Braking System Calculation help

[Nuggett]

Hello everyone! Here goes the first post!

I've been assigned to design this year's braking system for our car. I've grabbed the book Brake design and safety by Rudolf Rupert and I've been following his method, and so far it's ok. I've made an excel for all my calculations, however, there is one bit which I don't think it is well explained and I am getting a bit confused about it:

When determining the actual braking force (assume this case for the front) my main equation is:

Fx= (Nfront * Pedal Ratio * Fpedal * Apiston * Braking Factor * Front bias % * Rrotor) / (Amastercylinder * Rtire)

Now my question is: when the calliper has 2 pistons, will i multiply this by 2? and when 3, by 3 etc?

Also, for the braking factor, i know that it is about 0.35 for each pad, would that be 2*0.35 for one hub, so for the front in general that would be 4*.35 ?

I know this may sound a bit dumb but I am actually well confused about it! Also I can't find any info on it on the forum (i've used the search option etc!)

Thank you for your help!

Bests

Spyr

[Nuggett]

Hello everyone! Here goes the first post!

I've been assigned to design this year's braking system for our car. I've grabbed the book Brake design and safety by Rudolf Rupert and I've been following his method, and so far it's ok. I've made an excel for all my calculations, however, there is one bit which I don't think it is well explained and I am getting a bit confused about it:

When determining the actual braking force (assume this case for the front) my main equation is:

Fx= (Nfront * Pedal Ratio * Fpedal * Apiston * Braking Factor * Front bias % * Rrotor) / (Amastercylinder * Rtire)

Now my question is: when the calliper has 2 pistons, will i multiply this by 2? and when 3, by 3 etc?

Also, for the braking factor, i know that it is about 0.35 for each pad, would that be 2*0.35 for one hub, so for the front in general that would be 4*.35 ?

I know this may sound a bit dumb but I am actually well confused about it! Also I can't find any info on it on the forum (i've used the search option etc!)

Thank you for your help!

Bests

Spyr

[Pete Marsh]

Hi Spyr,
Check out the block on a incline definition of friction coefficient. See an area term anywhere?

There's your answer.

For the pistons question, a free body diagram will make it clear how much force is applied to the pads.

Pete

[Mike Cook]

Hey mr. nugget.

It might be easy if you break down each term in your equation. Basically you have

Pedal force (lb)
Multiplied by the pedal ratio (lb)
=
The force on the master cylinders (lb)
divided by the master cylinder area (in^2)
=
hydraulic pressure in brake system (lb/in^2)
Multiplied by the caliper piston area (in^2)
=
Force on rotor (lb)
Multiplied by coefficient of friction (unitless)
=
Drag force on rotor (lb)
Multiplied by radius of rotor (in)
=
Moment on tire (in-lb)
divided by tire diameter (in)
=
Force at contact patch (lb)

hope this helps mr nugget.

[Nuggett]

Looking into wilwood'd website:

An example of a 1 piston calliper:
http://www.wilwood.com/Caliper...=1&mtspec=&pistarea= (http://www.wilwood.com/Calipers/CaliperList.aspx?appl=&subname=&dustboot=&numofpist=1&mtspec=&pistarea=)

piston 1 dia: 1.62
piston area: pi*(1.62/2)^2 = 2.06

Now for a 2 piston calliper:
http://www.wilwood.com/Caliper...tspec=&pistarea=0.79 (http://www.wilwood.com/Calipers/CaliperList.aspx?appl=&subname=&dustboot=&numofpist=2&mtspec=&pistarea=0.79)

Pistons : 2
Piston 1 dia: 1"
Piston 2 dia: -

Piston area = pi*(1/2)^2 = 0.79

What's the point of the 2 pistons if the area's only calculated for one?

Here's another example of a calliper with 2 pistons, but it has a piston 2 diameter and the area is essentially doubled.

http://www.wilwood.com/Caliper...tspec=&pistarea=1.98 (http://www.wilwood.com/Calipers/CaliperList.aspx?appl=&subname=&dustboot=&numofpist=2&mtspec=&pistarea=1.98)
Shouldn't this be the case for before?

Thanks for your help

[Xeilos]

Logic and first principles will get you far my son.

Draw a FBD and apply the logic that is built into the gray matter between the ears.

[Buckingham]

You do need to multiply something by "2", but you seem to be getting hung up on which "2" makes sense.

Consider a single piston caliper (like on your car, 1 piston on one side, 0 pistons on the other side). It exerts a normal force "x" on the pad next to the piston. Assuming the caliper is infinitely rigid and the assembly is sufficiently floating, there is an equal force on the opposite pad. You need to multiply by "2" because there are "2" friction surfaces, one pad on each side of the rotor.

Now, take the caliper above and add an equally sized piston on the opposite side of the caliper to create a 2-piston caliper with opposing bores. Have you changed the normal force on each pad? Have you changed the number of friction surfaces? Has the total amount of clamping force the caliper exerts on the rotor changed?

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> What's the point of the 2 pistons if the area's only calculated for one? </div></BLOCKQUOTE>

See comment above about infinite rigidity and floating. On your single piston street car caliper, do the pads wear evenly?

[Nuggett]

I've found the answer to my question (which I don''t think I've formatted right ;p)

When you take the Piston area you need to assume half the caliper. That being said, a two-piston caliper, where the pistons are opposing, is considered as a one piston caliper in terms of the area. If you have two calipers on one side however (as was the case above in one of the links) then you do have to multiply the area by two.

Thanks for your help though!

[Der Krug]

Don't make one big equation. Have a spreadsheet with different parameters you can plug in/change frequently. If you more pistons your area changes.
Simple maths just work through and ask your professors for help.