OhioBob

Here’s another way of looking at the problem. Let’s go back to our 10,000 kg rocket traveling at an initial velocity 1000 m/s. Let’s first compute the initial momentum of the rocket, Pinitial = 10000*1000 = 10,000,000 kg·m/s Let’s say we now expel 1000 kg of reaction mass out the back of the rocket at 3000 m/s. Since the rocket is traveling +1000 m/s and the exhaust is traveling -3000 m/s relative to the rocket, the exhaust velocity is -2000 m/s as seen from a stationary reference frame. Therefore the momentum of the exhaust is, Pexhaust = 1000*(-2000) = -2,000,000 kg·m/s We know that linear momentum must be conserved, therefore the final momentum of the rocket plus the momentum of the exhaust must equal the initial momentum. Or, in other words, the final momentum of the rocket equals the initial momentum less that of the exhaust, Procket = 10,000,000 – (-2,000,000) = 12,000,000 kg·m/s We can therefore calculate the final velocity of the rocket, Vfinal = 12,000,000 / 9,000 = 1333.333 m/s We see that the Δv is 333.333 m/s. Let’s now compute the initial kinetic energy and the final kinetic energy, KEinitial = 0.5*10000*1000^2 = 5,000,000,000 J KEfinal = KEexhaust + KErocket KEexhaust = 0.5*1000*(-2000)^2 = 2,000,000,000 J KErocket = 0.5*9000*1333.333^2 = 8,000,000,000 J KEfinal = 10,000,000,000 J We see that the system gained 5,000,000,000 J of kinetic energy, which is the energy released by the burning of the propellant. (It is this energy that gives our reaction mass its 3000 m/s exhaust velocity.) Let’s now work the same problem but with an initial velocity of 2000 m/s. Pinitial = 10000*2000 = 20,000,000 kg·m/s Pexhaust = 1000*(-1000) = -1,000,000 kg·m/s Procket = 20,000,000 – (-1,000,000) = 21,000,000 kg·m/s Vfinal = 21,000,000 / 9,000 = 2333.333 m/s KEinitial = 0.5*10000*2000^2 = 20,000,000,000 J KEexhaust = 0.5*1000*(-1000)^2 = 500,000,000 J KErocket = 0.5*9000*2333.333^2 = 24,500,000,000 J KEfinal = 25,000,000,000 J Again we see that the Δv is 333.333 m/s, and again the energy gained by the system is 5,000,000,000 J. This is what we should expect because we’re burning the same amount of propellant, releasing the same amount of energy, and expelling the same amount of reaction mass at the same exhaust velocity. However, we see that the energy gained by the rocket, KErocket-KEintial, in the first case is 3,000,000,000 J, and in the second case it is 4,500,000,000 J. Also note that we don't have to consider the energy of the exhaust at all to compute the change in energy of the rocket. Computing the energy of the exhaust here is only a convenience to show that net energy change of the system is the same in both examples. All we need to know to compute the energy change of the rocket is its change in velocity, which we get from analyzing momentum. Here we calculated the momentum of the exhaust in order to determine the change in momentum of the rocket, but we normally don’t have to do that. Thrust gives us what we need to know because thrust is simply the rate of change of momentum. Note that the examples above include some mathematical simplifications. (For explanation, see hidden contents below.) This was done to demonstrate the basic concept without overcomplicating the math. In real life the math is a bit more complicated and the resulting numbers will be different, but the concept is the same. My previous examples are a better demonstration of the real life calculations.

I do make mistakes occasionally, so don't ever hesitate questioning me or calling me out on something. You just might be right.

This is why the change in kinetic energy of the rocket is greater in example #2 than in example #1. Thrust comes from the chemical energy released by burning, which is why relative exhaust gas velocity and thrust are constant. However, it is the transfer of the propellant's greater kinetic energy in example #2 that accounts for the rocket's greater increase in kinetic energy. This results just falls out of the calculations, as demonstrated in my examples. There is no need to add or fudge anything. (As Slashy said, "it will appear on its own like it's supposed to.")

Example #1 A rocket has an initial mass of 10,000 kg. Its engine produces a thrust of 100,000 N with a specific impulse of 350 s. When operating, the engine’s mass flow rate is, ṁ = 100000 / (350 * 9.80665) = 29.135 kg/s The rocket is traveling in gravity free space at an initial velocity of 1000 m/s. We burn the engine for 10 seconds, resulting in a final mass of, mf = 10000 – 29.135 * 10 = 9708.65 kg The change in velocity is, Δv = 350 * 9.80665 * LN( 10000 / 9708.65) = 101.5 m/s The final velocity is, vf = 1000 + 101.5 = 1101.5 m/s And the spacecraft’s change in kinetic energy is, ΔEk = 0.5 * 9708.65 * 1101.5² – 0.5 * 10000 * 1000² = 889,763,445 joules Example #2 A rocket has an initial mass of 10,000 kg. Its engine produces a thrust of 100,000 N with a specific impulse of 350 s. When operating, the engine’s mass flow rate is, ṁ = 100000 / (350 * 9.80665) = 29.135 kg/s The rocket is traveling in gravity free space at an initial velocity of 2000 m/s. We burn the engine for 10 seconds, resulting in a final mass of, mf = 10000 – 29.135 * 10 = 9708.65 kg The change in velocity is, Δv = 350 * 9.80665 * LN( 10000 / 9708.65) = 101.5 m/s The final velocity is, vf = 2000 + 101.5 = 2101.5 m/s And the spacecraft’s change in kinetic energy is, ΔEk = 0.5 * 9708.65 * 2101.5² – 0.5 * 10000 * 2000² = 1,438,166,420 joules @Geschosskopf, As you may have noticed, everything in these two examples is exactly the same except for the initial velocity of the rocket. The thrust is the same, the specific impulse is the same, the mass flow rate is the same, and the burn duration is the same. As you can see, the second example results in a much greater increase in kinetic energy than the first example. I didn’t require any “additional code” or perform any mathematical shenanigans to artificially add energy to the rocket. There is no extra thrust involved. I’m just performing straightforward calculations like we do in this forum all the time. If you think there is something missing from these calculations, what is it? What additional code would you add? If my method doesn’t take into account the Oberth effect, then why is the change in kinetic energy greater in the second example than in the first?

No, we are not getting extra thrust. Thrust is equal to the rate that mass is ejected out the back of the rocket times its velocity relative to the rocket. This relative velocity is constant regardless of the speed the rocket is traveling, therefore the thrust is constant (ignoring the possible effects of changing ambient air pressure).

No, we don't need extra code. We know the thrust of the engine and the mass of the vehicle, from which we compute the acceleration. We then integrate over time to get the change in velocity, from which we get the change in energy. We don't need to know anything about the exhaust stream and we don't need to add anything artificially. (edit) Note that all we need to know about the exhaust stream is provided to us by the thrust. Know the thrust and we can compute everything else. There is nothing more about the exhaust that we must consider.

The result of the Oberth effect is that a rocket gains more energy per unit of Δv when traveling faster than it does when traveling slower. That is, adding 1 m/s when traveling 2000 m/s adds more energy to the rocket than adding 1 m/s when traveling 1000 m/s. Change in velocity comes from acceleration * time, and acceleration comes from thrust / mass. As long as the physics model correctly computes thurst, mass, acceleration, time, and kinetic energy, the Oberth effect is there whether you want it to be or not. There is no need to add a separate algorithm to compute Oberth effect. In fact, we don't "compute" Oberth effect at all, it is simply a consequence of the equations that are fundamental to any proper physics model. Furthermore, Oberth effect doesn't care where the Δv comes from. 1 m/s is 1 m/s regardless of whether is comes from a Mainsail or an ion engine. It is true that propellant mass flow rate and specific impulse are components in the calculation of thrust. In KSP thrust is computed as follows, F = ṁ * Isp * go where ṁ is the propellant mass flow rate, Isp is the specific impulse, and go is the standard acceleration of gravity. In KSP, Isp is provided by a floatCurve in the engine's configuration file, where it is a function of ambient air pressure. So mass flow rate and specific impulse do have an effect on the Δv produced by a particular system, but to relate that to the Oberth effect is not correct.

Let's take a look at some sample numbers by considering a flight to Duna. In an orbit of 100 km, orbital velocity is 2246 m/s and escape velocity is 3177 m/s. Therefore, from an orbit of 100 km, we must increase our velocity 3177 - 2246 = 931 m/s just to escape Kerbin gravity. However, if we accelerate to exactly escape velocity we'll have no velocity left over after escaping. For a typical Hohmann transfer to Duna, we must be traveling about 900 m/s relative to Kerbin after escaping the Kerbin's gravity. This left over velocity is called hyperbolic excess velocity. Burning from an altitude of 100 km, we find that we must increase our velocity to 3302 m/s in order to have a hyperbolic excess velocity of 900 m/s (math provided on request). Therefore we must increase our velocity by 3302 - 3177 = 125 m/s above and beyond escape velocity, or we must provide a total Δv of 3302 - 2246 = 1056 m/s. We see that we obtain 900 m/s in hyperbolic excess velocity for only a 125 m/s expenditure in Δv. That's the Oberth effect at work. Now let's look at the numbers for an ejection from an orbit of 1000 km. Here orbital velocity is 1486 m/s, escape velocity is 2101 m/s, and, in order to produce a hyperbolic excess velocity of 900 m/s, we must attain a velocity of 2286 m/s. Therefore, just to escape we must provide a Δv of 2101 - 1486 = 615 m/s, and, to produce the extra 900 m/s, we must provide additional Δv of 2286 - 2101 = 185 m/s. The total ejection Δv is 2286 - 1486 = 800 m/s. So we see that takes less Δv to eject from a 1000 km orbit (800 m/s) than it does to eject from a 100 km orbit (1056 m/s). This is because we are higher up in the gravity well where it takes to less Δv to escape (615 m/s vs. 931 m/s). However, when we look at how much Δv is takes to produce the 900 m/s hyperbolic excess velocity, we see that it is less from the 100 km orbit than the 1000 km orbit (125 m/s vs. 185 m/s). This is because the Oberth effect is greater closer to the planet where the orbital velocity is higher. Therefore, the answer to your question is, for a hyperbolic excess velocity of 900 m/s (approximately that for a transfer to Duna), the difference in Oberth effect between a 100 km orbit and a 1000 km orbit is about 60 m/s. Of course you can see that that's not the total story. As we increase our orbital altitude, the Δv required to reach escape velocity decreases, and the Δv required to produce the hyperbolic excess velocity increases. If we were to plot total Δv versus altitude, we would see that the Δv initially decreases with increasing altitude, reaches a minimum, and then starts to increase with increasing altitude. The altitude at which the minimum occurs is called the gate orbit. The height of the gate orbit is different for every transfer because the hyperbolic excess velocity is different for every transfer. Of course, if you start out in a low orbit, you would never want to transfer to a higher orbit to perform the ejection. This is because it takes more Δv to raise the orbit than what you can potentially save on the ejection burn. In the example above, it takes 730 m/s to go from a 100 km orbit to a 1000 km orbit, while you save only 256 m/s on the ejection burn. If you are launching into a temporary parking orbit from which you plan to perform an ejection burn, it is almost always best in terms of Δv to make your parking orbit as low as possible. The only possible exception that I can think of is if you have a really long ejection burn, then there might be some benefit to launching into a higher and slower orbit.

There's no one size fits all answer to that question; the situation is more complicated than you might think. For every transfer there is an ideal orbit, called the gate orbit, from which the dV is minimum. Below is a thread from last year in which discussed this topic. You might find it interesting.

I agree, the payload should just be some inert mass. It should be up to the rocket designer to determine what propellant, engines, and stage configuration he wants to use. By the way, I was curious to see if I could beat bewing's cost per kilogram using a liquid fuelled first stage rather than cheap SRBs. I was successful, but not by a lot. 8444 launch cost + 340 LF cost = 8784 total cost / 6208 kg to Mun orbit = 1.415 funds/kg One of the things that ran my cost up was that I included four AV-R8 Winglets at a cost of 2560. With some practice I might be able to eliminate those, or replace them with something cheaper, to lower my cost further. (edit) Made some changes and got it down to: 6184 launch cost + 353 LF cost = 6537 total cost / 6129 kg to Mun orbit = 1.067 funds/kg

The real life rule of thumb is that we should give more Δv to the stages with the higher specific impulse. Liquid fueled engines in KSP generally don't differ all that much in Isp, so yeah, splitting it 50/50 is a pretty good way to go. In theory we should probably give a little bit more to the second stage, but @GoSlash27 make a good argument for why we might want to weigh it a little bit more toward the first stage. I generally don't look at the Δv distribution as much as the propellant distribution. I typically try to give my first stage twice as much propellant as my second stage. If I get that balanced about right, then the Δv should be well balanced also. If I'm using SRBs for the first stage, then I definitely like to give more Δv to the liquid stage, which we should because is has better Isp.

When I perform a two-burn strategy, I like to make the first burn just big enough to place the spacecraft in an orbit with a period of exactly 6 hours. This should take about 785 m/s and place the spacecraft is an orbit with an apoapsis about halfway to Mun. This orbit brings the spacecraft back to the exact same spot the next day to complete the burn. This makes it easy to plan things so that the final ejection burn occurs at the right time and at the right location. I simply perform the first burn at the exact spot and time of day that the final burn should occur, I just do it one day earlier. Fortunately I've never had a burn long enough that it couldn't be completed in two parts. Note that in 6 hours Kerbin revolves around the Sun about 0.85 degree. Therefore, it is necessary to perform the first burn at an ejection angle that is 0.85 degree less than the computed transfer ejection angle. This assures that when the spacecraft returns to the same spot 6 hours later, the ejection angle will be exactly correct.

Here are my rules of thumb for designing rockets:

I agree. Many of the early game goals and payloads don't require anything nearly as powerful as the Reliant or Swivel. I ended up creating my own engine to fill the need, the LV-T20 -- 120 kN max. thrust, 270/300 s ISP.

I've usually don't put much effort into naming things. Although I don't anymore, I once saved all my launch vehicles as subassemblies and simply designated them by how much payload they could lift to LKO. For example, I would simply name it LV-24 for "Launch vehicle-24 tons". Now I generally custom build my rockets for each individual payload (I follow a few simple rules of thumb for designing rockets that allows me the slap together a launch vehicle pretty quickly). Because of this, I don't even bother to name rockets, everything is named after the payload. In that case I generally give things pretty boring names that simply describe the destination and type of spacecraft. Examples: Mun Lander I, Duna Orbiter I, Eve Explorer I, etc. If I launch a second mission of the same type I simply increase the number, e.g. Mun Lander II. I often use the "Explorer" name, or something similar, when I have a multi-part craft, such as an orbiter and a lander. After the parts separate I usually rename them, i.e. Eve Explorer I might become Eve Orbiter I and Eve Lander I. For my manned missions I usually give them a bit more colorful names, though I haven't formalized any particular rules for doing so. I generally use whatever pops into my head, though it is almost always the name of a star or constellation. For space stations and bases I generally give them a Greek letter designation, e.g. Mun Base Alpha.

The two mods should be completely compatible without you having to do anything special. KScale64 uses SigmaDimensions to stretch the heights of the atmospheres by a factor of 1.33. It doesn't care if those atmospheres are stock or Realistic Atmospheres. If both KScale64 and RA are installed, you should get RA stretched by a factor of 1.33.

It's been done since the 1950s.

Yeah, what @Sharpy said. I just plugged the numbers into all three equations to show the internal consistency. You are correct, we would know the impulse and use it to compute Δv1 and Δv2.

I don't think this is right. I believe it should be, I = Δv2 m2 - Δv1 m1 Δv1 = – (I / 2) (1 + m2 / m1) / (m1 + m2) Δv2 = (I / 2) (1 + m1 / m2) / (m1 + m2) Let's work through an example from @SchweinAero's experiment. We have, m1 = 2000 kg Δv1 = -3 m/s m2 = 1000 kg Δv2 = 6 m/s Plugging these numbers into the equations we get, I = (6)(1000) - (-3)(2000) = 12000 Ns Δv1 = – (12000 / 2) (1 + 1000 / 2000) / (2000 + 1000) = -3 m/s Δv2 = (12000 / 2) (1 + 2000 / 1000) / (2000 + 1000) = 6 m/s (Earlier we talked about the impulse equaling 6000 Ns, but that was 6000 applied to each part for a total of 12000.)

Thanks, guys. I was just looking at the specs in the VAB, didn't realize the Wiki listed the impulse.

Good experiment. So it looks like each unit of "ejection force" is actually 10 Ns of impulse imparted to each half. That is, Δv = Impulse / mass = (600 * 10) / 1000 = 6 m/s Have you tried the same experiment with unequal masses, say 1000 kg on one side and 2000 kg on the other? I would think in that case the change in velocity of the 1000 kg mass would be 6 m/s and the change in velocity of the 2000 kg mass would be 3 m/s.

We really can't calculate it. The decoupler specs gives the ejection force, but what we really need to know is the impulse, i.e. force and duration.

Performing a Hohmann transfer to Duna from a solar orbit near Kerbin requires a Δv of about 800 to 1000 m/s. To this we must add the Δv that it takes to escape Kerbin in the first place, which is about 940 m/s. This gives a total of about 1740-1940 m/s. Performing the ejection burn from low Kerbin orbit requires a Δv of about 1050-1100 m/s. Therefore, you are costing yourself about 690-840 m/s by executing the maneuver the way that you are doing it.

I don't think so. The only changes that I made were to the atmosphere temperature and pressure curves, which shouldn't have any effect on the appearance. I specifically avoided changing to anything having to do with visual effects. My guess is that is more likely has something to do with Scatterer. What happens when you install SSRSS without Scatterer?

Realistic Atmospheres, version 1.2.2 When Realistic Atmospheres changed the heights of the atmospheres, it messed up some of the time warp altitude limits. This was particularly annoying at Eve, where we couldn't time warp past 10x if flying just above the atmosphere. I've corrected this to make sure we can time warp at least 50x when outside of a planetary atmosphere. Stock Duna also has different time warp limits than the other terrestrial planets and moons. Since I couldn't see a good reason for doing this, I changed it to be consistent with the others. In Jool's case I increased some of the time warp altitude limits to account for the added height of it's atmosphere. With Realistic Atmospheres 1.2.2 installed, the time warp altitude limits are: Warp Minimum Altitude Eve Kerbin Duna Laythe Jool 50x Top of atmosphere 100x 120 000 m 120 000 m 120 000 m 120 000 m 400 000 m 1 000x 240 000 m 240 000 m 240 000 m 240 000 m 500 000 m 10 000x 480 000 m 480 000 m 480 000 m 480 000 m 700 000 m 10 0000x 600 000 m 600 000 m 600 000 m 600 000 m 1 200 000 m I also adjusted the flying altitude threshold (border between "flying low" and "flying high") to account for the change in the heights of the atmospheres. To make it easy to remember, I made the new thresholds 25% of the atmosphere height, rounded off to the nearest kilometer. This keeps Kerbin's threshold at the familiar 18 km. For Jool I used 50% of the atmosphere height. With Realistic Atmospheres 1.2.2 installed, the flying altitude thresholds are: Flying Altitude Threshold Eve 14 000 m Kerbin 18 000 m Duna 18 000 m Laythe 15 000 m Jool 200 000 m