This is long, blame
@kbM3
Really though, I'm glad I wrote this as my view on the situation changed as a result of it
So, what we know(ish): 0.3g of road deceleration to achive 6.7 MPH/s
The vehicle has accelerometers as part of the ABS/stability system. It also has at least one axis of gyroscopic sensing for stability control (yaw). It also has wheel speed sensors on each hub for ABS/ traction control.
Out of all of these, the wheel speed sensors are the most non-drifting and should be fairly accurate (its used used for the speedometer), so wheel counts over time gives speed and change is wheel counts over time over time give acceleration. This works well for averaging also. So one might think Tesla is using that with a 6.7 MPH/s acceleration limit (over the actual sample time) for the hard braking threshold.
However, acceleration is also available and would require less code to implement, so what happens if they used that?
Well, it depends on if they adjust their reference frame to back out gravity. On a flat surface while stopped, gravity is tangential to the road, Az (acceleration vertically) is 1G, Ay (lateral/turning) is 0, and Ax (forward/backward) is 0.
If stopped on a 30 percent downhill slope, Az =cos(30)=0.866, Ay=0, Ax=0.5 (gravity is pulling the car forward).
An accelerometer without orientation correction would view this situation as the car decelerating. Why? Because the force is pulling the vehicle forward which has the same effect on the accelerometer as the car accelerating backwards.
This is F=ma and Newton's laws at work. The accelerometer itself is a mass connected via spring like load cells to its housing. The mass's inertia resists changes in velocity and this produces a displacement and force measured by the load cells when the device is accelerated.
When the car accelerates, the rear of the packaging the accelerometer is housed in pushes against the sensing element in the forward direction. This force is read as positive acceleration (normally forward speed increasing). When braking/ decelerating the load element pushes against the forward edge of the packaging and a negative acceleration is reported.
Now, back to the incline. Gravity is pulling on the entire car, including the load cell in the accelerometer. Since the car is stationary, the accelerometer housing is also stationary and the forward edge pushes back on the load cell. Net result, it looks like negative acceleration. And, from the car's point of view, it is producing a -0.5G acceleration to keep the car motionless in the presence of the 0.5G forward acceleration due to gravity. (Note, that this is the result of Accel=Force/mass, with force generated being equal to the gravitational force and no work being accomplished).
One might think the system should have a three axis gyroscope and transform (rotate) and subtract out the gravitational acceleration to provide a correct picture of what the car is doing, and for some applications that might be true (like derived motor power calculations).
However, let's take a step back and look at what the system is trying to accomplish. The vehicle's phyiscal interaction with the world comes down to four areas of rubber interacting with the road. These forces do depend on the accelerations as reported. A stopped car is not providing acceleration via the motor, but the tires are providing an acceleration to prevent the car from moving. This acceleration corresponds to a force that the tire must exert on the road. If we tilt the road steeper and steeper, eventually there will be an angle where the force due to the vehicle's mass times acceleration due to gravity in the forward direction (gravity * sine of the angle) is greater than the force the tires can provide via the coefficient of friction times the force normal to the road surface (acceleration due to gravity time vehicle mass times the cosine of the angle).
At this point the car is uncontrollable, the circle of friction has been exited and we're in the skid zone. Not good from a control point of view. Now, use magic to stop the car and back off the slope just untill the tires can hold it again. How much braking authority do we have? None, were we to start rolling, there is no margin to stop the car again. It doesn't matter how little of an acceleration we are requesting, the system physically can't provide it.
So, if you want to make a system that stays in control of the vehicle and the vehicle is limited by grip do you care most about :
A: the change in vehicle speed
B: the acceleration due to the motors/ brakes
C: the accelerations/ forces at the tires?
I would say C
A and B are different ways to measure the same thing (until grip is lost)
C is the real world limit of the system.
Another example of why C is the critical constraint is banked curves. A car on a flat surface can only generate a certain amount of lateral acceleration (rate of turn) based largely on tire grip. However, by banking the curve such that the slope matches the resultant vector of gravity and the lateral acceleration, the car can controllably handle much high speeds/ lateral accelerations/ turn rates irrespective of grip. To an accelerometer in the vehicle, this maneuver, rather than looking like a turn with lateral acceleration, would instead look like gravity were increasing as the turn, slope, and gravity combine to form a vector tangential to the road surface.
Now imagine the opposite situation, the problematic reverse banked turn. In this situation the curve is banked but the opposite way, now we lose grip due to slope and the latteral acceleration makes us lose more. To the accelerometer, the bank appears as additional lateral acceleration and gravity appears decreased, exactly what the tires are dealing with.
Okay, I've been typing for
over an hour almost two hours, so I'll try to wrap this up. If Tesla is dinging people when braking downhill at a lower level than when going uphill, they are not wrong in doing so. When headed downhill, there is less traction available for stopping due to gravity and the slope working against you. In the uphill case, they act to slow you, so less tire/ braking force is needed for an equivilent deceleration.
Further, if this is the case, the 0.3G figure is the driving parameter and the 6.7 MPH is an illustative value, but is only valid on a flat surface. The 0.3G is then really the coefficient of friction/ allowable tire force value the system uses as a do not exceed value.
So in the downhill case, one must start breaking earlier and with less force than if one were on a flat surface. Even though you may only be requesting 0.3G and the car is only slowing at 6.7 MPH, the tires are dealing with that plus the effect of the slope. The tires also having less of the car's mass to use toward friction, but that is not a huge factor. For a 10% grade (5.7 degree), forward acceleration due to gravity is 0.1G (same as the grade percentage) and normal force for the tires is only 0.5% less. Slowing the car at 0.3G in this situation requires 0.4G effective braking force (and that's what the accelerometer would report).
Final thought, think of regen, downshifting, or (worst case) riding the brakes down the side of the mountain. It requires continuous deceleration to keep the speed constant, and the tires are working even though the speed is not decreasing.
P.S. yes aerodynamics also figures in. In the mountains without brakes, your speed would stabilize once aerodynamic drag plus rolling resistance plus other losses equaled the force due to gravity with no dependence on tire friction. However, one cannot rely on calm air nor a head wind to stop the vehicle. Rather, to be conservative, one would plan for a tail wind. Also, aero force goes as the cube of speed, so less of a factor at lower velocities.