This sort of vehicle dynamics question best addressed by Racing Car Vehicle Dynamics
What follows is a basic discussion at the high school physics level. As you will see from the reference text, high school physics is insufficient to statically model the complete vehicle system. A dynamic model is required to agree with easily obtainable experimental data.
Assuming brake pads which are able to lock up the tires and ABS, and
ignoring downforce, is it all just down to the tires?
Weight of car - physics tells me that friction is proportional to the
normal force, so I figure this doesn't matter
Incorrect. Mass is the dominating factor.
Your critical equation here is:
d is your stopping distance and, as you can see, it is defined in terms of your acceleration
a (deceleration in your case), your initial velocity
v_0 and the time required to get to speed zero
F is the decelerating force of the entire vehicle and its braking system while
m is the mass of the vehicle.
For completeness: to solve for
t, you're best off using the energy equations. However, the above two equations are sufficient to do a numerical approximation in the spreadsheet of your choice.
So, all other things being equal, you can easily see that the mass of the vehicle dominates the entire system. If you change nothing else, a lighter vehicle stops faster (higher acceleration for the same force). Two identical cars with a different number of passengers will slow at different rates.
Experimental confirmation: All of this discussion is fun but solid physics should be confirmable (or falsifiable) using an experiment.
- A fully fueled car.
- A measured course for 60 mph braking tests (i.e., a straight track, a Brake Here sign and 10 meter markers).
- An additional load of 20 fifty pound bags of sand.
Perform ten braking tests from 60 mph without the additional load. Measure the distance required to stop on each run. Note that back-to-back runs will likely induce brake fade (the distance will increase with brake temperatures).
Refuel the car.
Add 10 fifty pound bags of sand to the vehicle (for a total of 500 additional pounds). Repeat the same braking tests from 60 mph on the same course. Note the stopping distances. Note that the distances required to stop are longer. Note that back-to-back runs are resulting in increasingly severe brake fade (i.e., those long stopping distances are getting longer much faster).
Refuel the car.
Add 10 more fifty pound bags of sand (for a total of 1000 additional pounds). Repeat the same braking tests. Note that stopping distances required are much longer. Note that back to back runs are resulting in brake fade much more quickly (possibly reaching a state of terrifying as the brakes seem to cease to function).
Returning to our original equations, we can see that as
m increases, there is a distinct increase in the distance
d and time
t required to stop. If
F were a pure kinetic or static frictional force, we would expect it to increase linearly with
m, resulting in nearly identical stopping distances.
As this is not the case at all, we can conclude that
F is not an idealized frictional force.
So assuming the same tires, and brakes that can hold the tires at that
maximum static force, I don't see why mass would increase braking
Yes, the situation would be very simple if a car were an ideal solid block from a physics textbook (i.e. a spherical chicken). In your comment, you're substituting
m in several places as if it were always the same on every wheel. This would be the case if we were talking about a physics problem. Unfortunately, an actual car is a hollow box sitting on springs, riding on round balloons made of rubber and steel.
When you mention the normal force, you're talking about the force of the tire patch on the road. This situation would only match the basic friction function if the vehicle were non-moving with locked brakes. There's an amount of force that's going to be required for me to push this heavy thing across the parking lot (with tires screeching the whole way, I assume). Heavy car => harder to push.
Unfortunately, none of the above is actually relevant.
The reality is that the normal force used for kinetic friction to stop the car isn't the tires. It's actually the force of the brake pads on the rotors (or drums but I'm assuming disk brakes for the sake of having something easy to point at). Their clamping force is what actually slows the wheel. That only results in slowing the vehicle if the tires are statically coupled to the surface. A wheel that stops spinning is skidding, not braking.
Suspension stiffness - this would lessen the front dipping, but does
that even affect braking distance?
Suspension rebound - a poor rebound
would possibly cause momentary airtime if the road is bumpy, would
this be substantial?
Chassis rigidity - would a chassis that warps a
bit more under heavy forces cause some tire skipping or something?
Anything that increases the tire contact patch will increase the ability of the road to induce a torque on the wheel (and vice versa). A big patch will grip the road well, keeping the wheel spinning and allowing the brakes to use more gripping force before the wheel stops, skidding the tires. A smaller patch grips poorly, resulting in a skid with much less gripping force.
This is where your ABS system attempts to optimize a bad situation: it tries to keep all the wheels from stopping by relaxing and re-gripping the clamping force of the brakes near the limit of traction. This modulation of the brakes is exactly what a skilled driver will attempt to duplicate.
Why is it worse to have the majority of the load on two tires instead
Again, the braking system is the totally reliant on the use of the brake disks and the clamping force that it can apply before the wheels lock. As the weight leaves the back tires, their contact patch approaches zero. As a result, the braking system cannot apply much clamping force before those wheels lock (i.e., they're now out of the picture).
However, shifting the weight forward has increased the contact patches of the front tires but they haven't doubled (this is a consequence of many things, including that each is an air-filled balloon that doesn't spread linearly with an increase in the weight it supports). As a result, each front tire is now receiving less than two times the torque that each tire was receiving before the weight transfer. Consequence: each front rotor can be gripped with a higher force but not double the force, resulting in reduced braking and longer stopping distances.
ABS quality - ignoring EBD, is there a substantial difference in the
effectiveness of ABS between car models?
This can really only be answered in terms of a specific situation. What is the scenario? What tires are in use? Snow vs. ice vs. sand vs. rain? Cold or hot tires?
In general, any modern ABS system is better than none. It is close to optimal when compared to the average driver, their situational awareness and their reaction times.
Can a very skilled driver brake better without ABS than they would
I firmly believe that Michael Schumacher could out-brake me on the track, in the same vehicle, no matter what ABS system I were to use.
Unless you're a champion Formula One driver (or an effective equivalent), I put it to you that this is a meaningless comparison when you talk about driving in the real world, on real roads with real people who aren't paying nearly as much attention as they should.