PakledHostage

I agree with Jarnis. Kerbin -> Mun slingshot towards Minimus -> Slingshot out of Kerbin influence to interplanetary trajectory is doable. It is also possible to fly a gravitationally assisted trajectory past the Mun to Minmus and back to Kerbin with very little expenditure of fuel. See the Mun/Minmus Gravity Slingshot Challenge for details. It is not possible to slingshot off Kerbin when starting from an orbit within Kerbin's SOI. Orbital slingshot manoeuvres work because they transfer momentum from the planet to the spacecraft. Kerbin is at rest relative to objects in orbit about it, and therefore Kerbin has no momentum to transfer.

I am no good at 3D modelling and I don’t know enough about Unity to make visual re-entry effects, but I did code a little module to compute the heating effects predicted by the method described in this thread. I have posted it over in the add-ons section of this forum. As was discussed upthread, I didn’t include radiation heat transfer in this calculation because the purpose of this exercise was to create a re-entry heating model that has somewhat realistic behaviour, with minimal computational overhead. Also, at least in the case of Earth re-entry, radiation heat transfer actually cools the re-entry vehicle rather than heats it. I’ve attached screenshots showing the raw numbers predicted by these calculations for a re-entry from low Kerbin orbit. As expected, re-entries from higher orbits (i.e. higher speed re-entries) and steeper re-entries result in more significant heating. But they only result in increased heating to a point. I also tried a high speed straight down re-entry from a 70000 km apoapsis, and although the shock temperatures reached upwards of 4500 °C, the heat shield temperature didn’t rise above about 1200 °C. This may actually make sense though, because the capsule was only exposed to maximum heat for a very short period of time. Much like in the case of fried ice cream or waving your hand through a flame, you can expose something to high temperatures for brief periods without causing significant heating. In the straight down re-entry case however, the resulting g-loading was off the scale. Maybe a proper re-entry model should also consider g-loading. A steeply re-entering spacecraft may not be destroyed by heat because it isn’t exposed long enough, but it would be destroyed by the extreme g forces. The_Duck kindly provided me with links showing the work that Nova has already done to develop a re-entry heating plug-in (those links are broken in the new forum), but Nova doesn’t seem to have finished that work and I suspect that he’s got other priorities now that he’s working with Squad. If anyone reading this wants to collaborate on developing a re-entry heat module, please send me a PM. (Or feel free to use the library that I posted in the add-ons area in your own project, if you like what I’ve done here but prefer to work alone.) PH.

A few months ago, I did some work developing a physics model for re-entry heat that would behave in a way that is representative of reality, while using minimal computational overhead. The thread can be found here. I finally got around to coding it, but I am no good at 3D modelling and I don’t know enough about Unity to make visual re-entry effects so I didn’t bother. Instead, I am posting the basic physics module here in case anyone wants to incorporate it into their own project. And if anyone reading this wants to collaborate on developing a re-entry heat module, please send me a PM. Note that this isn’t intended as a useable re-entry heat module so I am only posting the raw source code here. Feel free to incorporate it into your own projects though. Also I think a proper re-entry physics model should account for excessive deceleration as well. Maybe the spacecraft should be destroyed if the g-meter reaches the top of the red zone on the navball? I've included some screen shots below that were taken during a low-angle re-entry from low Kerbin orbit. The data displayed in the re-entry heat UI was calculated using the code library that I've attached to this post:

I agree that both have their advantages but I enjoy KSP more. Some people diminish KSP as a "lite" alternative to Orbiter, but the two are really quite different. Orbiter is a space simulator while KSP is really a physics sandbox. In KSP, we can number crunch to our heart's content. We are given enough information about the performance of our engines and the aerodynamic drag of our spacecraft that we can predict a lot of things very accurately if we want to. Add to that what we are able to figure out about Kerbin's atmosphere (and the Kerbalverse in general) through experiment, and we can do some pretty cool things. Further, the limited instruments in KSP require us to be creative if we want to navigate accurately. Unless you're using an addon tool such as MechJeb, you need to hand-fly your missions. Doing that accurately requires you to either be a pretty good "seat of the pants" pilot, or good at doing calculations. I get more satisfaction from playing KSP because it doesn't have all of the instrumentation and autopilots that are available in Orbiter. I know I am not alone.

Thanks everyone for the explanations! I only first tried out v0.16 late last week so I missed the initial discussions about the fuel bug. When I orbited my first rocket in v0.16, I ended up with more fuel than expected left over. I attributed the difference to overly conservative assumptions... Maybe they weren't so conservative after all? But something else I did notice during my own test firings was that the new Isp system seems to result in a variation in fuel flow vs. static pressure rather than thrust vs. static pressure (either could vary with static pressure to yield the Isp values quoted in the VAB). Intuitively, it makes more sense to me that thrust should vary with static air pressure due to back pressure on the engine's nozzle, while fuel flow would be constant for any given throttle setting. Can anyone confirm this, or is there another physical explanation for why it is the way that it is in the game?

Interesting reading: XKCD: Is there enough energy to move the entire current human population off-planet?

Great writeup! I Laughed, I Cried! It was better than Cats!

I didn't do just one orbit of Kerbol and I didn't stop at any of the bonus waypoints, but I'm going to submit an entry anyway. I setup my escape burn such that I'd have an apoapsis equal to Kerbin's orbital distance and a periapsis of 7.0855 million KM. This results in an orbit that is about 1 hour shorter than 2/3 of a Kerbin year. Three trips around this orbit finds you arriving at apoapsis just slightly ahead of Kerbin after 212 days in space, with Kerbin closing at about 1575 m/s. A few small orbital trim manoeuvres are all that are then needed to re-enter. These are my screenshots: Returning to Kerbin after 2 years away] Total delta-V expended from initial 103 km high parking orbit through re-entry ~1300 m/s.

Nice job. Looks better than my first attempt. I haven't done much in Kerbol orbit lately but back in January I had a bit of fun trying to plan my missions to minimise the fuel required. I eventually managed it with just a stock RCS tank and thrusters to get me home. I've been looking forward to when planets are added since that time. My first missions to those planets will be robotic missions that will study the planet's physical parameters and atmosphere before I send Kerballed missions there. Here are some screen shots from my January mission into Kerbol orbit with just stock RCS thrusters to get me home. (I’ll have to try it again with the new parts in v0.16): Orbital trim maneuver of +4.8 m/s upon crossing Kerbin\'s SOI into inter-planetary space resulted in this orbit about Kerbol Approaching Kerbin rendezvous after 212 days in space. I sized my orbit so that my spacecraft would complete 3 orbits in 5 hours less time than it took Kerbin to complete 2 orbits of Kerbol. This put me just slightly out in front of Kerbin as I neared Kerbol apoapsis, with Kerbin closing at aprox 1500 m/s. Jettisoning my trusty RCS tank and nozzles prior to re-entry

I'm probably one of the last of the "regulars" to try out v0.16 but I finally got to it this evening. And I figured that since so much has changed with the stock components, I would start by just trying to get to orbit. I didn't want to launch just any rocket though. I wanted to reach orbit on a fuel budget. I did some calculations and figured I could do it on 4 tanks with a two stage rocket, but it turned out that I was overly conservative in my assumptions. On my second mission, I replaced a full LFT in the first stage with a small LFT. Those 3-1/2 tanks were enough to get me into orbit and back to a landing on the KSP compound. I am sure someone can do better but I'll post my mission highlights anyway... These are my screenshots: Landed after 4 trips around Kerbin Edit: Fixed a broken link to an image

Thanks, UmbralRaptor, for confirming my and coolMathew\'s results through your own analysis and experimentation! CaptainArbitrary, does UmbralRaptor get a prize?

You\'re right, but you\'re kinda 'picking nits' by pointing it out... I chose to write gkerbin instead of g0 because I felt there could be an equal number of people who\'d be confused either way. The value used for g0 is, after all, arbitrary and was chosen merely as a convenient convention. There\'s nothing magic about it. And ultimately, the values are the same on Earth and Kerbin. If I made an error, it was writing the equation for Ve in terms of Isp. I should have written the equation for Ve in terms of the more fundamental mass flow rate of fuel and thrust. Isp is somewhat immaterial to this discussion anyway. But while I\'m here, I thought I\'d add another example calculation of delta-V, this time for a two stage rocket. For a multiple stage rocket, you just add the delta-V for each individual stage, treating the upper stages as payload for the lower stages. For the rocket shown below: Stage 1: fuelled mass = 8.35 tonnes mass of fuel burned = 1.05 tonnes Ve of LV-T30 engine = 5682 m/s (ignoring the effect of the small difference in fuel density between the two LFTs) delta-V of the lower stage is 763.6 m/s Stage 2: fuelled mass = 4.3 tonnes mass of fuel burned = 2.2 tonnes Ve of LV-909 engine = 5682 m/s delta-V of the upper stage is 4072.0 m/s Total: delta-V of the entire stack = 4836 m/s Interestingly, if you reverse the order of fuel tanks on the stack shown above (using the large tank on the lower stage), you get a total delta-V of only 4135 m/s (again, ignoring the effect of the small difference in fuel density between the two LFTs). That\'s enough of a difference that the first stack can comfortably make orbit while it may not even be possible to reach orbit with the second one!

You caught me. It was totally arbitrary... I used the last value that was still in my spreadsheet. I was too lazy to change it because I judged that it was sufficiently close to the Mun\'s SOI boundary that it wouldn\'t make much difference anyway. A further savings of one kerbalgram gets us closer to the easiest to accomplish (given our limited instruments) 'burn towards 90 degrees at 0 degrees elevation on the navball' method, but it still doesn\'t match even that less than optimal result. Maybe if the Mun was further from Keribn (so the SOI was bigger), we\'d do better with the bi-elliptic method than the 'burn towards 90 degrees at 0 degrees elevation' method. Maybe we should do a sample calculation for a similar approach to Minmus? I wonder if the results would be different there due to Minmus\' larger SOI? Unfortunately I won\'t have time to do it myself though. I\'m away on vacation for 2 weeks. I\'m probably going to miss all the fun of the v0.16 release. Work out the bugs for me, will you guys?

You calculate it. From the VAB specs for the LV-909 engine: Thrust = 50 kN at max throttle Fuel flow = 2 litre per second at max throttle Fuel density = (2.2 tonne/500 litre) = 4400 kg/m3 mass flow rate of fuel = 8.8 kg/s Isp = Thrust/(mass flow rate of fuel * gkerbin) = 579.4 seconds Effective exhaust velocity (Ve) = gkerbin * Isp = 5682 m/s Note 1: Because the fuel flow and thrust both vary linearly with throttle setting, the effective exhaust velocity for the LV-909 engine is the same for all throttle settings. Note 2: The effective exhaust velocity of the LV-T30 is the same as the LV-909 because the ratio of thrust to maximum fuel flow is the same for both engines. Note 3: I\'m using SI units for the values given in the VAB because it makes the results of these and other calculations match the actual performance of our in-game rockets. Note 4: The fuel density actually changes, depending on which fuel tank you use. I\'m using the fuel density for the FL-T500 fuel tank in these calculations.

Most scientific calculators have a 'ln' button and a 'log' button. In the case of those calculators, 'log' is log base 10, while 'ln' is log base 'e'. e is a mathematical constant much like its more famous cousin Pi. If you\'re doing this calculation using a scientific calculator, use the 'ln' button If you\'re doing this calculation using Microsoft Excel, type '=LOG(5.5/2.9)' in the cell. Note: Excel\'s log function returns the natural log. You\'d need to specify '=LOG(X,10)' if you wanted log base 10 of X.

The Wikipedia articles an the Tsiolkovsky rocket equation and Specific Impulse both give good background to your question plus the equations you\'re looking for, but here\'re a couple of sample calculations: Example 1: Q. What is the total delta-V of a small single stage stack having a parachute, MK1-Pod, decoupler, Large LFT and a LV-909 LFE? Spacecraft mass including fuel (Mfuelled) = 5.1 tonnes Fuel mass =2.2 tonnes Spacecraft empty mass (Mempty) = 2.9 tonnes LV-909 effective exhaust velocity (Ve) = 5682 m/s DeltaV = Ve*ln(Mfuelled/Mempty) DeltaV = 5682 * ln(5.1/2.9) = 3208 m/s Note that 'ln' is the natural logarithm. Example 2: Q. How much delta-V has been used up if the stack from Example 1 shows 480.0 kg fuel remaining when the tank is right-clicked with the mouse? First, the 480.0 'kg' value shown in the popup window (KSP v0.15) is mislabelled. It should be labelled as a unit of volume. The popup window shows 500 kg for a full tank but the specs for the large LFT in the VAB show it as having a volume of 500. The actual mass of fuel burned is 2.2 tonnes * (500-480)/500 = 0.088 tonnes. Spacecraft starting mass including fuel (Mfuelled) = 5.1 tonnes Mass of fuel burned =0.088 tonnes Spacecraft final mass (Mempty) = 5.012 tonnes LV-909 effective exhaust velocity (Ve) = 5682 m/s DeltaV = Ve*ln(Mfuelled/Mempty) DeltaV = 5682 * ln(5.1/5.012) = 98.9 m/s Edit: Added Example 2.

It was a long time ago, but does this count: Kerbals Walk on Mun The link above announced my first ever Mun landing. I flew all of my early missions to the Mun (Kerballo 8, 10, 11, etc) in real time because it increased the level of commitment. As I pointed out back in December, it would suck to spend 8-10 hours getting to the Mun only to crash or get stuck in Munar orbit because of lack of preparation (we didn\'t have persistence files back then, so there were no 'do overs'). I\'m no Neil Armstrong, so my heart was racing when I touched down sucessfully... This is my favorite screen shot from my Kerballo 11 mission:

I just did some number crunching and it looks like the bi-elliptic transfer can be more efficient than a simple Hohmann transfer, but the case I tried is not as good as any of the results on the leader board. My sample calculation predicts an efficiency that is roughly the same as burning normal to the velocity at 2150 km from the Mun to lower the hyperbolic periapsis to 3 km (my 'Test 2'). I imagine it would be hard to do much better because we\'re restricted to a maximum semi-major axis length by the Mun\'s SOI radius and a minimum semi-major axis length by the Mun\'s surface altitude. Here\'re the numbers: Bi-elliptic transfer insertion: Speed at hyperbolic periapsis at 901 km = 463.6 m/s Speed at periapsis of 901 km x 2138.5 km orbit = 283.6 m/s (requires -180.0 m/s delta-V) speed at apoapsis of 901 km x 2138.5 km orbit = 133.5 m/s speed at apoapsis of 3 km x 2138.5 km orbit = 66.7 m/s (requires - 66.8 m/s delta-V) Delta-V required to close down orbit to 3 km x 3 km circular orbit at periapsis = -202.0 m/s Total delta-V = 448.8 m/s Corresponds to ~183.6 kg fuel remaining Hohmann transfer insertion: Speed at hyperbolic periapsis at 901 km = 463.6 m/s Speed at apoapsis of 3 km x 901 km orbit = 135.7 m/s (requires -327.9 m/s delta-V) Delta-V required to close down orbit to 3 km x 3 km circular orbit at periapsis = -169.6 m/s Total delta-V = 497.5 m/s Corresponds to ~176.7 kg fuel remaining

I don\'t know much about bi-elliptic transfers either. I did read in the Wikipedia article on bi-elliptic transfers that: The reference given in the Wikipedia article for the above quote is: Vallado, David Anthony (2001). Fundamentals of Astrodynamics and Applications. Springer. p. 317. ISBN 0-7923-6903-3. At 1102 km for our initial semi-major axis and 203 km for our final semi-major axis, the ratio of final to initial semi-major axis is only 5.44:1. Much less than the 11.94:1 quoted above. It is probably worth doing a test to be sure, but maybe we can\'t do much better than the current record?

It is nice to see new members who share my enjoyment of pre-planning missions. Welcome to the forums! Can you provide more details about your planned trajectory? You seem to be using a hyperbolic Munar escape trajectory and then bleeding off the excess later in the flight. Have you tried planning your Munar escape burn using an elliptical trajectory with apoapsis outside the Mun\'s SOI, instead? I calculated (and later verified with an actual mission) that 185.1 m/s Delta-V was sufficient to transfer from a 50 km circular, prograde orbit about the Mun to a ~100 km orbit about Minmus. Delta-V for a typical plane change manoeuvre would add between 0 and about 30 m/s to that if done correctly, and depending on the timing of the window.

I ran four more tests, all of them burning towards 90 degrees (like in Tests 1, 2 and 3 and as shown in the image above), except I varied the angle of the heading reticle in the navball from 65 degrees above the horizon (retrograde) to 30 degrees below the horizon (roughly normal to the inital velocity vector). I\'ve plotted the data below: Test 4: - Burn toward 30 degrees above horizon starting at 2150 km altitude above Mun. - 224.2 kg fuel remaining after Pe lowered to 3.6 km. - 189.0 kg fuel remaining after closing down orbit to 2.9 km x 3.4 km - Net delta-V expended 410.6 m/s Test 5: - Burn toward 15 degrees above horizon starting at 2150 km altitude above Mun. - 227.2 kg fuel remaining after Pe lowered to 3.5 km. - 189.5 kg fuel remaining after closing down orbit to 2.9 km x 3.1 km - Net delta-V expended 407.1 m/s Test 6: - Burn toward 15 degrees below horizon starting at 2150 km altitude above Mun. - 227.3 kg fuel remaining after Pe lowered to 3.6 km. - 185.4 kg fuel remaining after closing down orbit to 3.2 km x 3.3 km - Net delta-V expended 435.8 m/s Test 7: - Burn toward 30 degrees below horizon starting at 2150 km altitude above Mun. - 225.2 kg fuel remaining after Pe lowered to 3.4 km. - 180.5 kg fuel remaining after closing down orbit to 2.9 km x 3.3 km - Net delta-V expended 470.2 m/s Note that burning normal to the spacecraft\'s velocity results in a burn that starts at about 35 degrees below the horizon and ends at about 20 degrees below the horizon. The variation during the burn is as a result of the spacecraft turning.

My plan is in line with yours, aceassasin. I will send robotic probes to orbit the planet and land on its surface before I send a Kerballed mission. I want to learn enough about the new planet and its orbit before sending Kerbals that I can build an efficient lander and maybe even work out a cycler trajectory for my kerballed flight. The cycler trajectory will give them a free trip home if their orbital insertion stage fails to ignite when they get to the new planet. I will probably also place the lander and booster for the return trip into orbit prior to launching the Kerbals on their historic mission. I want to be sure that they\'ve got a ticket home.

Having thought about it a bit more since yesterday’s post, I think that there’s some further testing that could be done to follow up on my three tests. First of all, Test 2 could be done again using MechJeb or by someone who’s a better pilot than me to determine if the difference between Test 2’s results and Test 3’s results are due to experimental error or are a real effect. And even if it isn’t a real effect, then it would be interesting to compare the two methods accurately using MechJeb. Maybe the method used in Test 3 is just more practical in that it is easier to achieve using the instruments that we’ve got available? Some more tests could also be done, burning at a range of angles in the plane of the hyperbolic trajectory, between retrograde and normal to the spacecraft’s velocity. If we had data from that range of tests, then we could plot it like I did in my post from last week. Maybe there’s an optimum point that hasn’t been found yet? I\'ll try to fit in a bit of experimenting. And finally, I agree that my description of Test 3’s burn direction wasn’t very good. I struggled to come up with something that wasn’t too wordy. I ended up leaving out a couple of critical ones. I should have just posted a picture: Better late than never… I can sketch up a diagram too if people think it would help.

I ran three tests from within the Mun\'s SOI this afternoon. None of them improves on the current records, but I\'m sharing them because they are potentially interesting: Test 1: - Burn retrograde at 2150 km altitude above Mun. - 209.2 kg fuel remaining after Pe lowered to 3.2 km. (delta-V expended = 271.3 m/s) - 179.1 kg fuel remaining after closing down orbit to 2.8 km x 3.2 km (delta-V expended = 208.8 m/s) - Net delta-V expended 480.1 m/s Test 2: - Burn normal to spacecraft\'s velocity (but in the Mun\'s orbital plane) at 2150 km altitude above Mun. - 224.5 kg fuel remaining after Pe lowered to 3.3 km. (delta-V expended = 168.0 m/s) - 179.8 kg fuel remaining after closing down orbit to 3.0 km x 3.4 km (delta-V expended = 307.2 m/s) - Net delta-V expended = 475.2 m/s Note: I am not using MechJeb so I didn\'t do a perfect job of holding a burn heading normal to the spacecraft\'s velocity. This may have affected the results of this test. Test 3: - Burn towards Mun\'s centre of gravity, in the plane normal to the view direction towards the Mun (i.e. towards 90 degrees on the navball) at 2150 km altitude above the Mun. - 227.9 kg fuel remaining after Pe lowered to 3.6 km. (delta-V expended = 145.3 m/s) - 188.4 kg fuel remaining after closing down orbit to 3.3 km x 3.5 km (delta-V expended = 269.5 m/s) - Net delta-V expended = 414.8 m/s I would summarise my observations from this and the other challenges that we\'ve all participated in as follows: An efficient transfer from Kerbin to Munar orbit and landing starts with a well planned TMI burn. If contingency for in-flight failure includes initial insertion into a free-return trajectory, then that free-return trajectory should have as low a Munar Pe as possible. If the Munar orbit insertion requires lowering the Munar Pe, then the burn to lower the Pe should be made towards the Mun\'s centre of mass, in the plane normal to the current view direction towards the Mun (i.e. heading of 90 degrees on the navball)1. The earlier this burn is made after entry into the Mun\'s SOI, the less fuel will be required. The Munar orbit insertion burn should occur as late as possible however, as this maximises the window during which a failure could occur and still allow the mission to take advantage of the Munar flyby abort trajectory. Due to the conflicting pressures of minimising fuel burn and maximising safety contingencies, a balance must be found. Presumably the MOI burn will occur within the Mun\'s SOI. The Munar orbit insertion burn should then occur as close as possible to Pe to maximise the benefits of the Oberth effect. Orbital trim manoeuvres should then be carried out as required to circularise the Munar orbit. The DOI burn should be done 180 degrees from the planned landing site, and Pe of the descent trajectory should be as close to the surface as is safe. Final approach to landing should be done using a reverse gravity turn. Any comments about or additions to the above? 1 I realise it is a bit of a stretch to draw this conclusion based on just three tests, but intuitively it makes sense that this is the most efficient means of lowering the hyperbolic trajectory\'s Pe. Maybe I\'ll try to find some time to do the analysis to verify my intuition. But for the time being, the experimental evidence suggests that it is an efficient method.

I got 'nerd sniped' by a friend last summer (v0.9 days). He waltzed into my office and told me about this game that had kinda crappy graphics, but that was lots of fun. The game has come a long way since then and I\'m still hooked.