Comments on brakes (1 Viewer)

[ntsqd]

Hydraulic Force Multiplication - How it works

Rather than trying to write this out by myself I'm going to start off by employing wiki since many folks have had a hand in writing theirs. Hopefully it will be more understandable to more people that way.

Cylinder C1 is one inch in radius, and cylinder C2 is ten inches in radius. If the force exerted on C1 is 10 lbf, the force exerted by C2 is 1000 lbf because C2 is a hundred times larger in area (S = πr²) as C1. The downside to this is that you have to move C1 a hundred inches to move C2 one inch. The most common use for this is the classical hydraulic jack where a pumping cylinder with a small diameter is connected to the lifting cylinder with a large diameter.

With the below graphic ignore the lower half, it has nothing to do with brake systems:

Notice that no mention is made of displaced volume. This is an important thing to keep in mind because while it isn't usually important to most hydraulic systems, it is important to brake systems. With such a system the trade-off is fluid volume for force. To get the needed force you have to use a small bore cylinder and move a lot of fluid volume, which means that the small bore piston has to move much further than does the large piston. So the small bore moves relatively a long ways with a small input force and the large piston moves a short distance with a lot of force.
The small bore cylinder is the master cylinder and the large bore cylinder is all of the caliper pistons acting as one (in parallel).

The ratio of the areas of the pistons is the hydraulic leverage. It is also the ratio that the relative travel of the pistons will be. In wiki's example the hydraulic leverage ratio is 100:1, so the force multiplication will also be 100:1, but so will the travel ratio - only in reverse. The small piston will move 100 times as far as the large piston.

Applying that to brakes means that if the brake system's hydraulic leverage ratio is 100:1 it means that for the caliper pistons to move 0.015" (about the width of 5-7 human hairs laid side by side) the m/c piston has to move 100 times that or 1.5"!

Since a typical air gap between the brake pads and rotors is about 0.015" it has used up 1.5" of the m/c piston's available travel just to push the pads into contact with the rotors!
Let's say that the m/c has a total travel of 2-1/4". That means that there is only 3/4" of travel left in the m/c to actually do the work of stopping. It also means that the driver has pushed the pedal some distance since there is a mechanical leverage built into the brake pedal. This ratio varies with application, it can be as low as 3:1 and as high as 7:1. Think about that a minute, even at 3:1 it means that the pedal has moved 4.5" just to move the brake pads into contact with the rotors, and at 7:1 the pedal has moved 10.5 inches!

The pedal's leverage ratio is multiplied by the hydraulic leverage ratio to get the overall force multiplication ratio. In the above examples that total force ratio ranges from 300:1 to 700:1! These ratios could can make a LOT of hydraulic pressure with that kind of ratio working for you. 10 pounds of force on the pedal would generate between 3000 psi and 7000 psi. The problem with this is that even at 3000 psi it is just about double what most brake systems have for a maximum, panic stop kind of pressure limit. Most brake systems do not operate at 100:1 hydraulic ratio. Depending on the type of booster they operate down to around 50:1 hydraulic ratio and ~3.5:1 mechanical ratio. This is done to reduce the total brake pedal travel. People don't like it when their brake pedal has to move several inches before anything happens.

 

[tunegoon]

Awesome thread, very good info ntsqd. I have also built many race cars and played around with bore diameters and leverage ratios to make drivers happy(happy driver=fast driver). Do you work for Willwood. I had their 13" rotors and 6 piston calipers front, 12" rotor 4 piston calipers rear on a 91 Mustang race car I had. Went through 3 MC's before I got the pedal feel I wanted with no brake booster. That car stopped well with no fade! Just bought my first Cruiser and I'm researching brakes that will work with 37" tires. It doesn't stop so well right now. I'll probably do the 4runner calipers. I just need to decide if rear drums are sufficient or if I need discs.

 

[ntsqd]

AFCO (afabcorp.com) offers a version of the metric GM caliper (D154 pad) with a 2-1/4" bore instead of the standard 2-1/2" bore. That is a 19% reduction in rear piston area. Comes at a price though, at the time of this posting they are listed at ~$85 each.

R/H p/n: [FONT=Arial, Helvetica][SIZE=-1]7241-9005
L/H p/n:[/SIZE][/FONT][FONT=Arial, Helvetica][SIZE=-1]7241-9006[/SIZE][/FONT]

Is this a custom caliper, or a rare OEM piece off some obscure vehicle like a... 78 olds starfire V8?

Another vendor is now offering this type of modified caliper. Speedway Motors is offering both a 2-3/4" (1/4" larger bore) and a 2-1/4" (1/4" smaller bore) caliper:
U.S. Brake GM Metric Calipers and Accessories

 

[ntsqd]

On using FJ80 M/C's with the 4Rnnr calipers

I posted the below in this thread: https://forum.ih8mud.com/60-series-wagons/408164-rear-drums-they-really-all-cool-2.html#post6455667

I finally got around to looking it up in my brake ratios spread sheet. The total leverage ratio is calculated by taking the ratio of the master's piston area divided into one side of the caliper's piston area, and multiplying that result by the brake pedal ratio. Since my 60 is too far away to measure I'm going to make the assumption that it's pedal ratio is the same as that of the pedal in my Xcab Mini. This may or may not be a safe assumption. I'll try to remember to measure the pedal tomorrow and adjust these numbers accordingly.

Stock FJ60 Calipers & M/C: 62.8:1
4Rnnr Calipers + FJ60 M/C: 70.4:1
4Rnnr Calipers + FJ80 M/C: 53.5:1

The bigger the number, the easier the brakes are to apply. That is, less pedal effort is needed to stop at some set rate. The smaller the number the more pedal effort is needed. People frequently confuse a system with a low total leverage number as having really good brakes. This is because the pedal feels very firm, which is a good thing. Unfortunately it also means that the effort level is quite high, which for most is a bad thing.

 

[Dynosoar]

Just to stir the pot....... The stock F60 MC is a 7/8" bore the FJ80 is a 1" bore. I decided to split the difference and I have been very happy with my brake feel, and pedal effort. I went with a '95 v6 4Runner MC it is 15/16" I don't know what ratio it would give but it feels and works great.

ntsqd, you want to run the ratio for me?

Dynosoar

 

[ntsqd]

Dynosoar,
A 15/16" M/C bore with the 4Rnnr calipers comes out to be: 61.3:1

 

[Dynosoar]

So at 61.3:1 I am pretty close to stock ratio. Thanks for running the numbers.


Dynosoar

 

[TobyB]

'I agree with Mace on this one. More surface area always = more friction.'

Actually, (3 years later) yes. BUT: increase the surface area with the same overall pressure on the pad,
and you end up with more surface area, but it's offset by less UNIT pressure, and so it balances out
exactly to the same pedal pressure.
So if you have a 2 square inch pad, cf of .5 and 1000 lbs, your unit pressure's 500 psi. 2 inches at 500 psi = 1000 pounds of stoppy force.
But if you switch to a 4 square inch pad and the same 1000 lbs, the unit pressure drops to 250 psi, but since area doubles, 250 x 4 still = 1000 lbs of stoppy....

I've actually done it with calipers that could handle several different
pad designs, and it's true. And yes, pad life usually goes up, and pad distress goes down.

As to using only one side of double fixed calipers- you can do that, or double the force on a slider
(as a slider piston provides force on BOTH sides piston)

t

 

[ntsqd]

The wilwood, and presumably industry std. (though I don't know that for sure) is to use only one side's piston area. Presumably it is less confusing?

I wish wilwood would've sent me to some of those SAE conferences on Brakes. I'm sure that there is a whole lot more, like the rest of the iceberg besides the tip, that I could've learned.

An esoteric discussion would be if the slightly greater area of the back of the piston bore of a sliding caliper (due to operational & seal clearances) more or less than offsets the sliding friction and results in the same pad pressure on the outside pad as that of the pad that the piston is directly in contact with.

 

[miker]

Another vendor is now offering this type of modified caliper. Speedway Motors is offering both a 2-3/4" (1/4" larger bore) and a 2-1/4" (1/4" smaller bore) caliper:
U.S. Brake GM Metric Calipers and Accessories

FYI, the 2 1/4" undersized piston version only looks to be offered in a right side caliper.

Willwood makes a GM Metric caliper with 2" piston that is listed as universal (for left or right side installation).
http://www.wilwood.com/Calipers/CaliperProd.aspx?itemno=120-9333

Sold by Summit:
http://www.summitracing.com/parts/WIL-120-9333/?rtype=10

 

[lehiguy]

More life to this thread;

About three years ago I put rather large GM calipers on the rear of my Cruiser (old solid axle Chevy 1/2 ton calipers, monsters) along with a 1" bore MC from a T-100 and front calipers from the same rig. I had to use a proportioning valve to tame the rears and it took pedal pumping due to the large amount of fluid reqired to actuate them. It didn't work as well as I wanted it to. Fast forward to now. I still have the same front calipers and MC form the T-100, but I switched rear calipers to the GM metrics from the front of an '88 Monte Carlo (fits about a thousand other rigs as well). Now there is no pumping required and I opened the proportioning valve all the way, meaning it could work without it. This combination, in my opinion, is just about perfect and the hardware is easy and cheap to get.

 

[ntsqd]

Started to post this in another thread and decided that it belonged here instead.

Hydraulics are relatively simple. It's all about force vs. area. If you have a piston in a bore connected with a fluid in a tube to a second piston in a second bore then by varying those two bore diameters you can multiply or divide the input force to become the output force. There is an inverse relationship between bore size and piston travel. All else being equal a smaller piston will travel further than a large piston to get the same output piston travel distance. This is because the fluid volume is fixed (or it had better be!). A large piston will move a small distance in it's bore when the fluid volume increases or decreases by a set amount. A small piston will move a lot further in it's bore when that same set volume is added to or subtracted from a small bore. The force multiplication or division is set by the ratio of the areas of the related bores. Push on a small piston to move a big piston and the force needed to move the small piston is low, but the distance traveled can be relatively large. That volume displaced by the small piston will move a big piston, but by a much smaller distance though with much greater force.

With brakes we're much more concerned about the force multiplication than we are with the piston travel distance. If you start thinking about needing to somehow compensate for piston travel you're off to a bad start. Better to totally ignore that aspect of the system as thinking about it will get you into trouble.

EDIT: Lost in my own thread, I covered this in post #61

 

[ntsqd]

Cf vs. Temp and brake fade

I believe that the friction does have nothing to do with surface area. But the coefficient of friction changes with not only the type of materials, but other factors such as temperature (see wikipedia article linked above). As the heat goes up, the coefficient of friction goes down. Thus hot brakes don't work as well as cold brakes, which explain brake fade going down steep hills, etc. So larger brakes don't work better because they have more friction, but because they can dissipate the heat better. There are also other factors such as wear, etc.

I see that I never addressed this. There is a definite relationship between Cf (Coefficient of Friction) and temperature, but the trend is not always for the Cf to go down as the temp goes up. It also does not explain brake fade. For a sampling of temperature vs. Cf look at wilwood's curves for their various pad compounds. http://www.wilwood.com/PDF/Flyers/fl227.pdf some pads are pretty flat in the upper temps, others do have a slight loss as the temperature increases, yet others fall off rapidly as the temp goes up.

Brake fade is caused by two distinctly different events. One is fluid boiling. Boiling fluid will cause a mushy feeling in the pedal as well as lack of braking performance. See my heat barrier piston illustration in this thread for how some racing calipers combat this. If you've ever experienced boiling brake fluid you should flush the brake system at first opportunity. Boiled brake fluid has changed and is no longer serviceable. Though it can wait until after you've changed your shorts.

The other way for fade to happen is the out-gassing of the pads from excessive heat. Usually, but not always, these gasses come from the solvents used in the production of the pad compound that have remained trapped in the pads. The gas creates a boundary layer between the pad and the rotor. As you might guess it's Cf is pretty low.........
The drilling of rotors was originally intended to remove any such gas as early disc brake pads really suffered from this problem. Not so much any more. The problem with drilled rotors is that they will crack prematurely starting at those holes. Slots were also devised to combat the out-gassing. They are still a stress riser so cracks will happen around them first, but they are less of a problem than are holes. An advantage of slots is that they have a slight cleaning effect on the pads.

 

[ntsqd]

Posted this in another thread and thought that I'd quote it here for my future ref if no one else's.

Disconnect the booster downstream of the vacuum check valve (booster side) and cap it off. That should render the booster null. If the check valve is still connected to the booster you'll have made the brakes firmer feeling than they really are because you'll be working against the booster too. Then try the brakes.

The calc is a simple one, total system leverage = pedal ratio * hydraulic ratio. Where people go astray is in calculating the caliper piston area. You're only interested in the area of one side of the caliper. You'll to either trust me on this or figure out why on your own. I'm not sure that I full understand the physics of it well enough to explain it, but that is how everyone in the industry does it.
Total ratio (TR) = (pedal pivot to center of foot pad dist./pedal pivot to push-pin dist.) * (one side piston(s) area/m-c bore area)
From this equation I hope that it can be seen that if you increase the caliper piston area and increase the m-c bore area by the same percentage of change that you've made no difference to the TR or how the pedal will feel.

The curve-ball is the booster. In a manual, unboosted brake system you want a TR somewhere above 93:1 and below 100:1 though some folks can tolerate or even like slightly more than 100:1 (well into mushy). Very, very few will like the actual stopping of below 93:1 even though the pedal feel is nice and firm and really inspires confidence in the brakes*. Ideal that most everyone will at least not complain about seems to hover very close to 95:1.
With a booster in the mix the TR can drop well into the 70's and maybe below that, I don't know. Boosters are too much of a wildcard to approach system design like this. Takes a more sophisticated approach and one that I'm not familiar with. The reason is that the valves in the booster can vary the boost. That's like having a variable ratio pedal. Two identical appearing boosters can have pedal feels that are totally different due to the 'tune' on the valve or valves. The venting valve isn't an all or nothing kind of valve, it vents a percentage based on how far the push-rod is moved. Move movement = more boost. My brakes education was more about race car systems and didn't deal with boosters at all so what I know about boosters is from what I can find (truthfully haven't looked very hard) online and in James Walker Jr.'s book on brakes about booster design.

*All too often on forums a good feeling pedal is equated to be good brakes. Those folks don't understand why the vehicle stops so poorly when the pedal feels so good. A good firm pedal that doesn't stop in a reasonable distance means that the TR is too low (or wrong/bad booster) so it doesn't generate enough line pressure. This is the devil in monkeying with brakes, a firm pedal is confidence inspiring, but the line btwn good and firm that stops well and good and firm that doesn't stop well can be very thin.

 

[Slow Left]

Thanks for the thread @ntsqd !