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Gary Novak
Background Second Proof Concise Math Math Explained Joule's Constant Potential Energy Contradictions
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Complete Math
Velocities
From Falling Object page. Symbols are given on Falling Object page. Formulas: v = gt
g = 9.81 m/s (rounded throughout) Numbers for each object: A - 4 kg dropped 1 m B - 1 kg dropped 4 m C - 4 kg dropped 1 sec D - 1 kg dropped 4 sec Time Dropping: A - 0.45152364098573 sec B - 0.90304728197146 sec C - 1 sec D - 4 sec A: t = [2(1m)/9.81]½ = 0.45152364098573 sec B: t = [2(4m)/9.81]½ = 0.90304728197146 sec Velocity Dropping: A - 4.42944691807002 m/sec B - 8.85889383614004 m/sec C - 9.8100000000000 m/sec D - 39.2400000000000 m/sec A: v = 9.81(0.45152364098573) = 4.42944691807002 m/s B: v = 9.81(0.90304728197146) = 8.85889383614004 m/s C: v = 9.81(1) = 9.8100000000000 m/sec D: v = 9.81(4) = 39.2400000000000 m/sec Constant C in burn time: A - 7783.22401633603 B - 7649.69262371151 C - 7783.22401633603 D - 7649.69262371151 velocity of rocket = -103ln(100mo - t) + c -c = velocity at t=0 c = 103ln(100m0) A: c = 103ln[100(24)] = 7783.22401633603 B: c = 103ln[100(21)] = 7649.69262371151 C: c = 103ln[100(24)] = 7783.22401633603 D: c = 103ln[100(21)] = 7649.69262371151 Total Time of Burn: A - 10.6071633272070 sec B - 18.5215158540212 sec C - 23.4288933861318 sec D - 80.8081749880098 sec 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: mp = mass of payload = 1 kg or 4 kg tp = time for payload including mass ratio integrated with time 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(100m0) Constant C in Mass Fraction: A - 3113.28960653441 B - 764.969262371151 C - 3113.28960653441 D - 764.969262371151 A: c = 100(4kg)ln[100(24kg)] = 3113.28960653441 B: c = 100(1kg)ln[100(21kg)] = 764.969262371151 C: c = 100(4kg)ln[100(24kg)] = 3113.28960653441 D: c = 100(1kg)ln[100(21kg)] = 764.969262371151 Time of Burn for Payload Only: A - 1.77177876722800 sec B - 0.885889383614004 sec C - 3.92400000000000 sec D - 3.92400000000000 sec 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: A/B – 2.00000000000000 C/D – 1.00000000000000 |
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