Gaarst

While torque depends on distance to centre of mass, concluding that putting reaction wheels further away from the CoM should be better is wrong, at least IRL, not sure how KSP handles this. Torque is defined by the application of an external force about a fixed centre, defined by: t = r x F Note that here, letters in bold are vectors, and "x" the cross product, in terms of magnitude, you'll get: t = r.F.sin(A) where A is the angle between r and F. r is the distance vectore from the centre of rotation to the point of application of the force, and F is the force vector itself. This means that for a given force vector you should put your RCS the furthest you can from your CoM for maximal torque, to get r as high as you can. For reaction wheels though this is different: reaction wheels use the principle of conservation of angular momentum. A mass starts spinning inside the wheel, and as no external force is applied the total angular momentum of the ship must be conserved, therefore the ship starts spinning in the opposite direction. Then, for maximum efficiency, the reaction wheel must be placed as close as possible to the natural centre of rotation of the ship: its centre of mass. Note that in KSP, reaction wheels are completely OP, and therefore physics show their limits pretty fast with those.

Be careful not to mix ignorance and stupidity. Ignorant people can be taught, truly stupid ones are often helpless. Though in the OP, the teacher insisting on his/her mistake was probably more a matter of pride than anything else, even unconsciously.

You can make multiple quicksaves by pressing Alt+F5 for saving, and Alt+F9 for loading one of your quicksaves.

The Earth is static in its own reference frame, that's it. A reference frame can't be absolutely static, because that would imply they are not all equivalent and this violates Einstein's first postulate of special relativity. Here, inertial is what is important. - - - Updated - - - A direct consequence of Einstein's first postulate is that, in an inertial frame, there is no way to say if you're moving or not without any interaction with another reference frame. And even then, the inertial frames' displacement is relative to each other. Speed and time dialtion don't matter either because as long as no external force is applied (ie: velocity changing), your reference frame is inertial.

The famous "Twin Paradox" is actually very straighforward to solve in special relatvity: the twin on the rocket ages slower. This is because usual postulates of special relativity are made under inertial frames. Even if you skip acceleration of the rocket, the one on the rocket is in two different inertial reference frames: the rocket going away from Earth, and the rocket going towards Earth. When this change of frame occur, the notion of simultaneity as it was in the first frame (here the twins at the start of the experiment) is lost and redefined according to the second frame. The best way to visualise this is using a Minkowski diagram Doing a little maths then concludes the resolution of the problem. You could also end up with the same result calculating proper time for both twins, and seeing that the stationary twin's proper time is greater. If you want to to consider continuous acceleration, then you'll have to use General relativity and time dilation under accelerations; and again the problem gets solved the same way.

You could calculate with the energy contained in the coke (kcal nutritional value) which, AFAIK, is basically how much energy you will get from drinking and digesting it. From then you could get dV, and all other stuff to make a rocket. I may do the maths tomorrow, if anyone else doesn't do it in the meantime

Do reverted test launches count ? If no, then none (I have 3 Kerbals stuck at Moho since quite a few years, but technically they are still alive). If yes, then probably hundreds, mostly because I don't bother landing test flights properly, I just crash them and revert to VAB.

Asking for a thread to be closed or moved is usually not appreciated by mods...

Approximately 31 million tons (or 10 000 Saturn V), but you won't get anywhere with 170 m/s dV It depends on the distance from the second rocket to the centre of mass of the first one. And ultimately the latter's density.

Use this formula to calculate the Isp of a stage with several engines: The fuel doesn't matter, all you need is the consumption rate or individual Isp. For TWR just divide thrust by mass*g

Oh ! Great, I might even see it from where I am (Scotland). But... Scottish weather...

I wonder how long this post will survive before it gets closed

Mine is somewhat beautiful (others said) but half the letters look exactly the same while the other half do not look like anything at all, so: pleasant to look at, unpleasant to read, is a good summary.

Jeb: Whackjob, like everyone else said Bob: regex, definitely Bill: sal_vager (although he does know what he's doing) Gene: RIC, Alshain Wernher: Nathankell, GoSlash27 It's kinda difficult to nominate only one or two persons per kerbal, there are so many awesome posters out there, and they do not all fit exactly to one kerbal in particular.

Actually, using a few tricks to make my calculator cooperate, I could get the final numbers. So, to put a 1 pound satellite in LEO with a coke rocket you'll need a rocket weighing around 10574 kg, made of approximately the same number of 2l coke bottles. And that is assuming all the structural parts of the rocket and the mentos used are massless.

1 pound = 450g Dry mass is then 500g, wet mass is 2500g. Taking K^2's ejection speed: v = 10 m/s Rearranging the rocket equation for n bottles gives us: M0 / M1 = edV / v M0 is the wet mass of n bottles plus one satellite, so: M0 = n * m0 + 0.450 M1 is the dry mass of n bottles plus one satellite, so: M1 = n * m1 + 0.450 But, as it turns out, the limit of M0 / M1 as n goes to infinity is equal to m0 / m1 = 2.055 / 0.055 = 37.4 In other words, even with no payload and infinitely many bottles, you couldn't get more than 36.2 m/s of dV with a single stage rocket. Now, consider many stages: The delta v of the k-th stage is given by: dV = 10 * ln(Mk,0 / Mk,1) Mk,0 is the wet mass of n bottles plus k - 1 stages above, so: Mk,0 = n * m0 + Sk-1 Mk,1 is the dry mass of n bottles plus k - 1 stages above, so: Mk,1 = n * m1 + Sk-1 Sk-1 is the mass of the k - 1 stages above the k-th stage. Then again, we can see that Mk,0 / Mk,1 tends to 37.4 as n goes to infinite. So, each stage of the rocket can only give 36.2 m/s of dV. If you're a good pilot, then you can get to LEO for 9000 m/s of dV. You'll then need 249 ideal stages to get to LEO. With each stage having infinitely many bottles, but more infinitely many than the previous one (this is actually possible, see transfinite numbers). Having infinitely many bottles on each stage is not very practical, so we won't use ideal stages anymore, but finite stages, which are much more convenient (you'll see that "convenient" is relative here) to work with. We decide we only need 34 m/s of dV per stage, we'll then need 265 stages to get to orbit. We already saw that the delta v of the k-th stage is : dV = 10 * ln(Mk,0 / Mk,1) We can consider Sk as a series (recall Sk is the total mass of the rocket at the k-th stage), then: Sk = ∑ (nk * m0) = m0 * ∑ nk S0 = 0.45 kg nk is the number of bottles needed in the k-th stage to achieve 34 m/s of dV. n1 = 0.45 * C = 32.02, C = (e3.4 - 1) / (m0 - m1 * e3.4) = 71.169 nk = C * Sk-1 Calculating the first few terms of nk and Sk shows us the problem: S3 = 1 427 066 kg In other words, only the first 3 stages of our coke rocket would weigh half a Saturn V. Considering the first few terms of Sk allows us to make the following approximation: Sk = S0 * e4.992k While my calculator doesn't want to give me a result for our 265 stages, it tells me that the 14 first stages would weigh half the mass of our Sun. Also, the first 20 stages would weigh more than our galaxy, the Milky Way; and the observable Universe is most likely somewhere around 25 stages. TL;DR: we would need a lot of bottles.

That could sound very wrong out of context...

If you have the soda effective exhaust velocity, getting the dV of a soda bottle is pretty straighforward. - - - Updated - - - So, using Mentos added into a coke bottle: We can get the exhaust velocity of the coke using the height of the coke "geyser": ve = sqrt(2gh) Assume h = 10m (pretty high for a Coke-Mentos reaction but still realistic), we know g = 9.81 m/s², then: ve = sqrt(2*9.81*10) = 14.0 m/s For Isp: Isp = ve / g = 1.42 s For dV: dV = ve*ln(m0/m) Consider a 2l bottle, then m0 = 2055 g and m = 55 g; dV = 14*ln(2055/55) = 50 m/s if all the coke is ejected at the same speed. The thrust is given by: F = dm * ve dm is the Coke flow rate in kg/s. Assuming the reaction lasts 5 secs, them dm = 0.4 kg/s F = 0.4 * 14 = 5.6 N

I knew it !

You're not the first one to report bugs with Kerbals. If you found something new, then report it to the bug tracker. Otherwise chances are the devs are already on it.

I would be wrong if I wrote W = m*g In my post, even if I should have mentioned it, I was talking about magnitude, not actual force vectors, so did the post I answered to. You're right about vectors though, I apologise if my post was not clear enough.

Well, not these ones. Maybe it was because of overheating but they did go *boom*

So I was going through the painful process of testing an install with different sets of mods to try to find the origin of a bug, and while monitoring a flight with the debug menu I saw this during reentry: This is the first time I actually notice that "blast awesomeness" thing in the debug menu, and besides making me laugh (well done Squad), that made me wonder what does this value actually correspond to. I did some quick investigation with another rocket (ie: blew it up with the launchpad for science) and realised there are different values for different part, but they do not seem very logical/intuitive: While it does seem to depend on parts, it does not depend on the part's size or mass as I would have imagined, larger parts do no have a larger awesomeness Some parts have 0 blast awesomeness, other 0.1 or 0.5, and I have yet to see values >0.5 I assume that this value defines what type or how big the explosion is but I don't really get how this actually works. I don't recall seeing it on the forums recently, so if anyone knows, I'll be happy to hear that explanation ! PS: it's these little things that make me love KSP

RSS, obviously. The launcher and insertion stage are made from SpaceY and SpaceY Expanded parts, with Interstellar Fuel Switch installed to change their colour patters. The blue engine flames are from etiher Real Fuels, Real Plume or directly from SpaceY, not really sure. The lander and command module are essentially stock parts redesigned by Ven's Stock Revamp. There also are two TAC Life Support cans on the lander. I did this mission kinda quickly and not on my main save because of several reasons, including a bugged save and missing plumes. I probably will make another one with more screenshots whenever I manage to solve these bugs.

Landed on the Moon is RSS for the first time