closette

You killed Kermans! You bastards!

You mean this chart? http://kerbalspaceprogram.com/forum/showthread.php/6664-Mini-challenge-max-altitude-with-this-supplied-spacecraft?p=100912&viewfull=1#post100912

Terminal speed and max Q are not the same thing at all. I think the OP was using the result that the most efficient ascent speed (at a given altitude) happens to be equal in magnitude to the local terminal speed, i.e. when the drag force and gravitational force are equal in magnitude. Unlike in the real world, if you follow the optimal ascent speed vs. altitude, the instant of "max Q" (maximum drag force) occurs soon after launch for most spacecraft in KSP, due to the weird mass-dependence in the drag model. As the rocket ascends, it not only encounters thinner air but also loses mass as fuel is used up. Even though the drag is proportional to speed-squared, those two effects more than offset that. (For my Goddard Challenge rocket when following this fuel-efficient ascent profile, max Q occurred at a mere 250 m altitude on the way up while at full thrust, and at about 8000 m when falling back down from 33000 m). Most craft will experience the greatest effect of drag (i.e. drag force per unit mass) on descent from orbit, around the 8-12km altitude mark, when their re-entry speeds get slowed from a few thousand m/s down to a few hundred very quickly.

See http://kerbalspaceprogram.com/forum/showthread.php/14273-Smallest-orbit

One small issue is that KSP currently displays apo- and periapsis as altitudes above the planet's surface, not as distances "r" from the planet's center. However, I believe that the orbital elements of spacecraft are in the persistent.sfs file so you can pull them from there to get the planet's mass, and then compare with the displayed apses values to get the body radius.

I wasn\'t trying the (rather pointless as described) challenge - I was just verifying that the edge of the atmosphere is where it has been since before version 0.16. Time to move on...

I just did this and verified a pre-0.16 result reported here: http://kerbalspaceprogram.com/forum/index.php?topic=5623.msg92331#msg92331 that the atmosphere ends at 69077.5 m above sea level. Screenshot attached to show x5 warp is possible below 69200 m. (I didn\'t hit F1 fast enough to catch it right after 69077.5 m, but you can try for yourself).

@PLB: Not sure where you get that from. 70 km is outside Kerbin\'s atmosphere so the drag is non-existent there, not 'least'. Try this experiment: establish a 72x72km orbit (or close thereto), then gradually retrofire at apoapsis to bring periapsis down to, say, 68km. Time-warp and note the altitude that you are thrown back into 1x speed. That\'s where Kerbin\'s atmosphere 'ends'.

Well I realize that everyone has moved on to 0.16 which doesn\'t work with the parts provided, but I fired up my precious copy of 0.15 and with a more carefully tuned strategy achieved: 1. 202.92 kg remaining to get into a 3.2x3.8km orbit (first screenshot attached, with the MunArch in the background). This time, after lowering Pe while far away, I was more careful in changing from hyperbolic to circular orbit. I took two orbits to do it so that I could have engines firing as close to periapsis as possible. This time I also took care to start the burn slightly before periapsis such that the periapsis point would always remain just in front of the spacecraft. If it advanced too far in front, I would stop firing briefly and coast a little. 2. Landed with 115.0 kg remaining (second screenshot attached). This would make a great separate challenge. I did some research on terminal descent onto the Moon from lunar orbit - plenty of papers but very few show the actual throttle profiles. But rather than 'stop, then drop', they seem to employ a 100% throttle braking burn to kill off most but not all of horizontal speed, then a slowly decreasing throttle for a gravity turn, possibly going to zero, followed by a brief throttle-up burst just before the vertical landing. This is what I did, and despite a 'bounce' on landing I got away with an intact spacecraft. I learned a lot from this challenge - looking forward to similar ones in 0.16 once the LFT fuel bug is fixed.

Not to mention the whole 'tilt over to 270 degrees heading' thing, which I\'ve seen a few people post that they do to get into orbit. There\'s a reason why KSC (in game and in Florida) are on the east coasts of their continents. Launching eastwards gives you a free boost in that direction due to the planet\'s rotation - 174.5 m/s in the case of Kerbin. In the game, click the 'Surface' indicator on the Navball while sitting on the launch pad and you\'ll see what your 'orbital' velocity is relative to Kerbin\'s center. Very few satellites are launched into a retrograde orbit (see http://en.wikipedia.org/wiki/Artificial_satellites_in_retrograde_orbit) since it would be an enormous waste of fuel to cancel out Earth\'s rotation.

Thanks for the kind words (this challenge was a team effort) and for revisiting the challenge - it\'s been ongoing in some form since 1919! http://en.wikipedia.org/wiki/Goddard_problem. I\'ve only just installed 0.16 (Mac users had to wait) but it certainly sounds like a fuel problem, and not a change to the atmospheric drag model which I would be very interested in! I\'ll check it out later on but it sounds like a 0.16.1 revision should bring the challenge back to 'normal' and restore honor and dignity to the leaderboard...

Hairline receding? How do you know without a 'before' picture? I guess we all have our own 'background stories' for them, which is a good reason for the developers to leave some things undefined. The fan fiction and role-playing forums seem to have taken off based on pure speculation and imagination filling in the blanks, which is fine (just not my thing). So I was surprised that the developers recently gave Kerbals a voice on video, and allowed us to see them without space suits in the 0.16 VAB, but I guess that more details will slowly be filled in as the game progresses towards crew selection. Yes, eventually having some flexibility with Kerbals\' features and names for users to choose, without explicitly acknowledging a gender binary, seems like a good way to go.

Personally, I would prefer to keep Kerbals unisexual (or asexual?), to keep focus on the space program, which I find a safe haven from imposition of gender stereotypes. (I wasn\'t too keen on that comic\'s first female character being referred to as 'fresh meat', by the way). In my own imagination, the Kerbals are grafted and grown like plants and later develop limbs the way that tadpoles develop into frogs. They are cute the way they are, and although I use male pronouns to describe them I don\'t feel there\'s a great need to add another gender (and if so, why stop at just two?). Weird for me to be defending the status quo. though!

Congratulations on reaching orbit. The free version you are using (0.13.3) has fewer features and does not limit fps as far as I know. The paid version has many more features which might well end up producing a slower fps overall, depending on your system. I keep both versions (and a couple of intermediate ones) on my laptop\'s hard drive depending on whether I want to show a quick demo or do something more in-depth with all the features. My advice - feel free to ignore - is to play the free version up to being able to complete a Mun mission, before going for the more fully featured version, which by then you will both want and appreciate all the more.

Well I guess I\'m one of the 'impatient' crowd as well (although I am a Mac user and have to wait for the Patcher to be fixed to even get 0.16, grrr...). If x2 warp is so bad, replace it with x3 Practically, leaving the game to run while you go and make tea / check email / whatever isn\'t the best solution since we\'ll either forget to check back in time or will be constantly checking 'are they there yet?'. Aerobraking is one of the cool features of the game and it sounds like it just got a bit more tedious in 0.16.

Actually the mass term in the drag equation means that a full and an empty tank should both fall together, accelerating at the same rate and reaching the same terminal speed. For a falling object the downward acceleration of a falling part in the direction of 'g' is given by accel = {Net Force}/ mass = {Weight - Drag}/mass = (mass x gravity - 1/2 x mass x maximum_drag x density x speed2) / mass = g - 1/2 x maximum_drag x density x speed2 ...so mass cancels out in both terms. Setting a=0 one can solve for the terminal speed: vT = Sqrt[2g/(maximum_drag x density)] Unlike in the real world where if you drop a full and an empty styrofoam cup off a balcony you\'ll easily see the difference in accelerations and terminal speeds. The atmospheric density on Kerbin for drag purposes is given by 0.009785 * exp(-altitude / 5000 m), so you can see that the local terminal speed at the surface for most parts (with maximum_drag = 0.2) is about 100 m/s for any mass.

I think the challenge is to distinguish that there were 5 distinct songs back-to-back, and to keep count while your mind is numb.

How about calculating (apoapsis - periapsis) / (apoapsis + periapsis)? Oh wait, that is equal to the orbital eccentricity! (Yes there may be other ways of describing how out-of-round an orbit is, but this is already a well-established, scale- and unit-independent measure of what you want). I vote e < 0.1 is 'practically circular'. That only just includes Mars, but excludes Mercury and Pluto in our solar system. Calculating differences in kilometers for their apses would be silly, for instance Jupiter and Uranus have very similar orbital shapes (as given by their eccentricities) but the (apoapsis - periapsis) distances are very different since the orbits are different sizes. However (remember that Kerbin is 600 km in radius), that would mean a 100x150km altitude orbit, eccentricity 0.034, would be 'practically circular', which some may disagree with. Can we have another poll please?

Bug report (sorry if this was already mentioned): For an 'inner' bi-elliptic transfer from low to high (or an outer transfer high to low), the first of the 'delta-V's is negative, which is fine since you are slowing down, but for the total delta-V budget, you should add the absolute values of all the individual delta-v impulses because each impulse uses fuel regardless of direction. To demonstrate, select 'Bi-elliptic', 'Kerbin', 'Low to High' and Lower Orbit Altitude: 300 km Higher Orbit Altitude: 2000 km Midcourse Impulse Altitude: 100km and you\'ll see that the 3 impulses are -128, 437, and 406 m/s, but the TOTAL is given as 715 m/s, which is misleading. It should be about 971 m/s. OK not a very practical example but this bug also shows up when I try to plan high-to-low transfers around the Mun and it spoils an otherwise useful tool. The V/Vc values also suffer from this bug. It might also affect the output under 'Bi-elliptic transfer analysis' as well. By the way, this is a great utility and I hope you can maintain it, and perhaps add Minmus at some point. I like that it works 'standalone' so that I don\'t have to fire up KSP and rely on a bunch of add-ons.

@PakledHostage: This result was bothering me for a while, since the large intermediate orbit seemed arbitrary, but just like a bi-elliptic transfer, the larger the better. For a very large intermediate ellipse you can do this transfer with a delta-V of only 354 m/s, except you can\'t because it would go outside the Mun\'s SOI. So for an ellipse which just fits inside the Mun\'s SOI (apoapsis at 2400 km) I calculated that you can get the delta-V down to 441 m/s if you wait long enough. I think that will save you an additional ~ 1 kerbalgram of fuel over PakledHostage\'s chosen transfer orbit. It may still be more efficient to lower the hyperbola\'s periapsis and periapsis speed from afar, with a back-and-inwards impulse as you carefully plotted in Reply 48. I also learned that apparently, for a 1-burn capture into a circular orbit from a hyperbola, the most efficient capture occurs when rpvinf2/GM equals exactly 2. I\'m waiting for my library to get an orbital mechanics book which I hope has the proof of this. In the meantime I\'ve been avoiding real work by tracking down papers such as the two attached. Figures 2 through 4 in each paper look very much like what we (sometimes) do on a Munar approach. Having finite thrust means that sometimes it takes a few orbits to circularize. Enjoy.

Please don\'t judge all scientists by a few bad apples. Being rude and defensive doesn\'t make them wrong and you correct of course, and I think you realize that. Nevertheless, NO SCIENTIST SHOULD BE OFFENDED WHEN THEIR FUNDAMENTAL ASSUMPTIONS ARE QUESTIONED. And it should not matter who is asking the question or how many letters anyone has after their name. That\'s how our understanding of the Universe itself 'evolves'. Having said that, science is a human activity, and has its flaws, such as tying either 'being first to discover' or 'confirming the status quo' to putting food on the table (which is what 'funding' ultimately boils down to). I have seen, and felt pressure from myself, examples of confirmation bias which really made me stop and think 'Is this worth confronting an egotistical supervisor to mention?'. Not a good situation, so I got out fast. I\'m sorry you were treated rudely and hope you can continue asking great questions. I don\'t know much about biology...but I do know that there are many questions about evolution that still need to be explored, and you have every right to ask away, as long as you keep an open mind to the responses.

Step 3: Get stranded in orbit around Kerbol for all time. Planets are moving targets.

PakledHostage, your delta-V numbers check out just fine. (I used the KSP Orbit Mechanic but it still took ~ 10 mins to check \'em). Nicely presented too. The Hohmann vs. bielliptic efficiency quoted is for circular to circular orbits, as we had in the Munar Descent challenge, where I believe the Hohmann method is still the most efficient since a 100 km altitude to ~ 0km altitude transfer has a radius ratio (after adding in the Mun\'s 200km radius) of only 300 km/ 200 km =1.5, clearly favoring Hohmann\'s 2-impulse method. For hyperbolic -> circular transfers though, PakledHostage has shown that a bi-elliptic 3-impulse maneuver is slightly better. I think this might be because a hyperbolic orbit close to periapsis looks only slightly flatter than a really, really big ellipse. I\'m sure this is a 'known' result but it\'s good to see here, and applicable to every Mun mission. On my challenge attempts, when I reached the hyperbola\'s periapsis, I have found it difficult to get the apoapsis down to even 40km in a single impulse. Now I can try it in 2 stages. It will also go faster because of time-warp at higher altitudes.

Not really mentioned as an option but I\'d prefer a Mars-like planet in a Venus-like orbit, i.e. closer to Kerbol than Kerbin. (Call it Menus?) That would make for a different trajectory plan from Kerbin, and there could be some interesting gravity-assist (aka slingshot) trajectories to outer planets once they get added in. Landing through a thin atmosphere would be fun - aerobraking, then parachutes and (perhaps) airbags to deploy.

Very nice PakledHostage. I\'ve been wrestling with the orbital equations off and on (hard to find some distraction-free time) but your experiments bear out a reasonable guess, that the optimum angle for an impulse to reduce periapsis is somewhere towards 'backwards and inwards'. Periapsis distance is given by a(1-e) where semi-major axis a is related directly to orbital energy and eccentricity e is related less simply to both energy and angular momentum. For a hyperbola a is negative and e is greater than 1. The optimum angle you found reduces the magnitude of a and reduces e at the same time.