift the same mass as a horizontal takeoff SSTO, so it is somewhat less efficient due to the extra mass. And while it is trivial to recover a vertical lift SSTO intact, it's not liable to end up on the runway at KSC at the end of the mission, whereas a (properly designed) spaceplane will do that every time, thus recovering all of your funds. Another difference I didn't mention is that a vertical lift SSTO has no abort options in the event of failure, so there's a safety risk to the payload that's not a problem for spaceplanes. A spaceplane can land somewhere safely in the event of a failed launch without destroying the cargo, whereas a vertical mass lifter will doom it's cargo if something goes wrong. I use both types, and specialize them to their ideal payloads. Kerbals and supplies go in spaceplanes, while structures go aboard vertical lift SSTOs. Best, -Slashy

The original question never specified how much payload we are talking about here, and that's the single most important thing. Also a disclaimer, I have zero experience at making "legit" Kerbin to Laythe SSTO spaceplanes. Just not something I ever personally had a need for. So whatever I propose should be evaluated with that fact in mind... I do have a good amount of experience at making efficient SSTO spaceplanes that are multi- hop capable and LV-N mass movers, which is really what we're talking about here. The requirement is for an SSTO spaceplane that can make 2 atmospheric burns to orbit from fuel alone with reserve and can carry a single LV-N with enough fuel/ oxidizer to make 5KM/sec DV in space. This is, IMO, easily do-able. A twin engine dual hop SSTO would have 2 turbojets, 1200 units of fuel, 28 "strake" wing panels, and would have the capacity to lift 9 tonnes of payload to orbit with it. That 9 tonnes would be the LV-N and 6.75 tonnes of fuel/ oxidizer/ tanks, which is way more than you'd need to move 25 tonnes of mass that much at a vacuum Isp of 800. So my preliminary design would be: 1 Mk 2 cockpit 3 short LF+O Mk2 sections 1 LV-N 4 Mk-1 jet fuel fuselage 2 turbojet 4 XM-G50 radial scoop intakes 2 shock cone intakes 28 "strake" wing panels Other assorted/ massless parts as necessary. Best, -Slashy

^ I agree with RIC on all his points. T/W is the most important thing on bodies with atmospheres. You want the velocity to match the increase in terminal velocity so that you are generating constant drag (in newtons) throughout the ascent, and this will naturally give you a huge variation based on t/w. On airless bodies OTOH, the characteristics of the ship don't matter. A gravity turn is all about balancing gravity losses against drag losses, but since there's no drag, there's only one ideal profile; get it going horizontal as fast as you can without hitting the terrain. For a generic profile on Kerbin, my profile very closely matches RICs. Boost phase: realistically, 2 Gs would be perfect, but since you naturally gain acceleration as the fuel burns off that's more engines than you need. It is more efficient to start with just enough engine to get 1 G and let the acceleration build. Vertical to 7KM. Transstage phase: This is a gradual reduction in thrust from 1G to .75G as the nose falls prograde from vertical to 45*. altitude/ pitch/ acceleration 7KM/ 90 / 1G 15KM/70 / .95G 25KM/ 45 / .75G Injection phase: 25KM/ 45 / .75G 35KM/ 25 / .5G Circularization phase: From here, we will maintain .5G acceleration and pitch to establish our orbit. There are 2 ways to do this: 1) The KSP way: rotate prograde. Establish the apoapsis, then throttle just enough to keep it until you get within 15 seconds of it. Throttle as necessary to keep the apoapsis 15 seconds ahead until periapsis is at orbital height. 2) The real world way: Establish the apoapsis, then pitch as necessary to maintain the apoapsis without throttling. When periapsis matches orbital height, cut the engine. This is an on-the fly piloting thing, so there's no standard profile for it. It's just a gradual transition from prograde to normal. Burning prograde causes your apoapsis to get higher and closer, while burning radial in makes it go lower and move farther away. I use the KSP method because... frankly, I'm lazy and why not make it as easy for myself as possible, since the equipment lets me do it. HTHs, -Slashy

Hahaha well.. none of that stuff applies to KSP and certainly not anything going on in this thread. You might want to pose that in the Science Labs forum. I do recall that NASA has a pretty good tutorial and interactive Java applet on the subject, though. http://www.grc.nasa.gov/WWW/k-12/airplane/nozzled.html Good search terms to start with are "CD nozzle" "DeLaval nozzle" "Convergent Divergent Nozzle" and "ConDi nozzle". I did some research on this subject a few months back on a subject completely unrelated to rockets, and there was lots of info out there. Best, -Slashy

I haven't seen it happen the second way, either... but that's not to say it doesn't. That first scenario happens all the time, tho'. Thanks and have a splendid weekend, -Slashy

My profile is outlined in detail here. It's not for FAR, tho'. That's not bad. Only 10% more fuel for a much easier maneuver that you cant (read: is much harder to) screw up. Depends on your definition of "bad". That probe normally escapes Kerbin's SOI entirely and completes both normalization *and* inclination burns on less fuel than the vertical munar intercept took. I personally would never waste 340 m/sec if I didn't have to. I don't know enough about FAR aerodynamics to answer this question, but "optimal" acceleration doesn't light you up like a Christmas tree on the way up, regardless of pressure gradients. Again, I don't know enough about FAR profiles to answer to this. Your issue wasn't with when you made the turn, but rather that you're generating so much drag due to the acceleration. If you're bleeding DV like a stuck pig due to drag losses, the vertical ascent is going to seem less wasteful than it actually is in comparison. Sure it was. You used the word "efficient" several times in the OP. Your other criteria are your personal preferences, so there are no wrong answers there. Those are subjective opinion rather than fact. If you wanna go vertical because you find it easier, that's your business. For the rest of us, most of us will probably prefer one of the other 2 methods because maintaining control during a gravity turn isn't a problem for us. If you go into advanced edit on the OP, it will give you the option to fix that. Best, -Slashy