Gaarst

Because using as strap-on boosters would be more complicated and useless. It would require more engines than it currently has and more complicated designs around them. Having liquid fueled boosters of this size is just throwing money away (LF engines are expensive) and the USSR was never fond of SRBs. What is hard to realise when only having KSP as a reference for rockets, is that in real life, fuel and oxidiser tanks are separated. You don't just have a bit of both here and there as in KSP, but you have one oxidiser tank, and below or above one "fuel" tank. The Proton is a special case. The core of the first stage is an oxidiser tank (NTO) while the 6 side tanks are fuel (UDMH). It was designed this way for a simple reason: to fit in the trains. 4.1m is about the largest diameter that the Soviet railroads could transport making stages with a greater diameter harder, and therefore more expensive to move to Baikonur (the Proton is built by Khrunichev which has its facilities near Moscow). It turns out that 4.1m is the diameter of the Proton core, having the fuel tanks mounted on the side allows the rocket to fit on the railway (the side tanks are transported separately and assembled at Baikonur where they don't have this problem) and therefore allows it to be bigger without increasing its diameter and having to rebuild half of Russia's railway.

If you have the apoapsis and periapsis, finding the SMA and eccentricity is rather easy. The semi-major axis is given by: a = (Rap + Rpe) / 2, where Rap and Rpe are the distances to the "center" of your orbit at apoapsis and perisapsis. Note that the apoapsis height and periapsis heights that the game gives you do not include the size of the body you're orbiting (they are measured from the surface rather than from the center); to obtain the proper distances, you need to add the radius of the body you are orbiting (600,000m for Kerbin, 261,600,000m for the Sun). From then, the eccentricity is given by: e = (Rap - Rpe) / (2a) The time to periapsis is trickier as it would be a value for a body in your target orbit, not the orbit itself (I'm realising I was wrong saying that the calculator would be enough for this and I apologise ). Forget all the above! The launch window calculator is helpful if you want to plan an encounter with a body (ie: be at a given position at a given time to encounter your target), but in your case it is not because it will not tell you the dV required to perfectly match your target orbit so the launch window planner will be utterly useless! So, now time for actually useful stuff. What you want is to combine an inclination change with a Hohmann transfer orbit. Note that for Hohmann transfer orbits, we will be assuming that your start and final (target) orbits are circular because otherwise the maths becomes horrendous (even more than what's below). So if your target orbit is highly elliptical you're basically screwed. The easiest way to do this is to use velocities vectors (they allow us to consider angle and magnitude tied together and avoid a lot of trouble with equations). A simple cosine rule shows that the result for a single maneuver is: , where vi and vf are the initial and final velocities and θ is the inclination change. Since we want to completely change orbits (ie: do a full Hohmann maneuver + inclination change) we need two of these equations, put together, giving the total delta-v required from a sun orbit: ; let's consider the two terms separately to understand the equation: the 1st term: all velocities are those at the point of the first maneuver, vi is the velocity of the initial orbit and vt,a is the velocity of the transfer orbit the 2nd term: all velocities are those at the point of the second maneuver, vt,b is the velocity of the transfer orbit and vf is the velocity of the final orbit θ is the inclination difference. The inclination change is split in the two maneuvers with a weighing coefficient s, this is so to do the transfer more efficiently s is found by solving a transcendental equation (which by definition you can't solve) but we can find an approximation when θ = π/2: Now, remember that we are starting from Kerbin, therefore, considering ΔvS is a bit wrong: the first maneuver will be done in Kerbin's reference frame, and the Δv required will therefore be different. Using the law of cosines (again!) we find that the excess hyperbolic velocity at Kerbin after the first maneuver is: , which is really familiar! Since we start in orbit of Kerbin, we can consider that vi = vK (Kerbin's orbital velocity about the Sun). v∞ is then equal to the Δv required for our first maneuver, about the Sun. Using a bit of orbital mechanics, we can now write the Δv for the first maneuver done starting in a circular orbit with radius r around Kerbin: , note that here we are using µK Kerbin's standard gravitational parameter which is different from the Sun's standard gravitational parameter denoted as µS (we will need it later). This first maneuver must be done in the direction of Kerbin's motion (pro- or retrograde) at an angle α relative to Kerbin's equatorial plane such that: Using the vis-viva equation and assuming that the target orbit is circular, we can express our different speeds (vt,a, vt,b and vf) in terms of distances to the Sun's center: Finally, after replacing values and rearranging a bit, we can express the total Δv needed for our mission: With: This isn't pretty but (hopefully) correct. Note that since we have assumed that most of our orbits were circular, the different orbital values you mentioned earlier are not important (you can fine tune your orbit after doing the bulk of the maneuver); the only important thing is that you have to do your maneuvers at the ascending or descending nodes of your target orbit, ie: eject when Kerbin crosses your destination orbit's plane.

It works in this case. Search a bit.

There is a bug report here. The issue seems to be linked with the highlighting FX, deactivating it in the options should remove the stutter. Squad should be aware of that and a fix should be in 1.2.1. (Info coming from the bug report itself)

Try it in flight. I'm using Real Fuels and the engineer report always says that the RCS are not fed by monopropellant; except they don't use monoprop anymore but hydrazine or whatever fuel. So even if they work in flight the engineer report says they don't. Remember to have ServiceModule type tanks for RCS though as they are pressure-fed and therefore require pressurised tanks.

Might be a bit severe for the poor moderator

Yeah you're right, there are not dots when you didn't comment and there are no new posts. Never really noticed.

Blue star = you commented on it and there are new posts Grey star = you commented on it and there are no new posts Blue dot = you did not comment on it and there are new posts Grey dot Nothing = you did not comment on it and there are no new posts That's how I got it so far.

The thread mentions "real life rocket"; I hereby suggest Elon Musk's delusional hallucination (a.k.a. ITS) to be banned from this thread.

I appear to be missing a post in a thread. When browsing activity, I see the thread with the little star meaning I commented on it, but there are no posts I wrote inside the thread. It might have been deleted by a mod, but I don't remember writing anything purge-worthy recently (and haven't got any notification, private or in the thread, about breaking rules. Has anyone seen this before? Just to know if it's a forum software bug.

Inb4 someone gets to orbit with a 100% decoupler rocket

Thank you for correcting me, I did mean accepted and not completed. If you did launch after you accepted the contract, then I'd guess it's that they need separate launches. Are you sure it isn't written in the very small lines at the bottom of the contract? (The ones you never read and realise too late that your LKO station needs to accomodate 1257 Kerbals)

For this kind of missions, I think you need at least one (maybe both) vessels to be launched after the contract was completed accepted. I don't know if they need to be launched separately, but I'd assume so. Otherwise, you could just abuse it: accept contract, undock and redock your Mun space station, and complete the contract in 2 minutes.

Yes. The Earth's escape velocity at its surface is 11.2km/s. Neglecting atmospheric effects and remembering that gravity is a conservative field this means that a craft going over 11.2km/s on the Earth's surface in any direction (though going downwards might be problematic) will eventually escape Earth's attraction. Since your VTOL can't magically reach 11.2km/s in an instant, you need to accelerate to this speed in a given time. Going vertically means that you will constantly be fighting gravity face off, and this will make you lose dV depending on the time you spend fighting it: gravity losses are non-conservative since they are not a function of path/position; imagine a craft with 1 million km/s of dV but a surface TWR of 0.99, even though you should have enough dV to escape Earth (and probably the local galaxy cluster at this point) you will never get off the ground. A craft with infinite TWR needs 11.2km/s of dV to escape Earth's gravity from its surface, a craft with <1 TWR needs infinite dV to escape Earth's gravity from its surface, your VTOL will stand somewhere in the middle. Note that you don't need to actually reach 11.2km/s at any point to escape Earth's gravity. If you go up steadily at say 10m/s, you will eventually reach a point where the escape velocity (which is inversely proportional to the square of your distance to Earth) becomes less than 10m/s and you will have effectively escaped Earth's gravity. This is also the reason why you don't need to break physics to escape a black hole's horizon, just rocket engines. Due to gravity losses I mentioned earlier and considering that you will spend a lot of time going up at a fixed speed (so needing continuous 9.81m/s2 of acceleration from your engines at the surface, a bit less going up) this method is very inefficient dV wise, which is why we don't do it in real life: we just accelerate a lot in LEO and follow a ballistic trajectory to wherever we want to end up. 5% of the speed of light is a lot. The Kerbal sun's escape velocity at its surface is about 100km/s, here we're talking 15000km/s, that's 150 times as fast. At this point you can just use a ruler to plan your trajectories. Planning an encounter going at this speed is fairly easy. Draw a straight line between you and where your target will be when you reach it (even though you're going very fast, stuff moves so you can't just shoot point blank at other planets, no matter how fast you go) and you're done. If you do have a craft able to reach 5% of c and stop afterwards, then planetary alignement is just a matter of how long your trip will be (and trying to avoid going through the Sun to reach a planet in conjunction). To calculate in which direction to go you need some basic maths. Assume your target is at a constant distance D from your departure point (unless you're aiming for another galaxy, the phase change between your planet and the target will be small in the time you need to reach it, so constant distance). To travel this distance D at 5% of c you will need a time t = 20*D/c. In this time the planet will have travelled a small fraction of its orbit, this displacement θ is equal to: θ = t/T*360° with T the target planet's orbital period. Therefore, you need to aim at the planet's current position plus an angle θ (relative to its orbital motion). Under the spoiler is the calculations done for the Kerbal system's planets. Note that: the angles are really small, so it was OK to assume constant distance. But don't assume they are zero! If you do, you'll miss each body by a few dozens of radii (except Jool, which you'll still miss by 3 radii though) going from 25 radii for Eve to 244 for Dres the further away a body is, the smaller the angle. This is because the orbital period of a body is proportional to its SMA to the power 3/2, therefore the period "rises" quicker than distance, and the angle is smaller

I meant the bug tracker. (If Laythe was gone said bug tracker would probably be DDoSed by countless issues) The bug tracker has switched to 1.2 since the version was released.

Did you even go there? It's literally at the exact same place it has ever been. Oh, and:

Scroll down a bit, Downloads, and click on the zip: RasterPropStuffStuffStuff.zip (not the source code, unless you want to have fun compiling the mod yourself).

Using a Latin non-English keyboard is not hard. Just remember to change your key bindings from Qwerty to Qwertz/Azerty or whatever (I just copy paste my settings.cfg file in my new installs out of laziness now). For the translation, I have no problem with English even though it's not my first language and I'm used to play KSP in English, so TBH even if there was a translation, I'd probably stick to English. Although I still think translations are a good idea, they open up the game to new people who may see English language as a barrier. The translation should be decent of course: not only in terms of languages (typos, grammar and such) but also for the meaning of words, to avoid losing the jokes (in part descriptions for example) or simply to keep the astronautical vocabulary accurate.

This could make a fun thread for the Forum Games. Edit: oh yeah, and I have stutter too. A lot worse than anything I ever experienced so far, even with modded installs.

Now I have it on all my saves. Every second, the game stops for a second. On any size of crafts. 1.2 was about to be the first decent release since Beta (didn't get to enjoy 1.1 before 1.1.3 because of random crashes) but actually no. It is currently heading the same way 1.0 and 1.1 did. I seriously hope the issue is on my end.

I'm back on this thread, on the other side this time. Unbelievably annoying stutter every couple seconds when in space/reentring with a 5 parts pod (just started a career). Only mod installed is KER. I experienced stutter before, but on heavily modded installs and never to that extent. My other 1.2 saves didn't have this problem.

1.1 had a pre-release too.

Yes, it's stable. A bit too much for me. Hopefully, 1.2.1 will come and fix this.

I use sarcasm when I can, but often not in a derogatory way. If what I wrote sounded condescending at some point, know that this was/is not my intent at all. The position of Kerbin in its galaxy is hard to define. Looking at the skybox, there does appear to be a direction in which the galaxy appears larger. Especially, there seem to be a lot of gas clouds, which could be nebulae; large concentrations of nebulae are typical of spiral galaxies bulges. Therefore, I'd situate the Kerbin system in a rather small spiral galaxy (everything is too small, so why not galaxies?), maybe a bit closer to the centre than the Earth system is (bulge appears "taller", so it could mean it's closer). About the dense planets, I don't know why they are so dense. Neither do I know why the Kerbin sun is so not-dense (we really need an antonym for dense), as smaller stars are typically denser than larger ones (larger are heavier, so hotter, so more radiative pressure, so things pushed outwards stronger, so less dense star, I assume). Being ignorant is no flaw. Being unwilling to learn is one. My guess is that Laythe used to be a Europa-style moon, with a big layer of ice coating it. For some reason, there was a gravitational disturbance which sent it flying close to Jool; very close. So close that the gravitational pull of Jool almost ripped it apart (tidal forces). This immense strain on the planet's core could have been sufficient to heat it enough to trigger intense volcanic activity, heating the planet. The heat could have been enough to melt its ice layer, releasing gasses that formed an atmosphere. The problem with this theory is that I have no idea of the actual number involved, especially if a major tidal event could have such effect. I know that tidal heating is a thing (tidal forces exert forces on a planets interior, heating it by friction) but I don't know if it can happen at such scales. Io is very similar to this case (tidal forces of Jupiter and the other 3 Galilean moons heat it, giving it an intense volcanism) but it is still far from being as hot as Laythe. Though Io's tidal heating is regular, while what I described is an exceptional event, this could be enough to justify the difference, but it also means that Laythe will eventually freeze again. (Note that tidal heating is the main argument for Europa having a liquid ocean below its surface, though we're still far from having a surface ocean and a thick atmosphere) Funny enough, I read it, and happened to use the same example in my explanation.

I had some free time.