Project

Study how the weight of the car affects braking. Ensure that our current braking system meets FSGP 2017 and 2018 regulations.

Relevant Regulations

Must be able to stop from speeds of 50 km/h or greater. For these calculations, we will look at speeds between 50km/h and 100km/h.
Average deceleration must be greater than 4.72 m/s^2
Hydraulic brake lines must be able to handle much greater pressure beyond what is needed.
Rear brake must be able to hold the car in place (ie force from one of the master cylinders) for a force up to 15% of the car's weight (varying between 400kg - 1000kg) with ballasted driver (~70kg)

Assumptions

Car is on level ground.
Friction between wet pavement and rubber tire.

Information (to be checked from current system)

Thickness of brake pad (>6 mm) - current is less than 5mm, need to buy new ones. New ones have 0.3 inches thick
Brake pad contact area (>600mm^2) - should be 2 inches area contact
Maximum pressure allowed in brake lines - assume lines meet max pressure

Relevant Equations From Unknown Source

See Updated Formula below for a better description.

Possibly this site - http://www.engineeringinspiration.co.uk/brakecalcs.html

See page for other calculations. I still need to double check that these apply to our car and can be properly used. Also, can use FBD to determine braking force for deceleration. Formulas for stopping distance also from above page.

Results

Weight of Car	Results	Rear Braking Required	Does our system work?
400kg
500kg
600kg
700kg
800kg
900kg
1000kg

Updated Formulae

The following equations were taken from the winning solar car and outlines what is required for our braking system

Deceleration at wheel lock up is

a = f(N/2)

a = acceleration

N = normal force on front or back of car

f = coefficient of friction on road and tire (0.8)

Normal force on the front and rear of the car is

Nf = [(CGf)W + hCG(Fb)]/WB

Nr = [(CGr)W - hCG(Fb)]/WB

Nf = normal force on front of car

Nr = Normal force on rear of car

CGf = distance from centre of gravity to rear wheel

CGr = distance from centre of gravity to front wheel

WB = length of the wheel base

hCG = height of the centre of gravity off the ground

Fb = mass of car multiplied by deceleration

W = car weight (N)

Torque of the brake when locking up occurs is given to be

T = (f)(N)r/2

f = coefficient of friction between tires and road

T = torque

r = tire radius (m)

Friction Force required on braking pad is

F = T/2R

F = friction force

R = radius from centre of wheel to point where brake pad contacts brake disc

Normal force required on brake pad

n = F/0.2

n = normal force on brake pad

0.2 = coefficient of friction of brake pad on steel brake disc (safe estimate)

The pressure required from the master cylinder than is

P = 4N/[pi(d^2)]

P = pressure in cylinder

d = diameter of piston in brake caliper

Lastly the force required in compressing the master cylinder is

Fp = P(pi/4)D^2

Fp = force compressing master cylinder (reasonable to expect 446N)

D = master cylinder diameter

For more details on how these equations were derived, or what each variable, refer to the Winning Solar Car Pg 241 - 253

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