How to Slow Down Faster: The Math behind Brakes. Part 5, An Example

brake2

Part 1 is here.

Part 2 is here.

Part 2b is here.

Part 3 is here.

Part 4 is here.

Part 4b is here.

     Now I’ll use these formulas to find out the brake forces (front and back) of a 1993-1995 FD3S rx7, and we’ll see how balanced the system is.  I couldn’t find the real numbers for some of the variables, so I’ll use some approximate numbers based on most automobiles.  The original formulas are here for reference.  This will obviously change the end result, but its just an exercise.  Here’s all the variables:

Front Caliper Piston Diameter (in): [1.42125 in]. (x2 since there are 2 pistons) (actual value)

Rear Caliper Piston Diameter (in): [1.42124 in]. (actual value)

Master Cylinder Piston Diameter (in): [15/16th in]. (actual value)

Booster Output Force (lb): [1000 lb] (estimated value)

Mu of Brake Pad (unitless): [0.50] (EBC Redstuffs) (actual value)

Front Rotor Diameter (in): [11.6 in]. (actual value)

Rear Rotor Diameter (in): [11.6 in]. (actual value)

Tire Rolling Radius (in): [12.4 in]. (stock wheels/tires) (actual value)

Total Weight (lb): [2800 + 150 lb] (stock R1 trim + driver’s weight) (actual value + estimated value)

Wheelbase (in): [95.5 in] (actual value)

Center of Gravity Height (in): [17 in] (actual value)

     I’ll start from the bottom, and work my way up.

     Divide the Caliper Piston Diameter(in) by 2, square it, then multiply by pi:

= [1.586 in^2] for the Front Inboard Caliper Piston Area  (in^2).

     Do the same with the Master Cylinder Piston Diameter(in):

= [.690 in^2] for the Master Cylinder Piston Area(in^2).

     Divide the Booster Output Force by the Master Cylinder Piston Area(in^2):

= [1449.275 Psi] for the Master Cylinder Pressure(Psi).

     Multiply the Master Cylinder Pressure(Psi) by the Inboard Caliper Piston Area(in^2) x 4 (due to floating calipers and 2 pistons per caliper):

= [9196.903 lb] for the Front Caliper Clamp Force(lb).

     Multiply the Caliper Clamp Force(lb) by the Mu of the Brake Pad(unitless):

= [4598.451 lb] for the Front Brake Pad Friction Force(lb).

     Divide the Rotor Diameter(in) by 2, and then subtract the Caliper Piston Diameter(s)(in) divided by 2 from that:

= [5.089 in] for the Rotor Effective Radius(in).

     Multiply the Brake Pad Friction Force(lb) by the Rotor Effective Radius(in) divided by 12:

= [1950.126 ft-lb] for the Wheel Torque(ft-lb).

     Divide the Wheel Torque(ft-lb) by the Rolling Radius(in) divided by 12:

= [1887.219 lb] for the Front Brake Force(lb).

     Do the same for the rear, and we get:

= [943.609 lb] for the Rear Brake Force(lb) (since the rear brakes are basically the same here with the same rotor diameter and piston size.  It just has half the number of pistons; 1 vs 2).

     So our static brake bias is 66:33 to a 50:50 weight biased machine.  Not so good, eh?

     Let’s solve for the Deceleration/Mu for both the front and rear without regard to tire grip:

     Multiply the Front Brake Force(lb) by 2:

= [3774.437 lb] for the Total Front Brake Force(lb)

     Divide the Total Front Brake Force(lb) by the Front Weight(lb) (which is 1/2 of our total weight; 1475 lb):

= [2.559 g] for the Front Coefficient of Friction(or deceleration.  they’re the same thing).

= [1.280 g] per tire.

     Do the same for the rear:

= [1887.219 lb] for the Total Rear Brake Force(lb)

= [1.279 g] for the Rear Coefficient of Friction.

= [.640 g] per tire.

     The front is pretty high, since an all-season tire can only grip up to [.9] or so gs while the rear tire’s grip is wasted due to such a low brake force.  But we haven’t factored in weight transfer, so lets do that next:

     The Weight Transfer formula’s here.

     I’ll be substituting the Deceleration with .9gs first, as a normal all-season tire, and then 1.2 gs for grippier performance tires.

     Multiply Total Vehicle Weight(lb) by Deceleration(gs) by Center of Gravity Height(in).  Then divide that by the Wheelbase(in):

= [472.618 lb] of Weight Transfer(lb).

     We then solve for Deceleration/Mu for the front and rear.

     Divide the Total Front Brake Force(lb) by the Front Weight(lb) [1475 lb] plus the Weight Transferred(lb) [472.618 lb]:

= [1.938 g] for the Front Coefficient of Friction.

= [.970 g] per tire.

     Divide the Total Rear Brake Force(lb) by the Rear Weight(lb) [1475 lb] minus the Weight Transferred(lb) [472.618 lb]:

= [1.883 g] for the Rear Coefficient of Friction.

= [.942 g] per tire.

     Ok that’s a lot better.  With a decent all-season tire operating at a maximum of .9gs, we have a pretty well balanced brake system (with a difference of .028 gs per tire).  The engineers at Mazda definately did their homework.  Now let’s try it again with a max tire grip and Deceleration/Mu of 1.2gs.

= [630.157 lb] of Weight Transfer(lb).

= [1.793 g] of Front Coefficient of Friction.

= [.897 g] per front tire.

= [2.234 g] of Rear Coefficient of Friction.

= [1.117 g] per rear tire.

     Notice here that as we increase the weight on the front tire through more weight transfer, our deceleration/grip is reduced (1.928 gs vs 1.793 gs).  That’s right.  This is proof that weight is NOT good for braking.  The more weight you add, the less the tires can decelerate.  But now we have a problem.  With a harder deceleration due to track/competition tires, we see that our brake balance is now unbalanced.  So how do we fix that?  The easiest way is to change the brake pads between the front and rear brakes.  That’s the easiest way, but racers don’t want to get worse brake pads to even out their brake balance since they’re probably running the grippiest brake pads out there.  So we’re left with increasing our rotor size (big brake kits), or/and increasing our caliper capacity (once again, bbks).  So for next time, let’s tune our brake system for our competition tires, and see if we can achieve an almost perfect brake balance like Mazda did for everyday driving.

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