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      09-22-2014, 10:49 PM   #36
Hujan
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Quote:
Originally Posted by SenorFunkyPants View Post
See
http://hyperphysics.phy-astr.gsu.edu/hbase/crstp.html

Its not that heavier cars stop in the same distance just that the brakes need to do more work to make them stop in the same distance.

Also
http://en.wikipedia.org/wiki/Braking_distance
That the brakes might be able to overcome the effect of additional weight in order to stop the car in the same distance as it would without the weight, does not mean that weight has no effect on stopping distance.

I sense that you and I actually agree: For the most part, the chief limiting factor on stopping distance is friction, as a factor of the tire (or "tyre") grip and the road surface. Most cars are built with brakes that have ample extra braking force, so that if you load the car with passengers or cargo, the braking distance will not suffer much if at all (assuming similar surfaces and tire conditions).

But as you pointed out, the brakes have to do more work to stop a car that weighs more. This work will translate to additional heat, which will cause the brakes to heat up faster than those same brakes on the same car but with less weight in that car. (The same would be true of the cars, tires, and weights were the same, but one car was traveling 80 miles an hour faster than the other car when the brakes were applied.) The extra heat starts to build up on the track and braking distances will start to grow.

It's probably true that additional weight will not increase stopping distance significantly if at all in a single-stop scenario, assuming dry tarmac and good tires. But this is because cars are built with ample braking capacity, not because weight has no impact on braking distance.

As the Wikipedia page you linked to said, the vehicle's mass is a critical component of the stopping-distance formula:

Quote:
The theoretical braking distance can be found by determining the work required to dissipate the vehicle's kinetic energy.

The kinetic energy E is given by the formula:

E=\frac{1}{2}mv^{2},

where m is the vehicle's mass and v is the speed at the start of braking.
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