Hi All,
I am doing some research into regenerative braking and was wondering if you guys could proof read my theory, and tell me if anything is missing! I'm assuming that everything is 100% efficient.
Basically, we have a biker on top of a hill (100m). The combined weight of the biker and the bike is 85 kg.
So, the Potential Energy (P.E.) of the biker is :
P.E.= M x g x h = 85 x 9.81 x 100 = 83385joules
Does that mean that this is the energy that a motor must overcome (provided no pedal assistance) to climb the 100m hill? Is this also the amount of energy that could be recaptured on the way back down?
So to raise or lower the biker by 1 metre using gravity, 834 joules must be spent.
Does this mean that for every 1 metre travelled in a second (i.e. travelling at 1m/s) the motor would regen 834watts per second? (With 100% efficiency)
If the motor is a 24v motor, is 24v the voltage directly given off in regen mode?
And if that is the case, then W= V x I , I = W/V = 834/24 = 34.74A
So the current given back is 34.74Amps?
And finally,
I'm not sure if any of the P.E. is used up in motion, as kenetic energy? Or is it that P.E. is transfered to K.E. and then into Electrical energy?
Thanks for your help guys!
James
I am doing some research into regenerative braking and was wondering if you guys could proof read my theory, and tell me if anything is missing! I'm assuming that everything is 100% efficient.
Basically, we have a biker on top of a hill (100m). The combined weight of the biker and the bike is 85 kg.
So, the Potential Energy (P.E.) of the biker is :
P.E.= M x g x h = 85 x 9.81 x 100 = 83385joules
Does that mean that this is the energy that a motor must overcome (provided no pedal assistance) to climb the 100m hill? Is this also the amount of energy that could be recaptured on the way back down?
So to raise or lower the biker by 1 metre using gravity, 834 joules must be spent.
Does this mean that for every 1 metre travelled in a second (i.e. travelling at 1m/s) the motor would regen 834watts per second? (With 100% efficiency)
If the motor is a 24v motor, is 24v the voltage directly given off in regen mode?
And if that is the case, then W= V x I , I = W/V = 834/24 = 34.74A
So the current given back is 34.74Amps?
And finally,
I'm not sure if any of the P.E. is used up in motion, as kenetic energy? Or is it that P.E. is transfered to K.E. and then into Electrical energy?
Thanks for your help guys!
James
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