BRAKING SYSTEM

Design report of braking

BRAKES

The design criterion for the brake system is that it must lock all four wheels and comply with all the rules. The brake system consists of four wheel solid disc brakes actuated by a

dual master cylinder. The brake pedal uses a 6.2:1 pedal ratio to multiply the input force applied by the driver.

A side by side dual master cylinder of 1 inch bore each was used to meet the requirement of having two independent front and rear brake circuits. We chose this setup over a

tandem master cylinder and proportioning valve because it was easier to install and adjustable. We are able to adjust the front brake pressure by using balancing bar fitted on dual master cylinder. Driver from dashboard can adjust this and hence balance brake for

different driving conditions.

The brake calipers chosen were single piston floating-type caliper of maruti 800. These calipers were used because they were found to apply enough pressure on the brake pads to lock up all four wheels as required by the rules. We will be using 25-8-12 atv tires. So,

the disc and caliper of maruti 800 can be easily fitted in these tires as their rim size is similar to maruti 800 rim size. This will make fabrication easier and more efficient. The

brake rotors used are of maruti 800 and are 8.5 inch in diameter. Dust cover will be provided to protect it from dust and mud. Brake lines and hoses will be also of maruti 800. DOT-4 brake fluid will be used for braking purposes. The brake rotors will be

mounted on the outboard side, which allows them to receive more cooling air than the inboard design.

Specifications

Pedal ratio = 6.2:1 Pedal size = 7.75”

Braking percentage = 66:34 Master cylinder type: dual master cylinder fabricated by using two master cylinders and a balance bar.

Master cylinder bore size = 1” Brake fluid: DOT-4

Brake lines: maruti 800 brake lines and hoses Caliper type: single piston floating-type calipers Caliper cup size = 1.5”

Rotor type: solid Rotor size = 8.5”

Rotor thickness = 10mm

CALCULATIONS

1. BRAKES APPLIED TO ALL FOUR WHEELS

θ = Angle of inclination of the plane with horizontal

W = Weight of vehicle = mg = 350g N. Rf, Rr = Reactions of the ground on the front and rear wheels respectively

f = Retardation of the vehicle ft = Retardation at which vehicle can topple b = Wheel base of the vehicle = 62 inches or 1.5748 m

h = Height of the center of mass of the vehicle from the inclined surface = 16 inches or 0.4064 m

l = Distance of the center of mass from the rear axle = 28.1 inches or 0.71374 m u = Coefficient of friction between the ground and the tires = 0.8 (Source: Advanced Vehicle Technology 2nd edition. - H. Heisler (2002))

s = Stopping distance t = Stopping time

So according to center of gravity, weight distribution when vehicle is at rest at θ = 0 is

Rf = 0.45W or 45% of W Rr = 0.55W or 55% of W

Now when brakes are applied weight distribution at θ = 0 changes to

Rf = 0.66W or 66% of W

Rr = 0.34W or 34% of W

Therefore, optimum braking percentage is 53:47. Weight transfer from rear to front is 21% of W, i.e. vehicle will see 21% more weight

than before on front wheels after braking.

Braking percentage can easily be adjusted by balance bar provided in dual master cylinder by the driver as per requirement.

f = 7.85m/s2

Rf = 2266.1N and Rr = 1167.4N. this will be the front wheel and rear wheel reactions after a hard braking on level track, producing a deceleration of 7.85m/s2

STOPPING DISTANCE

s = v^2/2f At 40 kph s = 7.86m

Time t = 1.41 seconds

TOPPLING AT ACCELERATION AND DECELERATION

When brakes are applied, due to deceleration weight transfer of vehicle occurs and it develops a tendency to topple about front wheels. The deceleration at which Rr becomes zero and vehicle starts toppling is given by

ft = g ( cos θ * (b-l)/h – sin θ)

The vehicle will have tendency to topple during braking while coming down of an inclination. For θ = 30,

ft = 13.1m/s2

And actual deceleration that will be produced at θ = 30, f = 1.89m/s2

This deceleration is very smaller than ft = 18.74m/s2 at θ = 30. Hence, vehicle will not topple at the event of braking.

When vehicle is accelerated, weight transfer of vehicle occurs and it develops a tendency to topple about rear wheels. The acceleration at which Rf becomes zero and vehicle starts

toppling is given by a = g ((l/h)* cos θ – sin θ)

The vehicle will have tendency to topple during braking while going up of an inclination. For inclination θ = 30

a = 10.01m/s2 This acceleration is far greater than our vehicle’s max achievable acceleration. So the

vehicle will not topple.

Determination of master cylinder and brake pedal

http://msis.jsc.nasa.gov/sections/section04.htm

Here we have to determine the values of pedal ratio and master cylinder bore size. Brake rotors and calipers are already decided. They are going to be of Maruti 800 car.

Pedal force = 400N (from above figure) Pedal ratio = p

Master cylinder bore size = dmc

Caliper cup size = 1.5” Weight of vehicle = 350g N

Radius of tire = 12.5” Coefficient of friction between road and tire = 0.8

Coefficient of friction between rotor and pads = 0.45 Diameter of rotor = 8.5” Effective radius of rotor = 3.5”

Assuming front axle and rear axle reactions are equal and the brake percentage required

is 50:50. This has been assumed just for calculations as any other assumptions will not affect the answer.

Torque on a single wheel, when skidding will start , T = 0.8*350g*12.5”*0.0254/4 = 218 Nm

Pedal force applied = 400N

Force applied on master cylinder = 200*p N Force taken by one master cylinder = 200*p N

Pressure generated in one master cylinder = Pr = 200*p/(pi*dmc^2)*4 Force applied on rotor = Pr*pi*(1.5^2)/4

Braking torque formed on rotor, BT = Pr*pi*(1.5^2)/4 * 0.45*2*3.5*0.0254 Nm = 36*p/ dmc ̂ 2 Nm

BT >= T P/dmc ̂ 2 >= 6.056

Master cylinder

bore size

Pedal ratio

¾” 3.5:1

7/8” 4.7:1

1” 6.1:1

1” 6.2:1

The bold numbers are the values which we decided to accommodate. BT = 223.2Nm

And this can increase as the pedal force will increase. Let length of pedal be now = 7.75” Mechanical advantage in pedal = 6.2:1

With balance bar being 1.25” above pivot point and rubber pad being 7.75” above pivot point, the pedal ratio becomes 6.2:1

Now with above values of master cylinder and pedal ratio and 66:34 braking percentage

Force applied on

pedal = 400N

Increased force

due to pedal ratio = 2480N

Force applied on front wheel master

cylinder = 0.66*400 = 1636.8N

Force applied on rear wheel master

cylinder = 0.34*400 = 843.2N

Pressure applied on front wheels = 3230266.115 N/m2

Pressure applied on rear wheels = 1664076.484N/m2

force applied on front wheel rotor = 3682.8N from both side of rotor

force applied on rear wheel rotor = 1897.2N from both side of rotor

Torque applied on front wheel = 294.66Nm

Torque applied on rear wheel = 151.8Nm

Pedal travel = 3.3”

Angular travel of pedal = 25.55

degrees