Myggen

Wow, thanks alot for explaining all that, it all makes more and more sence, still a way to go tho. Still some things i dont understand: 9.81: is this because of this? > the acceleration due to gravity at the Earth's surface, 9.81 m/s2?` If so, then isnt this the engines isp while flying upwards in earth gravity and atmosphere?, but in my project we are pushing at the moon in vacuum, would it then still be 9.81? And no, you wont cheat by ignore the weight of the fuel, because this is my plan: http://img.photobucket.com/albums/v209/total-uffe/Projekt%20Stop%20Maringnen_zps5lyer4ds.png All the fuel is in a ballon in a stationary orbit, also with a gravity manipulator thing attached so it wont fall down...and not have to think about when spending fuel the mass gets lower etc. But first i would really like to know the fuel needed, but need to know if i leave out the M0/M1? So i wont be needing the M0 and M1 thing, if i leave out the weight of the fuel...and the engines, if i modify one to have all twr needed? Then the twr to be able to stop the moon in one orbit 2,358,720 sec (60x60x24x27.3) No, but i have just checked it out. Its Force=Mass*acceration right? But then i guess i need to now the acceleration, and someone aka ChrisSpace came up with these numbers 0.0002416m/2, but why and how? and whats the /^2 mean? I hope i make myself understandable, im very tired atm , thanks again to far.

Hey all again, thanks for all the replies, and sorry for the late reply, but had alot of stuff to do. I look specially at this: But here i wanna go with the russian RD-170 with Isp on 338 https://en.wikipedia.org/wiki/RD-170 Also these i dont understand: You write M_1 and M_2, dont you mean M_0 and M_1? and then these: 0.13036= ln (M_1/M_2) 1.1392 = M_1/M_2 How to come up with those numbers? Can you explain a little more? i would like to know just a little about what im talking about, not standing there sounding more stupid than the project itself haha. Then this: Yea i would need some calculations of how much thrust i need to stop the moon in one orbit, how did you come up witht the thrust needed number? Also, shouldnt ^-4 be ^4? my calculator says syntax error using ^-4 Im only a 1. year student, we havent "learned" about atoms and such stuff yet haha. Another good idea, and i see the point, but i think it would be pretty hardcore calculations, involving grafs. But thanks alot so far, there is really something to think about, and i think i will consult the physic teacher tomorrow.

Hello all, im a electrician student, and in school we got the task (not having anything to do with electricity) to explain about something in 5-10 min, it could be hobbies..well, also something about electricity, anything goes. But, actually the night before i saw this video by Scott Manley: , and was instantly thinking about the video when we got the task telling about something, i just want to try it on the real moon (the one turning red in 1 hour time, im staying awake to see it atm, at about 04.30 am danish time ).Anyway, what does it take to stop the velocity of the REAL Moon??? My plan is, to leave out the factor of the forces of gravity in the calculations, by putting a gravitymanipulator on the surface of the moon (so it wont fall "down" when loosing speed). But, in the video, Scott Manley just show some number without explaining where he got em from and why. The calculation he show at 1:28, says (280x124 trillion) / (800x9.81) = ..... Well, i guess the (280x124 trillion) = (deltaV x mass of gilly), but then this / (800x9.81) I spoke to a friend who play this game alot more than i do, and he is talking about that the 800 might be isp of the engine used, but then we are in doubt of the 9.81? Also i wanna leave out the factor of the weight of all the fuel, as its stored in a big ballon in orbit, with a long hose connected to the engine(s) on the surface... I wanna use this rocket engine: https://en.wikipedia.org/wiki/RD-170 And i want to stop all velocity in one single orbit (27.3 days), not Scott Manley's 19 million years. So first im gonna figure out how much fuel is needed, then how many engines is needed to complete the stop in only 27.3 days. So far i have these notes: 1 orbit = 2,412,517.5 km / 27,3 / 24h / 60 min / 60 sec = 1022,8 m/s Moon mass = 7,349×10^22 kg > 7,349×10^19 t > 73.477.000.000.000.000.000 t Orbital period = 27.321582 d (27d 7h 43.1) delta v / masse = (1022 x 73.477.000.000.000.000.000) / (isp? x 9.81?) = Strongest rocket engine: Russian RD-170 (https://en.wikipedia.org/wiki/RD-170) Thrust (vac.) 7.887 MN (1,773,000 lbf) > ibf = pound force > 1 ibf = 4.448222 N (https://da.wikipedia.org/wiki/Newton_%28enhed%29) Thrust to weight ratio: 82 (TWR > wiki.kerbalspaceprogram.com/wiki/Thrust-to-weight_ratio) Any ideas/help would be very much appriciated, and thanks in advance

Well, its on steam for -40 %, thats cool, so i bought it, eventho i bought it here from the site also, when it was v 0.15 i think. Yea, i know i could get a free steam key, whatever , just want to support the developers i guess, and then i love a good deal too haha. I dont even know why i make this post, but thanks Squad for a great game that deserve lots of attention.

havent played this since last year, now i see its on steam, whats up with activation on steam. i bought the game long time ago.