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  • I have done just that on a large scale , I have created a wheel generator that only captures kinetic energy when vehicle system has a velocity over 30mph or 50kph. 1/2mv2 = for 10,000lb suv at 30mph = 44ft/sec or 13.3 m/sec for the us systemthis means 5,000 pounds x 44ft/sec x 44ft/sec = 5,280,000 poundals ft seconds and multiply that by 0.0421 to convert to watts is 497 kw but at highway speeds of 60 mph this number for the 10000lb SUV 38,720,000 poundal ft x 0.0421 = 1,6 MW of stored kinetic energy. So at highway speeds and multiple 12vdc alternator with built in IVR's internal voltage regulators. I have been able to capture 28Kw the extra rolling drag is 28,000 / 1,600,000 watts 1.7% extra drag so if it took 20Kw vdc to power electric vehicle. it would not be linear ...see more »I have done just that on a large scale , I have created a wheel generator that only captures kinetic energy when vehicle system has a velocity over 30mph or 50kph. 1/2mv2 = for 10,000lb suv at 30mph = 44ft/sec or 13.3 m/sec for the us systemthis means 5,000 pounds x 44ft/sec x 44ft/sec = 5,280,000 poundals ft seconds and multiply that by 0.0421 to convert to watts is 497 kw but at highway speeds of 60 mph this number for the 10000lb SUV 38,720,000 poundal ft x 0.0421 = 1,6 MW of stored kinetic energy. So at highway speeds and multiple 12vdc alternator with built in IVR's internal voltage regulators. I have been able to capture 28Kw the extra rolling drag is 28,000 / 1,600,000 watts 1.7% extra drag so if it took 20Kw vdc to power electric vehicle. it would not be linear drag it would be 28,000watts - (20,000wtsx 1.7%) = net gain of 27, 660 watts in kinetic energy harvest . now at lower velocities it is less efficient at the 30 mph speed the percentage of extra drag 28,000 watts/ 497,000 watts = 5.6% drag from wheel generator 20,000 - (20,000 x 5.6%) = 1,120 wawatts neededwe have powered mobile trailer refrigeration doing this. see http://4dde.com

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