alexmun

Sure! Yes, this angle only applies to the mid-course plane change trajectory and it is how many degrees before you get to your destination that you need to execute the plane change. It's not quite that simple. The ejection inclination is what you want the inclination of your hyperbolic ejection orbit to be after your maneuver. If you hover your mouse over the ejection delta-v value you'll get a tooltip with the prograde and normal components of the burn. The actual calculation requires you to know the velocity of your ejection orbit immediately after your burn, which I don't display. The formulas are: normal delta-v = ejection velocity * sin(ejection inclination) prograde delta-v = ejection velocity * cos(ejection inclination) - initial orbital velocity Yup! It's the angle between the velocity of your ship and the velocity of the target planet at intercept. Another way of looking at it is that it is the inclination of your orbital plane around the planet relative to the planet's orbital plane around the sun.

I don't have any plans to add a graphical plot of the transfer orbit at the moment. I know it would be useful, but I'm not convinced it's useful enough to justify the effort, at least for me. Thanks for the bug report, I'll work on getting that fixed.

The inclination you want is the ejection inclination. You would want to launch when the KSC is at the ejection angle relative to Kerbin's prograde vector (or actually, just slightly before). The delta-v of the burn from that orbit is a little more complicated than just the prograde component, unfortunately. If v(prograde) is the prograde component of the normal burn, v0 is your circular orbital velocity and i is the inclination, you can calculate the delta-v by: dv = (v(prograde) + v0) / cos(i) - v0 And you would want to time your burn for when you're passing over the equator in your inclined orbit, either the ascending node for positive inclinations or the descending node for negative inclinations. Overall the procedure is complicated enough that I don't think it's usually worth it.

FYI, Moho is by far the hardest planet to get to and MechJeb's calculations aren't perfect for getting there. If this is your first interplanetary mission, you might want to try going to Duna or Eve first.

Ok, the main thing is that a mid-course plane change transfer is optimal for transferring to Jool. For a mid-course plane change you need to add both the ejection delta-v and the inclination change delta-v to get the delta-v required to reach your target. In the 150km final orbit case your delta-v to reach jool is 1950m/s ejection delta-v + 138 m/s inclination change delta-v for a total of 2088m/s. In the fly-by case, your delta-v is 2003m/s ejection + 60m/s inclination change for a total of 2063m/s, which is 25m/s cheaper than the insertion case. The reason the inclination change is included is because without the inclination change you will never reach your target. The other thing to note is that for the fly-by case the point it picked is at the very top of the plot. That usually means there is a more optimal transfer with a longer time of flight. If you increase the time of flight to go up to 600 days in the advanced settings, you'll find a ballistic transfer with an ejection delta-v of 2047m/s which is even cheaper than the previous fly-by solution.

Running those settings for a day 1 plot gives 1996m/s for a fly-by vs 2002m/s for a transfer with insertion. Can you tell me what other parameters you were using that gave you inconsistent results?

The launch window planner doesn't know exactly where you are in your orbit around Kerbin because it doesn't have all of your orbital elements, just the altitude. So the figure you want to use to time your ejection burn is the ejection angle which is measured relative to the direction Kerbin is moving in its orbit around the sun. That's probably a little confusing, but you can get some good illustrations at Olex's Illustrated Interplanetary Guide or have a look at the tutorial Mr Shifty made.

It all depends on how expensive it is to do the plane change required to intercept your target. The two factors that go into determining how expensive the plane change is are: - The delta-v required for a plane change is directly related to your orbital velocity. So if you're in an eccentric orbit, the closer you are to apoapsis the cheaper it is. - The delta-v is also related to the sine of the half angle of the inclination change you need to make. Your inclination change is smallest if you change your plane 90 degrees before intercept and increases to 90 degrees if you wait until you're directly "under" your intercept point before making your plane change maneuver. So for a transfer to an outer planet the mid-course strategy works to find the point at which the decreasing delta-v because you are getting farther from the sun starts to get counteracted by the increasing delta-v as the angle you have to achieve increases towards 90 degrees. Now, when you're doing a ballistic transfer the raw delta-v for the inclination change is probably a lot higher than the inclination change delta-v for the mid-course strategy. For one thing, your orbital velocity around Kerbin is probably pretty high, and for another to shift your transfer orbit by a degree might take several degrees of inclination change to your ejection orbit (compare your ejection orbit inclination with your transfer orbit inclination to see what I mean). Both of those factors would seem to suggest that the mid-course plane change should always be better. However, there is one big factor in the ballistic strategy's favor and that is the law of cosines. What the law of cosines says is that if you combine two maneuvers into one you don't have to spend the sum of their delta-v's, instead it's related to the square root of the sum of their squares and the cosine of the angle between them (for a 90 degree angle, this is just the pythagorean theorem). So if we combine our ejection burn with the plane change burn the net delta-v becomes: sqrt(v0^2 + (v0 + dv[ejection])^2 - 2*v0*(v0 + dv[ejection])*cos(angle)). For a typical Duna transfer, the ejection delta-v without any plane change is about 1040 m/s. To do a ballistic transfer takes about a 2 degree ejection inclination in order to get about a 0.1 degree transfer inclination. Doing the plane change alone in a 100km Kerbin orbit (which has an orbital velocity of about 2250 m/s) would take about 80 m/s. Plugging into the formula we get sqrt(2250^2 + 3290^2 - 2 * 2250 * 3290 * cos(2)) = 1044 m/s. So by combining the maneuvers we get to do an 80 m/s plane change maneuver for only 4 m/s more than the ejection would otherwise take. TL;DR: It works out that for transfers outer planets with small inclination changes and for transfers to inner planets with moderate inclination changes the ballistic transfer tends to be cheaper because the benefits from combining the maneuvers outweigh the cost of doing the plane change at a less optimal time. However for transfers to outer planets with significant inclination changes or transfers to inner planets with extreme inclination changes the benefits of a cheaper plane change exceed the benefit from combining maneuvers.

Kerbal Engineer's flight engineer component will give you the angle to your target body.

It's generally most efficient not to try to match the inclination of your target during the transfer, but just to make your transfer orbit intersect the target orbit at the right time. Two ways of doing that are to include an inclination change component in your ejection burn (where you get to take advantage of the law of cosines which can reduce a 30m/s inclination change down to about 0.5m/s net) or to change your inclination mid-transfer like you're suggesting. The optimal point to do the inclination change mid-transfer varies depending on the eccentricity of your transfer orbit and the angle by which you need to change your inclination. I hacked my porkchop plotter a bit to explore the transfer to Minmus from a 100km Kerbin orbit and it looks like the two strategies are pretty much equivalent in terms of delta-v (about 1410m/s total depending on the window), but the window is much larger for the mid-course plane change strategy. The optimal time to do the plane change maneuver varied but was generally quite close to Minmus (around 5 degrees prior to intercept, for example). I think the simplest and most efficient transfer to Minmus would be to wait for a window where Minmus will intersects an ascending/descending node at the time of intercept. Then you don't have to do any plane change and you can get an intercept with a single maneuver node.

Good catch on the Minmus typo. It's fixed now. The Jool to Eeloo transfer is unusual, I think because they are in resonant orbits. If you zoom out (click Show Advanced Settings then set the time of flight between 1 and 4000 days) you can start to see the pattern, particularly on the ballistic transfer type. On the plane change or optimal transfer type there are some discontinuities caused by my assumption that you always want a counter-clockwise transfer orbit. I think there may be a small region near that boundary where a clockwise transfer orbit would be more efficient (but still very expensive). Ejection delta-v components are available if you hover over the delta-v value. I've added an underline to make that more obvious. And I've added some more sanity checking to the time of flight values.

Hmm, Chrome 28 is working fine for me. If you're still having problems, can you let me know what platform you're using (i.e. Windows/Linux/Mac/Android)? Also, if you could go to the Tools menu and click on Javascript Console and let me know any error messages that are displayed, that would be helpful. Oops, yeah that's a bug. It should be fixed now.

Moho intercepts are tricky and the standard Hohmann calculation fails pretty badly due to Moho's inclination, eccentricity, and orbital velocity. If you don't hit a good window, your intercept will come in at a large angle which means a lot of delta-v since you're so close to the sun. A good launch window can get you from LKO to 100km Moho orbit in ~5000m/s delta-v. This exact problem is one of the main reasons I made my launch window planner: http://alexmoon.github.io/ksp

Unfortunately that only works if the scale of the two axes is the same. I've thought of trying to round the scale of the y-axis to a harmonic of the x-axis to reduce the workload, but I haven't had a chance to try it yet.

So I finally got around to implementing these for you, tavert. I decided to just make the transfer time axis the default because I think it makes the periodic nature of the transfer windows a little bit more clear. It is more computationally intensive because I have to recalculate the position of the destination planet for every point instead of just once for every row, but the improvements I've made to the lambert solver recently help to offset the performance penalty.

Eeloo is sort of a weird case. Because it's orbit is so slow you can pretty much leave whenever you want as long as you pick a good arrival time back at Kerbin. I've just put up an update that includes advanced controls to change the scale of the x and y axes of the plot that might help you visualize it better.

I've put up a new version! In addition to an all new lambert solver on the back end, the main new feature is that it now calculates the prograde and normal components of your ejection maneuver. Just hover over the ejection delta-v value and a tooltip will pop up with the details. So if you use something like MechJeb's maneuver node editor you can now enter the values for the maneuver directly.

Ah, yes. I don't actually use the sphere of influence for anything so I had forgotten that. I treat the mass as the underlying value and the gravitational parameter as a (possibly less precise) derived value so I will use the mass where that makes sense such as in that equation. I think I got it from that same source. I don't remember if I validated it in-game or not, but I think I may have. When you write mods you have access to the innards of the game such as that value.

Very true. If you want a visualization of what a launch window actually looks like, you can use my web app: http://alexmoon.github.io/ksp

KSP uses the truncated gravitational constant, so that's where that comes from. I can't think of any place where I'm dividing two gravitational parameters off the top of my head, do you have an example? Generally for (simplified, 2-body) orbits all you care about is the gravitation parameter, not the actual masses.

I think it's going to depend on how much of a boost you're getting from the Mun and how close to the right direction your final ejection vector is. Ignoring the gravity assist for a second, if you're in a 100km orbit your orbital velocity is ~2246m/s. Escape velocity at that altitude is ~3176m/s giving you a delta-v of 930m/s for a minimal escape from Kerbin's SOI. Using my calculator (http://alexmoon.github.io/ksp) to calculate the solar orbit delta-v gives 854m/s (set initial orbit to 0 to get this value). So your total delta-v would be 1784m/s minus whatever boost you get from the Mun. Doing a straight transfer from LKO to Duna costs 1046m/s so you'd need to be saving 742m/s from your gravity assist for it to be cheaper. Transferring to Jool requires 2802m/s from solar orbit vs 1996m/s from LKO. Adding in the escape velocity delta-v gives 3732m/s meaning you'd need to save 3732 - 1996 = 1736m/s. The largest boost you can get from the Mun is roughly double its orbital velocity or 1083m/s so that's probably never going to be more efficient. Of course, if you could incorporate a Munar gravity assist into a single ejection burn from LKO and end up on the correct transfer trajectory that would be the most efficient, but calculating that would be a challenge!

Depending on your launch window it looks like it'll cost between 2500 and 3000 m/s without aerobraking. You can play with the various options using my calculator here: http://alexmoon.github.io/ksp

Yup, that's an orbit that passes very near the center of the sun. Basically a hyperbolic orbit that is nearly a straight line. And to answer your earlier question, yes the calculator handles hyperbolic orbits just fine.

More updates! Since tavert challenged my assumption that 90 degrees would be an optimal point to perform the plane change maneuver, I decided to implement another transfer type to search for the best time to do the plane change. Unfortunately, this is pretty slow because it has to try many different solutions for each point, however it does show that 90 degrees is usually not optimal although it is usually pretty close to the optimal burn time in terms of delta-v. The biggest difference in the first launch window for the outer planets is again Dres, which goes from 3611 m/s (including insertion) to 3537 m/s. Also, I have added support for Internet Explorer 10 so now all major browsers are supported.

90 degrees isn't really optimal in all cases. For example, you could set up a bi-elliptic transfer with an inclination change at apoapsis which would end up being cheaper (assuming the bi-elliptic transfer is cheaper than a two-impulse transfer in the first place). However, I think it should be optimal or close to optimal in nearly all reasonable cases since it minimizes the angle of the plane change required for intercept and it makes the math dramatically simpler because the plane change does not otherwise interfere with your transfer orbit (whereas at other angles a plane change will also change the radius of your intercept point). Stupid browser compatibility bugs! It should be fixed now.