Does the new Model S really get 400+ miles?

[comanchepilot]

@TIppy - I routinely see about those wh/mi myself. In my wife's old Blue 75D S - I'd generally average 263-270wh on the way from our house to LAX - which is 2000MSL to 100' MSL and 313-330 on the way back at 70-75 without traffic. That is for about 55 miles of actual driving - which equates to about 290-300wh/mi which is right in your ball park there.

I always got ALOT better mileage than my wife because my driving was much more anticipatory - so I could use regen much more efficiently - she is ALWAYS in the 350-360 range - because she uses the actual brake much much more often than I do. When we drive to our house in Tucson from near Ontario Calif I only have to stop once - and she ALWAYS has to stop twice in the 2020 Long Range plus - and we're usually at the same speed.

 

[Jughead135]

no one will ever drive it until it stops unless they're being followed by a diesel powered generator

 

[Jughead135]

Friction brakes kill efficiency. Regen had limits which are exacerbated by SoC, ambient temp, pack temp, and peak power demand.

Ah! So, you guys are talking about using (over-using) the brake pedal in lieu of allowing regen to do its thing...? I guess that would boil down to driving in such a way that requires friction brakes, vs leading stops / slow-downs with enough distance to use max regen?

Understood on the "limits" regen faces, but there's not much the driver can do with that--it is what it is.

 

[TIppy]

Help me understand please. I’m following the math and it makes perfect sense. I understand. But doesn’t that make the required power curve for against speed increases linear?

Power equals force times velocity, P = FV;

Drag (a force) is 0.5 * density * speed squared * Cd * Area. Here Cd = 0.24, but you also need the frontal area of the car. This is one of the forces captured in the coast down procedure used to build the second order equation for the road force discussed below.

This is how the force on the car changes with speed:

pounds = 33.40 + 0.5032 * mph + 0.0106 * mph * mph

The last term with the coefficient of 0.0106 says that the force has a component that changes with the speed squared (aerodynamic drag). In addition to this, you must multiply the force by the speed to get the required power, so the power required is a function of the speed cubed.

But to get the wh / mile you must divide the power (in watts) by the speed (in mph), so wh / mile is a force not power. That it's a force is also apparent because it is energy (wh) per distance (miles) and the definition of energy , E = F * d, gives F = E / d.

 

[quickstrike12]

@TIppy
Thank you for that. The math behind this topic fascinates me. Some of it is over my head but that’s ok. I like to learn.
Appreciate your time.

 

[TIppy]

@TIppy
Thank you for that. The math behind this topic fascinates me. Some of it is over my head but that’s ok. I like to learn.
Appreciate your time.

You are welcome, Happy to answer any other questions you have.