# Maths for capacitors and moving cars

1. Apr 25, 2010

### garryokeeffe

I've been working around with this little problem for some weeks now and I think I have most of it figured, but there are some points that keep confusing me, its long time since I finished college and I have forgotten most of the maths. So any pointers on mistakes or indeed if my maths is correct would be great.

Modern capacitors are massive ,i.e. 3000f @ 2.7V, and are used in the auto industry. I'm trying to figure out how far a car can be driven using only the power stored in the capacitors. To get a decent voltage the capacitors are put together in series, referred to as strings. I will assume a voltage of 400V that gives a 20f capacitor.

Assume the capacitor is fully charged.

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I think the formula I need are :

charge on the capacitor J = 0.5CV^2
C = capacitance, V = voltage
gives 1600000J (1.6x10^6)

energy to maintain a speed J = 5v^3t
v = speed of the car in m/s
t = the time the speed is maintained in s

force to accelerate the car F = ma (Newton)

acceleration a = (Vf-Vi)/t
Vf = final velocity
Vi = initial velocity
t = time to get to speed

distance traveled while accelerating
s = 0.5(Vf-Vi)t

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Lets say I want to accelerate the car (mass 900Kg) from 0-50kmph( 0-30mph) in 20 seconds and then maintain that speed for as long as possible, how long can I maintain the speed.

First the acceleration
final velocity = 50kph = 13.8m/s
a = 13.8/20 = 0.69m/s^2

required force F = ma

F = 900x0.69 = 625N.

In this time the car will have traveled
s = 0.5(Vf)t = 0.5x13.8x20 = 276m.

Energy used in the time
J = N.m = 625x276 = 172500J

The capacitor had 1600000J after the acceleration it will now have
1427500J

Assuming that the conversion is lossless the motors are perfect and that all the energy can be used then to calculate how long the car will travel I would used

J =5v^3t ( I got this from Wiki, its simplified and gives the energy to maintain a car speed)

rearrange to give time.
t = J/(5v^3) = (1427500)(5x13.7^3) = 108.6 seconds

Can anybody tell me if this is correct.

My next task is to take into account losses in the system and to work out when the motor will stop i.e. assume the motor can work down to 100V, for how long will the car be driven.

Any guidance ?