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FSAE Braking System: Preliminary Kinetics and Dynamic Dynami c Analysis Analysis « Previous / Next »
By Steven Holland / October 9, 2013 / FSAE / 2 comments
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Preface I wanted to give a brief background on how to approach the braking sys tem for an FSAE car. A thorough understanding of the physics in the brake system is essentially to component selection, the first thing the brake lead should should consider in his/her design. This was the method that my colleague, Daniel Stamper, and I took in the 2012-2013 2012-2013 year on the SDSU FSAE team. Please feel free to leave comments or your input on our approach. Note: the number here were only chosen for simplicity and don’t represent the actual values of our car or our analysis.
Simply put, an initial brake input force (force exerted by the driver) is converted through mechanical/hydraulic means to apply a much larger braking force on each of the 4 rotors. This converts the vehicles kinetic energy (speed) into heat. This is the basic principal that stops a car. The following will introduce the equations that will be involved in a brake system analysis and was adapted from the StopTech website at: http://stoptech.com/docs/media-center-documents/the-physics-of-braking-systems The focus of this document is to relate these equations in respects to brake lock, since this is an essential requirement that all FSAE cars must demonstrate at competition. The ultimate objective the year of our design was to minimize brake input force that resulted in brake lock. 3.
Assumptions In this analysis it is initially assumed that all parts are perfectly ridged, the fluids are incompressible, and heat loss is minimal. Essentially it is assumed that the forces are perfectly transferred in the processes outlined. Other notes and assumptions are referenced in the corresponding sections.
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Kinetics and Dynamic Analysis 1.
Brake Pedal The best place to begin in understanding the braking system is beginning with the drivers input force on the brake pedal. By summing the moments about the pivot tube, the force output can be found. F_out=F_in (d_1/d_2 )
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Bias Bar That force is then transformed via the bias bar (also called the balance bar), again by summing the moments the force going into the master cylinders can be determined. This bias bar is typically adjustable, so depending on the driver and driving conditions c an be fine-tuned. Note: It is important to consider that when the bias bar is actually adjusted, the angle between the master cylinder piston rods and the bias bar deviate from perfect perpendicularity. So further consideration should be taken when dramatic adjustments are required. F_out=F_in (L_1/L_total )
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Master Cylinders and Caliper Pistons Using Pascal’s Law, a relation is developed depending on the master cylinder bore size and the calipers piston diameter to equate the force acting on a single rotor. This can be calculated in the following manner for each rotor. You can see from this equation that in order to increase braking force, you can either decrease the master cylinder size, increase caliper piston area, or increase the pedal ratio. Note that increasing brake pad size does not increase braking force at the rotor, as the equations indicates, only an increase in piston area or number of pistons increases clamping force on the rotor. F_out=F_in ((2A_cal)/A_MC )
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Brake Pads The caliper’s piston is then pushed against the brake pad, where the actual braking force can be determined depending on the brake pads c oefficient of friction. Note: This friction c oefficient is often provided by the manufacture, and is highly dependent on temperature. It is assumed that the rotor and brake pad have properly been embedded. Brake pad embedding is where
F_out= μ F_in
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Braking Force That braking force is itself dependent on the radius of the piston in relation the diameter of the tire. A larger radius tire will require a higher brake torque, increasing rotor diameter increases brake torque produced by the cars brake system. Ideally, you want as large a brake rotor as possible, however we are limited by things such as rim size, suspension travel interference to name a few. It also adds the cars unsprung weight which is undesirable. F_out=F_in (r/R)
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Weight Transfer Here is where things start to get a bit tricky in the brake system. I’m sure all of you notice that when you stomp on the brakes in your own car, you have a tendentiously to move forward, this is weight transfer. So once the braking forces acting on each tire are found, you can solve for the deceleration using the basic kinematic equations F=ma. The weight transfer can then be found depending on that deceleration rate, the height of the c enter of gravity, the weight of the car, and the length of the wheel base. This new weight distribution is called the dynamic weight distribution, and is simply the static weight distribution including the effect of weight transfer WT=(W_t H a)/WB
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Brake Lock That weight transfer is added to the front wheels and subtracted from the rear wheels. Then the force required to lock the brakes can be determined by simply multiplying that weight times the static coefficient of friction of the tire on pavement. Any applied force below this value, the brakes do not lock, above this, the
conditions using high and low coefficient of frictions between the tires and pavement. This way, by adjusting the bias bar, we can achieve maximum braking performance in all driving conditions. F_lock=μ W
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The Problem Our focus is to ensure that all drivers are capable of achieving the brake force required to initiate brake lock, therefore a more thorough investigation should take place. The weight transfer phenomenon is dependent on the deceleration of the vehicle (w hich varies with applied brake force) and the braking force at brake lock must be equal to the resulting frictional force from weight transfer. It is also important to consider the fact that not all tires will lock at the same instant, and this too will vary the weight transfer. Since these calculations can be very tedious, a computer iterative process must be done to solve for the force of brake lock in a timely and accurately manner (will be discussed in a fore coming blog post). F_lock (a(F_brake))=F_brake (F_in)
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Tire Data
Note: the best fit polynomial below was chosen with the assumption that a single tire load will never exceed 350 lbs. since the equation diverges from tire data. Also, the frictional value derived from the tire data should only be taken tentatively until a correction factor I applied from actual testing conditions. (-7.5×10^(-008))W^3 +(4.891×10^(-005))W^2 + -0.01082W + 3.105
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Tags: ar13, brake, equations, FSAE, sdsu, system
2 comments
Jon Sullivan
September 14, 2014 at 9:12 pm
Your diagrams need to be redone. It should be noted that the opposing piston s hould not be counted in the MC and caliper piston forc e calculations. What is currently shown in the diagrams for “brake pads” and “Master Cylinders and Caliper Pistons” is that the total caliper piston area to MC area is 3:1, which with the 200 lb force being applied to the MC results in 600lbs of clamping force with the brake pad coefficient of .5 this results in 300 lbs of friction force. However the area ratio should only be 1.5:1 because the piston on the other side of the caliper is only providing a reaction force but allows the pistons to center about the rotors position. so the clamping force should only be 300 lbs and the total friction force due to the brake pads should be 150 lbs. If you placed both pistons on only one s ide of the caliper and a fixed pad on the other (floating caliper setup) then the numbers would hold true because the fixed pad would be providing the reaction forc e. If possible please change this because I can almost guarantee a FSAE team will improperly design their brake system based on this haha.
Steven Holland
September 14, 2014 at 9:50 pm
It has been a while since I looked at these equations, but I believe my depiction is still true. I
please if I do have it wrong, let me know. I know my old team is still running my numbers and would like to notify them if I’m wrong. Thanks for keeping me on my toes .
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