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I thought of a maths problem for myself to solve, as it seemed like a good problem to try to get my head around. My issue is that I got stuck at the first hurdle. I was reading up about the braking systems on f1 cars and it said they the drivers put around 1500N of force through the brake pedal, and the coefficient of friction between the brake pads was about 0.6 at optimum temperature. I would have thought that this would have then given the car a decelerative force of 900N. f1 cars weigh 700kg, so this would give a deceleration of 1.28ms^-2. This is very far from the real value, as f1 cars decelerate at up to 5g, which is 49ms^-2. Can anybody figure out where I have gone wrong as I have been thinking about it for well over an hour by now.

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  • $\begingroup$ Is it 1500N through the brake pedal or per pad? (You would be off by a factor or 10 instead of 40... :) ) $\endgroup$ – Matt Jun 27 at 21:52
  • $\begingroup$ 1500N through the brake pedal, I have also realised where I have gone wrong as I didn't take into account for the moments in the wheels. I am having another approach st it using the energy equations. $\endgroup$ – finlay morrison Jun 27 at 21:57
  • $\begingroup$ Have you included the brake disc diameter? $\endgroup$ – user207455 Jun 27 at 21:58
  • $\begingroup$ yea I have got to this point atm "braking power = (FxV)/D" where 'f' is the force at the pedal, 'x' is the diameter of the brake disk, 'v' is the velocity of the car and 'D' is the wheel diameter $\endgroup$ – finlay morrison Jun 27 at 22:05
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You are missing at least two things here.

First, the total braking force is not directly applied by the driver's foot. All modern road cars, and F1 cars, have a servo assisted braking system so the driver's foot controls the amount of braking force, but doesn't create all of it directly. But having said that, a total force of 1500N would be reasonable for a road car. I don't know about the specification for F1 cars.

Second, that force is then multiplied by the hydraulic braking system when it is applied to the wheels. A hydraulic (fluid filled) system transmits pressure, i.e. force/area, not force. If the area of the pistons in the "wheel cylinders" that operate the brake calipers is say 10 times as big as the area of the "master cylinder" which is operated by the drivers foot (boosted by the servo system mentioned earlier), the fluid pressure will be the same in all the cylinders, but the force acting on the brake shoes will be 10 times bigger because the area is 10 times bigger.

The downside of that is that the piston in the master cylinder (and the driver's foot) has to move 10 times as far as the brake pads, but that isn't a big problem because the pads themselves only move a fraction of a millimeter from "off" to "full on."

To summarize, the force decelerating the car can be much greater than the force the driver applies to the brake pedal.

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  • $\begingroup$ Your first point is a bit far fetched. I'm sure OP didn't think that a pilot could lift 150 kg in earth's gravity with the tip of his toe... It was just a shortcut. $\endgroup$ – Matt Jun 28 at 9:13
  • $\begingroup$ i understand that the force isnt all appled instantly, but i am just trying to get started by assuming it does, and as for assisted braking, formula 1 cars are not allowed any electronics controlling the brakes (apart from the energy recovery systems systems in the rear wheels). $\endgroup$ – finlay morrison Jul 3 at 13:35

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