[Bearleener]

Yeah, getting the power to the pavement is the main problem of any retrofit system.

Here's a back-o-the-envelope calculation for a compressed air / air motor regen system (all metric units here, makes it easier):

Let's assume we have a suitable air motor and the system is 100% efficient (yeah, right...). The popular LP gas conversion systems have cylindrical tanks that fit in the spare tire well. With my 185/60 R14 tires that's about 50 litres capacity to use as a compressed-air tank. The LP tanks usually run at about 8 bar (120 psi) pressure. That means I could store an energy of E= Volume*Pgage = (50/1000 m^3)*(8 bar * 101.325 kPa/bar) = 40.5 kJ.

My car weighs about m= 1100 kg. Set the air energy equal to the kinetic energy of the car: E = (1/2)*m*v^2 means I could accelerate from a stop to a speed of about v= 8.6 m/s or 31 km/h (19 mph) -- not bad! I could bump start from there.

Let's accelerate modestly and say we can live with a 10-second zero-to-31 km/h time. That's an average acceleration of a= v/t = 8.6 m/s / 10 s = 0.86 m/s^2. Newton told us that F= m*a = 1100 kg*0.86 m/s^2 = 946 N of force needed to accelerate the car. That's quite a lot. (By the way, the power is (at half of the 31 km/h terminal speed) P = F*v = 946 N * (8.6/2) m/s = 4 kW (5.5 hp).)

Suppose the air motor uses a roller to drive the tire or the pavement directly. This roller needs to achieve 946 N of traction. Using an optimistic coefficient of friction of 1.0, the roller would have to press down with a force of 946 N (about 95 kg or 210 lb) to avoid slipping. Too much! Very hard to do with a roller or tire. Back to the drawing board! For comparison, Mike's Insight (see 1st post of this thread) presses with 130 lb of force.

Okay, but maybe we can live with less power and only use it to boost acceleration over longer periods. Or figure out a way to couple it more directly to the driveline.

No wonder nobody offers such a system!