Brakes

02.1 / Theory

How pedal force becomes braking torque, where the car’s energy goes, and what limits its ability to stop.

Project methodology · this page: Understand and Model

  1. 01 Understand
  2. 02 Model
  3. 03 Inspect
  4. 04 Measure
  5. 05 Analyze
  6. 06 Improve
  7. 07 Test
  8. 08 Reflect
01 / UNDERSTAND

Brake Identity

What design am I using as the reference?

I’m starting with the early Spitfire’s Girling hydraulic system: disc brakes at the front and drum brakes at the rear.

The factory manual lists 9-inch front discs and 7-inch rear drums with 1¼-inch-wide linings. A separate handbrake linkage operates the rear brakes mechanically. These are reference specifications, not measurements of the parts currently on my car. [S1]

Before deciding what to change, I want to understand how the pedal, hydraulic system, friction surfaces, and tires work together. Identifying the installed components and checking their condition belongs in Characterization.

02 / UNDERSTAND

Brake Purpose

What has to happen for the car to slow down?

The brakes apply torque that opposes wheel rotation. The tires then transmit a braking force to the road, while friction at the discs and drums converts much of the car’s kinetic energy into heat. A useful system has to provide enough torque, remain controllable, and cope with repeated stops. [S3]

Inputs
  • Pedal force and travel
  • Vehicle speed and load
  • Tire grip
  • Initial brake temperature
Apply torque
↓
Dissipate energy
Outputs
  • Braking force at the road
  • Reduced vehicle speed
  • Heat in the brake components
  • Pedal feedback and wear
Two different limits: brake hardware determines the torque available at a wheel; the tire and road determine how much of that torque can be used before the wheel locks.
03 / UNDERSTAND

Operating Principles

What happens between pressing the pedal and slowing the car?

The pedal provides leverage, the master cylinder creates pressure, and the brakes at each wheel turn that pressure into friction torque.

01 Pedal leverage

The pedal acts as a lever. Its effective ratio compares the perpendicular moment arms of the foot force and the master-cylinder pushrod force. A larger ratio gives more pushrod force for the same effort, but it also requires more pedal movement for the same master-cylinder stroke. [S2]

02 Hydraulic pressure and displacement

Once the master cylinder closes its reservoir connection, piston movement displaces fluid and raises pressure. In the ideal model, that pressure change is transmitted through the connected fluid. A larger receiving piston produces more force at the same pressure. [S5]

A smaller master-cylinder bore raises pressure for a given pushrod force, but moves less fluid per inch of stroke. Pressure, fluid volume, and available pedal travel therefore have to be considered together.

03 Front disc brakes

The disc rotates with the wheel while the caliper holds pads against its two faces. The friction forces act at an effective radius and produce braking torque. In a fixed opposed-piston caliper, hydraulic pistons act on both sides of the disc. A floating caliper uses movement of its body to apply the opposite pad. [S3]

I have to define piston area carefully in a torque calculation. If I use the area on one side of a symmetric fixed caliper, I must include both friction faces. Adding every piston area and then doubling it again would overestimate torque.

04 Rear drum brakes

A wheel cylinder pushes the shoes outward against the rotating drum. Return springs draw them back when pressure is released. Depending on the shoe arrangement and direction of rotation, friction can help pull a shoe into contact. This self-energizing action means drum torque depends on shoe geometry as well as hydraulic force. [S4]

The disc-brake equation below cannot simply be reused for the rear drums. I would first need the wheel-cylinder dimensions, shoe layout, and lining characteristics.

05 Tire grip and brake balance

During braking, the front axle carries more vertical load and the rear carries less. The car’s total weight has not changed; the loads at the contact patches have. This changes how much braking force each axle can support. [S8]

Brake balance describes the front-to-rear share of braking force. Equal hydraulic pressure does not guarantee equal brake torque. Piston sizes, friction materials, effective radii, and drum geometry all matter. The ideal balance also changes with deceleration and loading.

06 Heat, fade, and fluid

A stop adds thermal energy faster than the brakes may be able to shed it. Repeated stops can raise temperatures until friction performance changes. Cooling and thermal capacity therefore matter even when the brakes can already generate enough torque to lock a wheel. [S9]

Fluid boiling is a separate problem: vapor is much more compressible than liquid, so pedal movement can compress bubbles instead of building pressure effectively. Friction fade and vapor-related loss of pedal firmness should not be treated as the same mechanism. [S10]

07 Release, clearance, and the handbrake

The brakes also have to release. The master cylinder must return far enough to restore its connection to the reservoir, and the wheel mechanisms need their intended clearances. Excessive clearance increases the fluid travel needed before contact; insufficient release can cause drag and heating.

The factory handbrake applies the rear brakes mechanically rather than through service-brake hydraulic pressure. Its linkage is a separate part of the system to identify and assess. [S1]

04 / MODEL

Mathematical Modeling

How can I connect the component dimensions to braking performance?

These models let me trace the relationships from pedal effort to pressure, wheel torque, load transfer, and energy. They also show which measurements I need before I can make useful predictions for the car.

All worked examples are illustrative. The example weight, bores, lever ratio, friction values, geometry, and temperatures are not measured Spitfire data or component recommendations. The models assume level-ground, straight-line braking unless stated otherwise. Force is in lbf, pressure in psi, and distances in feet or inches as labeled; g = 32.174 ft/s².
M.01

Pedal force multiplication

R = lfoot / lrod
Fmc = Fpedal R ηp

The pedal’s moment arms determine its leverage. The efficiency factor represents mechanical losses; this example includes no brake booster.

Example: 100 lbf at the pedal produces 450 lbf at the master-cylinder pushrod.

Variables & worked example

Variables

R
Effective pedal ratio.
l
Perpendicular moment arm of each force.
Fmc
Master-cylinder pushrod force, lbf.
ηp
Assumed pedal-mechanism efficiency.

Illustrative inputs

lfoot = 10 in; lrod = 2 in; Fpedal = 100 lbf; ηp = 0.90.

R = 10 / 2 = 5
Fmc = 100 × 5 × 0.90 = 450 lbf

From moment balance, the ideal forces satisfy Fpedallfoot = Fmclrod. Effective leverage can change through the stroke as the force angles change. [S2]

M.02

Pressure and fluid volume

Amc = πdmc² / 4
P = Fmc / Amc
V = Amcsmc

Bore size affects both pressure and displaced volume. More pressure from a smaller bore comes with more stroke for the same required volume.

Example: a 0.75-inch bore and 450 lbf pushrod force produce approximately 1,019 psi.

Variables & worked example

Variables

dmc
Master-cylinder bore, in.
Amc
Piston area, in².
P
Gauge pressure, psi.
smc
Effective piston stroke, in.
V
Displaced fluid volume, in³.

Illustrative inputs

dmc = 0.75 in; Fmc = 450 lbf; smc = 0.50 in.

Amc = π(0.75)² / 4 = 0.4418 in²
P = 450 / 0.4418 ≈ 1,019 psi
V = 0.4418 × 0.50 ≈ 0.221 in³

For incompressible fluid, displaced volume must supply piston movement throughout the connected circuit. Real systems also consume displacement through compliance and fluid compression. This is a single effective-piston model, not a full tandem-master-cylinder analysis. [S5]

M.03

Disc-brake torque

Nface = PAside
Tb = 2μpadPAsidereff

For a symmetric fixed caliper, each pad develops a friction force. Multiplying both friction forces by their effective radius gives torque at one wheel.

Example: approximately 5,040 lbf·in, or 420 lbf·ft, at one front disc.

Variables & worked example

Variables

Aside
Total piston area on ONE side of the fixed caliper, in².
Nface
Normal force on one pad, lbf.
μpad
Assumed pad/disc friction coefficient.
reff
Effective friction radius, in; not the outer disc radius.
Tb
Brake torque at one wheel, lbf·in.

Illustrative inputs

One 1.50-inch piston on each side; P = 1,018.59 psi from M.02; μpad = 0.40; reff = 3.50 in.

Aside = π(1.50)² / 4 = 1.7671 in²
Nface = 1,018.59 × 1.7671 = 1,800 lbf
Tb = 2 × 0.40 × 1,800 × 3.50
Tb = 5,040 lbf·in = 420 lbf·ft

The two pad-normal-force magnitudes sum to 3,600 lbf. The factor of two already accounts for both faces. This calculation assumes equal pressure, symmetric geometry, and constant friction. It does not model the rear drums.

M.04

Wheel torque and tire grip

Fbrake,w ≈ Tb / rtire
Fgrip,w ≈ μtireNw

The torque-based force must be compared with the available tire grip. In this simple rolling-wheel model, usable force cannot exceed the smaller of these two values.

Example: 5,040 lbf·in and an 11-inch tire radius correspond to about 458 lbf before the grip limit is applied.

Variables & worked example

Variables

rtire
Effective tire radius, in when torque is in lbf·in.
μtire
Assumed tire/road friction coefficient, separate from pad friction.
Nw
Dynamic normal load at this wheel, lbf.

Illustrative calculation

Fbrake,w = 5,040 / 11 ≈ 458 lbf
Froad,w ≲ min(Tb/rtire, μtireNw)

This approximation neglects wheel rotational inertia and detailed tire-slip behavior. An 11-inch tire radius is only an example. Wheel rim diameter is not tire outside diameter, so it cannot be used directly to obtain this radius.

M.05

Load transfer and balance

ΔN = W(a/g)(h/L)
Nf = Nf0 + ΔN
Nr = Nr0 − ΔN

Braking adds normal load to the front axle and removes the same amount from the rear. The braking force acting below the center of gravity creates a pitching moment.

Example: 225 lbf transfers forward, changing a 50/50 static load split to 65/35 at 0.70 g.

Variables & worked example

Variables

W
Total operating weight, lbf, including occupants and fluids.
a
Positive deceleration magnitude, ft/s².
h, L
Center-of-gravity height and wheelbase in the same units.
Nf0, Nr0
Static front and rear axle loads, lbf.

Illustrative inputs

W = 1,500 lbf; a/g = 0.70; h = 18 in; L = 84 in; initial axle loads = 750 lbf each.

ΔN = 1,500 × 0.70 × 18/84 = 225 lbf
Nf = 750 + 225 = 975 lbf
Nr = 750 − 225 = 525 lbf
βf,ideal = Nf/W = 0.65

Moment balance gives ΔN·L = ma·h, with m = W/g. This assumes negligible pitch acceleration and aerodynamic effects. The 65% front share is an idealized grip-matching result for equal tire friction coefficients, not a recommended brake-bias setting. [S8]

M.06

Energy removed in a stop

ΔE = ½m(vi² − vf²)
m = W/g

Translational kinetic energy grows with speed squared. Doubling the starting speed gives four times the energy to remove when stopping the same vehicle.

Example: stopping a 1,500 lbf vehicle from 60 mph removes about 180,500 ft·lbf, or 245 kJ, of translational energy.

Variables & worked example

Variables

m
Mass in slugs when using lbf, ft, and seconds.
vi, vf
Initial and final speeds, ft/s.
ΔE
Energy removed, ft·lbf.

Illustrative inputs

W = 1,500 lbf; vi = 60 mph; vf = 0.

60 mph × 5280/3600 = 88 ft/s
m = 1,500/32.174 ≈ 46.62 slugs
ΔE = ½ × 46.62 × 88²
ΔE ≈ 180,518 ft·lbf ≈ 244,750 J

This excludes rotating-component energy and changes in elevation. Drag and rolling resistance remove some energy during a real stop, so the brakes do not necessarily absorb all of this value. [S6]

M.07

Stopping distance

db = (vi² − vf²)/(2a)
dtotal = vitr + db

Constant deceleration gives a braking-distance estimate. Reaction distance is added separately because the car keeps moving before braking begins.

Example: 60–0 mph at a constant 0.70 g takes about 172 ft of braking, or 260 ft with an assumed one-second reaction time.

Variables & worked example

Variables

db
Braking distance, ft.
a
Positive deceleration magnitude, ft/s².
tr
Assumed reaction time, s.
dtotal
Reaction distance plus braking distance, ft.

Illustrative calculation

a = 0.70 × 32.174 = 22.522 ft/s²
db = 88²/(2 × 22.522) ≈ 171.9 ft
dreaction = 88 × 1.0 = 88 ft
dtotal ≈ 259.9 ft

The equation follows from vf² = vi² − 2ad. Under the same assumptions, 30–0 mph takes about 43 ft of braking. This does not establish the car’s achievable deceleration or include a gradual buildup of brake pressure. [S7]

M.08

Disc temperature rise

Qdisc = χβfΔE/2
ΔT = Qdisc/(mdcp)

A simple heat-capacity model estimates the average temperature rise of one front disc. The front axle’s energy share is divided between its two discs.

Example: an average rise of about 62°C, or 112°F, in one front disc during the modeled stop.

Variables & worked example

Variables

βf
Assumed front-axle share of dissipated energy.
χ
Fraction of front brake energy entering the discs.
md
Mass of ONE disc, kg.
cp
Assumed specific heat, J/(kg·K).

Illustrative inputs

ΔE = 244,750 J from M.06; βf = 0.65; χ = 0.90; md = 2.5 kg; cp = 460 J/(kg·K).

Qdisc = 0.90 × 0.65 × 244,750 / 2
Qdisc ≈ 71,589 J
ΔT = 71,589/(2.5 × 460) ≈ 62.3 K
Temperature rise ≈ 62.3°C = 112.1°F

This example uses SI units for heat capacity and assumes no cooling during the stop, uniform disc temperature, and constant energy split. It predicts neither surface-temperature peaks nor a safe operating limit. A temperature rise is added to the starting temperature; it is not the final temperature. [S11]

05 / UNDERSTAND

Design Development

What changed within the Spitfire’s brake design?

The factory manual distinguishes the earlier 12P calipers from the 14LF type introduced with the Spitfire Mk III. The broad disc-front, drum-rear arrangement remained, but that does not make the individual components identical. [S1]

For this project, the useful question is what each change affects. A different piston area changes hydraulic force and fluid demand. A different friction radius changes torque. A different disc mass or cooling arrangement changes thermal behavior. I need to follow those effects through the whole system before calling a component an improvement.

06 / MODEL

Design Assessment

What does this arrangement do well, and where are its limits?

I’m separating the ability to produce brake torque from pedal control, heat management, and reliability. A change can improve one of those without improving all of them.

A Useful characteristics
  • Pedal leverage and hydraulic area provide force multiplication without requiring large movement at each friction surface.
  • The front discs expose their friction surfaces to cooling air.
  • The rear drums provide a mechanical handbrake interface as well as service braking.
B Limits to evaluate
  • More brake torque does not create more tire grip.
  • More hydraulic leverage can require more fluid travel and pedal movement.
  • Changing one axle’s components can change balance and which tires reach their grip limit first.
  • Repeated-stop performance depends on heat capacity, cooling, and friction behavior at temperature.
  • Leakage, corrosion, wear, and restricted movement cannot be evaluated from the nominal design alone.
07 / MODEL

Factory Baseline

What is published, and what still needs measuring?

The factory reference and the worked-example inputs serve different purposes. I will compare the installed brakes with the reference, then replace the illustrative model inputs with measurements.

  • Published reference: Girling hydraulic service brakes, 9-inch front discs, and 7-inch rear drums with 1¼-inch-wide linings. [S1]
  • Parking brake: mechanical operation of the rear brakes. [S1]
  • Identification to confirm: caliper type, master cylinder, rear wheel cylinders, circuit arrangement, and any previous modifications.
  • Geometry to measure: pedal leverage, cylinder bores, disc dimensions, pad sweep, drum dimensions, and effective tire radius.
  • Vehicle inputs to establish: operating weight, axle loads, wheelbase, and center-of-gravity height or a clearly identified estimate.
  • Performance still unknown: line pressure, pedal travel, brake balance, deceleration, stopping distance, and temperature response.
08 / EVIDENCE

Sources & Uncertainty

What supports the theory, and what can it not tell me yet?

The factory manual supplies the historical reference. The technical sources explain the underlying mechanisms; the numerical examples are my illustrative calculations using the assumptions shown above.

  1. S1 Triumph workshop manual, Brakes . Factory brake data, printed page 3.201; master-cylinder operation and brake components in the following sections. Archived scan.
  2. S2 Wilwood, Pedal Ratios and How to Find Them . Pedal leverage and travel.
  3. S3 Akebono, Disc Brakes . Disc operation and fixed versus floating calipers.
  4. S4 Akebono, Drum Brakes . Shoe arrangements and self-energizing action.
  5. S5 OpenStax, Pascal’s Principle and Hydraulics . Pressure, piston area, and displacement.
  6. S6 OpenStax, Kinetic Energy and the Work-Energy Theorem . Energy model.
  7. S7 OpenStax, Motion with Constant Acceleration . Braking-distance model.
  8. S8 MotoSPEC, Weight Transfer Math Channel . Longitudinal load transfer. M.05 applies force and moment balance to the simplified car model.
  9. S9 Brembo, Brake Disc Overheating . Heat generation and cooling.
  10. S10 Brembo, Brake Fluid . Boiling and compressible vapor.
  11. S11 OpenStax, Heat Transfer, Specific Heat, and Calorimetry . Average temperature-rise model.
This page establishes the reference design and the models. It does not confirm the installed parts, their condition, or the car’s braking performance. Those questions come next through identification, measurement, and testing.