The Brake Bible
By Bill "BillaVista" Ansell
Photography: Bill Ansell
Copyright 2008 - Bill Ansell
Introduction
Why?
At first glance, it may seem that an article on brakes isn't a terribly sexy idea. After all, brakes don't make you go faster, look cooler or sound meaner. But, few things are less enjoyable than driving a rig with lousy brakes. Not only can it be stressful and needlessly hard work, it can also be dangerous. Good brakes can actually improve your overall abilities by allowing much better control of the rig at all times. Finally, with the recent dramatic increase in the popularity of "racing" forms of 'wheeling competition, good brakes have become much more important.
What ?
So, my aim with this article is to help you better understand brake systems and the factors affecting brake performance; to dispel the many common myths and misconceptions, and to address important off-road factors that you won't find covered by any other source.
By the end I hope that you will be able to:
At the end I will also showcase a kick-ass brake system modification from which virtually any rig will benefit.
That said, this article isn't a beginner's guide to brakes - it is assumed that you are already familiar with the basic components. It's also not aimed at maintenance - how to replace the brake pads on this or that model truck - there are many factory and aftermarket repair manuals that cover the topic well. Also, because this article is intended for off-road rigs, I won't be covering aerodynamic braking or down force. Finally, if for no other reason than I detest the damn things, I will not be covering drum brakes, but instead focusing on disc brake systems.
Actually, there are a number of good reasons why disc brakes are far superior to drums, including:
So if you still have drum brakes on the rear of your rig, there's plenty of good information in this article of interest to you, but I won't be covering drum brake improvements - you're far better off to convert to discs, whether you fabricate your own or use one of the many commercially available conversion kits. What I will cover, is how to adjust the rest of your system to accommodate newly installed discs in place of drums - but now I'm getting ahead of myself.
Table of Contents
Basic Brake Components
Each component in a properly functioning brake system must work in harmony with the other components for us to achieve maximum braking performance. We will cover each component; it's selection, modification, and integration into the 'system' as a whole. For now, the basic components of a 4-wheel disc-brake system are:
How Brakes Work
It seems pretty simple - the driver presses on the pedal and the rig stops. But there's quite a bit going on, and understanding it more fully (yes, including a little physics and math, sorry!) can really help us get the most out of our brake system.
Here's what's going on:
The driver presses on the brake pedal, actuating the piston in the master cylinder (MC). The piston in the master cylinder displaces the hydraulic fluid in the brake lines. Because the system is sealed, the displacement (movement) of the hydraulic fluid moves the piston(s) in the brake callipers. The moving callipers bring the brake pads into contact with the rotor. At this point, because there is no more movement possible in the system, pressure begins to build, and the pads are pressed harder and harder against the rotor, creating friction, and stopping the rig.
OPTIONAL GEEK TECH - MAY CONTAIN MATH AND/OR PHYSICS
Why does the rig stop?
The reason the rig stops is because of two important little pieces of physics - Newton's First Law, and the Law of Conservation of Energy.
Wha??
Newton's First Law, aka the Law of Inertia, states that:
The law of conservation of energy states that:
As we're driving down the road in a straight line at a constant speed the buggy has a constant velocity. In order to stop it, we must create an unbalanced force - that Newton's First Law.
The buggy driving down the road also develops energy in the form of kinetic energy. (Energy is the ability to do work, and kinetic energy is the energy developed by moving objects.) In order for the buggy to stop, this kinetic energy must be converted to another form (it cannot be destroyed) - that's the Law of Conservation of Energy.
In order to stop the buggy, then, the brakes must apply an unbalanced force to the buggy and convert the buggies Kinetic energy to some other form
The unbalanced force that stops the car is the increase in friction between the tires and the road that is created as the brakes slow the rotation of the wheels. When the brakes are applied, the friction between the pads and rotors also creates tremendous heat - converting the kinetic energy of the buggy to heat (and in some cases a little noise if the tires or brakes squeal.)
In summary, in order to stop a rig, the brakes must have three properties. They must:
Brake Torque
Brake torque in in-lbs (for each wheel) is the effective rotor radius in inches times clamping force times the coefficient of friction of the pad against the rotor. Brake torque is the force that actually decelerates the wheel and tire. There are two components - how hard the pads clamp the rotor (clamping force) and how far that clamping takes place from the center of the wheel hub. The larger the effective rotor radius, the further the clamping takes place from the wheel center, and the more torque generated by this longer "lever effect". This is very similar to the manner in which a longer handle on a ratchet generates more torque than a short handle (for the same input). To increase brake torque it is necessary to increase the hydraulic pressure, the calliper piston area, the coefficient of friction between pad & rotor, or the effective rotor diameter.
Clamping Force
The clamping force that a calliper exerts, measured in pounds, is the hydraulic pressure (in psi) multiplied by the total piston area of the calliper (in a fixed calliper) or two times the total piston area (in a floating calliper), in square inches. To increase the clamping force it is necessary to either increase the hydraulic pressure or the calliper piston area. Increasing the coefficient of friction will not increase clamping force.
Coefficient of Friction
he coefficient of friction between pad & rotor is an indication of the amount of friction between the two surfaces. The higher the coefficient, the greater the friction. Typical passenger car pad coefficients are in the neighbourhood of 0.3 to 0.4. Racing pads are in the 0.5 to 0.6 range. "Hard" pads have a lower coefficient but wear less, "soft" pads have a higher coefficient but can wear quickly. With most pads, the coefficient is temperature sensitive - which is why sometimes racers need to "warm up" the brakes before they work well, and also why most brakes will "fade" when they overheat - the coefficient of friction is reduced as the temperature rises. For more info in coefficient of friction, see section on pads.
Thermal Capacity
The brake rotors must be capable of absorbing the heat generated by the brakes as they convert the moving car's kinetic energy into heat. The amount of kinetic energy a car has (and, therefore, the amount of heat the rotors must be able to absorb) depends on the weight of the car and the square of the speed of the car. The rotor's ability to absorb this heat depends on its mass (weight), and on how well it cools. Exposed as they are to cooling airflow, this is one area where discs are superior to drums.
OPTIONAL GEEK TECH - MAY CONTAIN MATH AND/OR PHYSICS
The following equations and examples will help to clarify the concepts:
Brake Torque
Brake Torque Required is calculated as:
TBr = Friction Force on Tire x Rolling Radius of Tire
Where:
TBr = Brake Torque Required (in. lbs.)
Friction Force on Tire = Vertical Force on Tire x Grip
Grip = coefficient of friction between the tire and road
The tire's grip is difficult to measure, and can vary from 0.1 on wet ice to about 1.4 for a racing slick on a hot, dry track. If you don't have a value for your tires, use 1.0 as an average value.
The calculation of Friction Force on Tire is different for front and rear tires, takes into account weight transfer, and requires the calculation of vertical force on both tires first.
Front:
FF = µ Ff / 2
Where:
FF = Friction Force on Front Tire
µ = grip (use 1.0)
Ff = Vertical Force on both front tires
and Ff = Wc [1-(Xcg/l) + (µYcg/l)]
Where:
Ff = Vertical Force on both front tires
Wc = Weight of car (in lbs)
Xcg = Distance from front axle to car's center of gravity (in)
l = wheelbase (in)
µ = grip (use 1.0)
Ycg = height above ground of car's center of gravity
Rear:
FR = µ Fr / 2
Where:
FR = Friction Force on Rear Tire
µ = grip (use 1.0)
Fr = Vertical Force on both rear tires
And
Fr = Wc - Ff
Where:
Fr = Vertical Force on both rear tires
Wc = Weight of car (lbs)
Ff = Vertical Force on both front tires
By examining the above equations carefully, we can learn some valuable things. For example, note that the equation for Brake Torque Required:
Great, so we can calculate a theoretical value for how much brake torque our system needs to be able to deliver - but how do we determine what our brakes are capable of?
I'm glad you asked.
Clamping Force is calculated as:
CF = PM x AT
Where:
CF = Clamping Force (lbs)
PM = Maximum hydraulic pressure (psi)
AT = Total effective area of calliper's pistons (sq. in.) - for fixed callipers this is the actual area of the pistons, for floating callipers this is equal to 2 x the actual area of the pistons
Brake Torque Developed is calculated as:
TBd = CF(µL )Re
Where:
TBd = Brake Torque Developed (in-lbs)
CF = Clamping Force (lbs)
µL = Coefficient of friction between brake pads and rotors (use 0.3, manufacturer's specs, or estimate derived from the pad's DOT edge code (see section on pads))
Re = Effective rotor radius (in.) - measured from the center of the rotor to the canter of the brake pad.
Re = effective rotor radius
Ah-ha - now we're getting to some more familiar terms - pressure, piston size, size of brakes. Lets just dig a little deeper.
The maximum hydraulic pressure developed in your braking system can either be measured with an inline pressure gauge, or can be calculated as:
PM = Fp / Ap
Where:
PM = Maximum Pressure (psi)
Fp = Force on Master Cylinder piston (lbs) = Pedal Effort x Pedal Ratio (i.e. how hard the driver pushes the pedal multiplied by the pedal ratio.)
Ap = Area of the MC piston (sq. in.) = 0.785 x Dp^2 (Dp^2 = the MC piston diameter, in inches, squared)
This is a very important equation. Note how the maximum pressure determines the brake torque, and the maximum pressure is the force applied DIVIDED by the area of the MC piston. This means, all other things being equal, the bigger the MC piston, the LESS pressure developed, and the LESS brake torque generated. The trade-off is, the smaller the MC, the less fluid is displaced per inch of travel, and therefore the greater the pedal travel required - more on this later.
Thermal Capacity
As previously discussed, the brake rotors must be capable of absorbing the heat generated by the brakes as they convert the moving car's kinetic energy into heat.
The formula for the Kinetic Energy (K) in the moving car is:
K = (W * S^2) / 29.9
Where:
K = Kinetic Energy (ft. lbs.)
W = Weight of car (lbs.)
S = Speed of car (mph)
(Note: The root equation for kinetic energy is actually K = 1/2mv^2, the above version has a conversion factor included so we can use weight instead of mass and so the result is given in force units instead of energy units)
Nothing really surprising here - we know from instinct and experience that how much brake you need depends on how heavy the car is and how fast it's going. Note though, that speed is squared in the equation - meaning that as the speed increases, the kinetic energy developed goes up by the square of the speed increase - for example, if speed doubles, kinetic energy increases by a factor of 4. If speed triples - kinetic energy goes up by a factor of (3^2) or 9!! This is a bitch for race-car drivers but not so much trail rigs. However - if you're beginning to "race" your rock crawling buggy - be aware that your brake requirements are going to increase exponentially!
OK, so, moving rig has kinetic energy, kinetic energy must be turned into heat, rotors must absorb said heat. Big deal, so what?
Well, the equations that follow are used to calculate the temperature increase in the rotor for a given kinetic energy. Remembering that kinetic energy depends on weight and speed, they also explain in incontrovertible terms (big word - means you can't bloody argue with me about it!) exactly why "pinion brakes", flat out suck, and why Patooyee's rotors keep trying to melt off ! Remember - to add insult to injury "pinion brakes" are often used on Rockwell-axled rigs - which are big and heavy, and weight is a multiplying factor in the equation for Kinetic energy.
[As an aside - the previous discussion on brake torque also explains why some feel that pinion brakes work well "at slow speed". It's because the rotor is placed before the axle differential, meaning the pinion brake's brake torque is calculated as above and then multiplied by a factor equal to the axle ratio. This means, even with small callipers and rotors, they can develop tremendous brake torque. But - remember what we said about the requirements of a braking system - it must also have sufficient thermal capacity - and they simply do not. In fact, they're dangerously inadequate in this regard!]
The formula for temperature rise is:
TR = Kc / 77.8 * Wb
Where:
TR = Temperature rise (degrees F)
Kc = Kinetic energy change (from start of braking to end of braking) (ft. lbs.)
Wb = weight of all rotors (lbs)
And:
Kc = Kinetic energy change (ft. lbs.) = Kb - Ka
Where:
Kb = kinetic energy before stop
Ka = Kinetic energy after stop
Let's do an example, because math is fun!
Imagine I'm stopping my 5000lb buggy from 40 mph to a dead stop.
First, let's calculate the change in kinetic energy for a 5000 lb buggy from 40 mph to 0 mph - this remains the same regardless of the brake system.
Kb = Wc * S^2 / 29.9
= 5000 * (40)^2 / 29.9
= 267 559 ft lbs
Ka = 0
Kc = Kb - Ka
= 267 559 - 0
= 267 559
Now, with 3/4 ton truck disc brakes at each wheel, each rotor weighs about 21 lbs, for a total rotor weight of 84 lbs.
In this configuration, the temp rise of the rotor will be:
TR = Kc / 77.8 * Wb
= 267 559 / (77.8 * 84)
= 267 559 / 6535
= 41*F
If it were a hot day, after the stop my rotors could be at 141*F
Now, imagine I have pinion brakes with 2 small rotors, weight, say, 12 lbs each.
Now,
TR = Kc / 77.8 * Wb
= 267 559 / (77.8 * 24)
= 267 559 / 1867
= 143*F
On the same hot day, the rotors are now at 243*F!!
Make a few more stops, with insufficient time for the rotors to cool fully between them, and it's easy to see how you can seriously overheat the small pinion brake rotors!
OK, so much for physics and math. Let's move on now to the more practical side of how the braking system works.
Recall that brakes need to be able to do 3 things:
And they must do all three with good and consistent feel!
The following "flow-chart" presents a stylized summary of the brake system and its interrelated components / factors. Following it, we will examine each component in more detail:
The Pedal
As we have discussed, in order to apply the brakes the driver must be able to both move and pressurize the hydraulic fluid. That is, the master cylinder piston has the job of moving the brake fluid through the lines in order to bring the brake pads into contact with the rotor, and of pressurizing that fluid create clamping force.
The job of the pedal is to allow the driver to actuate the brakes, and also to multiply the driver's input through a lever effect called the "pedal ratio." In other words, the pedal transfers movement & force from the driver's foot to the master cylinder piston. This is a key concept - that brakes require both movement and force to operate, as the two are largely inversely proportional (that is: as one goes up the other goes down, and vice-versa) - we will return to this concept many times so keep it in mind.
The force the driver applies to the pedal when braking is called the "pedal effort". This amount of force, measured in pounds, is usually insufficient to operate the brakes - values ranging from 50-100 lbs are common. As such, pedals are designed to operate like a lever, multiplying this force before it is applied to the master cylinder piston. The amount by which any particular pedal design does so is called the "pedal ratio."
The pedal ratio is calculated as the distance between the pedal's pivot point and the centre of the foot pad, divided by the distance between the pivot point and the master cylinder push rod, as shown in the following diagrams.
Pedal Ratio = A / B
As can be seen from the diagram - the longer the pedal, the higher the pedal ratio, and the greater the mechanical advantage - meaning the lower the pedal effort required.
However, there are trade-offs. First, longer pedals are harder to fit - especially in cramped buggy cockpits. Also, the longer the pedal, the greater the pedal stroke required to move the master cylinder piston - i.e. the further the driver must press the pedal. The best design is a trade-off that will depend on many factors including desired pedal effort, master cylinder and calliper piston diameters, and the room available. As far as pedal effort is concerned, 75 lbs is common for high-performance race-car brakes, 100 lbs feels hard, and 50 lbs is common in power-assisted brakes. Common pedal ratios range from 3 or 4:1 for power brakes up to 6 or 7:1 for manual brakes.
Of course, if you are using a stock pedal, you are pretty much stuck with the ratio it has, and must design the rest of the system to accommodate.
Just remember - the longer the pedal, the greater the ratio, the less leg force required but the greater the pedal travel required. Longer pedals tend to deflect more under hard use, which can lead to a "spongy" feeling to the brakes. You should always strive to use a pedal that will go from fully retracted to fully applied in as short a distance as possible - usually less than about four inches. Any more than this, and there will not be sufficient travel in the drivers leg to fully apply the brakes under emergency conditions where travel is suddenly increased - such as during brake fade or when there is air in the fluid.
Keep in mind that total brake pedal travel includes not only the stroke of the master cylinder piston, but also includes all slop and deflection of all parts in the system.
Designing and fabricating a good pedal set up requires engineering far beyond the scope of this article - especially since the brakes are such a critical safety system on the rig.
My advice is to either incorporate a rugged stock pedal system, or use one of the pre-fab systems available (in a variety of ratios) from many aftermarket vendors. Choose a ratio that will work with the rest of your system and your personal preference for how the brakes will feel. Common ratios are 3:1 for power brakes and 5:1 for manual brakes - but, you may desire a more custom ratio - like 7:1 for manual brakes if using a large diameter master cylinder.
One final word on pedal effort:
You can design a brake system for any value of pedal effort (values ranging from 50-100 lbs are common), and the pedal effort will largely determine how the brakes "feel" to the driver (excepting problems like brake fade and fluid boil). For this reason, in the following discussion, assume, unless stated otherwise, that we are discussing the brake system with a "consistent and acceptable pedal effort". For example, if we say, changing to a larger bore master cylinder (all other things remaining the same) will decrease pedal travel but reduce braking power, we mean "with the same pedal effort." No matter what the system, or mod's we make to it, we always have the option of just increasing pedal effort, but this is rarely experienced by the driver as an acceptable solution. i.e. one "could" design a system with undersize callipers and an oversize MC that could still stop the rig effectively if only we would push on the pedal with 300lbs - but clearly this would not be suitable.
Brake Hydraulics
OK, now we're getting to the good stuff.
Recall that the brake's hydraulic system must supply movement and force. The movement must be enough to take up all slop, clearance, and deflection of parts as well as move the calliper pistons sufficiently to bring brake pads into firm contact with the rotors. The force must be enough to create enough friction between pad and rotor to stop the car.
It is the piston of the master cylinder that provides both the movement and the force, and the brake fluid that transmits both to the callipers.
Keep in mind, as you read this section, that the goal is to have a system that provides maximum force with small movement - i.e. we want to be able to brake hard without excessive pedal travel.
Force applied to the MC piston creates pressure in the brake fluid. The pressure is the force applied, DIVIDED by the area of the piston. Therefore, the smaller the master cylinder piston, the greater the pressure created.
System Pressure = Fp / Ap
Where:
Fp = Force on the piston (lbs.)
Ap = Area of the piston (sq. in.)
Example: Calculate system pressure for both a 1/2" diameter MC and a 1" diameter MC, given an input of 100 lbs.
Case one - 1/2" MC piston (d= 0.5):
P = Fp / Ap = 100 / pi * (d/2)^2 = 100 / 3.14 * (0.5/2)^2 = 510 psi
Case 2 - 1" MC piston(d=1.0):
P = Fp / Ap = 100 / pi * (d/2)^2 = 100 / 3.14 * (1.0/2)^2 = 127 psi
Recalling that the pressure created is a direct factor in how much clamping force and therefore brake torque is developed, it may seem that for the most powerful brakes, we would want to use a small MC piston.
But, the trade-off is the other component required of the brake system - namely movement. Because the fluid is incompressible, any movement in the MC piston translates into movement of the calliper pistons (excluding expansion of hoses and lines, which should be minimal in a properly working system). This movement in a hydraulic system is known as displacement and is calculated as the area of the MC piston multiplied by its stroke. Displacement is a volume, measured in cubic inches. Therefore, the smaller the master cylinder piston, the less displacement created.
Displacement (cubic inches) = Ap * Sp
Where:
Ap = Area of the piston (sq. in.)
Sp = Distance piston moves (stroke) (in.)
Example: Calculate displacement for both a ½" diameter MC and a 1" diameter MC, both with a stroke of 2 inches.
Case 1 - 1/2" MC piston (d= 0.5):
Displacement = Ap * Sp = 2 * [ pi * (d/2)^2] = 2 * [ 3.14 * (0.5/2)^2] = 0.39 cu. in.
Case 2 - 1" MC piston(d=1.0):
Displacement = Ap * Sp = 2 * [ pi * (d/2)^2] = 2 * [ 3.14 * (1.0/2)^2] = 1.57 cu. in.
Now we can clearly see that there is a trade-off between force and movement in selecting a MC piston size. The smaller the piston, the greater the pressure created but the less displacement produced (and therefore greater pedal travel required.)
So far, we have considered only the effect of the size of the MC piston, but of course in a brake system there are two pistons - the MC piston and the calliper piston (for calculations, the total area of all pistons in a multi-piston calliper act the same as a single piston of equivalent area in a single-piston calliper.)
There is, of course, a relationship between the pistons in the system that affects both force and movement.
Because the brake hydraulic system is a closed, sealed system, and brake fluid cannot be compressed, there is a law of hydraulics that we make use of to multiply force - that is, to apply more force at the callipers than the driver applies to the MC piston. It is quite simple, and quite possibly the most important concept in this entire article. It is this:
In a closed hydraulic system, pressure is equal over all surfaces of the containing system.
In our discussion of brake systems we will refer to the MC piston as the "input" piston and the calliper piston as the "output" piston.
The above law means that whatever pressure is created by the input piston is applied equally to the output piston. Because the output (calliper) piston is of much larger area than the input (MC) piston, this has the effect of multiplying force in the brake system.
The amount of force-multiplication thus achieved is known as the brake's "Hydraulic Ratio" Hydraulic Ratio can be calculated or expressed a number of ways. It is the ratio of fluid displacement by the master cylinder to fluid displaced in the calliper pistons. It is also equal to the ratio of force applied to the MC piston to the force generated by the calliper pistons. Hydraulic ratio is an important factor in the pedal effort equation, the higher the ratio, the less pedal effort is required (and the longer the pedal travel to achieve a given clamping force). The stiffer the calliper and the stiffer the pad, the higher the hydraulic ratio that can be employed.
For example, suppose we apply 100 lbs of force to a ½" diameter MC piston, we develop approx. 500 psi. This 500 psi acts evenly on all other surfaces in the system, including the calliper pistons. Suppose the calliper piston has a diameter of 3 inches. Multiplying our 500psi by the area of the calliper piston (~ 7 sq. in), we develop nearly 3500 pounds of clamping force at the brake pads.
The equation for this force-multiplying relationship is:
Fc = [Fmc * Ac] / Amc
Where:
Fc = Force at calliper piston (output) (lbs)
Fmc = Force on master cylinder piston(input) (lbs)
Ac = Area of calliper piston (sq. in.)
Amc = Area of master cylinder piston (sq. in.)
Brilliant! From the equation we can see that, for a given input force (pedal effort), in order to increase the force at the calliper (what we're after) we can either increase the area of the calliper piston(s) or decrease the area of the master cylinder piston (or both).
Of course, nothing is ever as easy as that! We can't forget the other factor - movement. Of course, there is once-again, a trade-off.
Unfortunately, both actions that increase force (good!) also increase pedal travel (bad!). That is, if we decrease the size of the MC piston we decrease its displacement - requiring us to increase its stroke to compensate - increasing pedal travel required. Similarly, as we increase the size of the calliper piston, we increase the clamping force, but we also increase the volume of fluid (the displacement) required to move the piston a given distance - the distance it must move to bring the pads into contact with the rotor - this also demands an increase in pedal travel.
The equation for relating piston movements is:
Mc = [Mmc * Amc] / Ac
Where:
Mc = Movement at calliper piston (in.)
Mmc = Movement of MC piston (in.)
Amc = Area of MC piston (sq. in.)
Ac = Area of calliper piston (sq. in.)
So, we can see that, in order to increase the movement of the calliper piston without affecting pedal travel (Mmc), we would have to either increase the size of the MC piston, or decrease the size of the calliper piston - both actions that will decrease force!
A simple example with which we are all intimately familiar, that illustrates the concepts of force and movement in a 2-piston hydraulic system, is that of the common floor jack.
Consider the jack's handle piston as the input piston (like the MC piston) and the piston that raises the lifting saddle as the output piston (like the calliper piston). Look at their relative sizes - the handle piston is tiny compared to the output piston - that's how you can lift a 5000lb car by applying only 50 lbs to the jack handle (multiplied, of course, by the length of the handle - just like pedal ratio in a brake system.) The trade off, of course, is movement - you have to pump that handle a lot of times (much input piston movement) to raise the car even just a little (output piston movement). Just as in brakes, movement and force are a trade-off. To produce great force without great input, you have to trade travel.
It's my considered opinion that this interrelationship between force and movements in a 2-piston hydraulic system such as brakes is not very well understood. Consider as evidence the following scenario - a guy swaps rear disc brakes in place of drums on the rear axle of his rig. The resulting brake performance is predictably inadequate - obviously the calliper pistons are much larger than the previous drum-brake wheel-cylinder pistons - meaning that more fluid must be displaced to actuate them. The common advice found is to "swap in a larger master cylinder." That's great from a movement perspective - the larger MC piston will displace grater fluid, moving the calliper pistons further for a given amount of pedal travel. BUT - we now know that the result will ALSO be a decrease in force at the calliper pistons (including the front callipers) than was achieved with the smaller MC.
So, what are we to do? We really have only two options - either careful compromise - or power assist. The beauty of power assist is that it allows us to substantially increase input force while maintaining pedal effort. This, in turn, enables the production of adequate force at the callipers while allowing the use of larger MC pistons and smaller calliper pistons, decreasing pedal travel. This is particularly important when you consider that most often, a single MC piston must actuate two callipers, (usually one for each of the front or rear wheels). That single MC piston must displace enough fluid to move two or more large-diameter pistons in the callipers. Making it large enough to do so without requiring way too much pedal travel means we compromise the force multiplication in the hydraulic system. The result is, we either accept sub-optimal force (and resulting brake torque), or we must make up for the lack of hydraulic force multiplication by increasing input force. And the only way to significantly increase input force without requiring much greater pedal effort is to increase pedal ratio or add power assist. Given that cockpit space availability place a very real upper limit on how much pedal ratio we can practically achieve - the answer becomes power assist.
Master Cylinders
The master cylinder is the heart of the brake system. Actuated by the pedal, its piston provides the force and the movement required to apply the brakes. When the pedal is released, an internal return spring returns the piston to its resting position.
Initially, as the pedal is pushed, the piston moves forward and fluid volume is displaced, taking up all clearances in the system. This fluid movement actuates the calliper pistons which extend and bring the brake pads into contact with the rotors. Because the fluid is incompressible, once the pads are in contact with the rotor, fluid movement stops and pressure rises. The harder the pedal is pushed, the greater the pressure achieved, the more brake torque is developed. How hard the driver can push the pedal, and therefore how much braking is achieved, is a function of input force (leg strength), combined with pedal ratio and any power assist.
The critical specs of a master cylinder are its bore (diameter of the piston) and stroke (how far the piston can travel - and therefore how much fluid it can displace when applied). Common values for bore range from 5/8" to 1-1/2", and stroke from 1" to 1.5". Matching both to the requirements of your system is the key to satisfactory performance. Remember that, for a given input (pedal effort) the smaller the bore, the more pressure is generated but the less fluid displaced. Similarly, the longer the stroke the more fluid is displaced, but the greater the pedal travel required. As with most things, the best result is achieved through carefully considered compromise.
The most common type of MC in use today is known as a "dual-tandem" master cylinder. It has separate circuits (pistons and outlets) for front and rear brakes ("dual"), with each circuit having its own independent reservoir. The reason for this is safety - the theory is: if one circuit (front or rear brakes) should fail, there remains the other separate, independent circuit. The two pistons are arranged inline in a single bore ("tandem"). The piston closest to the pedal is the "primary" piston, and the other is the "secondary" piston. Normally, the primary piston operates the font brakes and the secondary piston the rear. Fluid reservoirs may be integrated or remote (attached to the cylinder via hoses). The most common design has the reservoir attached directly to the cylinder.
Often, master cylinders are listed as being either for disc/drum brakes or for disc/disc brakes (I'm ignoring older drum/drum systems). Obviously a disc/disc MC is designed to be used with 4-wheel disc brakes, and obviously a disc/drum MC with disc/drum brakes. Assuming that nobody in their right mind would convert disc brakes to drums, that means the only remaining issue of concern would be using a disc/drum MC with disc/disc brakes. When choosing a master cylinder, traditionally, you may have been told that there are three differences between disc/disc and disc/drum masters that preclude the use of a disc/drum master with disc/disc brakes. They are:
Here's my take on these three "issues":
Reservoir Size
Disc brake reservoirs are larger than those for drum brakes. You will often see two reasons given for this:
It is my opinion that the second reason is the only real practical reason for the difference in size between disc and drum reservoirs. I say this because I have never seen a drum MC reservoir that was so small it didn't contain enough fluid to actually apply disc brakes - even though discs require more volume than drums.
With this in mind, we can see that, if we routinely check brake fluid levels as pads wear, reservoir size alone should not disqualify us from using a disc/drum MC with disc/disc brakes - especially if we're using a large MC with moderate or small callipers (e.g. 1-ton MC with 3/4-ton callipers).
Piston Size
It is true that disc brakes require both more pressure and more movement (volume) to operate than drum brakes. However, after reading this article, with your new encyclopedic knowledge of pressure and volume, you're going to be selecting a MC for your brakes based on its specs - not on what the counter guy says the OE application was. That is to say - given the right bore and stroke to match your system, there's no particular reason that, based solely on piston bore and stroke, an MC originally designed for disc/drums can't be used successfully on a disc/disc setup. That said, you will by now also understand that leaving in your old disc/drum MC after converting the rear to discs will not work well.
Built-in Valving
This one can be a deal-killer. If the MC in question was designed for disc/drums and has a built-in residual pressure valve, it will not be suitable for disc/disc brakes. See section below on valving for description of residual pressure valves. For now, the point is, know the MC in question and whether or not it has built-in valving (many do not). If it does, you would have to either modify it by removing the residual pressure valve to make is suitable for use with disc brakes, or choose a different master cylinder.
Power or Manual?
Again, you may see master cylinders marketed as being for manual or power brakes. This is based on two potential differences:
Case one may be true if all you are doing is swapping MC's. Normally, manual brakes use a 1" bore or less MC and power brakes 1" and up. Leaving everything else the same, and using a "power" MC (large bore) with manual brakes will normally result in a "hard pedal" - that is, the pedal effort required of the driver, without power assist, becomes too great. However, we already know that we can compensate by using a greater pedal ratio or larger piston callipers. Therefore, when overall system design is considered, and if MC specs are selected carefully to match the pedal ratio and calliper piston size, there is no technical reason that a larger "power" MC couldn't be used with manual brakes.
Case two is more of an issue. The piston hole is the are where the actuating rod (from the pedal or booster) contacts the MC piston. Normally it will be a "shallow" hole for a power brake MC as the power booster rod is captive between the booster and the MC. With manual brakes, the actuating rod is attached to the pedal which is obviously a lever. In this case the MC piston hole is normally deeper to prevent the actuating rod from falling out when the brakes are released - a very useful feature! However, if you must use a "shallow piston hole" MC with manual brakes - it may be possible to overcome the problem by appropriately limiting pedal travel (in the direction opposite to brake application) to prevent actuating rod fall-out.
Summary
There are normally differences between disc and drum master cylinders and between power and manual master cylinders, as follows:
Normally, the easiest route is to use an MC designed for the type of brakes you have. However, when total system design is considered, as long as the specific specs and design of the MC in question are well understood, a much greater range of interchange possibilities opens up. That said, there are a few time-honoured rules of thumb:
Effects of Pedal Ratio and Bore Size on Hydraulic Pressure Output
By Bill "BillaVista" Ansell
Photography: Bill Ansell
Copyright 2008 - Bill Ansell
Introduction
Why?
At first glance, it may seem that an article on brakes isn't a terribly sexy idea. After all, brakes don't make you go faster, look cooler or sound meaner. But, few things are less enjoyable than driving a rig with lousy brakes. Not only can it be stressful and needlessly hard work, it can also be dangerous. Good brakes can actually improve your overall abilities by allowing much better control of the rig at all times. Finally, with the recent dramatic increase in the popularity of "racing" forms of 'wheeling competition, good brakes have become much more important.
What ?
So, my aim with this article is to help you better understand brake systems and the factors affecting brake performance; to dispel the many common myths and misconceptions, and to address important off-road factors that you won't find covered by any other source.
By the end I hope that you will be able to:
- Understand how your brakes work, and why they perform the way they do.
- Accurately troubleshoot / identify causes of dissatisfaction with your brakes.
- Accurately describe the performance / feel of your brakes using proper terminology.
- Decide what modifications/components will work for you to help you achieve your objectives.
- Select appropriate components and understand how they interrelate and work together (and sometimes why they work against one-another).
- Be able to identify and avoid common misconceptions and bad advice
At the end I will also showcase a kick-ass brake system modification from which virtually any rig will benefit.
That said, this article isn't a beginner's guide to brakes - it is assumed that you are already familiar with the basic components. It's also not aimed at maintenance - how to replace the brake pads on this or that model truck - there are many factory and aftermarket repair manuals that cover the topic well. Also, because this article is intended for off-road rigs, I won't be covering aerodynamic braking or down force. Finally, if for no other reason than I detest the damn things, I will not be covering drum brakes, but instead focusing on disc brake systems.
Actually, there are a number of good reasons why disc brakes are far superior to drums, including:
- Better cooling - less likely to overheat and fade
- Less susceptible to contamination from mud and water
- Less maintenance
- Much easier maintenance
- Lighter
- More pad-to-rotor contact area
So if you still have drum brakes on the rear of your rig, there's plenty of good information in this article of interest to you, but I won't be covering drum brake improvements - you're far better off to convert to discs, whether you fabricate your own or use one of the many commercially available conversion kits. What I will cover, is how to adjust the rest of your system to accommodate newly installed discs in place of drums - but now I'm getting ahead of myself.
Table of Contents
- Basic Brake Components
- How Brakes Work
- Brake Torque
- Clamping Force
- Coefficient of Friction
- Thermal Capacity
- The Pedal
- Brake Hydraulics
- Master Cylinders
- Valving
- Brake Tubing and Hoses
- Brake Callipers
- Rotors
- Pads
- The Braking Requirements of Extreme Off-road Rigs
- Brake Performance and Limits
- Brake Fade and other complaints - Design Troubleshooting
- Brake System Design
- Maintenance Troubleshooting
- Review - VANCO Power Brake Supply Hydroboost System
Basic Brake Components
Each component in a properly functioning brake system must work in harmony with the other components for us to achieve maximum braking performance. We will cover each component; it's selection, modification, and integration into the 'system' as a whole. For now, the basic components of a 4-wheel disc-brake system are:
- A pedal mechanism - for the driver to actuate the brakes. Also acts as a lever that serves to multiply the driver's input via the "pedal ratio".
- A master cylinder (MC) & reservoir - contains hydraulic brake fluid to actuate the brake callipers and a piston to move and pressurize that fluid.
- Hydraulic lines - to convey the brake fluid to the brake callipers, and contain the hydraulic fluid, allowing it to be pressurized.
- Valves - To improve performance, a brake system may contain some of the following valves between the MC and the callipers: residual pressure valve, metering valve, combination valve, proportioning valve, or pressure limiting valve.
- Brake Callipers - located at each wheel, the brake callipers are hydraulically actuated clamps that clamp the brake pads against the rotors.
- Brake pads - located inside the callipers, the pads are the friction material that the callipers clamp against the rotors.
- Rotors - Bolted to the axle hub, the rotors slow and stop the rotation of the wheels when clamped by the callipers. They also absorb the heat created from the friction of the pads against the rotor.
How Brakes Work
It seems pretty simple - the driver presses on the pedal and the rig stops. But there's quite a bit going on, and understanding it more fully (yes, including a little physics and math, sorry!) can really help us get the most out of our brake system.
Here's what's going on:
The driver presses on the brake pedal, actuating the piston in the master cylinder (MC). The piston in the master cylinder displaces the hydraulic fluid in the brake lines. Because the system is sealed, the displacement (movement) of the hydraulic fluid moves the piston(s) in the brake callipers. The moving callipers bring the brake pads into contact with the rotor. At this point, because there is no more movement possible in the system, pressure begins to build, and the pads are pressed harder and harder against the rotor, creating friction, and stopping the rig.
OPTIONAL GEEK TECH - MAY CONTAIN MATH AND/OR PHYSICS
Why does the rig stop?
The reason the rig stops is because of two important little pieces of physics - Newton's First Law, and the Law of Conservation of Energy.
Wha??
Newton's First Law, aka the Law of Inertia, states that:
(*velocity just means the combination of speed and direction)
The law of conservation of energy states that:
To see how these laws of physics relate to brakes, let's examine them with an example - driving a buggy down the road at a constant velocity (constant speed in a straight line), and then stopping.
As we're driving down the road in a straight line at a constant speed the buggy has a constant velocity. In order to stop it, we must create an unbalanced force - that Newton's First Law.
The buggy driving down the road also develops energy in the form of kinetic energy. (Energy is the ability to do work, and kinetic energy is the energy developed by moving objects.) In order for the buggy to stop, this kinetic energy must be converted to another form (it cannot be destroyed) - that's the Law of Conservation of Energy.
In order to stop the buggy, then, the brakes must apply an unbalanced force to the buggy and convert the buggies Kinetic energy to some other form
The unbalanced force that stops the car is the increase in friction between the tires and the road that is created as the brakes slow the rotation of the wheels. When the brakes are applied, the friction between the pads and rotors also creates tremendous heat - converting the kinetic energy of the buggy to heat (and in some cases a little noise if the tires or brakes squeal.)
In summary, in order to stop a rig, the brakes must have three properties. They must:
- be able to apply a force to the rotor to decelerate the wheel's rotation so that friction is increased between tires and road and the vehicle slows/stops; this ability is described as the brake system's BRAKE TORQUE.
- be able to create enough friction between the pad and rotors to convert the vehicle's kinetic energy to heat; this is called CLAMPING FORCE; and
- be large and heavy enough (the rotors) to absorb that heat without damage; this is called THERMAL CAPACITY.
Brake Torque
Brake torque in in-lbs (for each wheel) is the effective rotor radius in inches times clamping force times the coefficient of friction of the pad against the rotor. Brake torque is the force that actually decelerates the wheel and tire. There are two components - how hard the pads clamp the rotor (clamping force) and how far that clamping takes place from the center of the wheel hub. The larger the effective rotor radius, the further the clamping takes place from the wheel center, and the more torque generated by this longer "lever effect". This is very similar to the manner in which a longer handle on a ratchet generates more torque than a short handle (for the same input). To increase brake torque it is necessary to increase the hydraulic pressure, the calliper piston area, the coefficient of friction between pad & rotor, or the effective rotor diameter.
Clamping Force
The clamping force that a calliper exerts, measured in pounds, is the hydraulic pressure (in psi) multiplied by the total piston area of the calliper (in a fixed calliper) or two times the total piston area (in a floating calliper), in square inches. To increase the clamping force it is necessary to either increase the hydraulic pressure or the calliper piston area. Increasing the coefficient of friction will not increase clamping force.
Coefficient of Friction
he coefficient of friction between pad & rotor is an indication of the amount of friction between the two surfaces. The higher the coefficient, the greater the friction. Typical passenger car pad coefficients are in the neighbourhood of 0.3 to 0.4. Racing pads are in the 0.5 to 0.6 range. "Hard" pads have a lower coefficient but wear less, "soft" pads have a higher coefficient but can wear quickly. With most pads, the coefficient is temperature sensitive - which is why sometimes racers need to "warm up" the brakes before they work well, and also why most brakes will "fade" when they overheat - the coefficient of friction is reduced as the temperature rises. For more info in coefficient of friction, see section on pads.
Thermal Capacity
The brake rotors must be capable of absorbing the heat generated by the brakes as they convert the moving car's kinetic energy into heat. The amount of kinetic energy a car has (and, therefore, the amount of heat the rotors must be able to absorb) depends on the weight of the car and the square of the speed of the car. The rotor's ability to absorb this heat depends on its mass (weight), and on how well it cools. Exposed as they are to cooling airflow, this is one area where discs are superior to drums.
OPTIONAL GEEK TECH - MAY CONTAIN MATH AND/OR PHYSICS
The following equations and examples will help to clarify the concepts:
Brake Torque
Brake Torque Required is calculated as:
TBr = Friction Force on Tire x Rolling Radius of Tire
Where:
TBr = Brake Torque Required (in. lbs.)
Friction Force on Tire = Vertical Force on Tire x Grip
Grip = coefficient of friction between the tire and road
The tire's grip is difficult to measure, and can vary from 0.1 on wet ice to about 1.4 for a racing slick on a hot, dry track. If you don't have a value for your tires, use 1.0 as an average value.
The calculation of Friction Force on Tire is different for front and rear tires, takes into account weight transfer, and requires the calculation of vertical force on both tires first.
Front:
FF = µ Ff / 2
Where:
FF = Friction Force on Front Tire
µ = grip (use 1.0)
Ff = Vertical Force on both front tires
and Ff = Wc [1-(Xcg/l) + (µYcg/l)]
Where:
Ff = Vertical Force on both front tires
Wc = Weight of car (in lbs)
Xcg = Distance from front axle to car's center of gravity (in)
l = wheelbase (in)
µ = grip (use 1.0)
Ycg = height above ground of car's center of gravity
Rear:
FR = µ Fr / 2
Where:
FR = Friction Force on Rear Tire
µ = grip (use 1.0)
Fr = Vertical Force on both rear tires
And
Fr = Wc - Ff
Where:
Fr = Vertical Force on both rear tires
Wc = Weight of car (lbs)
Ff = Vertical Force on both front tires
By examining the above equations carefully, we can learn some valuable things. For example, note that the equation for Brake Torque Required:
- does not involve vehicle speed in any way
- will vary for different tires and road (or trail) conditions
- does involve tire/wheel radius - an important point for those of us running large diameter tires (do you think the car guys (who pretty much developed all the braking components we will choose from) have ever done the calculations for a 44, 47, or god forbid, a 54" tire?)
- does involve the vehicle's weight, height, and wheelbase - confirming what we knew from instinct -that a lower, lighter car will be easier to stop (and again - important to us since we are generally much taller and heavier than the average car)
Great, so we can calculate a theoretical value for how much brake torque our system needs to be able to deliver - but how do we determine what our brakes are capable of?
I'm glad you asked.
Clamping Force is calculated as:
CF = PM x AT
Where:
CF = Clamping Force (lbs)
PM = Maximum hydraulic pressure (psi)
AT = Total effective area of calliper's pistons (sq. in.) - for fixed callipers this is the actual area of the pistons, for floating callipers this is equal to 2 x the actual area of the pistons
Brake Torque Developed is calculated as:
TBd = CF(µL )Re
Where:
TBd = Brake Torque Developed (in-lbs)
CF = Clamping Force (lbs)
µL = Coefficient of friction between brake pads and rotors (use 0.3, manufacturer's specs, or estimate derived from the pad's DOT edge code (see section on pads))
Re = Effective rotor radius (in.) - measured from the center of the rotor to the canter of the brake pad.
Re = effective rotor radius
Ah-ha - now we're getting to some more familiar terms - pressure, piston size, size of brakes. Lets just dig a little deeper.
The maximum hydraulic pressure developed in your braking system can either be measured with an inline pressure gauge, or can be calculated as:
PM = Fp / Ap
Where:
PM = Maximum Pressure (psi)
Fp = Force on Master Cylinder piston (lbs) = Pedal Effort x Pedal Ratio (i.e. how hard the driver pushes the pedal multiplied by the pedal ratio.)
Ap = Area of the MC piston (sq. in.) = 0.785 x Dp^2 (Dp^2 = the MC piston diameter, in inches, squared)
This is a very important equation. Note how the maximum pressure determines the brake torque, and the maximum pressure is the force applied DIVIDED by the area of the MC piston. This means, all other things being equal, the bigger the MC piston, the LESS pressure developed, and the LESS brake torque generated. The trade-off is, the smaller the MC, the less fluid is displaced per inch of travel, and therefore the greater the pedal travel required - more on this later.
Thermal Capacity
As previously discussed, the brake rotors must be capable of absorbing the heat generated by the brakes as they convert the moving car's kinetic energy into heat.
The formula for the Kinetic Energy (K) in the moving car is:
K = (W * S^2) / 29.9
Where:
K = Kinetic Energy (ft. lbs.)
W = Weight of car (lbs.)
S = Speed of car (mph)
(Note: The root equation for kinetic energy is actually K = 1/2mv^2, the above version has a conversion factor included so we can use weight instead of mass and so the result is given in force units instead of energy units)
Nothing really surprising here - we know from instinct and experience that how much brake you need depends on how heavy the car is and how fast it's going. Note though, that speed is squared in the equation - meaning that as the speed increases, the kinetic energy developed goes up by the square of the speed increase - for example, if speed doubles, kinetic energy increases by a factor of 4. If speed triples - kinetic energy goes up by a factor of (3^2) or 9!! This is a bitch for race-car drivers but not so much trail rigs. However - if you're beginning to "race" your rock crawling buggy - be aware that your brake requirements are going to increase exponentially!
OK, so, moving rig has kinetic energy, kinetic energy must be turned into heat, rotors must absorb said heat. Big deal, so what?
Well, the equations that follow are used to calculate the temperature increase in the rotor for a given kinetic energy. Remembering that kinetic energy depends on weight and speed, they also explain in incontrovertible terms (big word - means you can't bloody argue with me about it!) exactly why "pinion brakes", flat out suck, and why Patooyee's rotors keep trying to melt off ! Remember - to add insult to injury "pinion brakes" are often used on Rockwell-axled rigs - which are big and heavy, and weight is a multiplying factor in the equation for Kinetic energy.
[As an aside - the previous discussion on brake torque also explains why some feel that pinion brakes work well "at slow speed". It's because the rotor is placed before the axle differential, meaning the pinion brake's brake torque is calculated as above and then multiplied by a factor equal to the axle ratio. This means, even with small callipers and rotors, they can develop tremendous brake torque. But - remember what we said about the requirements of a braking system - it must also have sufficient thermal capacity - and they simply do not. In fact, they're dangerously inadequate in this regard!]
The formula for temperature rise is:
TR = Kc / 77.8 * Wb
Where:
TR = Temperature rise (degrees F)
Kc = Kinetic energy change (from start of braking to end of braking) (ft. lbs.)
Wb = weight of all rotors (lbs)
And:
Kc = Kinetic energy change (ft. lbs.) = Kb - Ka
Where:
Kb = kinetic energy before stop
Ka = Kinetic energy after stop
Let's do an example, because math is fun!
Imagine I'm stopping my 5000lb buggy from 40 mph to a dead stop.
First, let's calculate the change in kinetic energy for a 5000 lb buggy from 40 mph to 0 mph - this remains the same regardless of the brake system.
Kb = Wc * S^2 / 29.9
= 5000 * (40)^2 / 29.9
= 267 559 ft lbs
Ka = 0
Kc = Kb - Ka
= 267 559 - 0
= 267 559
Now, with 3/4 ton truck disc brakes at each wheel, each rotor weighs about 21 lbs, for a total rotor weight of 84 lbs.
In this configuration, the temp rise of the rotor will be:
TR = Kc / 77.8 * Wb
= 267 559 / (77.8 * 84)
= 267 559 / 6535
= 41*F
If it were a hot day, after the stop my rotors could be at 141*F
Now, imagine I have pinion brakes with 2 small rotors, weight, say, 12 lbs each.
Now,
TR = Kc / 77.8 * Wb
= 267 559 / (77.8 * 24)
= 267 559 / 1867
= 143*F
On the same hot day, the rotors are now at 243*F!!
Make a few more stops, with insufficient time for the rotors to cool fully between them, and it's easy to see how you can seriously overheat the small pinion brake rotors!
OK, so much for physics and math. Let's move on now to the more practical side of how the braking system works.
Recall that brakes need to be able to do 3 things:
- Develop enough clamping force to create enough friction between pad and rotor to convert vehicle's kinetic energy to heat.
- Develop enough brake torque to reach the limit of traction (lockup the tires) in all conditions, and
- Have enough mass to absorb the conversion of the rigs kinetic energy to heat without boiling the fluid, warping the rotors, cooking seals, etc.
And they must do all three with good and consistent feel!
The following "flow-chart" presents a stylized summary of the brake system and its interrelated components / factors. Following it, we will examine each component in more detail:
The Pedal
As we have discussed, in order to apply the brakes the driver must be able to both move and pressurize the hydraulic fluid. That is, the master cylinder piston has the job of moving the brake fluid through the lines in order to bring the brake pads into contact with the rotor, and of pressurizing that fluid create clamping force.
The job of the pedal is to allow the driver to actuate the brakes, and also to multiply the driver's input through a lever effect called the "pedal ratio." In other words, the pedal transfers movement & force from the driver's foot to the master cylinder piston. This is a key concept - that brakes require both movement and force to operate, as the two are largely inversely proportional (that is: as one goes up the other goes down, and vice-versa) - we will return to this concept many times so keep it in mind.
The force the driver applies to the pedal when braking is called the "pedal effort". This amount of force, measured in pounds, is usually insufficient to operate the brakes - values ranging from 50-100 lbs are common. As such, pedals are designed to operate like a lever, multiplying this force before it is applied to the master cylinder piston. The amount by which any particular pedal design does so is called the "pedal ratio."
The pedal ratio is calculated as the distance between the pedal's pivot point and the centre of the foot pad, divided by the distance between the pivot point and the master cylinder push rod, as shown in the following diagrams.
Pedal Ratio = A / B
As can be seen from the diagram - the longer the pedal, the higher the pedal ratio, and the greater the mechanical advantage - meaning the lower the pedal effort required.
However, there are trade-offs. First, longer pedals are harder to fit - especially in cramped buggy cockpits. Also, the longer the pedal, the greater the pedal stroke required to move the master cylinder piston - i.e. the further the driver must press the pedal. The best design is a trade-off that will depend on many factors including desired pedal effort, master cylinder and calliper piston diameters, and the room available. As far as pedal effort is concerned, 75 lbs is common for high-performance race-car brakes, 100 lbs feels hard, and 50 lbs is common in power-assisted brakes. Common pedal ratios range from 3 or 4:1 for power brakes up to 6 or 7:1 for manual brakes.
Of course, if you are using a stock pedal, you are pretty much stuck with the ratio it has, and must design the rest of the system to accommodate.
Just remember - the longer the pedal, the greater the ratio, the less leg force required but the greater the pedal travel required. Longer pedals tend to deflect more under hard use, which can lead to a "spongy" feeling to the brakes. You should always strive to use a pedal that will go from fully retracted to fully applied in as short a distance as possible - usually less than about four inches. Any more than this, and there will not be sufficient travel in the drivers leg to fully apply the brakes under emergency conditions where travel is suddenly increased - such as during brake fade or when there is air in the fluid.
Keep in mind that total brake pedal travel includes not only the stroke of the master cylinder piston, but also includes all slop and deflection of all parts in the system.
Designing and fabricating a good pedal set up requires engineering far beyond the scope of this article - especially since the brakes are such a critical safety system on the rig.
My advice is to either incorporate a rugged stock pedal system, or use one of the pre-fab systems available (in a variety of ratios) from many aftermarket vendors. Choose a ratio that will work with the rest of your system and your personal preference for how the brakes will feel. Common ratios are 3:1 for power brakes and 5:1 for manual brakes - but, you may desire a more custom ratio - like 7:1 for manual brakes if using a large diameter master cylinder.
One final word on pedal effort:
You can design a brake system for any value of pedal effort (values ranging from 50-100 lbs are common), and the pedal effort will largely determine how the brakes "feel" to the driver (excepting problems like brake fade and fluid boil). For this reason, in the following discussion, assume, unless stated otherwise, that we are discussing the brake system with a "consistent and acceptable pedal effort". For example, if we say, changing to a larger bore master cylinder (all other things remaining the same) will decrease pedal travel but reduce braking power, we mean "with the same pedal effort." No matter what the system, or mod's we make to it, we always have the option of just increasing pedal effort, but this is rarely experienced by the driver as an acceptable solution. i.e. one "could" design a system with undersize callipers and an oversize MC that could still stop the rig effectively if only we would push on the pedal with 300lbs - but clearly this would not be suitable.
Brake Hydraulics
OK, now we're getting to the good stuff.
Recall that the brake's hydraulic system must supply movement and force. The movement must be enough to take up all slop, clearance, and deflection of parts as well as move the calliper pistons sufficiently to bring brake pads into firm contact with the rotors. The force must be enough to create enough friction between pad and rotor to stop the car.
It is the piston of the master cylinder that provides both the movement and the force, and the brake fluid that transmits both to the callipers.
Keep in mind, as you read this section, that the goal is to have a system that provides maximum force with small movement - i.e. we want to be able to brake hard without excessive pedal travel.
Force applied to the MC piston creates pressure in the brake fluid. The pressure is the force applied, DIVIDED by the area of the piston. Therefore, the smaller the master cylinder piston, the greater the pressure created.
System Pressure = Fp / Ap
Where:
Fp = Force on the piston (lbs.)
Ap = Area of the piston (sq. in.)
Example: Calculate system pressure for both a 1/2" diameter MC and a 1" diameter MC, given an input of 100 lbs.
Case one - 1/2" MC piston (d= 0.5):
P = Fp / Ap = 100 / pi * (d/2)^2 = 100 / 3.14 * (0.5/2)^2 = 510 psi
Case 2 - 1" MC piston(d=1.0):
P = Fp / Ap = 100 / pi * (d/2)^2 = 100 / 3.14 * (1.0/2)^2 = 127 psi
Recalling that the pressure created is a direct factor in how much clamping force and therefore brake torque is developed, it may seem that for the most powerful brakes, we would want to use a small MC piston.
But, the trade-off is the other component required of the brake system - namely movement. Because the fluid is incompressible, any movement in the MC piston translates into movement of the calliper pistons (excluding expansion of hoses and lines, which should be minimal in a properly working system). This movement in a hydraulic system is known as displacement and is calculated as the area of the MC piston multiplied by its stroke. Displacement is a volume, measured in cubic inches. Therefore, the smaller the master cylinder piston, the less displacement created.
Displacement (cubic inches) = Ap * Sp
Where:
Ap = Area of the piston (sq. in.)
Sp = Distance piston moves (stroke) (in.)
Example: Calculate displacement for both a ½" diameter MC and a 1" diameter MC, both with a stroke of 2 inches.
Case 1 - 1/2" MC piston (d= 0.5):
Displacement = Ap * Sp = 2 * [ pi * (d/2)^2] = 2 * [ 3.14 * (0.5/2)^2] = 0.39 cu. in.
Case 2 - 1" MC piston(d=1.0):
Displacement = Ap * Sp = 2 * [ pi * (d/2)^2] = 2 * [ 3.14 * (1.0/2)^2] = 1.57 cu. in.
Now we can clearly see that there is a trade-off between force and movement in selecting a MC piston size. The smaller the piston, the greater the pressure created but the less displacement produced (and therefore greater pedal travel required.)
So far, we have considered only the effect of the size of the MC piston, but of course in a brake system there are two pistons - the MC piston and the calliper piston (for calculations, the total area of all pistons in a multi-piston calliper act the same as a single piston of equivalent area in a single-piston calliper.)
There is, of course, a relationship between the pistons in the system that affects both force and movement.
Because the brake hydraulic system is a closed, sealed system, and brake fluid cannot be compressed, there is a law of hydraulics that we make use of to multiply force - that is, to apply more force at the callipers than the driver applies to the MC piston. It is quite simple, and quite possibly the most important concept in this entire article. It is this:
In a closed hydraulic system, pressure is equal over all surfaces of the containing system.
In our discussion of brake systems we will refer to the MC piston as the "input" piston and the calliper piston as the "output" piston.
The above law means that whatever pressure is created by the input piston is applied equally to the output piston. Because the output (calliper) piston is of much larger area than the input (MC) piston, this has the effect of multiplying force in the brake system.
The amount of force-multiplication thus achieved is known as the brake's "Hydraulic Ratio" Hydraulic Ratio can be calculated or expressed a number of ways. It is the ratio of fluid displacement by the master cylinder to fluid displaced in the calliper pistons. It is also equal to the ratio of force applied to the MC piston to the force generated by the calliper pistons. Hydraulic ratio is an important factor in the pedal effort equation, the higher the ratio, the less pedal effort is required (and the longer the pedal travel to achieve a given clamping force). The stiffer the calliper and the stiffer the pad, the higher the hydraulic ratio that can be employed.
For example, suppose we apply 100 lbs of force to a ½" diameter MC piston, we develop approx. 500 psi. This 500 psi acts evenly on all other surfaces in the system, including the calliper pistons. Suppose the calliper piston has a diameter of 3 inches. Multiplying our 500psi by the area of the calliper piston (~ 7 sq. in), we develop nearly 3500 pounds of clamping force at the brake pads.
The equation for this force-multiplying relationship is:
Fc = [Fmc * Ac] / Amc
Where:
Fc = Force at calliper piston (output) (lbs)
Fmc = Force on master cylinder piston(input) (lbs)
Ac = Area of calliper piston (sq. in.)
Amc = Area of master cylinder piston (sq. in.)
Brilliant! From the equation we can see that, for a given input force (pedal effort), in order to increase the force at the calliper (what we're after) we can either increase the area of the calliper piston(s) or decrease the area of the master cylinder piston (or both).
Of course, nothing is ever as easy as that! We can't forget the other factor - movement. Of course, there is once-again, a trade-off.
Unfortunately, both actions that increase force (good!) also increase pedal travel (bad!). That is, if we decrease the size of the MC piston we decrease its displacement - requiring us to increase its stroke to compensate - increasing pedal travel required. Similarly, as we increase the size of the calliper piston, we increase the clamping force, but we also increase the volume of fluid (the displacement) required to move the piston a given distance - the distance it must move to bring the pads into contact with the rotor - this also demands an increase in pedal travel.
The equation for relating piston movements is:
Mc = [Mmc * Amc] / Ac
Where:
Mc = Movement at calliper piston (in.)
Mmc = Movement of MC piston (in.)
Amc = Area of MC piston (sq. in.)
Ac = Area of calliper piston (sq. in.)
So, we can see that, in order to increase the movement of the calliper piston without affecting pedal travel (Mmc), we would have to either increase the size of the MC piston, or decrease the size of the calliper piston - both actions that will decrease force!
A simple example with which we are all intimately familiar, that illustrates the concepts of force and movement in a 2-piston hydraulic system, is that of the common floor jack.
Consider the jack's handle piston as the input piston (like the MC piston) and the piston that raises the lifting saddle as the output piston (like the calliper piston). Look at their relative sizes - the handle piston is tiny compared to the output piston - that's how you can lift a 5000lb car by applying only 50 lbs to the jack handle (multiplied, of course, by the length of the handle - just like pedal ratio in a brake system.) The trade off, of course, is movement - you have to pump that handle a lot of times (much input piston movement) to raise the car even just a little (output piston movement). Just as in brakes, movement and force are a trade-off. To produce great force without great input, you have to trade travel.
It's my considered opinion that this interrelationship between force and movements in a 2-piston hydraulic system such as brakes is not very well understood. Consider as evidence the following scenario - a guy swaps rear disc brakes in place of drums on the rear axle of his rig. The resulting brake performance is predictably inadequate - obviously the calliper pistons are much larger than the previous drum-brake wheel-cylinder pistons - meaning that more fluid must be displaced to actuate them. The common advice found is to "swap in a larger master cylinder." That's great from a movement perspective - the larger MC piston will displace grater fluid, moving the calliper pistons further for a given amount of pedal travel. BUT - we now know that the result will ALSO be a decrease in force at the calliper pistons (including the front callipers) than was achieved with the smaller MC.
So, what are we to do? We really have only two options - either careful compromise - or power assist. The beauty of power assist is that it allows us to substantially increase input force while maintaining pedal effort. This, in turn, enables the production of adequate force at the callipers while allowing the use of larger MC pistons and smaller calliper pistons, decreasing pedal travel. This is particularly important when you consider that most often, a single MC piston must actuate two callipers, (usually one for each of the front or rear wheels). That single MC piston must displace enough fluid to move two or more large-diameter pistons in the callipers. Making it large enough to do so without requiring way too much pedal travel means we compromise the force multiplication in the hydraulic system. The result is, we either accept sub-optimal force (and resulting brake torque), or we must make up for the lack of hydraulic force multiplication by increasing input force. And the only way to significantly increase input force without requiring much greater pedal effort is to increase pedal ratio or add power assist. Given that cockpit space availability place a very real upper limit on how much pedal ratio we can practically achieve - the answer becomes power assist.
Master Cylinders
The master cylinder is the heart of the brake system. Actuated by the pedal, its piston provides the force and the movement required to apply the brakes. When the pedal is released, an internal return spring returns the piston to its resting position.
Initially, as the pedal is pushed, the piston moves forward and fluid volume is displaced, taking up all clearances in the system. This fluid movement actuates the calliper pistons which extend and bring the brake pads into contact with the rotors. Because the fluid is incompressible, once the pads are in contact with the rotor, fluid movement stops and pressure rises. The harder the pedal is pushed, the greater the pressure achieved, the more brake torque is developed. How hard the driver can push the pedal, and therefore how much braking is achieved, is a function of input force (leg strength), combined with pedal ratio and any power assist.
The critical specs of a master cylinder are its bore (diameter of the piston) and stroke (how far the piston can travel - and therefore how much fluid it can displace when applied). Common values for bore range from 5/8" to 1-1/2", and stroke from 1" to 1.5". Matching both to the requirements of your system is the key to satisfactory performance. Remember that, for a given input (pedal effort) the smaller the bore, the more pressure is generated but the less fluid displaced. Similarly, the longer the stroke the more fluid is displaced, but the greater the pedal travel required. As with most things, the best result is achieved through carefully considered compromise.
The most common type of MC in use today is known as a "dual-tandem" master cylinder. It has separate circuits (pistons and outlets) for front and rear brakes ("dual"), with each circuit having its own independent reservoir. The reason for this is safety - the theory is: if one circuit (front or rear brakes) should fail, there remains the other separate, independent circuit. The two pistons are arranged inline in a single bore ("tandem"). The piston closest to the pedal is the "primary" piston, and the other is the "secondary" piston. Normally, the primary piston operates the font brakes and the secondary piston the rear. Fluid reservoirs may be integrated or remote (attached to the cylinder via hoses). The most common design has the reservoir attached directly to the cylinder.
Often, master cylinders are listed as being either for disc/drum brakes or for disc/disc brakes (I'm ignoring older drum/drum systems). Obviously a disc/disc MC is designed to be used with 4-wheel disc brakes, and obviously a disc/drum MC with disc/drum brakes. Assuming that nobody in their right mind would convert disc brakes to drums, that means the only remaining issue of concern would be using a disc/drum MC with disc/disc brakes. When choosing a master cylinder, traditionally, you may have been told that there are three differences between disc/disc and disc/drum masters that preclude the use of a disc/drum master with disc/disc brakes. They are:
- Reservoir size
- Piston size (bore and stroke)
- Built-in valving.
Here's my take on these three "issues":
Reservoir Size
Disc brake reservoirs are larger than those for drum brakes. You will often see two reasons given for this:
- First, because the pistons in a disc brake calliper are MUCH larger than the tiny pistons in a drum brake wheel-cylinder, disc brakes require more fluid volume to be displaced than drum brakes - requiring a larger reserve of fluid for operation.
- Secondly, as disc brake pads wear, disc brake callipers are self adjusting. That is, the callipers only retract the piston just enough to prevent pad-to-rotor contact. Now, imagine you start with disc brake pads with ½" thick linings and you have a 4" diameter calliper piston. Every time you apply the brakes and the pads wear a little bit, the calliper retracts just a tiny bit less. By the time the pads wear to 25%, or 1/8" thick, the piston at rest will be .75" further out in its stroke than it was when the pads were new. That .75" behind the piston must be taken up by additional fluid - and in the case of a 4" diameter piston, the additional volume required is given by (pi[d/2]^2 * 0.75) or about 8 cubic inches. Multiplied by two callipers (one for each wheel) and that's 16 cubic inches of extra fluid reserve required to compensate for pad wear. That's much higher than the amount required to compensate for drum brake shoe wear. Therefore, disc brake reservoirs are larger than for drums because OEM designers must design a reservoir for disc brakes large enough that the brakes will still function even if Joe Public doesn't check the fluid or add a drop between new pads and completely worn out pads.
It is my opinion that the second reason is the only real practical reason for the difference in size between disc and drum reservoirs. I say this because I have never seen a drum MC reservoir that was so small it didn't contain enough fluid to actually apply disc brakes - even though discs require more volume than drums.
With this in mind, we can see that, if we routinely check brake fluid levels as pads wear, reservoir size alone should not disqualify us from using a disc/drum MC with disc/disc brakes - especially if we're using a large MC with moderate or small callipers (e.g. 1-ton MC with 3/4-ton callipers).
Piston Size
It is true that disc brakes require both more pressure and more movement (volume) to operate than drum brakes. However, after reading this article, with your new encyclopedic knowledge of pressure and volume, you're going to be selecting a MC for your brakes based on its specs - not on what the counter guy says the OE application was. That is to say - given the right bore and stroke to match your system, there's no particular reason that, based solely on piston bore and stroke, an MC originally designed for disc/drums can't be used successfully on a disc/disc setup. That said, you will by now also understand that leaving in your old disc/drum MC after converting the rear to discs will not work well.
Built-in Valving
This one can be a deal-killer. If the MC in question was designed for disc/drums and has a built-in residual pressure valve, it will not be suitable for disc/disc brakes. See section below on valving for description of residual pressure valves. For now, the point is, know the MC in question and whether or not it has built-in valving (many do not). If it does, you would have to either modify it by removing the residual pressure valve to make is suitable for use with disc brakes, or choose a different master cylinder.
Power or Manual?
Again, you may see master cylinders marketed as being for manual or power brakes. This is based on two potential differences:
- Piston size (bore)
- Depth of piston hole
Case one may be true if all you are doing is swapping MC's. Normally, manual brakes use a 1" bore or less MC and power brakes 1" and up. Leaving everything else the same, and using a "power" MC (large bore) with manual brakes will normally result in a "hard pedal" - that is, the pedal effort required of the driver, without power assist, becomes too great. However, we already know that we can compensate by using a greater pedal ratio or larger piston callipers. Therefore, when overall system design is considered, and if MC specs are selected carefully to match the pedal ratio and calliper piston size, there is no technical reason that a larger "power" MC couldn't be used with manual brakes.
Case two is more of an issue. The piston hole is the are where the actuating rod (from the pedal or booster) contacts the MC piston. Normally it will be a "shallow" hole for a power brake MC as the power booster rod is captive between the booster and the MC. With manual brakes, the actuating rod is attached to the pedal which is obviously a lever. In this case the MC piston hole is normally deeper to prevent the actuating rod from falling out when the brakes are released - a very useful feature! However, if you must use a "shallow piston hole" MC with manual brakes - it may be possible to overcome the problem by appropriately limiting pedal travel (in the direction opposite to brake application) to prevent actuating rod fall-out.
Summary
There are normally differences between disc and drum master cylinders and between power and manual master cylinders, as follows:
- Disc brake MC's normally have a longer stroke and larger reservoir than those for drums
- Drum brake MC's may have built-in residual pressure valves
- Power brake MC's normally have a larger bore and shallower piston hole than those for manual brakes.
Normally, the easiest route is to use an MC designed for the type of brakes you have. However, when total system design is considered, as long as the specific specs and design of the MC in question are well understood, a much greater range of interchange possibilities opens up. That said, there are a few time-honoured rules of thumb:
- Use a 1" or smaller bore MC for manual brakes
- Use 1" or larger bore MC for power brakes
- Manual brakes with 1" MC use a pedal ratio of approx. 6:1
- Power brakes with a 1-1/8" MC use a pedal ratio of approx 4:1
- Disc brakes require a minimum pressure of 800-1000 psi
- Drum brakes require a minimum pressure of 400-500 psi
Effects of Pedal Ratio and Bore Size on Hydraulic Pressure Output
Pedal Ratio | Master Cylinder Bore Size (in.) | Input Force (lbs) | Pressure Developed (psi) |
6:1 | 1 1/8 | 75 | 453 |
6:1 | 1 | 75 | 573 |
6:1 | 7/8 | 75 | 748 |