Well yes, but only if the area of the master was 1 square inch. A 1 1/8" master is a hair under a square inch, whereas a 7/8" diameter measures up to a mere 0.6 square inches (plus some small change). But to continue the math, your 600lbs input on a 0.6 square inch master will push out a hefty 1000psi. But if the pedal ratio was say 5:1 the output would drop to 833 psi. Seemingly small changes have huge impact. Is the fog clearer, or fogger!! You'd have been dead right if you'd said 600# entering the master though. Chris Sent from my SM-T515 using The H.A.M.B. mobile app
While small, the fluid compressibility and pipe expansion under pressure are non-zero factors. See http://www.engineeringinspiration.co.uk/brakecalcs.html Steel Pipe Expansion Fluid Compression So a larger diameter pipe would affect system pressure. Rubber lines have their own expansion factor, as do flexible stainless braided lines. I’d expect the typical rubber line to outweigh any effect from the hard line expanding under pressure, or fluid compressibility, but it all adds up. Sent from my iPad using The H.A.M.B. mobile app
I agree. When you stroke the master cylinder, a fixed amount of fluid is transferred. There shouldn't be a "delay" in building pressure due to line size. The only basic difference would be in the velocity of the brake fluid as it transfers to the brake cylinder. I suppose there is pressure loss due to increased velocity, but I would think it would be pretty small.
David, Doing all that math on a Saturday isn't going to happen. After you've calced it out, let us know what percentage of fluid is required to make up for the stretch of the 1/4" line over the 3/16" line.
It doesn't, kind of. It's a ratio thing. A rate, not an absolute measure. Remember the high-school question, which applies the greater pressure on the ground the elephants foot or the ladies stilleto heel?? Same thing. Eg 500lbs applied to a square inch is 500psi. 500lbs applied to 1/2 square inch is 1000psi. 500lbs applied to 2 square inches is 250psi. Chris Sent from my SM-T515 using The H.A.M.B. mobile app
Since it looks like we've got some experts here how about this question. S10 front discs with 2-1/2" Pistons. Jag IRS xjs rear discs with 1.685 Pistons. 15 in tires front and rear. Rears are 255. Don't remember the size of the front but they are what we might call normal size. The car is 15 miles from me now. The Jag pads are about half the size of the S10 pads. Unknown master cylinder with power brakes. Under normal circumstances all brakes fine. I don't think I have ever locked the brakes up no matter how hard I hit the pedal. Pedal ratio is 7 to 1. Unknown proportional valve but it's supposed to be for discs / discs. Should I buy a brake pressure gauge and fittings next?
A good question! I'll see if I can form an answer that illustrates the process. The example above.....100 psi on the pedal with a 6 to one pedal arm(s) ratio equals 600 psi applied to the MC piston.......there is a "trade off" so to speak. For simplicity, lets say the pedal moves 6" of travel, but with the pedal arm(s) ratio, that means the small arm acting on the MC piston only moves one inch. Okay, now let use an analogy. If you have a pry bar, say 7' feet long, and you want to lift a heavy object with it, you put the lifting end under the object, place a fulcrum 1' from that point and push down on the remaining 6' of prybar handle. The force you are exerting is multiplied by the longer arm of the pry bar 6 times, lifting the weight more easily in the process. A claw hammer does the same thing when pulling a nail. Now, in the case of the fluid under discussion, lets agree for this purpose that the MC piston is one square inch of total area (not diameter, but area....what the heck... imagine it is a square piston of 1"x1", a 'square inch') So back to our leverage example above. Your foot is on the 'prybar' (long arm of the pedal) and whatever movement and/or pressure you apply is multiplied by 6 at the other end of the prybar (short arm of the pedal). If a lever is applied to a load with the fulcrum (pivot point) midway of the length of the prybar, there is equal linear movement and equal pressure at both ends. It is only when the fulcrum is offset from the center of the prybar that multiplication of force occurs......but at the expense of distance moved by one end vs the other. Ray
OK , just read (about12 times ) "hydraulic principle aka pascals principle , somewhere in the ancient fog I remember this ( probably 60 years ago) . I knew this , I just couldn't get my head around it . I think its like some mathematical functions , you can't prove them you simply accept that its so , so thanks.
This is my take on how brakes work, I am happy for corrections. In a hydraulic system with no booster, we are not creating any extra PSI than the foot can apply. We are using leverage and hydraulic principles. In the above description, pressure was applied to the brake pedal (100psi) multiplied by leverage to give as 600psi. The piston on moved 1/6 the pedal distance. Now in a brake system we need movement and pressure. We we start pushing the brake pedal we are not using the full 100psi. What we require is enough fluid to be moved to allow all of the pistons in the brake system to move enough to make the shoes/pads contact the surface of the drum disk. The only pressure we need is to overcome the springs and seals. The next phase is the movement is relatively small but we need pressure to take out flex in lines, flex in calipers and initial compression of brake pad and shoe. Please note that without proportioning valve, all brakes lines will have the same pressure (in theory). Now we get to the serious side of braking, we a pushing down on the pedal with 100psi or more of force. As in the above example with 6 to 1 leverage and 1 square inch piston we have 600psi in all brake lines (Front, Back,Left, Right). We are also getting virtually no fluid flow because or the flex in the system has been taken up. Think if a 100 pound woman sits on a chair or a 500 pound man. The chair does not sink into the floor but it is harder to drag with the man. If I have a 2 square inch piston in the front I will have 600psi pushing on 2 square inches of piston for a force of 1200psi (large man). If on the rear I have a 1 square inch piston with 600psi I will have 600psi of force (small woman). The rear will have less braking force. This is why we can find a combination of lever ratio, master piston size, and caliper piston size to give us enough flow for feel and to work with enough pressure to stop. Steve from down under
@Budget36 A car accelerates with force applied to the footprint of the tyres This is measured as torque from the rotation centreline, and is a sum of the total torque values multiplied ] wheel torque factors wheel radius,then rear end ratio, Transmission ratio, and Engine torque [the source of this force] Braking is the same. Deceleration, is brake torque applied at the axle centreline. So wheel radius is factored, as well as disc/ drum diameter. Brake bias can be adjusted by 4 methods [4 if a race car balance bar is used] These methods are 1: Disc/Drum diameter. 2: Coefficient of friction [basically pad sizes or compounds] 3: Clamping pressure . Different caliper piston or wheel cylinder sizes. These methods ^^^^ use equal line pressure throughout the whole system [up to a set point of a proportioning valve] 4: A race car balance bar with 2 M/C's is the only method that proportions front and rear line psi. [this method is infinitely adjustable, and perfect for people out of their depth with calculations] Jag rear pistons are smaller so there is reduced clamping pressure, and combined with smaller pads would net front brake bias. When calculations are made for race cars , we factor 0.005" compression for the brake pads. Then use "fluid mechanics" to calculate pedal compression. [not to be confused with initial pedal travel] Sometimes we have had a situation where there is too much pedal travel on one particular M/C, [causing the balance bar to bind, and excessive pedal travel] So we reduce the pedal travel with a slightly larger M/C and adjust the balance bar further over to one side. Anybody that is building a Hot Rod with "Big and Littles" I would highly recommend using twin M/C's with a balance bar [eg: a Wilwood Pedal box] The combination piston area of the Fr and Rr M/C's should equal the piston area of a single M/c [As a starting point] Then mismatched parts or traction issues etc can be remedied easily. Ray Your foot uses lbs force on the pedal , not psi. [you can use a bathroom scale to measure this force] Measure how many square inches your foot actually is I did understand what you meant.........I'm just winding you up!
Well, I can partially answer this. Jag used the same brakes on pretty much everything in the '70s/80s and this particular system into the '90s on the XJS although they did move the rear calipers outboard something like 3 years before changing it all up. Jags that used these brakes are heavy cars (typically almost 4K lbs curb weight), and tend to be nose-heavy to boot. Total piston area (factory was 4-piston) per front caliper is a bit over 7 square inches compared to your roughly 4.9 inches so you have about 30% less clamping force all else being equal. The pad size on the Jag calipers is about 4" x 2.25" or 9 square inches. Probably not a lot of difference there. But your '49 is undoubtedly lighter and probably even has a better F/R weight bias, so my SWAG is you're probably pretty close on sizing. I will note that the factory Jag system used NO proportioning or residual valves of any kind, not even a pressure-differential warning switch. The front half of the master went directly to the fronts, same for the rears. And before everyone gets fixated on hydraulic pressure, we need to remember that brakes are simply a way to dissipate energy by converting it to heat. Swept area and how well the assembly sheds that heat is a big part of how well your brakes are going to work. A bigger swept area can do as well or better at lower pressure than a smaller area at a higher pressure.
Have been having fun with Goooogle today. These two online calculators may be helpful. https://www.tceperformanceproducts.com/bias-calculator/ https://www.tceperformanceproducts.com/dual-bias-calc/ Sent from my iPad using The H.A.M.B. mobile app
There was a mention about smaller rear discs on newer cars, my pickup has 4 wheel disc (03), I looked at them today and my eyeball says they sure look the same size all the way around. If they are different, it isn’t by inches
[/QUOTE] Ray Your foot uses lbs force on the pedal , not psi. [you can use a bathroom scale to measure this force] Measure how many square inches your foot actually is I did understand what you meant.........I'm just winding you up![/QUOTE] @Mimilan Thank you for the comment. Let start by saying "I do understand what you meant" However, if we accept as a given that we have 600 psi in the master cylinder produced by a one sq inch piston being driven by an arm with a six to one leverage ratio, it follows that one hundred pounds of pressure must have been applied at the arm's pressure point. So, whether that should be discounted as psi in favor another term is, I think, debatable. The debate being dependent on the area size of the point on the pedal arm where the pressure/force is being applied. If that is a one sq inch surface area (as it happens, the precise size of the pedal pad I was visualizing), would that not be 100 psi? Ray
When I made that comment, I specifically limited it cars as trucks were very likely to be different because the wide range of weight you'd find on the rear wheels.
I thought it might be helpful to sketch a basic diagram of a brake system. Maybe it will help as we throw around different terms. We apply a mechanical Force with our foot on the brake pedal (50 pounds). That force is increased through a lever (6:1 increase to 600 pounds). That force is applied to a piston in a master cylinder. That piston (1 sq in in this sample) presses on 1 sq in of hydraulic fluid. That creates a pressure in the fluid system. (300 psi in this example). That 300 psi fluid presses on a 4 inch square wheel cylinder. This provides 1200 pounds of mechanical force on the brake pads. BRAKE SYSTEM by HotRodWorks posted Jan 18, 2021 at 11:30 AM
It is not There is 2 factors here [force and distance] for the same volume. The smaller diameter M/C requires more piston travel so the pressure is higher [for the same input force from the pedal] The volume is limited by clamping pressure at the other end @Mimilan Correct! Force doesn't get converted to psi until "area" is factored in; For the benefit of other readers. 100 lbs force from the foot x 6:1 pedal ratio = 600 lbs force. [as Ray has stated] If applied to a 1" M/C [0.7854 sq inches area] [1" x 1" x 0.7854 =0.7854"] 600 lbs ÷ 0.7854 = 763.94 psi [line pressure] If applied to a 3/4" M/C [0.441 sq inches area] [0.75" x 0.75" x 0.7854 =0.441"] 600 lbs ÷ 0.441 = 1360.54 psi [it starts getting up there] The math then goes the opposite way at a wheel cylinder [it is multiplied instead of divided. So ^^^ 1360.54 x 3/4" wheel cylinder = 600lbs force [each cyinder] But it gets interesting with calipers A 2" piston caliper has 3.141 sq inches of area . This must be multiplied by 2 because a floating single caliper moves 2 directions A floating 2" caliper has the same swept volume as a 2 piston [opposite] 2"caliper. So the above caliper effectively has the swept area of 6.282" With the 3/4" M/C's 1360.54 psi pressure there is 8546.9 lbs clamping pressure. With the 1" M/C the clamping pressure is 4799 lbs clamping pressure.
There were some comments about using smaller diameter pipe to the rear. Another poster questioned if it would make a difference. My answer is a definite maybe. An easy way to show this is to hook up your garden hose with a T piece on the tap. When you turn the tap on you will get flow out of one side of the T very quickly and a second or so later you will get flow out of the hose. This is because it takes time for the fluid to get through the hose even if the hose is full of fluid. The next example would be to put a 50 foot 1/4 hose on one side of the T and a 50foot 1 inch hose on the other side of the T. The 1 inch will flow more due to less surface friction. If you put a stopper on the ends of both hoses the pressure would be the same in both. So as we are applying the brakes the shorter larger diameter hose will flow more fluid and move the piston earlier than the longer thinner hose. As to how many milliseconds and if you will notice I cannot say. The other thing with a disk drum setup is some of the force applied to the rear drums is overcoming the shoe springs. If you have changed a set of shoes you will notice that this can be a lot. Steve
Steve Smith published a book called "Advanced Race Car Suspension Development" It has a lot of formulas I stumbled upon the calculations when I inherited a work sheet with a Race-Car I purchase [last Century]
brake fluid doesn't flow through the pipes, your pads move a very small amount, there will be a tad of movement but nothing much. Force applied at one end is applies at the other without movement required (as I said a tad of movement but flow or movement isnt required to get the force at the other end)
That sounds like it may lack the inputs needed (like the links in post 47) if using more mundane parts...
In the initial part of braking the fluid that is displaced by the master cylinder flows into the caliper or slave cylinder. This has to be considered if we are looking at a braking system as a whole. This is why we need a master cylinder that will flow enough fluid to account for the displacement of the caliper pistons or slave cylinders. The amount that the master need to flow can be calculated by the diameter of all of the pistons and how much the require to move. Some calipers use the seals to retract the pad from the disk to reduce friction to a minimum. If the flow required is more than the master cylinder can displace you will never build up pressure. If the flow is only a fraction of what the master can flow you will end up with very short pedal movement which makes it hard to regulate braking. The goldilocks solution is to have a master that will flow enough for all of the cylinders with a bit in reserve. Have the right diameter to get enough pressure, be connected to the pedal with a ratio that gives a comfortable amount of travel plus leverage and have the ratio of piston area between the front and rear just right to give slightly more braking on the front. It so happens that there are some very smart people on this site that will be able to suggest a setup that from experience that will be close to the goldilocks formula. It is unlikely it will be perfect because they don't know you have a bung knee that cant push as hard, a sponge soft rubber pad on your brake pedal and happen to have 70% of the cars weight over the rear wheels. Steve
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