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About BattleReadyKen

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    Bottle Rocketeer

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  1. Neat link. In their example, they find the average mass of the rocket during flight and use it as a constant. However a lot of the aero terms will be hard to determine using KSP.
  2. Yea, any objective that is non-orbital would be a good application for early game contracts like those. That wasn't my main goal, but it's definitely a use.
  3. Ah yes, you are correct. That is one of the first major issues I was wrestling with, once I elevate myself above 70kM, the horizontal velocity is negligible. My first attempts at the game were to burn straight vertical with a payload orbital burn stage capable of > 2400 m/s dV, however this is significantly much less efficient and almost double what may be required at an acute tilt around 8000m (at least that's what I've heard is efficient). Assuming I launch my rocket (a much larger- multi-stage one), at a slight angle to distribute velocity in both horizontal and vertical directions, I imagine actually calculating the resultant horizontal and vertical velocity after the stage(s) fire would be quite difficult. Or maybe it just involves a basic trig manipulation? *Also, the formula a = Ve * ( BTOTAL / MFULL ) - g, 'g' would likely have to be used with a trig function with the angle of incline as the center of gravity is no longer exerting a force directly through the center of thrust. P.S. I still do full vertical payloads then massive burns at AP like an idiot sometimes hahaha! I like the general 1500-2000 dV rule of thumb, but would like more exact calculations if it is feasible to do so. Thanks.
  4. Ah that makes sense. NASA has it here that the thrust force, which factors into Isp, is based on pressure of the system. Thank you. This algorithm is still useful and I will probably compile it into a C++ executable as a side tool for myself in the future. If anyone stumbles upon this and would like a copy of the tool just let me know! I find this particularly useful for determining the minimum parts/mass required in the first initial stage(s) to leave Kerbin's atmosphere.
  5. Hey everyone, I recently have been sucked into this game, and I'm loving the math. My question is this simple, determine the altitude a rocket will achieve on full fuel burn of a single stage. I've done a lot of research and have come up with the following example problem to test my algorithm/process of calculation. Let me know what you guys think of below and what I'm missing or potentially a force I haven't considered into the calculation such as lift, as you'll see my answer is off by nearly 3,300m. (For the sake of simplicity the rocket travels straight up in a vertical dimension only.) Known Values of my Rocket: Full Mass [MFull] (Entire Rocket) : 7.5t (7,500kg) Empty Mass [MEmpty] (First Stage Depleted) : 4.5t (4,500kg) Fuel Mass [MFuel] (Both LQ and OX) : 3.0t (3,000kg) Isp [Isp] (Reliant Engine) : 265 sec Thrust [FT] (Thrust Force Atm.) : 205.2kN (205,200N) LQ Rate [BLQ] (Burn Rate of LQ) : 7.105 u/sec OX Rate [BOX] (Burn Rate of OX) : 8.684 u/sec LQ Volume [VLQ] (LQ Fuel 45% Mix) : 270 u OX Volume [VOX] (OX Fuel 55% Mix) : 330 u Known Values of Kerbin: Accel. Kerbin [g] (Accel. of Gravity) : 9.81 m/sec^2 First I will calculate the time required to burn through the fuel mixture. This time will be needed in the final calculation. Tburn =VLQ /BLQ =VOX / BOX << >> 270u / (7.105u/sec) = 38.0 sec Next I convert burn rate units from volume/sec to units of kg/sec. (I assume 1u = 5kg of both LQ and OX) BLQ_M = BLQ * (5kg/u) << >> (7.105u/sec) * (5kg/u) = 35.525 kg/sec BOX_M = BOX * (5kg/u) << >> (8.684u/sec) * (5kg/u) = 43.42 kg/sec BTOTAL= BLQ + BOX << >> 35.525kg/sec + 43.42kg/sec = 78.945 kg/sec (M *Dot = Mass Flow Rate) Determine effective exhaust velocity of rocket motor related to Specific Impulse and Gravity. (NASA Formula) Ve= Isp * g << >> 265sec * 9.81m/sec^2 = 2,599.65 m/s Determine acceleration of rocket (Found this formula on a physics forum, not sure if valid) a = Ve ( BTOTAL / MFULL ) - g << >> 2,599.65 m/s * (78.945kg/sec / 7,500kg) - 9.81m/s^2 = 17.554 m/s^2 Apply classical kinematic physics equation for displacement with acceleration. (Vertical Axis only...) deltaX = 0.5 * a * (Tburn^2) << >> 0.5 * 17.554m/s^2 * (38sec ^ 2) = 12,673.988m So in the end this calculation results in an effective altitude of 12,673.98 meters. If anything, I expect drag (if simulated) among other forces to take away from this value. Instead the opposite occurred, my actual test flight while holding steady to the center of the NavBall resulted in roughly 16,000 meters altitude at 38 seconds into flight (after stage finished burning). Any ideas?