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lolwhatitis

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lolwhatitis
·2 anni fa·discuss
Presuming an 80 kWh battery and 80% efficiency:

80 kWh / 0.8 = 100 kWh

To charge in nine minutes:

100 kWh * 60 min/hr / 9 min = 667 kW

A 400 V DC setup is common for this sort of application, so:

667 kW / 400 V = 1667 A

How physically large do the cables and related apparatus need to be in order to deliver this sort of current? What sort of training and personal protective equipment will people need in order to plug and unplug these cables? (Hint: Arc flashes are no joke!) What sort of service would you need to order from the electric company to be able to power just one of these installations?
lolwhatitis
·3 anni fa·discuss
Motorcyclist here. The physics are the same for bicycles and motorcycles, and yes, you have it correct.