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About Zhetaan

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  1. Zhetaan

    What parts produce these experiments?

    Are you using Nertea's stuff? There's a plant growth experiment on the greenhouse of the Station Parts Expansion. The Triple-Z Radio Telescope has a radio astronomy experiment that's actually called 'Radio Astronomy'. My only other guess on the source is that there may be something in DMagic's Orbital Science, but I'm not familiar enough with it to commit. Also, depending on how you arrange your install, it's possible that the experiment exists but the parts to perform it do not: science experiments, their applicable situations, and other bits are stored in a science definitions file that is separate from the part configuration file which tells a science part which experiments to perform. [x]Science! looks at both, so if you have a vestigial science definitions file somewhere, then it will see that the experiment is possible but will never see the means to perform it on the vessel.
  2. Zhetaan

    Ksp Delta-V Map

    It shows the average of a series of optimum Hohmann transfers. That means that the numbers will tend to be low. There are more efficient trajectories, but most of these are not direct transfers; they require increased flight time, gravity assists, or both. There are infinitely many less efficient trajectories. As @bewing said, you can be 100% inefficient.
  3. Zhetaan

    Science Basics

    But there is also a minimum Reputation requirement to activate that strategy, so for practical purposes, yes, you can, in the sense that you can have a public image that is so bad that you cannot sell that image for money--the extraordinary need of which, as I understand it, is why you would want to sell that Reputation in the first place. The only thing to do then is hope that you get a contract to test something on the Launch Pad.
  4. Zhetaan

    Science Basics

    Strategies, explained: The Finances, Science, and Public Relations Departments are interested in Funds, Science, and Reputation, respectively. They each have two long-term strategies that turn the other two currencies into the one that they're interested in. There is a required minimum Reputation and a set-up cost, so you have to have some of what you're converting to begin, and the conversion amount varies depending on how much you choose to commit (which is also limited by how much you've upgraded the Administration Building), but it can go as high as 100%, meaning that you can conceivably trade all of your gains in one thing for another. This makes sense in the late game when you've researched the entire tech tree and no longer have a use for all of those Science points you keep collecting, for example, but you may want to have a few million Funds to pay for your new mega-rocket that uses all of those new parts that you just finished unlocking. It's also useful if you keep killing your Kerbals and want a balm for the strikes against your Reputation, though I've found that bad Reputation is actually the easiest obstacle to overcome. Finances: Fundraising Campaign: Trades Reputation for Funds Patents Licensing: Trades Science for Funds Science: Unpaid Research Program: Trades Reputation for Science Outsourced R&D: Trades Funds for Science Public Relations: Appreciation Campaign: Trades Funds for Reputation Open-Source Research Program: Trades Science for Reputation There are two more long-term strategies (both from the Operations Department) that change the balance of launching and recovery: Aggressive Negotiations: Trades some Reputation for lower costs on launches, building repairs and upgrades, and unlocking technologies in the R&D building (not the nodes which require Science, but the individual parts which require Funds). This is applied for each discount, meaning that there's a Reputation cost for every new launch, every new technology you research, and so on. On the other hand, the discount is a percentage of the Funds cost (up to thirty (Wow!) percent for launches) and the Reputation cost is limited to six (that's six Reputation Points, not six percent), so if you're in the habit of buying expensive rockets or you want to unlock the expensive third-tier upgrades (or you need to fix the third-tier VAB since you just crashed the expensive rocket into it), this is potentially a lot of money saved for a comparatively minor strike on Reputation. Recovery Transponder Fitting: Trades the maximum recovery value for a higher minimum. This is the only Funds-for-Funds strategy; it mainly is useful if you tend to drop your rockets on the other side of Kerbin. If land near KSC, use a lot of disposable probes, or like to keep your vessels in the sky for many multiple missions, then this strategy is silly. If you want to land on the other side of the planet and get 25% recovery at the risk of only getting 88% when you're next to KSC, or if you just have too much money and need to invent nonsensical ways to spend it, then this strategy is for you. It may have a use in the early game when you may not be able to land planes in one piece or you don't have good landing control, but as soon as the value of the rocket begins to make this worthwhile, you've got both the skill and the technology to do better on your own. Besides, if you want nonsensical ways to spend money, why do something spectacular, such as launching a rocket full of Xenon tanks into the sun? Lastly, there are two immediate-return strategies, one in the Science Department and one in the Public Relations Department, that respectively trade some amount of your Science and your Reputation for Funds right away: Research Rights Sell-Out: Sells Science (up to 100% of it) for Funds Bail-Out Grant: Sells Reputation (again, up to 100% of it) for Funds These strategies appear to be intended for those who manage, either through bad luck or bad management, to bankrupt their space program and have no other way to make money (because, for example, they don't have enough to pay for the rocket that can complete the contract to get more money--Rockomax does not, apparently, accept credit). Granted that people in this situation must also have managed to do a few experiments or else remain in the public's good graces, so if you're completely out of Reputation, Science, and Funds, you're also out of luck. The other use of these that I can think of is that, for example, let's say that you've completed the tech tree and set up Patents Licensing for 100%. There's a Science set-up cost, but after that, there is still a lot of Science left over; Patents Licensing only works on new incoming Science, not what you already have. The Sell-Out in this case gives you useful money from the Science you still have left over. Long-term, the strategies you've activated will both lower your reputation, so it will be more important for you not to fail contracts or to kill Kerbals. One will help your Funds by giving you cheaper launches and facility upgrades, and the other will give you Science for everything you do well (new World's Firsts and completing contracts mostly). However, the amount of Reputation loss is balanced by a few things. One of these is that Reputation is actually fairly easy to get, provided you have the standard difficulty settings for your game. Another is that Reputation, unlike Science or Funds, is not strictly additive. By that I mean to say that when a contract promises a certain amount of Science or Funds, it pays that amount and your totals increase. Reputation, on the other hand, is limited to 1000, so as you get closer to that value, contracts pay out less of it. Lowering your Reputation with strategies, oddly enough, actually opens the way for you to get more of it from each contract.
  5. Zhetaan

    Science Basics

    Firstly, welcome to the forum; you've definitely come to the right place. That depends on the experiment. Here's the essential idea of how science works: First, experiments work in specific 'situations'. There are a number of these: landed, splashed, flying low, flying high, space low, and space high. Some experiments work in all of these situations, but some experiments only work in a few situations. On top of that, any, all, or none of these can be biome-specific, meaning that instead of a global result for the whole planet, you can get specific results for each biome. For example, the seismic accelerometer only works when 'landed'--it won't let you take a report otherwise--but it gives different reports for every biome. The infrared telescope only works when in space high, and it's global, so it only takes one result total for each planet. You can take EVA reports for each biome in space low but only one for all of Kerbin in space high. Also, it's not necessary to enter orbit: so long as the situation is correct, you can get the results, even on a suborbital hop where you clear the atmosphere for only ten seconds, or a hyperbolic flyby where you dip low enough to get the low space science and then then head back to Kerbin. Second, not every world has all of these situations. 'Flying' requires an atmosphere; 'splashed' requires an ocean (that may or may not have water in it; it just needs to be liquid), and so forth. Third, the altitude transitions for these situations vary with the location. For Kerbin, 'flying low' changes to 'flying high' at 18,000 metres; 'flying high' changes to 'space low' at the edge of the atmosphere (70,000 metres), and 'space low' changes to 'space high' at 250,000 metres. You can't get flying science when you're in space, for example, so that's a limit, but all that means is that there's a different range of experiments to do. Eventually, you leave Kerbin's sphere of influence, so I suppose that's the absolute maximum. There's no preferred order of experiments; the only thing to be really concerned about is that when you take crew reports, you have to gather the data from the pod, as well. I'm probably telling you something you already know, but the steps are: get out of the capsule, right-click on it (or the experiment), select 'Take Data', right-click on the capsule and select 'Store Data', and then get back in for the next report. You may also choose to just board the capsule after taking the data, which will also store the data in the correct place. Otherwise, the most recent report is stored in the--we'll call it the active data slot--and keeps a new one from being added without overwriting the active one. You have to pull the data from the active slot and put it into inactive storage in order to make room for a new report. As to why you have to get out of the capsule to do this, I have no idea. If you continue to get errors such as duplicate report errors, that only means that you've already visited that biome. Alternatively, it means that you've got a global experiment and can only take one for that situation, which is the same problem but with fewer total reports. Temperature, for example, is global unless you're landed, splashed, or flying low (two of these are not an issue at the Mun--I'll let you guess which ones). The same is true of crew reports. Pressure is global except on the surface--it doesn't do biome-specific science even flying low. In fact, the only experiments that are biome-specific in space low are EVA reports and gravity scans. You can tell whether an experiment is biome-specific in any situation by looking at the experiment result. If it says, for example, 'Temperature while Landed at Mun's Poles', then the biome name (Poles) tells you that it gets results by the biome. Obviously, 'landed' tells you the situation. If it says, instead, 'Temperature while In Space Near the Mun', then that's a global experiment ('space near' is the same as 'space low'). Anyway, to properly answer your question, preparing your orbiter is a matter of putting the experiments you want on it. I'd recommend using a service bay to keep them out of the wind on the way up, but that doesn't help you get science except indirectly. However, assuming that you've already gotten Kerbin's low and high space temperature and pressure results, then on the way down to the Mun with a thermometer and barometer, you'll get four experiments in high space: temperature, pressure, EVA report, and crew report. All of those experiments are global in high space. The change from high to low occurs at 60 km at the Mun, so 15 km is definitely close enough. There, temperature, pressure, and crew reports are all global, so you'll only get one each, which means that in all honesty, getting the most science entails hanging off the ladder and taking EVA reports in each new biome (and remembering to put the reports away to make room for new ones). In general, preparing your vessel is a matter of knowing what you intend to do with it. Don't put a telescope on your lander; it only works in orbit. Don't put seismometers on your orbiter; they only work when landed. Take a barometer (surprise!) but not an atmospheric analysis unit, unless you're both visiting a world with an atmosphere and intent on flying there (Jool and Eve can be tricky). I'd recommend a polar orbit in order to get all possible biomes, but if you want any more science, then you'll need to land (or take different experiments). As I said above, use a polar orbit. When you launch and turn east, you end up in what's called an equatorial orbit (so named because it's above the equator). If you were to turn north instead, you'd end up in a polar orbit, so named because it goes over the poles. The ISS is an intermediate case; but the general term for all orbits that are not equatorial is inclined orbit. Much like a ramp has a tilted surface that is called an incline, an inclined orbit is tilted with respect to the equatorial plane. Something of which to be cautious is that the Mun, being tidally locked to Kerbin, has an extremely slow rotation period of nearly six and one-half days, so it will take a while for the Mun to rotate under your vessel. Forty-five degrees of inclination will give access to most biomes in less time. To set up a polar orbit, the most straightforward way is to burn to transfer to the Mun as usual, but as soon as you cross into the Mun's sphere of influence, set up a node for an immediate normal burn ( on the navball). The time isn't too critical but you'll want to burn as soon as possible after entering the sphere. However, you won't get the most up-to-date orbit information until after you enter the Mun's sphere of influence, so don't set up the burn until after you've crossed over, either. Add enough normal burn to tilt the orbit as much as you like (or put it over the pole if you want to try for a polar orbit), add enough prograde or retrograde so that your periapsis is out of the ground and at the altitude you want (15 km), and then burn to capture when you get there as usual. The comm system is useful for other purposes, but for science it's really only good for transmitting science results back to Kerbin. If you intend to bring Jeb back, you don't need it because you can recover the experiments along with the pilot. Also, transmitted results tend to be worth less and there's a limit to how much can be transmitted, anyway (whether or not that makes sense: temperature and pressure are capped at 50%, which is weird, because those are just numbers). Eventually, you have to bring the science back. The Science Jr. is a nice experiment and gives a lot of science, but it's heavy and single-use. It can be restored by a scientist if you have one along, but not by a pilot. It's worth taking one or two to get the space high and low science, but unless you really need the science to unlock something, you may want to wait for a dedicated landing mission with a scientist in a two-seat vessel, in which case you can grab the space science on the way there.
  6. As I understand these sorts of missions, the mission is accomplished when a vessel with the correct specifications enters the correct orbit and stays there for ten seconds. As such, I don't think there is any specific prohibition against taking it on the nose of a manned rocket. Don't forget any antenna or other accoutrements that the mission might require--some of them are terribly specific. There should be a target orbit in the map view; the correct longitude of ascending node is wherever the ascending node is on that target. KSP doesn't provide a good way to know that orbital parameter. Zero longitude is a specific direction in the game, but since the skybox is essentially wallpaper painted on the face of infinity, it doesn't correspond to any actual celestial object. You'll either need to use a mod that gives you the information or else get comfortable with flying from map view. Your design approach is still correct; it's just that you need to add up a bit more delta-V at the top end, which of course will have exponential increases down the rocket in terms of necessary fuel. But delta-V adds: if you need, for example, one hundred for the probe release and five hundred for the rescue, then take six hundred for that part of the mission. The Mun transfer is going to cost the same no matter what you take; you'll just need to account for the extra mass. Of course, you'll need to take care because of drag for the launch from Kerbin, but I assume you know how to handle that. The trick for these sorts of missions is knowing which parts can be combined with other parts and which are best done independently; anything combined saves fuel. For example, it may pay off the most to capture with a perispsis close to the Mun and an apoapsis at the probe orbit; you simply leave the probe with enough fuel of its own to fix its orbit. Then you can continue to burn down for the rescue (assuming the rescue is closer to the Mun than the probe orbit) at the next periapsis and you won't need to carry any extra fuel beyond what you need to loft the probe to that point in the first place. This works because it's trivially easy to give a probe the hundred (or whatever) metres per second it will need to do its mission, as opposed to changing orbits with a more massive rocket up high where you don't get any free efficiency. But the best part is that so long as you burn at the near-Mun periapsis, it doesn't cost you extra fuel to drop off the probe; the only expense is time. As far as knowing what the delta-V costs will be if you've never been in the neighbourhood, you'll need either a map or a good calculator. But I got the impression that you can plan for a single mission just fine; it's the two-for-one that you've not yet done. There's no reason not to do so, especially if you can cut the probe mass to near-trivial levels. I caution you to remember the extra required delta-V for the return and the higher speed of moon return re-entry. Also double-check your extra seat; nothing inspires imaginative cursing like taking a rescue ship that has no room, but without your knowledge until you get there. Probe design is fairly straightforward: you need the core, a battery, antenna, and solar panel or four, plus fuel and an ant engine if you want them (hint: yes you do), and whatever else the mission specifies. Sometimes the mission asks for certain science experiments. Sometimes, it wants a docking port. Sometimes, it asks you to take certain other parts (like landing gear, for some reason I don't ken), so be sure that you add them. Then add a separator (not a decoupler), a fairing, and stick it on top of your rocket's nose. For these kinds of probes, you should keep the size down to .625-metre parts. Don't forget that a probe on your nose means that a parachute won't be. Take radials. Good luck!
  7. You have my apologies; I oversimplified. You're correct in that the period of the orbit cares not a whit about the location of the nodes--though the comparison between two scaled orbits does care about the location of periapsis insofar as that defines zero true anomaly. My intent in saying that was only to recall that for this specific problem, there is a special reason why the ascending node must occur at ±90° true anomaly: the Draim constellation constrains not only the inclination and eccentricity, but the argument of periapsis as well. This is the angle between the ascending node and the periapsis as measured on the orbital plane (as opposed to the longitude of ascending node and longitude of periapsis, which are measured on the equatorial reference plane). Specifically, all of the orbits in the tetrahedron have an argument of periapsis of either 90° or 270°, which puts the periapsis at right angles to the line of nodes with respect to the primary as the vertex. This is why, in this case, the line of nodes coincides with the latus rectum of the ellipse, but for any other argument of periapsis, this would not be so. Since your original question was not how to find the latus rectum, but how to find the correct point at which to change inclination, I was pointing out that although you have a general solution to scale orbits of constant eccentricity by their orbital period regardless of their orientation and can also find the latus rectum of any orbit with relative ease, the equation will only find the correct burn point if you begin with the correct argument of periapsis: 90°, 270°, or undefined (which means starting from equatorial orbit). Granted, I'm also aware that your entire purpose was to simulate the Draim constellation, so alternative arguments of periapsis were never in your plans. Perhaps I was being over-thorough ... you may have noticed that I have that habit.
  8. I did swap them; that was my error. The apoapsis sight angle will always be smaller because it is farther away. I fixed it, in any case. That will work well enough for most situations. It will give issues when you need to deal with either massive bodies orbiting your targets closely (because their sphere of influence will interfere with your desired orbit) or targets with low mass orbiting closely to their primaries (because the target's sphere of influence is too small to give you the correct orbit within the constraints). I know that there is no solution for Laythe that fits the parameters (Laythe is fairly large for a moon but not compared to Jool at that orbital distance), but Duna should be possible provided that you orbit outside of Ike's influence. The target value from the other thread is a semi-major axis of at least 7.25 times the planetary radius, (which is the value of 1 / tan 7.853°) so that's yet another verification method. I saw nothing constraining the maximum value of the semi-major axis except insofar as the constellation needs to remain in orbit. Oh, I don't doubt that one launch would work better for anything interplanetary, at least in transfer costs. That being said, you may still want to consider the flotilla alternative, or possibly a hybrid where you have one carrier to take the satellites into interplanetary space but then separate the satellites and set them to their own intercepts before the target encounter. That may not be necessary but it's probably worth investigating; thirty-three degrees inclination is expensive no matter how you look at it. As I mentioned above, Laythe's sphere of influence is too small and Laythe itself too large for the tetrahedron to work with 33° inclination and .28 eccentricity. Its sphere of influence is 7.45 times its radius, which fits one parameter, but that requires orbits to have eccentricity of no more than .027, which is too circular to be of help (the minimum for a Draim tetrahedron appears to be more than .1). I have seen references to some other solutions that work with five satellites, but I am not familiar enough with the details to tell you whether they would work in Laythe's sphere. Duna has a radius of 320,000 metres, so that would require a constellation semi-major axis of 2,320,093.7 metres. Eccentricity of .28 gives an apoapsis for that orbit of 2,969,719.9 metres. Ike's periapsis is 3,104,000 metres, and with its sphere of influence of 1,049,599 metres, that gives a potential closest approach of Ike's influence at 2,054,041 metres. More likely, that conflict would occur farther out (33° inclination gives a lot of extra room in 3-space), but it would still occur. You're right about the coverage gaps: the table-given SMA of 2,080,000 metres reflects the three-dimensional geometry (it's too large for anything coplanar), but even so, 2,080,000 / 320,000 = 6.5, or alternatively a sight angle of 8.746°, which is too close to Duna. On the other hand, if you go higher, then Ike's apoapsis is 3,296,000 metres, so it gives a maximal influence conflict at 4,345,599 metres. Let that be the periapsis and that gives the satellites a semi-major axis of 6,035,554.2 metres. You can get closer (again because of three dimensions) but that value guarantees no conflicts with Ike. It's nearly nineteen times the ratio of the radius to the SMA, or a sight angle of only 3.035°, but I see no reason why that wouldn't work. Satellites on that orbit would have an apoapsis of 7,725,509.4 metres, which is well within Duna's sphere of influence (nearly 48 million metres). But there are still coverage gaps: Ike gets in the way. That's probably why the author chose the lower orbit. On the gripping hand, the solution to that is simple: put a constellation around Ike. By definition, a body with one hundred percent coverage is transparent to the network--meaning that if you can see every point on a planet, then there's no reason why you can't see through to what's beyond that planet; it's only a question of whether you're looking. So long as your Ike satellites have transmitters that can see Duna, then you'll get full coverage without needing to send anything extra beyond the more powerful antennas--which you need to send anyway if you want that constellation to talk to Kerbin. If you put your Kerbin-capable relays around Duna and have Ike talk to them, then you still get full coverage without needing to do anything extra; the required minimum transmission distance to the relays is greater than the distance to Duna's surface. Yes; assuming that you hold eccentricity constant and travel from a specific true anomaly θa to a specific true anomaly θb on both orbits, the change in mean anomaly will be the same and the travel time will scale linearly with the period: t = MT / 2π. It's not enough that the craft travel through the same angle; the caveat is that the trip must begin and end at the same true anomaly on both orbits. Since you're going from 0° to 90° each time, it works. However, understand that this is a special case since both endpoints are identifiable as other orbital parameters (Pe and AN/DN).
  9. Without access to that particular paper, I am limited in what I can figure out. The radius and the gravity tend to increase and decrease together, but there is no requirement that they do so--planetary density can vary quite a lot. Saturn, for example, is less dense than water, because its volume belies the fact that there simply isn't that much mass to it. In other words, Saturn is our fluffiest planet. However, satellite coverage has more to do with the visible surface, which is of course related to radius, but it appears that the parameters for this tetrahedron are very flexible. Nevertheless, I note that the table of parameters in the original challenge post seems to preserve the angular relationship from Draim's work, which is why the satellites in KSP have the semi-major axis that they do. For a satellite orbiting Kerbin with a semi-major axis of 4,350,000 metres and an eccentricity of .28, the (true) apoapsis and periapsis are as follows: Ap = 5,568,000 metres Pe = 3,132,000 metres These were derived from the relation Ap = (1 + ε) * SMA and Pe = (1 - ε) * SMA. With the planetary radius of 600,000 metres, the angle between the satellite and a point on the circumference of the greatest-size cross section that faces the satellite (in other words, the farthest possible visible point) is given by: Sight angle Ap: arctan 600,000 / 5,568,000 = 6.15° Sight angle Pe: arctan 600,000 / 3,132,000 = 10.84° Using a similar derivation for satellites orbiting Earth: Earth's standard gravitational parameter is about 3.986004419×1014 m3/s2. Earth's average planetary radius is 6,371,000 metres. The required minimum orbital period for Earth of 27 hours is equal to 97,200 seconds. A 27-hour period gives a semi-major axis of = 45,691,651.57 metres, so: Ap = 58,485,314.01 metres Pe = 32,897,989.13 metres Sight angle Ap: arctan 6,371,000 / 58,485,314.01 = 6.22° Sight angle Pe: arctan 6,371,000 / 32,897,989.13 = 10.96° I think we have our relationship. So long as your sight angle is less than 6.22 degrees at apoapsis and 10.96 degrees at periapsis (smaller angles mean greater distance, which means better coverage), it can work. Since we're using the same eccentricity for everything, we can also find the semi-major axis directly: Hypothetical sight angle: arctan 6,371,000 / 45,691,651.57 = 7.938° To figure the minimum semi-major axis for a given body of known radius: SMA = rbody / tan 7.938° To check for Kerbin: 600,000 / tan 7.938° = 600,000 / .1394375 = 4,303,002 metres. This is lower than the given value for Kerbin because, I think, the original poster decided to be safe, but again, without having access to the paper, it is possible that there is a practical reason for the given figure. If you prefer to have that margin, then the hypothetical angle to use is actually 7.853°. This is in line with your supposition that you can scale the semi-major axis by the radius times a constant; I simply took it a step farther to determine both that 1 / tan 7.853° is the value of that constant, and why it appears. To check these figures, we can use the Mun (radius = 200,000 metres): 200,000 / tan 7.938° = 1,434,334.13 metres Alternatively, with the safer angle: 200,000 / tan 7.853° = 1,450,058.58 metres See whether you can get a good network there using that parameter, and if so, you'll have your solution. Edit: Alternatively, you can use this set of precalculated figures that I just found, of course, only after doing all of that: https://forum.kerbalspaceprogram.com/index.php?/topic/155188-orbital-parameters-for-a-tetrahedral-satellite-constellation/ Edit edit: Draim's original paper is behind a paywall, but he also got a patent for it, and because of the wonderful scope of U.S. patent law, the patent itself is available: Tetrahedral Multi-Satellite Continuous-Coverage Constellation Edit edit edit: And here's an interesting animation of satellite coverage using this model:
  10. That's correct; all of the anomalies are measured under the condition that periapsis equals zero (and apoapsis is π radians (or 180 degrees); the anomalies all converge there, too). And to answer the unspoken question: if it's a circular orbit, the anomalies don't exist--though the hypothetical orbit described by the mean anomaly describes the actual orbit, so you can use that and simplify calculation immensely. In Kepler's time, orbits were assumed to be circular, so any deviation from circularity was anomalous--hence the term anomaly. I use Character Map if I need it. However, there's no special requirement to use difficult-to-type characters in these equations; I could have solved for the true anomaly as x just as easily as θ. It's simply convention to use Greek letters because of either tradition or a need to use more letters than are available in the Latin alphabet. That being said, the conventions are fairly common, so here's a cheat sheet: ε - Epsilon, used for eccentricity; e is often used instead, but since e is also a mathematical quantity and E is used for eccentric anomaly, it can get confusing, so any time that I calculate eccentric anomaly, I make certain that I use E for the anomaly and ε for the eccentricity. You'll also find ε used for planetaery axial tilt; it's not an issue in KSP but you'll want to know that if you ever try your calculations on real planets. θ - Theta, often used instead of x when the unknown variable is an angle; it's also used for true anomaly since t is almost universally used for time and T is used in astronomy for orbital period. You'll encounter it a lot in place of x in trigonometric functions, and it is used for angular measure in polar coordinates (which use the (r, θ) form instead of the (x, y) of Cartesian coordinates). ν - Nu, an alternative way to show true anomaly. It looks like Latin v in Arial but its actual form is a bit more elaborate (it has serifs and everything!); needless to say, its similarity to Latin v means that it can be mistaken for velocity, so I prefer to use θ. Incidentally, Latin f is yet another way to show true anomaly, but f can also mean frequency, and it is used to denote focal length for a lens--this isn't a concern in KSP, but in the kind of astronomy that requires telescopes, it can be confusing, so again, I prefer θ. π - Pi, and I'm sure you know this one. ω - Omega, used in physics for angular velocity; the difference between this and θ is that θ describes a static angle, whereas ω describes rotational speed (usually in terms of radians per second). It's used in orbital mechanics as the argument of periapsis (the angle between the periapsis and the ascending node). Note that capital omega (Ω) is used for the longitude of the ascending node, so there are two different omega parameters in orbital mechanics. It turns out that there aren't enough letters even with Greek. μ - Mu (pronounced myoo) (or mew, as though it were a kitten), which is used in many different contexts. It is the symbol for the SI prefix micro-, which in astronomy is likely only to arise if, for some reason, you need to describe micro-light-years. Astronomy has to scale colossally, though, generally doesn't bother with prefices, and instead uses scientific notation almost exclusively. You are more likely to encounter μ as the standard gravitational parameter, which is a measure of the gravitational influence of a large body and is actually a very important value to know when calculating orbital period and the like. This parameter first arose as the proportionality constant in the relationship T2 ∝ a3, which you may recognise as Kepler's Third Law (as an aside, the symbol ∝ means is proportional to). When taking the proportionality constant into account in order to make an equation, it becomes T2 = ka3, where T is the orbital period, a is the semi-major axis, and k is constant for all orbits about a given body (but it is different for every body). The constant k relates to μ by this relationship: k is equal to 4π2 / μ. If that seems esoteric then know that the π factor appears because you're chasing around an ellipse; the angular velocity relates to the period because the period is the amount of time it takes to travel 2π radians (one full revolution) around the ellipse. Since the period is squared in the Third Law, the period in terms of angular velocity needs must also be squared: hence, 4π2. A more intuitive way to describe the Third Law is this: (T / 2π)2 = a3 / μ, in which it is more obvious that the entire period factor refers to time per revolution. Correct. Though if you are starting from apoapsis, then it may be simpler to subtract 7000.63 seconds from six hours to get time after apoapsis rather than time before periapsis. This will work around any body, and yes, the orbital period encodes the gravitational relationship between the primary and the orbit in its calculation; that's why an orbit of a given semi-major axis around a given body has the orbital period it does. Furthermore, it's why we need further calculation to locate things on that orbit: any orbit with that SMA about that body will have that orbital period, regardless of its shape, and the need to find the differences that relate to eccentricity is the reason that the concept of orbital anomaly was invented in the first place. However, you won't need the orbital period until you are ready to include the mean orbital motion to get time from the mean anomaly; all ellipses of the same eccentricity are similar, so the angles will be the same in any case. Bear in mind that you must know the eccentricity to get that far, but orbital eccentricity is trivial to calculate, especially if you have Kerbal Engineer to do it for you. However, if not (or if you'd like to do it yourself): ε = (ra - rp) / (ra + rp), where ra and rp are the radii of the apoapsis and periapsis to the centre of mass (meaning the planet's centre if you're orbiting a planet), respectively. Bear in mind that this requires the actual apsides, but KSP displays the altitudes above the surface. You'll need to add the planetary radius to get the correct value, but that radius is available both on the wiki and in the KSP Knowledge Base. For Kerbin, it's 600,000 metres. I think that was all of your questions. In any case, good luck.
  11. I can help you with some of the prediction. If you want to know when you'll be at 90 degrees true anomaly, then you need Kepler's laws; that's non-negotiable. I'll assume that you're starting from a true anomaly of zero; it's one element that is easy enough to see both in KSP's interface and in Kerbal Engineer. True anomaly, θ, is given by the equation: (1 - ε) tan2 (θ / 2) = (1 + ε) tan2 (E / 2) Where E is the eccentric anomaly and ε = .28, the eccentricity. In this case, we know that we want a ninety-degree (or π/2) true anomaly, so this equation is a matter of solving for E. (1 - .28) tan2 (π / 4) = (1 + .28) tan2 (E / 2) --- This is easy enough; tan of π/4 is 1 .72 = 1.28 tan2 (E / 2) .5625 = tan2 (E / 2) .75 = tan (E / 2) .643501 = (E / 2) 1.287 rad = E This is about 73.7 degrees. Once you have E, you need to solve Kepler's Equation for mean anomaly, M: M = E - ε sin E --- This is the easy way to do it; if you were solving for position instead of time, you'd have to solve the transcendental M = 1.287 - .28 sin 1.287 M = 1.287 - .28 (.96) M = 1.287 - .2688 M = 1.0182 rad This is about 58.3 degrees. Mean anomaly is the mean motion times time, and 2π radians divided by the period is the mean motion, n; you mentioned earlier that the period is twelve hours, so I'll use that. Twelve hours is 43,200 seconds, so: 2π / 43,200 = 1.45444 x 10-4 radians per second. 1.0182 rad = 1.45444 x 10-4 s-1 * t 7000.63 seconds = t Your time after periapsis passage at true anomaly of 90 degrees is 7000.63 seconds, or 1 hour, 56 minutes, and 40.63 seconds. The next question is whether you are certain that you want the true anomaly to be ninety degrees. θ of ninety degrees actually gives the location of the latus rectum, which is certainly useful for some purposes, but if you really want the location of the semi-minor axis, then you want E to be ninety degrees, not θ. Of course, that's even easier to solve; you can dispense with the true anomaly completely and can begin with Kepler's Equation. Edit: Oh, why not--let's do it: M = E - ε sin E --- sin of π/2 is 1 M = (π / 2) - .28 M = 1.2908 radians, approximately, or 74.0 degrees 1.2908 = 1.45444 x 10-4 s-1 * t 8874.87 seconds = t This gives a time after periapsis passage of 8874.87 seconds, or 2 hours, 27 minutes, and 54.86 seconds.
  12. Welcome to the forum, @Pryor! Stock KSP doesn't have a speed limit--though in practical terms, the program will crash before you can carry enough rocket fuel to get any appreciable fraction of lightspeed. You'll have to deal with mods. To be honest, what you want is something probably best served by a combination of KSP Interstellar and an outlying-stars pack powered by Kopernicus. KSP doesn't simulate relativity (rockets never normally approach lightspeed) so you're probably out of luck with respect to time dilation. You may want to play a bit with Real Solar System (also powered by Kopernicus) if you want the authentic go-to-α Cen experience, but I don't know whether any RSS-based planet-and-star packs include it.
  13. Zhetaan

    Multi stage help

    @Lankspace: Others have covered theory of construction well enough that I don't need to repeat them. In terms of theory of design, however, I can say that the overall goal is to get something into space such that it doesn't: cost too much, blow up on the way there, fall back down later, and fail the mission. Usually for real-life space programs, cost is the deciding factor. There are a lot of missions that don't fly because they cost too much. For KSP, the practical issue of getting something to space--and making it stay in space--intact is more important. The main difference between the two designs you gave as examples involves drag versus thrust: a rocket with boosters has more of both. Sometimes, that's needed, and the reason for that is because getting off the ground requires you to fly in what would otherwise be the most inefficient way possible. I will assume that you know about how to manoeuvre in space. If not, then please accept that you generally want to change your orbit by thrusting in the prograde/retrograde direction, with normal added if you need to change inclination. Radial burns (at right angles to both prograde and normal but meaning roughly toward or away from the planet) are the least efficient because they don't add much useful velocity to your rocket and don't add any inclination at all. There are a couple of very specific use cases for them but normally only as corrective manoeuvres. When you launch, however, your rocket begins by going straight up--which is to say, radial out. Because of both the atmosphere and the gravity, it is necessary to include a radial-out component in your launch burn. Obviously, radial-out keeps you from falling down to the surface, but it also gets you out of the atmosphere, which otherwise parasitically drains prograde velocity. Spaceplanes get this radial-out component from lift, which is what makes spaceplanes potentially so efficient, but wingless rockets cannot take advantage of that. However, with the right rocket design, you can immediately begin adding prograde velocity at launch in such a way that gravity and drag cancel the radial part at the correct rate to match the declining need for radial velocity--that is to say, you want gravity and drag to work against you so that when you don't have any radial velocity left, you don't need it anymore, either. That's the idea behind a gravity turn. Ultimately, your choice for a rocket design is tied to how well it will fly to orbit. 'Well' can be defined as a matter of cost, weight, thrust, or a combination of these and other factors. One thing to remember is that because of the way gravity works, you will only need to provide a specific value of thrust to counter it, so anything above that value is thrust that you can use for other things, such as going to orbit. Side boosters provide that extra thrust (which means, practically, that you can tip the rocket a bit more on your way to orbit), but that also has you going faster in the atmosphere which increases drag and heat and wastes fuel--there's also such a thing as too much thrust. On the other hand, a long inline rocket will fly very well through the atmosphere--provided that it is pointed in the same direction that it flies. Also, with less thrust generally available, a larger proportion of it must be used to counteract gravity. This requires a longer gravity turn and wastes fuel--there's such a thing as too little thrust, as well. You are always going to 'waste' fuel to fight gravity in any surface launch; the object is to find a design that wastes less of it.
  14. Zhetaan

    Departing Mun from Polar Orbit

    Correct. And giving credit where it's due, that is the same thing that @GoatRider said, but with a different mathematical definition. I will caution you that because it is a polar orbit, you will have some inclination with respect to Kerbin; that's unavoidable. However, as I said before, it's probably going to be less than a degree unless you're in a really high orbit.
  15. Zhetaan

    Departing Mun from Polar Orbit

    There is an efficient way to do it without needing to correct inclination. Mainly, you need to understand that the object of a good return is to leave the Mun in a direction that is parallel to the Mun's own retrograde vector. That's easy enough to see when everything is coplanar: burn on the prograde side to raise the apoapsis on the retrograde side so that you leave along a path that ultimately shoots out the back side of the Mun's sphere of influence. There is nuance to that in the form of ejection angles and other tricks, but that's the idea. It can be done in three dimensions as well, but there are, as these things go, fewer solutions. In other words, it will take time. The key is to wait until your polar orbit's axis points through Kerbin. Then eject when your rocket is above the prograde point of the Mun as normal. This will happen twice in any given Mun orbit (of Kerbin--not your rocket's orbit of the Mun). However, this isn't an interplanetary transfer window, either; there's a lot of room for error. Remember that your polar orbit of the Mun is from Kerbin's perspective equal to the Mun's orbit of Kerbin plus minor perturbations: everything is a matter of scale. Your polar orbit of the Mun, if you were to leave the Mun at the exact north pole of its sphere of influence, constitutes an inclination of a bit over eleven degrees with respect to Kerbin. Close polar orbit of the Mun is less than a degree. Imagine that, for example, you wanted to go from Jool to the Mun: that trajectory is essentially the same as going to Kerbin until you get very close to (or even within) Kerbin's sphere of influence. With scale in mind, you can get a satisfactory Mun return from quite a wide selection of axis angles.