racer8432955 said:
https://grassrootsmotorsports.com/forum/build-projects-and-project-cars/econoline-pick-up/256677/page1/.
When designing the brake system I believe I made a mistake, and am now unsure on how to correct it. Basically, I am not generating enough stopping power. I put a brake pressure gauge on each caliper and am getting readings between 600-725 psi. This seems to add up to my (bad) design using this math: 50lbs of leg force to a 6:1 pedal ratio= 300 psi. 300psi / 2 capliers = 150psi @ each caliper. 150 x 5 (bore area of a 2.5" piston - GM Metric Caliper)= 750 psi.
Driven and Triumph are right on the money.
also, you are missing some important steps and mixing up some units in the math:
1. Pedal Force x Pedal Ratio = MC Input Force, not a line pressure
50 lbs x 6:1 = 300 lbs input force to master
2. MC Input Force divided by MC Area = line pressure, which will be the same through the entire brake system. it is not split between calipers.
300 lbs / 0.7854 sq in = 382 psi
so, if you're seeing 700 psi on your gauges, and your pedal is really 6:1, you must be pushing the pedal with more than 50 lbs. Line pressure divided by pedal ratio = pedal force: 700 psi / 6 = 117 lbs. That's a lot of pedal force, but let's go with these values since they match your gauge measurements.
3. Line pressure x caliper piston area = caliper clamping force. going with the 2.5" piston diameter you stated
caliper piston area = (pi / 4) x diameter squared = 0.7854 x 2.5 x 2.5 = 4.91 square inches
then caliper clamping force = 700 psi x 4.91 in^2 = 3436 lbs of clamping force. but that's not *braking* force, because we haven't considered the pad friction, the radius of the brake rotor, or the radius of the tire.
4. clamping force x 2 x pad friction value = caliper friction force. since you said "rock auto cheapskate pads", lets go with a common pad friction value of 0.3. then:
3436 lbs clamping force x 2 x 0.3 = 2062 lbs of caliper friction force. but that's still not braking force, because we haven't considered radius of rotor or radius of tire.
5. caliper friction force x rotor effective radius (distance from center of hub to center of piston) = brake torque. Metric calipers are typically used on 10.5" diameter rotors, so the effective radius is about half of the rotor diameter minus half of the piston diameter, or 5.25" - 1.25" = 4"
so we are applying a friction force of 2062 lbs at a distance of 4" from hub center, which gives a brake torque of 8248 lb-in, or 687 lb-ft
6. brake torque divided by tire rolling radius = braking force on that tire. i'll assume your tires are 26" diameter, so their radius is 13". then
8248 lb-in / 13 in = 634 lb. that's the brake force you're applying at ONE front tire. since you've got 2 equal brakes on the front axle, your total front brake force is 1268 lbs for a line pressure of 700 psi, which was generated from a pedal force of around 117 lbs.
Rock Auto tells me Explorer (i chose model year 2000) rear caliper piston diameter is 48 mm = 1.89". and the rotor diameter is 12.19". so, if we assume the same pad friction and the same tire size, we can lump all the equations together and calculate the rear brake force as
700 psi x 0.7854 x 1.89 x 1.89 x 2 x 0.3 x (6.09 - 0.95) / 13 = 466 lb of stopping force for one rear brake = 932 lb stopping force on the rear axle.
so for the total vehicle you've got 1268 lb front plus 932 lb rear = 2200 lb brake force. how much does the truck weigh? if we assume none of the tires are skidding at that brake force, then your deceleration in G's is simply total brake force divided by total vehicle weight. if the vehicle weighed 3300 lb, you'd be pulling 0.66 G deceleration.
go to a smaller bore MC to generate more pressure ? Two problems with that, one I already tried it, and seat of the pants testing resulting in worse stopping power. And, I already have way too much pedal travel.
if seat of pants is correct, it is possible that you bottomed out your pedal travel before you generated maximum pressure.
The too much pedal travel problem could be solved with a larger bore MC, but that would give me less pressure (but more volume - moving fluid), correct ? So you can see I'm kinda stuck in the middle.
yes, a larger MC bore would take more force, but would move more volume.
I have hours and hours into brake bleeding and see no air bubbles. I'd like to think thats not the problem.
some of the details that might help:
- Manual brakes - 4 wheel disc
- GM MC (from a manual disc 70s Corvette)
- GM Metric calipers on front, Ford Exploder rear calipers on an 8.8 axle
- Brake Pads are close to disc and not retracting too far (visual inspection)
- Cheapskate rock auto calipers and brake pads
your question is how to get more stopping power? the easiest way is by increasing the friction of your pads. going from a friction value of 0.3 up to a 0.36 is a 20% increase. in our simple example, which does not account for things like whether or not friction is constant with temperature or pressure or speed (hint, it's not), going from 0.3 to 0.36 pad friction would take you from 2200 lb to 2640 lb total braking force, and your decel would go up by 20%, from 0.66 G to 0.8 G.
