Bugatti Chiron’s braking power

I’m a big fan of cars and the new Bugatti Chiron is remarkable as it claims to be able to accelerate to 400km/hr and decelerate back to a complete stop in a span of 42s. It takes the Chiron 32.6s to reach its top speed, but only 9.3s to come to a complete stop. I wanted to take a closer look at this deceleration.

The Chiron goes from a velocity 400km/hr to 0km/hr in 9.3s and 491m

400km/h=111m/s a=-11.9m/s via vf=vi+at, a=-12.5m/s when vf2=vi2+2aΔx is used. This suggests that the deceleration of the car is not constant. As the formula using distance suggests that the deceleration of the car is greater than that of the deceleration calculated using time, the car most likely has a higher deceleration initially at its initial top speed before decreasing, allowing the total distance covered during the deceleration to be smaller than had the car been decelerating at the constant speed that was suggested with the time given. For the rest of the equations I will be using a=-12.3m/s as that is the average between the two.

I wanted to see how much force the brake pads would have to exert on the given tires, the back and front tires are different sizes, in order to accommodate for this deceleration. I assumed that the mass of the tire was located in the tread of the tire and that the brake pads were a point force acting one third of the way out from the center of the wheel. I am also estimating the mass of the tires.

2 tire sizes: 285/30r20(front tire) has a radius to the outside of the tread of 340mm, a radius of 255mm to the inside of the tread, and a mass of 10.5kg. 355/25r21 has a radius to the outside of the tread of 356mm, a radius to the inside of the tread of 266mm, and mass of 11.0kg.

Front tire: a=r⍺, -12.3(m/s)/340mm=⍺=-36.2rad/s2. τ =I⍺, τ=rfsin(θ), I⍺=rfsin(θ). I’m going to assume that θ =90, leaving I⍺=rf. I =½ M(R12+R22)=½ (10.5kg)((0.340m)2+(0.255m)2)=0.948. r=0.340m/3 as it is applied one third of the distance away from the center of the wheel. I⍺=rf, (0.948)(-36.2rad/s2)/(0.340m/3)=F=302N

Back Tire: a=r⍺, -12.3(m/s)/356mm=⍺=-34.6rad/s2. τ =I⍺, τ=rfsin(θ), I⍺=rfsin(θ). I’m going to assume that θ =90, leaving I⍺=rf. I =½ M(R12+R22)=½ (10.5kg)((0.356m)2+(0.266m)2)=1.04. r=0.356m/3 as it is applied one third of the distance away from the center of the wheel. I⍺=rf, (1.04)(-34.6rad/s2)/(0.356m/3)=F=303N

What is truly remarkable about the physics is that despite the back tire being bigger and thus decelerating at a slower rate, the difference in radius accommodates for this, and ultimately the two sets of tire have approximately the same braking force. The problem with the calculations is that they do not accommodate for the momentum of the car as theoretically the cars could stop spinning and it would still continue moving until the friction force between the wheels and the ground caused the car to stop moving. These do however show the force required to stop the wheels moving at extraordinary speeds to come to a complete stop.