JtPB

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Everything posted by JtPB

  1. let O(eccentricity, semimajor, inclination, AN, AoP, MA) be a function that returns an orbit while those parameters are the six elements of it. let N(v(prograde, radial, normal)) be a function that returns a manuever node according to a given manuever vector. the product (N,O) returns the new trajectory of the spacecraft if the given node is activated at the given location in the given orbit. how do i know what the new trajectory is by activating the product?
  2. i asked your opinion about my way to find the exact(more or less) survey's place. it's like dropping navaid marks around when you receive the messege that says you pass the place while flying above. the marks helps you to navigate when dark and you dont see any recognizable .... around with all them pesky same hills or the flat planes, by vice versa. or its only me and my crappy machine which cant deal with high surface detail settings...
  3. i've already unlocked basic aerodynamics and landing gears... psst. you could get rover tech until then. and even without this you can navigate pretty nice with your jumping rocket and EVA RCS, which is impossible on kerbin...
  4. planting flags aroud the approx. location and dropping parts with high crash speed for surveys on water and then find it according to them... do you have a better way? (of course, all these are when you dont have rover parts yet)
  5. i want to build a spaceplane with the new parts and i figured out that i have to build massive wings for it. so i tried to copy and paste the patterns of the lift surfaces, but... well, i failed. how do i do it?
  6. what is the 5% exactly? 5% of what? So i need to put renewable resource? I used the Okto probe for my first satellite, thanks. Is 'staying stable' means the orbit or the orientation of the probe?
  7. 1. how close the current orbit of the satellite have to be to the specified orbit to complete the contract? 2. i putted batteries instead of solar panels cuze i didnt researched them yet. is it accepted as a "power" for the contract? 3. the first condition of the contract is "launch an unmanned satellite..." and its never green. when will it be accepted? its not clear how do i supposed to do that...
  8. but rockets will ran out fast & ion take so much electricity. i suggested the disadvantages: they couldnt make much at high altitude & when theres no atmosphere, that means that they wont help you when you want to launch the vessel into orbit...
  9. it would be nice to fly an airplane probe at atmospheres that doesnt consist of oxygen. there will be at least 2 types of propellors: large propellor: the one on top of helicopters, creates a tourqe that offends its rotation & spins the vessel if there's no contra-torque to resist it. small propellor: the one on tails of the above helicopters & on non-jet planes. doesnt create contra torque. there will be options to make them foldable and invert the rotation of the large one. they will be inefficient at low atmospheric pressure, that means high altitude. there will be need for foldable wings as well for launching them to other planets...
  10. because they simulate ramjet and scramjet, and those engines cant work when the plane does not move
  11. and so many big asteroids have some little astero-moons orbiting them, as well as ceres. so why not dres?
  12. note that this thing will need a massive amount of electricity, which cannot be obtained through foldable solar panels due to the wind speed. it will consume more power even than ion engine.
  13. say, you want to dock a rover to the the base but you didnt plan the precise position of the docking ports, but the rover has RCS system. or you want flip a rover & the RCS thrusters arent strong enough. i think its not considered as cheating if you turn it off back just when you finished the annoying fix, cuze we dont have cranes & such stuff in vanilla
  14. thermal jet engine is like LF jet engine, but it can operate anywhere that has an atmosphere. the principle of operation is as follows: 1. there's an intake(could be any of the stock intakes that already exists) that takes gases from the atmosphere 2. the engine uses a lot of electricity to heat the gases 3. the heated gases are emitted from the engine like any jet/rocket that engine could work on any planet that has an atmosphere(bases on Eve and refundable science probe for Jool), but one big disadvantages is the large amount of electricity it consume for doing that.
  15. so i saw there that rule of thumb at non-atmospheric landing is to keep my movement direction as parallel to the horizon as possible. and TWR of above 1.2 is enough. the only problem now is how do i land exactly at the certain area that i chose... yep, that what i did...
  16. i have the trajectories-prediction mod, which tells me where will i impact by considering the rotation of the body. also can i just burn in radial direction(away from the surface. yep i know this is bad idea for using in dV, but nonetheless) to land?
  17. didnt understand your 'suggested' procedure... we talking about the methods that start on the parking orbit around the moon and do not goes out of it...
  18. dont you think that we should give names for all geographic features around the kerbal solar system? lets start at kerbin, the africa-like continent which KSC is in will be called 'scottea', named after scott manly & the desert will be called 'danny desert'... move on with your creativity!
  19. i thought that the dV is independent on the deorbit strategy you will choose and is determined by this formula. the "high drop" method is more simple, however. Im not expert enough to use the other one... The black curve is the total delta-v spent with this landing strategy. It peaks at 400 km (2R, as you correctly found). Here's the corresponding graph for the better landing strategy which has been mentioned in this thread: first lower the periapsis to the surface, coast to the surface (i.e., periapsis) and then kill all your velocity. http://i.imgur.com/tqZVLb4.pngNote that both landing strategies assume that TWR is large enough that velocity changes are nearly instantaneous. If this isn't true, expect landing costs to be higher than those calculated here. Doing a "vertical drop" from high altitude is a bad idea since the higher you start the more gravity accelerates you on your way down, which is extra speed you have to kill to land. And for small altitudes the vertical drop is still not the best idea because it's more efficient to kill your orbital speed (which you have to do at some point either way) when you're as low as possible in the gravitational field (yes, the good ol' Oberth effect). can you define "large enough TWR"? and i dont see the graphs // i assumed that the dV is eqaul for both landing and taking-off with these methods to the same height of parking orbit. see the note i edited in the first post.
  20. the lander's impact velocity formula expands the case when an object is falling at constant field of gravity, which is constant acceleration. at the lander's case, you just calculate the work of the gravitional force along the height of the falling(energy of the lander at the beginning of falling, energy of the lander at surface) and the velocity at impact is calculated according to E/W = (mv^2)/2 of course, i knew that the lander cant braking its speed immediately(iassumed this only when it braking the orbit velocity, which is completly worse assumtion, but when it goes downwards it must slow down it's speed sometime, and i suppose that this dV is anyway equal to the impact velocity. or i really dont know how they calculated all those things...
  21. well i thought you woould guess the calculation procedure because its pretty simple: generaly minimizing the dV(H) function which is consist of the orbit velocity part and the velocity the lander will reach when it hit the surface when the orbit's velocity is after zeroing. surface speed is not included in the formula cuze simplicity reasons.(the surface velocity on the moon is negligible anyway, 4 m/s approx) anyway, thats what i did: (when 'u' is the gravitional parameter GM of the body, 'R' is the radius(perfect sphere assumed) of the body, and 'h' is the orbit's height) orbit velocity: V = sqrt(u/(R+h)) lander's impact velocity: Vi = sqrt(2u(1/R - 1/(R+h)) adding them together is: VL = sqrt(u/(R+h)) + sqrt(2u(1/R - 1/(R+h)) we can extract the u parameter which what will lead us to this: (u^2)(sqrt(1/(R+h)) + sqrt(2(1/R - 1/(R+h))) we will derive this thing, but before lets do some things to simplify the buisness: 1/(R+h) = x, 1/R = t leaving us with: sqrt(x) + sqrt(2)sqrt(t - x) //u^2 is omitted because the derivation that will be afterwards derivation(d/dx): 1/2sqrt(x) - sqrt(2)/2sqrt(t-x) transffering members & multiplaying by common denominator: 2sqrt(t-x) = 2sqrt(2)sqrt(x) reduction by 2: sqrt(t-x) = sqrt(2)sqrt(x) square raise: t-x = 2x -> t = 3x remember that t = 1/R and x = 1/(R+h), and the above equation becomes 1/R = 3/(R+h) multiply by common denominator again: R+h = 3R -> h = 2R
  22. as a part of my munar-mission planning, i wanted to know what was exactly the profile of the apollo missions to copy this(they planned the missions as fuel-cheap as possible, for sure. this will be a key sentence as you will see below). i assumed that they were at the moon-orbit that took the little dV for landing as possible, which according to my calculations is where H, the height of the orbit above the surface, is 2 times the radius of the body, independent on the body's mass, without considering the surface-rotation speed. but that wasnt the parking orbit in apollo program. the orbit was at 110 km above the surface, which is clearly isnt the diameter of the moon(2 times the radius). can anyone here explain this? //i think that problem is symmetical, means that the dV that was required for landing is equal to the dV for taking-off to that height of orbit...
  23. as said above, you cant do this. you need an amphibious rovboat. or you can just try this with airplane & refueling stations around the globe.