# Thread: Demand factor for multiple car charging station

1. Senior Member
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Originally Posted by Ingenieur
work (or energy depending on context) = F d = m a d
d = v t and 1/2 v = a t or a = 1/2 v/t
substituting
work = m 1/2 v/t v t = 1/2 m v^2
2 work / m = v^2, not work/d, that equals Force
Your derivation shows that absent outside forces, the work done on an object starting at rest will equal the resulting kinetic energy of the object. Which agrees with conservation of energy.

Air resistance is a dissipative force like friction, energy conservation arguments aren't going to work. The energy lost to drag F over a distance d is just F * d. And your initial observation that F varies as v2 means that the energy you need to overcome drag for a given distance also varies as v2. In other words, at speeds where the energy required to keep the car moving is dominated by air resistance, the range you'll have for a fixed size energy source varies as v2.

Cheers, Wayne

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Originally Posted by wwhitney
Your derivation shows that absent outside forces, the work done on an object starting at rest will equal the resulting kinetic energy of the object. Which agrees with conservation of energy.

Air resistance is a dissipative force like friction, energy conservation arguments aren't going to work. The energy lost to drag F over a distance d is just F * d. And your initial observation that F varies as v2 means that the energy you need to overcome drag for a given distance also varies as v2. In other words, at speeds where the energy required to keep the car moving is dominated by air resistance, the range you'll have for a fixed size energy source varies as v2.

Cheers, Wayne
energy conservation better work
you heat the air/body surface and friction tires, bearings, etc

the point is going down the road at a constant speed
engine T x tran ratio x diff ratio / tire rad
= road thrust = drag + friction
apply more T and you accelerate

if you increase speed/rpm by 10% fuel rate increases ~ the same
engine T ~ (V x P x eff) / (4 Pi)
V = displacement, P = mean eff press ~ comp ratio x atm press, eff ~ vol eff
mean eff press ~ charge volume ~ throttle
consider boost on an

engine speed is linear and a/f constant at ~13:1 (varies with comb mode)
as does T to offset friction + drag
rpm linear with air, and a/f is constant, and T ~ charge
consider turbo/supercharger boost, a psi is ~ 7% more T/P
basically increasing vol eff and hence mean eff press
so even though drag F ~ v^2 the additional fuel required is linear with vel

all this varies with speed, typ fric > drag if v< 70 mph
gearing, etc

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my car gets 27 mpg at 70 mph, 1 hr, 70 miles, 2.6 gal, 0.037 gpm
if sq with v at 140 mph it would get 6.75 mpg, 1 hr, 140 mile, 20.7 gal, 0.148 gpm
car would run dry in 50-55 min or so
from videos it actually gets 15 based on videos

According to studies backed by the department of energy, the average car will be at its advertised MPG at 55 mph. But as the speed increases:

- 3% less efficient at 60 mph
- 8% less efficient at 65 mph
- 17% less efficient at 70 mph
- 23% less efficient at 75 mph
- 28% less efficient at 80 mph

80/55^2 ~ 2.1, it does not use twice as much fuel
only ~ 30% more

4. This was supposed to be about battery chargers.

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