Meithan

If you're referring to the "epoch" in "Mean anomaly at epoch", then the epoch is the reference time at which the mean anomaly is specified. It's a way to specify where the planet or body was along its orbit at some specific time, since this information can't be derived from the other orbital elements. For KSP planetary orbital elements, I think the reference epoch is UT=0, and I'd imagine alexmoon used the same convention in his webapp.

Hi. I do have one request, Felsmak, if time permits. Could you do Dres and Eeloo, too? Preferably on the same scale as the inner system planets. I was making a "bodies of the Kerbol system" wallpaper with your images (with due credit, of course), and I feel bad leaving those two out. Thanks!

Wow! Amazing work, I love it.

Aw, that's too bad! I was thinking those would make a great wallpaper for a KSP fan like me.

Slightly off-topic, but if you think Wooks' journey is crazily amazing (I do), you should check out the videos by the dude who's walking to the Minecraft Far Lands. That's a place in a Minecraft map that's about 12.5 million meters from the spawn point. For comparison, Kerbin's circumference is about 3.8 million meters. Keep in mind that one can move at about 4-5 m/s on foot in Minecraft, and there's no handy 4x timewarp. Also, there are monsters by night.

Wow, loved these! Great work! Could you share higher resolution versions of the ones in the OP? Particularly those of the stock bodies.

Wow, my estimate is a bit off . How come it took so long? That's an average speed under 1 m/s! I haven't watched the series, though. Slow ocean crossings?

Yup, Kerbin days are now 6 hours long (Kerbin's rotation period) and Kerbin years 426 days (Kerbin days, that is) long (Kerbin's orbital period around Kerbol). That means that 1 Kerbin year is 106.5 Earth days long, or about 0.29 Earth years. Personally, since I started playing KSP I felt this is how the game should keep track of time. I'm glad Squad switched to this.

Wow, very impressive. Congratulations! How much in-game time did the trip take? From a quick back-of-the-envelope calculation, at 20 m/s I get about 60 hours, so about 10 (Kerbin) days.

I just finished going over through the whole voyage. Fantastic stuff, sir, I loved it! And as someone who's actually designing his own Jool 5 mission, I thank you for all the tips and insights. They'll be certainly put into practice soon.

Can't wait .

Nice to know! I'm glad everything's in good health. It's what, about 100 days form Mars Insertion?

Exactly. Check out the diagrams on Olex's interplanetary calculator. Basically you always want to leave along Kerbin's orbit prograde direction (for targets farther from the Sun than Kerbin) or retrograde direction (for targets closer to the Sun). The ejection angle tells where you should do the transfer burn so that you're moving in the right direction when you exit the current SOI.

I think so, yes.

Hi. First of all, I want to thank you deeply for writing this fantastic tool. I rely heavily on it for all my interplanetary missions. I just have a small request. Could the lower/upper delta-v limits for the porkchop plot be user-configurable (maybe under advanced settings)? Sometimes a few values have very large delta-v's, which makes the color map less useful for it becomes almost completely blue. Here's an example (Jool -> Kerbin transfer with midcourse correction and aerocapture at destination, departing from a 150 km orbit on Year 10 Day 1 at the earliest, Kerbin time): As you can see, the upper limit of the plot is huge. I'd be more useful if the user could (optionally) set the limits, and have any value below the minimum simply be blue and any value above the maximum be red (or whatever colors are at the ends of the colormap). Thanks!

I fully agree, Streetwind.

This may be slightly off-topic, but it seems SpaceX just won the injunction, which will prevent ULA from selling rockets to the U.S. Air Force based on russian-built engines: SpaceX says rocket purchases violate Russian sanctions, gets injunction [Ars Technica] Musk is playing the "our engines are U.S.-built" card amidst the climate of tension between the U.S. and Russia over the Ukraine affair, and it would seem that Rogozin's statement arrived at just the right time for the court to make up its mind over the injunction. All I can say is: well played, Mr. Musk, well played.

Thank you, Streetwind, that was extremely informative (and what a good memory!). Too bad the link is down; I would've loved to watch the press conference. If the actual video feed is so bad, perhaps SpaceX could make an animated video using the telemetry data? With a big "using actual flight data" notice, that would be great advertising.

Well that was some crappy video. Is it water approaching what can be "glanced" at the top of the frame, and then engine wash / water being kicked up before the end?

That's awesome, tavert, thanks! I'll check it out when I get a chance. I've also never coded in Julia; it sure looks interesting.

Awesome! Thanks for the update, Tech Support. I'll definitely be watching for the insertion burn next September!

Thanks for answering. Ah yes, I was concocting a "manual" solution very similar to what MechJeb does. I see that MJ doesn't use the GetUTforTrueAnomaly() method from the KSP API, but does the computation using its own functions involving the eccentric anomaly as intermediate step. I guess the GetUTforTrueAnomaly() method is bugged somewhat? I'll try playing around with the wrapAfterSeconds parameter, and see if that fixes it. If not. I think I'll just do the following, which is similar to what MJ is doing: [Pseudocode] [given a radius r and a minimum UT min_UT, as well as an Orbit instance to extract orbital elements] E = acos(1/e*(1-r/a)) MA1 = E - e*sin(E) MA2 = 2*pi - MA1 (or MA2 = -MA1, should be equivalent) n = sqrt(G*M/a^3) t1 = (MA1-M0)/n + t0 t2 = (MA2-M0)/n + t0 while (t1 < min_UT) { t1 += P; } while (t2 < min_UT) { t2 += P; } return min(t1,t2) I haven't had chance to test it out, but I think it should work. That's for elliptic orbits, but the computation for hyperbolic ones is similar. Edit: yup, the above definitely works . Thanks for the tip in the right direction, though. MJ is an impressive piece of software.

Hi. I'm trying to compute the next UT at which a vessel with a given Orbit will be at some altitude (specifically, I want to find the UT for the next atmospheric contact, assuming the periapsis is within the atmosphere of the primary). I've been using several methods of the KSP Orbit class but I get unreliable results: sometimes the UT is spot on, sometimes it's nonsensical, not even close to the actual atmospheric contact. This is an example of what I've tried: atmoTopR = vessel.mainBody.Radius + vessel.mainBody.maxAtmosphereAltitude; contactUT = orbit.GetUTforTrueAnomaly(orbit.TrueAnomalyAtRadius(atmoTopR),0.0); I don't really understand what the second parameter of the GetUTforTrueAnomaly() method of the orbit class is for. It's a double called "wrapAfterSeconds". It sounds like it's used to clamp the resulting UT to a certain maximum value so as to get a result in the current orbital cycle and not far off into the future. However, the UTs I've been getting are always within the current orbital cycle, so I've simply left its value at zero. I've checked manually the results of the TrueAnomalyAtRadius() method, and even when the resulting UT is incorrect, the calculated true anomaly is correct. So it seems it's a problem in the GetUTforTrueAnomaly() method. I understand there is some ambiguity in these matters since orbits are cyclic and have certain symmetries. I guess I could compute the UT myself by computing the eccentric and mean anomalies, obtaining a time, and then adjusting that into the smallest value in the future. Thoughts, anyone?

Here's my first try at an SSTO, using the new R.A.P.I.E.R. engine. After a few tries I managed to reach orbit and dock with my space station. Then I returned to Kerbin and landed at the KSC runway under powered flight.

This one's just FANTASTIC.