caliper orientation and brake bleeding

[satbat7]

To what extent does the orientation of the brake caliper on the disc have an effect? Is there a recommended orientation/position?

Can i bleed the brakes with the bleeding screw at the highest point first to remove the air and then set the caliper such that the bleeding screw is NOT the highest point?
This problem is due to the byke caliper i am using is made only for the right hand side of the byke.

thanks,

[drivetrainUW-Platt]

If you make your brake calipers easily removable once the wheels are off, the bleed skrews can be anywheres in theory.

[PedalOnTheRight]

In theory, your calipers can be mounted anywheres you want. It gets tricky to mount them on top or bottom on a typical setup due to the location of the control arms at the uprights.

Also, you can bleed the brakes with the bleed screw at the highest point and then move the brakes where you want them. HOWEVER, if you do this, it is MORE likely that you'll still have air in the system if you only bleed it before you put your brakes together. This is because at higher temps and pressures, you're surely to dislodge a bubble here or there in the lines. Therefore, I'd recommend bleeding them with the bleed screw at the highest point, putting the system together, running the car for a few hours, then bleeding again. Repeat this process again and you should be set.

[BStoney]

...I would also advise considering vehicle dynamics in your caliper positioning (depending upon the packaging constraints your uni has inparticular). It would be advised to place your caliper as close to the ground AND the longitudinal location of the CG as to decrease CG height as well as yaw inertia. Maybe a good idea to calculate the difference in heights/inertias based on different packaging options to justify your compromise.

My .02...

[C.Zinke]

You will reduce the reaction forces in the wheel bearings during braking if you put the caliper behind the upright

[GSpeedR]

Behind the upright will reduce vertical reaction loads in the wheel bearing, above the upright will reduce longitudinal loads in the bearing under braking. Ortiz mentioned dangers of oscillating loads if the overall reaction force changes directions under braking, but I don't think it's as big of a deal as hitting bumps and such.

[Bill Kunst]

Can anyone explain to me how the net reaction forces in the brake/bearing setup would be reduced by putting the brake in any specific location. My brain thinks in torque moments, and if it is at the same diameter, I just can't see how the forces would be reduced in any way.

Bill

[Chris Allbee]

I usually reduce reaction loads by lifting my foot off the brake pedal...

[Sooner_Electrical]

Don't get me wrong I am no mechanical expert but I am already logged in thought I would give it a guess. I figure the only way that the placement of the caliper position will affect loading in the bearing is if it were in front, the load would be the resultant reaction plus forces caused by weight of car as were rear mounting would be opposite direction. That is my best guess.

[GSpeedR]

Originally posted by Bill Kunst:
Can anyone explain to me how the net reaction forces in the brake/bearing setup would be reduced by putting the brake in any specific location. My brain thinks in torque moments, and if it is at the same diameter, I just can't see how the forces would be reduced in any way.

Bill

The unsprung mass must be decelerated (to change to a slower rotational speed) and thus there is a net change in momentum transferred to the caliper, producing a linear force in the direction of motion of the unsprung mass. Originally I figured this changed wheel loading, but it was pointed out that the caliper is connected to the rotor in parallel with the wheel bearing. So caliper loads change the wheel bearing loads.

[Kyle Walther]

doherty could you please explain how the wheel loads are not changed. every time i draw the fbd if the caliper is mounted in any position not parallel to the ground it creates a vertical force that has to be reacted through the chassis or through the wheel onto the ground.

{edit}

the more i think about it though it doesn't make sense that it would. i think i confused myself

[Bill Kunst]

So, what you are saying is:
front caliper is equal to load tansfered weight plus the reactive force of the brake torque.

rear caliper would be reactive torque - load transfered weight.

Now that I look at it this way, it makes sense. Now the question becomes, "how big of a force is the reactive braking force?" Also, how many teams mount their caliper so that it is parallel to the force of load transfer? My guess would be that most teams mount it in such a way as to minimize materials used in the upright, therfore reducing weight.

[Chris Allbee]

Kyle, your initial FBD is correct. A vertical force generated would have to be reacted through the chassis through the push/pull rod with a corresponding change in wheel load.

Just to get a better visual of this I ran an FEA on our full suspension model with a static torque on the rotor and a fixed caliper. This produced a measurable amount of vertical motion (not much, but it is there). This was JUST a static load though representing a constant 2g stop. Obviously the impulse of initial application will amplify this movement. Initial measurements indicated just over an eight of a pound though. Again, not much change, but it is there.

For another visual you could always jack up the rear of your car and replace the dampers and springs with bungees. Then rev the thing up, pull the clutch and slam on the brakes. Look for any wheel movement. This of course assumes you have outboard rear brakes.

[GSpeedR]

Originally posted by Chris Allbee:
Kyle, your initial FBD is correct. A vertical force generated would have to be reacted through the chassis through the push/pull rod with a corresponding change in wheel load.

Just to get a better visual of this I ran an FEA on our full suspension model with a static torque on the rotor and a fixed caliper. This produced a measurable amount of vertical motion (not much, but it is there). This was JUST a static load though representing a constant 2g stop. Obviously the impulse of initial application will amplify this movement. Initial measurements indicated just over an eight of a pound though. Again, not much change, but it is there.

For another visual you could always jack up the rear of your car and replace the dampers and springs with bungees. Then rev the thing up, pull the clutch and slam on the brakes. Look for any wheel movement. This of course assumes you have outboard rear brakes.

The momentum transfer at the rotor/caliper interface is an internal force. There is an equal and opposite force on the rotor (and hence the bearing). So for a trailing caliper, there is an upward force at the caliper and thus an upward force at the upright. There is also a downward force at the rotor and thus a downward force at the upright. They are in parallel and thus you simply change the load path through these components (as described by Mr. Ortiz) but not the total load and so nothing is transmitted to the tire. It is not an external force; if it was then yes any vertical force is reacted through the wheel to the ground.

Also, I had my bearing loads backwards before: the caliper should be below the wheel centerline to reduce the bearing reaction load under braking.

Kyle, if you include the reaction at the rotor then your FBD should work out fine.

[Kyle Walther]

yeah there is only one instant the braking loads in my mind can cause a wheel load change, and that would be if at a large braking accel you have some deformation in your a-arms changing the attitude of your outboard. thus changing the cross wheights if your left to right deformations were different.

i think that is what chris is seeing in the system FEA. and again that has nothing to do with caliper placement only A-arm geo

[Chris Allbee]

actually, my terminology was off. Bearing reactions were changing and I wrongly interpreted that as wheel reactions. My bad. More Red Bull, less sleep.