GoSlash27

MabDeno, You might be able to gain some performance by losing one of the tanks. Best, -Slashy

If I understand your current question, my math says that it is not more efficient at any distance to launch from Kerbin's surface at the north pole than it is to do so from a spot on Kerbin's orbit away from it's SOI. If you got the idea otherwise from me, it was a misunderstanding, as I was referring to LKO. And I will take this opportunity to stop responding to your posts. Cheers, -Slashy

Sorry, didn't see this when I posted my last. I don't know how being static on the North pole could ever be more efficient than floating in open space. I'd have to chuck some numbers at it and see if that's possible. Best, -Slashy

arkie87, Again, it's not escaping faster that does it, but rather burning while traveling at a higher velocity. So accepting your assertion that you understand the Oberth effect, you should have no problem understanding that 1) the charts are actually correct, 2) the cost of a 2 burn profile is more than just the DV to escape Kerbin's SOI, and 3) The penalty in terms of DV actually increases with distance to the target planet. This should set to rest all of the questions that you had. Best, -Slashy

^This. The Oberth effect has nothing to do with gravity wells. Looking at it from that perspective isn't stating the same thing in a different way. It's just that the relationship between velocity and kinetic energy isn't linear, so adding velocity when travelling rapidly adds more kinetic energy than adding velocity when travelling slowly. That's really it, period, full stop. The original question was why (or indeed if) it was possible to achieve a Jool intercept more cheaply from the bottom of Kerbin's gravity well than the top, and this is why. You have more kinetic energy at the time of firing down there, so you create more kinetic energy for the same fuel expenditure. Best, -Slashy

Arkie, The Oberth effect isn't about gravity wells absorbing energy. It's about the nonlinear relationship between velocity and kinetic energy. Kinetic energy is proportional to the square of velocity. adding velocity is (V+D)^2, or v^2+2VD+D^2. It's that "+2VD" part that increases your efficiency the faster you're going. Best, -Slashy

Starman, The "lost" energy you're talking about is going into potential energy relative to Kerbin. That energy does nothing to aid a trip to Jool, and is thus wasted. The important part is that adding velocity at a higher velocity creates more kinetic energy than adding velocity at a lower velocity. That's why it's cheaper to make a trip to Jool from LKO than it is to make it from Kerbin's orbit around Kerbol even though it's in the bottom of a gravity well. Kamuchi, Your practice of refueling at high orbit and then slingshotting is completely sound. Harder to match your launch windows, but your potential energy is returned to kinetic just before the burn and your burn is even more efficient because you're doing it at higher speed. Best, -Slashy

Sorry, wrong answer. You are too proud to accept instruction, so I can't help you (even though I want to and I know what's going on here). Good luck looking elsewhere and I honestly hope you find the answers you're looking for. -Slashy

*Grumble...* If you "already knew" this stuff, you wouldn't have this confusion. Forget what you think you "know" about the Oberth effect and orbital mechanics and pay attention, and *maybe* you'll actually understand it if somebody's willing to take the time to explain it. Honestly... I don't know why I'm willing to give you another shot at this point, but I am. Act like you don't know anything about this, and I'll help you. What don't you get? How getting out of the bottom of a gravity well can be cheaper than operating in free space?

You're not listening, you *don't* understand the concept, and I'm done explaining. Your defensiveness ain't my problem, so maybe Concentric will have better luck. Good luck, -Slashy

*Sigh* Shouldn't have bothered. I have tried to explain to you in detail why your underlying assumptions are incorrect and how the math really works. But you're convinced that everybody who came up with this stuff is wrong because it "doesn't make sense" to you. I'm not trying to "out-nerd you" but rather going out of my way to help you out by explaining it to you, but you don't want to listen, so have it your way. Not worth my time and not appreciated by you. Screwit, -Slashy

I've already shown you the math. 1/2M*(V2)^2- 1/2M*(V1)^2. Look... all you have to do is answer your thought experiment in practice. You have a suitable simulator right here. Put a ship in LKO and another in Kerbin's orbit away from Kerbol and try it yourself. It will tell you the same thing I'm telling you and I've already told you why it is so. If you stop thinking of transfers in terms of "DV" and start thinking of them in terms of "DEk" it will all make sense. Velocity <> kinetic energy. Best, -Slashy

Discussing this in another thread at this very moment! The reason the Oberth effect works is because orbital mechanics aren't velocity, but rather kinetic energy. We all think of it in terms of "ÃŽâ€V" in m/sec, but in reality they're "ÃŽâ€Ek" in terms of joules. Adding velocity imparts a change in energy proportional to the square of the velocity, not linear. How fast you're going when you impart the velocity has a profound effect on how much energy you add to your motion. If KSP had an "energy" gauge, we would all understand this implicitly and wouldn't even need to discuss it. You start on the pad with kinetic energy from rotation and orbit. You have potential energy in chemical form in fuel. Launching converts the chemical potential energy into kinetic energy, with losses due to gravity and drag. Your burns then convert chemical energy into kinetic energy and it all suddenly makes sense; converting energy is more efficient when you're going fast. Best, -Slashy

What you're saying is *completely* different from what I'm saying, which is why I get it and you don't. Transfers are kinetic energy. They're not velocity. Your explanation of the Oberth effect didn't mention kinetic energy at all, and that's how it works. Best, -Slashy

Shockingly, it's 1.1Km/sec cheaper to make the trip to Jool from the bottom of the gravity well at Kerbin than it is to do it from the opposite side of Kerbol. The reason is the extra kinetic energy you have at the time of burn from whipping around Kerbin, which makes your conversion of DV into Ek more efficient.

No it's not. You *think* you understand the Oberth effect, but you don't. I think that's the source of your confusion. Kinetic energy (which is what gets you places) is not measured in m/sec, it's measured in Joules. Kinetic energy is (M*V^2)/2. If you add 100 M/sec when you're going 2.3 Km/ sec, Your gain in kinetic energy is a whole heckuva lot more than if you add it when you're doing zero. Say you have a 1 T mass at rest. Kinetic energy zero. You burn for 100M/sec. Your final kinetic energy is 5 megajoules, so that's what you've added. Now say you do the same at an initial velocity of 2.3KM/sec. Starting Ek 2.645 GJ. Add a hundred M/sec and your new Ek is 2.880 GJ. Now you've added 235 MJ... with the same burn. It's because a rocket expends part of it's chemical energy spitting exhaust out the back. The faster you're going, the more of your chemical energy gets converted into kinetic energy instead of fire. Given that this is what's *actually* going on rather than what you've assumed, your follow-up question is based on a false assumption and therefore immaterial. Best, -Slashy

I wasn't interested in KSP for the longest time until I downloaded the demo version and went through the tutorials. Best, -Slashy

I think we're trying to The charts aren't incorrect, your assumptions about how engines convert chemical energy into kinetic energy are incorrect. The reason it's cheaper to hit a transfer from low orbit (and even cheaper from the pad) is "oberth rah rah rah". It's not a matter of "double- counting your burn", but rather the engines being more efficient when they're moving faster. They spend less of their chemical energy spitting exhaust out the back and more of it moving the vehicle forward. Your vehicle is moving faster at low orbit than high orbit, so your engines are more efficient down there. That's the Oberth effect. Best, -Slashy

Kraft, I like the video! Adding you to the leaderboard. Best, -Slashy

-Using valid aerodynamics with ion gliders, still out. Until it's proven possible with infinigliders, it's impossible for normal wings. We've barely been successful with pure wing SSTO ion gliders from Kerbin. Eve is far worse and nobody currently working the Kerbin effort would consider Eve plausible at this point. Using infiniglide with ion engines, I haven't had any luck. I suspect the required part count would be prohibitive. Using infiniglide with kraken drive, That one's confirmed for .25. KrakBadger 2.5 Using ladder lifters, that one's been confirmed many times over. Anything short of these is a mathematical impossibility. Best, -Slashy

It's because of the Oberth effect. Your engines make more useful energy when travelling at a high velocity than a low velocity, so they waste less DV on a transfer burn from low orbit than high orbit. Counter- intuitive, but true. Rest assured, the graphs are correct. Best, -Slashy

You, Sir, have given me an idea! Suppose you launch a simple satellite to a point just outside Kerbin's SoI but in the same orbit. Using that, you could do the standard maneuver node thing and it would tell you when the transfer window would be. You could actually launch the mission from the pad with that info and do a little correction burn. Best, -Slashy

arkie, Some helpful info on this technique: http://forum.kerbalspaceprogram.com/threads/80857-A-Delta-V-Map-for-simple-flight-paths Of particular interest is the comparison between both profiles in terms of DV. A trip to Duna would cost 1,060 m/sec overall with this method (a 17% increase) while a trip to Eeloo costs an additional 3,190 m/sec (40%). And this understates the disparity quite a bit since most people do their transfer from orbit rather than the pad. Simple flight paths are definitely more costly for more distant missions. IMO KSP really needs a built- in function to plan for transfer windows. Best, -Slashy

Wanderfound, Please remove my entry from this competition. I'm afraid I won't have the time and energy to participate in it properly. Sorry for the inconvenience! Best, -Slashy

Also FWIW the Structural D panel isn't worth the effort. I ran a head- to- head comparison between the strake and the structural D on an identical testbed and identical profile. I balanced the number of wing panels used to yield the same lift. The strake was far superior. It made it over the hump and got to nearly 33KM at 1.4Km/sec, while the structural D was only able to muster 28KM at 1Km/ sec flat. The strake also out-performed the structural D in time to climb at both 10KM and 20KM. So there's that... Best, -Slashy