Jump to content

Search the Community

Showing results for tags 'eccentricity'.

  • Search By Tags

    Type tags separated by commas.
  • Search By Author

Content Type


  • General
    • Announcements
    • Welcome Aboard
  • Kerbal Space Program 2
    • KSP2 Dev Updates
    • KSP2 Discussion
    • KSP2 Suggestions and Development Discussion
    • Challenges & Mission Ideas
    • The KSP2 Spacecraft Exchange
    • Mission Reports
    • KSP2 Prelaunch Archive
  • Kerbal Space Program 2 Gameplay & Technical Support
    • KSP2 Gameplay Questions and Tutorials
    • KSP2 Technical Support (PC, unmodded installs)
    • KSP2 Technical Support (PC, modded installs)
  • Kerbal Space Program 2 Mods
    • KSP2 Mod Discussions
    • KSP2 Mod Releases
    • KSP2 Mod Development
  • Kerbal Space Program 1
    • KSP1 The Daily Kerbal
    • KSP1 Discussion
    • KSP1 Suggestions & Development Discussion
    • KSP1 Challenges & Mission ideas
    • KSP1 The Spacecraft Exchange
    • KSP1 Mission Reports
    • KSP1 Gameplay and Technical Support
    • KSP1 Mods
    • KSP1 Expansions
  • Community
    • Science & Spaceflight
    • Kerbal Network
    • The Lounge
    • KSP Fan Works
  • International
    • International
  • KerbalEDU
    • 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 3 results

  1. eccentricity has units of degrees instead of no units in the celestial body information panes. https://imgur.com/a/9Jl8gOq
  2. Hi Everyone, I'm having geeky fun with my HP-50g calculator, keying in all sorts of equations found on the Wiki pages. I know almost nothing in algebra/mathematics, but isn't KSP a nice place to learn? I'm on Entroper's Basic Orbiting (Math) Wiki page and there's something I don't get. If I know the altitudes of my apo- and periapsis, I can solve for the orbital speeds at the apsides with the orbital speed equation velocity = body_radius * sqrt( 9.80665 / ( body_radius + altitude ) ) which would yield one velocity for the apo- and another for the periapsis. Right? Now, further down the tutorial page, there are equations useful to solve for numbers in elliptical orbits, and they use a variable l (lowercase L) for the angular momentum. It is said that "l = m * r * v at apoapsis or periapsis", where I understand that m means the orbiting vehicle's mass, r its orbit radius (body radius + altitude) and v its orbital velocity. It says "at apoapsis or periapsis", so I understand value of variable l should be equal whether it is calculated at apo- or periapsis, but, using the two values for v found using the velocity equation discussed above, I don't get equal numbers. Are they not supposed to be equal? What am I not understanding?
  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'.
  • Create New...