Hacker News new | ask | show | jobs
by fiddlerwoaroof 1997 days ago
Teslas turn off regenerative breaking if you charge too far, but the car tells you that pretty prominently: however, the car handles very differently without Regen brakes.
4 comments

This was the dumbest thing I experienced with my Model 3. I was expecting my regen brakes to work after I pulled out and I had to brake super hard after.

Why can’t Tesla just give us an option to keep it consistent? I guess adding video games is more important.

Tesla has been religious about never blending regen and mechanical braking. The accelerator only controls the motors, the brake pedal only controls the brakes. They’ve broken that somewhat now with one-pedal driving mode. The only way they could make regen feel consistent with a cold or full battery would be to blend in friction braking to make up the difference, which is just not something Tesla is into doing.

My BMW i3 did this really well, FWIW.

I mean, the solution is not to charge past 90%: for most practical purposes there’s no reason to go past 90% anyways.
Because that's not how batteries work.

Regen works because the batteries are being charged by the induction of the motors taking energy from the moving car. If the batteries are full they cannot take any more change and hence cannot draw power from the motion of the car.

What do you expect them to do? Dump the current on a giant resistor and hope it doesn't melt?

> Dump the current on a giant resistor and hope it doesn't melt?

That's called an induction brake and it's already widely used in trains and semi trucks. That said, you still need a friction brake at low speeds, since the induction braking force is proportional to velocity.

Simulate the same braking curve with the physical brakes as when regen is functioning?
I suppose that's one approach. Mechanical brakes work by converting the kinetic energy of the vehicle into heat via friction, it seems fitting that inductive brakes would dump energy to heat by running current over a resistor. Though under normal braking circumstances I don't know what percentage of the vehicle's energy is lost to the brakes themselves and what percentage is lost to the friction of the tire with the road.
Can anyone do the math and tell me how big that resistor would need to be? I’ve always wondered how many watts are being dissipated by my brake pads and rotors when I stop.
Assuming your Tesla is around 2000 kg and traveling at 30 meters/s, it's got about 900,000 J of kinetic energy. Stopping in 10s is some serious power that you'd need to throw off.
All of yhe 90 kW doesn't need to be radiated in real time though as it's ok for the brake elements to heat up.
> What do you expect them to do? Dump the current on a giant resistor and hope it doesn't melt?

You know...they could use it to boil water for a built in coffee/green tea maker.

Just a thought...

Can’t you also set the car to never charge to actual max, giving regen breaking some leeway? (If this is somehow a problem you run into often)
Yes, and that is the default. But people do charge to 100% occasionally for road trips (or for escaping a hurricane path, etc).
Annoyingly, newer Leafs don't let you limit how much you charge anymore, and the regenerative braking curve changes based on how much battery you have…
I guess that's easier and cheaper than adding a circuit to keep regenerative braking on, but dump the energy into a big resistor...
Aka rheostatic braking. Trains do this, but they have pretty large radiators for it.
that would be an unnecessarily complicated way of reimplementing friction brakes.
Most of the hardware is already there for the regenerative brakes, and it allows the driver's muscle memory with respect to braking pressure to still apply in that edge case. I'd be surprised if it were simpler to make the friction brake fallback behave identically to the regenerative system when the battery is full.

Maybe it's not worth the cost/complexity anyway, but the idea has merit somewhere in the design space.

If a Tesla would use rheostatic braking, I think it would require more cooling capacity than the cars are currently designed for. If the resistor couldn't be kept cool, then the car would have to revert back to friction brakes anyway. More cooling probably means bigger air intakes, which I think would run contrary to their aesthetic goals.
Rheostatic and friction brakes both basically convert kinetic energy into heat. Is there some fundamental reason why a resistor would be harder to cool appropriately (low thermal conductivity comes to mind, though I'm pretty sure there are high performance ceramics used for both brakes and resistors -- my heuristics aren't really good enough to make a good guess for this one)?
Well, friction brakes are cooled by the airflow in the wheel wells. Perhaps you could aircool the resistors in the same place. The friction brakes would probably still be there though; would there be space for both?
Shouldn't the resistor having to end up dissipating exactly as much heat as the friction brakes do, by conservation of energy? Is there a reason that it's inherently mechanically harder for the resistor to dissipate that heat?
not really, but you have to have the friction brakes anyway, so why add more parts?
and then you need extra engineering and production costs to cool those resistors :)