Yasmy

I run a little script to watch for changes in quicksave.sfs and persistant.sfs, and automatically commit the changes to a repository. It basically runs 'git commit -m "autosaving" in a loop at regular intervals. This way I can scroll back in time to pretty much any quicksave or persistence update. Very handy. Would highly recommend.

Without mods: 1) quicksave 2) open the quicksave.sfs file located at <kspdir>/saves/<savename>/quicksave.sfs (replace <kspdir> with your ksp installation directory and savename with your save game name) 3) search for the name of your vessel, or just search repeatedly for VESSEL 4) look at the ORBIT section. For example, I have one that says: { SMA = 70444089952.4041 ECC = 0.8432536741597 INC = 1.30267898784267 LPE = 343.656117208333 LAN = 25.096559254239 MNA = 1.6918185180101 EPH = 35523493.2772298 REF = 0 OBJ = 0 } So that vessel is on an orbit with a 1.3 degree inclination.

Right again, Red. This time I'll stop before doing the general Fourier series solution for arbitrary thrust profile. You would think I hadn't seen simple harmonic motion a hundred times in the past 25 years...

So Red Iron Crown said you can change your inclination without changing your specific orbital energy by burning continuously normal to your current prograde. I thought Red Iron Crown was wrong, but he is pretty much right, and I was wrong. Sorry about that. Let x be your initial prograde direction, and y be your initial normal direction. Burning normal to your velocity (vx, vy) means you burn in the direction (-vy,vx) or (vy,-vx): dvx/dt = - a vy(t) / sqrt(vx(t)2 + vy(t)2) dvy/dt = a vx(t) / sqrt(vx(t)2 + vy(t)2) A perfectly good solution to this system of equations is: vx(t) = v cos(a t / v) vy(t) = v sin(a t / v) So your total velocity (and thus specific orbital energy) is constant at all times: vx(t)2 + vy(t)2 = v2. Your delta-v is the time integral of your constant magnitude acceleration: sqrt((dvx/dt)2 + (dvy/dt)2) = a delta-v = a t To change your orbital inclination to phi degrees, wait until vx = v cos(phi * pi / 180). This happens when (a t / v) = phi * pi / 180. And we already determined that delta-v = a t. Thus delta-v = phi * pi/180 * v, exactly like maccollo found. So thanks for teaching me something, Red Iron Crown. The following is wrong, but retained for thread continuity: Note that I assumed constant thrust. If you can make the whole maneuver with essentially zero change in your thrust to weight, a continuous normal burn will maintain your specific orbital energy. In practice your thrust to weight increases (if you are not staging) while burning. So, if for the ease of it, you want to do a continuous normal burn plane change, keep your acceleration (thrust to weight) somewhat constant by decreasing your thrust over time.

Bingo, maccollo. In the limit that the angle dphi goes to zero, dV = 2 V sin(dphi/2) -> V dphi. Then integrating from 0 to final inclination phi gives ÃŽâ€V = V phi. That's in radians. In degrees, ÃŽâ€V = V phi (pi/180). For phi = 90 degrees, ÃŽâ€V = pi/2 V ~= 1.5708 V.

Deleted for wrongness. Details in my next reply.

Of course it does. If your velocity is v, and you burn by dv at a right angle, your new velocity is sqrt(v^2 + dv^2). Your new orbital energy is approximately 1/2 (m-dm) dv^2 greater. It's much smaller than for a prograde burn, where the change in energy is approxiamtely (m-dm) v dv, but it's not zero.

I edited my post to indicate that I was explicitly referring to the fact that he said that you could make a burn perpendicular to your velocity without making a kinetic energy change. I removed one sentence from the quote. Please see my quote in the new context. I did not speak one way or another about when you do or do not get the benefit of the Oberth effect. I'm not sure why you responded to me. But you are wrong that you don't get any Oberth effect when making inclination changes. You can only make very small inclination changes by burning perpendicular to your initial prograde. If you want to make a serious inclination change, without extending your orbit, you have to make your final speed equal to your initial speed. That means you have to burn somewhat along your initial retrograde direction. See Scott Manley's Orbital Mechanics tutorial on inclination changes. He goes through the trig to pick the correct burn direction for an inclination change which doesn't extend your orbit. It ain't perpendicular to your initial prograde. Orbitial Mechanics 2

I'm not sure this is universally true. At this point in the thread, such statements should be qualified or backed up with equations. So let's look at pure prograde/retrograde instantaneous burns, using dm fuel to change the rocket velocity by dv: Ki = 1/2 m v2 Kf = 1/2 (m-dm) (v+dv)2 Conservation of momentum for a prograde burn: m v = (m-dm) (v+dv) + dm (v - ve), where ve is the exhaust velocity g Isp. Simplifying gives dm = m dv / ve. (Note that integration of this equation gives the rocket equation: Delta-v = ve ln(mi/mf.) For a retrograde burn, just let dv be negative, but you must explicitly replace ve with -ve. So let's use dm = +/- m dv / ve, where the top sign is for prograde, the bottom for retrograde. So the change in kinetic energy is dK = Kf - Ki: dK = 1/2 (m-dm) (v+dv)2 - 1/2 m v2 dK = 1/2 m (1 -/+ dv/ve) (v + dv)2 - 1/2 m v2 dK = m v dv -/+ 1/2 m v2 dv/ve dK/dv = m v (1 -/+ v/ve) Note that this is the statement of the Oberth effect: The change in kinetic energy due to a burn is proportional to your velocity (or momentum): dK/dv = m * v * stuff. The bigger your velocity v, the bigger dK/dv. That's it. dK/dvpro = m v (1 - v/ve) dK/dvretro = m v (1 + v/ve) (dv is negative, so the change in kinetic energy dK = m v (1 + v/ve) dv is negative.) Conclusions: 1) You get a larger change in kinetic energy per delta-v spent in the retrograde direction than prograde direction. This is basically because for a retrograde burn the change in v2 is in the same direction as the change in m (both get smaller): (m-dm) * (v-dv)^2 vs (m-dm) * (v+dv)^2. 2) Here's a fun one: For prograde burns, if your rocket velocity is greater than your exhaust velocity, you actually lose kinetic energy. You gain speed, but (1 - v/ve) is less than zero, so you lose kinetic energy. Cute. 3) You do of course gain specific orbital energy, which is really what matters. When not burning, your mass is irrelevant, so it makes more sense to look at your energy divided by mass. dEpro = 1/2 (v+dv)2 - 1/2 v2 = 2 v |dv| + dv2 dEretro = 1/2 (v-|dv|)2 - 1/2 v2 = -2 v |dv| + dv2 Note that I've used -|dv| for retrograde because earlier I chose to let the sign of dv indicate prograde or retrograde. Here taking the sign out of dv makes it clear that the two changes in specific energy are different. Now clearly prograde burns add more specific orbital energy than retrograde burns subtract specific orbital energy. So, I guess I don't disagree with Kasuha so much as caution people to be specific. (No pun intended.)

This is incorrect. Any radial or normal burn (or mixed radial and normal) burn will change the magnitude of your velocity, thus your kinetic energy. It's plain old Pythagorus. Initially, K = 1/2 m v^2. Finally, K = 1/2 (m-dm) (v^2 + dv^2). So for a radial or normal or mixed radial/normal burn, dK = 1/2 (m-dm) (v^2 + dv^2) - 1/2 m v^2 = 1/2 (m - dm) dv^2 - 1/2 dm v^2 If you want to do a pure plane change maneuver, resulting in the same shape orbit but at a different inclination, you need to burn away some of your initial prograde velocity. Even then, your kinetic energy (and total energy) will be lower in the end, because you reduced your mass by the amount of fuel spent.

Right. On a delta-v map, from low orbit around one body to another body, you see an intercept/transfer burn and a circularization burn. Going in reverse, undoing the circularization burn is the new intercept/transfer burn, and reversing the intercept/transfer burn makes it a circularization burn.

Let's do an approximate calculation. It's only a loose approximation. Suppose you want to match Jool's p2 = 1.3 degrees inclination. You need about 965 m/s in-plane exit velocity. (This is the number you read off of a delta-v chart for Kerbin to Jool transfer). This in-plane delta-v is the horizontal component of your Kerbin exit velocity: v1 cos(p1) = 965 m/s. Now use tan(p2) = v1 sin(p1) / (vK + v1 cos(p1)) to find the out of plane delta-v: v1 sin(p1) = tan(p2) * (vK + v1 cos(p1)) = 233 m/s. Thus your Kerbin inclination p1 is arctan( v1 sin(p1) / v1 cos(p1) ) = arctan(233/965) = 13.6 degrees. And your exit velocity v1 is sqrt(965^2 + 233^2) = 993 m/s. This is where you get your savings. You combine your plane change with your transfer burn and it only cost an extra 993 - 965 = 28 m/s. To get your approximate burn delta-v at Kerbin, add about 20 m/s to this exit velocity (993 m/s). A handy delta-v map will show that it takes about 20 m/s to go from a Minmus transfer orbit to a Kerbin escape orbit. Add to this a burn at Minmus to change your inclination by 13.6 +/- 6 degrees. That will cost less than 25 m/s. External to Kerbin, a 1.3 degree plane change would cost something below that 233 m/s, depending on your current velocity. But it would still be more than 28 m/s + the Minmus plane change delta v. (It's sqrt(2*v2*(1-cos(1.3))), which drops as you get further from Kerbin.) Of course this was just an approximation. You may need a second inclination change burn if you are not on the line of nodes. And someone is going to come along and tell you to just drop some maneuver nodes and play around until you got it. And I wouldn't argue with them

Since you are going about 50 m/s at Minmus at Kerbin apoapsis, the maximum plane change delta-v is about 100 m/s. That's pretty cheap. So what can you get out of it in terms of a plane change relative to Kerbol? This requires just a little bit of trig. Suppose that just before exiting Kerbin's SOI, you are on an inclined orbit, inclined by angle p1, moving at v1 m/s. Just outside of Kerbin's SOI, you pick up Kerbin's orbital velocity, vK = 9284 m/s. Then your velocity relative to Kerbol is v2 = sqrt((vK + v1 cos(p1))^2 + (v1 sin(p1))^2) = sqrt(vK2 + v12 + 2 vK v1 cos(p1)). Your inclination relative to Kerbol given by is p2 = arctan( v1 sin(p1) / (vK + v1 cos(p1))). Whatever your inclination was inside of Kerbin's SOI, it will be greatly suppressed outside Kerbin's SOI. If you exit Kerbin at 1000 m/s at 90 degrees inclination, your final inclination is arctan(1000/9284) = 6.1 degrees. Typically though you would want to combine your plane change burn with your transfer burn, so you wouldn't want a 90 degree exit inclination. You may still need a second plane change burn on the line of nodes between your orbit and your target's orbit. There are definitely delta-v savings to be found by doing part of your plane change near Kerbin. How much depends on exactly where you are going, and when. Practically, you want to invert the equations above: You know the v2 ( = transfer v - kerbin escape v) and p2 you need (based on the many delta-v charts, or whatever), and want to know p1 and v1. (Note: if you leave Kerbin retrograde, replace v1 with -v1 everywhere above. This makes plane changes to the inner planets slightly cheaper than to the outer planets, per delta-V spent at Kerbin.)

You got it. Kerbin's gravitational parameter is mu = 3.5316e12 m^3 / s^2. The orbital velocity around Kerbin at 100 km is v = sqrt(mu / 700000) = 2246 m/s. The periapsis velocity on a return from the Mun is v = sqrt(mu * (2/7e5 - 2/(1.2e7 + 7e5)) = 3088 m/s. So there is 842 m/s of the escape from Kerbin circular orbit. Another ~190 puts you at ~1030 m/s. Seems about right for a burn from Kerbin orbit to Eve intercept. Maybe off by 10 m/s.

Yes. You are right that you need to consider Minmus's orbital plane, but you really don't have to worry about the delta-V. When you exit Minmus SOI on a trajectory to low Kerbin orbit, you will be going about 47 m/s relative to Kerbin. A 6 degree plane change should cost almost nothing: sqrt(2*v^2 (1 - cos(6*pi/180))) = 5 m/s. Like DMagic said above, you can use the cheap plane change cost at Minmus to reduce the future cost of a plane correction burn in interplanetary space.

Gilly -> Eve -> Kerbin takes about 10 m/s more delta-V than the approximately 500 m/s for Gilly -> Kerbin, if you ignore the plane change cost. Thus, if Kerbin will be near the line of nodes (ascending or descending) at Kerbin intercept, then it doesn't matter which way you go, because you won't need a big plane change burn later on. If you need to make a significant plane change to return to Kerbin, you may be able to save a lot of fuel by setting up a inclined pass by Eve. While still near Gilly, adjust your inclination so that you get a kick out of Eve's orbital plane towards Kerbin's orbital plane. I haven't done the math on this. Your mileage may vary.

Well, I've opened a can of worms before I was ready to write up the math. From low circular orbit around the Mun, for idealized, zero plane change transfers to planets on idealized circular orbits, the delta-V cost in meters per second is approximately: [table=width: 500] [tr] [td]Target[/td] [td]Kerbin Slingshot[/td] [td]Direct Interplanetary[/td] [/tr] [tr] [td]Moho[/td] [td]1135[/td] [td]1509[/td] [/tr] [tr] [td]Eve[/td] [td]462[/td] [td]348[/td] [/tr] [tr] [td]Duna[/td] [td]498[/td] [td]414[/td] [/tr] [tr] [td]Dres[/td] [td]987[/td] [td]1279[/td] [/tr] [tr] [td]Jool[/td] [td]1361[/td] [td]1841[/td] [/tr] [tr] [td]Eeloo[/td] [td]1520[/td] [td]2065[/td] [/tr] [/table] For Mun to Eve or Duna, it's best to go direct, but for Mun to the other planets you can save a few hundred meters per second delta-V. For low orbit around Minmus: [table=width: 500] [tr] [td]Target[/td] [td]Kerbin Slingshot[/td] [td]Direct Interplanetary[/td] [/tr] [tr] [td]Moho[/td] [td]943[/td] [td]1957[/td] [/tr] [tr] [td]Eve[/td] [td]270[/td] [td]440[/td] [/tr] [tr] [td]Duna[/td] [td]306[/td] [td]567[/td] [/tr] [tr] [td]Dres[/td] [td]795[/td] [td]1700[/td] [/tr] [tr] [td]Jool[/td] [td]1169[/td] [td]2318[/td] [/tr] [tr] [td]Eeloo[/td] [td]1328[/td] [td]2559[/td] [/tr] [/table] Thus for Minmus, you save anywhere from 170 m/s to 1200 m/s by sling-shotting around Kerbin. One day soon I'll get around to posting the equations and derivation, and the tables for other moons. All you need to derive these values is the vis-viva equation.

Note that if you are going from Minmus to Eve or Duna, it is cheapest to go directly interplanetary, without sling-shotting past Kerbin. If you are going to any other planet from Minmus, it is cheaper to drop into a low Kerbin slingshot. Sorry. The above statements are correct for the Mun, not for Minmus. See my post below. I've been meaning to write up my tutorial on when to slingshot around the parent planet vs. when to go direct vs. when to put the refueling station around the parent. Seems to come up a lot. (I like to know the delta-V and other practical factors, but my answer is put the refueling station where you will have the most fun with it.)

The difference in delta-v between low Minmus orbit and near Minmus SOI is so small that I would just put the station near Minmus, where the time between encounter/departure windows is much smaller, and time from the surface to the station and back is much smaller. Pick a minimum time warp you can deal with, and place the station just above the altitude cutoff for that time warp. 50× above 6 000 m 100× above 12 000 m 1 000× above 24 000 m 10 000× above 48 000 m 100 000× above 60 000 m

An option to consider is to make your drive section attach to the front and pull the rest of the vehicle. Put your engines out on arms so they don't roast the payload you drag behind. Payloads dragged behind are ideally self-stabilizing. Payloads pushed from behind are ideally unstable to wobbling.

Now the reason things look different from Kerbin then they really are when you show up at the Mun is simply because the Mun is moving faster than you are on its orbit. Suppose you show up on the near side of the Mun. While you are busy inserting yourself into a Munar orbit, the Mun is still moving around Kerbin faster than you are. Thus you end up losing ground and orbiting the Mun eastward (counter-clockwise from above).

First, the arrows you were talking about are sphere of influence change indicators. They appear at the end of a trajectory within one sphere. Note the color of the indicator is the color of the trajectory your are leaving. The easiest way to tell which way you are going to orbit after an encounter is change the conics mode to 0. The default mode is 3. Mods such as Enhanced Maneuver Node or PreciseNode (shown in the images below) allow you to change your conics draw mode without editing your config file. But if you are averse to mods, in your setting.cfg file in your KSP directory, just change CONIC_PATCH_DRAW_MODE = 3 to CONIC_PATCH_DRAW_MODE = 0. The first two images are the same encounter, but with conics mode changed from 3 to 0. In the second image you see the encounter relative to the moon, instead of relative to Kerbin. This way you can see the spacecraft will orbit the Mun counter-clockwise. Images three and four show an encounter that puts you into a clockwise orbit of the Mun (seen from above). Check out the purple trajectory around the Mun in image 4.

Nice analysis, Kasuha! Correct to about 1%, since he is almost, but not quite on a Kerbol escape trajectory. Beautiful back of the envelope calculation. For your rocket, with about 10.8 km/s delta-v left, you can circularize at about 1.6 Gm above the solar surface. As smidge204 says, use the vis-viva equation.

You are on the right track, Bane1998. You are correct in theory, but slightly incorrect in the KSP model. It is still better to take advantage of the Oberth effect in terms of delta-V savings, by doing your interplanetary burn near Kerbin such that your ejection angle is prograde or retrograde. But if you burn in any direction just enough to escape Kerbin's SOI, you will almost, but not exactly, be on Kerbin's orbit around Kerbol. Let's do the math. As usual, all we need is the vis-viva equation: v2 = mu (2/r - 1/a). Let rp = your periapsis around Kerbin (altitude + Kerbin radius) Let rs = the SOI radius Then the semi-major axis is a = (rp + rs)/2 Thus the velocity at the SOI of an orbit that just touches the SOI is v = sqrt( mu (2/rs - 2/(rs + rp))). Let's put in some numbers for Kerbin: [table=width: 500] [tr] [td]rp = 0 km[/td] [td]v = 0 m/s[/td] [/tr] [tr] [td]rp = 100 km altitude + 600 km Kerbin radius[/td] [td]v = 26.3 m/s[/td] [/tr] [tr] [td]rp = 200 km altitude + 600 km Kerbin radius[/td] [td]v = 28.1 m/s[/td] [/tr] [/table] So if you went straight up from the surface (but westward enough to cancel Kerbin's surface velocity) you would have a periapsis of zero, and you would be moving at zero relative to Kerbin at the SOI. However, if you start from orbit, and send your apoapsis exactly to the SOI, you will have some small velocity relative to Kerbin, perpendicular to the semi-major axis. Since your velocity relative to Kerbin is perpendicular to the semi-major axis, it is actually better to exit the SOI either at Kerbin noon or midnight, if your plan is to exit the Kerbin SOI, and then do your interplanetary burn. But we're talking about at most about 60 m/s difference for exiting on opposite sides of Kerbin. Eve injection will cost between approximately 60 and 120 m/s (ignoring plane change) based on where you exit the SOI. I'd call that significant. On the other hand, when I go to Jool, I go in style, and 60 m/s is entirely negligible. As one final complication, if you really care about the nitty-gritty details, you will almost always exit the SOI at a different radius around Kerbol than Kerbin's radius (by at most +/- Kerbin's SOI). This has an additional effect on the interplanetary burn delta-v. Instead of using Kerbin's orbital radius in the transfer calculation, use the spacecraft's Kerbol orbital radius.

Honestly, I don't think that the question you asked is really one you need to ask for mission planning. Like I said in the previous post, periapsis adjustment at the target, while you are far from the target SOI, is cheap. Generally all you need to do any orbital calculations is the vis-viva equation. I won't go over how you use it to calculate delta-v for going from one body to another here. There are tutorials around for that. Or check out en.wikipedia.org/wiki/Hohmann_transfer_orbit. But I will go ahead and answer your original question anyway. First you need to know the velocity (both speed and direction) of your spacecraft and the target planet at the SOI boundary. Next, when entering a planetary body's SOI, you have to subtract the body's orbital velocity from the spacecraft velocity, to get the spacecraft velocity relative to the planet. Using the planet's average orbital speed will not work unless the planet is in a perfectly circular, zero inclination orbit. You will have to calculate the planet's trajectory from its orbital elements. Now the question becomes: Given a velocity v and radius r of an orbit, what is the periapsis p? The standard way to approach such a calculation is to use conservation of energy and conservation of angular momentum. Instead I'll use the vis-viva equation (which has conservation of energy implicit in it) and conservation of angular momentum. Use the vis-viva equation at the SOI boundary to find your semi-major axis: 1/a = 2/r - v2/mu. Note that r is just the SOI radius. Use the vis-viva equation again at periapsis: vp2 = mu(2/p - 1/a) = mu(2/p - 2/r) + v2 Now use conservation of angular momentum to eliminate the periapsis velocity, vp: At periapsis, the magnitude of the angular momentum is just L = m vpp, since at periapsis, the velocity is perpendicular to the radius. At the SOI boundary, the magnitude of the angular momentum is L = m v r sin(theta), where theta is the angle between the line from the planet to the spacecraft and the velocity vector. Conservation of angular momentum says these two values are equal: m vpp = m v r sin(theta). Substitute vp = v r sin(theta) / p into the previous equation: vp2 = mu(2/p - 2/r) + v2 (v r sin(theta) / p)2 = mu(2/p - 2/r) + v2. Now multiply both sides by p2 and you get a quadratic equation for the periapsis p. Solve the quadratic equation. There are two roots: the periapsis and the apoapsis. Since you are on a hyperbolic orbit, the apoapsis will be meaningless. It should be obvious which root is the periasis. (Hint: It's the positive root.) Now the important thing to note in all of this, ie, the reason why I said this is pretty much useless to almost everyone, is that tiny changes in velocity while you are far form the target planet's SOI will have a huge effect on the velocity at the SOI. Specifically, it will have a huge effect on the sin(theta) term, the angle between the velocity and the line from the planet to the spacecraft.