Brake caliper position - galileo
Most makers seem to put front calipers ahead of the axle, but some place them behind: there must be a difference in the direction of the forces acting on the caliper (and thus suspension) when brakes are applied, does this have significant effects or is position purely a matter of convenience for servicing?
Brake caliper position - OldSock
I'd assume that the retarding torque provided by the steering knuckle would be independent of the caliper mounting position, by virtue of circular symmetry about the stub axle. Its effects on the suspension components would therefore be identical.

It's probably easier to cool a forward-mounted caliper.

Edited by OldSock on 29/06/2009 at 17:52

Brake caliper position - Number_Cruncher
This question came up quite recently;

Brake caliper position - OldSock

N_C, I'm a bit confused.

If I jack up the front wheel on a car such that there is no radial load on the wheel bearing (other than the weight of the rim and tyre), I would still expect the brakes to 'work' - even with no means for the braking force to be 'reacted' at the stub axle....
Brake caliper position - Number_Cruncher
The angular position of the caliper on the disc makes no difference to the braking force produced at all. For one brake / wheel;

BF = N * C * mu * R_eff / R_rolling

BF - Braking Force (Newtons)
N - Number of friction surfaces (2 in a standard disc brake)
C - clamping force (Newtons) [brake fluid pressure * piston area]
mu - co-efficient of friction between pad and disc
R_eff - effective radius of brake pad contact (metres)
R_rolling - rolling radius of wheel

There's absolutely nothing in there about the angular position of the caliper.

However, what does change during braking is the total load borne by the wheel bearing. If the caliper is ahead, braking force adds to the load borne by the bearing, and hence, the bearing and its housing, and therefore probably the brake disc itself needs to be bigger - it's a vicious spiral, and all contributes to both total mass, and unsprung mass.

Determining R_eff is a bit of fun. For most practical purposes, you can take it to be the mean radius of the active surface of the disc (i.e., the outer diameter plus the inner diameter divided by four), and is independant of brake pad shape. To evaluate it properly, you need to perform an integral over the area of the brake pad - the result is different, but, not by enough to trouble anyone but the most pedantic!

Brake caliper position - OldSock

... If the caliper is ahead, braking force adds to the load borne by the bearing....

So does the converse hold - for a 'trailing' caliper, does the braking force subtract from the load borne by the bearing? If so, wouldn't the front end of a car so equipped rise under braking - a result of the reduced weight acting perpedicularly to the stub axle?

Can't say I've ever seen that happening!
Brake caliper position - Number_Cruncher
>>does the braking force subtract from the load borne by the bearing?


>>If so, wouldn't the front end of a car so equipped rise under braking


This has nothing to do with what happens to the sprung mass of the car. Nothing to do with the nose dipping under braking. The forces resulting from braking are all contained within the hub (see the equilibrium arguments posted seperately), except for the braking torque which becomes a braking force at the road surface.

Brake caliper position - Number_Cruncher
>>even with no means for the braking force to be 'reacted' at the stub axle....

OK, imagine that you have a long, quite flexible, driveshaft which is spinning. On the far, unsupported end of it is a disc which is running between the pads of a caliper. The caliper squeezes - without any support, the disc would just turn or twist out of the caliper, move aside, and continue to spin.

There needs to be a bearing there to react that tangential force which acts on the disc, with an equal and opposite force.

In the case of a caliper on the front of the disc, during a brake application, the brake pads would tend to rotate with the disc, and so, the force acting on the pads is in the downwards direction. The force provided by the pads, acting on the disc at this location is, for force equilibrium, upwards. The force provided by the bearing, acting on the centre of the disc is, for the force equilibrium of the disc, is downwards.

The force on the rotating assembly caused by the reaction of the road is upwards. The force, provided by the bearing, for the force equilibrium of the rotating assembly, is downwards.

So, the braking forces in this example produce a force which must be reacted by the bearing which is in the same direction as the force which supports the car, whereas, if you put the caliper on the rear of the disc, you get some beneficial cancellation of these forces.

Brake caliper position - OldSock

Ah, that makes a bit more sense!

I've always struggled a bit with rotational mechanics - particularly with 'frames of reference' :-)
Brake caliper position - galileo
NC, many thanks for your usual full and lucid explanation - my initial question arose because I have just traded my Corolla (front location) for a Hyundai i30, which has rear located calipers and I wondered why there was a difference.

Brake caliper position - pmh2
Discussed recently with a good explanation by number_cruncher



Edited by pmh2 on 29/06/2009 at 18:24