What Line Pressure Do Gti Brakes Run At?

[Rich_p 23]

Need to know as a friend is trying to borrow an ex F1 bias valve from the factory for me

[Rich_p 23]

I really thought a techy person might know this.

[C_W 3 1 Cars]

Depends on how hard you press the brakes surely?

[Rich_p 23]

I guess even the max possible pressure would help.

 

They normally set up a graph based on what pressure they run at certain points when the pedal is pressed and then they optimize the valve for that.

Just need to make sure it’s going to be compatible that’s all as they may run their cars at a lot higher pressure.

[welshpug 1,621]

doesn't matter if you run at a higher pressure or not, a Bias valve is a bias valve, they're pretty much all the same bar the fitting and method of adjustment.

[Rob Thomson 6]

Quite simple to work out...

 

Maximum force you can exert on the pedal = 100kg = 200lbs

Leverage ratio of pedal box = 5:1

Assistance provided by servo = 14 x 0.8 x Pi x 4.5^2 = 700 lbs

MC piston area = Pi x 0.4^2 = 0.5 sq.in

 

Brake hydraulic pressure = (700 + 200 x 5)/0.5 = 3400 lb/sq.in

 

Ok, there's a few assumptions in there but that's probably a safe estimate of the maximum pressure you could ever possibly see.

 

Regardless, I thought F1 cars had proper bias pedal boxes rather than bias valves...?

[Rob Thomson 6]

You can also work it out the other way, to calculate the pressure needed to lock a wheel...

 

205 weight = 900kg = 2000 lbs

 

Assume under heavy braking 75% of that weight carried by front wheels.

 

Therefore, load through each front wheel = 2000 * 0.75 * 0.5 = 750 lbs

 

Assuming dry tarmac and therefore co-efficient of friction of 1, braking force required to lock wheel is the same.

 

Rolling radius of wheel = 15/2 + (195/25) x 0.5 = ~12 ins.

 

Effective brake disc diameter = ~4.5 ins.

 

Therefore, force required at caliper to lock wheel is 750 x (12/4.5) = 2000 lbs.

 

Due to the mechanical advantage of a sliding one-pot caliper, only half that force needs to be generated by the pads = 1000 lbs.

 

BUT, the co-efficient of friction of pads on disc is only ~0.35.

 

Therefore, the force that needs to be generated by the hydraulics = 1000/0.35 = ~3000 lbs.

 

Caliper piston dia = 50mm = 2 ins.

Caliper piston area = Pi x 1^2 = ~3 sq.ins

 

Hydraulic pressure required = 3000/3 = 1000 lb/sq.ins

 

 

That's MUCH less than the maximum theoretical pressure that you could produce by really stamping on the pedal, but gives you a good idea of the pressure that you might generate regularly during hard driving.

 

 

I'm bored....

[miamichris 9]

^^^^ that's some technical stuff goin on there Rob, I got lost halfway through!

 

and yes, you must be bored!