If overall efficiency is what you see reflected in the EPA mileage numbers (or similar European tests), in what scenario would a lower efficiency transmission (lower Pout/Pin over some range of operating points?) lead to or even allow better overall efficiency?

▲

taeric 2 days ago | parent [-]

I'm not sure I understand the question? You get the engine to hit its sweet spot and to hold it there. That gain outweighs the loss from the new transmission.

This would be similar to hitting the optimal torque point. The idea there would be that you can get out of the acceleration phase faster, so that you can transition to a more efficient gear to maintain the speed for longer.

The wikipedia looks to cover this well. One of the cites is specific on the efficiency of the CVT. I think I overstated how much higher the loss is, so maybe that is confusing things? I thought it was 10-20, but the cite on the page shows it solidly around 10.

▲

klaff 2 days ago | parent [-]

Ah, thank you, I see the argument. I'm still a bit skeptical though when many non-CVT automatics are 6 or 8-speed models that there would be much "sweet spot" benefit left by being able to make relatively small changes in engine speed. Like many things probably depends on exactly what comparison is being made.

▲

taeric 2 days ago | parent [-]

I mean, we are comparing 6-8 to effectively infinite? Question then comes into just how much is lost by being outside of the sweet spot for an engine. Which, I agree that I would not have thought it would be that big of a deal. Just going off modestly paying attention in my car, I see massive benefits keeping the car around 300 rpms lower.

Interestingly, my car gets better gas mileage around the 40ish speeds than I do at full highway speeds. That somewhat surprises me. It is very dependent on not having a heavy foot, of course.

▲

klaff a day ago | parent | next [-]

Best efficiency speed can be understood by considering impact of fixed losses like HVAC, lights, computing, which consume more energy the longer you drive (so the total energy for those would be minimized by going faster) and the aerodynamic losses in which force goes with speed squared, power w/ speed cubed, energy over a fixed distance back to squared because at higher speeds time is reduced (this component is theoretically minimized by going very slowly). In between is rolling resistance which requires a fixed energy per distance, so it doesn't care about speed.

For EVs, the drivetrain efficiency is so high that it's variability with operating point doesn't affect this calculation much, and so the most efficient speed of an EV is around the speed at which the fixed losses equal the aero ones. This will vary greatly with environmental conditions since AC or heating load can be large in hot or cold conditions but at the right temperature will go to near zero.

In ICE cars, the drivetrain efficiency is much lower and so the drivetrain efficiencies are a much more significant part of the optimization problem, but the basic physics of the aerodynamics are the same.

The model I used to use in my head is that for an ICE, the most efficient operating point is probably around the lowest speed the car can operate in the highest gear, so maybe around 40 mph / 60 km/h? Obviously a rough heuristic though.

▲

vel0city 2 days ago | parent | prev [-]

Generally speaking your car will get better fuel economy at 40mph than it will at 60 or 70mph. Your car gets so much more wind resistance the faster you go.

The main reason why city mileage is usually lower is because of all the stopping.

	▲	taeric a day ago \| parent [-]
		I think I just mentally had this idea shifted up about 10-15 mph. I'd expected 55 to be the sweet spot, for some reason.