OhioBob

In most cases, a few days one side or the other of the optimum departure isn't going to make a big difference. The big exception I think is Moho, which has an orbital period of just 102 days. Let's say we miss the ideal ejection time by 10 days, then Moho has moved 35 degrees from the optimal phase angle. This is quite significant and will result in a suboptimal transfer, likely costing several hundred m/s in ÃŽâ€v. Don't forget that it's just not the ejection burn that matters, it's the orbit insertion on the other end as well (unless you're doing an aerocapture). The penalty that you pay for a suboptimal transfer may come in the orbit insertion rather than the ejection burn. One technique that can be used to control the timing of the burn is to adjust the orbital period so that you arrive at the right place at the right time. What I've done is to identify the location on my orbit where the ejection burn will take place several orbits from now*. When I reach that spot I compare my orbital period to the "time to ejection burn". I then perform a short burn to change my orbital period so that the number of orbits remaining until the ejection burn is some whole integer. In this way I'll be exactly where I need to be when it comes time to perform the ejection burn. * One tricky part of this is that you must recognize that the 'angle to prograde/retrograde' that you want for your burn several days from now is not the current angle. Let's say we're orbiting Kerbin and we want to perform an ejection burn 20 days from now that has an angle to prograde of 100 degrees. In the 20 days between now and then, Kerbin will revolve 17 degrees around the sun. Therefore, the location of the maneuver node at the present time is 100 - 17 = 83 degrees to prograde. - - - Updated - - - the more I love it. At least when I rearrange it to: Vbo2 = Vesc2 + V∞2 This method is so much more direct and easier than the other methods I've used to figure out IP transfer costs. Srsly, thank you for bringing this to my attention. Yes, I too find it very useful. It, along with the vis-viva equation and Tsiolkovsky rocket equation, is one of my most frequently used equations.

Another potentially big problem with performing the transfer from a high orbit is the timing. The burn has to be performed when the 'angle to prograde' is correct. Minmus has an orbital period of over 56 days. When the ideal time for the ejection burn comes, we could be on the opposite side of Kerbin and 28 days away from where we need to be. Being off by that much means we're likely to miss the launch window altogether. And even if we're off by only a day or two, it still means we're performing a less than optimal transfer. - - - Updated - - - I agree that's likely the root of the misunderstanding.

Hyperbolic excess velocity, as I explained it, it the theoretical velocity that a spacecraft would retain at an infinite distance (i.e. the real life definition). As you correctly describe, this isn't completely relevant to KSP because of the use of patched conics. Likewise, the mathematically derived definition of escape velocity isn't applicable to KSP either. To "escape" in KSP we just have to raise the apoapsis of the orbit until it touches the sphere of influence, thus it's possible to escape with what is technically an elliptical orbit. In KSP, the residual velocity that a spacecraft retains after crossing the SOI is greater than the calculated value of V∞. The "effective V∞" is simply the spacecraft's velocity relative to Kerbin as it crosses the SOI. Similarly, "effective Vesc" is the velocity needed to raise the apoapsis to exactly the radius of the SOI. The relative velocity that we require at the SOI is the difference between the orbital velocities of Kerbin and our planned interplanetary transfer orbit. - - - Updated - - - One advantage of performing the burn from a high orbit is that the spacecraft will traverse a much smaller sweep angle in the same unit of time. This means longer burns are possible from a high orbit with fewer detrimental effects, such as loss of ejection accuracy. It's therefore possible to use a smaller engine and lower TWR from high orbit.

That's the same method of Isp calculation that Streetwind gave, which is incorrect. Specific impulse is given by, Isp = F / (á¹Â*go) When we have two different engines with different Isp, we must set F equal to the total thrust and á¹ equal to the total rate of mass flow. That is, Isp = ΣF / (Σá¹Â*go) Mass flow rate is found by simply rearranging the Isp equation, á¹ = F / (Isp*go) Substituting the above for á¹Â, and cancelling out go, we get Isp = ΣF / Σ(F/Isp) On the other hand, your (and Streetwind's) equation can be written as, Isp = Σ(F*Isp) / ΣF Let's take your example but with a small change. Since the Isp you have chosen are so similar, the two methods while yield nearly the same number. Instead of 305 s for the Isp of the boosters, let's call it 250 s. Your method yields, Isp = (1500*310+430*250)/1930 = 296.63 s The correct method yields, Isp = 1930/(1500/310+430/250) = 294.27 s The difference is not huge, but it is certainly significant.

The Mobile Lab itself must be occupied by two Kerbals for it to work. However, the Mobile Lab is not a command pod and cannot control your vehicle. You must also have a command pod or probe core. I'm not certain if you can move your crew from the command pod to the Mobile Lab, leaving the command pod unoccupied, and have your Mobile Lab work (assuming no probe core). Even if you can do that, I wouldn't recommend it. Having your command pod unoccupied will mean you'll have no way to control your vehicle (you'd be adrift). I'd recommend a three-kerbal crew so you can move two to the Mobile Lab and leave one in the command pod. Of course if you have a probe core then you don't need a crewed command pod.

Nice explanation, LethalDose. It is from this same equation that we derive hyperbolic excess velocity. (Note that "mu" is the Greek letter, μ, where μ = GM.) If r1 and v1 are the distance and velocity of the spacecraft when it is near the planet, and r2 and v2 are the distance and velocity of the spacecraft when it is far from the planet, then, since energy is constant, we have v12/2 - μ/r1 = v22/2 - μ/r2 To obtain the hyperbolic excess velocity we want r2 to be some distance far from the planet. As r2 approaches infinity, -μ/r2 goes to zero, therefore we have v12/2 - μ/r1 = v22/2 Multiplying by 2 we get v12 - 2μ/r1 = v22 Escape velocity occurs when we give the spacecraft just enough kinetic energy to overcome the negative gravitational potential energy, that is Vesc2 = 2μ/r Therefore we can substitute as follows v12 - Vesc2 = v22 Since v1 is the velocity near the planet we can call this the burnout velocity. And since v2 is the velocity at some infinite distance, it represents the hyperbolic excess velocity. Thus the final equation is Vbo2 - Vesc2 = V∞2

That's an excellent point. It should take only about 160-170 m/s to escape Minmus and establish an orbit with a periapsis close to Kerbin. Let's say we end up in a 46,400 x 100 km orbit; when we reach periapsis we'll be traveling 3153 m/s. However, the burnout velocity required to produce our transfer orbit to Jool is no different than we calculated before. At an altitude of 100 km we still require Vbo equal to 4232 m/s. Therefore, the ÃŽâ€v of our ejection burn at periapsis is only 1079 m/s.

The solar orbit that one requires to reach a particular planet at some particular launch window is largely defined by the hyperbolic excess velocity of the escape trajectory, i.e. the velocity leftover after the spacecraft reaches an infinite (hypothetically) distance from Kerbin. Excess hyperbolic velocity is given by the equation, V∞2 = Vbo2 - Vesc2 where V∞ is hyperbolic excess velocity, Vbo is the burnout velocity, and Vesc in the escape velocity. In your example we have a ÃŽâ€v of 1986 m/s from a 100 km orbit. At 100 km orbital velocity is, Vorb = SQRT(GM/r) = SQRT(3.5316E+12 / 700000) = 2246.1 m/s And escape velocity is, Vesc = SQRT(2GM/r) = SQRT(2 * 3.5316E+12 / 700000) = 3176.5 m/s The burnout velocity is the orbital velocity plus the ÃŽâ€v, Vbo = Vorb + ÃŽâ€v = 2246.1 + 1986 = 4232.1 m/s Therefore V∞ equals, V∞ = SQRT(Vbo2 - Vesc2) = SQRT(4231.12 - 3176.52) = 2795.0 m/s If we now move out to an orbit of 47,000 km, V∞ remains (approximately) the same; only Vorb and Vesc change. We now have, Vorb = SQRT(3.5316E+12 / 47000000) = 274.1 m/s Vesc = SQRT(2 * 3.5316E+12 / 47000000) = 387.7 m/s The burnout velocity is, Vbo = SQRT(V∞2 + Vesc2) = SQRT(2795.02 + 387.72) = 2821.8 m/s And the required ÃŽâ€v is, ÃŽâ€v = Vbo - Vorb = 2821.8 - 274.1 = 2547.7 m/s As you can see, the above isn't exactly the same as the number that you obtained, but it clearly confirms that it does indeed cost more ÃŽâ€v to launch from the vicinity of Minmus than it does from low Kerbin orbit. The burnout velocity out near Minmus is considerably less but, since you're starting orbital velocity is so much lower, the ÃŽâ€v is more.

It's not possible to have anything close to a "pure Hohmann transfer" to Moho unless you intercept Moho when it is at one of it's nodes. These ideal alignments happen only rarely. What you did was to manufacture the conditions for a Hohmann transfer by using very costly means. I don't consider what you did to be a Hohmann transfer at all. A perfect Hohmann transfer to Moho looks like this: Apoapsis is the launch point and is located at Kerbin, periapsis is the intercept point and is located 180o opposite the launch point. Since the launch point is Kerbin, and since Kerbin lies in the ecliptic plane, the launch point must be one of the nodes of the transfer orbit. The other node must be 180o opposite, which is the intended intercept point. Therefore, intercept must take place when Moho is at one of its nodes. A true Hohmann transfer is possible only when Moho will be at one of its nodes at the time of intercept, and when that node is 180o opposite of the launch point. These conditions are very rare. It's probably never possible to perform an absolutely perfect Hohmann transfer, but with the right conditions you should be able to get close. You just have to be patient and way for those right conditions to arrive.

It's my understanding that they use a pressurized bladder system; however, I'm not exactly sure how it is configured. You surmise that the fuel is inside the bladder with the pressure squeezing the outside of it. I envision a system that would have a gas pressurized bladder expanding into the tank as the liquid is drawn down. (edit) Here you go... http://www.space1.com/Artifacts/Apollo_Artifacts/SM_RCS_Tank/sm_rcs_tank.html Quote: "Helium pressurizing gas pushed against flexible bladders in the fuel and oxidizer tanks to force the weightless propellants out of the tanks." The link also included a diagram of the tank construction. As I guessed, the gas inflates the bladder and it expands into the tank.

I don't know anything about that. If somebody wants to write an old aero mod, then go for it. However, I think retaining an old aero option in v1.0 is a really bad idea.

In that case I say no. The current model has no basis in reality. I see no reason to keep around something that was just a place holder put into an early version of the game.

What is #lolaero?

I like collecting science. I prefer that to having to complete a bunch of contracts to collect funds. The next time I start a new game it will likely be a Science game rather than a Career game. But, hey, that's just me. BTW, it's would have or would've, not would of.

That's where you have to use something like a ÃŽâ€v map to plan ahead to see what the next maneuver will be.

I outline my design method in the following post: http://forum.kerbalspaceprogram.com/threads/107763-Designing-Launch-Vehicles?p=1678756#post1678756 The guidelines I give are for stock aero. If you use NEAR/FAR, then I would disregard. For very small launchers, say those for payloads of 5-10 t, I rarely get payload fractions greater than about 0.14. Almost all of my mid-sized launchers give a payload fraction right around 0.16. For my larger designs, say for payloads ≥75 t, the payload fraction is usually about 0.18. Of course these payloads fractions are the maximum that can be achieved. I'm often lifting a payload that is a little smaller than what a particular rocket can theoretically lift. Cost varies considerably depending largely upon how many solids I use. My cost can vary anywhere from about 1500 to 3000 funds per tonne of payload. The median cost for my mid to heavy launch vehicles is about 2100 per tonne of payload (not counting the cost of the payload itself). I prefer smaller rockets because very large ones can become difficult and sluggish to control. The biggest launcher that I've saved as a stock design is capable of lifting 93 t. Anything bigger than that and I start using multiple launches and in orbit assembly and refueling. - - - Updated - - - I generally like to design my upper stage TWR based on the ÃŽâ€v of my ejection burn. I try to limit the duration of my burn to about 5 minutes so that my sweep angle is not too large. If I have to deliver only 1000 m/s ÃŽâ€v, then my average TWR can be at low as 0.34. On the other hand, if I have to deliver 3000 m/s ÃŽâ€v, then my average TWR must be 1.02. It can be hard to obtain a TWR that high with big payloads using LV-N engines, so I've been know to break my rule on occasions. As Yakky said, when you have a high ÃŽâ€v and a low TWR, you can always do multi burns through successive periapsis passages. - - - Updated - - - That's close but incorrect. The steps should be: 1.) Sum up the thrust of all your meta-stage's engines 2.) Calculate Thrust / Isp for every engine 3.) Sum up the results of 2. 4.) Divide the result of 1. by the result of 3. In other words, Isp = ΣF / (ΣṠ* g)

With the current engines in the game, surface Isp on Eve would be atrocious if the curve were allowed to extend all the way to 5 atm. Take a LV-T45 for instance... it's Isp at 5 atm would be only 120 s. However, those engines are configured for use on Kerbin, not Eve. By having special purpose, low expansion ratio engines designed specifically for Eve, a respectable Isp could still be achieved, perhaps only 5-10% lower than the sea level performance of current engines. Of course, such an engine would have very poor vacuum performance. For example, an engine designed for Eve might have an Isp of 270 s at 5 atm, but it's vacuum Isp would barely be any better, say 280 s. These engines would only be used during the early stages of ascent. Once the atmospheric pressure drops to close to 1 atm, you'd stage and switch to using the current engines.

Sometimes yes, sometimes no. What you want is the encounter that has the smallest relative velocity between the spacecraft and the planet/moon. One way to do this is to make the angle formed by the crossing of the orbits as small as possible. Generally, if the planet/moon is in the part of its orbit where it has passed periapsis and is approaching apoapsis, then you want to encounter the planet/moon before your spacecraft reaches apoapsis (called a Type I transfer). If the planet/moon is in the part of its orbit where it has passed apoapsis and is approaching periapsis, then you want to encounter the planet/moon after your spacecraft has passed apoapsis (called a Type II transfer). The reason for this is that if both are approaching apoapsis, then both have a positive flight path angle. And if both have past apoapsis and are approaching periapsis, then both have a negative flight path angle. By keeping the flight path angles both positive or both negative you reduce the relative angle and, therefore, the relative velocity. If the target planet/moon has a perfectly circular orbit (such as Mun), then it makes little difference whether you use a type I or type II transfer, the magnitude of the relative velocity will be the same, it will just have different direction. Of course a type I trajectory always has a shorter travel time than a type II.

50,774.6

I just kept burning retrograde until I killed nearly all my velocity (both horizontal and vertical). I then free fell from about 7-8 km AGL until I started a final breaking burn at about 2.5-3 km AGL. The stage I used for final breaking had a TWR of about 3.5. This higher TWR stage was also used at the very end of the suicide burn, killing the last 250-300 m/s. Using this method my total landing ÃŽâ€v was about 3400 m/s. My ascent/reorbit ÃŽâ€v was right around 2640 m/s, for a total of 6040 m/s. I'm sure that if I used a higher TWR for the majority of the suicide burn, and started from a lower altitude, I could get down to your 5900 m/s or maybe even a little less.

I just successfully completed my first manned launch from Eve. It took what essentially amounted to eight stages, giving me a total ÃŽâ€v of about 12.8 km/s. In simulations I determined it could achieve Eve orbit from an elevation as low as 500 m, though I launched it from about 900 m. From that elevation I got into a 120 km orbit with 200 m/s to spare. All I launched was a Kerbal in a command seat and my total liftoff mass was 83 tonnes (the landed mass was 91 tonnes). The total delivered to orbit was about 320 kg. I also did an unmanned practice launch using a smaller design (also capable of 12.8 km/s). The unmanned version delivered a little over 200 kg to orbit with a launch mass of 51 tonnes. In both cases I'm getting a payload fraction of only about 0.004. There are probably others who have done better than that, but the point is, it takes a huge vehicle to launch a small mass from Eve. Each additional m/s of ÃŽâ€v is very, very hard to attain. (edited to add) The 12.8 km/s ÃŽâ€v is based on vacuum Isp. It's more like 12.1 km/s in actual practice.

Important point. Performing a suicide burn at periapsis of an elliptical orbit is really no different in terms of ÃŽâ€v than first circularizing and then performing a suicide burn. The only difference is that in the first case you burn right through the point where the orbit is circularized, while in the second case there is a pause in the action. The advantage or circularizing first, as Warzouz said, is that you can then select any site along your orbital path to land. In the other case you're pretty much committed to landing at the site of the initial periapsis. I think that's true of everyplace but Tylo. Orbital velocity around Tylo is much higher than any other airless body, which means that the duration of the suicide burn is much longer than other places. Unless you have a very high TWR, this can result in a considerable loss of altitude during the burn. The initial altitude must be high enough to account for this. For my one and only Tylo landing, the stage I used for the suicide burn had an initial TWR of only about 1. I started the burn at an altitude of about 47 km and was down to about 8 km when the burn ended. If I had to do it again, I'd consider increasing the TWR.

I generally classify as "light" any launcher made exclusively of 1.25 m parts. As soon as I start adding 2.5 m parts I classify it as "medium". When I start adding 3.75 m parts it becomes "heavy". I consider "super-heavy" to be launchers that start adding additional 3.75 m parts attached radially. When you say "typical mass" do you mean the total launch mass or the mass of the payload (i.e. lift capacity). Although there are no precise lines of demarcation, my light launch vehicles generally have lift capacities of 5 tonnes or less; my medium launchers lift between 6 and 30 tonnes, my heavy launchers between 30 and 80 tonnes, and my super-heavy launchers >80 tonnes. In regard to total launch mass, light is approximately <40 tonnes, medium 40-200 tonnes, heavy 200-450 tonnes, and super-heavy >450 tonnes.

I use the TT-38K whenever I can because it is cheaper and lighter than the others. In cases where I need more room, or I want to extend my footprint for greater stability, I use the TT70. In the past I used the Hydraulic Detachment Manifold for heavy attachments, but I've found that using the TT-38K or TT70 with Separatrons and extra struts is usually lighter and can even be cheaper in some cases. (I think the mass of the Hydraulic Detachment Manifold needs to be rebalanced to make it more viable.) I've never used the Structural Pylon.

My launch vehicles I save as subassemblies and I'm not too original at naming them. I just give them a number designation to signify their lift capacity. For example, my 30-tonne launch vehicle is simply named LV-30. I'm not much more creative when it comes to naming my robotic payloads. I typically just name them to indicate the destination and mission type, i.e. Mun Rover, Minmus Lander, Duna Orbiter, etc. I also include a Roman numeral (or a number/letter combination) to indicate the number in a series, or the generation of spacecraft. My manned missions is where I use more colorful names. I generally name them after constellations or stars. I haven't done much with stations and bases other than to deploy them just to fulfill contracts. I generally name stations using Greek letters, i.e. Alpha Station, while bases I name after the planet or moon, i.e. Eve Base I. I've been thinking of trying to come up with a more formalized naming system for all my payloads, but I haven't settled on anything specific yet.