Brakes
24th of July, 2018
The brake system design and analysis can be divided in different steps.
Rules requirements 2019.
Necessary braking force.
Calculate the necessary brake line presure. Analytical part
Calculate the brake pedal dimensions.
Component selection.
First draft of geometry design.
Physical part
FEM analysis and geometry optimization.
Final design.
Rules requirements 2019
The formula SAE rules 2019 describe the specifications for the construction of the brake system. Where:
The car must be equipped with a brake system that acts and lock on all four wheels and is
operated by a single control.
There are two independent brake circuits are required.
For each brake circuit a separate reservoir (OEM-Style) must be present.
In case of failure of a brake circuit two wheels must still be able to brake.
A single brake acting on a limited-slip differential is acceptable.
Brake-by-wire system is prohibited.
Unarmored plastic brake lines are prohibited.
The brake systems must be protected withscatter shields from failure of the drive train or from
minor collisions.
The brake system will be dynamically tested and must demosntrate the capability of locking all
four wheels and stopping the vehicle in a straight line at the end of an acceleration run specified
by the brake inspectors (50km/h).
A brake pedal over-travel switch must be installed on the car (BOTS). This switch must be
installed so that in the event of brake system failure such that the brake pedals overtravels, the
switch will be activated and will stop the engine from running. This switch must kill the ignition
and cut the power to any electrical fuel pumps. Repeated actuation of the switch must not
restore power to these components, and it must be designed so that the driver cannot reset it.
The switch must be implemented with analog components, and not through recourse to
programmable logig controlers, engine control units, or similar functioning digital controllers.
The car must be equipped with a red brake light of at least 15 watts, or equivalent, clearly visible
from the rear. If a LED brake light is used, it must be clearly visible in every bright sunlight.
This light must be fixed between the wheel centerline and below the driver’s shoulder level
vertically aproximatelly on vehicle centerline laterally.
In side view any portion of the brake system tha is mounted on the sprung part of the car must
not project below the lower surface of the chassis.
Fasteners in the brake system are critical fasteners (SAE grade 5, hex head or hexagonal recessed
drive, Nylon lock nuts).
Must be fabricated from steel or aluminum or machined from steel, aluminum or titanium.
Designed to withstand a force of 2000N without any faillure of the brake system or pedal box.
The analytical part can be explaned in a better way considering the following diagram:
•Braking force at •Braking force at •Master cylinder
•Vehicle Brake line pressure displacement/ Pedal Force
Wheel disk
force
Each section of the calculation is explaned in detail in the following parts.
Necessary braking force
The first step in the braking system design is to analized the forces acting on the vehicle at the moment
of braking, by this way the necessary braking forrce for each wheel can be found.
𝑅
𝑊⁄ (𝑎)
ℎ2 𝑔
ℎ1
𝑅𝑟𝑟 𝑚𝑔 𝑅𝑟𝑓
𝐹𝑏𝑟 𝐹𝑏𝑓
𝑊𝑓 𝑊𝑟
Where:
h1 CG height. W Vehicle weight.
h2 Air resistance height. Ra Air resistance.
L Wheel base. Rrf Rolling resistance front.
l1 Distance from CG to front axle.
Rrr Rolling resistance rear.
l2 Distance from CG to rear axle.
Fbf Front braking force. Wf Front Weight.
Fbr Rear braking force. Wr Rear Weight.
During braking a weight transfer from the rear axle to the front axle is happening. Due to the geometry
of the Formula SAE frontal area centroid is closer to the CG height, for this reason the approximation of
h=h1=h2 can be made in order to simplify the analysis.
Front tire
The rear weight during braking can be calculated analyzing the moments action on the front wheel.
𝑊
∑ 𝑀𝐹𝑡 = 𝑊𝑟 𝐿 − 𝑊𝑙1 + 𝑎(ℎ) − 𝑅𝑎 (ℎ) = 0
𝑔
𝑊
𝑊𝑟 𝐿 = 𝑊𝑙1 − 𝑎(ℎ) + 𝑅𝑎 (ℎ)
𝑔
1 𝑊
𝑊𝑟 = 𝐿 [𝑊𝑙1 − ℎ ( 𝑔 𝑎 − 𝑅𝑎 )] (1)
Rear tire
Similar to the rear weight the front weight during barking can be calculated analyzing the rear tire.
𝑊
∑ 𝑀𝑅𝑡 = −𝑊𝑓 𝐿 + 𝑊𝑙2 + 𝑎(ℎ) − 𝑅𝑎 (ℎ) = 0
𝑔
𝑊
𝑊𝑓 𝐿 = 𝑊𝑙2 + 𝑎(ℎ) − 𝑅𝑎 (ℎ)
𝑔
1 𝑊
𝑊𝑓 = 𝐿 [𝑊𝑙2 + ℎ ( 𝑔 𝑎 − 𝑅𝑎 )] (2)
X Direction
Analyzing the Sum of the forces in the longitudinal direction we have:
𝑊
∑ 𝐹𝑥 = 𝐹𝑏𝑓 + 𝐹𝑏𝑟 + 𝑅𝑟 + 𝑅𝑎 − 𝑎=0
𝑔
𝑊
[ 𝑔 𝑎 − 𝑅𝑎 ] = 𝐹𝑏𝑓 + 𝐹𝑏𝑟 + 𝑅𝑟 (3)
Substituting equation 3 in the equations 1 and 2 we have :
1
−𝑊𝑓 = 𝐿
[𝑊𝑙2 + ℎ(𝐹𝑏𝑓 + 𝐹𝑏𝑟 + 𝑓𝑟𝑤 )] (4)
1
−𝑊𝑟 = [𝑊𝑙1 + ℎ(𝐹𝑏𝑓 + 𝐹𝑏𝑟 + 𝑓𝑟𝑤 )] (5)
𝐿
For a 4 wheel vehicle the maximum braking force is happening when the 4 wheels reach the friction
limit, this means that the wheels are close to the slip point at the same time, this defined the maximum
braking force as µW.
-----------------------------------
𝐹𝑏𝑟 + 𝐹𝑏𝑓 = 𝐹𝑏𝑚𝑎𝑥 = 𝜇𝑊 … . (6)
𝐹𝑏𝑓𝑚𝑎𝑥 = 𝜇𝑊𝑓
𝐹𝑏𝑟𝑚𝑎𝑥 = 𝜇𝑊𝑟
𝜇
𝜇𝑊𝑓 = [𝑊𝑙2 + (𝜇𝑊 + 𝑓𝑟 𝑊)]
𝐿
𝜇𝑊
𝑏𝑓𝑚𝑎𝑥 = [𝑙 + ℎ(𝜇 + 𝑓𝑟 )] … . (7)
𝐿 2
De manera similar:
𝜇𝑊
𝑏𝑟𝑚𝑎𝑥 = [𝑙 − ℎ(𝜇 + 𝑓𝑟 )] … . (8)
𝐿 1
Diego parte 2
--------------------------------
The front and rear braking force is not equal due to the weight transfer, at the maximum brake
condition the brake ratio is:
𝑘𝑏𝑓 𝐹𝑏𝑓𝑚𝑎𝑥 𝑙2 + ℎ(𝜇 + 𝑓𝑟)
= =
𝑘𝑏𝑟 𝐹𝑏𝑟𝑚𝑎𝑥 𝑙1 − ℎ(𝜇 + 𝑓𝑟)
Braking torque
Ivonne 2 (Diagrama de llanta)
According the image we can define the necessary braking torque as follow:
∑ 𝑀𝑇 =
http://www.engineeringinspiration.co.uk/brakecalcs.html
https://content.sciendo.com/view/journals/afe/16/4/article-p196.xml
http://www.mscsoftware.com/Free-Trial/ADM743/index.html
http://www.multibody.net/teaching/dissertations/multibody-analysis-
of-the-formula-sae-car-mg0712/
