# How much braking force do cars generate?

I've been trying to find actual figures for the braking power of cars (in order to determine things like stopping time). As I understand it this comes in two parts: the maximum force that the brakes can apply on the tires, and the maximum force the tires can apply on the road.

I'd love to find figures for particular cars (or brake systems or tires) but I would be fine with any figures for actual cars under reasonable conditions. It's easy to find math problems of this sort online but I don't expect their figures to have much in common with the real world.

If someone could point me in the right direction (car data sheets, third party testing, government regulations, etc.) I'd be very much obliged.

• It depends on many factors like road and tyre texture, tyre width (relative to car weight), and further on the condition of the tyre; its temperature, level of wear, humidity, pressure etc. In general on normal vehicles you can't really expect more traction force, than the force on the tyres themself by the car's weight, which would give you max. 1G of deceleration.
– Bart
Jan 21 '18 at 18:11
• Agreed - just been driving on fresh snow and braking distance is normal * 20 so easier to plan and use engine braking... Jan 21 '18 at 19:49
• Reviews will often have stopping distance Jan 21 '18 at 20:25
• @SolarMike I'll take any numbers you can dig up, but ideally I'd like numbers for dry asphalt. Jan 21 '18 at 20:50

To elaborate on my comment; obviously generated braking force is defined by how hard you push the braking pedal. The required force to lock the wheels (given that ABS is disabled) depends on how much traction the tyre has. Let's say our tyre has a constant friction coefficient of 1. That's a simplified situation where the tyre has as much traction force, as there is force on the tyre by the vehicle's weight.

This allows you to decelerate(or accelerate) with a maximum of 1G or 9.81m/s2, before your tyres start slipping. That means that you'd get to a stop in 2.83s when travelling 100km/h. You can see from this figure, that most supersports cars are limited in their 0-100 time by the tyres. They use as wide tyres as they can on the driven wheels, to maximise traction. Heating up the tyres and lowering their pressure, and a slick surface on dry hot asphalt further increases their traction. But it doesn't allow you to accelerate to 2G or something.(except for crazy top fuel dragsters) But that's why i'm rather sceptical about Tesla's promised 0-100 figures for the announced roadster.

In reality, a friction coefficient is not close to constant and way more complex. In reality, your tyres are always slipping, which is the main reason they wear out. It gets worse in corners, acceleration, braking etc. but it's always present in a certain amount.

Also, traction sharply decreases once the wheel starts slipping by a certain amount. I believe that was usually 20% or something. You only have kinetic friction at that point, which is always lower than static friction. That's the reason ABS works, it constantly regains static friction by releasing the brakes briefly. I hope this gives you an better understanding of how and why tyres work.

• Thank you, this is great. Do you have any recommendations/resources I could use to get an idea of when you can exceed 1g and what can decrease your ability to get to 1g? It sounds interesting and applicable. Jan 22 '18 at 21:55
• @Charles It's been a while since I last calculated with tyres and i can't remember the literature i used, but I recommend searching for 'white papers' published by tyre manufacturers like Vredestein, Goodyear, Michelin etc. White papers generally quickly and clearly explain how stuff works without tiring you(lol) with all the complex stuff. Bosch also might've published stuff about the subject. Japanese companies are generally exceptionally generous with white papers about their technology.
– Bart
Jan 23 '18 at 16:15

braking systems in cars can be rated in terms of braking horsepower capacity, which is the net frictional force exerted on the brake disc or drum times the rotational speed of the disc or drum at the point of contact of that force. When described this way, car brakes can dissipate hundreds of horsepower in a hard stop.

I do not know if there are any mandated braking standards based on braking horsepower ratings, but as a practical matter the brakes in a car can be considered adequate if they are capable of locking up the wheels at any forward speed the car can achieve (in the absence of any ABS system) using the driver's muscle forces. This would guarantee that the braking system can exert the optimum braking force, defined as that generated just before rolling friction gives way to sliding friction.

• Thank you. What can be said of the maximum force of the tires on the road before skidding? Jan 21 '18 at 20:41
• the retarding force of the tire's rolling contact point with the pavement times the speed of that contact point will be exactly equal to the same product for the brakes as outlined above- that is, the tire horsepower equals the brake horsepower. Jan 22 '18 at 0:49
• They're equal unless the tires lock, so I'm interested in figuring out how much force the tires can take before locking. I assume that under most conditions the tires, not the brakes, are the limiting factor when braking hard. Jan 22 '18 at 3:28
• that's right, and tire performance is very hard to accurately model. Jan 22 '18 at 3:36
• ...but there has to be literature on this topic that one could research! Jan 23 '18 at 7:07