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I think you have the concept; but I prefer to look at it as a function of time; telecom.A one kilogram missile moving 19.81m/sec will rise 20 meters. From d = ½ v²/a. This will take 2.019 second. From d = ½ at². A kilogram mass applies 9.81 newtons of force. This is 9.81 newton applied for 2.019275 seconds. = 19.81 N seconds; N*sForm a chain of twenty one kilogram masses; they are one meter apart, 20 meters high. This vertical chain could be dropped one meter. The original configuration can be restored if only one kilogram is raised 20 meters. Connect this 20 kilogram chain to a 80 kilogram flywheel. It will take 1.009637 seconds for the chain to drop one meter while it spins the wheel. But this is 20 kilograms dropping one meter for (20 kg * 9.81 N/kg) 196.2 N applied for 1.009637 seconds for 198.09 N seconds.The output momentum is ten times that of the input momentum. The output momentum is 100 kilograms moving 1.9809 m/sec for 198.09 kg*m/sec. The input momentum was 1 kg moving 19.81 m/sec for 19.81 kg*m/sec. The input energy is 196.2 joules. The output energy is 19,620 joules. In the 20 kilogram; twenty meter; vertical chain arrangement one kilogram applies its 9.81 newtons for 1.0096 seconds 20 times before it needs to be returned to its original vertical position at the top of the chain. The other 19 kilograms in the chain always assist in the application of force but each one kilogram applies its 9.81 N for 1.0096 seconds 20 times. The descending mass is 9.81 N applied for 20.19274 seconds. This is 198.09 Ns. The ascending mass is only 9.81 N applied for 2.0193 second for 19.809 Ns. This time difference multiplies the momentum by 10 times and the energy by 100 times.
Stack 20 dominoes on top of each other on their sides. Hold the second domino from the bottom and slide out the bottom domino. Place the removed domino on the top of the stack.We now have the starting position; of 20 dominoes that are at a distance of one domino width from the table. We can lower the stack of twenty dominoes by one domino width so that the stack again rests on the table. We can remove the bottom domino again; and place it on the top of the stack. So by raising one domino twenty domino widths we can continually restore the original configuration. (This is quite a trick with twenty; but works easily with six)All the one kilogram masses in the chain drop one meter; but only one needs to be raised 20 meters to restore the driving chain (stack) to its original configuration. It takes a whole lot less momentum to raise one kg twenty meters than to raise 20 kg one meter. It takes 19.81 (19.81 m/sec * 1 kg) units of momentum to raise one kilogram 20 meters: and 88.59 (4.429 m/sec * 20 kg) units of momentum to raise 20 kilograms one meter. 19.81 / 88.59 = 22.36%You are actually using 19.81 units of momentum to make 88.59. The momentum increase is 88.59 / 19.81 447%. When you transfer all the motion of the 88.59 units of momentum to one kilogram you get (½ * 1 kg * 88.59 m/sec * 88.59 m/sec) 3924 joules of energy: for an energy increase from 196.2 J to 3924 joules; 2,000%. If you attach the chain to a 80 kilogram flywheel you are then using 19.81 units of momentum to make 198.1.
Is the only way to capture this excess energy by sending the projectile upward?And then using it as a potential energy?