Weight transfer is a big reason adjustable brake proportioning valves should be used to fi
Gil Acheson, Denver, CO: I've read Car Craft from cover to cover since its inception, and I race a SuperFormance Cobra with Wilwood brakes. Now that I've established my "street creds," let me ask a question: On page 44 of the "What's Your Problem" column in the Mar. '10 issue, you are discussing proportioning valves, front and rear adjustments, and so on. You state that a "taller tire requires less brake pressure than a shorter tire." Please help me understand this, because my intuition would tell me the opposite would be true due to the higher leverage or turning moment of the taller tire exceeding that for the shorter tire, therefore requiring more pressure to equal the braking power.
Jeff Smith: My first editor, Rick Voegelin, used to tell me, "Doctors bury their mistakes; we publish ours." Both of you are very much correct. Let's get into why this is, and then I'll explain where my statement originated and how my experience set me up.
In the HP Books Brake Handbook by Fed Puhn, which is now out of print, there is a section on brake system design that lists a formula:
New pedal effort = old pedal effort x (new tire radius/old tire radius)
The values for this formula include pedal effort in pounds, and tire radius in inches. But to be completely accurate, these tire numbers must be the loaded tire radius. We'll look at the difference between the tire diameter of 26 inches and the new tire diameter of 28 inches.
New pedal effort = 100 pounds x (14 inches/13 inches)
New pedal effort = 100 pounds x 1.077
New pedal effort = 107.7 pounds
This means that a taller tire requires greater pedal effort to generate the same brake torque. Conversely, a shorter tire will require less pedal effort to achieve the same brake torque. So you are correct, Gil, in your assumption that the taller tire creates greater leverage. This is true for front or rear brakes.
My statement originated from an experience many years ago testing a friend's Chevelle in which we had tested his brakes with several aggressive stops and set the brake balance with an adjustable proportioning valve. Then he installed a set of taller drag tires on the rear of the car, and on his first aggressive stop, the rear tires locked up and the car spun. Luckily, he didn't hit anything, but it was plenty scary. I assumed the taller tires required less brake pressure. The reality was that the combination of a more aggressive use of the brakes combined with more weight transfer that was partially due to the taller tires (that created a greater static rake that moved weight to the front) all contributed to a greater transfer of dynamic weight onto the front wheels and off of the rear wheels. With reduced load on the rear tires but the same hydraulic pressure to the rear, the rear brakes locked up prematurely. To fix the problem, we reduced the rear brake pressure with the adjustable proportioning valve and the car stopped fine with no drama after that. While I thought that reducing the pressure was because of the taller tire, it really was because of the greater weight transfer. It's a small point, but it led me to an incorrect assumption.
As an interesting aside, there was also a formula in the book for calculating the equivalent amount of horsepower it takes to stop a vehicle of a given weight at a given speed and deceleration rate. For fun, I used a 3,500-pound car decelerating at 1 g from 60 to 0 mph and the formula says it would take the equivalent of 563 hp to pull it off. This gives you an idea of how much power it takes to stop a heavy muscle car from even a relatively mild 60 mph.
ARP recently invited us to its open house. Among the 50-plus cars that showed up were this tribute Penske Camaro and the Poteet and Main streamliner that has already run 436 mph. They're shooting for 500! Also worth noting: This is ARP's main production shop and the floor is this clean all the time, not just for the show.