Car weight and starting/stopping distance.

[Fuel]

We all know the the lighter a car is, ceteris parabis (center of gravity, tires, ect.), the faster it will stop or accelerate (assuming the car has the power to test the friction on the tires). But is this a linear relationship?

If the x-axis is the weight of the car and y-axis is 60-0mph starting/stopping distance, what would the graph look like, generally speaking?

I realize that when you lighten a car, you must adjust certain things such as the suspension. Let’s try to assume that all settings are optimal or assume such things as no suspension at all. I want to try and keep this question as simple as possible without dealing with the nagging details.

Thanks.

[Desmostylus]

The weight doesn’t really affect the stopping distance.

Say you’ve got a car with ABS, so the brakes will always be working at their best. Throw a few bags of sand in the car (distributing them so as not to affect the CG). The extra mass of the sandbags is exactly offset by the extra downforce on the tyres that the sandbags create.

The brakes will get hotter with the sandbags on board, but the stopping distance will be the same.

Of, course, you’ll eventually reach a point where the brakes can’t handle the load, and you need to put in bigger brakes.

[Jinx]

If the OP assumes that the braking force is constant (and initial velocity is a constant so we can compare apples to apples) then it boils down to simply a matter of F=ma where: m is the mass of the car. As m decreases, the rate of negative acceleration (deceleration). And, as m decreases, in short, we’d find this would result in a smaller and smaller stopping distance.
(Also we are ignoring drag, etc…which would cause your actual mileage to vary.)

However, this isn’t a perfect world, and braking forces are not constant…and there is drag…but at least there’s Wahlgreens!

• Jinx

[Jinx]

*Originally posted by Jinx *
…As m decreases, the rate of negative acceleration (deceleration). And, as m decreases, in short, we’d find this would result in a smaller and smaller stopping distance…

Oops! Me postus expidous! I forgot a critical word! I meant to say: …As m decreases, the rate of negative acceleration (deceleration) increases. So, as m decreases ** and acceleration increases**, in short, we’d find this would result in a smaller and smaller stopping distance…

I “Jinxed” myself!

• Jinx

[Desmostylus]

*Originally posted by Jinx *
If the OP assumes that the braking force is constant (and initial velocity is a constant so we can compare apples to apples) then it boils down to simply a matter of F=ma where: m is the mass of the car.

Why assume that braking force is constant? It isn’t what really happens, and it also leads to the wrong conclusion.

Leonardo da Vinci was the first to notice this - “friction produces double the amount of effort if the weight be doubled”.

[Jinx]

Desmostylus, I do not dispute your initial answer. But, I think the OP was trying to boil the problem down to its simplest form. Beyond your mention of DaVinci’s observations, rolling friction alone is whole different animal, and wind resistance can be a parabolic function above a minimum speed… But, the OP doesn’t need to go study aerodynamics just to get a feel for the most basic answer relating mass to stopping distance…as opposed to relating mass to how hard (or soft) the brake pedal must be exerted… Or, we could roll down a window and change the parameters yet again! - Jinx

[Desmostylus]

Wind resistance doesn’t enter into it. Nor does the pressure on the brake pedal, provided that you can supply sufficient pressure to lock the tyres, and that’s not that hard to do in a modern, well designed car.

Rolling resistance does affect things somewhat, you’d need to compensate for the extra weight by increasing the tyre pressures.

[Princhester]

Jinx I think Desmostylus is on the right track as far as the OP is concerned. Note that the OP says “assuming the car has the power to test the friction on the tires”. So (given that the friction on the tires is going to vary with weight) the power (on acceleration) or the braking force (on deceleration) is going to have to vary to test the friction on the tires.

[CookingWithGas]

A semi with a fully loaded trailer has a shorter stopping distance than one with an empty trailer. (Not sure why although it’s reasonable that the braking system is optimized for a full load.)

[Fuel]

*Originally posted by Desmostylus *
**The weight doesn’t really affect the stopping distance.

Say you’ve got a car with ABS, so the brakes will always be working at their best. Throw a few bags of sand in the car (distributing them so as not to affect the CG). The extra mass of the sandbags is exactly offset by the extra downforce on the tyres that the sandbags create.

The brakes will get hotter with the sandbags on board, but the stopping distance will be the same.

Of, course, you’ll eventually reach a point where the brakes can’t handle the load, and you need to put in bigger brakes. **

Holy crap, duh, my whole question was about something I already knew.

You can’t pull any more than a G on a flat surface, or incline/decline for that matter. the laws of physics do not allow any more, unless you start getting into adhesion (like at a good drag strip). so, given a set of good soft racing tires, you will not be able to brake or accelerate any faster than the tires will let you, no matter the weight. the tires are the deciding factor, nothing else.

sorry i wasted a whole post like this. but maybe someone learned something from this post!?

[scotth]

There is an advantage to less weight, but you have to adjust some other parameters to get that advantage.

With lower weight, softer tire compounds can be used. Shorter stopping distances could then be achieved. But, it would be a direct result of the softer tires.

In any vehicle that can generate aerodynamic downforce, lower weight will reduce braking distance. Again, another special case.