Anti-dive and Inboard brakes.

[Caito]

Hi guys, I need some enlightment here. I'm havng trouble understanding how brake position (inboard/outboard) affects antidive.

As I read on Tires Suspension and Handling (or the wiki, too:Anti's ) you get the instant center as viewed from the side, get a line from the tire contact pantch and join them, then you do some force analysis.


The thing is that that is valid for outboard brakes, for inboard brakes you need to draw a line from wheel centerline to the instant center. Why??? It's driving me nuts .


Thank you for your help!

Bye.

PS:

Some reference pic:

[thisisatest]

for inboard brakes, the caliper is attached to the chassis. you hit the brakes, the tire tries to stick to the ground, chassis moves forward, wheel moves backward. it's all nice and neat.
for outboard brakes, the caliper is on the hub, and the hub rotates around the instant center as the suspension goes up n down. so you have to account for the torque that is being applied to the hub.
i'm pretty sure that's right. i would have thought that for outboard brakes, there are two separate equations to deal with (the axle being pushed back, and the hub being rotated), but i guess it can be consolidated into calculating the single force at the contact patch...

[Caito]

But as I see it, those forces are internal and should cancel out.




If we imagine the car going to the right.

When you brake the caliper generate a force going down. This could be seen as a force down, plus a torque in the wheel.


The caliper gets the same force up. Which could be seen as a force down and a torque in the upright.

The force going down to the wheel transfers to the bearings, to the hub and it's cancelled with the one for the caliper.

Still, as you said, I have that torque in the hub which I wouldn't if the brakes where inboard. That torque would be clockwise as seen in the image.

Now we have a similar situation with inboard brakes:



The car is going to the right.

As brakes are applied you get a force pointing down on the disc, and one pointing up in the caliper.

The force at the disc can be imagined as a torque and a force on the center of the disc.

This force would cancel up with one in the caliper giving a net moment at the chassis, halfway between the calipper and the center of the disc. Is this right?


Anyway, supposing what I said is correct, I still can't get why with inboard brakes the brake force is applied at the wheel centerline.

[bettonracing]

With outboard brakes the caliper loads travel through the upright & the arms. With inboard brakes the braking loads travel through the axle - approximated (rightly or wrongly) at the wheel center.

Regards,

Kurt

[Caito]

But still, in both cases, the braking force is at the contact patch, which we may assume below the wheel centerline at floor level.

So why does the outboard brake lets you move that force at wheel centerline?


Sorry that I don't get it :p

[mep]

Hi Caito
nice question, you got me thinking for a couple minutes. Now here is my explanation:
When the brakes are outboard the tire and the upright are rigidly fixed together trough the brake pad/disc and can be considered as a solid part (more or less). So the braking force and the torque will be transmitted over the upright into the suspension arms.

When the brakes are inboard there will be an axle running from the diff to the rim which transmits the torque to the chassis. Also when there are no outboard brakes there is no connection between rim and upright which can transmit torque so there is only the braking force left which has to be carried. This force will be transmitted by the central bearing to the upright and into the suspension arms.

So what happens is a split of torque and the force causing the difference. In first case both is carried by the suspension arms in second case just the force has to be carried while the torque gets carried by the axle.

Do you need a drawing?
I made some pencil drawing but it didn't turned out to be nice.

[Belatti]

Right, the same happens with antisquat in a rear suspension when you have RWD vs. FWD. The reaction loads the wishbones, thus affect chassis.

[Caito]

Thank you very much. Then the next question follows:

With inboard brakes the brake force is carried to the suspension to the chassis.

But there's a torque on the chassis itself due to the caliper, isn't it?

As the you brake the caliper and disc get opposing forces which cancel out, but you get torque on the driveshaft and torque on the chassis.. is this ok?


bye!

thank again

[ysyy88]

Hi guys, I need some enlightment here. I'm havng trouble understanding how brake position (inboard/outboard) affects antidive.

As I read on Tires Suspension and Handling (or the wiki, too:Anti's ) you get the instant center as viewed from the side, get a line from the tire contact pantch and join them, then you do some force analysis.


The thing is that that is valid for outboard brakes, for inboard brakes you need to draw a line from wheel centerline to the instant center. Why??? It's driving me nuts .


Thank you for your help!

Bye.

PS:

Some reference pic:

i think, the line should always be from the "tire print point" to IC, no matter how the tire brakes.

[GSpeedR]

i think, the line should always be from the "tire print point" to IC, no matter how the tire brakes.

If you create a free body diagram of the tire and wheel, neglecting wheel torque, you'll find that lateral tire force is reacted at the wheel bearing. Including wheel torque of outboard brakes and differentials allows one to replace the force and moment with a force located at the tire patch. It depends on whether the chassis is reacting the torque or the suspension. Gillespie has a nice derivation in Fundamentals of Vehicle Dynamics.

[mep]

Oh I forgot this topic good that it pops up again.

Thank you very much. Then the next question follows:

With inboard brakes the brake force is carried to the suspension to the chassis.

But there's a torque on the chassis itself due to the caliper, isn't it?

Yep.

As the you brake the caliper and disc get opposing forces which cancel out, but you get torque on the driveshaft and torque on the chassis.. is this ok?

So far yes.
Actually it's a pretty good question.
What happens with the torque once it is reacted into the chassis?
I must say once again I am a bit disappointed by the forum. I wonder why nobody else tries to answer the question. I must admit you got me confused, too.

When I follow your way of thinking there will be torque acting on the rear of the chassis (inboard brakes) trying to rotate it so that the front dives down. There will also be torque acting on the front of the chassis because it doesn't matter if the brakes are inboard or outboard because in both cases torque is reacted to the chassis. Just in one case it goes over the axle and in the other over the suspension. So those two torques would tilt the chassis down + the mass of the car. With this we would end up with twice the amount of dive so this simply can't be true.

I guess that is what you where heading to?
I think the theory above is wrong.
The torque acting on the diff/chassis is nothing else than an internal load. It is just required to make the system stiff but it does not increase tilting of the chassis.
So when you think about the torque on the chassis we can assume a model of suspension/chassis as a rigid body. By this we get trough a static analysies the external forces acting on the car and no incluence of the torque on the diff.

The difference to the "anti model" is that we take a closer look at the suspension and try to figure out how it is compressed. Important here is the force going over the suspension arms. With tilted arms there must be a reaction force in the pushrod which pushes the car body up.

So far my understanding of it.

[PlatinumZealot]

When considering the chassis as one the braking torque is not an internal load because this force cannot travel through the wheel bearings (which the brake disk is mounted on). It has to travel through the calliper bracket - no other place.

This torque is causing the car to dive. And in a steady state, it is cancelled out by the reaction from the front suspension.

The internal torque caused by this external load, is reacted by the actual strength and rigidity of all the bodies that connect the brake pad to the chassis of the car, to the front suspension. Like a long chain of bodies reacting the stresses put through them. Just my reasoning though.