Gary Novak
Background
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▼▼▼   Concise Math
The proof that there are no errors in the math is that the ratio for the first two tests is 2.000000, while the ratio of the second two tests is 1.000000. Any error would replace the zeros with other numbers. To determine the amount of energy in the falling objects, a mathematical model of a rocket can be used to add the energy. The total energetics can be calculated without doing an experiment. The rocket is constant powered, as rockets usually are. This means fuel is used at a constant rate. For this reason, burn time can replace actual fuel use. In other words, a two second burn uses twice as much fuel as a one second burn. Therefore, the method is to determine the time of burn required to give each of the objects their final velocities. The burn time is then adjusted for the payload only by subtracting the proportion of the mass due to the rocket. The rocket mass changes as fuel is burned; so an average mass is determined through calculus techniques. The Rocket Equations: x' = rate of change for x = dx/dt (' is rate of change) ∫ 10³/(100mo - t)dt = -10³ln(100mo - t) + c (-c is quantity at T = 0) burn time = t = mo/0.01 - [(c-v)/10³]ex ex is inverse of natural log for the preceding quantity. (That means, when you get the preceding quantity established, hit inverse of natural log on the calculator.) mass ratio averaged for loss of mass = -100mpln(100mo - t) + c -c is quantity at t = 0 Example: Constant C in burn time: Total Time of Burn: 24kg/0.01 - [(7,783.224016-4.429446918)/103]ex = Fraction for Payload: -100mpln(100mo - t) + c c (- quantity at t = 0) = 100(4kg)ln[100(24kg)] = 3,113.289607 Explanation: Numbers for each object: Time Dropping: A: t = [2(1m)/9.81]½ = 0.45152364098573 sec Velocity Dropping: A: v = 9.81(0.45152364098573) = 4.42944691807002 m/s Constant C in burn time: velocity of rocket = A: c = 103ln[100(24)] = 7783.22401633603 Total Time of Burn: t = mo/0.01 - [(c-v)/103]ex (ex is inverse natural log for preceding quantity.) A: 24kg/0.01 - [(7783.22401633603-4.42944691807002)/103]ex = 10.6071633272070 sec B: 21kg/0.01 - [(7649.69262371151-8.85889383614004)/103]ex = 18.5215158540212 sec C: 24kg/0.01 - [(7783.22401633603-9.81)/103]ex = 23.4288933861318 sec D: 21kg/0.01 - [(7649.69262371151-39.24)/103]ex = 80.8081749880098 sec Fraction for payload: equals test mass divided by average total mass (loses 0.01kg/s). The mass ratio integrated with time is this: ∫ mp/(mo - 0.01t) dt = -100mpln(100mo - t) + c -c is quantity at t = 0 c = 100mpln(100mo) Constant C in Mass Fraction: A: c = 100(4kg)ln[100(24kg)] = 3113.28960653441 Time of Burn for Payload Only: A: t = -100(4kg)ln[100(24kg) - 10.6071633272070] + 3113.28960653441 = 1.77177876722800 sec B: t = -100(1kg)ln[100(21kg) - 18.5215158540212] + 764.969262371151 = 0.885889383614004 sec C: t = -100(4kg)ln[100(24kg - 23.4288933861318] + 3113.28960653441 = 3.92400000000000 sec D: t = -100(1kg)ln[100(21kg - 80.8081749880098] + 764.969262371151 = 3.92400000000000 sec Ratios: |