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Help Me Decide Between Long Range and Performance
- Thread starter NinjaFoot
- Start date
. Not surprisingly, the new M3P is slower than the old M3P because of the wheel/tire combo. I have no doubt that if you put 18s on the new M3P, the old M3P, and LR AWD w/Boost and did a roll race at 60mph, they would all be equal. I test drove both LR AWD and Performance and there was no question I wanted the Performance. The acceleration of the Performance was literally the only selling point for me getting a Tesla anyway. It’s almost as quick as my GT-R but is silent and doesn’t draw any attention. I don’t take long trips in my Model 3 so the added range of the LR was not nearly as important to me as having a quicker car.
MagnusMako
Member
The guy on Engineering Explained on Youtube actually released a great video on Youtube today talking about Tesla and just this particular topic.
Here's the video that everyone should watch:
I just watched that earlier today and it's an excellent video. It also makes sense that as long as your brakes are powerful enough to lock up your wheels (necessitating ABS activation), then the limiting factor on braking performance is basically tire grip. If the tires on the P and non-P models were equal (they aren't from the factory), they'd probably brake exactly the same (at least the P and LR would since they weigh the same...SR+ would beat them, and it may already beat them in braking stock). As others have said, the advantage to bigger/performance brakes is that they can maintain performance longer under stress (ie, when driving on a track). There is no difference for road use. It's all in the tires.
The SR would brake the same. This is an acceleration calculation. It takes mass into account. The force would be different but the accel is the same.
SR has the same wheels and brakes as the LR but it weighs less due to the smaller battery pack, so it should be able to stop more quickly.
F = ma. The calculation goes around the acceleration (a). So the amount of force required to stop the vehicle is the same as the amount of force required to accelerate. Being lighter it requires less force and less energy to stop an SR than a LR but the whole argument around tires is independent from the mass of the vehicle. The tires are the constraining factor. So no, being lighter does not affect the acceleration/deceleration, only the amount of energy required to do so.
Mahamilto
Member
Assuming the tires\friction is the limiting factor, and is constant between the heavier and lighter car, the lighter car still would stop in a shorter distance because of exactly what you just said. The friction isn’t zero... the inertia you are overcoming is inherently less because of having less mass, so if the force is a constant between the two calculations you are still looking at a = F/m... a smaller denominator means greater acceleration (or deceleration) and the F is the sum of the forces.
Mahamilto
Member
Time = (vf- vo)/a... implying a larger acceleration results in less time to stop.
d = (initial velocity x time) + (at^2)/2.... so a smaller time means smaller displacement (distance) to stop.
Well that’s just it. Friction is NOT constant. Also the force in each case is different. The acceleration is the same which is why weight is not a factor. This cease to be true if you are skidding in which case the weight does play a role (lighter better). But under “ABS” braking, the tire quality is all that matters.
Re calculate the equation of motion. Look at the video. Weight is not a factor.
Mahamilto
Member
brain fart. The friction isn’t the same. The coefficient of friction is (the tire compound)... the friction itself is different because friction is in and of itself calculated on the basis of the normal force on the object; which is relative to its weight (mass x gravity; gravity being constant). Thus negating mass from the equation of displacement.
That's super counterintuitive but yeah essentially, the "braking force" is proportional to the friction coefficient which varies linearly with mass so it cancels out. Disclosure: I am a Ph.D. in Physics.
Mahamilto
Member
However
“For rubber tyres on cars, the coefficient of friction (μ) decreases as the mass of the car increases. Additionally, μ depends on whether the wheels are locked or rolling during the braking, and a few more parameters such as rubber temperature (increases during the braking) and speed”
I’d need to dive further into how much the coefficient of friction decreases based on mass so it stands to reason the car with less mass may still stop shorter...
“For rubber tyres on cars, the coefficient of friction (μ) decreases as the mass of the car increases. Additionally, μ depends on whether the wheels are locked or rolling during the braking, and a few more parameters such as rubber temperature (increases during the braking) and speed”
I’d need to dive further into how much the coefficient of friction decreases based on mass so it stands to reason the car with less mass may still stop shorter...
Mahamilto
Member
Not a PhD. Just a physics minor with an MD. Haven’t visited classical physics in awhile; all I deal with now is radiology physics for boards purposes. Very different side of physics from classical kinetics.
Mahamilto
Member
So, as I admittedly don’t know the variation of change (decrease it seems based on what I’m reading) of the coefficient of friction for standard air filled tires based on the load of the tires themselves, care to comment more?
Seems this would bring back the possibility that mass does indeed play some roll in stopping distance due to the intrinsic property of the tire itself.
I mean weight does change things because 18-wheelers' braking distance is much longer because they are heavier. But they are like 10x+ heavier, not a couple hundred pounds heavier. So the bottom line is for consumer streetcars, tire quality (and air pressure) is the most important factor*, in which case weight is not. In the case of SR vs LR, it would measure I think. Difference may be an inch or two if anything.
Think of it this way: in driving classes the napkin calculations around following distance are never mentioning weight of vehicles except where large trucks (And trains) are concerned.
* braking from the same starting velocity
Think of it this way: in driving classes the napkin calculations around following distance are never mentioning weight of vehicles except where large trucks (And trains) are concerned.
* braking from the same starting velocity
Mahamilto
Member
I hear you, just wondering the relationship of mass and the coefficient of friction of the tire itself; is this linear? Exponential? If its linear and 1:1 proportional then the loss of coefficient of friction due to mass/load on the tire itself would negate the argument that weight doesn’t impact stopping distance.
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