How to calculate power required by vehicles

[Phrak]

sophiecentaur. Sorry, by variable brake peddle pressure I was referring to, in rough terms to the variable power presented to the conversion circuit. The actual pressure has nothing to do with it. In effect, your foot pressure is just a mean to tell the braking circuitry how much power needs to be absorbed to reduce the vehicle's kinetic energy at any given time.

likephysics. Good thinking! I don't know if it's a viable idea, but I've only been thinking in the electronics box. Maybe you found a way around it.

Say the flywheel can quickly absorb a vehicle's kinetic energy during a braking episode. Now you may not have to dump it so quickly into electrical storage. The longer the time you have to do this, the smaller the currents that are involved. The smaller the currents, the more efficient, and the smaller, simpler and lighter the electronics that is required to do this.

 

[sophiecentaur]

It's amazing to think that, in a very few years, we are all likely to be driving vehicles which use something like we're discussing. Something else to go wrong, I suppose!

I am reassured by my (sailing) boat engine (diesel) which is 35ys old, has a flywheel that you can hardly lift and one bit of electronics - the rev counter. It seems to be refusing to die. (Refuses to start, too, on occasions)

 

[mheslep]

And in the end, the bottom line is not BTU comparison but cost.

In which case the energy cost for an electric mile is well established to be 3-4X less expensive than a gasoline powered mile, given the average efficiency of the US gasoline vehicle and the current price of gasoline.

 

[mheslep]

Someone here at the forum pointed out to me that there is development on fast charging LiPo batteries:..

Almost, that was Lithium Ion Iron Phosphate, LiFePO, that has demonstrated high power, not the polymer version of Li ion which is inexpensive but lower power as mentioned above.

 

[mheslep]

Adding a bit more to your fine response...

At some point air resistance dominates rolling friction.

Yes, break even (drag vs rolling resistance ) is about 40 km/h (25mi/h) in a sedan, and interestingly only 10 km/h for a bicyclist.

That's because it is quadratic with velocity. A quick measure of air resistance is given by the Bernoulli pressure 1/2*(rho)*(v^2). For 40 m/sec this gives approx 1000 N/m^2. Take this over the cross section of the vehicle (approx 1 sq meter) at the velocity in question and you get a power of 40 kW or about 50 horsepower.

Yes, just so for 1 M^2 at 40m/s = 144 km/h = ~90 mi/h, moving right along. The average car effective drag area is actually a bit lower, Cd*A = ~0.8 M^2, and at the highway speed of 60 mi/h the tractive load for air drag alone is about 18HP. The rolling resistance coeffient for a sedan is about Crr=0.01. So for a one ton vehicle the rolling resistance force is MgCrr = 1000kg x 10 x 0.01 = 100N, a constant over any speed. The power load for rolling resistance is also F x V, so at 60 mi/h we have 100N * 26.6M/s = 2.66kW = 3.6HP. Total tractive load, drag and RR, at 60 mi/h level road, constant speed, 1 ton sedan is ~22HP. Another metric is energy cost per distance, in this case that is 0.27 kWh per mile ( about 3 cents per mile at my electric rates).

Streamlining helps only up to a point. That's because the friction of sliding sideways through the air eventually equals the amount of friction of butting headlong into the air. This point occurs for a cylinder approx 40 times its diameter in length. In the above example, then for a cylindrical car 40 meters long and 1 meter in diameter, you would have 100 horsepower consisting of 50 horsepower butt-end friction and 50 horsepower irreducible sliding friction.

True, though there's a 10x gap between the Cd of the average sedan (0.3) and the average airplane (0.03)

From D. McCay's excellent reference on this subject.
The energy cost per 100 km here includes a engine efficiency of 0.25. To get actual tractive load divide by 4.

http://www.inference.phy.cam.ac.uk/withouthotair/cA/page_254.shtml