# Vehicle Braking Figures

1. Jun 16, 2015

### joesoap

Hi, Im looking for some help with a project I am doing where I am designing a braking system for a machine.

The machine can travel up to a maximum speed of 12khm and weighs 5000kg.

I need to stop the vehicle in approx. 6 meters/1.2m/s deceleration, using two drum brakes which are operated with hydraulic pressure.

I have calculated the required braking force
F = Ma
6000 = 5000x1.2

Is there an equation to calculate the size of brake drum that I need and the amount of hyd pressure I need to operate the brake?

Last edited by a moderator: Jun 16, 2015
2. Jun 16, 2015

### joesoap

I also need an equation for the what force would make the wheel lock.

I am using a coefficient of friction between pad and drum - 0.3
between floor and wheel 0.8

I just need basic figures really.

3. Jun 16, 2015

### CWatters

khm ?
If you meant 12 kilometres per hour then a stopping distance of 6m isn't equivalent to a deceleration of 1.2 m/s/s.

4. Jun 16, 2015

### CWatters

I believe these two things are related. eg Bigger drums allow lower hydraulic pressures.

I suspect you may have to find empirical data (for example how much power can brakes of a certain size dissipate while still having a reasonable lifetime). That may dictate the minimum size of drum/pad. From that and the friction coefficient (and any leverage in the system) you should be able to work out the pressure required.

5. Jun 16, 2015

### joesoap

They are related, that's why I asked if anyone knew the equations.

12 kph stopped in 6m is 1.2 ms

6. Jun 16, 2015

### Staff: Mentor

Welcome to the PF, Joe.

You should try to be a bit more careful when typing in units -- it's very confusing trying to read some of what you typed. Acceleration has units of m/s^2, for example.

Anyway, this article should get you going. It covers the hydraulics and friction/leverage in the 2nd page...

http://auto.howstuffworks.com/auto-parts/brakes/brake-types/brake.htm

7. Jun 17, 2015

### CWatters

What I meant was that the equation for braking force is likely to have (at least) two unknowns...

The dimensions (eg diameter of drum)
The hydraulic pressure.

One equation with two unknowns can't be solved so eventually you will need more information such as empirical data that relates power or wear to the dimensions/diameter of the drum or pads. I had a look for that yesterday but only found this from the 1970's...

http://deepblue.lib.umich.edu/bitstream/handle/2027.42/1354/13894?sequence=2

Page 8 onwards appears to have some of the equations you need but I haven't read it all. For example see Eqn 5 and 6 page 8.

8. Jun 17, 2015

### CWatters

Perhaps this also helps.. With reference to the diagram..

The braking torque Ti is

uNRd ........................................(1)

Where
u = coefficient of friction (you said 0.3)
N = Normal force (depends on brake pressure and area, unknown)
Rd = Radius of the drum (unknown).

The required Braking Torque at each wheel is

maRw/4 ...................................(2)

where
m = Mass of vehicle
a = Deceleration of vehicle
Rw = Radius of Wheel
(m is divided by 4 because there are four wheels)

Equate (1) and (2)

uNRd = maRw/4

This can be rearranged to give essentially the same equation as Eqn5 in the reference in #7. Note they summed over 4 wheels and used weight instead of mass.

There are two unknowns as I predicted. The radius of the drum and the Normal force (Hydraulic pressure)

Edit: If you choose a "practical" drum diameter and pad area you can solve for the hydraulic pressure needed.

Last edited: Jun 17, 2015