Calculatus eliminatus:
[edited]
[The acceleration for a 6 second 1/4 mile is 73.33 ft/sec^2, or about 2.27 g., from x=.5A t^2]
Take [their advertised] 1000 Hp over 6 seconds to determine the energy (or work over 1/4 mile) to determine that the [force on the accelerated mass] is 2500 lbs.
[1000 hp x 550 x 6s = 3,300,000 lb-ft, or 1,243 Whr.]
[Divide the 2500 lb force by the acceleration to get the mass = 34.1 and the bike weight is ~1100 lbs.]
Equate the kinetic energy, [1243 Whr] to the capacitor energy, 1/2 CV^2, using 3p = 6000F and 6000/210s = 28.6 F for C, it looks like they are assuming a voltage drop of only 9.4 V in the pack during the 6 second burn.
[edited]
[The acceleration for a 6 second 1/4 mile is 73.33 ft/sec^2, or about 2.27 g., from x=.5A t^2]
Take [their advertised] 1000 Hp over 6 seconds to determine the energy (or work over 1/4 mile) to determine that the [force on the accelerated mass] is 2500 lbs.
[1000 hp x 550 x 6s = 3,300,000 lb-ft, or 1,243 Whr.]
[Divide the 2500 lb force by the acceleration to get the mass = 34.1 and the bike weight is ~1100 lbs.]
Equate the kinetic energy, [1243 Whr] to the capacitor energy, 1/2 CV^2, using 3p = 6000F and 6000/210s = 28.6 F for C, it looks like they are assuming a voltage drop of only 9.4 V in the pack during the 6 second burn.