Jump to content

Search the Community

Showing results for tags 'inclination'.

  • Search By Tags

    Type tags separated by commas.
  • Search By Author

Content Type


  • General
    • Announcements
    • Welcome Aboard
  • Kerbal Space Program 2
    • KSP 2 Discussion
    • KSP 2 Suggestions & Development Discussion
    • KSP 2 Dev Diaries
    • Show and Tell
  • Kerbal Space Program
    • The Daily Kerbal
    • KSP Discussion
    • KSP Suggestions & Development Discussion
    • Challenges & Mission ideas
    • The Spacecraft Exchange
    • KSP Fan Works
  • Community
    • Player Spotlight
    • Science & Spaceflight
    • Kerbal Network
    • The Lounge
  • Gameplay and Technical Support
    • Gameplay Questions and Tutorials
    • Technical Support (PC, unmodded installs)
    • Technical Support (PC, modded installs)
    • Technical Support (Console)
  • Add-ons
    • Add-on Discussions
    • Add-on Releases
    • Add-on Development
  • Making History Expansion
    • Making History Missions
    • Making History Discussion
    • Making History Support
  • Breaking Ground Expansion
    • Breaking Ground Discussion
    • Breaking Ground Support
  • International
    • International
  • KerbalEDU Forums
    • KerbalEDU
    • KerbalEDU Website

Find results in...

Find results that contain...

Date Created

  • Start


Last Updated

  • Start


Filter by number of...


  • Start



Website URL



About me



Found 9 results

  1. I am not sure of how to match my own orbit's inclination with the orbit that is shown on the map by a contract. I had to send a satellite into a specific orbit which was shown on the map. I had no problems with the apoapsis and periapsis (except for matching the exact spot for the ap and pe, I just eyeballed it, would appreciate a hint here as well) but when it came to matching up the inclination I was really struggling. The contract said the desired orbit had an inclination of 170 or so degrees but relative to what? I also had the argument of the periapsis and longitude of ascending node, just didn't know what they meant (still don't know). If I aligned the map/ planet with the orbit it looked like about 290 degrees. If I launched and put the navball on 170 I got nowhere. Can someone tell me how I can launch at the correct angle or sort of figure it out? And also how to adjust my orbit too once in the air, like you would do for a moon (using maneuver nodes doesn't work because I can't set an orbit as a target so I'm not sure how much normal of delta-v to add/ subtract)? In the end I just had to eyeball things by aligning the planet with the orbit and stuff like that which was fine but still way to much hassle in comparison to other things in this game. Also if you add normal or anti-normal the contract orbit tells you your ascending and descending nodes but maneuver nodes would be useful in this situation. I have Kerbal Engineer installed so I think there has to be a better way of doing these sorts of missions just not sure how. Would MechJeb help in this situation? Thanks in advance for any answers although I'll probably thank you again later
  2. Hi folks! As far as I've looked, I couldn't find anything like this, so here goes... I've tried to create an Excel workbook to help you calculate how much Δv a craft needs for different manoeuvres, including going to a stationary or resonant orbit, hohmann transfers and inclination changes. I'm Dutch so my OS (and by extension, Excel) is Dutch as well, but it should work fine on any other language. Here are some screenshots of the first release: https://imgur.com/a/AMqom Time for the first release! Hope you guys like it! I strongly advise you to read the instructions. Since my native tongue is Dutch I hope the translations make sense. If you have any advice for me to make things clearer, please don't hesitate to tell me. LINK! Changelist: How to use the sheets: This workbook is most suitable for, though not limited to, both stock and OPM installations of KSP. Calculate stationary/resonant orbit Use this sheet to calculate at which altitude your craft/satellite/etc. needs to be at, when you want it to stay above a specific point on the surface of the celestial it's orbiting. You can also calculate resonant orbits. In combination with the other calculations, it can help you figure out how much Δv your craft needs. Δv Hohmann Manoeuvre Use this calculation when you want to know how much Δv your craft needs to change orbit altitudes. This can come in very handy when building your craft. Δv Change inclination (e≈0) Use this calculation when you want to know how much Δv your craft needs to change its inclination. This can come in very handy when building your craft. However, this calculation method is only valid if your orbit is (close to) circular. If the eccentricity is too much, use the next sheet. Δv Change inclination (e<0) Use this calculation when you want to know how much Δv your craft needs to change its inclination AND it has an orbital eccentricity that's not (close to) 0. This can come in very handy when building your craft. (You can still use this calculation when the eccentricity is 0, but it’s not worth the hassle.) Free Sheet You are free to use this sheet any way you wish; to store your own values and/or make your own calculations. Things to take into account when working with these formulas - For the inclination change calculations, you need to use your judgement to decide whether to use the e≈0, or e<0 variant. - All the calculations assume you wish you change only that for which the calculation is suitable, nothing else. If you wish to change apsis and inclination at the same time, you’re out of luck, because that’s not included in this workbook and you can’t simply add up the results of formulas. It’s way more complicated than that. Disclaimer This workbook can help you out in KSP (and maybe some other games/simulators). Keep in mind though that as thorough and accurate as I have tried to be, errors may still have slipped in. I (the creator of this workbook) will not be held responsible for ruined games. Having said that, of course I would be sorry if something went wrong because of an error in this workbook. Would you kindly point it out to me in this thread. I’ve deliberately put a lock on all the worksheets (except the ‘free’ one), so the cells containing essential information and formulas are protected from accidentally selecting and changing them. I’ve also provided the password for each of them, in case you do want to change something. If you’d like to see anything added and/or changed, please ask me in this thread. I can and will not make any promises, except for taking a look. Please don’t spread this workbook elsewhere without consulting me first, thank you very much!
  3. For just about every journeys there are questions that need to be made. Just about all of us have taken off from Kerbin with 10k dV in the Moho's orbital direction and finding out, when we go close to Moho that we did not have enough fuel left to circularize (or some other nice oversight like pointing the solar panel in the direction of the sun for 6 months). Can we sit down with a spreadsheet and make decisions that factor the choices that are presented. I decided to basically write these posts because of the eccentricity thread in order to illustrate what the real value of eccentricity is when all is said and done. To make the answer short, energy is much more important, but eccentricity gives us on the fly information. For example if you are circularizing from an eccentric orbit close to Pe or Apo (whichever is the burn point) and your delta-e/t is too low compared to time to pe/apo, this informs you that you need to increase thrust or should have carried a more powerful engine. Another situation is during reentry from distal targets, the delta-e/t tells you how rapidly or effective your entry-theta was. If your ship is overheating and your e is not likely to approach zero at some point during reentry then you probably should have used a bigger shield(or kept some retro fuel) and choosen a steeper entry angle (lower no-ATM Pe). The eccentricity argument has an effective range of 0.0005 to 1.0000 below or above which are meaningless in the game. In this case an eccentricty of 0.5 = 0.4995 to 0.5005. This differs from other stats such as dV which are accurate over a 100,000 fold range in the game, altitudes are accurate to >10 decimal places. IOW, the values used to derive e are much more precise than e itself. Does it make a difference, yes and no, theoretically if you had a TWR = infinity, an exactly angle to prograde for a perfect burn (with perfect thruster control) there is no wasted dV and you end up intersecting the minimum orbit of the target planet. In reality the dV calculated at best puts the craft in a range were RCS thrusters in the departure orbit can be used to get within about 5000 meters of the perfect arrival orbit (on a good day). However knowing energy makes some logical sense of what is going on, for example by the Oberth effect works, why burn from low orbit, why use kicks on lowTWR craft in low orbits (versus spiralling away from the celestial). When we are using e for on-the-fly decision making accuracy is not really an issue, however in the formulation of travel strategies we do want to use as accurate as possible starting information. So what about everything else? The procedure is this. Step one. For a target planet orbital ap _and_ pe (meaning two parallel analyses), assign a departure and arrival altitudes relative to kerbol, transform to radius, derive a. Step two. Assign u/a (Escape energy) and u/2a (SKE at a) Step three. Assign SPE changes from kerbin-to-a and from a-to-target. Step four. Assign deltaSPE changes (changes in Kinetic energy) from a to kerbin or target. Step five. Calculate SKE at kerbin or target. Step six. determine dV required to achieve kerbin or target orbits without entering kerbin or targets SOI. Step seven. determine the SKE at planets SOI entry or exit based on step six. Step eight. Add this to planets minimum orbit radius escape energy, this give energy to reach minimum orbit around the planet and free fall to planet. Step nine. Convert this to dV required to free fall from minimum stable orbit Step ten. Subtract the circular orbital velocity from freefall at minimum orbit dV requirement. Step eleven. Add the two dV (kerbin and target planet) together and get total dV. At 6 specific points in the 11 step process unique energy parameters were used to derive decision making information. The table below compares the Total dV (m/s) cost of intersecting orbits (values rounded for clarity) and also compares to inclination dV performed in circumkerbol orbit. Planet Target Intrcpt δV inclination at Apo at Pe dV at a Moho 4724 4001 723.1 1818 (Depart from Kerbin at the Kerbin-Moho inclination node closest to Moho's Apo, inclination nodes are priorities) Eve 2911 2913 002 400 (Eve's orbital inclination nodes are priority) Duna 1928 3009 1081 78 (Depart from Kerbin close to Duna's Pe) Dres 2819 4837 2081 466 (Depart from kerbin closest to Dres's Pe, inclination nodes should be also considered) Jool 5202 5686 484.0 79 (more analysis of satellites requires) Eeloo 3416 3449 32.7 386 (Eeloo's orbital inclination nodes are the priority). As we can see above the analysis is devoid of any consideration of the e parameter, although it is easily obtained from the information we have. How can we get those pesky inclination nodes. One way is to place a satellite in a Kerbinesce orbit at theta = 2/3 pi and 4/3 pi relative to kerbin (in the same orbit as Kerbin but at maximum distance. Then target a planet, the nodes will show up also relative to kerbin. Such satellites can have a dual function since one can also place a deep space array on the satellite. That allows communication to objects that current orbit is on the other side of Kerbol. [Another set of energy and dV calculations that involve the equation The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. in which we create an orbit a which is 2/3 or 4/3 the period of kerbin by generating a periapsis or apoapsis (respectively) whose average with kerbin orbit gives us a]. Disclaimers. Above assumes two sets of reciprocal process, launch/ascent & descent/land and transfer commence and complete that have a boundary at the minimum stable orbit of both planets. 1. There is a simpler mechanic, single step "star trek" mechanic, in which your ship has so much power that you travel from departure x,y,z, vx, vy, vz to arrival x, y, z, vx, vy, vz before either of which have significantly changed position (arrival point and departure points are oriented to each other). Given that human life could never survive the dV/dt and the dV & TWR required do not exist this scenario can be disregarded and physically impossible and should not be considered. Deep space probes such as New Horizons need not depart into a circular orbit. That if they reach an angle to prograde of theta>090' while still in upper atmosphere while at the same time burning the dV required for a highly eccentric orbit does not require circularization and may use less dV. In such a burn the base assumption is that the best delta-SKE on hill sphere exit is obtained from the lowest altitude burn even if that burn starts at a suborbital trajectory with pe always below safe circular orbit altitude. To convert this to KSP the ascent angle is say inclination is at >15' and you are at alt=45 you simply can burn to Eeloo Apo intersect as you are crossing some theta~90 and have a large plume of overheated gas momentarily burning through the fairings before they are deployed. 2. KSP provides perfect examples, kerbin is in a zero inclination, zero eccentricity orbit about kerbol and as it so happens it common departure planet or arrival planet for most of the transfers. The argument of periapsis differs for all other planets so you are not going to be able to match a Apo/Pe angle of a planetary departure with 180 degree Pe/Apo of a planetary arrival except from Kerbin because kerbins orbit a=Pe=Apo. This has some relevance for the Eeloo, Jool transfer which is damn cheap relative to traveling back to kerbin and traveling to Jool (given some dV spent on plane matching). 3. The smallest-sweep area stable orbit about a planet may be unpreferential. Jool being the example. This infers there is complex decision making involved in getting to a system moon in which the planet is not the target. mechanical thermodynamics can be used to burn less than the amount needed to circularize at the planets rMin, that dV would be used to circularize at an apo that intersects the moons orbit. In the case of Jool, all three inner planets have a=pe=apo, so this is not too much of a problem. In these instances you want to compare intersecting the moons orbit directly (using a different planetary Rx,orbit) versus a hohmann transfer to 200k Jool-altitude and a partial circularization burn to intersect the target orbit and circularize. 4. If we make the assumption that inclination nodes are approximate to r = a (semi-major axis), that the dV required for inclination burn (see table) is low enough not to be a priority. In these cases we can, if we desire burn at a bearing above or below the departure planets equitorial plane on depart to send the inclination node to r = a and get rid of some inclination. In comparing the table below the difference between an Kerbin-Moho transfer Pe-target and Apo-target is delta-dV = 724 but the inclination change dV averages at 1814. Therefore its simply intelligent to set a priority on changing planes over departing theta = Moho's apo theta (fortunately Moho-apo is relatively close to the inclination node). The cost of changing inclination at kerbol is reduced by 100s of dV. The same logic is also true for Eve, and Eeloo. For the other planets a departure window closest to the target planets periapsis is a better choice than choosing a departure window closest to an inclination node. 5. Entry burns particularly on planets like Jool need transfers that seldomly overlap with their pe or Apo, consequently there is a triangulation between time to get good window for efficient inclination change, or close to Jool theta. In other instances like Moho, which is so small oberth effect is minimal, free burn times at pe near a kerbin inclination node is going to occur separately than the moho circularization burn. 6. Depending of kerbol relative altitude of the target the true burn altitude is different from the planets altitude. Our burn starts 670,000 meters closer but a maximum efficiency burns leaves kerbin's SOI at the moment of crafts circumkerbol Apo or Pe (depending on an interior or exterior target). We always want the exit trajectory to be parallel to kerbins path of travel even if the line is not identical with Kerbin, otherwise predicting intercept could be off and correcting dV would be required. This occurs both on kerbin exit and on target arrival. For example a the flat part of the escape curve to moho should generally be at a final angle to prograde ever so slightly more than 180 at kerbin SOI otherwise the Apo for the circumkerbol orbit will occur in the future. This means that the numbers for apo and pe differ slightly relative to the calculation. If the target was exactly one SOI in front or behind Kerbin, the difference would be zero, on an interstellar trajectory that the difference is nominal, from Kerbin 670,000 radius is 0.99995 that of the calculated. On such a trajectory 670000 = 85000000 sin theta, translates to an angle to prograde of is 180.45'.
  4. Hi guys, I'm doing some sat contracts (in RSS) and need some advice. I'm talking, of course, about launching into the appropriate inclination. I know there have been threads about this topic before, but I did pretty extensive searching and couldn't find what I was looking for, so here goes. Since I'm using RSS, I'm launching from Cape Canaveral AFS in Florida, which is at 28.6 degrees. The orbit I'm going for is a tundra orbit 71,000x460km, inclined at 110 degrees. The orbit appears to cross the equator about half way between Ap and Pe. I've noticed though, that the ascending/descending nodes slide around and change value as I time warp on the pad. As a full day passes on the pad, the nodes swing back and forth along the target orbit (~45 degrees), and vary between 88 and 145 degrees. I looked up a launch azimuth calculator and it gave me ~202.9 as a heading, which sounds right to me. I rotated the rocket in the VAB 115 degrees so I'd only have to do a little yaw correction achieve the proper heading. Does all this sound right? I assumed it would make sense to launch when the nodes read 110/-110, but when I tried it, I was off by more than 15 degrees. I then tried launching when they read 138/-138 (110+28.6), but I was off by 40 degrees. So does anyone know when I should actually launch? Am I not taking travel distance during launch into account? How should I calculate this?
  5. Could we finally get inclination data in map view and tracking station? Why was it never added? I do remember people asked about it very often.
  6. Hello, kind sirs and ladies. Does anybody know of any way to launch your vessel into orbit in a pre-calculated, non equitorial, non polar orbit? You see, I'm running Real Solas System, but the reason I chose not to submit this topic into the mods' discussions, is because this question is one about the technicals of ksp. The problem is, the moon in the RSS setup is rotating in a highly inclined orbit. So, to reach it, the most efficient way is to launch into an inclined orbit to begin with. Thus, here's the question: Is there any way for me to pre-calculate the inclination of the orbit I'm gonna launch into, without randomly pointing towards different directions during ascend and watching what the orbit is doing? Any help or advice is strongly appreciated. Cheers!
  7. Hey all, So I recently posted a thread about trans-Mars injections, and some of the responses really got me thinking. Thanks to you all, I've finally learned that transfer orbits do not have to be in the same plane as either the origin or target, so long as you can put the AN/DN at the intercept point. For some reason, this topic seems to be rarely discussed here or in almost all the articles about interplanetary spaceflight I could find. I feel like I've discovered a well kept secret that holds the key to efficient transfers. Anyway, in fiddling around with a maneuver node at the appropriate launch window, I can see how important it is to be as close to the correct ejection inclination (as shown by Precise Node) as possible. Adding a normal/antinormal component to the ejection burn can help a bit, but not by much without adding a hefty amount of dV. So, as in almost every other aspect of spaceflight, launching into the appropriate inclination is crucial. I figured out a pretty good method for placing the LAN where my ejection burn with take place, but I have no idea what inclination to go for. My relative inclination to target is 1.6 degrees, but Transfer Window Planner says I need an ejection inclination of 0.1 degrees. How can I figure out what inclination to launch into in order the get the desired ejection inclination? Am I correct that it will NOT work to simply launch into a 0.1 inclination orbit? Thanks!
  8. Hi I just completed my first contract to insert a satellite into a polar orbit around the sun (apoapsis 82.2Mkm, periapsis 24.8Mkm). This was actually quite difficult as it took approximately 20k delta v just to do the inclination change at the ascending node from Kerbin's orbit to the required orbit of 90 degrees. Normally I try and do inclination changes near the apoapsis but in this instance the required orbit was rather elliptical with the apoapsis being near the southern pole. Any tips on how to make inclination changes more reasonable? (assume I'm launching a new craft from Kerbin). Thanks Stuart
  9. Hey guys, haven't posted in a long time cause I've been completely revamping how I play KSP. Got RSS, Real Fuels and basically every realism mod except RO itself. Anyway, I've got a contract to launch a satellite into equatorial orbit. I've launched other satellites already, including a two polar sats, but this is the first equatorial sat I've tried to launch. In this career game, I'm only launching from Cape Canaveral or Vandenberg, so I'm going to have to make a quite significant plane change maneuver. Minimum inclination I can seem to get from the Cape is around 27 degrees, maybe 26.x if I'm lucky. So here's my question. At the contract orbit (~1,336 km), it takes about 3200-3300 dV to zero my inclination. I know that it is sometimes more efficient to push my apoapsis on the ascending/descending node way out before doing the plane change, so I thought I'd give that a try. I'm wondering what the appropriate AP would be though, or if it would even be worth this extra maneuver. How big of a plane change do I need to be making to make this technique wothwhile? Also, how do I go about calculating the ideal AP for a given plane change? Thanks!
  • Create New...