Supernovy

Also, backspace returns the focus to the current craft. You can't focus on other crafts, however.

I thought most of the delay was caused by initialising physics. If that was true, then there would be a correlation between part count and loading time. I may have to investigate this myself. On topic, I'm sure the space centre is pretty well optimised already, but perhaps you could go to lower resolution textures like you can do with parts.

You get something like 80 percent if you return one, and due to the diminishing returns it takes about 4 experiments returned to max out the science. The way I did it was to take a measurement, process, clean out experiment, and repeat until I could transmit any more. I think 2 returns would be worth more than 1 return and 1 transmit, even with the processing boost. Only Kerbin, Mun and Minmus have biomes in 0.23. Although, water and land surface readings count as two different situations - relevant for Laythe.

Moved to the Spaceport forum. I believe there is already a Crafts category.

Welcome to the community, EricLimaB!

There's always persistence file editing. I don't know exactly how you'd do it, but look for the "ACTIONS" modules and look at other craft that have action groups, you should be able to figure out how to hack together a solution. Keep backups, though.

Often for interplanetary slingshots, I don't wait for things to line up, I go for it, and just keep waiting until I can find a low Δv manoeuvre to get an intercept with the next planet. This ends up taking quite a long time, but very little Δv, I once got to Duna with a probe only designed to escape Kerbin.

Moved to Gameplay Questions and Tutorials.

Hi, GiantZombies, and welcome to the community! There are some great tutorials for getting started in modding in this thread, including some stuff on Blender. I'm sure you can also ask questions in the Modelling and Texturing Discussion forum or Add-on Development forum and get helpful input from the rest of the community. I personally recommend learning how to use Model Nodes and config editing. I also recommend starting small, with fuel tanks or structural parts before moving on to more complicated things. Again, welcome to the community!

Moved to The Forum Forum. Attachments do not work. They have been disabled since the forum moved to vBulletin. If you want to put something in your post, you'll have to use an external host such as imgur for images (e.g. [noparse][/noparse]) or dropbox for other files. Attachments on old posts moved over from the old forum won't work either.

For a circular orbit, r is identical to a (the semimajor axis of a circle is its radius). So v2 = GM(2/r - 1/r) => v2 = GM((2 - 1)/r) => v2 = GM/r v2 = 6.67×10−11 * 1.2244127×1023 / (700 000 + 120 000) v2 = 8.1717302×1012 / 820 000 v2 = 9.9656 v = 3157 ms-1 EDIT: Gah! Ninja'd! The difference between my and Rhomphaia's answer can be attributed to different precisions and using different orbit altitudes.

Which answers my first question. Which answers my second question. And this is a bonus question answered. Thank you very much, alexmun and metaphor. I think I can set this thread to "answered" now. This is a valuable piece of knowledge added to the Kerbal knowledge base - when and how to save Δv on plane changes by raising apoapsis and doing the change there. In summary, for less than 38.9 degrees, don't raise apoapsis. For 60 degrees or more, raise it as much as possible. In between those, raise it as given by the second equation. If aerobraking, these angles become 19.2 degrees and 28.96 degrees respectively, and in between use the third equation. I'm very surprised to see that it ends up independent of μ - but then again, so is the phase angle equation. I suppose the dependence on orbital velocity gets eaten up by the plane change term.

So I suppose I was asking the wrong question? Is there no case for which a higher apoapsis results in less Δv than than a lower apoapsis (aside from your case) for the overall manoeuvre? Also thanks alexmun and Kosmo-not for the math. Geschosskopf: In the case where you're going from an inclined orbit to equatorial, you'd put the apoapsis at either the ascending or descending node. For going into a more inclined orbit, it doesn't really matter I believe.

I'm using "apoapsis plane change manoeuvre" to refer to the three burn manoeuvre (starting from a circular orbit) of increasing apoapsis, performing a plane change at apoapsis, then re-circularising at periapsis. If I'm correct, the total Δv for this manoeuvre is: ΔvT = 2[(μ(2/a1 - 1/a2))1/2 - (μ/a1)1/2] + (μ(2/(2a2-a1)-1/a2)1/2(sin(Δi)/sin(90-(Δi/2))) Whereas the Δv for a simple plane change is: Δv = (μ/a1)1/2(sin(Δi)/sin(90-(Δi/2))) μ is the gravitational parameter, a1 is the circular orbit SMA, a2 is the elliptical orbit SMA and Δi is the desired change in inclination (in degrees here). If either of these equations are wrong or not fully simplified, please correct me. I can post the derivation if needed. I am interested in finding the "break even point" - the a2 for a constant Δi, a1 and μ at which the Δv for the apoapsis plane change manoeuvre equals the Δv for the simple plane change. I hope to use this to find a simple (enough) equation that determines a good a2 for a given Δi, a1 and μ. So far, I've figured out that the difference between the simple plane change Δv and the apoapsis plane change manoeuvre Δv must be equal to twice the prograde burn Δv (the total Δv of the apoapsis raising burn and the recircularisation). (μ/a1)1/2(sin(Δi)/sin(90-(Δi/2))) = 2[(μ(2/a1 - 1/a2))1/2 - (μ/a1)1/2] + (μ(2/(2a2-a1)-1/a2)1/2(sin(Δi)/sin(90-(Δi/2))) => (μ/a1)1/2(sin(Δi)/sin(90-(Δi/2))) - (μ(2/(2a2-a1)-1/a2)1/2(sin(Δi)/sin(90-(Δi/2))) = 2[(μ(2/a1 - 1/a2))1/2 - (μ/a1)1/2] Any help in simplifying or solving this expression would be greatly appreciated.

From what I've seen in MCE, the long-term payouts that you can get are capped, e.g. you get 500 K per day for a set amount of days, which adds up to a total comparable to a single other satellite mission. If there was a similar thing in stock, time warping would only get you the reward faster in real-world time, and perhaps your reputation would decay over that time too.

Deleting your settings.cfg file and letting the game generate a new one next time you started it would work, but it would also get rid of your other settings (e.g. Graphics). You could possibly note down, or copy out all of the non-keybinding settings, delete, restart, and paste them back in.

Welcome to the community, Pinoy Kerbonaut! As an astronomy enthusiast, what do you think about the recent supernova in the Cigar galaxy? It's the closest one we've had for years, and hopefully since it's a type Ia, we can get more accurate distance measurements to the galaxy.

Welcome to the community, Comma_dose. If you were wondering why it took your thread a while to show up, it's because the first five posts of a user must be approved by a moderator as a spam prevention measure. I look forward to seeing your work in Live from Mission Control!

Because FΔt = mΔv. Force is essentially change in momentum over time, the longer you apply a force the more velocity you get out of it. This is because the acceleration caused by the force occurs over a longer period, and Δv=aΔt.

I think I see what you're saying - a Jool escape has to require less Δv than a Jool escape and transkerbin orbit, due to the effort of lowering solar periapsis (and these will be the same Δv as circularising from a parabolic Jool orbit and circularising at Jool from a transjool trajectory respectively). I believe the maps simply don't show escape and transfer as separate for non-Kerbin bodies, because the usefulness of that would be limited. Otherwise, you are correct. FAKEEDIT: Actually I think I've got it now This should actually be It takes 2630 m/s to leave Jool's SOI and intercept Kerbin. It takes 965 m/s to go from an encounter to an escape trajectory of Kerbin. It takes 950 m/s to circularise to LKO.

If you think about it, staying on prograde while it changes will get you the most velocity increase for your burn, but staying pointing the same way will change the direction of the velocity slightly as well as increasing speed. I'd say compare the manoeuvre node orbit to your actual orbit - if they diverge significantly, point at the node marker, but if they're almost the same, point prograde. Also, Welcome to the community!

Also, on some Δv maps it lists escape and intercept separately, so a Jool encounter from LKO on that map would be 950 + 965 = 1915 ms-1.

I'd like the wormholes to be in orbits of increasing difficulty, so that you'd have to preform (for example) solar orbit rendezvous in order to do interstellar travel. My reasoning is that rendezvous is one of the hardest things you can do in game, so makes sense to be the barrier to unlocking an entirely new system. This way, you'd get plopped in some orbit on the other side. You'd have to travel from the outside in, as opposed to starting from Kerbin and going out. I think each wormhole would go to a different system, and that speed would be conserved as you go through them. Ship speed that is, I don't want you to come out at a wonky angle because the two ends of the wormhole were at different inclination orbits. Also, I like wormholes.

As far as I know, the 1 atmosphere value is the lowest the specific impulse can go. It's not extrapolated past that. In the above example, the engine would still be at 320s on Eve.