don't forget that at F1 speeds a good portion of the braking force comes from aerodynamic drag, so there's quite a bit of energy leaving there as well, though that's speed dependent.

xpensive wrote:It was obviously a rethorical question, there is no chance of employing a 600 kW MGU-K, the point is that today's braking distances are way to short to make anything useful of this idea, far too much power over too short time.

It's not that ludicrous; Toyota's LMP1 car can harvest at ~360 kW from front and rear by using supercapacitors, which are more amenable to large currents than batteries.

And it's not like the energy they store is negligible. I wouldn't say that they're not making anything useful of it.

But how about this, let's talk numbers. First, that 1849 kW is peak, not average. Assuming a 700 kg car, 304 km/h down to 67 km/h amounts 850 kW average and 2.374 MJ. If peak is 1849 kW, then a significant amount of time is spent with braking power below that average. This makes sense. As speed decreases, so does downforce and with it the amount of braking force that can be applied.

If we take the above, that means MGU-K is picking up 14% of the available energy. Except that's not entirely true because MGU-K only harvests from the rear wheels. Let's account for aerodynamic drag and front wheel work. Let's assume front wheels are responsible for 65% of the work done, which if you don't think is reasonable, feel free to dispute. But that's 65% of total braking work done by the wheels. What about drag? They say that an F1 car makes 1g of braking at top speed. Let's assume the car makes 1g of drag at 304 km/h. Using the same weight assumption as above, 700 kg, that makes drag force at 304 km/h is equal to about 7 kN

Now let's use the equation F = .5 * rho *Cd * A * V^2. Since Cd and A are invariant, let's combine them into one variable, CdA, and now we can solve for it if we assume rho = 1.225. We get CdA = 1.571. So now we have drag force as a function of speed. Let's integrate between 67 and 304 km/h. This tells us that the energy dissipated by aero drag is about 0.191 MJ.

Now let's combine everything. Total work done by brakes is 2.374 - .191 = 2.184 MJ. Multiply that by .35 to get .764 MJ. MGU-K can theoretically harvest 44% of the energy available to it. That doesn't sound so bad, does it?