maltesh

Easiest way (Assuming your internet speeds and data caps are fast and loose enough to do so) is to log into the store and download a clean copy of the game. Perhaps save the .zip file on your hard drive somewhere to unleash it again should you get the desire. Another option would be to sort your parts folder by Date Modified. If you didn't carry forth any of your mods from the demo version, all the stock part folders will have the earliest date.

With regards to the adjective for Minmus, I Invariably wind up using "Minmal."

Not even quite that. 100km LKO's about 2250. Burn up to 5000 m/s in LKO, pointed along the direction of Kerbin's travel, and you'll escape Kerbol. Made some mistakes on ascent, so only barely had enough fuel to hit the escape trajectory. Engines on full when burning, all stock. A better pilot than I could probably have flown it without ASAS. I'm sure smaller escape craft than this one are possible. A better ranking criterion probably would be maximum hyperbolic velocity excess, or the related quantity, specific orbital energy.

In the free version, Kerbol had no collision surface, and was treated as a gravitational point in space. You could pick up speeds in excess of several thousand kilometers per second by passing near its center, and floating point errors compounded by the Oberth effect made it possible to carry huge amounts of that velocity back out. In the paid version,there's an invisible solid surface at an altitude of about 4500km above Kerbol's defined surface. Nearly all objects that smack into that will do so at a couple hundred kilometers per second, and be destroyed. Scraping (or hitting) Kerbol at over 200km/s is probably only barely in the realm of possible now. Though if the Kraken is slain, a bi-elliptic sundive into a sunscraping powered gravitational slingshot would be a pretty good way to carry very large velocities into the interstellar void.

Sometimes, you've just got to dump things into the sun. A bi-elliptic transfer sundive can cost less than half the delta-V of a hohmann transfer sundive, at the cost of taking much, much, much longer. If your Munlander has a reasonable amount of fuel on landing, it's probably capable of a bi-elliptic sundive.

Getting back to Kerbin can be difficult, but not impossible... unless you don't have any fuel. If you don't have any fuel, and didn't plan your departure very carefully, the most likely result is /extremely/ long periods between passes through the Kerbin SOI, on the order of decades to centuries, and and a decent chance of ultimately being thrown by one of these rare passes onto an escape trajectory of the Kerbol system entirely.

Someone in my Google+ stream pointed me at the Kerbal Space Program SomethingAwful thread, and I downloaded v. 0.8 soon afterwards.

There are four bodies in the game. Have you orbited Kerbol and returned to Kerbin? Have you sundived?

It's not possible. Any object in orbit around Kerbin with a period equal to that of the Mun has to have a semi-major axis the same as that of the mun. In the KSP physics simulation (and in general, in reality, with a few significant exceptions) any object whose orbit is entirely inside the Mun's orbit will have a period shorter than that of the Mun, and will drift (or race) out of the line of Kerbin-Mun alignment.

As a fellow Munar Circumnavigator, I'd say you don't want to hold the I key down the whole time. The mun going to unexpectedly throw significant drops in front of you as you drive through the Munar night, and a flipped vehicle can be pretty inconvenient if your last Quicksave wasn't recent. I'd also recommend dropping stations and waypoints every 30 degrees of longitude (about every 100km) , giving you something to drive towards so you don't drift too far off course, and natural stopping points when you want to go and do something else.

Mechjeb does indeed do an inclination change to get to Minmus, if necessary. Mechjeb does it at a high altitude , when the spacecraft is moving much slower, and thus the fuel cost to change inclination is lower (generally halfway to two-thirds of the way there, in my expeirience, I presume it's actually doing the math for minimal fuel costs.) After all, you don't have to match Minmus' orbital inclination to reach Minmus, you only have to go through Minmus' orbital plane at the place and time when Minmus is nearby. If you want to return from the Mun just using Mechjeb, get into orbit, then use its Orbital Operations panel to raise your apoapsis to a point higher than the radius of the Mun's SOI (2200km or so.) That will get you out of the Mun's SOI, and allow you to then land on Kerbin. It's not the most efficient way to return from the Mun, however. A very efficient way involves burning directly from the Mun's surface (or very low Munar Orbit) into an orbital path that is hyperbolic, and puts you at or near the center of the trailing edge of the SOI moving fast enough to cancel most of the Mun's 545 m/s of velocity. If you're in orbit of Kerbin at 11400km altitude, and moving slower than 176 m/s in any direction, your orbit will intersect Kerbin, as shown in this video.

You can, however, put three equal-mass objects into a stable Figure-Eight Orbit, and even put that figure-eight in orbit around a star. Wouldn't work in the KSP physics simulation, of course.

Does "Moving craft" include rovers?

Yes, the Oberth Effect is in the game. If the Oberth Effect doesn't work, then the orbital equations don't work, because you're calculating kinetic energy wrong. The parking orbit is the problem. You generally spend a significant amount of fuel circularizing at the intermediate parking orbit, and then more fuel burning out of it. For what it's worth, here's what I got. Planet V at 9.79 Gm semimajor axis. (Venus analog) Planet M at 20.7 Gm semimajor axis. (Mars analog) Planet J at 70.7 Gm. (Jupiter analog) For the direct Hohmann transfer, assuming you aim it to dump you directly in the destination SOI.. Total Delta-V from 100km Kerbin orbit: Planet V: 1476 m/s. Arrival relative velocity: 856 m/s. Planet M: 1058 m/s. Arrival relative velocity: 821 m/s Planet J: 1947.9 M/s. Arrival relative velocity: 1758 m/s. For the method posted above Planet V. Intermediate Orbit SMA 15.6 Gm (yes, this works out to be higher than Kerbin, though planet V is lower) From 100km Kerbin orbit to Intermediate orbit: 788.6 m/s. Circularizing at Intermediate Orbit: 297.5 m/s. Transfer from Intermediate Orbit to Planet V: 1050.6 m/s. Total Delta-V: 2293.0 m/s. Relative velocity at arrival 1,180.4 m/s. Planet M. Intermediate Orbit SMA 13.0 GM (yes, this works out to be lower than Kerbin, though Planet M is higher) Transfer from 100km Kerbin Orbit to Intermediate Orbit: 931.4 m/s. Circularizing at Intermediate Orbit: 94.4 m/s. Transfer from Intermediate Orbit to Planet M: 1012.7 m/s. Total Delta-V: 2038.5 m/s. Relative velocity at arrival 902.3 m/s. Planet J. Intermediate orbit SMA 44.5 Gm. Transfer from 100km Kerbin Orbit to Intermediate Orbit: 1620.9 m/s. Circularizing at Intermediate Orbit: 1620.0 m/s. Transfer from Intermediate Orbit to Planet J 554.2 m/s. Total Delta-V:3785.2 m/s. Relative Velocity, 493.2 m/s. So for Planet V and Planet M, Direct Hohmann results in lower delta-V (817 m/s less for V, 980 m/s less for M), and lower arrival velocity (324 m/s less for V, 81 m/s less for M) over the method in the powerpoint. For Planet J, the trip still takes less delta-V on the direct hohmann. ( 1837 m/s less) The arrival velocity is higher than for the powerpoint method( 1264.8 m/s higher.), but not enough to offset the delta-V settings....if you were to come to a dead stop directly at the SOI edge. At the surface of the planet, thanks to the Oberth Effect, the difference between the two planet-relative velocities will be smaller. Much smaller, if J is a gas giant. And you get the best efficiency by braking at the last minute, rather than coming to a dead stop at the SOI edge, if your destination is the planet. In addition, if planet J is a gas giant you can aerobrake to shed velocity, as Kosmo-not mentions, instead of burning fuel.

Having digested the method fully now, I'll have to agree with Kosmo-Not on this one. Going to an intermediate parking orbit to make angle-measurement easier can pose significant extra fuel costs in Kerbol orbit, depending on how far away your parking orbit is. Still, the idea of using the navball as a protractor gives the option of using trigonometry to measure the current phase angle on a ship in interplanetary space using the angle between the nadir pole on the navball and the KSC marker, and the ship's and Kerbin's semi-major axes..

Holy crap. Using the navball to measure angles. That's brilliant.

If the orbits are slightly elliptical, the information in the post will get you close to the destination object. If the orbits are more than slightly elliptical, the mathematics is going to get pretty hairy, and likely beyond the scope of something you'd post in a thread like this one. You'd really want a tool built into the game, or an app that could read the persistence file, for something like that.

My own superwide 3-Man lander, sitting on a 20-degree slope. It has more than three times the amount of RCS I actually needed to land it, and a bit more fuel than I needed. I could have, and probably should have raised the legs higher.

Do it in one trip. Build a spacecraft that can transport all five satellites, and deploy them one at a time Take it up to that 100km orbit. It will have a period of 32 minutes, 39 seconds. 1. Deploy the first satellite. 2. Burn prograde to boost your apoapsis to 285.255 km (delta-V, about 127 m/s). Period will be about 39 minutes 18 seconds, or 20% longer than the 100km orbit's period. 3. Go around the new orbit once. 4. When you come back to 100km, circularize again. You are now 72 degrees behind the first satellite. Drop the second satellite. Repeat steps 2,3, and 4 for the third, fourth and fifth satellites, then head back to Kerbin. It should be noted though, that five satellites evenly spaced in a circular 100km altitude orbit over Kerbin are too close to the planet for any one of the satellites to have line of sight to any other of the satellites in the set. Kerbin's edge is 31 degrees below the horizontal at an altitude of 100km, and the angle to the next satellite in a 5-sat constellation is 36 degrees below horizontal.

As Kosmo-Not implies, but I thought someone should explicitly state, you got really lucky here, and hit a mathematical coincidence. You can't generally substitute cos(x) for cos-1(x). They're inverse functions of one another, and it's analagous to deciding to take the square root of a number instead of squaring it. For certain values of x, you'll get close enough to the right answer, but for most values of x, the answer you'll get will be very, very wrong.

I'm fairly certain he is Scott Manley.

Wiki says 2868.4 km altitude, though I remember calculations that placed it a little higher. Fairly easy way to measure it in the game (assuming you're playing the paid version) Get into orbit. Any orbit that doesn't dump you in the atmosphere will do. Check your apoapsis and periapsis times on the map screen. Burn until they're exactly three hours apart (Kerbin's sidereal rotation period is 6 hours). The altitude for stationary orbit will be the average of your apoapsis and periapsis altitudes. (All orbits around a body with the same period have the same semi-major axis.) An the speed you have when you cross that altitude will be the circular orbit speed for that altitude. (All orbits around a body with the same semi-major axis have the same specific orbital energy, and thus, the same speed for all orbital altitudes that they share.) .It just won't be in the proper direction to make it circular.

Getting into Kerbol orbit is fairly easy. You only need to be moving about 90 m/s faster in LKO to get into Kerbol orbit than you do to get to the Mun, and only about 10 m's faster than you need to be going to get to Minmus. The thing with using the Moons as gravitational slingshots is that it is extremely easy to spend more fuel aiming and pointing a slingshot than it would potentially save you, and the game currently doesn't have the flight-planning instrumentation to help you make these decisions without doing the math yourself. Also, if you're aiming for a gravity slingshot around the moons themselves, you're not taking full advantage of the far more easily predictable Oberth Effect by burning directly into your Kerbol orbit directly LKO. There's a reason why NASA doesn't slingshot interplanetary probes by the Moon, and NASA's dealing with far tighter fuel constraints and higher velocities than KSP is.

If you go the other way, and circularize yes, you'll hit the hypothetical Far Planet in up to half a period of the planet at 20million km after reaching that altitude.. The problem is going the other way. Let's assume you do the (relatively) sane thing, and Hohmann transfer out from Kerbin's orbit to the 20 millon Km planet's orbit, then reverse course and circularize. and attempt to come into it head-on. (Reversing course at Kerbin's orbit is going to significantly more fuel-hungry) When you arrive out threre, you'll be moving at 6.9km/s. To reverse course, you'll have to kill all that velocity, then pour on another 7.6 km/s to circularize. So there's an extra expenditure of 14.5 km/s delta-V. Going the normal way, You'd be moving at 6.9 km/s at apoapsis, and only have to boost up another 1.5 km/s to circularize. And if you waited for a launch window, the Far planet will be nearby, and you might not even have to circularize. So, by reversing course at the Far planet's orbit, you spend an extra 13 km/s of delta-V over if you'd gone the right way. That is a lot of extra fuel you had to bring along. Also, if you've decided to come into the planet head-on, you'll be moving at more than 15 km/s relative to the planet when you enter the planet's SOI. Kerbin's atmosphere could probably eat that. A hypothetical Far Planet's atmosphere may not be so accommodating.

Most players do aim a launch window when heading for the Mun or Minmus, using the Munrise/Minmusrise burn method. The reason why looking for launch windows is so important is because the "launch at any old time, push my apoapsis out to the planet's distance, circularize there and wait to catch up" method can easily take hours at 100,000x for planets as close as 20 million km from Kerbol, because you're attempting to catch up to an object dozens of millions of kilometers ahead at a relative velocity of maybe 20 meters/second. And then once you get there, you still have to find the planetary SOI. At the very least, eyeballing the proper position for the launch window determined by the equations will put you a heck of a lot closer to your eventual target.