OhioBob

Thanks for letting me know. I searched for any mention of this issue but couldn't find it. Sorry for being repetitive. Hopefully Squad will fix it.

I use something very similar, though I've put Kerbin, Mun and Minmus all on one sheet. I find it a great way to see what science is completed and what is still available. There is enough science in the Kerbin-Mun-Minmus system to complete the tech tree, so I don't bother having a check list for the planets. Once I complete the tech tree, I generally stop charting my science.

I like planning missions and designing vehicles to complete them. I usually like to incorporate contracts into my missions, though they often have addition self-defined objectives. I also enjoy flying the missions, but I like the planning/designing a little more. I find it very satisfying to be able to complete a mission and to have a vehicle perform according to the plan. I do everything with rockets; no air or spaceplanes.

Bumping this thread in the hope that Squad addresses this issue before the release of version 1.0.

There are actually two common usages of the term ÃŽâ€V. There is the one that everyone here has given, i.e. the potential change in velocity that your propulsion system can provide. But it is also the change in velocity needed to produce a change in orbit, or to complete a particular maneuver. For example, let's say you are in a 100 km circular orbit around Kerbin and you want to raise your apoapsis to 1000 km. That maneuver is going to require a change in velocity, or a ÃŽâ€V, of 403 m/s. To complete a mission you need to identify all the maneuvers you'll have to perform and add up the ÃŽâ€V for all those maneuvers to produce a ÃŽâ€V budget. You then have to design your spacecraft to make certain it can produce a ÃŽâ€V equal to or greater than your ÃŽâ€V budget (likely a little bit more to play it safe).

I usually start right at 5 km (using stock aero).

That works for Kerbin because Kerbin's orbit is circular. It's not so simple when both orbits have a significant eccentricity.

I'm glad I could help. I stopped searching when I reached the end of year 1000. There may be closer/farther distances in later years, but I figured it would take a seriously addicted individual to play the game that far into the future. (In fact, I know Jool and Eeloo have closer encounters in the second millennia but I had to stop somewhere.) I put all the orbital parameters into an Excel spreadsheet and computed the planet positions vs. time, from which I calculated the separation distances. I then just searched for the minimums and maximums.

2 + 4 + 4 + 16 + 10 + 16 + 0 + 2.5 + 2.5 = 57 / 20 = 2.85 It looks like I need to start doing some return missions.

Minimum (km) Maximum (km) Moho Eve 3,546,900 16,117,600 Moho Kerbin 7,289,400 19,913,500 Moho Duna 14,641,100 26,768,900 Moho Dres 30,052,400 51,629,500 Moho Jool 60,949,000 76,594,200 Moho Eeloo 60,715,300 119,514,300 Eve Kerbin 3,668,900 23,530,800 Eve Duna 9,792,200 31,665,400 Eve Dres 25,237,900 56,462,800 Eve Jool 55,583,200 81,962,500 Eve Eeloo 56,927,000 123,317,100 Kerbin Duna 6,069,300 35,383,000 Kerbin Dres 21,402,600 60,320,600 Kerbin Jool 51,735,000 85,811,900 Kerbin Eeloo 53,183,400 127,081,700 Duna Dres 13,732,500 68,079,200 Duna Jool 44,584,600 92,947,000 Duna Eeloo 45,095,900 135,224,000 Dres Jool 24,543,300 113,355,000 Dres Eeloo 28,733,500 151,339,300 Jool Eeloo 11,617,100 169,943,700 It sounds like what you want is the minimum/maximum distance between the orbits of the planets, which would be the theoretical minimum/maximum distance between the planets. I'm sure that can be determined mathematically, but that's not what I did. I projected the motions of all the planets over a 1000-year period and searched for the minimum and maximum true separations. These results are tabulated above.

I agree that this is really the only reason to use separators. And even then it really only makes sense if both parts are to perform propulsive maneuvers after separation. In that case you don't want either part carrying needless mass. On the other hand, if one of the parts will not be performing any further maneuvers, then there is no harm in it retaining the mass of a used decoupler.

I'm not adding much new here, but... First, you need to budget some ÃŽâ€V for your deorbit burn, say 50 m/s. As long as you're using plenty of parachutes and landing at a low elevation, you shouldn't need much ÃŽâ€V for landing. I generally budget about 50 m/s just in case I have to do a little breaking just before touchdown. If you plan to land at a high elevation, then you'll likely need more. How much more, however, I can't tell you because I don't have much experience with that. I budget about 1350 m/s for launching into a low orbit of around 50 km. I haven't spent much effort trying to optimize my ascent. I've found the following works good enough: burn vertically until my velocity is 200 m/s, pitch over to 45o and burn until my velocity is 400 m/s, pitch over horizontally and continue burning until my velocity is 800 m/s, cut the engine and coast up to apoapsis, and, at apoapsis, burn to circularize the orbit. As others have said, if you must rendezvous and dock with another vehicle in orbit, you'll have to add ÃŽâ€V for that. I think 100 m/s + RCS should be enough in most cases (assuming you're good at rendezvous). Of course that assumes you're already in the correct plane. Chances are you're going to be somewhat out of plane and will have to make a correction. A budget of 100 m/s will allow a 6o plane change (hopefully you're not off by more than that). Adding another 100 m/s margin isn't a bad idea, which get's us up to a total of 1750 m/s for everything. Others have suggested as high as 2000 m/s. That number should be more than enough, but, for your first attempt, better to have too much margin than to come up short. You can always get more miserly with your ÃŽâ€V budget after you've gained some experience.

That's not entirely correct, Isp is not constant. In real life the only constant is the fuel flow. As a rocket rises through the atmosphere and the ambient pressure drops, the thrust increases. Since Isp = thrust / (fuel flow * go), as the thrust increases, so does the Isp. The exhaust velocity doesn't change (maybe that's what you're thinking of), but the Isp does because of the changing pressure difference at the nozzle exit. - - - Updated - - - That's not true (aerospikes are not correctly depicted in KSP). The nozzles of engines are optimized for the ambient conditions at which must operate, i.e. low altitude (near sea level pressure) or high altitude (near vacuum). At low altitudes, nozzles have small expansion ratios, and at high altitude they have large expansion ratios. An aerospike effectively acts like a variable expansion ratio nozzle engine, obtaining a sea level performance similar to a small expansion engine, and a high altitude performance similar to a large expansion ratio engine. In a specific environment, however, aerospikes tend to perform a little worse than a fixed nozzle engine that is optimized for that specific condition. Here are some hypothetical Isp numbers to illustrate this point: Small expansion ratio: 260 s sea level, 300 s vacuum High expansion ratio: 200 s sea level, 330 s vacuum Aerospike: 250 s sea level, 320 s vacuum

For Moho the boundary is 80 km. The minimum for 50X is 30 km. 50 km is minimum for 100X.

First off, I'd replace the part that says "the point at which there is no atmospheric drag." Although the atmosphere gets thinner and thinner with increasing altitude, atmospheric drag is still a problem even up to hundreds of kilometers. In the U.S., the maximum lift capacity of a rocket is often stated as the amount of payload it can deliver to a 185 km orbit (100 nautical miles), and the minimum altitude for a short-term parking orbit seems to be 167 km (90 nautical miles). For any orbit that must last days or longer, I rarely see anything with a perigee below 200 km. You can choose your own words, but I'd revise it to say something like, "...the point at which the atmosphere is thin enough to allow a stable orbit, is at about 200 kilometers."

Interesting, I did not know that. A good piece of trivia to log away for future reference.

One thing you might consider changing is the part that says "the point at which there is no atmospheric drag, is at 100 kilometers". The 100-kilometer definition is that accepted by the Fédération Aéronautique Internationale (FAI), which is an international standard setting and record-keeping body for aeronautics and astronautics. For example, the X-15 and SpaceShipOne pilots that exceeded the 100 km limit were considered to have entered space and were deemed astronauts. However these were suborbital flights where the velocity achieved was much slower than needed to orbit. For orbital flight the velocity required is so high that even at 100 km the atmosphere is still too thick. I don't know what the bare minimum is to be able to complete a full orbit, but I know for sure that very short-term orbits can be achieved at an attitude of 90 nautical miles (103.6 statute miles, 166.7 km). For example, the last three Apollo missions to the Moon launched into parking orbits of 90 miles, but they only completed 1.5 orbits before heading out to the Moon. Geostationary satellites will also briefly park in 90-mile orbits before being transferred out to geostationary distance. The 90 nautical mile altitude is the minimum that I've seen used in actual practice. Even at that altitude the orbit cannot be maintained for long periods before atmospheric drag will cause it to decay. When an orbit must last a number of weeks, altitudes in the 200-300 km range are typical. And for very long durations, such as manned space stations, the orbits are more like 300-400 km.

My comment in which I quoted 1100 m/s was in reference to the opening post. The OP wrote "my engine must fire continuously for 60 minutes to get to Duna", which is a reference to the ejection burn only. Obviously ejection and insertion are two separate burns that would not be continuous.

Are you sure you're looking a ejection ÃŽâ€V and not total ÃŽâ€V? Ejection ÃŽâ€V is that needed to depart Kerbin orbit on a trajectory to Duna. Insertion ÃŽâ€V is that needed to enter orbit around Duna. Total ÃŽâ€V is the sum of ejection ÃŽâ€V and insertion ÃŽâ€V. It is the ejection ÃŽâ€V that is typically about 1050-1100 m/s. Total ÃŽâ€V is usually about 1600-1700 m/s. If 1600 m/s is the ejection ÃŽâ€V, then it's possible you may have selected a less than ideal date for launch. The application will give you the lowest ÃŽâ€V launch that lies between the earliest and latest departure dates. I generally search for the most ideal launch window in about a 2- or 3-year range, which is long enough to search a complete synodic period. If you limit the departure dates to a smaller range, you may miss out on the most ideal launch windows.

1050-1100 m/s is pretty normal for a Duna transfer from a low Kerbin orbit of about 70-75 km. If you haven't used it before you might consider planning your transfers using the following: Launch Window Planner for Kerbal Space Program

If we assume a normal Duna transfer having a ÃŽâ€V of about 1100 m/s, then he's trying to push about 210 tonnes for every one LV-N engine (a TWR of 0.03).

It is best to perform a course correction while you're some distance out. When you are still a long way from the planet it is really hard to have fine control because the slightly velocity change has a big effect. I usually just try to get close, primarily focusing more on being in the correct plane than I am about getting the altitude right. Once crossing into the planet's SOI a final course correction can be made. At this time you can fine tune the altitude to be just what you need. I suggest you take a look at the following. It gives equations for calculating the ÃŽâ€V of altitude changes, plane changes, and combination altitude/plane changes. If your answer isn't there then you can always ask a follow-up question. http://www.braeunig.us/space/orbmech.htm#maneuver

The OP appears to be AWOL.

Making it even more confusing is that in real life there are different ways in which altitude can be expressed. Sometimes altitude is expressed as the height above an assumed spherical Earth, usually with a diameter equal to Earth's equatorial diameter. Other times it is the height above a reference ellipsoid. One has to be careful to use the correct coordinate system. From what I can tell, there is no polar flattening KSP. All the bodies are treated as spheres, i.e. the sea level radius on Kerbin is the same across the entire globe. This isn't true on Earth.

I aim for ~75 km unless it is something that I plan to dock with. If I plan to dock with the ship then I aim for ~90 km, which allows me to launch the second ship into a ~75 km orbit and play catch up.