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Showing results for tags 'rocket equation'.
I have had KSP for the PS4 a total of 4 days now, and it's complexity and intricacy exceeds what I prepared for. I am happy in this, but mortified as well. I have started a science save file, and have only researched 4 additional groups including: basic rocketry, engineering 101, general rocketry, and survivability. With the new parts acquired from researching these groups I built a simple, 3 stage ship, and have decided to try to achieve LKO. To do so efficiently, I delved into the KSP wikis and forums to find the relevant information. I understand to a certain point the rocket equation, or the "*dudes name I can't spell*'s equation". Anyways, I found the ISP of the rockets used in each stage, and I wasn't going for complete accuracy at first, using the ATM ISP for every equation done. Then, I found the total mass of each individual stage, as well as the dry mass of the stages by themselves. In order to do so I had to take the first to stages off, and measure the total and dry masses for the last stage, then add the second stage and subtract the total mass of both stages my the total mass of the last stage to get the total mass of the second stage. Then I proceeded to do so with the first stage attached as well. I found both total and dry this way for the first and second stages. Now that I have the ISP, M_FULL, and M_EMPTY, I figured I could calculate and add the answers together to find my overall delta-v, granted I SHOULD have more than I calculate based on the idea that my ISP for each equation will be the atmospheric ISP. My work: (weights are rounded) Stage 1 (final stage): pod, fuel tank, and 'reliant' thruster. - Dv = 265 * 9.8 * In (4/3) = 747 Stage 2 (second stage): 2 fuel tanks and 'reliant' thruster. - Dv = 265 * 9.8 * In (3.5/1.5) = 2,200 Stage 3 (first stage) 2 BACC thrusters, and 1 'hammer' thruster, with added radiator panels. Dv = 520 * 9.8 * In (19.5/4.5) = 7,472 I am now realizing that each stage must also push the weight of the stages above them, meaning I must carry the mass of those stages into that equation, adding to the total and dry mass of the stage being figured. If this is not so and I have created a false solution, please let me know, thank you.
WARNING, Math ahead. So I was trying to find an intuitive way of approximating the reverse rocket equation by hand (without havinng to developp an exponential) : M = m.exp(DV/ve) And came up with that : (1) M1 = m.(DV/ve) As the mass of fuel required to propel the playload mass up by a desired DV, with an exhaust velocity of ve (conservation of momentum). (2) Mn+1 = (Mn/2).(DV/ve) As the mass of fuel required to carry the mass of fuel Mn while accelerating (by summing over a triangle, one can see you only need to propel up to Dv/2 on average). Edit : here is exactly where the above formula comes from : If DV < 2.ve, then the sum (Mn)n>=0with M0=m is convergent and easy to compute by hand. Its sum is : (3) M = m. (2ve + DV)/(2ve - DV) Which is not consistent with the correct reverse rocket equation. Close enough for small DeltaV : if you take r = M/m, and z = DV/ve, we find r(z) = (2+7)/(2-z) = 1 + z + z2/2 + z3/4 + o(z3) Instead of r(z) = exp(z) = 1 + z + z2/2 + z3/6 + o(z3) For r > e, (z > 1), at interstage mass ratio that are not unseen, the approximation gets significantly off (in a conservative way at least). https://prnt.sc/jnq406 But more importantly, I don't know why I don't end up getting the correct formula, altough I strongly suspect step (2) to be at fault, and I'd like to be enlighten.